Degree of $W(x)$ is 2015. $W(x)=\frac{1}{x}$ for $x=1,2,...,2016$. Calculate $W(2017)$.
I think that $W(2017)=0$ but how to prove this? I thought of $W(x)=\frac{1-(x-1)(x-2)...(x-2016)}{x}$, but I don't know how to factor $x$ out of counter.
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I do not think the $W$ that you defined is a polynomial. – Aniruddha Deshmukh Oct 24 '17 at 18:07
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1Consider $xW(x)-1$. – Angina Seng Oct 24 '17 at 18:09
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@AniruddhaDeshmukh: That relation holds only for a particular set of $x$, not all $x$. – Alex R. Oct 24 '17 at 18:09
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Same type of problem here and here. – StubbornAtom Oct 24 '17 at 18:10
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Also: https://math.stackexchange.com/q/1379483/42969 – Martin R Oct 24 '17 at 18:11
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Using my "hopefully induction" I say that $W(2017)=0$ :)) – Raffaele Oct 24 '17 at 18:56
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Suppose the leading coefficient of $W(x)$ is $1$. Note that the degree of $W(x)$ is $2015$ and so the degree of $xW(x)-1$ is $2016$. Now $xW(x)-1$ has roots $1,2,\cdots,2016$. Then $$ xW(x)-1\equiv (x-1)(x-2)\cdots(x-2016). $$ So $$ 2017W(2017)-1=2016\cdot2015\cdot\cdots1=2016! $$ and hence $$ W(2017)=\frac{2016!+1}{2017}. $$
xpaul
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