0
$\begingroup$

enter image description here

I am having troubles with proving the above identity. Can I please ask for someone's help? I have spent more than two days on this question and I am kind of exhausted. So I would really appreciate if someone could just provide a full proof. Thanks so much for your help.

Note: In the above identity, $e_j$ denotes the elementary symmetric functions and $P_k$ denotes the power sum symmetric functions ($P_k=\alpha_1^k+\alpha_2^k+\alpha_3^k+...+\alpha_n^k$).

For any power series $f(t)=c_{1}t+c_{2}t^2+...$ with zero constant term, $exp(f(t))$ is defined by the following:

enter image description here

1 Answers 1

1

Let's see if you can get it with these two hints:

Left side: factor using Vieta's formulas.

Right side: rewrite using the Mercator series.

Alright, I'll spoil the exercise.

Left side is

$$ \begin{array}{ll} (1+tx_1)(1+tx_2)\cdots(1+tx_n) & = \phantom{+} 1 \\ & \phantom{=} +(x_1+\cdots+x_n)t \\ & \phantom{=} +(x_1x_2+\cdots+x_{n-1}x_n)t^2 \\ & \phantom{=} \qquad\qquad\quad \vdots \\ & \phantom{=} +(x_1x_2\cdots x_n)t^n. \end{array} $$

Right side is

$$ \begin{array}{l} \displaystyle \phantom{=} \exp\left(\frac{x_1+\cdots+x_n}{1} \, t \,-\, \frac{x_1^2+\cdots+x_n^2}{2} \, t^2 \,+\, \frac{x_1^3+\cdots+x_n^3}{3} \, t^3 \,-\, \cdots\right) \\[5pt] \displaystyle =\exp\left[\left(\frac{x_1t}{1}-\frac{(x_1t)^2}{2}+\frac{(x_1t)^2}{3}-\cdots\right) + \cdots + \left(\frac{x_nt}{1}-\frac{(x_nt)^2}{2}+\frac{(x_nt)^2}{3}-\cdots\right) \right] \\[5pt] \displaystyle = \exp\left(\frac{x_1t}{1}-\frac{(x_1t)^2}{2}+\frac{(x_1t)^2}{3}-\cdots\right) \,\cdots\, \exp\left(\frac{x_nt}{1}-\frac{(x_nt)^2}{2}+\frac{(x_nt)^2}{3}-\cdots\right) \\[5pt] \displaystyle =\exp\!\big(\!\ln(1+tx_1)\big) \,\cdots\, \exp\!\big(\!\ln(1+tx_n)\big) \\[5pt] \displaystyle =(1+tx_1)\cdots(1+tx_n). \end{array} $$

  • 0
    Maybe it is related to what I have seen in this link: https://holdenlee.wordpress.com/2010/11/22/newton-sums/ But in that page I did not see the desired conclusion using exponential generating functions and I have not found any way to manipulate what is proved to get it myself. Could you please give a full proof? – 2017-02-26
  • 0
    I tried to expand the equation. But so far, I still have no idea about how to use them and get the conclusion... – 2017-02-26
  • 0
    @Y.X. Let's do some special cases. First, say $n=3$; can you factor $$ 1+(x_1+x_2+x_3)t+(x_1x_2+x_2x_3+x_3x_1)t^2+(x_1x_2x_3)t^3 $$ using Vieta's formulas? Second, can you prove the equality when there is only one variable ($n=1$)? In that case the left side is just $1+tx$. – 2017-02-26
  • 0
    I have tried using the Vieta's formulas. Using the identity "$x_{1}x_{2}...x_{n}=(-1)^n*(a_{0}/a_{n})$" (from Wiki), I seemed to get $s_{3}^2=-1$ which is absurd. I really do not know how to factor that. May I please ask for some more help? – 2017-02-26
  • 0
    Sorry for not understanding your approach. I have spent more than two days on this problem. So I would really appreciate if you could just provide a full proof. – 2017-02-26
  • 0
    I am able to prove the case when n=1, it is trivially by expand the RHS and compare the coefficients on $t$ and note that they are same. – 2017-02-26
  • 0
    @Y.X. It is not entirely trivial with $n=1$, because you still need to invoke the fact the RHS is the Taylor series for $\ln(1+x)$. Surely if you can work out the RHS for $n=1$, then you can work out the RHS for *all* values of $n$? And maybe let's try factoring the LHS when $n=2$, since it has fewer variables. You should be able to factor $1+(x_1+x_2)t+(x_1x_2)t^2$. – 2017-02-26
  • 0
    If I have get it correctly, in this polynomial $a_{2}=x_{1}x_{2}$ $a_{1}=x_{1}+x_{2}$ $a_{0}=1$,thus by Vieta's formula we can get$ t_{1}+t_{2}=-a_{1}/a_{2}=-(x_{1}+x_{2})/x_{1}x_{2}$ and $t_{1}t_{2}=a_{0}/a_{2}$,so we should be able to solve for$ t_{1} $and $t_{2}$ and factorize it, but without solving for roots what should we do? – 2017-02-26
  • 0
    @Y.X. You are missing the forest for the trees. I was not intending for you to merely plug things into formulas to get the factorization, I was hoping for you to see the **pattern** inherent in Vieta's formulas so you would understand what's going on. Anyway, I did the whole problem for you in an edit fifteen minutes ago. – 2017-02-26
  • 0
    Thanks a lot! Now I think I have understand it. Moreover, I am thinking about a quite related question about where does this identity can used for expressing **every symmetric polynomial**. Could you please have a look to this question http://math.stackexchange.com/questions/2162002/what-is-the-condition-for-any-symmetric-polynomial-in-rx-can-be-expressed-as? – 2017-02-26