# Question:False result even though having done the right manipulations (Expanding Wronskian of Heun Functions into a series)

## Question:False result even though having done the right manipulations (Expanding Wronskian of Heun Functions into a series)

Maple 16

Hello,

I have the following problem:

My function is defined by the determinant of 2 Heun functions

If I plot the phase I get something which looks quite what I'm looking for.

To get a better result I thought I would manually carry out the Wronskian as far as possible...

Doing some manipulations I get another form of the Wronskian which in fact should give the same result...

the problem is it doesnt :-(

I've added the spreadsheet. Does anyone know where the problem lies?

Thx Chris

Wronski-Determinant.mw

If I calculate the Wronskian directly over the implemented Heun functions I think I get the right result...

 (1)

 (2)

 (3)

 (4)

Now as mentioned before I tried to calculate the wronskian as far as possible to maybe get a better result.

For this i introduce the first Heun function g:=HeunG(c,q,a,b,g,d,t)=Sum(c_n*t^n)

The other Heun function over the Singularity t=1 shall be h:=Sum(d_m*(1-t)^m)

evaluating at t=1/2

 (5)

the wronskian becomes...

Wronskian(fg,fh)=f^2*Wronskian(g,h)=f^2*(g*h'-g'*h)

g*h'=-Sum(n=0 .. infinty) Sum(m=1 .. infinity) c_n * d_m * m * 2^-(n+m-1)

Now I introduce a new summationindex k=n+m-1 and the sum becomes

g*h'=-Sum(n=0 .. infinty) Sum(k=n .. infinity) c_n * d_(k-n+1) * (k-n+1) * 2^-(k)

Using the Cauchy Product

g*h'=-Sum(k=0 .. infinity) Sum(n=0 bis k) c_n * d_(k-n+1)*(k-n+1)*2^(-k)

the other term g'*h is analogous to the first and except for an overall sign the result is quite the same

So:

g'*h = Sum(k=0 .. infinity) Sum(n=0 .. k) c_(k-n+1) * d_n * (k-n+1)*2^(-k)

and therefore

f^2*(g*h'-g'*h) = -2^(-I*p)*Sum(k=0 .. infinity) Sum(n=0 .. k) (c_n*d_(k-n+1)+c_(k-n+1)*d_n)*(k-n+1)*2^(-k)

The recursion relations of the heun functions are:

f_n * c_(n+1) + g_n * c_n + h_n * c_(n-1) = 0

with

f_n=c*(n+1)*(n+g)               (here c is the singularity as above)

g_n=-(q+n(n-1+g)(1+c)+n(c*d+a+b+1-g-d))

h_n=(n-1+a)(n-1+b)

Finally I tried to implement it in maple:

Now with c_n=r(n) and d_n=o(n)

 (6)

 (7)

Checking the correctness of the sequence by expanding the implemented heunfunction into a series

 (8)

 (9)

Analogous:

 (10)

 (11)

So they should be right

For the wronskian should therefore hold:(considering the factor exp(-I*p*ln(2))=2^(-I*p)

Which is obviously not the same.
I just don't know where the problem arises...