Converegence Problem for forward and backward propogation

  • Thread starter aashish.v
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  • #1
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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.
 

Answers and Replies

  • #2
80
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Are you sure that you typed the function correctly?
 
  • #3
13
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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)
 
  • #5
Ray Vickson
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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
 
  • #6
13
0
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
[itex]C.λ^n[/itex] 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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