How to solve a recursion relation with a constant using hints?

  • Context: Graduate 
  • Thread starter Thread starter spaghetti3451
  • Start date Start date
  • Tags Tags
    Recursion Relation
Click For Summary

Discussion Overview

The discussion centers around solving a recursion relation of the form ##y_{k}=k(2j-k+1)y_{k-1}##. Participants explore methods to derive a closed-form expression for ##y_{k}##, specifically aiming to show that ##y_{k}=\frac{k!(2j)!}{(2j-k)!}##. The context involves mathematical reasoning related to recursion and factorials.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the recursion relation and requests hints for solving it.
  • Another participant suggests a product form for the ratio of successive terms in the sequence.
  • A later post clarifies the initial condition ##y_{0}=1## and derives an expression for ##y_k## through a series of transformations involving products of integers and terms dependent on ##j##.
  • The final expression for ##y_k## is presented as ##(k!) \frac{(2j)!}{(2j-k)!}##.

Areas of Agreement / Disagreement

Participants appear to build on each other's contributions without explicit disagreement, leading to a derived expression. However, the initial request for hints indicates that the process may still involve exploration and refinement.

Contextual Notes

The discussion does not address potential limitations or assumptions underlying the recursion relation or the derived expressions.

spaghetti3451
Messages
1,311
Reaction score
31
I have the recursion relation ##y_{k}=k(2j-k+1)y_{k-1}##

and I would like to solve it to obtain ##y_{k}=\frac{k!(2j)!}{(2j-k)!}##.

Can you provide some hints on how I might proceed?

P.S.: ##j## is a constant.
 
Mathematics news on Phys.org
## \frac{y_k}{y_1} = \prod_{i = 2}^k \frac{y_i}{y_{i-1}} ##
 
I forgot to mention that ##y_{0}=1##.

All right, then, we have

##\frac{y_k}{y_0} = \prod_{i = 1}^k \frac{y_i}{y_{i-1}}##

##y_k = \prod_{i=1}^{k} i(2j-i+1)##

##y_k = \bigg(\prod_{i=1}^{k} i\bigg) \bigg(\prod_{i=1}^{k} (2j-i+1)\bigg)##

##y_k = (k!) \bigg(\prod_{i=1}^{k} (2j-(i-1))\bigg)##

##y_k = (k!) \frac{(2j)!}{(2j-k)!}##

Thanks!
 
You're welcome
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K