2
$\begingroup$

$f(x)=c_{2014}x^{2014}+c_{2013}x^{2013}+\dots+c_1x+c_0$ 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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.