Taylor Polynomials and Series

What is the Taylor polynomial?

$$T_{n,a}(f(x))=\sum_{r=0}^{n}\frac{f^{r}\left(a\right)}{r!}\left(x-a\right)^{r}$$

with $f^0(x)=f(x)$

What’s the remainder of a Taylor polynomial?

$$R_{n,a}f(x)=f(x)-T_{n,a}f(x)$$

or

$$R_{n,a}f(x)=T_{n,x}f\left(x\right)\left.-T_{n,a}f(x\right)$$

If we define $h(t)=T_{n,t}f(x)$ then

$$R_{n,a}f(x)=h(x)-h(a)$$

If the first n+1 derivatives of $f$ exist on an open neighbourhood containing a and x. Then what does $\frac{d}{dt} T_{n,t}f(x)$ =?

$$\frac{(x-t)^n}{n!}f^{(n+1)}(t)$$

Proof:

Differentiating each term of $T_{n,t}f(x)$ leads to cancelations leaving only the $f^{(n+1)}$ term.

State and prove Taylor’s Theorem with Cauchy’s form of the error?

Assume the first n+1 derivatives of f exist on an open neighbourhood containing a and x. Then $R_{n,a}f(x)=\frac{f^{(n+1)}\left(c\right)}{n!}\left(x-c\right)^{n}\left(x-a\right)$ for some c between a and x.

Proof:

Assume the first $n+1$ derivatives of $f$ exist on an open interval containing $a$ and $x$.

Let $h(t)=T_{n,t}f(x)$. Hence, we have $R_{n,a}f(x)=f(x)-T_{n,a}f(x)=T_{n,x}f(x)-T_{n,a}f(x)=h(x)-h(a)$.

By the Mean Value Theorem, for some $c$ between $a$ and $x$, $\frac{h(x)-h(a)}{x-a}=h'(c)$. By results in the course, $h'(c)=\frac{(x-c)^n}{n!}f^{(n+1)}(c)$. Therefore $R_{n,a}f(x)=(x-a)h'(c)=\frac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a)$.

State and prove Taylors Theorem with Lagrange’s error?

Assume the first n+1 derivatives of f exist on an open neighbourhood containing a and x. Then $R_{n,a}f(x)=\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}\left(x-a\right)^{n+1}$ for some c between a and x.

Proof:

Assume the first $n+1$ derivatives of $f$ exist on an open interval containing $a$ and $x$.

Let $h(t)=T_{n,t}f(x)$, so $R_{n,a}f(x)=h(x)-h(a)$.

Apply Cauchy’s Mean Value Theorem to $h$ and a function $g$ (differentiable on $(a,x)$ and continuous on $[a,x]$) chosen so that $g'(t)=(x-t)^n$.

Then for some $c$ between $a$ and $x$, $\frac{R_{n,a}f(x)}{g(x)-g(a)}=\frac{h'(c)}{g'(c)}$ by Cauchy’s M.V.T.

$h'(c)=\frac{(x-c)^n}{n!}f^{(n+1)}(c)$, so $\frac{h'(c)}{g'(c)}=\frac{f^{(n+1)}(c)}{n!}$.

Also, $g(x)-g(a)=\int_a^x (x-t)^n\,dt=\frac{(x-a)^{n+1}}{n+1}$ .

Therefore $R_{n,a}f(x)=\frac{f^{(n+1)}(c)}{n!}\frac{(x-a)^{n+1}}{n+1}=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$.

What’s the Taylor series of f at a?

Assume that all derivatives of f exist in some neighbourhood of $a\in \mathbb{R}$. The Taylor series is

$$\sum_{r=0}^{\infty}\frac{f^{\left(r\right)}\left(a\right)}{r!}\left(x-a\right)^{r}$$

What two questions must be asked about a Taylor series?

1. Does it converge

2. If it does, does it converge to $f(x)$

A Taylor Series converging doesn’t automatically mean it converges to the function.

What does the radius of convergence tell us for a Taylor series?

There is some $R\ge0$ such that:

• if ∣x−a∣<R, the series converges;

• if∣x−a∣>R, the series diverges;

• if$∣x−a∣=R$, the series must be checked separately.

When does a Taylor series converge to $f(x)$?

If $\lim_{n\to\infty}R_{n,a}f\left(x\right)=0$.

To prove it converges to the original function use an error formula (usually Lagrange’s) and bound it and show it tends to 0 as n tends to infinity.