How to solve simultaneous trigonometric equation with summation from n=1 to inf. ?

I am trying to solve the set of simultaneous equations (shown in file attached, see bottom). x and y are known. For a particular value of psi3, I have to obtain psi1 and psi2 from these simultaneous equations? I wonder if a combination of 'symsum' and 'solve' command will give a solution as the series is not convergent. In that case, how many terms should I take in the series? Never encountered such a set of equations. Any kind of help is appreciated.Thanks for kindness.

Answer 1

[Roger Stafford]

I think you will find that if x satisfies the inequalities 0<x<1, your infinite series will in fact be convergent. Otherwise they are divergent and the problem has no meaning.

It is doubtful that 'symsum' can find a symbolic sum for the series, or that 'solve' could find a symbolic solution even if 'symsum' succeeds.

However, provided again that 0<x<1, you should be able to numerically evaluate the sum of each of the eight different infinite series which are involved in these expressions for any given x and y to a sufficient accuracy. Then solving for psi1 and psi2 in terms of psi3 is a trivial linear problem in two unknowns.

For example, one of the eight series is

 sum from n equals 1 to inf of:
    cos(n*pi)*cos(n*pi*y)*sinh(n*pi*x)/sinh(n*pi)

The cosine factors are bounded, so the magnitude of successive terms in this series decreases pretty much according to the factor

   exp(-(1-x)*pi)

Therefore it should be possible to approximate it with sufficient accuracy with a reasonably finite number of terms (unless x is too close to 1.)

[Roger Stafford]

Added advice: To avoid premature overflow in the numerator and denominator of

   sinh(n*pi*x)/sinh(n*pi)

you can express it as

   exp(-n*(1-x)*pi)*(1-exp(-2*n*pi*x))/(1-exp(-2*n*pi))

Similar advice applies to the other seven series.

[Roger Stafford]

All eight of those series are still convergent with x = 5e-6 or even 1e-6. However, with x that close to zero, four of them require that many more terms in the series be included in the summation before an accurate value is attained. I conducted an experiment on my computer with y set to zero so that the series you mentioned would then be sinh(n*pi*(1-x))/sinh(n*pi) which must be expressed as exp(-n*pi*x)*(1-exp(-2*n*pi*(1-x)))/(1-exp(-2*n*pi)) to avoid overflow. The result was that for x = 5e-6 it took 3e+6 summations to achieve convergence and with x = 1e-6 it required 1.2e+7 summations. Here are the results:

 x = 5e-6;
 n = 3000000:-1:1;
 s = sum(exp(-n*pi*x).*(1-exp(-2*n*pi*(1-x)))./(1-exp(-2*n*pi)))
 ans = 63661.47723800815

 x = 1e-6;
 n = 12000000:-1:1;
 s = sum(exp(-n*pi*x).*(1-exp(-2*n*pi*(1-x)))./(1-exp(-2*n*pi)))
 ans = 318309.3861840408

Note that this summation is presumably being performed backwards from the tail of the series to the beginning so as to allow the smaller terms at the tail to have a cumulative effect before being overwhelmed by the large value of the sum. If necessary, you could use a backwards for-loop instead to ensure that the order of summation is backwards. I am not certain of the way the 'sum' function works in this respect.

[Vikas Manhotra]

I am able to do it now. I appreciate your help. Thank you very much. This website is awesome.