Finding a particular solution when there are infinitely many

Question

Ikhsan 2013-3-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/65430-finding-a-particular-solution-when-there-are-infinitely-many

I need to find a particular solution for a specific range.

The solution I got using the solve method led me to a 0.

>> a1 = 5;
>> a2 = 10;
>> b = 13;
>> c = 1*10^-3;
>> d = 0.57 * 10^-3;
>> e = 0.57 * 10^-3;
>> d1 = d/c;
>> e1 = e/c;
>> f = 0.02 * 10^-3;
>> g = d/c+e/c-1;
>> h = acos(1 - f/(2*g*c));
>> n = 6*(10^6)*g^2+3*(10^6)*g*6894.75729;
>> syms m; 
k = @(a,m) a ./ (b .* ((c)^2)*n) - sin(m) .* ((cos (h) ./ cos (m))-1).^1.5;
solve(m);
>> solve(m)
ans =
0

However, the range of values where my solution lies is 0.35< m < 0.41 as shown in the photo. which isn't 0. http://i.imgur.com/RrMO1Ro.jpg

Previous related question I asked which was solved for some history. http://www.mathworks.com/matlabcentral/answers/64353-error-using-char-how-do-i-solve-a-transcendental-equation-for-given-variables

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Ikhsan 2013-3-1

在 MATLAB Online 中打开

I tried using this. http://people.clarkson.edu/~wwilcox/ES100/eqsolve.htm "Here are the steps to find a solution numerically:

Convert the equation to a function consisting of everything moved to the left-hand side, e.g. func2(x) = sin(x)-1/2. Create this function in the MATLAB editor, save to the Current Directory, and make certain the current directory is in the path (File / Set Path). For example:

function out = func2(x) out = sin(x) - 1/2; end

Plot this function over the range of interest to see where the solutions are, e.g.:

>> clear, x = - 4:0.01:4; plot(x,func2(x))

Use one of the following formats to find the solution:

>> fzero('func2',3) or >> fzero(@func2,3) or >> fzero('func2', [2,3])

MATLAB finds the value of x nearest 3 that gives func2 = 0. Notice (“whos” or Workspace) that the result is double, so no conversions are necessary for subsequent numeric calculations."

However I received this error.

ans = 
      @(a,m)a./(b.*((c)^2)*l)-sin(m).*((cos(h)./cos(m))-1).^1.5
>> m = - 4:0.01:4;  plot(m,func2(m))
??? Error using ==> plot
Conversion to double from function_handle is not possible.

Juan Camilo Medina 2013-3-1

编辑：Juan Camilo Medina 2013-3-5

在 MATLAB Online 中打开

Matlab cannot plot func2 because you haven't define a, in other word, you are trying to do a 2D plot of a multivarible equation without the necessary input arguments.

You need to define what a is, let's say you picked a to be 5 then try:

m = - 4:0.01:4; plot(m,func2(5,m))

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Answer 1

Walter Roberson 2013-3-1

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https://ww2.mathworks.cn/matlabcentral/answers/65430-finding-a-particular-solution-when-there-are-infinitely-many#answer_76970

Do not apply solve() to a function handle. Apply fsolve() to function handles, or apply solve() to symbolic expressions.

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Walter Roberson 2013-3-5

Which MATLAB version are you using?

Ikhsan 2013-3-6

MATLAB R2011b(7.13.0.564), using my school's computer.

My goal is to retrieve the value of 'm' for a given value of 'a', in a specified range of 'm'.

I thought that this would be the way to do it.

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Answer 2

Juan Camilo Medina 2013-3-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/65430-finding-a-particular-solution-when-there-are-infinitely-many#answer_77039

在 MATLAB Online 中打开

your syntax is wrong, instead of using

solve(m)

try

    syms m a;
    k = @(a,m)a./(b.*((c)^2)*n)-sin(m).*((cos(h)./cos(m))-1).^1.5;
    solve(k(a,m)==0,m)

But it also appears that your equation is ill-posed. If you look at the picture you posted, they are plotting E vs. a, and not m. It's rather strange that an equation with a bunch of sines and cosines gives you a straight line.

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Juan Camilo Medina 2013-3-5

I think your problem is not how you solve it, but your equation instead. It seems to be ill-posed. Check your problem statement.

Ikhsan 2013-3-6

编辑：Ikhsan 2013-3-6

You might be right. As stated above, I'm intending to find the value of 'm' for a given value of 'a', in a specified range of 'm'.

The equation involves the solution to quintics. However the solutions(range) i'm interested in only lies in the range of the graph as shown in the picture.

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