how to code equation

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Cesar Cardenas 2023-2-28

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https://ww2.mathworks.cn/matlabcentral/answers/1920610-how-to-code-equation

编辑： Voss 2023-2-28

Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);

Hello, is this code right? not sure, any help will be appreciated thanks.

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Les Beckham 2023-2-28

It looks like it might be right, but that depends on the size and class of all of the variables; are they scalars, vectors, 2d arrays, something else, integers, complex, double. The details matter.

Did you try to run the code? If so, what happened? If you got an error message, cut and paste all of the text of the error message into an edit of your question, or a new comment.

Since we can't see your Matlab workspace, we can't answer your question until you share more information.

Cesar Cardenas 2023-2-28

在 MATLAB Online 中打开

I have not run the code, but I see this on the workspace "parse error at EOL"

Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);

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Answer 1

Voss 2023-2-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1920610-how-to-code-equation#answer_1182005

编辑：Voss 2023-2-28

在 MATLAB Online 中打开

Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);
%                                                                      ^^ you need another closing parenthesis here
%    ^       ^  ^       ^  ^                        ^ extraneous parentheses (that don't affect the operation)
%     ^^^^^ should be 0.332

After correcting those things:

Nu = 0.332.*Pr.^(1/3).*(rho.*Uinf.*x/mu).^(1/2).*(1 -(x0/x).^(3/4)).^(-1/3);

And some of those / may have to be ./, depending on whether x and/or mu are scalars. Doesn't hurt to just use ./