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The degree of the polynomial $\ W(x) $ is equal $\ 2015$. Knowing that $\ W(n)= \frac{1}{n}$ for $\ n = 1,2,...,2016$ calculate $\ W(2017)$. Please help

math.trouble
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    Consider the polynomial $f(x) = xW(x) - 1$. What is $f(n)$ for $n=1,2,3,\ldots$ ? Use this to factor $f(x)$. – Winther Sep 18 '17 at 09:01
  • Raffaele, Can you explain why $\ W(2017)=0$? – math.trouble Sep 18 '17 at 10:56
  • A question like this can have just ONE answer $$W(2017)=0$$ Edit. I left the page 2 hours ago without refreshing it and I did not see the answer below. Anyway: my thought has been the following. The inventor of the question put $2015$ as often happens in context question, but I was convinced that the answer would be the same for a degree like, let's say, $3$. I made the algebra (in a less sophisticated way wrt Julian) and discovered the answer is zero. Call it HOPEFULLY INDUCTION, but it worked :) – Raffaele Sep 18 '17 at 10:59
  • $$W_3(x)=a x^3+b x^2+c x+d$$ Plug the values $$\left{ \begin{array}{l} a+b+c+d=1 \ 8 a+4 b+2 c+d=\frac{1}{2} \ 27 a+9 b+3 c+d=\frac{1}{3} \ 64 a+16 b+4 c+d=\frac{1}{4} \ \end{array} \right.$$ Get the actual $3$rd degree polynomial $$W_3(x)=-\frac{x^3}{24}+\frac{5 x^2}{12}-\frac{35 x}{24}+\frac{25}{12}$$ Evaluate $W(5)$ and get $0$ as result – Raffaele Sep 18 '17 at 11:04

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The polynomial of degree $2015$ is uniquely determined by the 2016 predescribed values. Since $$W(x)=\frac{1-\prod_{n=1}^{2016}\big(\frac{x}{n}-1\big)}{x}$$ has the correct values for $n=1,\dots,2016$, $W$ has to be of this form. Also note that the given formula is indeed a polynomial since the constant term in the numerator vanishes. Thus we can conclude that $W(2017)=0$.

Julian
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