# Find the following finite sum

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

• 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. – Gerry Myerson Aug 15 at 5:03
• @GerryMyerson yes solutions for $f(x)=0$ – user593646 Aug 15 at 5:49
• Answer given is 1 – 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.

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

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