Plotting the sum of series
up vote
0
down vote
favorite
I use this code and i don't know what it needs to work for my problem:
syms x k t
for t=0:10
num=((-1)^k)/k
t1=sin(8*3.1415*k*t)
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
end
Plot(x)
On the x axis I show time and on the y axis I show the function over four periods.
When I try to run the code I get the following error:
Undefined function 'symsum' for input arguments of type 'double'.
Maybe I can't use symsum with my argument type, but is there another function I can use? Sum also didn't work:
Error using sum Dimension argument must be a positive integer scalar within indexing range.
matlab plot matlab-figure matlab-guide
add a comment |
up vote
0
down vote
favorite
I use this code and i don't know what it needs to work for my problem:
syms x k t
for t=0:10
num=((-1)^k)/k
t1=sin(8*3.1415*k*t)
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
end
Plot(x)
On the x axis I show time and on the y axis I show the function over four periods.
When I try to run the code I get the following error:
Undefined function 'symsum' for input arguments of type 'double'.
Maybe I can't use symsum with my argument type, but is there another function I can use? Sum also didn't work:
Error using sum Dimension argument must be a positive integer scalar within indexing range.
matlab plot matlab-figure matlab-guide
Does N=2 in the equation so that thesymsum
ranges from 1 to 2?
– Banghua Zhao
Nov 11 at 0:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I use this code and i don't know what it needs to work for my problem:
syms x k t
for t=0:10
num=((-1)^k)/k
t1=sin(8*3.1415*k*t)
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
end
Plot(x)
On the x axis I show time and on the y axis I show the function over four periods.
When I try to run the code I get the following error:
Undefined function 'symsum' for input arguments of type 'double'.
Maybe I can't use symsum with my argument type, but is there another function I can use? Sum also didn't work:
Error using sum Dimension argument must be a positive integer scalar within indexing range.
matlab plot matlab-figure matlab-guide
I use this code and i don't know what it needs to work for my problem:
syms x k t
for t=0:10
num=((-1)^k)/k
t1=sin(8*3.1415*k*t)
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
end
Plot(x)
On the x axis I show time and on the y axis I show the function over four periods.
When I try to run the code I get the following error:
Undefined function 'symsum' for input arguments of type 'double'.
Maybe I can't use symsum with my argument type, but is there another function I can use? Sum also didn't work:
Error using sum Dimension argument must be a positive integer scalar within indexing range.
matlab plot matlab-figure matlab-guide
matlab plot matlab-figure matlab-guide
edited Nov 11 at 10:10
SecretAgentMan
508113
508113
asked Nov 11 at 0:23
Alex Mihai
32
32
Does N=2 in the equation so that thesymsum
ranges from 1 to 2?
– Banghua Zhao
Nov 11 at 0:52
add a comment |
Does N=2 in the equation so that thesymsum
ranges from 1 to 2?
– Banghua Zhao
Nov 11 at 0:52
Does N=2 in the equation so that the
symsum
ranges from 1 to 2?– Banghua Zhao
Nov 11 at 0:52
Does N=2 in the equation so that the
symsum
ranges from 1 to 2?– Banghua Zhao
Nov 11 at 0:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Since you want to plot x(t)
, you need to use plot(t,x)
where t
and x
are vectors.
Instead of using for t=0:10
, just let t=0:10
and calculate the corresponding x
.
Also, the symbolic variable is just k
.
syms k
t=0:10;
num=((-1)^k)/k;
t1=sin(8*3.1415*k*t);
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
plot(t,x)
It is noted that if you let t=0:10
, then the sin(8*k*pi*t)
will always be 0 since t
is a vector of the integer from 0 to 10. The result of x(t)
will be 5:
Output when t=0:10
:
As you can see, the value of x(t)
is very close to each other. Theoretically, they should all be 5. But there is some numerical approximation which leads to the small error.
You probably want non-integer t
. Here is a output when t=0:0.1:10
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since you want to plot x(t)
, you need to use plot(t,x)
where t
and x
are vectors.
Instead of using for t=0:10
, just let t=0:10
and calculate the corresponding x
.
Also, the symbolic variable is just k
.
syms k
t=0:10;
num=((-1)^k)/k;
t1=sin(8*3.1415*k*t);
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
plot(t,x)
It is noted that if you let t=0:10
, then the sin(8*k*pi*t)
will always be 0 since t
is a vector of the integer from 0 to 10. The result of x(t)
will be 5:
Output when t=0:10
:
As you can see, the value of x(t)
is very close to each other. Theoretically, they should all be 5. But there is some numerical approximation which leads to the small error.
You probably want non-integer t
. Here is a output when t=0:0.1:10
add a comment |
up vote
0
down vote
accepted
Since you want to plot x(t)
, you need to use plot(t,x)
where t
and x
are vectors.
Instead of using for t=0:10
, just let t=0:10
and calculate the corresponding x
.
Also, the symbolic variable is just k
.
syms k
t=0:10;
num=((-1)^k)/k;
t1=sin(8*3.1415*k*t);
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
plot(t,x)
It is noted that if you let t=0:10
, then the sin(8*k*pi*t)
will always be 0 since t
is a vector of the integer from 0 to 10. The result of x(t)
will be 5:
Output when t=0:10
:
As you can see, the value of x(t)
is very close to each other. Theoretically, they should all be 5. But there is some numerical approximation which leads to the small error.
You probably want non-integer t
. Here is a output when t=0:0.1:10
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since you want to plot x(t)
, you need to use plot(t,x)
where t
and x
are vectors.
Instead of using for t=0:10
, just let t=0:10
and calculate the corresponding x
.
Also, the symbolic variable is just k
.
syms k
t=0:10;
num=((-1)^k)/k;
t1=sin(8*3.1415*k*t);
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
plot(t,x)
It is noted that if you let t=0:10
, then the sin(8*k*pi*t)
will always be 0 since t
is a vector of the integer from 0 to 10. The result of x(t)
will be 5:
Output when t=0:10
:
As you can see, the value of x(t)
is very close to each other. Theoretically, they should all be 5. But there is some numerical approximation which leads to the small error.
You probably want non-integer t
. Here is a output when t=0:0.1:10
Since you want to plot x(t)
, you need to use plot(t,x)
where t
and x
are vectors.
Instead of using for t=0:10
, just let t=0:10
and calculate the corresponding x
.
Also, the symbolic variable is just k
.
syms k
t=0:10;
num=((-1)^k)/k;
t1=sin(8*3.1415*k*t);
S1=symsum((num*t1),k,1,2);
x=0.5-((1/3.1415)*S1);
plot(t,x)
It is noted that if you let t=0:10
, then the sin(8*k*pi*t)
will always be 0 since t
is a vector of the integer from 0 to 10. The result of x(t)
will be 5:
Output when t=0:10
:
As you can see, the value of x(t)
is very close to each other. Theoretically, they should all be 5. But there is some numerical approximation which leads to the small error.
You probably want non-integer t
. Here is a output when t=0:0.1:10
answered Nov 11 at 4:33
Banghua Zhao
1,095617
1,095617
add a comment |
add a comment |
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Does N=2 in the equation so that the
symsum
ranges from 1 to 2?– Banghua Zhao
Nov 11 at 0:52