Taylor and Maclaurin Series

In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function

f

and the series converges on some interval, how do we prove that the series actually converges to

f ?

Overview of Taylor/Maclaurin Series

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series [link] is a representation for

f

In addition, we would like the first derivative of the power series to equal

f^{'} (a)

Continuing in this way, we look for coefficients c_n such that all the derivatives of the power series [link] will agree with all the corresponding derivatives of

f

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from [link] that power series representations are unique. Therefore, if a function

f

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.

Taylor Polynomials

is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of

f

We now provide a formal definition of Taylor and Maclaurin polynomials for a function

f .

We now show how to use this definition to find several Taylor polynomials for

f (x) = ln x

Finding Maclaurin Polynomials

For each of the following functions, find formulas for the Maclaurin polynomials $p_{0}, p_{1}, p_{2}$

and $p_{3} .$

Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of $p_{0}, p_{1}, p_{2}$

and $p_{3}$

with $f .$

$f (x) = e^{x}$
$f (x) = sin x$
$f (x) = cos x$

Since $f (x) = e^{x},$
we know that
$f (x) = f^{'} (x) = f ″ (x) = ⋯ = f^{(n)} (x) = e^{x}$
for all positive integers n. Therefore,

$f (0) = f^{'} (0) = f ″ (0) = ⋯ = f^{(n)} (0) = 1$

for all positive integers n. Therefore, we have

$\begin{matrix} p_{0} (x) & = & f (0) = 1, \\ p_{1} (x) & = & f (0) + f^{'} (0) x = 1 + x, \\ p_{2} (x) & = & f (0) + f^{'} (0) x + \frac{f ″ (0)}{2!} x^{2} = 1 + x + \frac{1}{2} x^{2}, \\ p_{3} (x) & = & f (0) + f^{'} (0) x + \frac{f ″ (0)}{2} x^{2} + \frac{f ‴ (0)}{3!} x^{3} \\ = & 1 + x + \frac{1}{2} x^{2} + \frac{1}{3!} x^{3}, \\ p_{n} (x) & = & f (0) + f^{'} (0) x + \frac{f ″ (0)}{2} x^{2} + \frac{f ‴ (0)}{3!} x^{3} + ⋯ + \frac{f^{(n)} (0)}{n!} x^{n} \\ = & 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{n}}{n!} \\ = & \sum_{k = 0}^{n} \frac{x^{k}}{k!} . \end{matrix}$

The function and the first three Maclaurin polynomials are shown in [link].
For $f (x) = sin x,$
the values of the function and its first four derivatives at
$x = 0$
are given as follows:

$\begin{matrix} f (x) & = & sin x & f (0) & = & 0 \\ f^{'} (x) & = & cos x & f^{'} (0) & = & 1 \\ f ″ (x) & = & − sin x & f ″ (0) & = & 0 \\ f ‴ (x) & = & − cos x & f ‴ (0) & = & −1 \\ f^{(4)} (x) & = & sin x & f^{(4)} (0) & = & 0. \end{matrix}$

Since the fourth derivative is
$sin x,$
the pattern repeats. That is,
$f^{(2 m)} (0) = 0$
and
$f^{(2 m + 1)} (0) = {(−1)}^{m}$
for
$m \geq 0 .$
Thus, we have

$\begin{matrix} p_{0} (x) = 0, \\ p_{1} (x) = 0 + x = x, \\ p_{2} (x) = 0 + x + 0 = x, \\ p_{3} (x) = 0 + x + 0 - \frac{1}{3!} x^{3} = x - \frac{x^{3}}{3!}, \\ p_{4} (x) = 0 + x + 0 - \frac{1}{3!} x^{3} + 0 = x - \frac{x^{3}}{3!}, \\ p_{5} (x) = 0 + x + 0 - \frac{1}{3!} x^{3} + 0 + \frac{1}{5!} x^{5} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!}, \end{matrix}$

and for
$m \geq 0,$

$\begin{matrix} p_{2 m + 1} (x) & = p_{2 m + 2} (x) \\ = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - ⋯ + {(−1)}^{m} \frac{x^{2 m + 1}}{(2 m + 1)!} \\ = \sum_{k = 0}^{m} {(−1)}^{k} \frac{x^{2 k + 1}}{(2 k + 1)!} . \end{matrix}$

Graphs of the function and its Maclaurin polynomials are shown in [link].
For $f (x) = cos x,$
the values of the function and its first four derivatives at
$x = 0$
are given as follows:

$\begin{matrix} f (x) & = & cos x & f (0) & = & 1 \\ f^{'} (x) & = & − sin x & f^{'} (0) & = & 0 \\ f ″ (x) & = & − cos x & f ″ (0) & = & −1 \\ f ‴ (x) & = & sin x & f ‴ (0) & = & 0 \\ f^{(4)} (x) & = & cos x & f^{(4)} (0) & = & 1. \end{matrix}$

Since the fourth derivative is
$sin x,$
the pattern repeats. In other words,
$f^{(2 m)} (0) = {(−1)}^{m}$
and
$f^{(2 m + 1)} = 0$
for
$m \geq 0 .$
Therefore,

$\begin{matrix} p_{0} (x) = 1, \\ p_{1} (x) = 1 + 0 = 1, \\ p_{2} (x) = 1 + 0 - \frac{1}{2!} x^{2} = 1 - \frac{x^{2}}{2!}, \\ p_{3} (x) = 1 + 0 - \frac{1}{2!} x^{2} + 0 = 1 - \frac{x^{2}}{2!}, \\ p_{4} (x) = 1 + 0 - \frac{1}{2!} x^{2} + 0 + \frac{1}{4!} x^{4} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!}, \\ p_{5} (x) = 1 + 0 - \frac{1}{2!} x^{2} + 0 + \frac{1}{4!} x^{4} + 0 = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!}, \end{matrix}$

and for
$n \geq 0,$

$\begin{matrix} p_{2 m} (x) & = p_{2 m + 1} (x) \\ = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - ⋯ + {(−1)}^{m} \frac{x^{2 m}}{(2 m)!} \\ = \sum_{k = 0}^{m} {(−1)}^{k} \frac{x^{2 k}}{(2 k)!} . \end{matrix}$

Graphs of the function and the Maclaurin polynomials appear in [link].

Taylor’s Theorem with Remainder

at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials

{p_{n}}

converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to

f .

we need the remainder R_n to converge to zero. To determine if R_n converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.

is differentiable on an interval I containing a and x, then by the Mean Value Theorem there exists a real number c between a and x such that

f (x) - f (a) = f^{'} (c) (x - a) .

is n times differentiable on an interval I containing a and x, then the nth remainder R_n satisfies

for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder R_n. If we happen to know that

\| f^{(n + 1)} (x) \|

We now state Taylor’s theorem, which provides the formal relationship between a function

f

This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for

f

Proof

We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t), it is a differentiable function. Also, g is zero at

t = a

Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that

g^{'} (c) = 0 .

By Rolle’s theorem, we conclude that there exists a number c between a and x such that

g^{'} (c) = 0 .

Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by

n + 1,

as desired. From this fact, it follows that if there exists M such that

\| f^{(n + 1)} (x) \| \leq M

Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of

f (x) = \sqrt[3]{x}

and determine how accurate these approximations are at estimating

\sqrt[3]{11} .

Using Linear and Quadratic Approximations to Estimate Function Values

Consider the function $f (x) = \sqrt[3]{x} .$

Find the first and second Taylor polynomials for $f$
at
$x = 8 .$
Use a graphing utility to compare these polynomials with
$f$
near
$x = 8 .$
Use these two polynomials to estimate $\sqrt[3]{11} .$
Use Taylor’s theorem to bound the error.

For $f (x) = \sqrt[3]{x},$
the values of the function and its first two derivatives at
$x = 8$
are as follows:

$\begin{matrix} f (x) & = & \sqrt[3]{x} & f (8) & = & 2 \\ f^{'} (x) & = & \frac{1}{3 x^{2 / 3}} & f^{'} (8) & = & \frac{1}{12} \\ f ″ (x) & = & \frac{−2}{9 x^{5 / 3}} & f ″ (8) & = & - \frac{1}{144} . \end{matrix}$

Thus, the first and second Taylor polynomials at
$x = 8$
are given by

$\begin{matrix} p_{1} (x) & = & f (8) + f^{'} (8) (x - 8) \\ = & 2 + \frac{1}{12} (x - 8) \\ p_{2} (x) & = & f (8) + f^{'} (8) (x - 8) + \frac{f ″ (8)}{2!} {(x - 8)}^{2} \\ = & 2 + \frac{1}{12} (x - 8) - \frac{1}{288} {(x - 8)}^{2} . \end{matrix}$

The function and the Taylor polynomials are shown in [link].
Using the first Taylor polynomial at $x = 8,$
we can estimate

$\sqrt[3]{11} \approx p_{1} (11) = 2 + \frac{1}{12} (11 - 8) = 2.25 .$

Using the second Taylor polynomial at
$x = 8,$
we obtain

$\sqrt[3]{11} \approx p_{2} (11) = 2 + \frac{1}{12} (11 - 8) - \frac{1}{288} {(11 - 8)}^{2} = 2.21875 .$
By [link], there exists a c in the interval $(8, 11)$
such that the remainder when approximating
$\sqrt[3]{11}$
by the first Taylor polynomial satisfies

$R_{1} (11) = \frac{f ″ (c)}{2!} {(11 - 8)}^{2} .$

We do not know the exact value of c, so we find an upper bound on
$R_{1} (11)$
by determining the maximum value of
$f ″$
on the interval
$(8, 11) .$
Since
$f ″ (x) = - \frac{2}{9 x^{5 / 3}},$
the largest value for
$\| f ″ (x) \|$
on that interval occurs at
$x = 8 .$
Using the fact that
$f ″ (8) = - \frac{1}{144},$
we obtain

$\| R_{1} (11) \| \leq \frac{1}{144 \cdot 2!} {(11 - 8)}^{2} = 0.03125 .$

Similarly, to estimate
$R_{2} (11),$
we use the fact that

$R_{2} (11) = \frac{f ‴ (c)}{3!} {(11 - 8)}^{3} .$

Since
$f ‴ (x) = \frac{10}{27 x^{8 / 3}},$
the maximum value of
$f ‴$
on the interval
$(8, 11)$
is
$f ‴ (8) \approx 0.0014468 .$
Therefore, we have

$\| R_{2} (11) \| \leq \frac{0.0011468}{3!} {(11 - 8)}^{3} \approx 0.0065104 .$

Representing Functions with Taylor and Maclaurin Series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

We know that the Taylor series found in this example converges on the interval

(0, 2),

We consider this question in more generality in a moment, but for this example, we can answer this question by writing

can be represented by the geometric series

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

Therefore, the Taylor series found in [link] does converge to

f (x) = \frac{1}{x}

We now consider the more general question: if a Taylor series for a function

f

converges on some interval, how can we determine if it actually converges to

f ?

To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for

f

at a, the nth partial sum is given by the nth Taylor polynomial p_n. Therefore, to determine if the Taylor series converges to

f,

and show that these series converge to the corresponding functions for all real numbers by proving that the remainders

R_{n} (x) \to 0

Finding Maclaurin Series

For each of the following functions, find the Maclaurin series and its interval of convergence. Use [link] to prove that the Maclaurin series for $f$

converges to $f$

on that interval.

e^x
$sin x$

Using the nth Maclaurin polynomial for e^x found in [link]a., we find that the Maclaurin series for e^x is given by

$\sum_{n = 0}^{\infty} \frac{x^{n}}{n!} .$

To determine the interval of convergence, we use the ratio test. Since

$\frac{\| a_{n + 1} \|}{\| a_{n} \|} = \frac{{\| x \|}^{n + 1}}{(n + 1)!} \cdot \frac{n!}{{\| x \|}^{n}} = \frac{\| x \|}{n + 1},$

we have

$lim_{n \to \infty} \frac{\| a_{n + 1} \|}{\| a_{n} \|} = lim_{n \to \infty} \frac{\| x \|}{n + 1} = 0$

for all x. Therefore, the series converges absolutely for all x, and thus, the interval of convergence is
$(− \infty, \infty) .$
To show that the series converges to e^x for all x, we use the fact that
$f^{(n)} (x) = e^{x}$
for all
$n \geq 0$
and e^x is an increasing function on
$(− \infty, \infty) .$
Therefore, for any real number b, the maximum value of e^x for all
$\| x \| \leq b$
is e^b. Thus,

$\| R_{n} (x) \| \leq \frac{e^{b}}{(n + 1)!} {\| x \|}^{n + 1} .$

Since we just showed that

$\sum_{n = 0}^{\infty} \frac{\| x {\|}^{n}}{n!}$

converges for all x, by the divergence test, we know that

$lim_{n \to \infty} \frac{{\| x \|}^{n + 1}}{(n + 1)!} = 0$

for any real number x. By combining this fact with the squeeze theorem, the result is
$lim_{n \to \infty} R_{n} (x) = 0 .$
Using the nth Maclaurin polynomial for $sin x$
found in [link]b., we find that the Maclaurin series for
$sin x$
is given by

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

In order to apply the ratio test, consider

$\frac{\| a_{n + 1} \|}{\| a_{n} \|} = \frac{{\| x \|}^{2 n + 3}}{(2 n + 3)!} \cdot \frac{(2 n + 1)!}{{\| x \|}^{2 n + 1}} = \frac{{\| x \|}^{2}}{(2 n + 3) (2 n + 2)} .$

Since

$lim_{n \to \infty} \frac{{\| x \|}^{2}}{(2 n + 3) (2 n + 2)} = 0$

for all x, we obtain the interval of convergence as
$(− \infty, \infty) .$
To show that the Maclaurin series converges to
$sin x,$
look at
$R_{n} (x) .$
For each x there exists a real number c between 0 and x such that

$R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} x^{n + 1} .$

Since
$\| f^{(n + 1)} (c) \| \leq 1$
for all integers n and all real numbers c, we have

$\| R_{n} (x) \| \leq \frac{{\| x \|}^{n + 1}}{(n + 1)!}$

for all real numbers x. Using the same idea as in part a., the result is
$lim_{n \to \infty} R_{n} (x) = 0$
for all x, and therefore, the Maclaurin series for
$sin x$
converges to
$sin x$
for all real x.

Key Concepts

Key Equations

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

f (x) = 1 + x + x^{2}

at $a = 1$

f (x) = 1 + x + x^{2}

at $a = −1$

f (−1) = 1; f^{'} (−1) = −1; f ″ (−1) = 2; f (x) = 1 - (x + 1) + {(x + 1)}^{2}

f (x) = cos (2 x)

at $a = π$

f (x) = sin (2 x)

at $a = \frac{π}{2}$

f^{'} (x) = 2 cos (2 x); f ″ (x) = −4 sin (2 x); p_{2} (x) = −2 (x - \frac{π}{2})

f (x) = \sqrt{x}

at $a = 4$

f (x) = ln x

at $a = 1$

f^{'} (x) = \frac{1}{x}; f ″ (x) = - \frac{1}{x^{2}}; p_{2} (x) = 0 + (x - 1) - \frac{1}{2} {(x - 1)}^{2}

f (x) = \frac{1}{x}

at $a = 1$

f (x) = e^{x}

at $a = 1$

p_{2} (x) = e + e (x - 1) + \frac{e}{2} {(x - 1)}^{2}

In the following exercises, verify that the given choice of n in the remainder estimate $\| R_{n} \| \leq \frac{M}{(n + 1)!} {(x - a)}^{n + 1},$

where M is the maximum value of $\| f^{(n + 1)} (z) \|$

on the interval between a and the indicated point, yields $\| R_{n} \| \leq \frac{1}{1000} .$

Find the value of the Taylor polynomial p_n of $f$

at the indicated point.

[T] $\sqrt{10}; a = 9, n = 3$

[T] ${(28)}^{1 / 3}; a = 27, n = 1$

\frac{d^{2}}{d x^{2}} x^{1 / 3} = - \frac{2}{9 x^{5 / 3}} \geq −0.00092 …

when $x \geq 28$

so the remainder estimate applies to the linear approximation $x^{1 / 3} \approx p_{1} (27) = 3 + \frac{x - 27}{27},$

which gives ${(28)}^{1 / 3} \approx 3 + \frac{1}{27} = 3. \bar{037},$

while ${(28)}^{1 / 3} \approx 3.03658 .$

[T] $sin (6); a = 2 π, n = 5$

[T] e²; $a = 0, n = 9$

Using the estimate $\frac{2^{10}}{10!} < 0.000283$

we can use the Taylor expansion of order 9 to estimate e^x at $x = 2 .$

as $e^{2} \approx p_{9} (2) = 1 + 2 + \frac{2^{2}}{2} + \frac{2^{3}}{6} + ⋯ + \frac{2^{9}}{9!} = 7.3887 …$

whereas $e^{2} \approx 7.3891 .$

[T] $cos (\frac{π}{5}); a = 0, n = 4$

[T] $ln (2); a = 1, n = 1000$

Since $\frac{d^{n}}{d x^{n}} (ln x) = {(−1)}^{n - 1} \frac{(n - 1)!}{x^{n}}, R_{1000} \approx \frac{1}{1001} .$

One has $p_{1000} (1) = \sum_{n = 1}^{1000} \frac{{(−1)}^{n - 1}}{n} \approx 0.6936$

whereas $ln (2) \approx 0.6931 ⋯ .$

Integrate the approximation $sin t \approx t - \frac{t^{3}}{6} + \frac{t^{5}}{120} - \frac{t^{7}}{5040}$

evaluated at πt to approximate $\int_{0}^{1} \frac{sin π t}{π t} d t .$

Integrate the approximation $e^{x} \approx 1 + x + \frac{x^{2}}{2} + ⋯ + \frac{x^{6}}{720}$

evaluated at −x² to approximate $\int_{0}^{1} e^{− x^{2}} d x .$

\int_{0}^{1} (1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \frac{x^{8}}{24} - \frac{x^{10}}{120} + \frac{x^{12}}{720}) d x

= 1 - \frac{1^{3}}{3} + \frac{1^{5}}{10} - \frac{1^{7}}{42} + \frac{1^{9}}{9 \cdot 24} - \frac{1^{11}}{120 \cdot 11} + \frac{1^{13}}{720 \cdot 13} \approx 0.74683

whereas $\int_{0}^{1} e^{− x^{2}} d x \approx 0.74682 .$

In the following exercises, find the smallest value of n such that the remainder estimate $\| R_{n} \| \leq \frac{M}{(n + 1)!} {(x - a)}^{n + 1},$

where M is the maximum value of $\| f^{(n + 1)} (z) \|$

on the interval between a and the indicated point, yields $\| R_{n} \| \leq \frac{1}{1000}$

on the indicated interval.

f (x) = sin x

on $[− π, π], a = 0$

f (x) = cos x

on $[- \frac{π}{2}, \frac{π}{2}], a = 0$

Since $f^{(n + 1)} (z)$

is $sin z$

or $cos z,$

we have $M = 1 .$

Since $\| x - 0 \| \leq \frac{π}{2},$

we seek the smallest n such that $\frac{π^{n + 1}}{2^{n + 1} (n + 1)!} \leq 0.001 .$

The smallest such value is $n = 7 .$

The remainder estimate is $R_{7} \leq 0.00092 .$

f (x) = e^{−2 x}

on $[−1, 1], a = 0$

f (x) = e^{− x}

on $[−3, 3], a = 0$

Since $f^{(n + 1)} (z) = ± e^{− z}$

one has $M = e^{3} .$

Since $\| x - 0 \| \leq 3,$

one seeks the smallest n such that $\frac{3^{n + 1} e^{3}}{(n + 1)!} \leq 0.001 .$

The smallest such value is $n = 14 .$

The remainder estimate is $R_{14} \leq 0.000220 .$

In the following exercises, the maximum of the right-hand side of the remainder estimate $\| R_{1} \| \leq \frac{max \| f ″ (z) \|}{2} R^{2}$

on $[a - R, a + R]$

occurs at a or $a \pm R .$

Estimate the maximum value of R such that $\frac{max \| f ″ (z) \|}{2} R^{2} \leq 0.1$

on $[a - R, a + R]$

by plotting this maximum as a function of R.

[T] e^x approximated by $1 + x, a = 0$

[T] $sin x$

approximated by x, $a = 0$

This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.5966, 0.2).

Since $sin x$

is increasing for small x and since $si n ″ x = − sin x,$

the estimate applies whenever $R^{2} sin (R) \leq 0.2,$

which applies up to $R = 0.596 .$

[T] $ln x$

approximated by $x - 1, a = 1$

[T] $cos x$

approximated by $1, a = 0$

This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.44720, 0.2).

Since the second derivative of $cos x$

is $− cos x$

and since $cos x$

is decreasing away from $x = 0,$

the estimate applies when $R^{2} cos R \leq 0.2$

or $R \leq 0.447 .$

In the following exercises, find the Taylor series of the given function centered at the indicated point.

x^{4}

at $a = −1$

1 + x + x^{2} + x^{3}

at $a = −1$

{(x + 1)}^{3} - 2 {(x + 1)}^{2} + 2 (x + 1)

sin x

at $a = π$

cos x

at $a = 2 π$

Values of derivatives are the same as for $x = 0$

so $cos x = {\sum_{n = 0}^{\infty} (−1)}^{n} \frac{{(x - 2 π)}^{2 n}}{(2 n)!}$

sin x

at $x = \frac{π}{2}$

cos x

at $x = \frac{π}{2}$

cos (\frac{π}{2}) = 0, − sin (\frac{π}{2}) = −1

so $cos x = \sum_{n = 0}^{\infty} {(−1)}^{n + 1} \frac{{(x - \frac{π}{2})}^{2 n + 1}}{(2 n + 1)!},$

which is also $− cos (x - \frac{π}{2}) .$

e^{x}

at $a = −1$

e^{x}

at $a = 1$

The derivatives are $f^{(n)} (1) = e$

so $e^{x} = e \sum_{n = 0}^{\infty} \frac{{(x - 1)}^{n}}{n!} .$

\frac{1}{{(x - 1)}^{2}}

at $a = 0$

(Hint: Differentiate $\frac{1}{1 - x} .)$

\frac{1}{{(x - 1)}^{3}}

at $a = 0$

\frac{1}{{(x - 1)}^{3}} = − (\frac{1}{2}) \frac{d^{2}}{d x^{2}} \frac{1}{1 - x} = − \sum_{n = 0}^{\infty} (\frac{(n + 2) (n + 1) x^{n}}{2})

F (x) = \int_{0}^{x} cos (\sqrt{t}) d t; f (t) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{n}}{(2 n)!}

at $a = 0$

(Note: $f$

is the Taylor series of $cos (\sqrt{t}) .)$

In the following exercises, compute the Taylor series of each function around $x = 1 .$

f (x) = 2 - x

2 - x = 1 - (x - 1)

f (x) = x^{3}

f (x) = {(x - 2)}^{2}

{((x - 1) - 1)}^{2} = {(x - 1)}^{2} - 2 (x - 1) + 1

f (x) = ln x

f (x) = \frac{1}{x}

\frac{1}{1 - (1 - x)} = \sum_{n = 0}^{\infty} {(−1)}^{n} {(x - 1)}^{n}

f (x) = \frac{1}{2 x - x^{2}}

f (x) = \frac{x}{4 x - 2 x^{2} - 1}

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

f (x) = e^{− x}

f (x) = e^{2 x}

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

[T] In the following exercises, identify the value of x such that the given series $\sum_{n = 0}^{\infty} a_{n}$

is the value of the Maclaurin series of $f (x)$

at $x .$

Approximate the value of $f (x)$

using $S_{10} = \sum_{n = 0}^{10} a_{n} .$

\sum_{n = 0}^{\infty} \frac{1}{n!}

\sum_{n = 0}^{\infty} \frac{2^{n}}{n!}

x = e^{2}; S_{10} = \frac{34,913}{4725} \approx 7.3889947

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} {(2 π)}^{2 n}}{(2 n)!}

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} {(2 π)}^{2 n + 1}}{(2 n + 1)!}

sin (2 π) = 0; S_{10} = 8.27 \times 10^{−5}

The following exercises make use of the functions $S_{5} (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120}$

and $C_{4} (x) = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24}$

on $[− π, π] .$

[T] Plot ${sin}^{2} x - {(S_{5} (x))}^{2}$

on $[− π, π] .$

Compare the maximum difference with the square of the Taylor remainder estimate for $sin x .$

[T] Plot ${cos}^{2} x - {(C_{4} (x))}^{2}$

on $[− π, π] .$

Compare the maximum difference with the square of the Taylor remainder estimate for $cos x .$

This graph has a concave up curve that is symmetrical about the y axis. The lowest point of the graph is the origin with the rest of the curve above the x-axis.

The difference is small on the interior of the interval but approaches $1$

near the endpoints. The remainder estimate is $\| R_{4} \| = \frac{π^{5}}{120} \approx 2.552 .$

[T] Plot $\| 2 S_{5} (x) C_{4} (x) - sin (2 x) \|$

on $[− π, π] .$

[T] Compare $\frac{S_{5} (x)}{C_{4} (x)}$

on $[−1, 1]$

to $tan x .$

Compare this with the Taylor remainder estimate for the approximation of $tan x$

by $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} .$

This graph has two curves. The solid curve is very flat and close to the x-axis. It passes through the origin. The second curve, a broken line, is concave down and symmetrical about the y-axis. It is very close to the x-axis between -3 and 3.

The difference is on the order of $10^{−4}$

on $[−1, 1]$

while the Taylor approximation error is around $0.1$

near $\pm 1 .$

The top curve is a plot of ${tan}^{2} x - {(\frac{S_{5} (x)}{C_{4} (x)})}^{2}$

and the lower dashed plot shows $t^{2} - {(\frac{S_{5}}{C_{4}})}^{2} .$

[T] Plot $e^{x} - e_{4} (x)$

where $e_{4} (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24}$

on $[0, 2] .$

Compare the maximum error with the Taylor remainder estimate.

(Taylor approximations and root finding.) Recall that Newton’s method $x_{n + 1} = x_{n} - \frac{f (x_{n})}{f' (x_{n})}$

approximates solutions of $f (x) = 0$

near the input $x_{0} .$

If $f$
and
$g$
are inverse functions, explain why a solution of
$g (x) = a$
is the value
$f (a) of f .$
Let $p_{N} (x)$
be the
$N th$
degree Maclaurin polynomial of
$e^{x} .$
Use Newton’s method to approximate solutions of
$p_{N} (x) - 2 = 0$
for
$N = 4, 5, 6 .$
Explain why the approximate roots of $p_{N} (x) - 2 = 0$
are approximate values of
$ln (2) .$

a. Answers will vary. b. The following are the $x_{n}$

values after $10$

iterations of Newton’s method to approximation a root of $p_{N} (x) - 2 = 0 :$

for $N = 4, x = 0.6939...;$

for $N = 5, x = 0.6932...;$

for $N = 6, x = 0.69315...; .$

(Note: $ln (2) = 0.69314 ...)$

c. Answers will vary.

In the following exercises, use the fact that if $q (x) = \sum_{n = 1}^{\infty} a_{n} {(x - c)}^{n}$

converges in an interval containing $c,$

then $lim_{x \to c} q (x) = a_{0}^{}$

to evaluate each limit using Taylor series.

lim_{x \to 0} \frac{cos x - 1}{x^{2}}

lim_{x \to 0} \frac{ln (1 - x^{2})}{x^{2}}

\frac{ln (1 - x^{2})}{x^{2}} \to − 1

lim_{x \to 0} \frac{e^{x^{2}} - x^{2} - 1}{x^{4}}

lim_{x \to 0^{+}} \frac{cos (\sqrt{x}) - 1}{2 x}

\frac{cos (\sqrt{x}) - 1}{2 x} \approx \frac{(1 - \frac{x}{2} + \frac{x^{2}}{4!} - ⋯) - 1}{2 x} \to - \frac{1}{4}

Taylor and Maclaurin Series

Overview of Taylor/Maclaurin Series

Taylor Polynomials

Taylor’s Theorem with Remainder

Proof

Representing Functions with Taylor and Maclaurin Series

Key Concepts

Key Equations

Glossary