Expanding $$a(x-r_1)(x-r_2)\cdots (x-r_n)$$ should give $$ax^n-a(r_1+r_2+\cdots r_n)x^{n-1}+a(r_1 r_2+r_1 r_3+\cdots r_{n-1}r_n)x^{n-2}+\cdots (-1)^{n}ar_1 r_2\cdots r_n$$ but I fail to prove it. I was only able to do cases $n=1$, $n=2$ and $n=3$ (and even the result for $n=3$ seems a bit different): $$\begin{align*}a(x-r_1)&=ax-ar_1\\ a(x-r_1)(x-r_2)&=(ax-ar_1)(x-r_2)\\&=ax^2-axr_2-axr_1+axr_1 r_2\\&=ax^2-a(r_1+r_2)x+ar_1 r_2 x\\a(x-r_1)(x-r_2)(x-r_3)&=(ax^2-a(r_1 +r_2)x+ar_1 r_2 x)(x-r_3)\\&=ax^3-a(r_1 +r_2)x^2+ar_1 r_2 x^2-ar_3 x^2+ar_3 (r_1 +r_2)x-ar_1 r_2 r_3x\\&=ax^3-x^2(a(r_1 +r_2)-ar_1 r_2+ar_3)+x(ar_3 (r_1+r_2)-ar_1 r_2 r_3)\\&=ax^3-a(r_1+r_2-r_1r_2+r_3)x^2+a(r_1r_3+r_2r_3-r_1r_2r_3)x\end{align*}$$ Could someone help me to prove the general case for all $n$?

Edit: There's an error in my computation.

  • 1
    $\begingroup$ There is an error in your calculation for n=3 where in the very first equality in the RHS there should be simply $ar_{1} r_{2}$ $\endgroup$ – Uday Khanna Jun 11 '20 at 11:31
  • $\begingroup$ Hint: Have you ever read about elementary symmetric polynomials? $\endgroup$ – IMOPUTFIE Jun 11 '20 at 11:32

Let us first examine some examples to guess the general case:$$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$$ $$(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3+r_2r_3)x-r_1r_2r_3.$$So, we can guess the following identity:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$Let us prove the claim by induction.

The base case is trivial. So, let us assume that the claim is correct for $n=m$, that is,$$\prod_{i=1}^m(x-r_i)=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k}.$$ So, we need to prove the claim for $n=m+1$ as follows.$$\prod_{i=1}^{m+1}(x-r_i)=\left ( \prod_{i=1}^m(x-r_i) \right ) \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $$=\left (\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k} \right ) \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $${=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} -\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}}$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=0}^{m-1}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=1}^{m}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^{k-1}r_{j_1} ... r_{j_{k-1}}r_{m+1}x^{m-(k-1)}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )\tag{*}\label{*}$$ $$= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k \left ( \sum_{1 \le j_1 \lt ... \lt j_k \le m}r_{j_1} ... r_{j_k}+ r_{j_1} ... r_{j_{k-1}}r_{m+1} \right ) x^{(m+1)-k} \right )$$ $${= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}r_{j_1} ... r_{j_k} x^{(m+1)-k} \right )}$$ $$=\sum_{k=0}^{m+1} \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}(-1)^k r_{j_1} ... r_{j_k} x^{(m+1)-k}.\tag{**}\label{**}$$Thus, by induction we proved that for any natural number $n$ the following identity holds:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$


\ref{*} is followed from the following property of summation:$$\sum_{i=m}^nA_i=\sum_{i=m+1}^{n+1}A_{i-1}.$$ \ref{**} is followed from considering the fact that for any fixed $k$ one can decompose the sum $\sum_{1 \le j_1 \lt ... \lt j_k \le m+1} r_{j_1} ... r_{j_k}$ into two sums: (i) the sum of terms not containing $r_{j_{m+1}}$, that is, $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_k}$, and (ii) the sum of terms containing $r_{j_{m+1}}$, that is $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_{k-1}}r_{j_{m+1}}$.


Let's drop the $a$ as it is common everywhere

Now, without expanding, let's analyze:

How many ways to make an $x^n$ term? Only 1 way (taking $x$ from each bracket)

How many ways to make an $x^{n-1}$ term? You can take $n-1 $ powers of $x$ and one other term from any bracket . Thus the coefficient is $-r_1-r_2...$

Continue as above...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.