Taylor Series

## Taylor series

Taylor's theorem.

1. If $f\left(x\right)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots$ then $a l = 1 l ! d l f d x l | x = 0 .$
2. If $f:\left[a,b\right]\to ℝ$ and $N\in {ℤ}_{>0}$ and ${f}^{\left(N\right)}:\left[a,b\right]\to ℝ$ is continuous and ${f}^{\left(N+1\right)}:\left(a,b\right)\to ℝ$ exists then there exists $c\in \left(a,b\right)$ such that $f b = f a + f ′ a b - a + 1 2 ! f ″ a b - a 2 + ⋯ + 1 N ! f N a b - a N + 1 N + 1 ! f N + 1 c b - a N + 1 .$

Remark. The following results follow from (a) after a tiny bit of algebra. $If f a + x = a 0 + a 1 x + a 2 x 2 + ⋯ then a l = 1 l ! d l f d x l | x = 0 .$ $If f z = a 0 + a 1 z - a + a 2 z - a 2 + ⋯ then a l = 1 l ! d l f d z l | z = a .$

Remark. The special case $N=1$ is the mean value theorem. The special case $N=1$ and $f\left(a\right)=f\left(b\right)$ is Rolle's theorem. The last term in $f a + f ′ a b - a + 1 2 ! f ″ a b - a 2 + ⋯ + 1 N ! f N a b - a N + 1 N + 1 ! f N + 1 c b - a N + 1$ is Lagrange's form of the remainder.

 Proof of Taylor's theorem, part (a). Assume $f\left(x\right)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots =\sum _{i=0}^{\infty }{a}_{i}{x}^{i}$ To show: ${a}_{l}=\left({\frac{{d}^{l}f}{d{x}^{l}}|}_{x=0}\right)$ for $l\in {ℤ}_{>0}$. We have $d l f d x l = d l d x l ∑ i = 0 ∞ a i x i = ∑ i = l ∞ i - l - 1 i - l - 2 ⋯ i - 1 i a i x i - l = ∑ i = l ∞ ∏ j = 1 l i - l - j a i x i - l ,$ by repeated differentiation. So $d l f d x l | x = 0 = ∑ i = l ∞ ∏ j = 1 l i - l - j a i x i - l | x = 0 = ∏ j = 1 l l - l - j a l = l ! a l .$ Solving for ${a}_{l}$ givs the required result.proof changed

 Proof of the mean value theorem. To show: there exists $c\in \left(a,b\right)$ such that $f b = f a + f ′ c b - a .$ First do the case: $f\left(a\right)=f\left(b\right)$. Then, since $\left[a,b\right]$ is compact and connected and $f:\left[a,b\right]\to ℝ$ is continuous, $f a b is compact and connected.$ So $f\left(\left[a,b\right]\right)$ is a closed interval. So $f\left(\left[a,b\right]\right)=\left[\mathrm{min},\mathrm{max}\right]$ for some $\mathrm{min},\mathrm{max}\in ℝ$. So there exists $c\in \left(a,b\right)$ such that $f\left(c\right)=\mathrm{max}$. Then, if $\epsilon \in {ℝ}_{>0}$ is small enough $f\left(c+\epsilon \right)\le f\left(c\right)$ and $f\left(c-\epsilon \right)\le f\left(c\right)$. So $f\prime \left(c\right)\ge 0$ and $f\prime \left(c\right)\le 0$. So $f\prime \left(c\right)=0$. Next do the case $f\left(a\right)\ne f\left(b\right)$. Let $g\left(x\right)=-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)\left(x-a\right)+f\left(x\right)$. So $g\left(a\right)=f\left(a\right)$ and $g\left(b\right)=f\left(a\right)$ and $g\prime \left(x\right)=-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)+f\prime \left(x\right).$ So by the first case, there exists $c\in \left(a,b\right)$ with $g\prime \left(c\right)=0$. So $f\prime \left(c\right)-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)=0$. So $f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ and $f\left(b\right)=f\left(a\right)+f\prime \left(c\right)\left(b-a\right)$.