Question about diff function

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Francisco
Francisco 2014-12-9
编辑： Star Strider 2014-12-9
I have defined a function as follows
function A = A(V, h)
A = V/h + 2*sqrt(V*pi*h)
end
In another function, I try to call it:
h = h - A (V, h) / diff (A (V, h), h);
But I get the error
Error using / Matrix dimensions must agree.
I am at a loss. Can someone give me a hand?
  1 个评论
Matt J
Matt J 2014-12-9
It is dangerous to use "A" as both a variable name and the name of your function.

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回答（3 个）

Star Strider
Star Strider 2014-12-9
To take a numerical derivative most accurately, use the gradient function.
  2 个评论
Francisco
Francisco 2014-12-9
I was looking for an analytic derivative.
Star Strider
Star Strider 2014-12-9
编辑：Star Strider 2014-12-9
That requires the Symbolic Math Toolbox.
EDIT —
If you are satisfied with an approximation, you can do a numeric version of the definition of the derivative as an anonymous funciton:
dfdx = @(f,x) (f(x+1E-8)-f(x))./1E-8; % Conventional Derivative
A = @(V,h) V./h + 2*sqrt(V*pi*h);
V = 3;
h = linspace(0.1,10);
dAdh = dfdx(@(h) A(V,h), h);
figure(1)
plot(h, A(V,h), '-b')
hold on
plot(h, dAdh, '-r')
hold off
legend('A(V,h)', 'dA(V,h)/dh')
You may want to experiment with the 1E-8 value (this is the value that most often ‘works’ for me). At the very least, it will provide you with something to experiment with.

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Roger Stafford
Roger Stafford 2014-12-9
h = h - (V/h+2*sqrt(V*pi*h))/(-V/h^2+V*pi/sqrt(V*pi*h));
(Maybe I am old-fashioned, but with elementary functions like this, it's easier to use calculus than to have to mess around with the 'diff' function in my opinion.)
Azzi Abdelmalek
Azzi Abdelmalek 2014-12-9
a=[ 2 4 8]
b=diff(a)
b= [4-2 8-4]
a is 1x3 and b is 1x2
  2 个评论
Francisco
Francisco 2014-12-9
I thought diff was http://www.mathworks.com/help/symbolic/diff.html, as in, I was trying to differentiate A. How can I achieve this, then?
Azzi Abdelmalek
Azzi Abdelmalek 2014-12-9
B=A(V,h)
C=diff (A (V, h), h);
h = h - B(h+1:end)/C ;

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