Finding the root of a function
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I'm trying to find x in the following equation using the fzero function.
I'm not sure whether I should use int or integral, and my code doesn't work. Any help?
My code:
N = 100;
S1 = 0;
S2 = 0;
for n = 0:N
S1 = S1 + ((-1)^n/((n + 1/2)*pi)^4*tanh((n + 1/2)*pi*(1+x+x.^2)/2));
S2 = S2 + (1/((n + 1/2)*pi)^5*tanh((n + 1/2)*pi*(1+x+x.^2)/2));
end
cs = 1/2 - 4./(1+x+x.^2).*S1;
cp = 1/3 - 4./(1+x+x.^2).*S2;
syms x
fun = @(x) cs./cp;
q = int(fun,0,x)
gx = @(x) q - 0.0062;
x0 = [0 10];
x = fzero(gx,x0);
The error:
Error using subsindex
Function 'subsindex' is not defined for values of class 'function_handle'.
Error in line 69
q = int(fun,0,x)
2 个评论
Geoff Hayes
2017-12-11
Lilach - please clarify what you mean by your code is not working. Is there an error? If so, please copy and paste the full error message here. Or, are you not getting the expected answer?
Daenerys
2017-12-11
采纳的回答
David Goodmanson
2017-12-12
编辑：David Goodmanson
2017-12-12
Hi Lilach,
After moving the syms x statement to the top so that the code runs, it is not so clear that it is going to get there. Here is a slightly different method.
A = .0062;
N = 100;
% integrate in z from 0 to x, subtract A, find root
p = fzero(@(x) integral(@(z) ratfun(z,N),0,x)-A, [0 .1])
% take a look at the integrand
z = 0:.001:1;
plot(z,ratfun(z,N))
function y = ratfun(x,N)
S1 = 0;
S2 = 0;
for n = 0:N
S1 = S1 + ((-1)^n/((n + 1/2)*pi)^4*tanh((n + 1/2)*pi*(1+x+x.^2)/2));
S2 = S2 + ( 1/((n + 1/2)*pi)^5*tanh((n + 1/2)*pi*(1+x+x.^2)/2));
end
cs = 1/2 - (4./(1+x+x.^2)).*S1;
cp = 1/3 - (4./(1+x+x.^2)).*S2;
y = cs./cp;
end
The result is p = 0.004647. From the plot, the integrand starts out at about 1.34, and the value of the integral is so small, .0062, that you would not expect the integrand to change much from 1.34. So you would expect the rectangle result p = .0062/1.34 which is pretty close.
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