1

I'm working on a problem that resulted in the following recurrence (which I think is wrong, but that's not the question here):

$$ (n-i)*2i*Z_i = n(n-1) + (n-i)(i+1)Z_{i+1} + (n-i)(i-1)Z_{i-1} $$ Subject to $Z_n = 0$.

The solution for $i=1$ to the above recurrence is said to be $$ Z_1=(n-1)^2 $$ Is there a quick way I can find if this solution is indeed correct?

user5965026
  • 677
  • If you were to post the complete question, it would be great. – AryanSonwatikar May 08 '20 at 15:08
  • @AryanSonwatikar The complete question is extremely long and I don't think it's relevant for answering this question, which is just about this recurrence relation and whether $Z_1$ satisfies that. – user5965026 May 08 '20 at 15:13
  • Well, I'm not too well versed in recurrence relations, but here goes. Observe that if you put$(n-i)iZ_i=\alpha_i$ you get a nice little recurrence $2\alpha_i=n(n-1) +\alpha_{i+1} +\alpha_{i-1}$. Put $i+1$ in place of $i$ and subtract. Then you get the characteristic equation $(x-1)^3=0$. And now I'M stumped. I haven't been taught such recurrences. See if it help you. – AryanSonwatikar May 08 '20 at 15:34

1 Answers1

1

This is a second order recurrence. With a single final condition, the solution is indeterminate.

Assuming you fix this, a possible check is to solve the recurrence explicitly with small $n$, say $3,4,5$, and check if the solutions match $(n-1)^2$.