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Active Calculus 2nd Ed

Matthew Boelkins

Section 8.5 Finding and using Taylor series

Subsection 8.5.1 Introduction

So far, we have focused our attention on a collection of five basic functions functions — \(\frac{1}{1-x}\text{,}\) \(\ln(1+x)\text{,}\) \(\sin(x)\text{,}\) \(\cos(x)\text{,}\) and \(e^x\) — and their Taylor series centered at \(a = 0\text{.}\) One of the reasons we were able to find these Taylor series is the patterns that arise in the derivatives of each of these functions. While we can always use Definition 8.4.1 to find the first few terms of the Taylor series, for most functions it is much more challenging to find a pattern among the various derivatives that allows us to state the general \(n^{\text{th}}\) term of the series.
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In Preview Activity 8.5.1, we explore this issue and investigate a different approach to finding the Taylor series of a given function.
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Preview Activity 8.5.1.
Let
\begin{equation*} g(x) = \frac{1}{1+x^2}\text{.} \end{equation*}
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(a)
Note that we can write \(g(x) = (1+x^2)^{-1}\text{.}\) Determine \(g'(x)\text{,}\) \(g''(x)\text{,}\) and \(g'''(x)\text{.}\)
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(b)
In (a), you found that
\begin{equation*} g'''(x) = \frac{24x - 24x^3}{(1+x^2)^4}\text{.} \end{equation*}
What derivative rules would be needed to find \(g^{(4)}(x)\text{?}\)
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(c)
Find \(g(0)\text{,}\) \(g'(0)\text{,}\) \(g''(0)\text{,}\) and \(g'''(0)\text{,}\) and use the results to determine \(T_3(x)\text{.}\)
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(d)
Since the derivatives of \(g\) become so complicated, we consider a different approach to finding the Taylor series centered at \(a = 0\) for \(g(x) = \frac{1}{1+x^2}\text{.}\) We take advantage of the fact that \(f(x) = \frac{1}{1-x}\) has a Taylor series expansion we already know, and observe that \(f\) has an algebraic structure similar to \(g\text{.}\)
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We introduce a new variable, \(u\text{,}\) and recall that
\begin{equation} f(u) = \frac{1}{1-u} = 1 + u + u^2 + u^3 + \cdots + u^n + \cdots\tag{8.5.1} \end{equation}
for \(|u| \lt 1\text{.}\)
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Let \(u = -x^2\text{.}\) What equation results from evaluating \(f(-x^2)\) in Equation (8.5.1)?
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(e)
How does our work in (d) compare to the formula you found for \(P_3(x)\) and suggest the Taylor series centered at \(a = 0\) for \(g(x) = \frac{1}{1+x^2}\text{?}\) For what values of \(x\) do you expect this series will converge? Why?
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Subsection 8.5.2 Using substitution and algebra to find new Taylor series expressions

The substitution technique we used in Preview Activity 8.5.1 can be used to find the Taylor series for any function whose structure is similar to that of a Taylor series we already know.
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Example 8.5.1.

Find the Taylor series expansion for \(g(x) = x^4 \cos(x^3)\) and determine the set of all \(x\)-values for which the series converges.
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Solution.
Because \(\cos(x^3)\) is part of the function whose Taylor series we seek, we start with the familiar series for \(\cos(x)\text{.}\) We know that
\begin{equation} f(u) = \cos(u) = 1 - \frac{1}{2!}u^2 + \frac{1}{4!}u^4 - \cdots + (-1)^n \frac{1}{(2n)!} u^{2n} + \cdots\tag{8.5.2} \end{equation}
for all real numbers \(u\text{.}\)
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By letting \(u = x^3\) in Equation (8.5.2), it follows that
\begin{equation*} \cos(x^3) = 1 - \frac{1}{2!}(x^3)^2 + \frac{1}{4!}(x^3)^4 - \cdots + (-1)^n \frac{1}{(2n)!} (x^3)^{2n} + \cdots\text{,} \end{equation*}
so
\begin{equation} \cos(x^3) = 1 - \frac{1}{2!}x^6 + \frac{1}{4!}x^{12} - \cdots + (-1)^n \frac{1}{(2n)!} x^{6n} + \cdots\text{.}\tag{8.5.3} \end{equation}
Then, multiplying both sides of Equation (8.5.3) by \(x^4\text{,}\) we find that
\begin{equation*} g(x) = x^4 \cos(x^3) = x^4 - \frac{1}{2!}x^{10} + \frac{1}{4!}x^{16} - \cdots + (-1)^n \frac{1}{(2n)!} x^{6n+4} + \cdots\text{.} \end{equation*}
Since Equation (8.5.2) is valid for every real number \(u\text{,}\) letting \(u = x^3\) tells us that Equation (8.5.3) is valid for every real number \(x\text{.}\) The Ratio Test can be used to show that multiplying every term of a Taylor series by the same power of \(x\) does not change the set of \(x\)-values for which the series converges, so the Taylor series for \(g(x)\) also converges for every value of \(x\text{.}\)
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Because the approaches in Preview Activity 8.5.1 and Example 8.5.1 each require us to use a known Taylor series, we restate the Taylor series we’ve established so far for \(5\) important functions in Table 8.5.2. These \(5\) Taylor series will also be valuable when we soon learn another approach for finding series representations of other more complicated functions.
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Important Taylor series representations.

Table 8.5.2. Taylor series and the open intervals of \(x\)-values where they converge for \(5\) important functions.
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\(f(x)\) Taylor series, \(T_f\text{,}\) centered at \(0\) \(f(x) = T_f(x)\)
\(\frac{1}{1-x}\) \(\sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \cdots\) if \(|x| \lt 1\)
\(\ln(1+x)\) \(\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^k}{k} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots\) if \(|x| \lt 1\)
\(\sin(x)\) \(\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots\) for all real \(x\)
\(\cos(x)\) \(\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!} = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \cdots\) for all real \(x\)
\(e^x\) \(\sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots\) for all real \(x\)
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In Activity 8.5.2, we find several Taylor series using substitution and algebraic techniques.
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Activity 8.5.2.
Use known Taylor series and substitution and algebraic techniques to find a Taylor series representation for each of the following functions. In addition, state the interval of \(x\)-values for which you expect each Taylor series to converge.
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(e)
\(r(x) = \dfrac{e^{3x}-1}{3x}\)
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Subsection 8.5.3 Differentiating and integrating Taylor series

In Chapter 5, we discussed the challenge posed by definite integrals such as
\begin{equation*} \int_0^1 \sin(x^2) \, dx\text{.} \end{equation*}
Because we are unable to find a simple algebraic antiderivative for the function \(\sin(x^2)\text{,}\) we cannot use the First Fundamental Theorem of Calculus to evaluate the integral exactly. We learned in Section 5.2 that the Second Fundamental Theorem of Calculus provides us with an antiderivative of a given function by using an integral function: one antiderivative of \(f(x) = \sin(x^2)\) is
\begin{equation*} F(x) = \int_0^x \sin(u^2) \, du\text{.} \end{equation*}
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Our recent work with Taylor series now suggests another way to find an antiderivative \(F(x)\) for \(f(x) = \sin(x^2)\text{,}\) and this approach also provides new options for finding additional Taylor series. In Activity 8.5.2, as part of our work in finding a Taylor series for \(x^3 \sin(x^2)\) we found that
\begin{equation} \sin(x^2) = x^2 - \frac{1}{3!}x^6 + \frac{1}{5!}x^{10} - \cdots + (-1)^{n-1} \frac{1}{(2n-1)!} x^{4n-2} + \cdots\tag{8.5.4} \end{equation}
and that this representation of \(\sin(x^2)\) is valid for every value of \(x\text{.}\) This infinite series representation suggests that we could find an antiderivative \(F(x)\) of \(\sin(x^2)\) by using Equation (8.5.4) to actually evaluate the integral \(\int_0^x \sin(u^2) \, du\text{.}\) Doing so, we see
\begin{align*} F(x) \amp= \int_0^x \sin(u^2) \, du\\ \amp= \int_0^x \left( u^2 - \frac{1}{3!}u^6 + \frac{1}{5!}u^{10} + \cdots + (-1)^{n+1} \frac{1}{(2n-1)!} u^{4n-2} + \cdots \right) \, du\\ \amp= \left. \frac{1}{3}u^3 - \frac{1}{7 \cdot 3!} u^7 + \frac{1}{11 \cdot 5!}u^{11} + \cdots + \frac{(-1)^{n+1}}{(4n-1)(2n-1)!} u^{4n-1} + \cdots \right|_0^x\\ \amp= \left( \frac{1}{3}x^3 - \frac{1}{7 \cdot 3!} x^7 + \frac{1}{11 \cdot 5!}x^{11} + \cdots + \frac{(-1)^{n+1}}{(4n-1)(2n-1)!} x^{4n-1} + \cdots \right) - 0\\ \amp= \frac{1}{3}x^3 - \frac{1}{7 \cdot 3!} x^7 + \frac{1}{11 \cdot 5!}x^{11} + \cdots + \frac{(-1)^{n+1}}{(4n-1)(2n-1)!} x^{4n-1} + \cdots \end{align*}
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Hence we have found that
\begin{equation*} F(x) = \frac{1}{3}x^3 - \frac{1}{7 \cdot 3!} x^7 + \frac{1}{11 \cdot 5!}x^{11} + \cdots + \frac{(-1)^{n+1}}{(4n-1)(2n-1)!} x^{4n-1} + \cdots \end{equation*}
is an antiderivative of \(f(x) = \sin(x^2)\text{.}\)
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It turns out that integrating a Taylor series has no effect on the open
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It is possible for the convergence status at the endpoints of the interval to change, but we are normally not concerned with those specific \(x\)-values.
interval of convergence of a Taylor series, nor does differentiating such a series. This fact is stated formally in a result called The Power Series Differentiation and Integration Theorem. A power series is any series of the form
\begin{equation*} f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k\text{.} \end{equation*}
Every Taylor series is a power series, and a famous result called Borel’s Theorem tells us that every power series is in fact the Taylor series of a related function.
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The Power Series Differentiation and Integration Theorem.

If \(f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k\) converges on the open interval \(|x-a| \lt R\) for a positive real number \(R\text{,}\) then
\begin{equation*} f'(x) = \sum_{k=1}^{\infty} k \cdot c_k (x-a)^{k-1} \end{equation*}
and
\begin{equation*} \int f(x) \, dx = C + \sum_{k=0}^{\infty} \frac{c_k}{k+1} (x-a)^{k+1} \end{equation*}
and both of these series converge on the same open interval, \(|x-a| \lt R\text{.}\)
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Note how the index of representation of \(f'(x)\) in sigma notation changes. This is because if we originally have
\begin{equation*} f(x) = c_0 + c_1(x-1) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots + c_k(x-a)^k + \cdots \end{equation*}
and we differentiate, we get
\begin{equation*} f'(x) = c_1 + 2c_2(x-a) + 3c_3(x_a)^2 + \cdots + k \cdot c_k(x-a)^{k-1} + \cdots \end{equation*}
and so to have sigma notation for \(f'(x)\text{,}\) we start with \(k=1\) to accommodate the fact that the general term is \(k \cdot c_k(x-a)^{k-1}\) in order to ensure all powers of \(x\) are nonnegative.
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Furthermore, this result shows that differentiating or integrating a power series has no effect on radius of convergence of the power series. Stated more informally, the Power Series Differentiation and Integration Theorem tells us that when it comes to differentiating or integrating a Taylor series, we can do so just as if they were finite polynomials: we can differentiate or integrate the Taylor series term-wise following the Power Rule for differentiating or integrating \(x^n\text{.}\) Moreover, doing so doesn’t change the open interval on which the Taylor series converges
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Differentiating or integrating can change the convergence status at the endpoints of the interval, but we again will not concern ourselves with that issue .
. Since polynomials are the easiest of all functions to differentiate and integrate, we can now find many more Taylor series of interesting functions, and even find Taylor series of some familiar functions in different ways.
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Example 8.5.3.

Use the familiar Taylor series for \(g(x) = \frac{1}{1-x}\) to develop the Taylor series for \(h(x) = \ln(1+x)\) using the Power Series Differentiation and Integration Theorem.
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Solution.
In Example 8.2.5, we developed the Taylor polynomials centered at \(a = 0\) for \(h(x) = \ln(1+x)\text{,}\) using the definition. And in Example 8.4.2, we considered the Taylor series for \(h(x) = \ln(1+x)\) that we deduced from patterns in the Taylor polynomials.
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Here, we take a different approach to finding the Taylor series for \(h(x) = \ln(1+x)\) that starts with the familiar geometric series expansion of \(g(x) = \frac{1}{1-x}\text{.}\)
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The key insight in this approach is the fact that
\begin{equation*} \frac{d}{dx}\left[ \ln(1+x) \right] = \frac{1}{1+x}\text{.} \end{equation*}
We thus first find a Taylor series representation for \(\frac{1}{1+x}\text{,}\) and then integrate to find the Taylor series for \(\ln(1+x)\text{.}\)
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We know that for \(|x| \lt 1\text{,}\)
\begin{equation*} g(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots + x^n + \cdots\text{.} \end{equation*}
Using the variable \(u\) instead (in anticipation of a change of variables), we have
\begin{equation*} g(u) = \frac{1}{1-u} = 1 + u + u^2 + u^3 + \cdots + u^n + \cdots\text{.} \end{equation*}
Next, letting \(u = -x\text{,}\) it follows that
\begin{align*} g(-x) = \frac{1}{1+x} \amp= 1 + (-x) + (-x)^2 + (-x)^3 + \cdots + (-x)^n + \cdots \\ \amp= 1 - x + x^2 - x^3 + \cdots + (-1)^n x^n + \cdots\text{,} \end{align*}
which also converges for \(|x| \lt 1\text{.}\)
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Now, since \(\frac{1}{1+x}\) is the derivative of \(\ln(1+x)\) and \(\ln(1) = 0\text{,}\) we can use the Second FTC to write
\begin{equation*} \ln(1 + x) = \int_0^x \frac{1}{1+t} \, dt\text{,} \end{equation*}
which combined with our earlier series representation for \(\frac{1}{1+x}\) shows that
\begin{align*} \ln(1 + x) \amp= \int_0^x \frac{1}{1+t} \, dt\\ \amp= \int_0^x 1 - t + t^2 - t^3 + \cdots \, dt\\ \amp= \left. t - \frac{1}{2}t^2 + \frac{1}{3} t^3 - \frac{1}{4}t^4 + \cdots \right|_0^x\\ \amp= x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \cdots \end{align*}
and this series is guaranteed to converge on the same open interval as the series we started with (\(-1 \lt x \lt 1\)). This is also the same series we found for \(h(x) = \ln(1+x)\) working directly from the definition of a Taylor polynomial in Example 8.2.5.
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Activity 8.5.3.
In this activity we determine the Taylor series expansion for \(\sin(x)\) in a different way and then also find the Taylor series for \(\arctan(x)\text{.}\)
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(a)
To find the Taylor series for \(\sin(x)\) without using the definition, we consider the function \(f(x) = \cos(x)\) and recall that \(f'(x) = -\sin(x)\text{.}\) Then we use the Power Series Differentiation Theorem on the known series for \(f(x) = \cos(x)\) to find the Taylor series for \(f'(x) = -\sin(x)\text{.}\)
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(i)
Consider \(f(x) = \cos(x)\) and recall that its Taylor series in sigma notation is
\begin{equation*} f(x) = \cos(x) = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!}\text{.} \end{equation*}
Write out the first four nonzero terms of the Taylor series in expanded form.
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(ii)
Using the expanded form of the Taylor series for \(f(x) = \cos(x)\text{,}\) apply the Power Series Differentiation Theorem to find the Taylor series for \(f'(x) = -\sin(x)\) in expanded form, being sure to simplify each constant coefficient in the expansion. Show the first three nonzero terms.
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(iii)
Now recall the series expansion of \(\cos(x)\) in sigma notation. What is the derivative of the general term of the series,
\begin{equation*} \frac{d}{dx}\left[ (-1)^{k} \frac{x^{2k}}{(2k)!} \right]\text{,} \end{equation*}
in its most simplified form?
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(iv)
Finally, state the Taylor series for \(f'(x) = -\sin(x)\) in sigma notation. Be sure to decide whether the series should be indexed starting with \(k = 0\) or \(k = 1\) and to justify your choice.
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(v)
How can you now use your work in part (iv) to find the Taylor series for \(\sin(x)\text{?}\)
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(b)
Consider the function \(g(x) = \arctan(x)\text{.}\)
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(ii)
Recall that in Preview Activity 8.5.1, we found the Taylor series expansion for
\begin{equation*} f(x) = \frac{1}{1+x^2}\text{.} \end{equation*}
State the Taylor series for \(f(x)\) that you found in the preview activity.
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(iii)
Explain why we can think of \(g(x)\) as being given by
\begin{equation*} g(x) = \int_0^x f(u) \, du\text{.} \end{equation*}
Use this relationship to find a power series expansion for \(g(x)\text{.}\)
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(iv)
For what interval of \(x\)-values is the Taylor series for \(g(x) = \arctan(x)\) guaranteed to converge?
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Activity 8.5.4.
In this activity we will determine the Taylor series expansion for the famous error function,
\begin{equation*} \erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt\text{,} \end{equation*}
that arises in probability theory and other important applications.
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(a)
Use the Taylor series for \(e^{x}\) to find the Taylor series for \(e^{-t^2}\text{.}\)
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(b)
Next, evaluate the integral \(\int_0^x e^{-t^2} \, dt\) by replacing \(e^{-t^2}\) with its Taylor series.
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(c)
Use your work in (b) to state the Taylor series for \(\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt\)
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(d)
For what interval of \(x\)-values will the Taylor series for \(\erf(x)\) converge? Why?
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(e)
In probability theory, \(\erf(x)\) is important because of its connection to the normal distribution, which is represented by a bell curve. Indeed, \(\erf(x)\) represents the fraction of a normally distributed characteristic in a population that lies between \(0\) and \(x\text{.}\) How can you use your result in (c) to estimate \(\erf(0.5)\text{?}\)
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Near the end of Section 8.4, we noted the important big-picture perspective that for familiar basic functions that are infinitely differentiable, such as \(f(x) = \sin(x)\text{,}\) not only can we find the function’s Taylor series and determine the open interval of \(x\)-values for which the Taylor series converges, but the Taylor series converges to the function itself. Furthermore, we noted that these representations play a key role in how computers provide decimal approximations to quantities such as \(e^1\text{,}\) which can be represented as
\begin{equation*} e^1 = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots\text{.} \end{equation*}
Our most recent work with Taylor series shows that the news is better still: now we can easily represent even more complicated functions with Taylor series, such as \(f(x) = e^{-x^2}\) and \(g(x) = \sin(x^2)\text{,}\) and determine their antiderivatives using their infinite Taylor series and treating that representation just like a polynomial. In Section 8.6, we investigate further how certain infinite series of numbers can be easily and accurately approximated using partial sums that result from evaluating Taylor series, and how we can even quantify how much error is present in certain partial sum approximations.
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Subsection 8.5.4 Summary

  • Through substitution, we can use any known Taylor series to find the Taylor series of a related function. For example, we know that if \(|u| \lt 1\text{,}\)
    \begin{equation*} \ln(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \frac{1}{4}u^4 + \cdots \end{equation*}
    Letting \(u = 4x^2\text{,}\) it follows that
    \begin{equation*} \ln(1+4x^2) = 4x^2 - \frac{1}{2} \cdot 16x^4 + \frac{1}{3} \cdot 64x^6 - \frac{1}{4} \cdot 256x^8 + \cdots \end{equation*}
    and that this representation converges if \(|4x^2| \lt 1\text{,}\) so for \(x\) such that \(|x| \lt \frac{1}{2}\text{.}\)
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  • The Power Series Differentation and Integration Theorem tells us that we can differentiate or integrate a Taylor series in the natural way and that doing so has essentially no impact on the set of \(x\)-values for which the series converges. For instance, we might note that if we first found the Taylor series for \(\sin(x)\text{,}\) which is
    \begin{equation*} \sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \cdots \end{equation*}
    and converges for every value of \(x\text{,}\) it follows by differentating that
    \begin{align*} \cos(x) \amp= \frac{d}{dx}[\sin(x)]\\ \amp= \frac{d}{dx}[x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \cdots]\\ \amp= 1 - \frac{1}{3!} \cdot 3x^2 + \frac{1}{5!} \cdot 5x^4 - \frac{1}{7!} \cdot 7x^6 + \cdots\\ \amp= 1 - \frac{1}{2!}x^2 + \frac{1}{4!} x^4 - \frac{1}{6!} x^6 + \cdots \end{align*}
    which is precisely the Taylor series for \(\cos(x)\) that we found by taking derivatives and applying Definition 8.4.1.
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Exercises 8.5.5 Exercises

1.

If \(y = \sum_{k=0}^\infty \left(k+1\right)x^{k+3}\) then \(y^\prime = \sum_{k=0}^\infty\)
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2.

(a) Part 1.
Note that when the constant in the denominator is not 1, we still can use the geometric series, but we have to multiply by a form of 1 that helps us. We’ll use \(\frac{\frac{1}{4}}{\frac{1}{4}}\text{.}\)
\begin{equation*} \frac{1}{(4+x)}=\frac{\frac{1}{4}}{\frac{1}{4}(4+x)}=\frac{\frac{1}{4}}{1+\frac{1}{4}x}=\frac{1}{4}\cdot \frac{1}{1+\frac{1}{4}x} \end{equation*}
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And now we can use the geometric series expression to get
\begin{equation*} \frac{1}{(4+x)}=\frac{1}{4}\cdot \frac{1}{1+\frac{1}{4}x}=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{-1}{4}x\right)^n=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{-1}{4}\right)^nx^n \end{equation*}
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Bringing the coefficient inside the sum means that
\begin{equation*} \frac{1}{(4+x)}=\sum_{n=0}^{\infty}\frac{(-1)^n}{4^{n+1}}x^n \end{equation*}
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Differentiate both sides of the equation above and use the result to express the following function as a power series centered at \(x=0\text{.}\)
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\begin{equation*} f(x) = \frac{1}{(4+x)^2} \end{equation*}
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\(f(x) = \displaystyle \sum\limits_{n=1}^{\infty}\)
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(b) Part 2.
Use your answer above and more differentiation to now express the following function as a power series centered at \(x=0\text{.}\)
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\begin{equation*} g(x) = \frac{1}{(4+x)^3} \end{equation*}
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\(g(x) = \displaystyle \sum\limits_{n=2}^{\infty}\)
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(c) Part 3.
Use your answers above to now express the function as a power series centered at \(x=0\text{.}\)
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\begin{equation*} h(x) = \frac{x^2}{(4+x)^3} \end{equation*}
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\(h(x) = \displaystyle \sum\limits_{n=2}^{\infty}\)
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3.

For the following indefinite integral, find the full power series centered at \(t=0\) and then give the first 5 nonzero terms of the power series and the open interval of convergence.
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\begin{equation*} f(t) = \int \frac{t}{1+t^{5}}\ dt \end{equation*}
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\(f(t) = C + \displaystyle\sum\limits_{n=0}^{\infty}\)
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\(f(t) = C +\) \(+\) \(+\) \(+\) \(+\) \(+ \cdots\)
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The open interval of convergence is: (Give your answer in .)
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4.

For the following indefinite integral, find the full power series centered at \(x=0\) and then give the first 5 nonzero terms of the power series and the open interval of convergence.
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\begin{equation*} f(x) = \int x^2 \ln(1+x)\ dx \end{equation*}
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\(f(x) = C + \displaystyle\sum\limits_{n=1}^{\infty}\)
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\(f(x) = C +\) \(+\) \(+\) \(+\) \(+\) \(+ \cdots\)
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The open interval of convergence is: (Give your answer in .)
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5.

You can interpret “power series” in the following problem statement as “Taylor series”.
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Represent the function \(\displaystyle \frac{7}{(1 - 10 x)}\) as a power series \(\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n\)
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\(c_0 =\)
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\(c_1 =\)
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\(c_2 =\)
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\(c_3 =\)
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\(c_4 =\)
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Find the radius of convergence \(R =\) .
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6.

You can interpret “power series” in the following problem statement as “Taylor series”. You might also want to recall the result of Activity 8.5.3 where we found the series expansion for \(\arctan(x)\text{.}\)
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The function \(f(x) = 8 x \arctan (5 x)\) is represented as a power series
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\(\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n .\)
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Find the first few coefficients in the power series.
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\(c_0 =\)
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\(c_1 =\)
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\(c_2 =\)
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\(c_3 =\)
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\(c_4 =\)
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Find the radius of convergence \(R\) of the series.
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\(R =\)
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7.

In this exercise we find the Taylor series representation for two famous functions, the Fresnel integral functions
\begin{equation*} C(x) = \int_0^x \cos(t^2) \, dt \end{equation*}
and
\begin{equation*} S(x) = \int_0^x \sin(t^2) \, dt\text{.} \end{equation*}
The Fresnel integral functions are important in optics and are used in the design of Fresnel lenses such as those found in lighthouses along the Lake Michigan shore.
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  1. Use the Taylor series for \(\cos(x)\) to find the Taylor series for \(\cos(t^2)\) and hence write \(C(x)\) as a Taylor series.
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  2. For what interval of \(x\)-values will the Taylor series for \(C(x)\) converge? Why?
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  3. Apply your result from (a) to estimate \(C(0.5)\) to within \(0.001\text{.}\)
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  4. Similarly, use the Taylor series for \(\sin(x)\) to find the Taylor series for \(\sin(t^2)\) and hence write \(S(x)\) as a Taylor series.
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  5. For what interval of \(x\)-values will the Taylor series for \(S(x)\) converge? Why?
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  6. Apply your result from (d) to estimate \(S(0.8)\) to within \(0.001\text{.}\)
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8.

The fact that we can differentiate or integrate a Taylor series reveals other important ways we can think about functions such as \(e^x\text{.}\)
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  1. Perhaps the most important property of the function \(h(x) = e^x\) is that \(h'(x) = e^x\text{;}\) that is, the function \(e^x\) is its own derivative. Suppose that we didn’t yet know the coefficients of the Taylor series expansion for \(e^x\text{,}\) so we just said
    \begin{equation} e^x = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n + \cdots\text{.}\tag{8.5.5} \end{equation}
    Let \(x = 0\) in Equation (8.5.5); what does this tell us about the value of \(a_0\text{?}\)
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  2. Take the derivative of both sides of Equation (8.5.5) and call your resulting equation for \(e^x\) “Equation 2”. Why do Equation (8.5.5) and Equation 2 together tell us that \(a_1 = a_0\text{?}\) Combine this observation with your conclusion in (b) and note that you now know the numerical value of both \(a_0\) and \(a_1\text{.}\)
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  3. Why do Equation (8.5.5) and Equation 2 together tell us that \(a_2 = \frac{1}{2}a_1\text{?}\)
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  4. Continue reasoning similarly to find the value of \(a_3\text{,}\) \(a_4\text{,}\) and \(a_5\text{.}\) What do you observe?
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9.

Taylor series also provide an alternate way to evaluate indeterminate limits.
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  1. Find the Taylor series for \(\sin(2t)\) and use it to evaluate the indeterminate limit given by
    \begin{equation*} \lim_{t \to 0} \frac{\sin(2t)}{t}\text{.} \end{equation*}
    Compare your result to what follows from L’Hôpital’s Rule (see Section 3.2 as needed).
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  2. Consider the indeterminate limit given by
    \begin{equation*} \lim_{x \to 0} \frac{e^x - 1 - x}{\cos(2x)-1}\text{.} \end{equation*}
    Find the Taylor series representations for \(f(x) = e^x - 1 - x\) and \(g(x) = \cos(2x)-1\) and use them to evaluate the given limit. How does your result compare to using L’Hôpital’s Rule?
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