A Taylor series uses the derivatives of a function at one point to build a power series. Its finite versions, called Taylor polynomials, give practical approximations near that point.

Recovering coefficients from derivatives

Suppose a function has a power-series representation centered at $a$:

$$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n =c_0+c_1(x-a)+c_2(x-a)^2+\cdots.$$

Setting $x=a$ removes every positive power of $(x-a)$, so

$$c_0=f(a).$$

Differentiate once:

$$f'(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots.$$

At $x=a$, this gives $c_1=f'(a)$. Differentiating twice gives

$$f''(a)=2!c_2,$$

and continuing the pattern gives

$$f^{(n)}(a)=n!c_n.$$

Therefore every coefficient is forced by the derivatives of $f$ at the center:

$$\boxed{c_n=\frac{f^{(n)}(a)}{n!}}.$$

This also gives a quick way to read derivatives from a known power series: the coefficient of $(x-a)^n$ is $f^{(n)}(a)/n!$.

Example: reading derivatives from a power series

From the geometric-series calculation in the previous section,

$$\frac{3}{2+x^2} =\sum_{n=0}^{\infty}\frac{3(-1)^n}{2^{n+1}}x^{2n}, \qquad |x|<\sqrt2.$$

To find $f^{(10)}(0)$, look for the $x^{10}$ term. Since $2n=10$, it comes from $n=5$, and its coefficient is

$$\frac{3(-1)^5}{2^6}=-\frac{3}{64}.$$

The coefficient of $x^{10}$ is also $f^{(10)}(0)/10!$, so

$$\boxed{f^{(10)}(0)=10!\left(-\frac{3}{64}\right)}.$$

Only even powers appear in this series. There is no $x^9$ term, so its coefficient is $0$ and

$$\boxed{f^{(9)}(0)=0}.$$

Taylor and Maclaurin series

The Taylor series of $f$ centered at $a$ is

$$\boxed{ f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n }$$

where the equality is valid at the values of $x$ for which the series actually converges to $f(x)$. We use the conventions

$$f^{(0)}(a)=f(a), \qquad 0!=1, \qquad (x-a)^0=1.$$

A Taylor series centered at $0$ is called a Maclaurin series:

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n.$$

The degree-$k$ Taylor polynomial keeps only the terms through degree $k$:

$$\boxed{ T_k(x)=\sum_{n=0}^{k} \frac{f^{(n)}(a)}{n!}(x-a)^n }.$$

It is a finite polynomial, so it approximates the function without requiring an infinite sum.

Three standard Maclaurin series

Every derivative of $e^x$ is $e^x$, and each derivative equals $1$ at $0$. Therefore,

$$\boxed{e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$

The derivatives of $\sin x$ cycle through

$$\sin x,\quad \cos x,\quad-\sin x,\quad-\cos x.$$

At $0$, their values cycle through $0,1,0,-1$. Only the odd powers remain:

$$\boxed{ \sin x=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} } =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$

For example, the degree-$4$ Maclaurin polynomial has no fourth-degree term, so

$$T_4(x)=x-\frac{x^3}{3!}.$$

The derivatives of $\cos x$ give the same alternating pattern on even powers:

$$\boxed{ \cos x=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} } =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots.$$

These three series converge to their functions for every real $x$.

The Taylor remainder

A Taylor polynomial is usually an approximation, not the whole function. The difference

$$\boxed{R_k(x)=f(x)-T_k(x)}$$

is the $k$th Taylor remainder. Rearranging gives

$$f(x)=T_k(x)+R_k(x).$$

For a fixed $x$, the Taylor series equals $f(x)$ when

$$\lim_{k\to\infty}R_k(x)=0.$$

The remainder is therefore the part that must disappear as more Taylor terms are included.

Taylor's inequality

Taylor's inequality gives an upper bound on the approximation error. Suppose

$$|f^{(k+1)}(u)|\le M$$

for every $u$ between the center $a$ and the point $x$. Then

$$\boxed{ |R_k(x)| \le \frac{M}{(k+1)!}|x-a|^{k+1} }.$$

The bound becomes smaller when $x$ is closer to the center or when the factorial in the denominator grows faster than the derivative bound.

Example: approximating $e^x$ on $(-1,1)$

The degree-$4$ Maclaurin polynomial for $e^x$ is

$$T_4(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}.$$

Every derivative of $e^x$ is $e^x$. On $(-1,1)$,

$$|f^{(5)}(x)|=e^x\le e,$$

so we may use $M=e$. Taylor's inequality gives

$$\begin{aligned} |R_4(x)| &\le \frac{e}{5!}|x|^5\\ &<\frac{e}{120}\\ &\approx0.023. \end{aligned}$$

Thus, throughout $(-1,1)$,

$$\boxed{|e^x-T_4(x)|<0.023}.$$

The number $0.023$ is a guaranteed worst-case error bound on that interval. The actual error may be smaller.