Recently I've asked a question related with the Taylor Series of the Lagrange Basis functions. In the comments someone wrote that the Taylor series of a polynomial is the polynomial. I know that eventually the derivatives will be zero...but I cannot see why we get the same polynomial.
$$T_{a}\{f\}(x) = \sum_{k=0}^{\infty} { \frac{f^{(k)}(a)(x-a)^k}{k!} } $$
if we take $f(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0} $
now $$ f^{(0)}(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0} = \sum_{j=0}^{n}{a_{j}x^{j}} $$ $$ f^{(1)}(x) = na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_{1}= \sum_{j=1}^{n}{ja_{j}x^{j-1}} $$ $$ f^{(2)}(x) = n(n-1)a_{n}x^{n-2}+(n-1)(n-2)a_{n-1}x^{n-3}+\cdots+2a_{2} =\sum_{j=2}^{n}{j(j-1)a_{j}x^{j-2}} $$
$$ \cdots $$
$$ f^{(n)}(x) = n(n-1)(n-2)\cdots(2)(1)a_{n} = \sum_{j=n}^{n}{j(j-1)(j-2) \cdots (j-n+1 ) a_{j}x^{j-n}} $$
maybe we can put this in the infinite sum and see what happens
$$ f^{(n+1)}(x) = 0 $$
Also we can use the Binomial Theorem: $$(x+(-a))^{k} = \sum_{i=0}^{k}{ \frac{k!}{(k-i)!(i)!}x^{k-i}(-a)^{i} }$$
These are my questions:
1.Do you know how to prove that?
2.what book do you recommend where they do the proof? Thank you!
Taylor's Theorem says that if $f$ has derivatives $f^{(1)},f^{(2)},...,f^{(n)},f^{(n+1)}$ defined on an interval $[a,x]$ then
$$f(x) - T_{a}^{n}\{f\}(x) = \frac{f^{(n+1)}(\lambda)}{(n+1)!}(x-a)^{(n+1)}$$
for some $\lambda \in (a,x)$ and where
$$T_{a}^{n}\{f\}(x) = \sum_{k=0}^{n} { \frac{f^{(k)}(a)(x-a)^k}{k!} } $$
if we choose $f(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\ \ $ a polynomial of degree $n$... note that $f^{(n+1)}(x) = 0$. For example $h(x)=x^2$ has derivatives $h^{(1)}(x)=2x$, $h^{(2)}(x)=2$ and finally $h^{(3)}(x) =0$
Then
$$f(x) - T_{a}^{n}\{f\}(x) = \frac{f^{(n+1)}(\lambda)}{(n+1)!}(x-a)^{(n+1)} = \frac{0}{(n+1)!}(x-a)^{(n+1)} = 0$$ and this implies that
$$ T_{a}^{n}\{f\}(x) = f(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$
Thus the Taylor Series of a polynomial is the polynomial.