Converegence Problem for forward and backward propogation

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The discussion revolves around proving the convergence and divergence of the function J(n-1) + J(n+1) = 2nJ(n). Participants are attempting to demonstrate that the function diverges in the forward direction while converging in the backward direction. A transformation of the function is provided to facilitate this proof, leading to the equations J(n) = 2(n-1)J(n-1) - J(n-2) for forward propagation and J(n) = 2(n+1)J(n+1) - J(n+2) for backward propagation. The challenge lies in showing that the solution takes the form C.λ^n, with the complication that λ appears to be a function of n. The discussion emphasizes the need for clear work to support the proof.
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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.
 
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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)
 
aashish.v said:
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
 
Ray Vickson said:
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.
 
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