Taylor Theorem

Topics: Calculus - Taylor Polynomial

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Let $f:[a,x]\to \mathbb{R}$ be a Function $C_{n+1}$ (i.e. continuous $n+1$ times; differentiable $n+1$ times).

We define the remainder as:

$$R_{n,a}(x):=f(x)-\left(f(a)+f^{\prime }(a)(x-a)+\cdots +\frac{f^n(a)(x-a{)}^n}{n!}\right)$$

That is, the remainder is the function minus its degree $n$ Taylor polynomial.

Furthermore, we can represent the remainder in the following forms:

1. Lagrange form: for some $t\in (a,x),$

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

1. Cauchy form: for some $t\in (a,x)$,

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

1. Integral form:

$$R_{n,a}(x)={\int }_a^x\frac{f^{n+1}(t)}{n!}(x-t{)}^{n}dt$$

Example with application

We’ll use a neat example to show not only how the remainder can be used, but also what it could be useful for.

Let $f(x)=e^x$. Then, we have that the Taylor Polynomial of $f$ of degree $1$ around $a=0$ is:

$$P_f(1,0)=1+x$$

…which approximates $e^x$ to a certain degree around $0$, of course. We can calculate the difference between this polynomial and actual $f$ by using the remainder:

$$R_f(1,0)=e^x-P_f(1,0)$$

$$R_f(1,0)=e^x-(1+x)$$

Then, we can isolate $e^x$:

$$e^x=(1+x)+R_f(1,0)$$

…rewrite the remainder in its integral form:

$$e^x=(1+x)+{\int }_0^x{e}^t(x-t)dt$$

…and when setting $x=1$, we can approximate the value of $e$:

$$e^1=2+{\int }_0^1{e}^t(1-t)dt$$