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Intuitively, I would expect the Taylor expansion around $x_0$ of a polynomial in $(x-x_0)$ to be identical to the polynomial. However, I cannot seem to show that/whether this is the case:

For a finite power series $f(x) = \sum_{i=0}^{n} a_i (x-x_0)^i$ the $k$-th derivative is given by $$\frac{d^k f(x)}{dx^k} = \sum_{j=k}^n \frac{j!}{(j-k)!} a_j (x-x_0)^{j-k}$$

To get the Taylor series, I would write $$ F(x) = \sum_{i=0}^{\infty} \frac{(x-x_0)^i}{i!} \frac{d^i f(x_0)}{dx^i} $$ and substitute the expression for the $i$-th derivative of $f(x)$: $$ F(x) = \sum_{i=0}^{\infty} \sum_{j=i}^{n} \frac{j!}{(j-i)!i!} (x_0 - x_0)^{j-i} a_j (x-x_0)^i $$

I am confused by the $(x_0 - x_0)$ factor that appears and the binomial coefficient that shows up. Is my intuition wrong?

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  • $\begingroup$ What do you get if you differentiate an $n$'th degree polynomial $n+1$ times? $\endgroup$ – David Mitra Feb 13 '14 at 13:32
  • $\begingroup$ Right, it becomes zero, so the sum in the expression for $F(x)$ does not need to be evaluated for $i\to\infty$, but just for $i$ up to $n$. $\endgroup$ – Marijnn Feb 13 '14 at 13:37
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The following is an algebraic approach that may not be particularly relevant in your context. We first show that any polynomial $P(x)$ can be expressed as $Q(x-x_0)$, where $Q(t)$ is a polynomial. It is enough to prove the result for $P(x)=x^n$.

We prove that $x^n$ is a polynomial in $x-x_0$ by induction on $n$. This is obviously true if $n=0$. Suppose it is true for $x^k$, for a particular $k$. We show it is true for $x^{k+1}$.

Note that $x^{k+1}=(x-x_0)x_k+x_0x^k$. By the induction hypothesis, $x_k$ is a polynomial in $x-x_0$, and therefore so is $(x-x_0)x^k+x_0x^k$.

Finally, note that the Taylor series for $Q(x-x_0)$ is just $Q(x-x_0)$. This can be done by writing $Q(x-x_0)$ as $a_0+a_1(x-x_0)+\cdots +a_n(x-x_0)^n$ and computing the Taylor coefficients.

Remark: Or else one can use some analysis. The Taylor expansion of $P(x)$ about $x-x_0$ is a polynomial $Q(x-x_0)$ in $x-x_0$. It converges to $P(x)$ for all real $x$. So the polynomial functions $P(x)$ and $Q(x-x_0)$ are the same function. But two polynomials with real coefficients that determine the same fnction are the same polynomial.

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