I've recently learnt Taylor series, and am trying to prove very simple lemma by induction.
I need to prove that every polynomial $Q(x)$ can be written as
$Q(x)=\sum_{k=0} ^{n} c_k (x-a)^k$
When $n$ is the polynomial degree, $a$ is constant and $(c_k)_{k=0} ^m$ is a sequence of real numbers.
The base case is: $n=0: Q(x)=c_{k}={\displaystyle \sum_{k=0}^{0}c_{k}(x-a)^{k}}$
Induction step:
$\sum_{k=0}^{n}c_{k}(x-a)^{k}+bx^{n+1}=\sum_{k=0}^{n}c_{k}(x-a)^{k}+b((x-a)+a)^{n+1}$
When $b$ is the n+1 coefficient of the given polynomial.
Am I missing something?
Thanks!
Your idea is correct, but some details are missing.
In the case $n=0$, you should say that $Q(x)$ is constant; it always takes a value $M$. So, take $c_0=M$.
Assume that the statement is true when $\deg Q(x)=n$. Let $Q(x)$ be a polynomial whose degree is $n+1$. Then $Q(x)$ can be written as $P(x)+\alpha x^{n+1}$ for some polynomial $P(x)$ such that $\deg P(x)=n$ and some $\alpha\ne0$. So, by the induction hypothesis, $P(x)$ can be written as $\sum_{k=0}^nc_k(x-a)^k$. Besides,\begin{align}\alpha x^{n+1}&=\alpha\bigl(a+(x-a)\bigr)^n\\&=\alpha\sum_{k=0}^{n+1}\binom{n+1}ka^{n+1-k}(x-a)^k.\end{align}Therefore, if$$d_k=\begin{cases}c_k+\alpha\binom{n+1}ka^{n+1-k}&\text{ if }k<n+1\\\alpha&\text{ if }k=n+1,\end{cases}$$then $Q(x)=\sum_{k=0}^{n+1} d_k(x-a)^k$.