Numerical solution for trascendental equations

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Giacomo Db 2022-2-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1648630-numerical-solution-for-trascendental-equations

评论： Giacomo Db 2022-2-14

采纳的回答： Walter Roberson

i would like to solve the following trascendental eqn using the following code:

syms omega

syms a

omega = sqrt(a^(2) - 1) - a*exp(-(2*pi - asin(1/a))/(omega));

z=solve(omega)

Eventually i would like to plot (omega,a)

but i get the Warning: The solutions are parametrized by the symbols:

k = Z_ minus {(log((a^2 - 1)^(1/2)/a)*I)/(2*PI)}

do you have any suggestion to solve this type of eqns?

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Paul 2022-2-13

在 MATLAB Online 中打开

Running the code here exactly as it was presented yields a different result.

syms omega

syms a

omega = sqrt(a^(2) - 1) - a*exp(-(2*pi - asin(1/a))/(omega));

z = solve(omega)

z =

As best I can tell, what this code is really doing is

clear omega

syms omega

eq1 = sqrt(a^(2) - 1) - a*exp(-(2*pi - asin(1/a))/(omega));

z = solve(eq1,omega) % solve the equation eq1 == 0

z =

subs(eq1,omega,z)

ans =

0

Are you sure about this line:

omega = sqrt(a^(2) - 1) - a*exp(-(2*pi - asin(1/a))/(omega));

Just asking because it looks peculilar to have omega on the LHS and the RHS of an assignment.

Giacomo Db 2022-2-14

thanks

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Answer 1

Walter Roberson 2022-2-13

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https://ww2.mathworks.cn/matlabcentral/answers/1648630-numerical-solution-for-trascendental-equations#answer_894925

If your values are confined to real then as far as I can see at this time of night, substitute 0 for k in the answer you get.

MATLAB computes a general solution. However because asin is periodic, a periodic solution has to be given. That leads to the general solution having a term of + 2*pi*1i*k, with integer k, except that there are particular a values that the term is not valid for.

Notice that the term is complex valued when k is not 0. So the immediate instinct would be to say "Oh, k has to be zero then, to avoid the complex values."

But... the expression involved itself has complex terms, and also involves log of a value that could be negative, which would generate a complex value. So hypothetically there might be a range of integer k values and a values such that there are an even number of 1i, leading to a real-valued result.

My analysis suggests that there are no real solutions at all for k ~= 0, and that for k=0 there are solutions only for a>1. But it might be worth rechecking non-zero k, as my late-night analysis for that case might not have been thorough enough.