how to solve non-linear transcendental equation with many roots

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David Zhang
David Zhang 2013-2-13
Hi All,
I have to solve the following equation
x-(2*(sinh(x)*((cos(x)).^2+cos(x)-1)+sin(x)*(cosh(x)-1)))/((1-cos(x)*cosh(x))) =0.
It is a nonlinear with many roots, so is there any function in matlab can do it? the function fzero only return a root near a initial root value, which is not a automatic task.
Thank you.

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回答（3 个）

Walter Roberson
Walter Roberson 2013-2-13
I suspect not, but I do not have the Symbolic Toolbox to test with. Using a different symbolic package, I find that there are an infinite number of roots, some real and some imaginary. There is no nice analytic expression for the roots.
One of the basic items repeated over and over in the forms is the root of cos(x) = x, which has no analytic solution.
The generalized solution for reals has three forms; the overall solution is the union of {0} and these three forms and three forms for the imaginary solutions. The three real forms are:
.7429621155 + 1.655668423 * B + 6.283185308 * Z
-0.7429621155 + 4.627516885 * B + 6.283185308 * Z
3.141592654 * B + 6.283185308 * Z
where B must be either 0 or 1, and Z is any arbitrary integer.
The 6.28<etc> is obviously 2*Pi, and the 3.14<etc> is obviously Pi, but the other numeric values have no obvious interpretation.
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Walter Roberson
Walter Roberson 2013-2-14
It is possible that the above solution is incorrect; I am working through some difficulties I found with the other package.
Walter Roberson
Walter Roberson 2013-2-14
I have confirmed that the solution I showed in my Answer is incorrect, but I do not have anything worked out yet as to what the correct general solution should be.

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Youssef Khmou
Youssef Khmou 2013-2-13
编辑：Youssef Khmou 2013-2-13
Hi, there many ways to get solution,
You use a function Handle :
>>f=@(x) x-(2*(sinh(x)*((cos(x))^2+cos(x)-1)+sin(x)*(cosh(x)-1)))/(1-cos(x)*cosh(x))
>>y=fsolve(f,10e-3)
'fsolve' starts at x0=10e-3 and tries to solve the equation , you can verify the solution :
>>f(y)
-2.0334e-009
This helps ?
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David Zhang
David Zhang 2013-2-14
Yes, it has infinite number of real roots. What i need is the first several results, such as the first 50 roots.

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Youssef Khmou
Youssef Khmou 2013-2-13
2nd answer :
You can use " solve" with symbolic variable :
syms x
y=x-(2*(sinh(x)*((cos(x))^2+cos(x)-1)+sin(x)*(cosh(x)-1)))/((1-cos(x)*cosh(x)))
z=solve(y) .
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David Zhang
David Zhang 2013-2-14
I guess i have got the results. pls see the following code,
clear clc close all
xx=0:0.001:100; for i=1:length(xx) x=xx(i); fx(i)=(2*(sinh(x)*((cos(x)).^2+cos(x)-1)+sin(x)*(cosh(x)-1)))/((1-cos(x)*cosh(x))); flinerx(i)=x; end
plot(xx, fx); hold on plot(xx, flinerx, 'r') axis([0, max(xx), -max(xx), max(xx)])
options=optimset('Display','iter','TolFun',1e-10); % Option to display output
xguess = [2, 5,12, 18, 24, 30, 37, 43, 49, 55.5, 62, 68, 75, 81, 87, 93.5, 99.5]; for i=1:length(xguess) % Root_results(i)=fzero(@myfun,xguess(i)); Root_results(i,1)=fsolve(@myfun,xguess(i),options); end
fsolve is much better than fzero in finding the root, so i finally use fsolve instead of fzero.

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