Possible definitions of exponential function











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I was wondering how many definitions of exponential functions can we think of. The basic ones could be:



$$e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$$
also
$$e^x:=lim_{ntoinfty}bigg(1+frac{x}{n}bigg)^n$$
or this one:
Define $e^x:mathbb{R}rightarrowmathbb{R}\$ as unique function satisfying:
begin{align}
e^xgeq x+1\
forall x,yinmathbb{R}:e^{x+y}=e^xe^y
end{align}

Can anyone come up with something unusual? (Possibly with some explanation or references).










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    Related : math.stackexchange.com/questions/833962/…
    – Arnaud D.
    Nov 7 at 13:35






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    continued fraction: en.wikipedia.org/wiki/…
    – R zu
    2 days ago






  • 1




    Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
    – Paramanand Singh
    2 days ago






  • 1




    I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
    – Hagen von Eitzen
    2 days ago






  • 2




    @Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
    – user3641
    2 days ago

















up vote
30
down vote

favorite
11












I was wondering how many definitions of exponential functions can we think of. The basic ones could be:



$$e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$$
also
$$e^x:=lim_{ntoinfty}bigg(1+frac{x}{n}bigg)^n$$
or this one:
Define $e^x:mathbb{R}rightarrowmathbb{R}\$ as unique function satisfying:
begin{align}
e^xgeq x+1\
forall x,yinmathbb{R}:e^{x+y}=e^xe^y
end{align}

Can anyone come up with something unusual? (Possibly with some explanation or references).










share|cite|improve this question




















  • 5




    Related : math.stackexchange.com/questions/833962/…
    – Arnaud D.
    Nov 7 at 13:35






  • 1




    continued fraction: en.wikipedia.org/wiki/…
    – R zu
    2 days ago






  • 1




    Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
    – Paramanand Singh
    2 days ago






  • 1




    I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
    – Hagen von Eitzen
    2 days ago






  • 2




    @Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
    – user3641
    2 days ago















up vote
30
down vote

favorite
11









up vote
30
down vote

favorite
11






11





I was wondering how many definitions of exponential functions can we think of. The basic ones could be:



$$e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$$
also
$$e^x:=lim_{ntoinfty}bigg(1+frac{x}{n}bigg)^n$$
or this one:
Define $e^x:mathbb{R}rightarrowmathbb{R}\$ as unique function satisfying:
begin{align}
e^xgeq x+1\
forall x,yinmathbb{R}:e^{x+y}=e^xe^y
end{align}

Can anyone come up with something unusual? (Possibly with some explanation or references).










share|cite|improve this question















I was wondering how many definitions of exponential functions can we think of. The basic ones could be:



$$e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$$
also
$$e^x:=lim_{ntoinfty}bigg(1+frac{x}{n}bigg)^n$$
or this one:
Define $e^x:mathbb{R}rightarrowmathbb{R}\$ as unique function satisfying:
begin{align}
e^xgeq x+1\
forall x,yinmathbb{R}:e^{x+y}=e^xe^y
end{align}

Can anyone come up with something unusual? (Possibly with some explanation or references).







real-analysis definition big-list






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edited Nov 7 at 13:25


























community wiki





3 revs
Michal Dvořák









  • 5




    Related : math.stackexchange.com/questions/833962/…
    – Arnaud D.
    Nov 7 at 13:35






  • 1




    continued fraction: en.wikipedia.org/wiki/…
    – R zu
    2 days ago






  • 1




    Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
    – Paramanand Singh
    2 days ago






  • 1




    I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
    – Hagen von Eitzen
    2 days ago






  • 2




    @Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
    – user3641
    2 days ago
















  • 5




    Related : math.stackexchange.com/questions/833962/…
    – Arnaud D.
    Nov 7 at 13:35






  • 1




    continued fraction: en.wikipedia.org/wiki/…
    – R zu
    2 days ago






  • 1




    Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
    – Paramanand Singh
    2 days ago






  • 1




    I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
    – Hagen von Eitzen
    2 days ago






  • 2




    @Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
    – user3641
    2 days ago










5




5




Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
Nov 7 at 13:35




Related : math.stackexchange.com/questions/833962/…
– Arnaud D.
Nov 7 at 13:35




1




1




continued fraction: en.wikipedia.org/wiki/…
– R zu
2 days ago




continued fraction: en.wikipedia.org/wiki/…
– R zu
2 days ago




1




1




Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
– Paramanand Singh
2 days ago




Your last definition is tricky and the existence of such a function is usually proved by showing that this this definition leads to other more suitable ones.
– Paramanand Singh
2 days ago




1




1




I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
– Hagen von Eitzen
2 days ago




I was not aware that your last one could actually used as a definition. And I am impressed because I have the habit of making my mathSE answers involving the exponential depend solely on these two properties!
– Hagen von Eitzen
2 days ago




2




2




@Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
– user3641
2 days ago






@Aloizio I'm not sure Community Wiki was warranted here. If a question is valuable enough that you believe it belongs on the site, chances are you don’t need it to be community wiki! [...] Instead, strive for quality. If you’re unsure a certain question class belongs on the site, don’t tolerate the worst examples — demand that these questions be awesome. [...] questions rarely, if ever, need community wiki.
– user3641
2 days ago












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The exponential function is the unique solution of the initial value problem



$y'(x)=y(x) , quad y(0)=1$.






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    up vote
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    down vote













    We can also define $e^x$ as follows:




    • the inverse function of $ln x$, defining $ln x$ independently as follows


    $$ln x := int_1^x frac{dt}{t}$$




    • the unique solution to IVP $f'(x)=f(x)$ with $f(0)=1$ which existence is guaranteed by Existence and Uniqueness” Theorems for first order IVP






    share|cite|improve this answer



















    • 2




      Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
      – j4nd3r53n
      Nov 7 at 15:47








    • 8




      And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
      – Daniel Schepler
      Nov 7 at 17:28










    • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
      – gimusi
      Nov 7 at 18:19






    • 1




      @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
      – Daniel Schepler
      Nov 7 at 18:23










    • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
      – Paramanand Singh
      2 days ago


















    up vote
    11
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    Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.



    First define it for the natural numbers:



    Define $e^2 = e times e$, $e^3 = e times e times e $, etc.



    Now define it for other integers:



    $e^0 = 1$, $e^{-n} = frac{1}{e^n}$, etc.



    Now for other rational numbers (getting a bit harder):



    $e^{frac{p}{q}} = sqrt[q]{e^p}$



    Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.



    This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.






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    • 1




      Can you elaborate your answer with something concrete?
      – Michal Dvořák
      Nov 7 at 13:28










    • I expanded my answer.
      – badjohn
      Nov 7 at 13:39






    • 1




      Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
      – badjohn
      Nov 7 at 15:43








    • 3




      You might just define $x^y$ at this point.
      – Ori Gurel-Gurevich
      Nov 7 at 16:04






    • 1




      True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
      – badjohn
      Nov 7 at 16:31




















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    Throw $n$ balls into $n$ bins uniformly at random, and take $n to infty$. Define $frac{1}{e}$ to be the limiting fraction of empty bins.



    A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-frac{1}{e}$ to be the fraction of distance covered after one unit of time.



    Given positive $x$, consider a set of independent Bernoulli random variables with $sum_{i=1}^n p_i = x$. As $n to infty$ and $max_i p_i to 0$, define $e^{-x}$ to be the probability that all are zero.






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      If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{nrightarrow infty}left[frac{n!}{n^nsqrt{(2n+frac13pi}}right]^{frac xn}$






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        You can also define the exponential function like this:
        $$
        e^x = lim_{ntoinfty} frac{f_n(x)}{f_n(-x)}
        $$

        where
        $$
        f_n(x) = sum_{j=0}^n frac{(2n-j)!}{j!(n-j)!}x^j
        $$






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        • 3




          I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
          – Michal Dvořák
          2 days ago


















        up vote
        4
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        Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define
        $$
        frac{1}{e}=lim_{ntoinfty}frac{D_n}{n!}.
        $$






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          up vote
          4
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          EDIT: thanks to @HagenvonEitzen's comment.



          I've often wondered if the following is sufficient for a general power function:



          $$f(x+y) = f(x)f(y)$$



          And then:



          $$f(1) = e$$



          for the base.



          Think this is probably similar to @badjohn's answer.



          EDIT: thanks to @CarstenS and @R



          Turns out we must also demand that $f(x)$ is continuous or measurable.






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          • 6




            Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
            – Carsten S
            Nov 7 at 16:44








          • 1




            @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
            – R..
            Nov 8 at 2:28






          • 2




            @R.., yes I should have written "not without additional conditions".
            – Carsten S
            2 days ago






          • 2




            $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
            – Hagen von Eitzen
            2 days ago


















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          $e^x := cos(-ix) + i sin(-ix)$



          (See Euler's formula)






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          • Is it possible to define the sine and cosine of a complex number without using exponential?
            – Surb
            Nov 7 at 23:08










          • That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
            – Ben Jones
            Nov 7 at 23:47






          • 7




            Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
            – weee
            Nov 7 at 23:48






          • 3




            So we keep running in circles and going "@weee..."? =)
            – user21820
            2 days ago






          • 5




            @weee Where else would one run when dealing with sin and cos? :)
            – Hagen von Eitzen
            2 days ago


















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          2
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          Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:



          Suppose we have $frac{dy}{dx}=ky.$ And we have $y(0)=1$.



          By Euler's Method:
          Pick some small $Delta x$. Then Let:
          $$x_{n+1}=x_n+Delta x$$
          $$y_{n+1}=y_n+frac{dy}{dx}Delta x=y_n+(ky_n)Delta x$$



          With some rearranging:
          $$y_{n+1}=y_n(1+kDelta x)implies y_{n+p}=y_n(1+kDelta x)^p$$
          $$x_{n+p}=x_n+pDelta x $$



          Now we have that $x_0$=0 and $y_0=1$. So:
          $$x_{n+p}=(0)+pDelta x$$
          $$y_{n+p}=(1)(1+kDelta x)^p$$



          Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $Delta x$ to be $z/p$.



          Then letting $n=0$,



          $$y_p=(1+kfrac{z}{p})^p$$



          By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.






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            $f(x) = e^x > 0$ is such a function that



            $$ intlimits_{1}^{f(x)} tfrac{1}{u} , mathrm{d}u = x$$






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              The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one.
              It is the Lie group exponential map of the latter group.






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                Definition of $e$:



                $$lim_{hrightarrow 0} frac{e^h - 1}{h} = 1.$$



                Define exponents as a supremum of a set of a real number to rational powers.






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                • 2




                  This needs a bit more elaboration as to how you define arbitrary real exponents.
                  – Paramanand Singh
                  2 days ago






                • 4




                  Also, this seems to define $e$ but not $xmapsto e^x$
                  – Surb
                  2 days ago


















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                0
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                Similar things have been said, but not in that way:



                $exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(mathbb R) to C^{k}(mathbb R)$ for some $k in [0,infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and ${exp}$ is a somewhat "canonical" basis for that space.)






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                  If we equip the one-dimensional manifold $(0,infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function.
                  This is the unique metric that makes the exponential function an exponential map.






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                    up vote
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                    The exponential function is the unique solution of the initial value problem



                    $y'(x)=y(x) , quad y(0)=1$.






                    share|cite|improve this answer

























                      up vote
                      45
                      down vote













                      The exponential function is the unique solution of the initial value problem



                      $y'(x)=y(x) , quad y(0)=1$.






                      share|cite|improve this answer























                        up vote
                        45
                        down vote










                        up vote
                        45
                        down vote









                        The exponential function is the unique solution of the initial value problem



                        $y'(x)=y(x) , quad y(0)=1$.






                        share|cite|improve this answer












                        The exponential function is the unique solution of the initial value problem



                        $y'(x)=y(x) , quad y(0)=1$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 7 at 13:24









                        Fred

                        41.2k1641




                        41.2k1641






















                            up vote
                            25
                            down vote













                            We can also define $e^x$ as follows:




                            • the inverse function of $ln x$, defining $ln x$ independently as follows


                            $$ln x := int_1^x frac{dt}{t}$$




                            • the unique solution to IVP $f'(x)=f(x)$ with $f(0)=1$ which existence is guaranteed by Existence and Uniqueness” Theorems for first order IVP






                            share|cite|improve this answer



















                            • 2




                              Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                              – j4nd3r53n
                              Nov 7 at 15:47








                            • 8




                              And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                              – Daniel Schepler
                              Nov 7 at 17:28










                            • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                              – gimusi
                              Nov 7 at 18:19






                            • 1




                              @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                              – Daniel Schepler
                              Nov 7 at 18:23










                            • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                              – Paramanand Singh
                              2 days ago















                            up vote
                            25
                            down vote













                            We can also define $e^x$ as follows:




                            • the inverse function of $ln x$, defining $ln x$ independently as follows


                            $$ln x := int_1^x frac{dt}{t}$$




                            • the unique solution to IVP $f'(x)=f(x)$ with $f(0)=1$ which existence is guaranteed by Existence and Uniqueness” Theorems for first order IVP






                            share|cite|improve this answer



















                            • 2




                              Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                              – j4nd3r53n
                              Nov 7 at 15:47








                            • 8




                              And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                              – Daniel Schepler
                              Nov 7 at 17:28










                            • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                              – gimusi
                              Nov 7 at 18:19






                            • 1




                              @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                              – Daniel Schepler
                              Nov 7 at 18:23










                            • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                              – Paramanand Singh
                              2 days ago













                            up vote
                            25
                            down vote










                            up vote
                            25
                            down vote









                            We can also define $e^x$ as follows:




                            • the inverse function of $ln x$, defining $ln x$ independently as follows


                            $$ln x := int_1^x frac{dt}{t}$$




                            • the unique solution to IVP $f'(x)=f(x)$ with $f(0)=1$ which existence is guaranteed by Existence and Uniqueness” Theorems for first order IVP






                            share|cite|improve this answer














                            We can also define $e^x$ as follows:




                            • the inverse function of $ln x$, defining $ln x$ independently as follows


                            $$ln x := int_1^x frac{dt}{t}$$




                            • the unique solution to IVP $f'(x)=f(x)$ with $f(0)=1$ which existence is guaranteed by Existence and Uniqueness” Theorems for first order IVP







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 2 days ago

























                            answered Nov 7 at 13:24









                            gimusi

                            83.2k74292




                            83.2k74292








                            • 2




                              Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                              – j4nd3r53n
                              Nov 7 at 15:47








                            • 8




                              And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                              – Daniel Schepler
                              Nov 7 at 17:28










                            • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                              – gimusi
                              Nov 7 at 18:19






                            • 1




                              @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                              – Daniel Schepler
                              Nov 7 at 18:23










                            • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                              – Paramanand Singh
                              2 days ago














                            • 2




                              Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                              – j4nd3r53n
                              Nov 7 at 15:47








                            • 8




                              And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                              – Daniel Schepler
                              Nov 7 at 17:28










                            • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                              – gimusi
                              Nov 7 at 18:19






                            • 1




                              @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                              – Daniel Schepler
                              Nov 7 at 18:23










                            • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                              – Paramanand Singh
                              2 days ago








                            2




                            2




                            Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                            – j4nd3r53n
                            Nov 7 at 15:47






                            Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp colon (mathbb{R},+) rightarrow (mathbb{R},cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) cdot exp(r_2)$ and $exp(0) = 1$
                            – j4nd3r53n
                            Nov 7 at 15:47






                            8




                            8




                            And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                            – Daniel Schepler
                            Nov 7 at 17:28




                            And a good candidate for the definition of $ln x$ would be $ln x := int_1^x frac{dt}{t}$.
                            – Daniel Schepler
                            Nov 7 at 17:28












                            @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                            – gimusi
                            Nov 7 at 18:19




                            @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers.
                            – gimusi
                            Nov 7 at 18:19




                            1




                            1




                            @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                            – Daniel Schepler
                            Nov 7 at 18:23




                            @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer.
                            – Daniel Schepler
                            Nov 7 at 18:23












                            @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                            – Paramanand Singh
                            2 days ago




                            @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions.
                            – Paramanand Singh
                            2 days ago










                            up vote
                            11
                            down vote













                            Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.



                            First define it for the natural numbers:



                            Define $e^2 = e times e$, $e^3 = e times e times e $, etc.



                            Now define it for other integers:



                            $e^0 = 1$, $e^{-n} = frac{1}{e^n}$, etc.



                            Now for other rational numbers (getting a bit harder):



                            $e^{frac{p}{q}} = sqrt[q]{e^p}$



                            Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.



                            This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.






                            share|cite|improve this answer



















                            • 1




                              Can you elaborate your answer with something concrete?
                              – Michal Dvořák
                              Nov 7 at 13:28










                            • I expanded my answer.
                              – badjohn
                              Nov 7 at 13:39






                            • 1




                              Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                              – badjohn
                              Nov 7 at 15:43








                            • 3




                              You might just define $x^y$ at this point.
                              – Ori Gurel-Gurevich
                              Nov 7 at 16:04






                            • 1




                              True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                              – badjohn
                              Nov 7 at 16:31

















                            up vote
                            11
                            down vote













                            Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.



                            First define it for the natural numbers:



                            Define $e^2 = e times e$, $e^3 = e times e times e $, etc.



                            Now define it for other integers:



                            $e^0 = 1$, $e^{-n} = frac{1}{e^n}$, etc.



                            Now for other rational numbers (getting a bit harder):



                            $e^{frac{p}{q}} = sqrt[q]{e^p}$



                            Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.



                            This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.






                            share|cite|improve this answer



















                            • 1




                              Can you elaborate your answer with something concrete?
                              – Michal Dvořák
                              Nov 7 at 13:28










                            • I expanded my answer.
                              – badjohn
                              Nov 7 at 13:39






                            • 1




                              Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                              – badjohn
                              Nov 7 at 15:43








                            • 3




                              You might just define $x^y$ at this point.
                              – Ori Gurel-Gurevich
                              Nov 7 at 16:04






                            • 1




                              True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                              – badjohn
                              Nov 7 at 16:31















                            up vote
                            11
                            down vote










                            up vote
                            11
                            down vote









                            Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.



                            First define it for the natural numbers:



                            Define $e^2 = e times e$, $e^3 = e times e times e $, etc.



                            Now define it for other integers:



                            $e^0 = 1$, $e^{-n} = frac{1}{e^n}$, etc.



                            Now for other rational numbers (getting a bit harder):



                            $e^{frac{p}{q}} = sqrt[q]{e^p}$



                            Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.



                            This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.






                            share|cite|improve this answer














                            Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.



                            First define it for the natural numbers:



                            Define $e^2 = e times e$, $e^3 = e times e times e $, etc.



                            Now define it for other integers:



                            $e^0 = 1$, $e^{-n} = frac{1}{e^n}$, etc.



                            Now for other rational numbers (getting a bit harder):



                            $e^{frac{p}{q}} = sqrt[q]{e^p}$



                            Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.



                            This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 7 at 13:39

























                            answered Nov 7 at 13:27









                            badjohn

                            3,8341619




                            3,8341619








                            • 1




                              Can you elaborate your answer with something concrete?
                              – Michal Dvořák
                              Nov 7 at 13:28










                            • I expanded my answer.
                              – badjohn
                              Nov 7 at 13:39






                            • 1




                              Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                              – badjohn
                              Nov 7 at 15:43








                            • 3




                              You might just define $x^y$ at this point.
                              – Ori Gurel-Gurevich
                              Nov 7 at 16:04






                            • 1




                              True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                              – badjohn
                              Nov 7 at 16:31
















                            • 1




                              Can you elaborate your answer with something concrete?
                              – Michal Dvořák
                              Nov 7 at 13:28










                            • I expanded my answer.
                              – badjohn
                              Nov 7 at 13:39






                            • 1




                              Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                              – badjohn
                              Nov 7 at 15:43








                            • 3




                              You might just define $x^y$ at this point.
                              – Ori Gurel-Gurevich
                              Nov 7 at 16:04






                            • 1




                              True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                              – badjohn
                              Nov 7 at 16:31










                            1




                            1




                            Can you elaborate your answer with something concrete?
                            – Michal Dvořák
                            Nov 7 at 13:28




                            Can you elaborate your answer with something concrete?
                            – Michal Dvořák
                            Nov 7 at 13:28












                            I expanded my answer.
                            – badjohn
                            Nov 7 at 13:39




                            I expanded my answer.
                            – badjohn
                            Nov 7 at 13:39




                            1




                            1




                            Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                            – badjohn
                            Nov 7 at 15:43






                            Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$.
                            – badjohn
                            Nov 7 at 15:43






                            3




                            3




                            You might just define $x^y$ at this point.
                            – Ori Gurel-Gurevich
                            Nov 7 at 16:04




                            You might just define $x^y$ at this point.
                            – Ori Gurel-Gurevich
                            Nov 7 at 16:04




                            1




                            1




                            True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                            – badjohn
                            Nov 7 at 16:31






                            True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$.
                            – badjohn
                            Nov 7 at 16:31












                            up vote
                            6
                            down vote













                            Throw $n$ balls into $n$ bins uniformly at random, and take $n to infty$. Define $frac{1}{e}$ to be the limiting fraction of empty bins.



                            A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-frac{1}{e}$ to be the fraction of distance covered after one unit of time.



                            Given positive $x$, consider a set of independent Bernoulli random variables with $sum_{i=1}^n p_i = x$. As $n to infty$ and $max_i p_i to 0$, define $e^{-x}$ to be the probability that all are zero.






                            share|cite|improve this answer

























                              up vote
                              6
                              down vote













                              Throw $n$ balls into $n$ bins uniformly at random, and take $n to infty$. Define $frac{1}{e}$ to be the limiting fraction of empty bins.



                              A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-frac{1}{e}$ to be the fraction of distance covered after one unit of time.



                              Given positive $x$, consider a set of independent Bernoulli random variables with $sum_{i=1}^n p_i = x$. As $n to infty$ and $max_i p_i to 0$, define $e^{-x}$ to be the probability that all are zero.






                              share|cite|improve this answer























                                up vote
                                6
                                down vote










                                up vote
                                6
                                down vote









                                Throw $n$ balls into $n$ bins uniformly at random, and take $n to infty$. Define $frac{1}{e}$ to be the limiting fraction of empty bins.



                                A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-frac{1}{e}$ to be the fraction of distance covered after one unit of time.



                                Given positive $x$, consider a set of independent Bernoulli random variables with $sum_{i=1}^n p_i = x$. As $n to infty$ and $max_i p_i to 0$, define $e^{-x}$ to be the probability that all are zero.






                                share|cite|improve this answer












                                Throw $n$ balls into $n$ bins uniformly at random, and take $n to infty$. Define $frac{1}{e}$ to be the limiting fraction of empty bins.



                                A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-frac{1}{e}$ to be the fraction of distance covered after one unit of time.



                                Given positive $x$, consider a set of independent Bernoulli random variables with $sum_{i=1}^n p_i = x$. As $n to infty$ and $max_i p_i to 0$, define $e^{-x}$ to be the probability that all are zero.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 2 days ago









                                usul

                                1,5131421




                                1,5131421






















                                    up vote
                                    5
                                    down vote













                                    If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{nrightarrow infty}left[frac{n!}{n^nsqrt{(2n+frac13pi}}right]^{frac xn}$






                                    share|cite|improve this answer



























                                      up vote
                                      5
                                      down vote













                                      If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{nrightarrow infty}left[frac{n!}{n^nsqrt{(2n+frac13pi}}right]^{frac xn}$






                                      share|cite|improve this answer

























                                        up vote
                                        5
                                        down vote










                                        up vote
                                        5
                                        down vote









                                        If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{nrightarrow infty}left[frac{n!}{n^nsqrt{(2n+frac13pi}}right]^{frac xn}$






                                        share|cite|improve this answer














                                        If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{nrightarrow infty}left[frac{n!}{n^nsqrt{(2n+frac13pi}}right]^{frac xn}$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 7 at 22:31

























                                        answered Nov 7 at 22:04









                                        Acccumulation

                                        6,4022616




                                        6,4022616






















                                            up vote
                                            4
                                            down vote













                                            You can also define the exponential function like this:
                                            $$
                                            e^x = lim_{ntoinfty} frac{f_n(x)}{f_n(-x)}
                                            $$

                                            where
                                            $$
                                            f_n(x) = sum_{j=0}^n frac{(2n-j)!}{j!(n-j)!}x^j
                                            $$






                                            share|cite|improve this answer



















                                            • 3




                                              I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                              – Michal Dvořák
                                              2 days ago















                                            up vote
                                            4
                                            down vote













                                            You can also define the exponential function like this:
                                            $$
                                            e^x = lim_{ntoinfty} frac{f_n(x)}{f_n(-x)}
                                            $$

                                            where
                                            $$
                                            f_n(x) = sum_{j=0}^n frac{(2n-j)!}{j!(n-j)!}x^j
                                            $$






                                            share|cite|improve this answer



















                                            • 3




                                              I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                              – Michal Dvořák
                                              2 days ago













                                            up vote
                                            4
                                            down vote










                                            up vote
                                            4
                                            down vote









                                            You can also define the exponential function like this:
                                            $$
                                            e^x = lim_{ntoinfty} frac{f_n(x)}{f_n(-x)}
                                            $$

                                            where
                                            $$
                                            f_n(x) = sum_{j=0}^n frac{(2n-j)!}{j!(n-j)!}x^j
                                            $$






                                            share|cite|improve this answer














                                            You can also define the exponential function like this:
                                            $$
                                            e^x = lim_{ntoinfty} frac{f_n(x)}{f_n(-x)}
                                            $$

                                            where
                                            $$
                                            f_n(x) = sum_{j=0}^n frac{(2n-j)!}{j!(n-j)!}x^j
                                            $$







                                            share|cite|improve this answer














                                            share|cite|improve this answer



                                            share|cite|improve this answer








                                            edited 2 days ago









                                            Dominique

                                            29627




                                            29627










                                            answered 2 days ago









                                            Glen O

                                            8,8481528




                                            8,8481528








                                            • 3




                                              I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                              – Michal Dvořák
                                              2 days ago














                                            • 3




                                              I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                              – Michal Dvořák
                                              2 days ago








                                            3




                                            3




                                            I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                            – Michal Dvořák
                                            2 days ago




                                            I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions?
                                            – Michal Dvořák
                                            2 days ago










                                            up vote
                                            4
                                            down vote













                                            Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define
                                            $$
                                            frac{1}{e}=lim_{ntoinfty}frac{D_n}{n!}.
                                            $$






                                            share|cite|improve this answer



























                                              up vote
                                              4
                                              down vote













                                              Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define
                                              $$
                                              frac{1}{e}=lim_{ntoinfty}frac{D_n}{n!}.
                                              $$






                                              share|cite|improve this answer

























                                                up vote
                                                4
                                                down vote










                                                up vote
                                                4
                                                down vote









                                                Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define
                                                $$
                                                frac{1}{e}=lim_{ntoinfty}frac{D_n}{n!}.
                                                $$






                                                share|cite|improve this answer














                                                Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define
                                                $$
                                                frac{1}{e}=lim_{ntoinfty}frac{D_n}{n!}.
                                                $$







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                answered 2 days ago


























                                                community wiki





                                                Foobaz John























                                                    up vote
                                                    4
                                                    down vote













                                                    EDIT: thanks to @HagenvonEitzen's comment.



                                                    I've often wondered if the following is sufficient for a general power function:



                                                    $$f(x+y) = f(x)f(y)$$



                                                    And then:



                                                    $$f(1) = e$$



                                                    for the base.



                                                    Think this is probably similar to @badjohn's answer.



                                                    EDIT: thanks to @CarstenS and @R



                                                    Turns out we must also demand that $f(x)$ is continuous or measurable.






                                                    share|cite|improve this answer



















                                                    • 6




                                                      Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                      – Carsten S
                                                      Nov 7 at 16:44








                                                    • 1




                                                      @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                      – R..
                                                      Nov 8 at 2:28






                                                    • 2




                                                      @R.., yes I should have written "not without additional conditions".
                                                      – Carsten S
                                                      2 days ago






                                                    • 2




                                                      $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                      – Hagen von Eitzen
                                                      2 days ago















                                                    up vote
                                                    4
                                                    down vote













                                                    EDIT: thanks to @HagenvonEitzen's comment.



                                                    I've often wondered if the following is sufficient for a general power function:



                                                    $$f(x+y) = f(x)f(y)$$



                                                    And then:



                                                    $$f(1) = e$$



                                                    for the base.



                                                    Think this is probably similar to @badjohn's answer.



                                                    EDIT: thanks to @CarstenS and @R



                                                    Turns out we must also demand that $f(x)$ is continuous or measurable.






                                                    share|cite|improve this answer



















                                                    • 6




                                                      Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                      – Carsten S
                                                      Nov 7 at 16:44








                                                    • 1




                                                      @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                      – R..
                                                      Nov 8 at 2:28






                                                    • 2




                                                      @R.., yes I should have written "not without additional conditions".
                                                      – Carsten S
                                                      2 days ago






                                                    • 2




                                                      $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                      – Hagen von Eitzen
                                                      2 days ago













                                                    up vote
                                                    4
                                                    down vote










                                                    up vote
                                                    4
                                                    down vote









                                                    EDIT: thanks to @HagenvonEitzen's comment.



                                                    I've often wondered if the following is sufficient for a general power function:



                                                    $$f(x+y) = f(x)f(y)$$



                                                    And then:



                                                    $$f(1) = e$$



                                                    for the base.



                                                    Think this is probably similar to @badjohn's answer.



                                                    EDIT: thanks to @CarstenS and @R



                                                    Turns out we must also demand that $f(x)$ is continuous or measurable.






                                                    share|cite|improve this answer














                                                    EDIT: thanks to @HagenvonEitzen's comment.



                                                    I've often wondered if the following is sufficient for a general power function:



                                                    $$f(x+y) = f(x)f(y)$$



                                                    And then:



                                                    $$f(1) = e$$



                                                    for the base.



                                                    Think this is probably similar to @badjohn's answer.



                                                    EDIT: thanks to @CarstenS and @R



                                                    Turns out we must also demand that $f(x)$ is continuous or measurable.







                                                    share|cite|improve this answer














                                                    share|cite|improve this answer



                                                    share|cite|improve this answer








                                                    edited 8 hours ago

























                                                    answered Nov 7 at 15:37









                                                    Paul Evans

                                                    23137




                                                    23137








                                                    • 6




                                                      Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                      – Carsten S
                                                      Nov 7 at 16:44








                                                    • 1




                                                      @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                      – R..
                                                      Nov 8 at 2:28






                                                    • 2




                                                      @R.., yes I should have written "not without additional conditions".
                                                      – Carsten S
                                                      2 days ago






                                                    • 2




                                                      $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                      – Hagen von Eitzen
                                                      2 days ago














                                                    • 6




                                                      Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                      – Carsten S
                                                      Nov 7 at 16:44








                                                    • 1




                                                      @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                      – R..
                                                      Nov 8 at 2:28






                                                    • 2




                                                      @R.., yes I should have written "not without additional conditions".
                                                      – Carsten S
                                                      2 days ago






                                                    • 2




                                                      $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                      – Hagen von Eitzen
                                                      2 days ago








                                                    6




                                                    6




                                                    Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                    – Carsten S
                                                    Nov 7 at 16:44






                                                    Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = exp(T(x))$.
                                                    – Carsten S
                                                    Nov 7 at 16:44






                                                    1




                                                    1




                                                    @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                    – R..
                                                    Nov 8 at 2:28




                                                    @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity).
                                                    – R..
                                                    Nov 8 at 2:28




                                                    2




                                                    2




                                                    @R.., yes I should have written "not without additional conditions".
                                                    – Carsten S
                                                    2 days ago




                                                    @R.., yes I should have written "not without additional conditions".
                                                    – Carsten S
                                                    2 days ago




                                                    2




                                                    2




                                                    $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                    – Hagen von Eitzen
                                                    2 days ago




                                                    $f(-x)=frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero
                                                    – Hagen von Eitzen
                                                    2 days ago










                                                    up vote
                                                    2
                                                    down vote













                                                    $e^x := cos(-ix) + i sin(-ix)$



                                                    (See Euler's formula)






                                                    share|cite|improve this answer










                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                                    • Is it possible to define the sine and cosine of a complex number without using exponential?
                                                      – Surb
                                                      Nov 7 at 23:08










                                                    • That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                      – Ben Jones
                                                      Nov 7 at 23:47






                                                    • 7




                                                      Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                      – weee
                                                      Nov 7 at 23:48






                                                    • 3




                                                      So we keep running in circles and going "@weee..."? =)
                                                      – user21820
                                                      2 days ago






                                                    • 5




                                                      @weee Where else would one run when dealing with sin and cos? :)
                                                      – Hagen von Eitzen
                                                      2 days ago















                                                    up vote
                                                    2
                                                    down vote













                                                    $e^x := cos(-ix) + i sin(-ix)$



                                                    (See Euler's formula)






                                                    share|cite|improve this answer










                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.


















                                                    • Is it possible to define the sine and cosine of a complex number without using exponential?
                                                      – Surb
                                                      Nov 7 at 23:08










                                                    • That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                      – Ben Jones
                                                      Nov 7 at 23:47






                                                    • 7




                                                      Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                      – weee
                                                      Nov 7 at 23:48






                                                    • 3




                                                      So we keep running in circles and going "@weee..."? =)
                                                      – user21820
                                                      2 days ago






                                                    • 5




                                                      @weee Where else would one run when dealing with sin and cos? :)
                                                      – Hagen von Eitzen
                                                      2 days ago













                                                    up vote
                                                    2
                                                    down vote










                                                    up vote
                                                    2
                                                    down vote









                                                    $e^x := cos(-ix) + i sin(-ix)$



                                                    (See Euler's formula)






                                                    share|cite|improve this answer










                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    $e^x := cos(-ix) + i sin(-ix)$



                                                    (See Euler's formula)







                                                    share|cite|improve this answer










                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    share|cite|improve this answer



                                                    share|cite|improve this answer








                                                    edited Nov 7 at 17:33





















                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    answered Nov 7 at 16:06









                                                    Ben Jones

                                                    1374




                                                    1374




                                                    New contributor




                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.





                                                    New contributor





                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    Ben Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.












                                                    • Is it possible to define the sine and cosine of a complex number without using exponential?
                                                      – Surb
                                                      Nov 7 at 23:08










                                                    • That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                      – Ben Jones
                                                      Nov 7 at 23:47






                                                    • 7




                                                      Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                      – weee
                                                      Nov 7 at 23:48






                                                    • 3




                                                      So we keep running in circles and going "@weee..."? =)
                                                      – user21820
                                                      2 days ago






                                                    • 5




                                                      @weee Where else would one run when dealing with sin and cos? :)
                                                      – Hagen von Eitzen
                                                      2 days ago


















                                                    • Is it possible to define the sine and cosine of a complex number without using exponential?
                                                      – Surb
                                                      Nov 7 at 23:08










                                                    • That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                      – Ben Jones
                                                      Nov 7 at 23:47






                                                    • 7




                                                      Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                      – weee
                                                      Nov 7 at 23:48






                                                    • 3




                                                      So we keep running in circles and going "@weee..."? =)
                                                      – user21820
                                                      2 days ago






                                                    • 5




                                                      @weee Where else would one run when dealing with sin and cos? :)
                                                      – Hagen von Eitzen
                                                      2 days ago
















                                                    Is it possible to define the sine and cosine of a complex number without using exponential?
                                                    – Surb
                                                    Nov 7 at 23:08




                                                    Is it possible to define the sine and cosine of a complex number without using exponential?
                                                    – Surb
                                                    Nov 7 at 23:08












                                                    That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                    – Ben Jones
                                                    Nov 7 at 23:47




                                                    That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions?
                                                    – Ben Jones
                                                    Nov 7 at 23:47




                                                    7




                                                    7




                                                    Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                    – weee
                                                    Nov 7 at 23:48




                                                    Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles...
                                                    – weee
                                                    Nov 7 at 23:48




                                                    3




                                                    3




                                                    So we keep running in circles and going "@weee..."? =)
                                                    – user21820
                                                    2 days ago




                                                    So we keep running in circles and going "@weee..."? =)
                                                    – user21820
                                                    2 days ago




                                                    5




                                                    5




                                                    @weee Where else would one run when dealing with sin and cos? :)
                                                    – Hagen von Eitzen
                                                    2 days ago




                                                    @weee Where else would one run when dealing with sin and cos? :)
                                                    – Hagen von Eitzen
                                                    2 days ago










                                                    up vote
                                                    2
                                                    down vote













                                                    Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:



                                                    Suppose we have $frac{dy}{dx}=ky.$ And we have $y(0)=1$.



                                                    By Euler's Method:
                                                    Pick some small $Delta x$. Then Let:
                                                    $$x_{n+1}=x_n+Delta x$$
                                                    $$y_{n+1}=y_n+frac{dy}{dx}Delta x=y_n+(ky_n)Delta x$$



                                                    With some rearranging:
                                                    $$y_{n+1}=y_n(1+kDelta x)implies y_{n+p}=y_n(1+kDelta x)^p$$
                                                    $$x_{n+p}=x_n+pDelta x $$



                                                    Now we have that $x_0$=0 and $y_0=1$. So:
                                                    $$x_{n+p}=(0)+pDelta x$$
                                                    $$y_{n+p}=(1)(1+kDelta x)^p$$



                                                    Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $Delta x$ to be $z/p$.



                                                    Then letting $n=0$,



                                                    $$y_p=(1+kfrac{z}{p})^p$$



                                                    By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.






                                                    share|cite|improve this answer

























                                                      up vote
                                                      2
                                                      down vote













                                                      Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:



                                                      Suppose we have $frac{dy}{dx}=ky.$ And we have $y(0)=1$.



                                                      By Euler's Method:
                                                      Pick some small $Delta x$. Then Let:
                                                      $$x_{n+1}=x_n+Delta x$$
                                                      $$y_{n+1}=y_n+frac{dy}{dx}Delta x=y_n+(ky_n)Delta x$$



                                                      With some rearranging:
                                                      $$y_{n+1}=y_n(1+kDelta x)implies y_{n+p}=y_n(1+kDelta x)^p$$
                                                      $$x_{n+p}=x_n+pDelta x $$



                                                      Now we have that $x_0$=0 and $y_0=1$. So:
                                                      $$x_{n+p}=(0)+pDelta x$$
                                                      $$y_{n+p}=(1)(1+kDelta x)^p$$



                                                      Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $Delta x$ to be $z/p$.



                                                      Then letting $n=0$,



                                                      $$y_p=(1+kfrac{z}{p})^p$$



                                                      By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.






                                                      share|cite|improve this answer























                                                        up vote
                                                        2
                                                        down vote










                                                        up vote
                                                        2
                                                        down vote









                                                        Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:



                                                        Suppose we have $frac{dy}{dx}=ky.$ And we have $y(0)=1$.



                                                        By Euler's Method:
                                                        Pick some small $Delta x$. Then Let:
                                                        $$x_{n+1}=x_n+Delta x$$
                                                        $$y_{n+1}=y_n+frac{dy}{dx}Delta x=y_n+(ky_n)Delta x$$



                                                        With some rearranging:
                                                        $$y_{n+1}=y_n(1+kDelta x)implies y_{n+p}=y_n(1+kDelta x)^p$$
                                                        $$x_{n+p}=x_n+pDelta x $$



                                                        Now we have that $x_0$=0 and $y_0=1$. So:
                                                        $$x_{n+p}=(0)+pDelta x$$
                                                        $$y_{n+p}=(1)(1+kDelta x)^p$$



                                                        Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $Delta x$ to be $z/p$.



                                                        Then letting $n=0$,



                                                        $$y_p=(1+kfrac{z}{p})^p$$



                                                        By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.






                                                        share|cite|improve this answer












                                                        Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:



                                                        Suppose we have $frac{dy}{dx}=ky.$ And we have $y(0)=1$.



                                                        By Euler's Method:
                                                        Pick some small $Delta x$. Then Let:
                                                        $$x_{n+1}=x_n+Delta x$$
                                                        $$y_{n+1}=y_n+frac{dy}{dx}Delta x=y_n+(ky_n)Delta x$$



                                                        With some rearranging:
                                                        $$y_{n+1}=y_n(1+kDelta x)implies y_{n+p}=y_n(1+kDelta x)^p$$
                                                        $$x_{n+p}=x_n+pDelta x $$



                                                        Now we have that $x_0$=0 and $y_0=1$. So:
                                                        $$x_{n+p}=(0)+pDelta x$$
                                                        $$y_{n+p}=(1)(1+kDelta x)^p$$



                                                        Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $Delta x$ to be $z/p$.



                                                        Then letting $n=0$,



                                                        $$y_p=(1+kfrac{z}{p})^p$$



                                                        By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered Nov 7 at 17:35









                                                        TurlocTheRed

                                                        42718




                                                        42718






















                                                            up vote
                                                            1
                                                            down vote













                                                            $f(x) = e^x > 0$ is such a function that



                                                            $$ intlimits_{1}^{f(x)} tfrac{1}{u} , mathrm{d}u = x$$






                                                            share|cite|improve this answer

























                                                              up vote
                                                              1
                                                              down vote













                                                              $f(x) = e^x > 0$ is such a function that



                                                              $$ intlimits_{1}^{f(x)} tfrac{1}{u} , mathrm{d}u = x$$






                                                              share|cite|improve this answer























                                                                up vote
                                                                1
                                                                down vote










                                                                up vote
                                                                1
                                                                down vote









                                                                $f(x) = e^x > 0$ is such a function that



                                                                $$ intlimits_{1}^{f(x)} tfrac{1}{u} , mathrm{d}u = x$$






                                                                share|cite|improve this answer












                                                                $f(x) = e^x > 0$ is such a function that



                                                                $$ intlimits_{1}^{f(x)} tfrac{1}{u} , mathrm{d}u = x$$







                                                                share|cite|improve this answer












                                                                share|cite|improve this answer



                                                                share|cite|improve this answer










                                                                answered Nov 8 at 3:00









                                                                robert bristow-johnson

                                                                204117




                                                                204117






















                                                                    up vote
                                                                    1
                                                                    down vote













                                                                    The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one.
                                                                    It is the Lie group exponential map of the latter group.






                                                                    share|cite|improve this answer



























                                                                      up vote
                                                                      1
                                                                      down vote













                                                                      The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one.
                                                                      It is the Lie group exponential map of the latter group.






                                                                      share|cite|improve this answer

























                                                                        up vote
                                                                        1
                                                                        down vote










                                                                        up vote
                                                                        1
                                                                        down vote









                                                                        The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one.
                                                                        It is the Lie group exponential map of the latter group.






                                                                        share|cite|improve this answer














                                                                        The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one.
                                                                        It is the Lie group exponential map of the latter group.







                                                                        share|cite|improve this answer














                                                                        share|cite|improve this answer



                                                                        share|cite|improve this answer








                                                                        answered 14 hours ago


























                                                                        community wiki





                                                                        Joonas Ilmavirta























                                                                            up vote
                                                                            0
                                                                            down vote













                                                                            Definition of $e$:



                                                                            $$lim_{hrightarrow 0} frac{e^h - 1}{h} = 1.$$



                                                                            Define exponents as a supremum of a set of a real number to rational powers.






                                                                            share|cite|improve this answer



















                                                                            • 2




                                                                              This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                              – Paramanand Singh
                                                                              2 days ago






                                                                            • 4




                                                                              Also, this seems to define $e$ but not $xmapsto e^x$
                                                                              – Surb
                                                                              2 days ago















                                                                            up vote
                                                                            0
                                                                            down vote













                                                                            Definition of $e$:



                                                                            $$lim_{hrightarrow 0} frac{e^h - 1}{h} = 1.$$



                                                                            Define exponents as a supremum of a set of a real number to rational powers.






                                                                            share|cite|improve this answer



















                                                                            • 2




                                                                              This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                              – Paramanand Singh
                                                                              2 days ago






                                                                            • 4




                                                                              Also, this seems to define $e$ but not $xmapsto e^x$
                                                                              – Surb
                                                                              2 days ago













                                                                            up vote
                                                                            0
                                                                            down vote










                                                                            up vote
                                                                            0
                                                                            down vote









                                                                            Definition of $e$:



                                                                            $$lim_{hrightarrow 0} frac{e^h - 1}{h} = 1.$$



                                                                            Define exponents as a supremum of a set of a real number to rational powers.






                                                                            share|cite|improve this answer














                                                                            Definition of $e$:



                                                                            $$lim_{hrightarrow 0} frac{e^h - 1}{h} = 1.$$



                                                                            Define exponents as a supremum of a set of a real number to rational powers.







                                                                            share|cite|improve this answer














                                                                            share|cite|improve this answer



                                                                            share|cite|improve this answer








                                                                            edited 2 days ago

























                                                                            answered 2 days ago









                                                                            PiKindOfGuy

                                                                            1077




                                                                            1077








                                                                            • 2




                                                                              This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                              – Paramanand Singh
                                                                              2 days ago






                                                                            • 4




                                                                              Also, this seems to define $e$ but not $xmapsto e^x$
                                                                              – Surb
                                                                              2 days ago














                                                                            • 2




                                                                              This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                              – Paramanand Singh
                                                                              2 days ago






                                                                            • 4




                                                                              Also, this seems to define $e$ but not $xmapsto e^x$
                                                                              – Surb
                                                                              2 days ago








                                                                            2




                                                                            2




                                                                            This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                            – Paramanand Singh
                                                                            2 days ago




                                                                            This needs a bit more elaboration as to how you define arbitrary real exponents.
                                                                            – Paramanand Singh
                                                                            2 days ago




                                                                            4




                                                                            4




                                                                            Also, this seems to define $e$ but not $xmapsto e^x$
                                                                            – Surb
                                                                            2 days ago




                                                                            Also, this seems to define $e$ but not $xmapsto e^x$
                                                                            – Surb
                                                                            2 days ago










                                                                            up vote
                                                                            0
                                                                            down vote













                                                                            Similar things have been said, but not in that way:



                                                                            $exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(mathbb R) to C^{k}(mathbb R)$ for some $k in [0,infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and ${exp}$ is a somewhat "canonical" basis for that space.)






                                                                            share|cite|improve this answer



























                                                                              up vote
                                                                              0
                                                                              down vote













                                                                              Similar things have been said, but not in that way:



                                                                              $exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(mathbb R) to C^{k}(mathbb R)$ for some $k in [0,infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and ${exp}$ is a somewhat "canonical" basis for that space.)






                                                                              share|cite|improve this answer

























                                                                                up vote
                                                                                0
                                                                                down vote










                                                                                up vote
                                                                                0
                                                                                down vote









                                                                                Similar things have been said, but not in that way:



                                                                                $exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(mathbb R) to C^{k}(mathbb R)$ for some $k in [0,infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and ${exp}$ is a somewhat "canonical" basis for that space.)






                                                                                share|cite|improve this answer














                                                                                Similar things have been said, but not in that way:



                                                                                $exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(mathbb R) to C^{k}(mathbb R)$ for some $k in [0,infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and ${exp}$ is a somewhat "canonical" basis for that space.)







                                                                                share|cite|improve this answer














                                                                                share|cite|improve this answer



                                                                                share|cite|improve this answer








                                                                                answered yesterday


























                                                                                community wiki





                                                                                Max























                                                                                    up vote
                                                                                    0
                                                                                    down vote













                                                                                    If we equip the one-dimensional manifold $(0,infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function.
                                                                                    This is the unique metric that makes the exponential function an exponential map.






                                                                                    share|cite|improve this answer



























                                                                                      up vote
                                                                                      0
                                                                                      down vote













                                                                                      If we equip the one-dimensional manifold $(0,infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function.
                                                                                      This is the unique metric that makes the exponential function an exponential map.






                                                                                      share|cite|improve this answer

























                                                                                        up vote
                                                                                        0
                                                                                        down vote










                                                                                        up vote
                                                                                        0
                                                                                        down vote









                                                                                        If we equip the one-dimensional manifold $(0,infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function.
                                                                                        This is the unique metric that makes the exponential function an exponential map.






                                                                                        share|cite|improve this answer














                                                                                        If we equip the one-dimensional manifold $(0,infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function.
                                                                                        This is the unique metric that makes the exponential function an exponential map.







                                                                                        share|cite|improve this answer














                                                                                        share|cite|improve this answer



                                                                                        share|cite|improve this answer








                                                                                        answered 13 hours ago


























                                                                                        community wiki





                                                                                        Joonas Ilmavirta


















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