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How to find the $n$-th degree polynomial function, $f:F\rightarrow F$, using the following?
Given the function $f:F\rightarrow F$ defined by:
$$\forall x\in F,f(x)=x^2$$
I have to find the $n$-th degree polynomial function, $f:F\rightarrow F$, by using the following formula:
$$f(x)=\sum_{i=0}^{n-1}a_ix^i$$
and $a_n=1$, $a_i=0$, for $i<n$
How to calculate $a_n$?
A:
The standard technique would be to take the derivative and "fix" the $a_n$ term. I'll give a hint, but for actual calculation, you need some results from complex analysis.
Let $g(x)=\sum_{i=0}^{n-1}a_ix^i$. Now consider $h(x)=g(x)+g(x-1)+g(x-2)+\cdots+g(x-n+1)$. It will follow from your assumptions that $h(x)=f(x)$. Now, $h'(x) = g'(x) = g'(x-1) + g'(x-2) + \cdots + g'(x-n+1)$, and $g'(x)=2xg(x)$. This means that $h'(x)=2xf(x)$. Since $h(0)=g(0)+g(1)+g(2)+\cdots+g(n-1)=\sum_{i=0}^{n-1} a_i = f(0)$, it follows that $h(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$. But $a_n=1$, so this lets you calculate the rest of the $a_i$.
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