Taylor proof from MacLaurin polynomial

Question

My book provides a full proof of MacLaurin theorem by induction, then assert that the theorem can be generalized for each $x_0$ (Taylor theorem), because any polynomial:

$$P_n(x) = {a_n}x^n + a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+...+a_0$$

can be rewritten with center $x_0$ as:

$$P_n(x) = {b_n}(x-x_0)^n + b_{n-1}(x-x_0)^{n-1}+ b_{n-2}(x-x_0)^{n-2}+...+b_0$$

with $b$ that depends from $a$.

But how can I prove always exist a $b $ that depends from $a$ that allows to rewrite the MacLaurin polynomial to Taylor polynomial?

Answer 1

From $$P_n(x) = {a_n}x^n + a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+...+a_0 $$ and $$P_n(x) = {b_n}(x-x_0)^n + b_{n-1}(x-x_0)^{n-1}+ b_{n-2}(x-x_0)^{n-2}+...+b_0 $$ put $x+x_0$ for $x$ to get $$P_n(x+x_0) = {a_n}(x+x_0)^n + a_{n-1}(x+x_0)^{n-1}+ a_{n-2}(x+x_0)^{n-2}+...+a_0 $$ and $$P_n(x+x_0) = {b_n}x^n + b_{n-1}x^{n-1}+ b_{n-2}x^{n-2}+...+b_0 $$

Use the binomial theorem in the first to get

$\begin{array}\\ P_n(x+x_0) &=\sum_{k=0}^n a_k(x+x_0)^k\\ &=\sum_{k=0}^n a_k\sum_{j=0}^k \binom{k}{j}x^j x_0^{k-j}\\ &=\sum_{j=0}^n\sum_{k=j}^n a_k \binom{k}{j}x^j x_0^{k-j}\\ &=\sum_{j=0}^nx^j\sum_{k=j}^n a_k \binom{k}{j} x_0^{k-j}\\ \end{array} $

This gives

$b_j = \sum_{k=j}^n a_k \binom{k}{j} x_0^{k-j} $.

Answer 2

For any $x$ and any $x_0$ let $y_{x,x_0}=x-x_0.$ So $x=y_{x,x_0}+x_0.$ So $$P(x)=a_0+a_1x+a_2x^2+...=a_0+a_1(y_{x,x_0}+x_0)+a_1(y_{x,x_0}+x_0)^2+...$$

Expand each of the terms in the above line and collect the result into powers of $y_{x,x_0}$ and we have $$P(x)=b_0+b_1y_{x,x_0}+b_2y_{x,x_0}^2+...$$ where $b_0,b_1,b_2,...$ are constants that are expressible in terms of $x_0$ and $a_0,a_1,a_2,...$ and do not depend on $x.$ Now since $y_{x,x_0}=x-x_0$ we have $$P(x)=b_0+b_1(x-x_0)+b_2(x-x_0)^2+...$$ for EVERY $x.$

This was known long before Taylor and MacLaurin.