M344: Calculus III
Section 11.10
Taylor and Maclaurin Series¶
In the last two sections we've seen how certain functions can be represented by power series, at least on portions of their domains. The fundamental series that we've used to represent the functions in the previous sections is the geometric series.
At this point it is natural to ask if a general function $f$ has a power series representation centered at a certain point $x_0$. Under certain conditions, the answer to that question is yes!
Let's investigate.
Suppose $f$ has a convergent power series representation centered at $x_0$,
$$
f(x) = \sum_{n=0}^\infty c_n\, \big( x - x_0 \big)^n = c_0 + c_1 \big(x-x_0\big) + c_2 \big(x - x_0\big)^2 + c_3 \big( x - x_0\big)^3 + \cdots.
$$
We will know the power series representation of $f$ if we can determine all of the coefficients $c_n$.
Video lecture
We have proved the following theorem.
1. Theorem¶
If $f$ has a power series representation centered at $x_0$,
$$
f(x) = \sum_{n=0}^\infty \, c_n\, \big(x - x_0 \big)^n, \quad \big\vert x - x_0 \big\vert < R,
$$
then its coefficients are given by
$$
c_n = \frac{f^{(n)}(x_0)}{n!}.
$$
Note: This theorem implicitly states that for a function $f$ to have a power series representation at $x_0$, then $f$ must be infinitely differentiable at $x_0$. If a function $f$ has a power series representation at a point $x_0$, then $f$ is said to be analytic at $x_0$.
2. Definition¶
Suppose $f$ has a power series representation at $x_0$. The series
$$
T(x) = \sum_{n=0}^\infty\, \frac{f^{(n)}(x_0)}{n!}\, \big(x - x_0\big)^n
$$
is called the Taylor series of $f$ (centered) at $x_0$.
When $x_0 = 0$, this series is called the Taylor-Maclaurin series, or simply Maclaurin series, of $f$.
Here is a great motivating video from our friend Grant at 3 Blue 1 Brown.
Motivating video
3. Example¶
Supposing the function $f(x) = e^x$ has a power series representation at $x_0 = 0$, find its Taylor-Maclaurin series.
Check your answer
$$ e^x = \sum_{n=0}^\infty \frac{1}{n!}\, x^n $$
Video solution
In the preceding theorem and example we have shown that if a function $f$ has a power series representation at $x_0$, then that series is the Taylor series of $f$ at $x_0$. But how can we be sure that a function indeed has a power series representation at a point $x_0$?
Suppose $T$ is the Taylor series for $f$ at $x_0$, and suppose that $f$ equals its Taylor series in an interval $\vert x - x_0 \vert < R$. Then the partial sums of $T$ approximate $f$ near $x_0$,
$$
f(x) \approx T_k (x) = \sum_{n=0}^k \, \frac{f^{(n)}(x_0)}{n!}\, \big( x - x_0\big)^n.
$$
We define the $k^\text{th}$ Remainder of $T$ to be the series
$$
R_k(x) = \sum_{n = k+1}^\infty\, \frac{f^{(n)}(x_0)}{n!}\, \big( x - x_0\big)^n.
$$
If $f$ indeed equals $T$ in an interval containing $x_0$, then $f(x) = T_k(x) + R_k(x)$ for all $k$ on that interval, and
$$
\lim_{k\to \infty} R_k(x) = \lim_{k\to infty} \big[ f(x) - T_k(x) \big] = f(x) - \lim_{k\to\infty} T_k(x) = f(x) - T(x) = 0.
$$
We obtain the following theorem.
4. Theorem¶
If $f(x) = T_k(x) + R_k(x)$, where $T_k$ is the $k^\text{th}$-degree Taylor polynomial of $f$ at $x_0$, and
$$
\lim_{k\to\infty} R_k(x) = 0,
$$
for $\vert x - x_0 \vert < R$, then $f$ is equal to the sum of its Taylor series on the interval $\vert x - x_0\vert < R$.
We state the following important theorem without complete proof. (For now.)
5. Theorem (Taylor's Inequality)¶
If $\vert f^{(k+1)}(x) \vert \leq M$ for $\vert x - x_0 \vert \leq \delta$, then the remainder $R_k(x)$ of the Taylor series satisfies
$$
\big\vert R_k(x) \big\vert \leq \frac{M}{(k+1)!}\;\big\vert x - x_0\big\vert^{\,k+1}
$$
for $\vert x - x_0\vert \leq \delta$.
Partial video proof
6. Example¶
Show that $f(x) = e^x$ equals the sum of its Taylor-Maclaurin series on $\mathbb{R}$.
Video solution
7. Example¶
Find the Taylor-Maclaurin series for $f(x) = \cos x$.
Video solution
8. Example¶
Find the Taylor series for $f(x) = \sin x$ centered at $x_0 = \tfrac{\pi}{2}$, two ways.
Video solution, one way
Video solution, another way
We wish to find a formula for the Taylor-Maclaurin series of a binomial function, $f(x) = \big(1 + x\big)^r$, for any real number $r$.
Video solution
This leads us directly to the following theorem.
9. Theorem (Binomial Series)¶
If $r$ is any real number and $\vert x \vert < 1$, then
$$
\big(1 + x\big)^r = \sum_{n=0}^\infty \, {r \choose n}\;x^n.
$$
10. Example¶
Find the Taylor-Maclaurin series for $f(x) = \dfrac{1}{\sqrt{4-x\,}}$.
Video solution
11. Table of Famous Series¶
We summarize some of the work done over the past few sections of the notes in a table. The following series are fundamental, and you can usually just use them without re-deriving them. However, you must know how to derive each of these series in case you are asked to do so.
Table of fundamental Taylor-Maclaurin series
| Function | Series Expansion | Radius of Convergence |
|---|---|---|
| $\dfrac{1}{1-x}$ | $\sum\limits_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots $ | $R =1$ |
| $e^x$ | $\sum\limits_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$ | $R = \infty$ |
| $\sin x$ | $\sum\limits_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$ | $R = \infty$ |
| $\cos x$ | $\sum\limits_{n=0}^{\infty} (-1)^n\dfrac{x^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots$ | $R = \infty$ |
| $\arctan x$ | $\sum\limits_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{2n+1} = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dfrac{x^7}{7} + \cdots$ | $R = 1$ |
| $\ln (1+x) $ | $\sum\limits_{n=1}^{\infty} (-1)^{n-1}\dfrac{x^{n}}{n} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots$ | $R = 1$ |
| $(1+x)^k$ | $\sum\limits_{n=0}^{\infty} \binom{k}{n}x^n = 1 + kx + \dfrac{k(k-1)}{2!}x^2 + \dfrac{k(k-1)(k-2)}{3!}x^3 + \cdots$ | $R = 1$ |
We end this section of the notes by revisiting an infamous integral.
12. Example¶
Use a power series to evaluate
$$
\int e^{-x^2}\, dx.
$$
Video solution
Discussion¶
Questions? You can ask them here.