墨卡托級數




在數學內，墨卡托級數(Mercator series)或者牛頓-墨卡托級數(Newton–Mercator series)是一個自然對數的泰勒級數：


ln⁡(1+x)=x−x22+x33−x44+⋯.{displaystyle ln(1+x);=;x,-,{frac {x^{2}}{2}},+,{frac {x^{3}}{3}},-,{frac {x^{4}}{4}},+,cdots .}{displaystyle ln(1+x);=;x,-,{frac {x^{2}}{2}},+,{frac {x^{3}}{3}},-,{frac {x^{4}}{4}},+,cdots .}

使用大寫sigma表示則為


ln⁡(1+x)=∑n=1∞(−1)n+1nxn.{displaystyle ln(1+x);=;sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n}}x^{n}.}{displaystyle ln(1+x);=;sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n}}x^{n}.}

當 −1 < x ≤ 1時，此級數收斂於自然對數(加了1)。




目录






  • 1 歷史


  • 2 推導


  • 3 特例


  • 4 複數級數


  • 5 參考資料





歷史


這級數被尼古拉斯·墨卡托，牛頓和Gregory Saint-Vincent分別獨立發現。首先被墨卡托出版於其1668年時的著作Logarithmo-technica。



推導


這級數可以由泰勒公式導出，藉由不斷地計算第n次ln x在x = 1時的微分，一開始是


ddxln⁡x=1x.{displaystyle {frac {d}{dx}}ln x={frac {1}{x}}.}{displaystyle {frac {d}{dx}}ln x={frac {1}{x}}.}

或者，我們可以從有限的等比數列開始(t ≠ −1)


1−t+t2−⋯+(−t)n−1=1−(−t)n1+t{displaystyle 1-t+t^{2}-cdots +(-t)^{n-1}={frac {1-(-t)^{n}}{1+t}}}{displaystyle 1-t+t^{2}-cdots +(-t)^{n-1}={frac {1-(-t)^{n}}{1+t}}}

這可以導出


11+t=1−t+t2−⋯+(−t)n−1+(−t)n1+t.{displaystyle {frac {1}{1+t}}=1-t+t^{2}-cdots +(-t)^{n-1}+{frac {(-t)^{n}}{1+t}}.}{displaystyle {frac {1}{1+t}}=1-t+t^{2}-cdots +(-t)^{n-1}+{frac {(-t)^{n}}{1+t}}.}

然後得到


∫0xdt1+t=∫0x(1−t+t2−⋯+(−t)n−1+(−t)n1+t)dt{displaystyle int _{0}^{x}{frac {dt}{1+t}}=int _{0}^{x}left(1-t+t^{2}-cdots +(-t)^{n-1}+{frac {(-t)^{n}}{1+t}}right),dt}{displaystyle int _{0}^{x}{frac {dt}{1+t}}=int _{0}^{x}left(1-t+t^{2}-cdots +(-t)^{n-1}+{frac {(-t)^{n}}{1+t}}right),dt}

接著逐項積分，


ln⁡(1+x)=x−x22+x33−⋯+(−1)n−1xnn+(−1)n∫0xtn1+tdt.{displaystyle ln(1+x)=x-{frac {x^{2}}{2}}+{frac {x^{3}}{3}}-cdots +(-1)^{n-1}{frac {x^{n}}{n}}+(-1)^{n}int _{0}^{x}{frac {t^{n}}{1+t}},dt.}{displaystyle ln(1+x)=x-{frac {x^{2}}{2}}+{frac {x^{3}}{3}}-cdots +(-1)^{n-1}{frac {x^{n}}{n}}+(-1)^{n}int _{0}^{x}{frac {t^{n}}{1+t}},dt.}

若−1 < x ≤ 1，餘項會在n→∞{displaystyle nto infty }n to infty時趨近於零。




這個表示法可以重複積分k次，得到


−xAk(x)+Bk(x)ln⁡(1+x)=∑n=1∞(−1)n−1xn+kn(n+1)⋯(n+k),{displaystyle -xA_{k}(x)+B_{k}(x)ln(1+x)=sum _{n=1}^{infty }(-1)^{n-1}{frac {x^{n+k}}{n(n+1)cdots (n+k)}},}{displaystyle -xA_{k}(x)+B_{k}(x)ln(1+x)=sum _{n=1}^{infty }(-1)^{n-1}{frac {x^{n+k}}{n(n+1)cdots (n+k)}},}



這裡的


Ak(x)=1k!∑m=0k(km)xm∑l=1k−m(−x)l−1l{displaystyle A_{k}(x)={frac {1}{k!}}sum _{m=0}^{k}{k choose m}x^{m}sum _{l=1}^{k-m}{frac {(-x)^{l-1}}{l}}}{displaystyle A_{k}(x)={frac {1}{k!}}sum _{m=0}^{k}{k choose m}x^{m}sum _{l=1}^{k-m}{frac {(-x)^{l-1}}{l}}}

和


Bk(x)=1k!(1+x)k{displaystyle B_{k}(x)={frac {1}{k!}}(1+x)^{k}}{displaystyle B_{k}(x)={frac {1}{k!}}(1+x)^{k}}

都是x的多項式。[1]



特例


令墨卡托級數裡面的x = 1，則我們會得到交錯調和級數


∑k=1∞(−1)k+1k=ln⁡2.{displaystyle sum _{k=1}^{infty }{frac {(-1)^{k+1}}{k}}=ln 2.}{displaystyle sum _{k=1}^{infty }{frac {(-1)^{k+1}}{k}}=ln 2.}


複數級數


下面的複數冪級數


z−z22+z33−z44+⋯{displaystyle z,-,{frac {z^{2}}{2}},+,{frac {z^{3}}{3}},-,{frac {z^{4}}{4}},+,cdots }{displaystyle z,-,{frac {z^{2}}{2}},+,{frac {z^{3}}{3}},-,{frac {z^{4}}{4}},+,cdots }

是ln(1 + z)的泰勒級數，這裡ln代表複對數(complex logarithm)的 主要分支(principal branch)。這個級數收斂於一個開放的單位圓盤 |z| < 1 以及圓 |z| = 1 ， z = -1除外 (根據阿貝爾判別法)，而且這裡的收斂對每個半徑小於一的圓盤是一致的 。



參考資料





  1. ^ Medina, Luis A.; Moll, Victor H.; Rowland, Eric S. Iterated primitives of logarithmic powers. 2009. arXiv:0911.1325. 




  • 埃里克·韦斯坦因. Mercator Series. MathWorld. 

  • Eriksson, Larsson & Wahde. Matematisk analys med tillämpningar, part 3. Gothenburg 2002. p. 10.


  • Some Contemporaries of Descartes, Fermat, Pascal and Huygens from A Short Account of the History of Mathematics (4th edition, 1908) by W. W. Rouse Ball




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