Taylor Series Preview

From Taylor Polynomials to an Infinite Series

Taylor Bounds showed how the derivative of order $n+1$ controls the error in a degree-$n$ Taylor polynomial. The next question is what happens as the degree increases. If we keep adding terms, do the resulting polynomials approach the function? Sometimes they do, but having derivatives of every order does not by itself settle the question.

Fix a center $a$, and suppose $f$ has derivatives of every order on an open interval containing $a$. Its degree-$n$ Taylor polynomial is

T_{n,f,a}(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.

The corresponding Taylor series is the infinite series suggested by these polynomials. At a particular $x$, its partial sums are exactly $T_{n,f,a}(x)$. Thus, saying that the Taylor series converges to $f(x)$ means that these polynomial values approach $f(x)$ as $n$ increases.

Definition: The Taylor series of $f$ at $a$ is $$ \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ At a point $x$, its partial sum through degree $n$ is $T_{n,f,a}(x)$. The series represents $f$ at $x$ if these partial sums converge to $f(x)$.

This definition separates two questions that are easy to confuse. First, does the series of Taylor coefficients converge at $x$? Second, if it does, is its sum $f(x)$? Taylor’s Theorem addresses the second question through the remainder. For $x\ne a$, that theorem gives a point $c_n$ between $a$ and $x$ such that

f(x)-T_{n,f,a}(x) = \frac{f^{(n+1)}(c_n)}{(n+1)!}(x-a)^{n+1}.

The point $c_n$ can depend on $n$. To prove that the Taylor polynomials approach $f(x)$, it is enough to show that the remainder on the left tends to zero. The next criterion makes this precise.

A Remainder Criterion for Recovering the Function

Theorem (Taylor-Series Recovery from a Remainder Bound): Let $a$ and $x$ lie in an open interval $I$, and suppose $f:I\to\mathbb{R}$ has derivatives of every order on $I$. If $$ \lim_{n\to\infty}\frac{|f^{(n+1)}(c_n)|\,|x-a|^{n+1}}{(n+1)!}=0 $$ for points $c_n$ between $a$ and $x$ supplied by Taylor’s Theorem, then $T_{n,f,a}(x)\to f(x)$. In particular, it is sufficient that there are numbers $M_n\geq0$ such that $|f^{(n+1)}(t)|\leq M_n$ throughout the segment between $a$ and $x$, and $$ \lim_{n\to\infty}\frac{M_n|x-a|^{n+1}}{(n+1)!}=0. $$

Proof. If $x=a$, then $T_{n,f,a}(a)=f(a)$ for every $n$, so the conclusion holds. Suppose $x\ne a$. Taylor’s Theorem gives, for each $n$, a point $c_n$ between $a$ and $x$ for which

\left|f(x)-T_{n,f,a}(x)\right| = \frac{|f^{(n+1)}(c_n)|\,|x-a|^{n+1}}{(n+1)!}.

Under the first hypothesis, the right-hand side tends to zero. Therefore the difference between $f(x)$ and the $n$th Taylor polynomial tends to zero, which is exactly $T_{n,f,a}(x)\to f(x)$. Under the stated derivative bound, $|f^{(n+1)}(c_n)|\leq M_n$, so

0\leq\left|f(x)-T_{n,f,a}(x)\right| \leq\frac{M_n|x-a|^{n+1}}{(n+1)!}.

The assumed limit of the upper bound again forces the remainder to tend to zero. $\square$

The criterion is about the remainder, not merely about the existence of each derivative. A convenient way to use it is to find bounds on all the relevant derivatives and then check whether the factorial in the denominator makes the resulting error tend to zero.

Worked Example: Recovering the Exponential from Its Taylor Series

Let $f(x)=e^x$ and take $a=0$. Every derivative of $f$ is $e^x$, so $f^{(k)}(0)=1$ for every nonnegative integer $k$. The Taylor polynomial is $$ T_{n,f,0}(x)=\sum_{k=0}^{n}\frac{x^k}{k!}. $$ Fix any real $x$. On the segment between $0$ and $x$, $e^t\leq e^{|x|}$. Thus the Lagrange remainder satisfies $$ \left|e^x-T_{n,f,0}(x)\right| \leq \frac{e^{|x|}|x|^{n+1}}{(n+1)!}. $$ For $x\ne0$, set $b_n=|x|^{n+1}/(n+1)!$. Then $$ \frac{b_{n+1}}{b_n}=\frac{|x|}{n+2}. $$ For all sufficiently large $n$, this ratio is at most $1/2$. Each subsequent term is then at most half the preceding term, so $b_n\to0$. If $x=0$, the remainder is zero for every $n$. The recovery criterion therefore shows that the Taylor polynomials converge to $e^x$ for every real $x$.

A Useful Uniform Derivative Bound

The recovery criterion becomes especially convenient when the derivatives grow no faster than a factorial times a fixed geometric factor. The following result gives a direct condition that works for all points sufficiently close to the center.

Theorem (Factorial Derivative Bound Gives Local Taylor Recovery): Let $I$ be an open interval containing $a$, and suppose $f:I\to\mathbb{R}$ has derivatives of every order on $I$. Let $C\geq0$ and $R>0$, and suppose that for every integer $n\geq0$, $$ |f^{(n+1)}(t)|\leq \frac{C(n+1)!}{R^{n+1}} $$ for every $t$ between $a$ and $x$. If $|x-a|<R$, then $T_{n,f,a}(x)\to f(x)$.

Proof. The Lagrange remainder gives, for each $n$, a point $c_n$ between $a$ and $x$ with

\left|f(x)-T_{n,f,a}(x)\right| = \frac{|f^{(n+1)}(c_n)|\,|x-a|^{n+1}}{(n+1)!}.

The derivative hypothesis applies at $c_n$. Substitution and cancellation of the positive factorial give

\left|f(x)-T_{n,f,a}(x)\right| \leq C\left(\frac{|x-a|}{R}\right)^{n+1}.

Write $q=|x-a|/R$. The hypothesis $|x-a|<R$ gives $0\leq q<1$. If $q=0$, then $x=a$ and every Taylor polynomial equals $f(a)$. If $0<q<1$, the powers $q^{n+1}$ tend to zero. Hence the remainder tends to zero in either case, proving that $T_{n,f,a}(x)\to f(x)$. $\square$

Worked Example: A Local Taylor Expansion for Sine

Let $f(x)=\sin x$ and $a=0$. The derivatives cycle through $\sin x$, $\cos x$, $-\sin x$, and $-\cos x$, so $|f^{(k)}(t)|\leq1$ for every order $k$ and every real $t$. The factorial derivative bound holds with $C=1$ and $R=1$, since $$ 1\leq(n+1)! $$ for every $n\geq0$. The theorem therefore guarantees that the Taylor polynomials converge to $\sin x$ when $|x|<1$.

For instance, the first nonzero terms at the center give $T_{3,f,0}(x)=x-x^3/6$. At $x=1/2$, $$ T_{3,f,0}\left(\frac12\right) =\frac12-\frac{(1/2)^3}{6} =\frac12-\frac1{48} =\frac{23}{48}. $$ The Taylor remainder bound using $|f^{(4)}(t)|\leq1$ gives $$ \left|\sin\left(\frac12\right)-\frac{23}{48}\right| \leq\frac{(1/2)^4}{4!} =\frac1{384}. $$ This is a finite-degree estimate. The recovery theorem adds that, as the degree increases, the corresponding Taylor polynomials approach the sine value.

Smoothness Does Not Guarantee Taylor-Series Equality

The recovery results require control of the remainders as the degree increases. Without such control, even a function with derivatives of every order can fail to equal its Taylor series away from the center. The following example shows why the distinction matters.

Worked Example: A Smooth Function with a Zero Taylor Series

Define $$ f(x)= \begin{cases} e^{-1/x^2},&x\ne0,\\ 0,&x=0. \end{cases} $$ For $x\ne0$, put $u=1/x$. Differentiating a polynomial in $u$ times $e^{-u^2}$ produces another polynomial in $u$ times $e^{-u^2}$, because $$ \frac{d}{dx}\left(P(1/x)e^{-1/x^2}\right) = \left(-u^2P'(u)+2u^3P(u)\right)e^{-u^2}. $$ The expression in parentheses is a polynomial in $u$. Inductively, for each $n\geq0$, the $n$th derivative on $x\ne0$ has the form $$ f^{(n)}(x)=P_n(1/x)e^{-1/x^2} $$ for a polynomial $P_n$.

As $x\to0$, $|1/x|\to\infty$. Exponential growth dominates every positive power, so any polynomial in $1/x$, multiplied by $e^{-1/x^2}$, tends to zero. This shows that each displayed derivative tends to zero as $x\to0$. To verify differentiability at zero at each stage, suppose the $n$th derivative is defined there to be zero. Its difference quotient is $$ \frac{f^{(n)}(h)-f^{(n)}(0)}{h} = \frac{P_n(1/h)}{h}e^{-1/h^2}, \qquad h\ne0. $$ The factor $P_n(1/h)/h$ is also a polynomial in $1/h$, so the same exponential domination shows that this quotient tends to zero. Thus the next derivative exists at zero and equals zero. Starting with $f(0)=0$, induction proves that $f$ has derivatives of every order at zero and $$ f^{(n)}(0)=0 $$ for every $n\geq0$.

Consequently, every Taylor coefficient at zero is zero, and every Taylor polynomial $T_{n,f,0}$ is the zero polynomial. But if $x\ne0$, then $e^{-1/x^2}>0$, so $f(x)>0$. The Taylor polynomials therefore converge to zero, not to $f(x)$, at every nonzero $x$. This function is infinitely differentiable, but its Taylor series at zero does not represent it away from zero.

What the Preview Establishes

A Taylor series is built from the Taylor polynomials, but its relationship to the original function is determined by the remainder. Taylor’s Theorem supplies a precise error formula; a derivative bound can then prove that this error tends to zero. When that happens, the polynomial approximations recover the function at the target.

The converse should not be assumed: knowing every derivative at the center does not ensure that the Taylor series represents the function nearby. In the smooth example above, all the coefficients vanish even though the function is positive away from zero. The important question is not simply whether derivatives exist, but whether their growth permits the Taylor remainder to vanish.

These results are a preview of later applications. They explain why Taylor expansions can be powerful approximations, while also identifying the additional estimate needed before treating an infinite Taylor series as the function itself.

Check Your Understanding

Use the remainder criterion and the examples to check your understanding.

What does it mean for the Taylor series at $a$ to represent $f$ at a particular point $x$?
Why does a remainder bound tending to zero imply that the Taylor polynomials converge to $f(x)$?
In the factorial derivative bound theorem, what role does the condition $|x-a|<R$ play?
Why does the exponential example satisfy the recovery criterion for every fixed real $x$?
How can a function have derivatives of every order at zero while its Taylor series there fails to represent it at nonzero points?

What You'll Learn