of the form:
\(\sum_{n=0}a_n(x-c)^n\)
Power series are all smooth. That is, they are infinitely differentiable.
\(f(x)\) can be estimated at point \(c\) by identifying its repeated differentials at point \(c\).
The coefficients of an infinate number of polynomials at point \(c\) allow this.
\(f(x)=\sum_{i=0}^{\infty }a_i(x-c)^i\)
\(f'(x)=\sum_{i=1}^{\infty }a_i(x-c)^{i-1}i\)
\(f''(x)=\sum_{i=2}^{\infty }a_i(x-c)^{i-2}i(i-1)\)
\(f^j(x)=\sum_{i=j}^{\infty }a_i(x-c)^{i-j}\dfrac{i!}{(i-j)!}\)
For \(x=c\) only the first term in the series is non-zero.
\(f^j(c)=\sum_{i=j}^{\infty }a_i(c-c)^{i-j}\dfrac{i!}{(i-j)!}\)
\(f^j(c)=a_ij!\)
So:
\(a_j=\dfrac{f^j(c)}{j!}\)
So:
\(f(x)=\sum_{i=0}^\infty (x-c)^i \dfrac{f^i(c)}{i!}\)
If \(x=c\) then the power series will be equal to \(a_0\).
For other values the power series may not converge.
Radius of convergence:
\(\dfrac{1}{R}={\lim \sup}_{n_\rightarrow \infty} (|a_n|^{\dfrac{1}{n}})\)
A Taylor series around \(c=0\).
\(f(x)=\sum_{i=0}^\infty (x-c)^i \dfrac{f^i(c)}{i!}\)
\(f(x)=\sum_{i=0}^\infty (x)^i \dfrac{f^i(0)}{i!}\)
For example, for:
\(f(x)=(1-x)^{-1}\)
\(f^i(0)=i!\)
So, around \(x=0\):
\(f(x)=\sum_{i=0}^\infty (x)^i\)
(root test, direct comparison test, rate of convergence, radius of convergence)