$f(x)=c_{2014}x^{2014}+c_{2013}x^{2013}+\dots+c_1x+c_0$ has 2014 roots $a_1,\dots,a_{2014}$ and $g(x)=c_{2014}x^{2013}+c_{2013}x^{2012}+\dots+c_1 $. Given that $c_{2014}=2014$ and $f '(x)$ is the derivative of $f(x)$, find the sum $\sum_{n=1}^{2014}\frac{g(a_n)}{f '(a_n)}$.

$f(x)=2014(x-a_1)(x-a_2)\dots (x-a_{2014})$

$f'(x)=2014^2x^{2013}+2013\cdot c_{2013}x^{2012}+\dots+c_1$

is there any relation between $f(a_n)$, $g(a_n)$ and $f'(a_n)$ or do you need different approach to solve? Edit As Gerry suggested now I have $\frac{-c_0}{2014}\left(\frac1{a_1\prod_{i\neq 1} (a_1-a_i)}+\frac1{a_2\prod_{i\neq 2} (a_2-a_i)}+\dots+\frac1{a_{2014}\prod_{i\neq2014} (a_{2014}-a_i)}\right)$

Would be helpful if someone tell me what to do next

  • $\begingroup$ When you write that $f(x)$ has $2014$ solutions, do you mean that the equation $f(x)=0$ has $2014$ solutions? If so, please edit the body of your question to reflect this. $\endgroup$ – Gerry Myerson Aug 15 at 5:03
  • $\begingroup$ @GerryMyerson yes solutions for $f(x)=0$ $\endgroup$ – user593646 Aug 15 at 5:49
  • $\begingroup$ Answer given is 1 $\endgroup$ – user593646 Aug 15 at 5:52

Here's what you need:

$g(x)=(f(x)-c_0)/x$, and $f(a_n)=0$, so $g(a_n)=-c_0/a_n$.
Also, $c_0=2014a_1a_2\cdots a_{2014}$.
Also, $f'(x)=2014\sum_n\prod_{i\ne n}(x-a_i)$, so $f'(a_n)=2014\prod_{i\ne n}(a_n-a_i)$.

Now you have to put those all together, and hit it with a dose of algebra.

  • $\begingroup$ I also got to this point. How to continue after this. $\endgroup$ – Sonal_sqrt Aug 15 at 8:16
  • $\begingroup$ @Sonal, let's wait and see how OP gets along. $\endgroup$ – Gerry Myerson Aug 15 at 8:17
  • $\begingroup$ That's really helpful but I am having trouble solving after that and also if some $a_n$ is $0$ what would we do $\endgroup$ – user593646 Aug 15 at 8:39
  • 1
    $\begingroup$ Show me what progress you have made, user5. $\endgroup$ – Gerry Myerson Aug 15 at 8:44
  • $\begingroup$ I have edited it now in the question $\endgroup$ – user593646 Aug 15 at 9:37

Here I provide a simple method which involves complex analysis for $c_0\neq 0$.

The sum implicitly implies all zeros of $f(z)$ are simple. Using residue theorem, the sum you want to find is equal to $$\frac{1}{2\pi i}\oint_C \frac{g(z)}{f(z)}dz$$ where C is circle given by $Re^{i t}$, and $R$ is large enough such that all zeros of $f(z)$ are inside C.

Note that $g(z)=\frac{f(z)-c_0}{z}$ for $z\neq 0$, so $$ \oint_C \frac{g(z)}{f(z)}dz=\oint_C \left(\frac{1}{z}-c_0\frac{1}{zf(z)}\right)dz$$ As $$\oint_C \frac{1}{z}dz=2\pi i$$ and $$\oint_C \frac{1}{zf(z)}dz=0$$ The result follows immediately.

  • $\begingroup$ Thanks I appreciate it but can we do it without complex analysis $\endgroup$ – user593646 Aug 15 at 9:35
  • $\begingroup$ Gerry Myerson's approach should work. $\endgroup$ – DragunityMAX Aug 15 at 9:39
  • $\begingroup$ Shouldn't $c_0/f(z)$ in the integrand be $c_0/(zf(z))$? $\endgroup$ – Gerry Myerson Aug 15 at 12:12
  • $\begingroup$ Yeah, thank for correction. $\endgroup$ – DragunityMAX Aug 15 at 12:51

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