Taylor Series

From Power Series to Taylor Series

The Binomial Series tutorial identified a function with a power series by using a coefficient recurrence and a differential equation. There is another way to determine the coefficients: if a function is represented by a power series near its center, repeated differentiation recovers each coefficient. This leads to the Taylor series, whose coefficients are defined directly from the derivatives of a function.

The definition applies to any function with sufficiently many derivatives at the chosen center. It does not, by itself, promise that the resulting series converges, or that its sum equals the function. We will first establish how Taylor coefficients relate to an existing power-series representation, then examine examples and an important limitation.

Definition: Suppose $f$ has derivatives of every nonnegative integer order on an open interval containing $a$. The Taylor series of $f$ centered at $a$ is $$ \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n. $$ Here $f^{(0)}=f$, and $0!=1$. The number $a$ is the center of the series.

The first terms are $f(a)$, $f'(a)(x-a)$, and $f''(a)(x-a)^2/2!$. Thus the coefficient of $(x-a)^n$ is the $n$th derivative at the center divided by $n!$. The factorial is necessary because differentiating $(x-a)^n$ exactly $n$ times produces the factor $n!$.

Derivatives Recover the Coefficients

Suppose a power series centered at $a$ has positive radius of convergence $R$, and let $f$ be its sum for $|x-a|<R$. The term-by-term differentiation theorem for power series allows us to differentiate the series repeatedly inside its radius. Evaluating the $n$th derivative at $a$ then isolates its $n$th coefficient.

Theorem (Taylor Coefficients of a Power-Series Sum): Suppose $$ f(x)=\sum_{k=0}^{\infty}c_k(x-a)^k $$ for $|x-a|<R$, where $R>0$. Then, for every integer $n\geq0$, $$ f^{(n)}(a)=n!c_n. $$ Consequently, the Taylor series of $f$ centered at $a$ is the original power series.

Proof. By the term-by-term differentiation theorem for power series, we may differentiate the series $n$ times at points inside its radius. After $n$ differentiations, each term with index $k<n$ has become zero, the term with index $k=n$ is the constant $n!c_n$, and each term with index $k>n$ contains a positive power of $x-a$. In detail, the differentiated series is

f^{(n)}(x) =\sum_{k=n}^{\infty} \frac{k!}{(k-n)!}c_k(x-a)^{k-n}, \qquad |x-a|<R.

At $x=a$, the $k=n$ term is $n!c_n$. For $k>n$, the exponent $k-n$ is positive, so the term is zero at $x=a$. Therefore $f^{(n)}(a)=n!c_n$, as claimed. This also shows that the coefficient of $(x-a)^n$ is $f^{(n)}(a)/n!$, so the Taylor series agrees term by term with the given power series. $\square$

This gives a uniqueness result: if two power series centered at the same point represent the same function on a neighborhood of that point, their corresponding coefficients must agree. The reason is that both coefficients of degree $n$ must equal the same derivative divided by $n!$.

Corollary (Uniqueness of Power-Series Coefficients): Suppose two power series centered at $a$ converge to the same function on an open interval containing $a$. Then their coefficients are equal at every index.

Proof. Let the two series have coefficients $c_n$ and $d_n$, respectively. Since both sum to the same function $f$ near $a$, the theorem gives $n!c_n=f^{(n)}(a)=n!d_n$ for every $n\geq0$. Since $n!\neq0$, it follows that $c_n=d_n$ for every $n$. $\square$

Worked Examples of Taylor Series

Worked Example: The Exponential Centered at an Arbitrary Point

Let $f(x)=E(x)$, the exponential function defined earlier in this course, and choose any real center $a$. The identity $E'(x)=E(x)$ implies, by repeated differentiation, that $f^{(n)}(a)=E(a)$ for every $n\geq0$. Its Taylor series is therefore

\sum_{n=0}^{\infty}\frac{E(a)}{n!}(x-a)^n =E(a)\sum_{n=0}^{\infty}\frac{(x-a)^n}{n!}.

The exponential series converges for every real input, and the multiplication law for $E$ gives $E(x)=E(a)E(x-a)$. Consequently, this Taylor series converges to $E(x)$ for every real $x$. For example, centered at $a=1$, its first terms are $E(1)+E(1)(x-1)+E(1)(x-1)^2/2!+\cdots$.

Worked Example: The Reciprocal Function Centered at 2

Take $f(x)=1/x$ and $a=2$. Repeated differentiation gives $f^{(n)}(x)=(-1)^n n!x^{-n-1}$. Indeed, this holds for $n=0$; differentiating the expression for order $n$ gives $(-1)^n n!(-n-1)x^{-n-2}=(-1)^{n+1}(n+1)!x^{-n-2}$, which is the expression for order $n+1$. At the center,

\frac{f^{(n)}(2)}{n!} =\frac{(-1)^n}{2^{n+1}}.

Thus the Taylor series centered at $2$ is

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

To identify its sum, set $u=-(x-2)/2$. When $|x-2|<2$, we have $|u|<1$, so the geometric series gives $\sum_{n=0}^{\infty}u^n=1/(1-u)$. Therefore

\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(x-2)^n =\frac12\sum_{n=0}^{\infty}\left(-\frac{x-2}{2}\right)^n =\frac{1/2}{1+(x-2)/2} =\frac1x.

The series represents $1/x$ for $|x-2|<2$, its interval of convergence around the center. The coefficient formula also makes clear why the radius is $2$: the series is a geometric series in $-(x-2)/2$.

Worked Example: The Square Root Centered at 4

For $x$ near $4$, write $\sqrt{x}=2\left(1+(x-4)/4\right)^{1/2}$. The Binomial Series theorem, with $\alpha=1/2$, applies when $|x-4|/4<1$. It gives the Taylor series

\sqrt{x} =2\sum_{n=0}^{\infty}\binom{1/2}{n} \left(\frac{x-4}{4}\right)^n, \qquad |x-4|<4.

The first coefficients can also be checked directly from derivatives. We have $f(4)=2$, $f'(x)=1/(2\sqrt{x})$, and $f''(x)=-1/(4x^{3/2})$. Hence $f'(4)=1/4$ and $f''(4)/2!=-1/64$. The next binomial coefficients give

\sqrt{x} =2+\frac{x-4}{4}-\frac{(x-4)^2}{64} +\frac{(x-4)^3}{512}-\cdots, \qquad |x-4|<4.

For example, the coefficient of $(x-4)^3$ is $2\binom{1/2}{3}/4^3=2(1/16)/64=1/512$. Because this is a power series representation of $\sqrt{x}$ near $4$, the Taylor Coefficients Theorem ensures that its coefficients are precisely the derivatives at $4$ divided by the corresponding factorials.

A Taylor Series Need Not Represent Its Function

The preceding examples all have Taylor series that equal the function in a neighborhood of the center. It is important not to mistake that pattern for a general theorem about infinitely differentiable functions. The definition uses derivatives at one point; it imposes no condition ensuring that the resulting series converges to the function at nearby points.

Worked Example: A Smooth Function with a Zero Taylor Series

Define $g(0)=0$ and $g(x)=E(-1/x^2)$ for $x\neq0$. This function is infinitely differentiable, but every derivative at zero is zero. To see why, put $u=1/x$. For $x\neq0$, successive derivatives have the form $g^{(m)}(x)=P_m(1/x)E(-1/x^2)$, where each $P_m$ is a polynomial. The claim holds initially with $P_0(u)=1$. If it holds for $m$, differentiation shows it holds for $m+1$, with $P_{m+1}(u)=-u^2P_m'(u)+2u^3P_m(u)$, also a polynomial.

For every positive integer $N$ and $t>0$, the exponential series gives $E(t)\geq t^N/N!$, since its terms are nonnegative. The multiplication law $E(t)E(-t)=E(0)=1$ then yields $0<E(-t)\leq N!/t^N$. Taking $t=1/x^2$ shows that $E(-1/x^2)$ tends to zero faster than any fixed power of $|x|$: for any integer $q\geq0$, choose $N$ large enough that $2N>q$. Then

|x|^{-q}E(-1/x^2) \leq N!|x|^{2N-q} \longrightarrow 0 \qquad(x\to0).

Since a polynomial in $1/x$ is bounded in magnitude by a constant times some power of $|x|^{-1}$ near zero, this estimate implies $P_m(1/x)E(-1/x^2)\to0$ for every $m$. It also implies $g^{(m)}(x)/x\to0$ as $x\to0$. Inductively, each derivative extends continuously to zero with value zero, and its derivative at zero is zero by the difference quotient. Thus $g^{(m)}(0)=0$ for every $m$.

The Taylor series of $g$ centered at zero is therefore the zero series. But $g(x)=E(-1/x^2)>0$ for every $x\neq0$. Its Taylor series converges to zero, not to $g(x)$, at every nonzero $x$. This example shows that having derivatives of every order is not enough to guarantee equality with the Taylor series.

What the Taylor Series Tells Us

A Taylor series is determined by the derivatives of a function at its center. If the function is already known to be represented by a power series near that center, the Taylor Coefficients Theorem identifies the two series, and uniqueness prevents a second set of coefficients from representing the same function there. This is a powerful way to find or verify coefficients.

The converse requires care. Knowing all derivatives at a point specifies a Taylor series, but does not establish that the series converges, or that its sum equals the function. The flat-function example has derivatives of every order and a convergent Taylor series, yet the series fails to represent the function away from its center. Equality must be justified by additional information about the function or the series; it does not follow from the definition alone.

Takeaway: The Taylor series centered at $a$ is $$ \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n. $$ For a function represented by a power series near $a$, its coefficient of degree $n$ is $f^{(n)}(a)/n!$. An infinitely differentiable function, however, need not equal the sum of its Taylor series.

Check Your Understanding

Use the definition, the coefficient theorem, and the examples to answer the following questions.

What is the coefficient of $(x-a)^n$ in the Taylor series of $f$ centered at $a$?
If $f(x)=\sum_{k=0}^{\infty}c_k(x-a)^k$ near $a$, what is $f^{(n)}(a)$?
Why must two power series centered at $a$ have identical coefficients if they represent the same function near $a$?
For the reciprocal example, what condition on $x$ makes the geometric-series representation valid?
What does the smooth function $g(x)=E(-1/x^2)$, with $g(0)=0$, demonstrate about Taylor series?

What You'll Learn