# Converegence Problem for forward and backward propogation

The function is J(n-1)+J(n+1)=2nJ(n)

I need to prove that it diverges in forward direction but converges in backward direction.
I am unable to find any method, kindly suggest.

## Answers and Replies

Are you sure that you typed the function correctly?

Yes.

for forward direction we can re-write it this way..
J(n)=2*(n-1)*J(n-1)-J(n-2)

and for backward propogation...
J(n)=2*(n+1)*J(n+1)-J(n+2)

Ray Vickson
Science Advisor
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The function is J(n-1)+J(n+1)=2nJ(n)

I need to prove that it diverges in forward direction but converges in backward direction.
I am unable to find any method, kindly suggest.

You need to show your work.

RGV

You need to show your work.

RGV

The typical approach shown in text is to show that if the solution of such function truns out to be in form of
$C.λ^n$ then we can say that such recursive function diverges, I have tried the same approach for the problem but the λ I am getting turns out to be function of n.

I can scan and upload my work if you wish.