Converegence Problem for forward and backward propogation

[aashish.v]

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.

 

[camjohn]

Are you sure that you typed the function correctly?

 

[aashish.v]

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 propagation...
J(n)=2*(n+1)*J(n+1)-J(n+2)

 

[aashish.v]

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e9oo6gajcl6

view this to avoid any ambiguity

 

[Ray Vickson]

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

 

[aashish.v]

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.