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The degree of the polynomial $W(x)$ is $2015$
$W(n) = \frac{1}{n}$ for $ n \in \{1,2,3,...,2016\}$

Calculate $W(2017)$.

I came to the conclusion $f(x) = 1 - xW(x)$, after checking all the $n$ and the degrees of the polynomials. $f(x) = a(x-1)(x-2)(x-3)...(x-2016)$ so $W(x) = \frac{1-f(x)}{x}$ After seeing an similar question I know that $a = \frac{1}{2016!}$ but how to prove it?

VereX
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2 Answers2

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To prove this, note that the constant term of $$f(x)=1-xW(x)$$ is $1$, so that $f(0)=1$, and therefore $1/a=(-1)^{2016}2016!$.

Angina Seng
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You need $1-f(0)$ to be $0$, because otherwise $\frac{1-f(x)}{x}$ would blow up at $0$ and then it certainly wouldn't be a polynomial function.

So $f(0) = a\cdot 2016!$ must be $1$.

hmakholm left over Monica
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