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How to properly create function for integration?

조회 수: 24(최근 30일)
Kaleem Graham
Kaleem Graham 2021년 9월 19일 22:40
댓글: Star Strider 2021년 9월 21일 11:15
Hi i'm trying to create functions that I can then use to compute a definite integral, here are the functions:
\\
Integral I'm trying to compute:
The Error i'm getting:
My Attempt:
syms theta_func
z_func = @(theta_func) exp(1i*pi*cos(theta_func));
E_func = @(theta_func) sin(theta_func).*(z_func(theta_func)-1).*((z_func(theta_func).^3)-1);
E_func = @(theta_func) abs(E_func(theta_func))^2;
expr = @(theta_func) E_func(theta_func).*sin(theta_func);
q = 2*pi*int(expr,theta,0,pi);
  댓글 수: 1
Jan
Jan 2021년 9월 20일 15:46
Prefer to post code as text, not as screenshot.
You show your apporach. Does it work? We cannot try this by our own, because it is a screenshot only. Most of the readers will not take the time to type this code again, most of all because it is not clear, how this code is related to your question.

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답변(1개)

Star Strider
Star Strider 2021년 9월 20일 16:07
There are several problems with the posted code image.
First, ‘z’ is not defined anywhere, and it is not an argument to ‘E_func’, and the second ‘E_func’ line should be:
E_func = @(theta_func) abs(E_func(theta_func)).^2;
It may also not be necessary to use the Symbolic Math Toolbox for this, anyway.
.
  댓글 수: 3
Star Strider
Star Strider 2021년 9월 21일 11:15
The way the code is written means that the Symbolic Math Toolbox functions are not appropriate.
Try this —
z_func = @(theta_func) exp(1i*pi*cos(theta_func));
E_func = @(theta_func) sin(theta_func).*(z_func(theta_func)-1).*((z_func(theta_func).^3)-1);
E_func = @(theta_func) abs(E_func(theta_func)).^2;
expr = @(theta_func) E_func(theta_func).*sin(theta_func);
q = 2*pi*integral(expr,0,pi) % Use 'integral' For Numeric Integration
q = 20.6011
.

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