8. Taylor Approximation

Possible to approximate function around point via tangent:

$$f(x)\approx f(x_0)+f^{\prime }(x_0)(x-x_0)=P_1(x)$$

Higher degrees $\to$ better approximation (degree $n$ $⟹$ matches up to $n$-th derivative):

$$\begin{array}{rl}f(x)& =f(x_0)+f^{\prime }(x_0)(x-x_0)+\cdots +\frac{1}{n!}{f}^{(n)}(x_0)(x-x_0{)}^n\\ & +\frac{1}{(n+1)!}{f}^{(n+1)}(c)(x-x_0{)}^{n+1}\\ & =\sum _{k=0}^n{f}^{(k)}(x_0)\frac{1}{k!}(x-x_0{)}^k+\frac{1}{(n+1)!}{f}^{(n+1)}(c)(x-x_0{)}^{n+1}\\ & =P_n(x)+R_n(x)\end{array}$$

If we take the $i$-th derivative at $x_0$:

• All higher-order terms get killed by $x-x_0$ $=0$ at $x_0$
• All lower-order terms get killed by the power rule
• The $i$-th term evaluates to $\frac{1}{i!}{f}^{(i)}i!(x-x_0{)}^0$ $=f^{(i)}$
  • $\frac{1}{i!}$ is there to cancel out $\frac{d^i}{d{x}^i}(x-x_0{)}^i=i!$

Error: $R_n$ is what we miss. For any $x$, there exists some $c\in (x_0,x)$ that makes the equality hold.

Taylor series: sum into infinity

Whether converges: only if the derivatives grow slower than the $1/(n+1)!$. Otherwise, error doesn’t go down in infinity. Ex. $ln(x)$: $\frac{1}{x},-\frac{1}{x^2},\frac{2}{x^3},-\frac{6}{x^4},\frac{24}{x^5},\dots$ $⟹$ $f^{(n)}(x)=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$

Multiplication of Series

Theorem

Consider two power series

$$f(x)=\sum _{n=0}^{\infty }{a}_n{x}^n,g(x)=\sum _{n=0}^{\infty }{b}_n{x}^n,$$

where $x$ lies in the convergence interval of both series. Then

$$f(x)\cdot g(x)=\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{a}_k{b}_{n-k}\right)x^n.$$

Intuition: