Please someone can help how to plot the following numerical
integration with respect to "z". Here "z" varies from "0" to "1". The
integral need to be done by numerical integration method.

 採用された回答

Star Strider
Star Strider 2020 年 10 月 23 日

0 投票

Try this:
zv = linspace(0, 1);
for k = 1:numel(zv)
z = zv(k);
u_int(k) = integral(@(u) sqrt(u)/((exp(u)/z)-1) + z/(1-z), 0, Inf, 'ArrayValued',1);
end
figure
semilogy(zv, u_int)
grid
xlabel('z')
ylabel('Integrated Value')
.

6 件のコメント

P Rakesh Kumar Dora
P Rakesh Kumar Dora 2020 年 10 月 23 日
Thanks. Here can we change the step size of "z" in your code?
Star Strider
Star Strider 2020 年 10 月 23 日
It can be whatever you want. This plots 100 points, so the step size is about 0.01. If you want more or fewer, use the third argument to linspace to define how many.
Example —
zv = linspace(0, 1, 20);
creates ‘zv’ as a 20-element vector and would result in a step size of about 0.05. The number of elements can be anything the computer memory permits.
P Rakesh Kumar Dora
P Rakesh Kumar Dora 2020 年 10 月 23 日
Ok, Thanks. Do you know how to do this in mathmaticae.Actually i am trying by using
"Nintegrate" function in mathematica , but unable to get the final answer.
Star Strider
Star Strider 2020 年 10 月 23 日
My pleasure!
I do not have Mathematica and have no experience with it. MATLAB does essentially everything I need to do.
If my Answer helped you solve your problem, please Accept it!
.
P Rakesh Kumar Dora
P Rakesh Kumar Dora 2020 年 10 月 23 日
Please can you tell how to locate precise value of the integrand corresponding
to precise value of "z" in the plot?
Star Strider
Star Strider 2020 年 10 月 23 日
The ‘z’ values are the elements of the ‘zv’ vector. They correspond to the elements of ‘u_int’. So ‘z=0’ corresponds to ‘u_int(1)’, ‘z=1’ to ‘u_int(end)’, with end depending on how many elements of ‘u_int’ you choose to calculate.

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