Looking for some help with a proof by iduction. Im looking to proove the following summation holds true:

$$\frac{\langle W \vert (C)^{N} \vert V \rangle}{\langle W \vert \vert V \rangle}=\frac{\langle W \vert (D+E)^{N} \vert V \rangle}{\langle W \vert \vert V \rangle}= \sum_{p=0}^N\frac{p(2N-1-p)!}{N!(N-p)!}\frac{\beta^{-p-1}-\alpha^{-p-1}}{\beta^{-1}-\alpha^{-1}}$$

The problem is related to the following rules: $$DE= D+E=C $$

$$D\vert V \rangle= \frac{1}{\beta} \vert V \rangle $$

$$\langle W\vert E= \frac{1}{\alpha}\langle W \vert $$

Now using the base case of N=1 yields to, $\frac{1}{\alpha}+\frac{1}{\beta}$ which I have shown but struggling with the hypothesis step.

Any help would be most appreciated, many thanks.

  • $\begingroup$ Hard to attack, given that you don't even define what appears on the left hand side. $\endgroup$ Jan 16, 2013 at 13:53
  • $\begingroup$ @MatemáticosChibchas sorry about that forgot to add in the relations. $\endgroup$
    – user24930
    Jan 16, 2013 at 14:08

1 Answer 1


Here's a lead:

So $C^N$ is a sum over all $N$-length words in $\{D,E\}$. If I've interpreted your rules correctly, the only way to simplify words ending in $E$ is to use the rule $DE = D + E$. This allows one (in theory) to simplify any word coming from $C^N$ into a sum of words ending in $D$ and perhaps a word consisting of some power of $E$.

The practical induction will be a mess, but here's a thought: (to do the induction it suffices to look at $C^{N} E$) \begin{align*} C^N E = C^{N-1}(D + E)E = C^{N-1} DE + C^{N-1} E^2 = C^N + C^{N-1} E^2 \\ = C^N + C^{N-1}E + C^{N-2}E^3 = \cdots = \sum_{i = 0}^{N-1} C^{N-i} E^i + E^{N+1} \end{align*} if I'm not mistaken.

Now concentrate on how to reduce the terms in the sum: for instance, $C E^{N-1}$ equals $(D + E) E^{N-1} = C E^{N-2} + E^N = C E^{N-3} + E^{N-1} + E^N = \cdots = C + \sum_{j = 2}^N E^j$. Try to use this form to come up with a closed form for the summands $C^{N-i} E^i$ (I would just hit both sides with a factor of $C$ and see what comes out).

  • $\begingroup$ Many thanks for the post, ive been told it may be more sensible to consider a binomial expansion? $\endgroup$
    – user24930
    Feb 9, 2013 at 12:50

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