Wichita State University Logo

Math 555: Differential Equations

5.1 Taylor Series, Part II

5.1.12 Taylor Series ¶

A Taylor Series is a series with positive integer powers of an independent variable $x$ in the definition of the terms of the series

$$\displaystyle\sum_{k=0}^{\infty} a_k(x-x_0)^k.$$
The partial sums

$$\begin{align*} s_0(x) &= a_0 \\ \\ s_1(x) &= a_0 + a_1(x - x_0) = \color{#307fe2}{(a_0 - a_1x_0)} + \color{#ec008c}{a_1}x \\ \\ s_2(x) &= a_0 + a_1(x + x_0) + a_2(x-x_0)^2 \\ &= \color{#307fe2}{(a_0 - a_1x_0 + a_2x_0^2)} + \color{#ec008c}{(a_1 - 2a_2x_0)}x + \color{#d17300}{a_2}x^2 \\ \\ s_3(x) &= a_0 + a_1(x + x_0) + a_2(x-x_0)^2 + a_3(x-x_0)^3 \\ &= \color{#307fe2}{(a_0 - a_1x_0 + a_2x_0^2 - a_3x_0^3)} + \color{#ec008c}{(a_1 - 2a_2x_0 + 3a_3x_0^2)}x + \color{#d17300}{(a_2 - 3a_3x_0)}x^2 + \color{#f3ad1c}{a_3}x^3 \\ &\ddots \\ s_n(x) &= a_0 + a_1(x + x_0) + a_2(x-x_0)^2 +\ \cdots\ + a_n(x-x_0)^n. \\ \end{align*}$$
are called Taylor polynomials and $s_n(x)$ is a Taylor polynomial of degree $n$. The right shift $x_0$ is called the base point . One gets a better view of the Taylor Series and Taylor polynomial if one first looks at examples where the base point $x_0=0$

$$\displaystyle\sum_{k=0}^{\infty} a_k(x-0)^k = \displaystyle\sum_{k=0}^{\infty} a_kx^k.$$
Such a Taylor series is called a Maclaurin series . The Taylor polynomials in a Maclaurin series are simpler

$$\begin{align*} s_0(x) &= a_0 \\ \\ s_1(x) &= a_0 + a_1x \\ \\ s_2(x) &= a_0 + a_1x + a_2^2 \\ \\ s_3(x) &= a_0 + a_1x + a_2x + a_3x^3 \\ &\ddots \\ s_n(x) &= a_0 + a_1x + a_2x +\ \cdots\ + a_nx^n. \\ \end{align*}$$

Each Taylor polynomial $s_n(x)$ is a function of one real variable; its domain is the entire real line and its codomain is the entire real line. They have all the properties of polynomials that we have come to expect from polynomials. Evaluating a polynomial for any real number $x$ is well-understood, which is why Taylor polynomials are such a useful tool.

Exercise 5.1.6 ¶

List the first 4 Taylor polynomials of the Taylor series

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(x+1)^k}{2^k}.$$

View Solution

Notice that the index starts at $1$ instead of $0$. This is allowed.

$$\begin{align*} s_1(x) &= \dfrac{(x+1)^1}{2^1} = \dfrac{x+1}{2} = \dfrac{x}{2} + \dfrac{1}{2}, \\ \\ s_2(x) &= \dfrac{(x+1)^1}{2^1} + \dfrac{(x+1)^2}{2^2} = \dfrac{x+1}{2} + \dfrac{(x+1)^2}{4} = \dfrac{x+1}{2} + \dfrac{x^2 + 2x + 1}{4} \\ \\ &= \dfrac{x}{2} + \dfrac{1}{2} + \dfrac{x^2}{4} + \dfrac{x}{2} + \dfrac{1}{4} = \dfrac{x^2}{4} + x + \dfrac{3}{4}, \\ \\ s_3(x) &= \dfrac{(x+1)^1}{2^1} + \dfrac{(x+1)^2}{2^2} + \dfrac{(x+1)^3}{2^3} = \dfrac{x+1}{2} + \dfrac{x^2 + 2x + 1}{4} + \dfrac{x^3 + 3x^2 + 3x + 1}{8} \\ \\ &= \dfrac{x}{2} + \dfrac{1}{2} + \dfrac{x^2}{4} + \dfrac{x}{2} + \dfrac{1}{4} + \dfrac{x^3}{8} + \dfrac{3x^2}{8} + \dfrac{3x}{8} + \dfrac{1}{8}, \\ \\ &= \dfrac{x^3}{8} + \dfrac{5x^2}{8} + \dfrac{11x}{8} + \dfrac{7}{8} \\ \\ s_4(x) &= \dfrac{(x+1)^1}{2^1} + \dfrac{(x+1)^2}{2^2} + \dfrac{(x+1)^3}{2^3} \dfrac{(x+1)^4}{2^4} \\ \\ &= \dfrac{x+1}{2} + \dfrac{x^2 + 2x + 1}{4} + \dfrac{x^3 + 3x^2 + 3x + 1}{8} + \dfrac{x^4 + 4x^3 + 6x^2 + 4x + 1}{16} \\ \\ &= \dfrac{x^4}{16} + \dfrac{3x^3}{8} + x^2 + \dfrac{13x}{8} + \dfrac{15}{16}. \end{align*}$$

Evaluate these Taylor polynomials at $x=-2,\ -1,\ 0,\ 1$.

View Solution

$$\begin{align*} s_1(-2) &= \dfrac{(-2+1)}{2} = -\dfrac{1}{2}, \\ \\ s_1(-1) &= \dfrac{(-1+1)}{2} = 0, \\ \\ s_1(0) &= \dfrac{(0+1)}{2} = \dfrac{1}{2}, \\ \\ s_1(1) &= \dfrac{(1+1)}{2} = 1, \\ \\ s_2(-2) &= s_1(-2) + \dfrac{(-2+1)^2}{4} = -\dfrac{1}{2} + \dfrac{1}{4} = -\dfrac{1}{4}, \\ \\ s_2(-1) &= s_1(-1) + \dfrac{(-1+1)^2}{4} = 0 + 0 = 0, \\ \\ s_2(0) &= s_1(0) + \dfrac{(0+1)^2}{4} = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}, \\ \\ s_2(1) &= s_1(1) + \dfrac{(1+1)^2}{4} = 1 + 1 = 2, \\ \\ s_3(-2) &= s_2(-2) + \dfrac{(-2+1)^3}{8} = -\dfrac{1}{4} - \dfrac{1}{8} = -\dfrac{3}{8}, \\ \\ s_3(-1) &= s_2(-1) + \dfrac{(-1+1)^3}{8} = 0 + 0 = 0,\\ \\ s_3(0) &= s_2(0) + \dfrac{(0+1)^3}{8} = \dfrac{3}{4} + \dfrac{1}{8} = \dfrac{7}{8}, \\ \\ s_3(1) &= s_2(1) + \dfrac{(1+1)^3}{8} = 2 + 1 = 3, \\ \\ s_4(-2) &= s_3(-2) + \dfrac{(-2+1)^4}{16} = -\dfrac{3}{8} + \dfrac{1}{16} = -\dfrac{5}{16}, \\ \\ s_4(-1) &= s_3(-1) + \dfrac{(-1+1)^4}{16} = 0 + 0 = 0,\\ \\ s_4(0) &= s_3(0) + \dfrac{(0+1)^4}{16} = \dfrac{7}{8} + \dfrac{1}{16} = \dfrac{15}{16}, \\ \\ s_4(1) &= s_3(1) + \dfrac{(1+1)^4}{16} = 3 + 1 = 4. \end{align*}$$

Graph these first four Taylor polynomials over the interval $[-4,2]$, on a single plot.

View Solution

Note the shared behavior near the base point $x = -1$.

5.1.13 Evaluating the Taylor Series ¶

We can evaluate the Taylor Series at every point as well. This can be more difficult if we don't recognize the series. For example, to evaluate the series

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(x+1)^k}{k2^k}$$
at $x=-1$ we proceed as always for algebraic expressions; we replace the variable $x$ with the value $-1$ to obtain

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(-1+1)^k}{2^k}.$$
Then we proceed to evaluate the series by computing the limit of the sequence of partial sums.

$$\begin{align*} \displaystyle\sum_{k=1}^{\infty}\dfrac{(-1 + 1)^k}{2^k} &= \displaystyle\sum_{k=1}^{\infty}\dfrac{0^k}{2^k} \\ \\ s_1(-1) &= 0 \\ \\ s_2(-1) &= 0 + 0 \\ &\ddots \\ s_n(-1) &= 0 + 0 +\ \cdots\ + 0 \\ \\ \displaystyle\lim_{n\to\infty} 0 &= 0. \end{align*}$$
If we evaluate the series at $x=0$, then obtain the series

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(0+1)^k}{2^k} = \displaystyle\sum_{k=1}^{\infty}\dfrac{1}{2^k}.$$
This is similar to the geometric series with $r=\frac{1}{2}$ and $a=1$ so

$$\begin{align*} \displaystyle\sum_{k=1}^{\infty}\dfrac{(0+1)^k}{2^k} &= \displaystyle\sum_{k=1}^{\infty}\dfrac{1}{2}^k \\ \\ &= \displaystyle\sum_{k=0}^{\infty}\dfrac{1}{2}^k - \dfrac{1}{2}^0 \\ \\ &= \dfrac{1}{1-\frac{1}{2}} - 1 = 2 - 1 = 1. \end{align*}$$
Notice that we added $0$ to the expression in the form $\frac{1}{2}^0 - \frac{1}{2}^0$ so that we could recognize the value of the series.

It seems that for at least some real numbers $x$, the series converges to a finite limit. We need to determine all of the real numbers $x$ for which the series converges to a finite limit.

5.1.14 Absolute Convergence ¶

We can apply the ratio test to our series. We compute the limit

$$\displaystyle\lim_{k\to\infty} \left|\dfrac{a_{k+1}}{a_k}\right|.$$
Where this limit is less than $1$, the series converges absolutely. Where the limit is greater than $1$ or does not exist, the series diverges. The test is inconclusive for values of $x$ in which the limit equals $1$. Each of the terms of our series is given by

$$a_k = \dfrac{(x+1)^k}{2^k}.$$
and

$$\left|a_k\right| = \left|\dfrac{(x+1)^k}{2^k}\right| = \dfrac{|x+1|^k}{2^k}.$$
Thus for our series the limit becomes

$$\displaystyle\lim_{k\to\infty} \left|\dfrac{a_{k+1}}{a_k}\right| = \displaystyle\lim_{k\to\infty} \dfrac{\dfrac{|x+1|^{k+1}}{2^{k+1}}}{\dfrac{|x+1|^k}{2^k}} = \displaystyle\lim_{k\to\infty} \dfrac{|x+1|^{k+1}}{2^{k+1}}\cdot\dfrac{2^k}{|x+1|^k} = \displaystyle\lim_{k\to\infty} \dfrac{|x+1|}{2}.$$
Notice that the last expression has no instances of the index $k$ so the limit is the limit of a constant expression. We need this expression to be less than 1 to pass the Ratio Test .

$$\begin{align*} \displaystyle\lim_{k\to\infty} \dfrac{|x+1|}{2} &< 1 \\ \\ \dfrac{|x+1|}{2} &< 1 \\ \\ |x+1| &< 2 \\ \\ -2 < x+1 &< 2 \\ \\ -3 < x &< 1. \end{align*}$$
This means that for any real value $x$ in the interval $(-3,1) = (-1-2 < x < -1+2)$, the series

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(x+1)^k}{2^k}$$
converges absolutely. We need to evaluate the series at $x=-3$ and $x=1$. At $x=1$ we have

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(1+1)^k}{2^k} = \displaystyle\sum_{k=1}^{\infty} 1 = \infty.$$
At $x=-3\ $ we have the series

$$\displaystyle\sum_{k=1}^{\infty}\dfrac{(-3+1)^k}{2^k} = \displaystyle\sum_{k=1}^{\infty} (-1)^k,$$
which diverges. We can create a function $f:(-3,1)\to\mathbb{R}$ that assigns the value

$$y = f(x) = \displaystyle\sum_{k=1}^{\infty}\dfrac{(x+1)^k}{2^k}$$
for every $x\in(-3,1)$. For any value $x \le -3$ or $x \ge 1$ the series diverges. We call the function $f$ analytic because the function $f$ has an absolutely convergent Taylor series expression on its domain $(-3,1)$.

5.1.15 Radius of Convergence ¶

Recall that in our example the interval of absolute convergence is

$$|x+1| < 2.$$
If we consider $x$ to be a complex number instead of a real number, then this equation expresses a disk on the complex plane. This is because the absolute value of a complex number $x = a + bi$ is $|x| = \sqrt{a^2 + b^2}$. Thus our equation becomes

$$\begin{align*} \sqrt{(a+1)^2 + b^2} &< 2 \\ \\ (a+1)^2 + b^2 &< 4; \end{align*}$$
the equation of a circle on the plane of radius $2$ and centered at $(-1,0)$.
Disk of Convergence

Disk of Convergence for $f(x) = \displaystyle\sum_{k=1}^{\infty}\dfrac{(x+1)^k}{2^k}$

The radius of this disk of values is called the radius of convergence . The intersection of this disk with the real line, $(-3,1)$ is called the interval of convergence .

5.1.16 Analytic Functions ¶

The term analytic is used in a few different ways to describe functions:

A function is called analytic at the point $x_0$, if $f(x_0)$ has an absolutely convergent power (Taylor) series.
A function is called analytic on an interval $(a,b)$ , if it is analytic at every point $x\in(a,b)$.
A function is called an analytic function , if it is analytic at every point in the domain of $f$.

When solving differential equations, we need our functions to be analytic to guarantee the existence and uniqueness of a solution.

Additionally, we need to understand the relationship between functions we have learned about previously and analytic functions. The functions we have studied in our previous mathematics courses are called elementary functions:

trigonometric functions and their inverses
exponential functions and their inverses
polynomials
rational functions
algebraic combinations of these functions

All of the above are elementary . Most of these functions are analytic functions on their domains.

5.1.17 Analytic Solutions ¶

There are several important differential equations that do not have elementary solutions.

Our theorems guarantee us that a family of solutions exist and that our initial value problems have a unique solution.
However, we cannot describe these solutions using any of the functions we have learned about thus far in our mathematical education.
We will be forced to represent our solutions as an infinite series of some kind.

To provide some context for how we think about using analytical solutions of differential equations, we consider how we employ rational and irrational numbers. The rational numbers are like our elementary functions.

$\sqrt{2}$ is not rational
$\pi$ is transcendental
$e$ is transcendental

A transcendental number is one that cannot be expressed as a root of a (finite) polynomial with rational coefficients. All numbers that are roots of such polynomials are called algebraic .

How do we perform computations with these numbers?

We write down a rational approximation to our irrational number; we write down the first several decimals of our irrational number. We cannot write down all of them. We truncate our decimal expansion so that the errors in our computations are small enough to be acceptable to our application.

How do we perform computations with a power series?

The solutions to many important differential equations have an analytic power series solution. We cannot write down all the terms in the series. We truncate our power series expansion so that the errors in our computations using the Taylor polynomial approximation are small enough to be acceptable to our application.

But first we must find the power series expansion of our solution.

5.1.18 Calculus of Analytic Functions ¶

If $f$ is an analytic function how do we find the power series expansion (representation) of $f$?

What functions have a power series expansion?
What functions are analytic?
How do we find radius of convergence of this function?
How do we the interval of convergence of this function?

The easiest way to compute the power series at $x_0=0$, that is the Maclaurin series, of a function is to use Taylor's Theorem .

Taylor's Theorem ¶

If $f$ has a convergent power series representation (expansion) at $a$, that is if

$$f(x) = \displaystyle\sum_{k=0}^{\infty} a_k(x-a)^k,\qquad |x-a|<\rho,$$
then the coefficients have the form

$$a_k = \dfrac{f^{(k)}(a)}{k!};$$
and the power series becomes

$$f(x) = \displaystyle\sum_{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k,\qquad |x-a|<\rho.$$
Here $f^{(0)}(a)$, the zeroth derivative of $f$ at $a$ is just $f(a)$. If $a=0$, then we have the Maclaurin series whose form is

$$f(x) = \displaystyle\sum_{k=0}^{\infty}\dfrac{f^{(k)}(0)}{k!}x^k,\qquad |x|<\rho.$$

Exercise 5.1.6 ¶

Consider the polynomial $p(x) = 3x^2 - 7x + 13$.

View Solution

Exercise 5.1.7 ¶

Consider the function $f(x) = e^x$.

View Solution

The limit is zero because regardless of the value of $|x|$, because the limit as $k\to\infty$ is much larger. The limit is zero for every real number $x$; and $0\ <\ 1$ for all real numbers $x$, so the radius of convergence is $\infty$.

The interval of convergence is $(0 - \infty, 0 + \infty) = (-\infty,\infty)$.

5.1.19 Computing the Derivative of an Analytic Function ¶

If $f$ is an analytic function with absolutely convergent powers series

$$f(x) = \displaystyle\sum_{k=0}^{\infty} a_k x^k,$$
then the derivative can be easily computed term by term .

$$\begin{align*} f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 +\ \cdots\ + a_nx^n +\ \cdots \\ \\ f'(x) &= 0 + a_1 + 2a_2x + 3a_3x^2 +\ \cdots\ + na_nx^{n-1} +\ \cdots \\ \\ &= \displaystyle\sum_{k=1}^{\infty} ka_nx^{k-1} \\ \\ f''(x) &= 0 + 0 + 2a_2 + 6a_3x +\ \cdots\ + n(n-1)a_nx^{n-2} +\ \cdots \\ \\ &= \displaystyle\sum_{k=2}^{\infty} k(k-1)a_nx^{n-2} \\ \\ f'''(x) &= 0 + 0 + 0 + 6a_3 +\ \cdots\ + n(n-1)(n-2)a_nx^{n-3} \\ \\ &= \displaystyle\sum_{k=3}^{\infty} k(k-1)(k-2)a_nx^{n-3} \\ &\ddots \\ f^{(n)}(x) &= 0 + 0 + 0 +\ \cdots\ + n!a_n +\ \cdots \\ \\ &= \displaystyle\sum_{k=n}^{\infty} k(k-1)\cdots(k-n)x^{k-n} \\ \end{align*}$$
Notice that the derivative of a constant is $0$, not a lower power of $x$ so the index of the power series must start at a higher positive integer with each derivative.

If a series is conditionally convergent then the derivatives cannot be calculated using this method. This only works if the series is absolutely convergent. That is why we will always require analytic functions and analytic solutions.

Exercise 5.1.8 ¶

Consider the analytic function $f(x) = e^x$. Differentiate the Maclaurin series term by term to obtain the derivative of the exponential function.

View Solution

The power series is $f(x) = \displaystyle\sum_{k=0}^{\infty} \dfrac{x^k}{k!}$. Differentiating term by term

$$\begin{align*} f'(x) &= \displaystyle\sum_{k=1}^{\infty} \dfrac{kx^{k-1}}{k!} \\ \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{x^{k-1}}{(k-1)!} \\ \\ &= 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3} +\ \cdots\ + \dfrac{x^k}{k!} +\ \cdots \\ \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{x^k}{k!} \\ \end{align*}$$
We can also substitute $k=j+1$ into the expression $\displaystyle\sum_{k=1}^{\infty} \dfrac{x^{k-1}}{(k-1)!}$

$$\begin{align*} f'(x) &= \displaystyle\sum_{k=1}^{\infty} \dfrac{x^{k-1}}{(k-1)!} \\ \\ &= \displaystyle\sum_{j+1=1}^{\infty} \dfrac{x^{j+1-1}}{(j+1-1)!} \\ \\ &= \displaystyle\sum_{j=1}^{\infty} \dfrac{x^{j}}{j!} \\ \\ &= \displaystyle\sum_{k=1}^{\infty} \dfrac{x^{k}}{k!} = f(x) \end{align*}$$
The last substitution of $j=k$ is possible because the index variable is a dummy variable .

Your use of this self-initiated mediated course material is subject to our Creative Commons License .

Creative Commons Attribution-NonCommercial-ShareAlike 4.0

Creative Commons Logo - Black
Attribution
You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.

Creative Commons Logo - Black
Noncommercial
You may not use the material for commercial purposes.

Share Alike
You are free to share, copy and redistribute the material in any medium or format. If you adapt, remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.