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Let $f$ be a polynomial having $n$ distinct real roots: $$f(x)=(x-x_1)(x-x_2)\dots(x-x_n)$$
Prove that $$\frac{1}{f(x)}=\sum_{k=1}^n \frac{1}{f'(x_k)(x-x_k)}, \: \forall x \in \mathbb{R} - \{x_1,x_2,\dots,x_n \} $$

I don't know much about partial fractions, but this looks very much like them. I tried to use induction for this, but I couldn't really make the jump from $n-1$ to $n$ and I honestly wouldn't have been satisfied even if I had solved it that way, because I would really like to see where this formula comes from. I'm sorry if this is actually very easy, I simply don't know how to approach it

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    $\begingroup$ HINT: first try to find an expresson for the derivative of $f$ in terms of its zeroes $\endgroup$ – M. Van Mar 20 '18 at 9:13
  • $\begingroup$ How familiar are you with complex analysis? $\endgroup$ – user228113 Mar 20 '18 at 9:28
  • $\begingroup$ I don't know anything about it yet. $\endgroup$ – AndrewC Mar 20 '18 at 9:38
  • $\begingroup$ @G.Sassatelli: complex analysis is not required at all. $\endgroup$ – Yves Daoust Mar 20 '18 at 9:47
  • $\begingroup$ @YvesDaoust It might be unnecessary, but you can indeed use the fact that you are eliminating the poles of a haolomorphic function on $\Bbb C\setminus\{x_1,\cdots,x_n\}$ which goes to zero as $z\to\infty$. Which also accounts for the need of all the roots to be there and to be distinct. $\endgroup$ – user228113 Mar 20 '18 at 10:27
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Have you tried decomposing $\frac{1}{f(x)}$ into partial fractions. By that i mean writing $\frac{1}{f(x)} = \frac{A_1}{x-x_1}+...+\frac{A_n}{x-x_n}$ and solving for each $A_i$? Then i think the first fraction would look something like $\frac{1}{(x-x_1)(x_1-x_2)(x_1-x_3)...(x_1-x_n)}$. Then calculating $f'(x_1)$ should get you the result you want.

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  • $\begingroup$ Yes, I tried that, but how to I prove that $A_1$ equals that, for example? $\endgroup$ – AndrewC Mar 20 '18 at 9:16
  • $\begingroup$ Multiplying the equation by $(x-x_1)$ and then realizing that substituting $x_1$ into the equation would "kill of" the other fractions leaving $A_1=\frac{1}{(x_1-x_2)...(x_1-x_n)} $ $\endgroup$ – Mrtny Mar 20 '18 at 9:22
  • $\begingroup$ Oh, silly me! You are right, thank you very much! $\endgroup$ – AndrewC Mar 20 '18 at 9:24
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    $\begingroup$ Regarding the calculation of $f'(x_1)$, i guess you could be a bit fancy and see that $\frac{1}{(x-x_1)(x_1-x_2)(x_1-x_3)...(x_1-x_n)}=\lim_{a \rightarrow {x_1}} \frac{1}{(x-x_1)(a-x_2)(a-x_3)...(a-x_n)}$ and by expandig the fraction by $(a-x_1)$ you get $lim_{a \rightarrow {x_1}} \frac{(a-x_1)}{(x-x_1)(a-x_1)(a-x_2)(a-x_3)...(a-x_n)}$. And this is $\frac{1}{(x-x_1)f'(x_1)}$. $\endgroup$ – Mrtny Mar 20 '18 at 9:30
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Let $s(x)$ be the summation and notice that

$$s(x)f(x)$$ is the Lagrangian interpolation polynomial* on the points $\left(x_k,y_k\right)$ where $$y_k:=\frac{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}{f'(x_k)}.$$

But we precisely have

$$f'(x_k)=\prod\limits_{\substack{i\ne k}}(x_k-x_i)$$

so that $$s(x)f(x)$$ is identically $1$.


*Identify with

$$\sum_{k} y_k\frac{\prod\limits_{\substack{i\ne k}}(x-x_i)}{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}.$$

For conciseness the ranges $1, n$ have been left implicit.

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$1/(x-x1)(x-x2)..(x-xk)..=A/(x-x1)+B/(x-x2) ...K/(x-xk)..$

So When you multiply both sides by denominator you get, $1=A(x-x2)..(x-xk..(x-xn)+B(x-x1)(x-x3)..(x-xk)..(x-xn) + K(x-x1)..(x-(k-1))(x-(k+1)..(x-xn)$

To find Constant K substitute $x=xk$

After cancelling all terms you end up with, $K=1/(xk-x1)..(xk-(k-1))(xk-(k+1)..(xk-xn)$

We can express $f(x)$ as follows,

$f(x)=(x-xk)*[(x-x1)(x-x2)..(x-(k-1))(x-(k+1)..(x-xn)]$

say

$g(x)=[(x-x1)(x-x2)..(x-(k-1))(x-(k+1)..(x-xn)]$

(also note that $K=1/g(xk)$)

Now $f(x)=(x-xk)*g(x)$

$f'(x)=(x-xk)g'(x)+g(x)*1$

when $x=xk$ is substituted, $f'(xk)=g(xk)$

So K th term is $1/(f'(xk)*(x-xk)$

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