8. Taylor’s formula and approximations

Using Lagrange theorem to approximate values

Recall that Lagrange Theorem states that for every function differentiable in an interval $[x_0, x+h]$ , there exists $c$ in this interval, such that

$\frac{f(x_0+h)-f(x_0)}{h}=f'(c),$

thus

$f(x_0+h)=f(x_0)+hf'(c),$

but for small $h$ we can approximate $f'(c)\simeq f'(x_0)$ and get the value $f(x_0+h)$ approximated by the values of $f$ and $f'$ at $x_0$ , so:

$f(x_0+h)\simeq f(x_0)+hf'(x_0).$

Taylor’s Theorem

Taylor’s Theorem is a very powerful generalization of Lagrange Theorem. It gives us a possibility to approximate a function with polynomials.

Let $k\in\mathbb{N}$ and $f\colon\mathbb{R}\to\mathbb{R}$ has all derivatives, up to $(k+1)$ -th. Let $x_0\in\mathbb{R}$ , then:

$f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f'''(x_0)}{3!}(x-x_0)^3+$

$+\ldots+\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+\frac{f^{(k+1)}(\theta)}{(k+1)!}(x-x_0)^{k+1},$

where $\theta$ is some number between $x$ and $x_0$ .

Sum $f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f'''(x_0)}{3!}(x-x_0)^3+$ $\ldots+\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ will be called $k$ -th order Taylor polynomial of $f$ in point $x_0$ , and $R_k(x)=\frac{f^{(k+1)}(\theta)}{(k+1)!}(x-x_0)^{k+1}$ is the $k$ -th remainder term.

Calculating approximations

Taylor’s Theorem enables calculating some approximate values of a function in a given point. E.g., $\cos 0.01$ . Its second order Taylor polynomial in $0$ is $1+0+\frac{-1}{2!}x^2$ , which for $x=0.01$ is $1-\frac{0,0001}{2}=0,99995$ and it is $\cos 0.01$ up to the remainder term, so the possible error is not greater than $\left|\frac{-\sin\theta}{3!}(0,01)^3\right|\leq\frac{0,000001}{6}$ .

Taylor series

Therefore, if on a given interval $R_n(x)\to 0$ , then we can write $f$ as a sum of a series. E.g. for $e^x$ $R_n$ is convergent on the whole real line, so the Taylor series of $e^x$ in point $1$ (it is easy because $e^{(k)}(x)=e^x$ ) is: