How to solve transcendental equation?

x1=0.1717 - 0.0638i

how to find x : tan(x)/x = x1

x is a complex number

Answer 1

[Walter Roberson]

f(x) == c implies that f(x) - c = 0. You can construct the expression f(x)-c and use it with a root-finding function such as fsolve. Just be sure to give an initial point that is complex valued.

You cannot use fzero for this purpose however. fzero cannot deal with complex values.

The method I gave is not necessarily the best one. You can often do better by splitting up the input x into a vector of two real values. In the calculation you would reconstruct the complex value by using complex() or by (real + 1i*imaginary). You would calculate with that. Then you would use real() and imag() to split it into two real values and return them.

fzero cannot deal with this situation because it can only deal with one real input (and no complex component) but fsolve can deal with it.

Splitting into real and imaginary components is mathematically often easier for root finders to deal with.

[SHANKAR AENAGANDULA]

The modified equation is like this:

Can you help me how to solve this type of transcendental equations.

y = (1+c)*z - c

r(k) = (sqrt(y) - i*tanh(x(k)*sqrt(y)) / (sqrt(y) + i*tanh(x(k)*sqrt(y))

where k=1,2

r(1),r(2),x(1),x(2) and c is known values

r(1),r(2) are complex values

how to find z?

z is a complex number.

[SHANKAR AENAGANDULA]

How to choose initial point, different initial points getting different output.

[Walter Roberson]

y = @(z) (1+c)*z - c
eqn = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z)))) /  (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z))) - r

z0 = something
z = fsolve(eqn, z0)

Or better,

y = @(z) (1+c)*z - c
eqnz = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z)))) /  (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z))) - r
eqn = @(zv) [real(eqnz(zv(1)+1i*zv(2))), imag(eqnz(zv(1)+1i*zv(2)))]

zv0 = [real guess component, imaginary guess component]
zv = fsolve(eqn, zv0)
z = zv(1) + 1i*zv(2);

You can then

funr = @(zr, zi) real(eqnz(zr+1i*zi))
funi = @(zr, zi) imag(eqnz(zr+1i*zi))
fsurf(funr)
hold on
fsurf(funi)
hold off

to plot the real and imaginary components to try to get an idea of where the roots are (they are where funr and funi are both zero)

[SHANKAR AENAGANDULA]

c=0.7835;
r = [0.7443+0.0616i,0.755+0.0525i];
x=[0.7995,0.91368];
y = @(z) (1+c)*z - c
eqnz = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z))) /  (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z))) - r(k)
eqn = @(zv) [real(eqnz(zv(1)+1i*zv(2))), imag(eqnz(zv(1)+1i*zv(2)))]
zv0 = [1,1]
zv = fsolve(eqn, zv0)
z = zv(1) + 1i*zv(2);

getting error as:

Error: File: Transcendental_3.m Line: 5 Column: 101

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

[Walter Roberson]

Your posted code had

r(k) = (sqrt(y) - i*tanh(x(k)*sqrt(y)) /  (sqrt(y) + i*tanh(x(k)*sqrt(y))
 1 0   1    2 1         2 3 2     3 21    2    3 2         3 4 3     4 32

The number above is the number of active bracket levels "after" the character it is below. So at the end of the line, there are two unclosed brackets.

I copied your line; you will need to figure out how to fix it.

The line you will be wanting will be of the form

 eqnz = @(z) EXPRESSION_IN_Z - r(k)

[SHANKAR AENAGANDULA]

i understood, rectified it, but one error showing like this:

Array indices must be positive integers or logical values.

Error in Transcendental_3>@(z)(sqrt(y(z)))-1i*tanh(x(k)*sqrt(y(z)))/(sqrt(y(z))+1i*tanh(x(k)*sqrt(y(z))))-r(k)

Error in Transcendental_3>@(zv)[real(eqnz(zv(1)+1i*zv(2))),imag(eqnz(zv(1)+1i*zv(2)))]

Error in fsolve (line 242)

fuser = feval(funfcn{3},x,varargin{:});

Error in Transcendental_3 (line 8)

zv = fsolve(eqn, zv0)

Caused by:

Failure in initial objective function evaluation. FSOLVE cannot continue.

The code is:

c=0.7835;
r = [0.7443+0.0616i 0.755+0.0525i];
x=[0.7995 0.91368];
y = @(z) (1+c)*z - c
eqnz = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z))) /  (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z)))) - r(k)
eqn = @(zv) [real(eqnz(zv(1)+1i*zv(2))), imag(eqnz(zv(1)+1i*zv(2)))]
zv0 = [1,1];
zv = fsolve(eqn, zv0)
z = zv(1) + 1i*zv(2);

[Walter Roberson]

What is your current value for k ? When I set k=1 then the code runs.

[SHANKAR AENAGANDULA]

k is a vector

we need to utilize both k=1 and k=2 gives two eqations. Using this two equations optimize the z value.

x(1), x(2), r(1), r(2) and c values are given in the above code.

[Walter Roberson]

c=0.7835;
r = [0.7443+0.0616i 0.755+0.0525i];
x=[0.7995 0.91368];

k = 1:2;

y = @(z) (1+c)*z - c
eqnz = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z))) /  (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z)))) - r(k)
eqn = @(zv) [real(eqnz(zv(1)+1i*zv(2))), imag(eqnz(zv(1)+1i*zv(2)))]
zv0 = [1,1];
[zv,fval] = fsolve(eqn, zv0);

z = zv(1) + 1i*zv(2)
fval

No error message, but also not a good fit.

we need to utilize both k=1 and k=2 gives two eqations.

Not really. It gives four equations: the real parts and imaginary parts of each of the two (total four values) all have to be solved to be 0. You would be trying to use a single pair of inputs (real and imaginary z) to solve four equations. You are not likely to get good results.

[SHANKAR AENAGANDULA]

IT gives two equations only, z (Real & Imaginary) is same for both equations.

Not really. It gives four equations: the real parts and imaginary parts of each of the two (total four values) all have to be solved to be 0. You would be trying to use a single pair of inputs (real and imaginary z) to solve four equations. You are not likely to get good results.

[SHANKAR AENAGANDULA]

1. c=0.7835;

2. r = [0.7443+0.0616i 0.755+0.0525i];

3. x=[0.7995 0.91368];

4. k = 1:2;

5. y = @(z) (1+c)*z - c

6. eqnz = @(z) (sqrt(y(z))) - 1i*tanh(x(k)*sqrt(y(z))) / (sqrt(y(z)) + 1i*tanh(x(k)*sqrt(y(z)))) - r(k)

7. eqn = @(zv) [real(eqnz(zv(1)+1i*zv(2))), imag(eqnz(zv(1)+1i*zv(2)))]

8. zv0 = [1,1];

9. [zv,fval] = fsolve(eqn, zv0);

10. z = zv(1) + 1i*zv(2)

11. fval

Can you help me, Line-7 in above code, what it is doing?

[Walter Roberson]

Suppose that I asked you to solve for the root of a complex-valued equation, so f(z) = 0 is being looked for. Is there any f(z) = 0 such that real(f(z)) ~= 0 -- is it possible for a complex value to equal 0 when its real part is non-zero? Is there any f(z) = 0 such that imag(f(z)) ~= 0 -- is it possible for a complex value to equal 0 when its imaginary portion does not equal 0?

So if you have f(z) = 0 and f(z) is complex valued is that really just a single equation, or is it really the pair of equations [real(f(z)) = 0, imag(f(z)) = 0] ?

[Walter Roberson]

My tests show that there are no z that satisfy both equations simultaneously.

If you plot the surface defined by abs() of the difference of the two equations, then that surface would be 0 at the point where the two equations are equal. If there are solutions to both equations simultaneously, then both equations would be 0 at the points of solution, and so both equations would be equal to each other at those points. Therefore any solutions to the two equations being equal must be a subset of the solutions to abs() of their difference being 0.

When you look at the plot of the difference, you can find three areas where the two equations might be zero. Using approximations of them from the graph as seeds, we can find that the two equations are equal at only three locations that are approximately

-7.434567016 - 10.84030833*I
-2.971210845 + 5.866746641*1i
 2.388387229 - 2.524573595*1i

However, at those locations, the equations are not especially close to zero, being (in order)

 1.665792984 - 4.553523447*I
 0.7774697970 + 2.885234165*I
 1.376375848 - 1.569379529*I

Again, for there to be simultaneous solutions to the two equations, the difference between the two equations would have to equal 0 and the individual equations would have to be 0 at the location.

Actually now that I have extended the search to +/- 50 Pi I see there might be some additional points where the two equations are equal. However:

If you plot abs() of each of the equations individually, you will see that they each have a single point where they might be 0, and in both cases those single points are "near" to 0. With the two equations each having only a single solution, then for the two equations to have simultaneous solutions, the single solution to each of them would have to be the same. But they are not. The first one has a root at 0.9658822712 + 0.5494187667*1i and the second one has a root at 1.018072372 + 0.5528635790*1i -- the roots are close to each other, but not the same place.

There is no simultaneous solution; if there were then it would have to be among the three possibilities listed above (we can constrain to those by approximate range from graphing.)

The surfaces here were generated by splitting the real and imaginary parts of z, z = zr + 1i*zi and doing a 2D plot varying zr independently of zi .

No solution.