I was calculating something and it seemed like any polynomial of the form $$f(x)=x^n+bx^{n-1}+cx^{n-2}+...+d$$ has this property. $$f(x+v)=\frac{\frac{d^{n}}{dv^{n}}f(v)}{(n)!}x^n+\frac{\frac{d^{n-1}}{dv^{n-1}}f(v)}{(n-1)!}x^{n-1}+\frac{\frac{d^{n-2}}{dv^{n-2}}f(v)}{(n-2)!}x^{n-2}...+\frac{f(v)}{0!}$$ Is this true, and if so, why is it true?
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2This is the Taylor series expansion around $v$. – Milten Oct 27 '19 at 17:03
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The solution set is clearly a vector space, so we need only check the case $f(x)=x^n$, which follows by the binomial theorem.
J.G.
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$$f(x)=x^n+bx^{n-1}+cx^{n-2}+...+d$$ Since you want to express $f(x+v)$ as a polynomial in standard form, first write out the taylor series of $f(x)$ around $v$ : $$f(x) = \frac{f^{(n)}(v)}{(n)!}(x-v)^n+\frac{f^{(n-1)}(v)}{(n-1)!}(x-v)^{n-1}+\frac{f^{(n-2)}(v)}{(n-2)!}(x-v)^{n-2}...+\frac{f(v)}{0!}$$
Above replace $x$ by $x+v$.
AgentS
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