Walkthrough

The core idea: matching a curve locally

Suppose you have a complicated function $f(x)$, but you only know what is happening at a single center point $x = a$. You know its value $f(a)$, its slope $f'(a)$, its concavity $f''(a)$, and perhaps higher derivatives.

Can we build a simple polynomial that matches the behavior of $f(x)$ near $a$?

A tangent line matches the value and the first derivative at $a$. If we add a quadratic term, we can match the second derivative as well. By adding terms of higher degree, we can match as many derivatives at $a$ as we like. The resulting polynomial bends and curves to mimic $f(x)$ around the center.

Why factorials appear in the formula

To see how derivatives determine the coefficients, suppose first that $f$ has a power-series representation near $a$:

This is a temporary assumption for deriving the coefficients. A smooth function need not equal its Taylor series; we will return to the remainder condition below.

If we plug in $x = a$, every term with $(x-a)$ vanishes, leaving just:

Now differentiate the series term by term:

Evaluating at $x = a$ isolates the next coefficient: $c_1 = f'(a)$.

Differentiate a second time:

Plugging in $x = a$ gives $f''(a) = 2! c_2$, so $c_2 = \frac{f''(a)}{2!}$.

Every time you take another derivative, the power rule brings down the next integer exponent. By the $n$th derivative, that power has contributed a factor of $n!$:

This formula means every coefficient is locked in place by a derivative of $f$ evaluated at the single point $a$.

Taylor polynomials and the Taylor series

Using these coefficients, the Taylor series associated with $f$ and centered at $a$ is:

This series is built from the derivatives of $f$ at one point, but it equals $f(x)$ only where its remainder approaches zero. When the center is set to $a = 0$, it is called a Maclaurin series. For example, because every derivative of $e^x$ is $e^x$ and $e^0 = 1$, its Maclaurin series is:

In practical calculations, we cannot sum infinitely many terms. Instead, we stop at a finite degree $k$ to form the degree-$k$ Taylor polynomial:

$T_k(x)$ is a standard polynomial that serves as a practical, computable approximation of $f(x)$ close to $a$.

Controlling the error: Taylor's inequality

An approximation is only useful if we know how accurate it is. We define the discrepancy between the true function and the degree-$k$ polynomial as the remainder:

The Taylor series converges to $f(x)$ if and only if this remainder approaches zero as the degree $k$ grows.

To determine how large this error can be, Taylor's inequality looks at the very next derivative, $f^{(k+1)}$. If we can find an upper bound $M$ such that $|f^{(k+1)}(u)| \le M$ for all points $u$ between $a$ and $x$, then:

For a fixed degree $k$, this bound shows why Taylor approximation is local: the factor $|x-a|^{k+1}$ grows as $x$ moves away from the center. Increasing $k$ can improve the estimate only when the corresponding derivative bounds $M_k$ remain controlled well enough for

to approach $0$. Smoothness alone does not guarantee that this happens.

Recovering coefficients from derivatives

A Taylor series uses the derivatives of a function at a single point to construct a power series. Truncating this series to a finite degree yields a Taylor polynomial, which provides a practical polynomial approximation of the function near that center point.

To see how derivatives determine the terms of a power series, suppose a function $f$ is represented by a power series centered at $a$:

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

Evaluating both sides at $x=a$ eliminates every term containing a positive power of $(x-a)$, isolating the constant term:

$$c_0=f(a).$$

Differentiating term by term gives

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

Evaluating at $x=a$ leaves only $c_1=f'(a)$. Differentiating a second time and evaluating at $x=a$ gives

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

which implies $c_2=f''(a)/2!$. Repeating this differentiation step $n$ times yields

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

Solving for $c_n$ shows that each coefficient is uniquely determined by the $n$th derivative of $f$ at the center:

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

This relationship also provides a direct way to read derivatives from an already known power series: the coefficient of $(x-a)^n$ is always $f^{(n)}(a)/n!$.

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 $f^{(10)}(0)$, locate the coefficient of the $x^{10}$ term. Setting the exponent $2n=10$ requires $n=5$, which gives the coefficient

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

Because this coefficient must equal $f^{(10)}(0)/10!$, equating the two gives

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

Since only even powers appear in this series, the coefficient of $x^9$ is $0$, which immediately implies

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

Taylor and Maclaurin series

Substituting the formula for $c_n$ into the general power series gives the Taylor series of $f$ centered at $a$:

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

This equality is valid for all values of $x$ where the series converges to $f(x)$. The definition relies on the standard conventions

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

When the center is chosen as $a=0$, the series is called a Maclaurin series:

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

Keeping only the terms through degree $k$ yields the degree-$k$ Taylor polynomial:

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

Because $T_k(x)$ is a finite polynomial rather than an infinite sum, it serves as a computable approximation of $f(x)$ near $a$.

Three standard Maclaurin series

Every derivative of $e^x$ is $e^x$, and evaluating each derivative at $0$ gives $1$. Consequently,

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

The derivatives of $\sin x$ cycle in a repeating pattern:

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

At $x=0$, these derivative values cycle through $0,1,0,-1$. The even powers drop out because their coefficients are zero, leaving only odd powers with alternating signs:

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

Because the fourth-degree coefficient is $0$, the degree-$4$ Maclaurin polynomial contains no fourth-degree term and matches the degree-$3$ polynomial:

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

Applying the same process to $\cos x$ produces the alternating pattern on even powers:

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

Each of these three series converges to its corresponding function for every real $x$.

The Taylor remainder and approximation error

Because a Taylor polynomial $T_k(x)$ is an approximation, the difference between the true function value and the polynomial is quantified by the $k$th Taylor remainder:

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

Rearranging shows how the function splits into the polynomial approximation and its error term:

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

For a given $x$, the Taylor series converges to $f(x)$ precisely when the remainder vanishes as the degree increases:

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

Thus, confirming that a Taylor series represents $f(x)$ requires verifying that this error term tends to zero.

Taylor's inequality

Taylor's inequality provides an explicit upper bound on the size of the remainder. Suppose that the $(k+1)$th derivative is bounded by a constant $M$ such that

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

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

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

This upper bound decreases as $x$ moves closer to $a$, or as the factorial $(k+1)!$ in the denominator outgrows the derivative bound $M$.

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

Consider approximating $e^x$ on $(-1,1)$ using its degree-$4$ Maclaurin polynomial:

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

The error depends on the next derivative, $f^{(5)}(x)$. Because every derivative of $e^x$ is $e^x$, and $e^x$ increases across $(-1,1)$,

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

allowing us to choose $M=e$. Applying Taylor's inequality with $k=4$, center $a=0$, and $|x|<1$ yields

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

Therefore, for all $x$ in $(-1,1)$,

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

The value $0.023$ is a guaranteed upper bound on the approximation error across the interval; the true discrepancy at any specific point may be smaller.

Convergence of the standard series everywhere

Writing down Taylor coefficients does not automatically prove that the resulting series equals the original function. The missing step is to show that the remainder tends to zero.

For $f(x)=\sin x$, every derivative is one of $\pm\sin x$ or $\pm\cos x$. Therefore,

$$|f^{(k+1)}(u)|\le1$$

for every real $u$. Taylor's inequality then gives

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

For each fixed real $x$, the factorial in the denominator eventually grows faster than the fixed power in the numerator, so

$$\frac{|x|^{k+1}}{(k+1)!}\longrightarrow0.$$

The Squeeze Theorem now forces $R_k(x)\to0$. This proves that the sine Maclaurin series represents $\sin x$ for every real $x$. The cosine argument is identical because its derivatives have the same absolute bound.

The same idea works for $e^x$, but its derivatives are not globally bounded by one. Instead, fix any $r>0$ and consider $|x|<r$. Since every derivative of $e^x$ is $e^x$,

$$\left|\frac{d^{k+1}}{dx^{k+1}}e^x\right| =e^x\le e^r.$$

If

$$T_k(x)=\sum_{n=0}^{k}\frac{x^n}{n!},$$

then Taylor's inequality gives

$$|e^x-T_k(x)| \le\frac{e^r|x|^{k+1}}{(k+1)!} \le\frac{e^r r^{k+1}}{(k+1)!} \longrightarrow0.$$

Thus the series equals $e^x$ whenever $|x|<r$. Because $r$ can be chosen larger than any fixed $|x|$, the identity holds for every real $x$.

Small-angle approximations and error bounds

The first nonzero terms of the sine and cosine series give the familiar small-angle approximations

$$\sin x\approx x, \qquad \cos x\approx1-\frac{x^2}{2}.$$

Here $x$ must be measured in radians. Taylor's inequality tells us exactly how much error these shortcuts can introduce.

For sine, the first- and second-order Maclaurin polynomials are both $x$ because the quadratic coefficient is zero:

$$T_1(x)=T_2(x)=x.$$

Since $f^{(3)}(x)=-\cos x$ and $|f^{(3)}(x)|\le1$,

$$\boxed{|\sin x-x|\le\frac{|x|^3}{3!}}.$$

For example, if $|x|\le\frac12$ radians, then

$$|\sin x-x| \le\frac16\left(\frac12\right)^3 =\frac1{48} \approx0.0208.$$

For cosine, the second- and third-order polynomials are both $1-x^2/2$ because the cubic coefficient is zero:

$$T_2(x)=T_3(x)=1-\frac{x^2}{2}.$$

Using the fourth derivative gives

$$\boxed{ \left|\cos x-\left(1-\frac{x^2}{2}\right)\right| \le\frac{|x|^4}{4!} }.$$

Therefore, when $|x|\le\frac12$,

$$\left|\cos x-\left(1-\frac{x^2}{2}\right)\right| \le\frac1{384} \approx0.00260.$$

The power of $|x|$ explains why the approximation improves rapidly near the center, while the factorial controls how quickly higher-order Taylor errors shrink.

Building a logarithm series from the geometric series

Instead of taking successive derivatives, some power series are easier to derive by integrating or differentiating a known series. Start with the geometric-series identity and rewrite $1/x$ around the center $x=1$:

$$\begin{aligned} \frac1x &=\frac1{1-(1-x)}\\ &=\sum_{n=0}^{\infty}(1-x)^n\\ &=\sum_{n=0}^{\infty}(-1)^n(x-1)^n, \qquad |x-1|<1. \end{aligned}$$

Because $(\ln x)'=1/x$, integrate the series term by term inside its radius of convergence:

$$\ln x =C+\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}(x-1)^{n+1}.$$

Reindexing gives

$$\ln x =C+\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n.$$

Substituting $x=1$ makes every series term vanish, so $C=\ln1=0$. Hence

$$\boxed{ \ln x =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n }, \qquad |x-1|<1.$$

Replacing $x$ by $x+1$ shifts the center to zero:

$$\boxed{ \ln(1+x) =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n }, \qquad |x|<1.$$

These conditions describe the open radius interval. As with any power series, endpoints must be tested separately.

The generalized binomial series

For a positive integer $b$, the ordinary binomial theorem is the finite expansion

$$(1+x)^b=\sum_{n=0}^{b}\binom{b}{n}x^n,$$

where

$$\binom{b}{n} =\frac{b!}{(b-n)!n!}$$

is read “$b$ choose $n$.” It counts the number of ways to select $n$ elements from a set of $b$ elements. For example, $\binom42=6$.

Now let $b$ be any real number and set $f(x)=(1+x)^b$. Repeated differentiation gives

$$f^{(n)}(x) =b(b-1)\cdots(b-n+1)(1+x)^{b-n}.$$

Evaluating at the Maclaurin center produces

$$f^{(n)}(0)=b(b-1)\cdots(b-n+1).$$

This motivates the generalized definition

$$\boxed{ \binom{b}{n} :=\frac{b(b-1)\cdots(b-n+1)}{n!} }.$$

Using the standard convention $\binom b0=1$, the Taylor series becomes

$$\boxed{ (1+x)^b =\sum_{n=0}^{\infty}\binom{b}{n}x^n }, \qquad |x|<1.$$

When $b$ is a nonnegative integer, the factors eventually include zero, so the infinite series terminates and reduces to the ordinary finite binomial expansion.

Evaluating limits with series

Power series can evaluate an indeterminate limit by exposing the lowest-degree term that survives cancellation. Consider

$$\lim_{x\to0} \frac{\ln(1+x)-\sin x+\frac{1}{2}x^2}{x^3}.$$

Near $x=0$,

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$

and

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$

The linear terms cancel, and the added $x^2/2$ cancels the quadratic term from $\ln(1+x)$. The first surviving power is cubic:

$$\begin{aligned} \ln(1+x)-\sin x+\frac{1}{2}x^2 &=\left(\frac13+\frac1{3!}\right)x^3 +\text{terms of degree at least }4\\ &=\frac12x^3+\text{terms of degree at least }4. \end{aligned}$$

After division by $x^3$, every remaining positive power of $x$ tends to zero. Therefore,

$$\boxed{ \lim_{x\to0} \frac{\ln(1+x)-\sin x+\frac{1}{2}x^2}{x^3} =\frac{1}{2} }.$$

Integrating a function without an elementary antiderivative

The function $e^{-x^2}$ has no elementary antiderivative, but its Maclaurin series can be integrated term by term. Substituting $-x^2$ into the exponential series gives

$$e^{-x^2} =\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{2n}.$$

Integrating each term produces a power-series antiderivative valid for every real $x$:

$$\boxed{ \int e^{-x^2}\,dx =C+\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)}x^{2n+1} }.$$

For the definite integral on $[0,1]$, every power of $x$ vanishes at the lower endpoint, so

$$\int_0^1 e^{-x^2}\,dx =\sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)}.$$

Keeping the first three terms gives

$$\int_0^1 e^{-x^2}\,dx \approx 1-\frac{1}{3}+\frac{1}{10} =\frac{23}{30}.$$

This is an alternating series whose term magnitudes decrease. The alternating-series remainder estimate bounds the error by the first omitted term:

$$\left| \int_0^1 e^{-x^2}\,dx-\frac{23}{30} \right| \le \frac{1}{3!\cdot7} =\frac{1}{42}.$$

Turning a Taylor error bound into an integral error bound

A pointwise approximation can also approximate a definite integral. The key is to integrate an absolute error bound, rather than assume positive and negative errors will cancel.

Let

$$f(x)=e^{\sin x}, \qquad T_2(x)=1+x+\frac{x^2}{2}$$

be the function and its degree-two Maclaurin polynomial. The coefficients follow from

$$\begin{aligned} f(0)&=1,\\ f'(x)&=e^{\sin x}\cos x,&f'(0)&=1,\\ f''(x)&=e^{\sin x}(\cos^2x-\sin x),&f''(0)&=1. \end{aligned}$$

Suppose a constant $M$ satisfies $|f^{(3)}(x)|\le M$ on $[-1,1]$. Taylor's inequality gives

$$|f(x)-T_2(x)|\le\frac{M}{3!}|x|^3 \qquad(-1\le x\le1).$$

Use linearity first, then the inequality $|\int g|\le\int|g|$:

$$\begin{aligned} \left|\int_{-1}^{1}f(x)\,dx-\int_{-1}^{1}T_2(x)\,dx\right| &=\left|\int_{-1}^{1}[f(x)-T_2(x)]\,dx\right|\\ &\le\int_{-1}^{1}|f(x)-T_2(x)|\,dx\\ &\le\frac{M}{3!}\int_{-1}^{1}|x|^3\,dx\\ &=\frac{2M}{3!}\int_0^1x^3\,dx\\ &=\frac{2M}{3!}\left[\frac{x^4}{4}\right]_0^1\\ &=\boxed{\frac{2M}{4!}=\frac{M}{12}}. \end{aligned}$$

The factor $2$ comes from the evenness of $|x|^3$. Keep the absolute value until splitting at zero: integrating $x^3$ over the symmetric interval would give zero, but that is not a bound on the error's magnitude.

The polynomial integral is easy to evaluate:

$$\int_{-1}^{1}\left(1+x+\frac{x^2}{2}\right)\,dx =2+0+\frac13=\frac73.$$

Therefore

$$\boxed{\left|\int_{-1}^{1}e^{\sin x}\,dx-\frac73\right|\le\frac{M}{12}.}$$

This method needs only a bound on the next derivative, not an elementary antiderivative of $f$ or an infinite series for the integral.

When can the bound be strict?

The chain above proves a non-strict inequality. For this particular function, we can also justify the strict bound. Differentiating once more gives

$$f^{(3)}(x)=-e^{\sin x}\sin x\cos x\,(\sin x+3).$$

This derivative is continuous, equals zero at $0$, and is not identically zero on $[-1,1]$, so any valid $M$ is positive. On a sufficiently small interval around zero, continuity gives $|f^{(3)}(x)|<M/2$. Applying Taylor's inequality on that smaller interval improves the local error bound to $(M/2)|x|^3/3!$.

Thus the integrand bound is strictly smaller than $M|x|^3/3!$ on an interval of positive length and no larger elsewhere. Its integral is strictly smaller than $M/12$, giving

$$\boxed{\left|\int_{-1}^{1}e^{\sin x}\,dx-\frac73\right|<\frac{M}{12}.}$$

Smooth does not always mean analytic

A function is smooth when it has derivatives of every order. It is analytic at a point when its Taylor series there equals the function on some interval around that point. Smoothness is necessary for analyticity, but it is not sufficient.

Consider

$$f(x)= \begin{cases} e^{-1/x^2},&x\ne0,\\ 0,&x=0. \end{cases}$$

The exponential approaches zero faster than every power of $x$ as $x\to0$. One can show that $f$ is smooth and that

$$f^{(n)}(0)=0 \qquad\text{for every }n\ge0.$$

Its Maclaurin series is therefore the zero series:

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

However, $e^{-1/x^2}>0$ whenever $x\ne0$. The Taylor series matches $f$ at the center but not on any interval around it, so $f$ is smooth at $0$ but not analytic there.

Differentiating a known power series

Known series can also generate new identities by differentiation. Starting with

$$\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n, \qquad |r|<1,$$

differentiate term by term:

$$\frac{1}{(1-r)^2} =\sum_{n=1}^{\infty}n r^{n-1}.$$

Substituting $r=2x$ gives

$$\boxed{ \frac{1}{(1-2x)^2} =\sum_{n=1}^{\infty}n(2x)^{n-1} }, \qquad |x|<\frac12.$$