Taylor polynomial

Our objective is to approximate a function $f(x)$ about an interior point, say $x_0$. We show that their exists a polynomial, called the Taylor polynomial, $P_n(x)$: $f(x)=P_n(x)+R_n(x)$ Let $\delta=x-x_0$. If $f(x)$ is differentiable at $x_0$ then the equation of the tangent: $y=f(x_0) + \delta f^\prime(x_0)$ is a linear approximation of $f(x)$: $f(x)\approx f(x_0) + \delta f^\prime(x_0)$ A better approximation than a linear can be attained if we start with the following identity: $f(x)=f(x_0)+\int_{x_0}^x f^\prime(t)$ Let $\mathrm{d}t=-\mathrm{d}(x-t)$ (just a trick) and apply integration by parts to the integral: \begin{align*} \int_{x_0}^x f^\prime(t) \mathrm{d}t &= - \int_{x_0}^x f^\prime(t) \mathrm{d}(x-t) \\ &=\left[f^\prime(t)(x-t)\right]_{x_0}^x+\int_{x_0}^x f^{\prime\prime}(t)(x-t) \mathrm{d}t \\ &=\delta f^\prime(x_0) +\int_{x_0}^x f^{\prime\prime}(t)(x-t) \mathrm{d}t \end{align*} This results in: $f(x)=f(x_0)+\delta f^\prime(x_0) + \int_{x0}^x f''(t)(x-t) \mathrm{d}t$

If $f(x)$ is twice differentiable we can continue: \begin{align*} \int_{x0}^x f''(t)(x-t) dt&=-\left[f''(t)\frac{(x-t)^2}{2!}\right]_{x_0}^x+\int_{x0}^x f'''(t)\frac{(x-t)^2}{2!} dt \\ &=\frac{\delta^2}{2!}f''(x_0)+\int_{x0}^x f'''(t)\frac{(x-t)^2}{2!} dt \end{align*} This leads to the following identity: $f(x)=f(x_0)+\delta f'(x_0) + \frac{\delta^2}{2!}f''(x_0)+\int_{x0}^x f'''(t)\frac{(x-t)^2}{2!} dt$ As long as $f(x)$ is differentiable we can continue and achieve the following result: $f(x)=\sum \limits_{k=0}^n \frac{\delta^k}{k!}f^k(x_0)+R_n(x)$ with $R_n(x)=\int_{x0}^x f^{n+1}(t)\frac{(x-t)^n}{n!} dt$

Taylor Polynomial

$P_n(x)=\sum \limits_{k=0}^n \frac{\delta^k}{k!}f^k(x_0), \; \delta=x-x_0$

Taylor series

If for $f(x)$ we have $\lim_{n\rightarrow \infty} R_n(x)=0$ and also $P_n(x)$ converges then we can write $f(x)$ as a Taylor series: $f(x)=\sum \limits_{k=0}^\infty \frac{\delta^k}{k!}f^k(x_0), \; \delta=x-x_0$ We will now derive two forms of the remainder so that we are better able to determine whether this remainder converges. The first form is called Lagrange and the second Cauchy.

Lagrange form of the remainder

Suppose $\delta>0$ and $x=x_o+\delta$, let $f^{n+1}$ be continuous on $[x_0,x]$. Since $f^{n+1}$ is continuous it follows that $f^{n+1}$ takes a minimum $m$ and a maximum $M$ on $[x_0,x]$. Therefore $\int_{x0}^x m\frac{(x-t)^n}{n!} dt \leq \int_{x0}^x f^{n+1}(t)\frac{(x-t)^n}{n!} dt \leq \int_{x0}^x M\frac{(x-t)^n}{n!} dt$ Notice: $\int_{x0}^x \frac{(x-t)^n}{n!} dt = \left[\frac{-(x-t)^{n+1}}{(n+1)!}\right]_{x_0}^x=\frac{\delta^{n+1}}{(n+1)!}$ Substitute this result in the previous equation to get: $m\frac{\delta^{n+1}}{(n+1)!} dt\leq \int_{x0}^x f^{n+1}(t)\frac{(x-t)^n}{n!} dt \leq M\frac{\delta^{n+1}}{(n+1)!}$ So, $R_n(\delta)=\frac{\delta^{n+1}}{(n+1)!}\lambda \;, \lambda \in [m,M]$ From the mean value theorem follows that fn+1 takes on each value between $m$ and $M$ on $(x_0,x)$, so there exists a $\xi \in (x0,x)$ such that $\lambda=f^{n+1}(\xi)$ this finally gives: $R_n(\delta)=\frac{\delta^{n+1}}{(n+1)!}f^{n+1}(\xi) \;, \xi \in (x_0,x)$

Cauchy remainder

Let $F(t)$ defined by: $F'(t)=f^{n+1}(t)\frac{(x-t)^n}{n!}, \; x=x_0+\delta$ this gives: $R_n(\delta)=F(x)-F(x_0)$ then apply mean value theorem: $R_n(\delta)=\delta F'(\xi) \;, \xi \in (x_0,x)$ Substitute the definition for $F'(t)$: $R_n(\delta)=\delta f^{n+1}(\xi)\frac{(x-\xi)^n}{n!}$ Notice $\xi=x_0+\phi\delta,\;0< \phi <1$ and $x-\xi=\delta(1-\phi)$: $R_n(\delta)=f^{n+1}(\xi)\frac{\delta^{n+1}}{n!}(1-\phi)^n \;, 0< \phi <1$