# Proving the generating functions version of Newton's identities

We are asked to prove the identity in the generating functions version in the Expressing elementary symmetric polynomials in terms of power sums section of the Wikipedia page on Newton's identities

The following is what I am thinking about now:

I have not even seen generating functions before, and since $\infty$ is not a number, I do not know that if it suffices to prove that for any $n$ in N every coefficient of $T^n$ consists. If so, I think maybe it can be done using induction.

I have not familiar with the exponential generating functions so I am afraid that there are some useful properties of this function which I do not know. If so, could someone tell me what property of this function should I use?

Seeing this question:Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I guess maybe there is an algorithm to translate a given recurrence into generating functions.

But until now I have not found a track to prove it, Could someone please prove it or give some hints? Thanks in advance!

EDIT: The proofs in the link

https://yakovenko.files.wordpress.com/2011/04/lecture5symmetry.pdf

are understandable and in generating functions version.

But they are not in exponential generating functions version. I do know that there is a method to translate ordinary generating functions into exponential version. But I cannot understand the principle behind the method and I am afraid that I cannot use that. So is it possible to modify the proofs above to give the desired conclusion? Any hint will be appreciate? Thanks!

• math.stackexchange.com/a/2100857/155436 – Count Iblis Feb 25 '17 at 2:10
• @CountIblis Sorry I have no back groud of exponential generating function as well as log generating function. May I ask what should I learn about from this answer? – PropositionX Feb 25 '17 at 2:14

What you need here is the following statement: the OGF of the cycle index $Z(P_k)$ of the unlabeled set operator $\mathfrak{P}_{=k}$ is given by

$$Z(P_k) = [w^k] \exp\left(\sum_{l\ge 1} (-1)^{l-1} a_l \frac{w^l}{l}\right).$$

This is known as the exponential formula. To see this consider a repertoire of variables $\sum_q X_q$ and substitute the corresponding power sum

$$a_l = \sum_q X_q^l$$

into the formula to get

$$Z(P_k)\left(\sum X_q\right) = [w^k] \exp\left(\sum_{l\ge 1} (-1)^{l-1} \sum_q X_q^l \frac{w^l}{l}\right) \\ = [w^k] \prod_q \exp\left(\sum_{l\ge 1} (-1)^{l-1} X_q^l \frac{w^l}{l}\right)\\ \\ = [w^k] \prod_q \exp\log(1 + w X_q) = [w^k] \prod_q (1 + w X_q)$$

This is

$$\bbox[5px,border:2px solid #00A000]{ \sum_{Q\subseteq X, |Q|=k} \prod_{X_q\in Q} X_q}$$

by inspection and we have expressed the elementary polynomials in terms of power sums.

Readings. A closely related identity was discussed at this MSE link and the proof was given at this MSE link II. An alternate proof due to Lovasz is included among the links at Wikipedia on cycle indices.

Example. The cycle index $Z(P_6)$ for the operator $\mathfrak{P}_{=6}$ is given by

$$Z(P_6) = {\frac {{a_{{1}}}^{6}}{720}}-1/48\,a_{{2}}{a_{{1}}}^{4} \\+1/18\,a_{{3}}{a_{{1}}}^{3}+1/16\,{a_{{1}}}^{2}{a_{{2}}}^{2} -1/8\,a_{{4}}{a_{{1}}}^{2}-1/6\,a_{{1}}a_{{2}}a_{{3}} \\+1/5\,a_{{5}}a_{{1}}-1/48\,{a_{{2}}}^{3} +1/8\,a_{{2}}a_{{4}}+1/18\,{a_{{3}}}^{2}-1/6\,a_{{6}}.$$

• I really appreciate your answer. But your answer is far more beyond my knowledge. I have posted another question which is more specific (math.stackexchange.com/questions/2161737/…). Thanks so much. – PropositionX Feb 26 '17 at 3:00
• Looks like it has been closed. I for one understood your question immediately on seeing the Wikipedia entry. – Marko Riedel Feb 26 '17 at 22:34