Error in using Inline function and fzero

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Venkatesh M Deshpande 2014-3-20

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https://ww2.mathworks.cn/matlabcentral/answers/122458-error-in-using-inline-function-and-fzero

评论： Venkatesh M Deshpande 2014-3-23

Hello, Can anybody help me? I want to figure out where I am doing a mistake in program. The only variable is X and Cp is an input.

Cp                          = double(zeros(49792,1));
P                           = size(x.Time_Series);
Q                           = P(1,1);
Cp(1,1)= 0.126826003999670;
Cp(2,1)= 0.325943036562508;
for k=3:Q;
    k
    hx = inline( '12.175*(X - 2*Cp(k-1,1) + Cp(k-2,1)) + 13.525*((X - Cp(k-1,1))* abs(X - Cp(k-1,1))) + X - SINGLE_FINAL_TIME_SERIES_OF_TAPS_NEAR_DOMINANT_OPENING(k,1)','X',Cp);
    hp = fzero (hx,0,Cp,SINGLE_FINAL_TIME_SERIES_OF_TAPS_NEAR_DOMINANT_OPENING(k,1));
    Cp(k) = hp(1);
end
I am getting this error
Error: The expression to the left of the equals sign is not a valid target for an
assignment.

If in inline function, I remove Cp which is given after X (at the very last near the bracket), then it shows error: too many inputs to inline function. Pls help. thanks.

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采纳的回答

Star Strider 2014-3-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/122458-error-in-using-inline-function-and-fzero#answer_129411

It doesn’t like the ‘==’ in SINGLE_FINAL_TIME_SERIES_OF_TAPS_NEAR_DOMINANT_OPENING(k,1)==0.

Unlike anonymous functions that use workspace variables, inline functions ignore anything that isn’t in their argument list. It has no idea what SINGLE_FINAL_TIME_SERIES_OF_TAPS_NEAR_DOMINANT_OPENING(k,1) or k are.

Also, fzero only accepts single-variable functions.

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Star Strider 2014-3-22

编辑：Star Strider 2014-3-22

在 MATLAB Online 中打开

Does this produce the result you want? The plot looks strange, but then I don’t know what you’re researching. (I plotted it to see if it looked reasonable.)

SINGLE_FINAL = [1.88848000000000 1.74232000000000 1.54512000000000 1.46624000000000 1.20408000000000 0.874640000000000 0.570720000000000 1.13216000000000 1.31544000000000 1.43608000000000 1.63328000000000 1.42912000000000 1.22032000000000 1.49640000000000 1.63328000000000 1.71680000000000 1.51728000000000 0.965120000000000 0.468640000000000 0.740080000000000];
Cp(1)= 0.126826003999670;
Cp(2)= 0.325943036562508;
% hx = inline( '12.175*(X - 2*Cp(k-1,1) + Cp(k-2,1)) + 13.525*((X - Cp(k-1,1))* abs(X - Cp(k-1,1))) + X - SINGLE_FINAL_TIME_SERIES_OF_TAPS_NEAR_DOMINANT_OPENING(k,1)','X',Cp);
% CALLING ‘f1’:
%   Define cp = [Cp(k-1); Cp(k-2)]
%   f1(x,cp,SF(k))
f1 = @(X,cp,SFTSOTNDO) 12.175 .* (X - 2 .* cp(1) + cp(2)) + 13.525 .* ((X - cp(1)) .* abs(X - cp(1))) + X - SFTSOTNDO;
for k = 3:length(SINGLE_FINAL)
    cp = [Cp(k-1); Cp(k-2)];
    sf = SINGLE_FINAL(k);
    X = fzero(@(x)(f1(x,cp,sf)),1);
    Cp(k) = X;
end
figure(1)
plot(SINGLE_FINAL, Cp)
grid
xlabel('SINGLE FINAL')
ylabel('Cp')

I kept your inline function in for reference, but commented it out.

I also vectorised f1 because that generally avoids problems.

Star Strider 2014-3-23

My pleasure!

Wow! And a tribute, too! Thank you!

I have a background in physical chemistry and am an instrument-rated private pilot, so I'm slightly familiar with what you are doing. I do not have sufficient background in fluid dynamics to understand it in any detail, though.

Venkatesh M Deshpande 2014-3-23