TBIL-Cal Taylor Series (PS4)

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Section 9.4 Taylor Series (PS4)

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Learning Outcomes

Determine a Taylor or Maclaurin series for a function.

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Subsection 9.4.1 Activities

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Activity 9.4.1.

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The following tasks will help us find a mechanism to produce a power series given information about its derivatives.

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(a)

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Find the 2nd derivative of

x^{2} .

$2 x$
$2$
$4 x$
$4$

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(b)

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Find the 3rd derivative of

x^{3} .

$2$
$3 x$
$6$
$12 x$

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(c)

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Find the 4th derivative of

x^{4} .

$18$
$24$
$32$
$64$

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(d)

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Based on these results, which of the following should always equal the

n

th derivative of

x^{n}

with respect to

x ?

$n$
$n^{2}$
$n!$
$n^{n}$

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Activity 9.4.2.

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Let’s use derivatives to rediscover the sequence

a_{n}

which gives a power series representation for

e^{x} .

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(a)

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Let’s say that

e^{x} = \sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} \dots .

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What must

a_{0}

be to satisfy

e^{0} = 1 ?

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(b)

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Then,

\frac{d}{d x} [e^{x}] = e^{x} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} \dots .

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What must

a_{1}

be to also satisfy

e^{0} = 1 ?

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(c)

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Then,

\frac{d^{2}}{d x^{2}} [e^{x}] = e^{x} = 2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots .

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What must

a_{2}

be to also satisfy

e^{0} = 1 ?

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(d)

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Then,

\frac{d^{3}}{d x^{3}} [e^{x}] = e^{x} = 6 a_{3} + 24 a_{4} x + \dots .

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What must

a_{3}

be to also satisfy

e^{0} = 1 ?

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(e)

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So this

6 a_{3}

term was obtained from the fact that the

3

rd derivative of

x^{3}

3! = 6 .

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So finally, we may skip ahead to the

n

th derivative:

\frac{d^{n}}{d x^{n}} [e^{x}] = e^{x} = n! \cdot a_{n} + (n + 1)! \cdot a_{n + 1} \cdot x + \dots .

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What must

a_{n}

be to also satisfy

e^{0} = 1 ?

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(f)

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This reveals the power series we previously found for

e^{x} :

e^{x} = \sum_{n = 0}^{\infty} a_{n} x^{n} = \sum_{n = 0}^{\infty} \frac{1}{n!} x^{n} .

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So in general, if

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots,

then

\frac{d^{n}}{d x^{n}} [f (x)] = f^{(n)} (x) = n! \cdot a_{n} + (n + 1)! \cdot a_{n + 1} \cdot x + \dots .

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What must

a_{n}

be to produce the correct value for

f^{(n)} (0) ?

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Fact 9.4.3.

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f (x)

can be written as a power series, then there is a real number

c

such that

\begin{aligned} f (x) & = \sum_{n = 0}^{\infty} \frac{f^{(n)} (c)}{n!} (x - c)^{n} \\ = f (c) + f^{'} (c) (x - c) + \frac{f^{''} (c)}{2!} (x - c)^{2} + \frac{f^{(3)} (c)}{3!} (x - c)^{3} + \dots \end{aligned}

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on some interval centered at

x = c .

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In fact, the functions that can be represented as power series are exactly those functions which are infinitely differentiable on some open interval.

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Definition 9.4.4.

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The Taylor series generated by

f (x)

and centered at

x = c

is given by

\begin{aligned} f (x) & = \sum_{n = 0}^{\infty} \frac{f^{(n)} (c)}{n!} (x - c)^{n} \\ = f (c) + f^{'} (c) (x - c) + \frac{f^{''} (c)}{2!} (x - c)^{2} + \frac{f^{(3)} (c)}{3!} (x - c)^{3} + \dots \end{aligned}

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with an interval of convergence determinable by series convergence rules.

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When

c = 0,

\begin{aligned} f (x) & = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \\ = f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \frac{f^{(3)} (0)}{3!} x^{3} + \dots \end{aligned}

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is called the Maclaurin series generated by

f .

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Activity 9.4.5.

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Observe that

f (x) = \sin (x)

is a function such that:

\begin{array}{cccccccc} f (0) & f^{'} (0) & f^{″} (0) & f^{(3)} (0) & f^{(4)} (0) & f^{(5)} (0) & f^{(6)} (0) & f^{(7)} (0) \\ \sin (0) & \cos (0) & - \sin (0) & - \cos (0) & \sin (0) & \cos (0) & - \sin (0) & - \cos (0) \\ 0 & 1 & 0 & - 1 & 0 & 1 & 0 & - 1 \end{array}

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(a)

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Given the zeros appearing for every even derivative above, which of these is a valid simplification of the Maclarin series

\sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

for

\sin (x) ?

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

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(b)

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Now consider the following consolidated chart:

\begin{array}{cccccccc} f^{(1)} (0) & f^{(3)} (0) & f^{(5)} (0) & f^{(7)} (0) \\ \cos (0) & - \cos (0) & \cos (0) & - \cos (0) \\ 1 & - 1 & 1 & - 1 \end{array}

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Which formula yields these alternating

1

s and

- 1

s appearing for

f^{(2 n + 1)} (0) ?

$f^{(2 n + 1)} (0) = (- 1)^{n}$
$f^{(2 n + 1)} (0) = (- 1)^{n + 1}$
$f^{(2 n + 1)} (0) = (- 1)^{2 n}$
$f^{(2 n + 1)} (0) = (- 1)^{2 n + 1}$

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Fact 9.4.6.

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The power series we’ve introduced for each of the following functions are in fact their Maclaurin series (Taylor series centered at

0

\begin{aligned} \frac{1}{1 - x} = \sum_{n = 0}^{\infty} \frac{n!}{n!} x^{n} & = 1 + x + x^{2} + x^{3} + \dots \\ e^{x} = \sum_{n = 0}^{\infty} \frac{1}{n!} x^{n} & = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots \\ \cos (x) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n)!} x^{2 n} & = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \frac{x^{6}}{720} + \dots \\ \sin (x) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n + 1)!} x^{2 n + 1} & = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \frac{x^{7}}{5040} + \dots \end{aligned}

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Definition 9.4.7.

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For a function

f (x)

with a Taylor series centered at

x = c,

\begin{aligned} f (x) & \approx T_{k} (x) \\ = \sum_{n = 0}^{k} \frac{f^{(n)} (c)}{n!} (x - c)^{n} \\ = f (c) + f^{'} (c) (x - c) + \frac{f^{''} (c)}{2!} (x - c)^{2} + \dots + \frac{f^{(k)} (c)}{k!} (x - c)^{k} \end{aligned}

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where

T_{k} (x)

is called the

k^{t h}

degree Taylor polynomial generated by

f

and centered at

x = c .

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The

k^{t h}

degree Taylor polynomial can be seen as the “best” polynomial of degree

k

or less for approximating

f (x)

for values close to

x = c .

Note that the

1^{s t}

degree Taylor polynomial is also known as the linearization of

f .

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Activity 9.4.8.

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Let

f (x)

be a function such that:

\begin{array}{ccccccc} f (4) & f^{'} (4) & f^{″} (4) & f^{‴} (4) & f^{(4)} (4) & f^{(5)} (4) & f^{(6)} (4) \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 \end{array}

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(a)

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Find a Taylor polynomial for

f (x)

centered at

x = 4

of degree

3 .

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(b)

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Using the table above, find a general closed form for

f^{(n)} (4) .

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(c)

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Use (b) to find a Taylor series for

f (x)

centered at

x = 4 .

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Activity 9.4.9.

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Let

f (x)

be a function such that:

\begin{array}{ccccccc} f (- 2) & f^{'} (- 2) & f^{″} (- 2) & f^{‴} (- 2) & f^{(4)} (- 2) & f^{(5)} (- 2) & f^{(6)} (- 2) \\ 0 & 2 & - 16 & 54 & - 128 & 250 & - 432 \end{array}

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(a)

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Find a Taylor polynomial for

f (x)

centered at

x = - 2

of degree

3 .

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(b)

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Using the table above, find a general closed form for

f^{(n)} (- 2) .

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(c)

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Use (b) to find a Taylor series for

f (x)

centered at

x = - 2 .

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Remark 9.4.10.

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You might have seen

\sqrt{- 1}

written as

i,

and know that

z

is a complex number if

z = a + b i

for some real numbers

a

and

b .

Note that

i^{2} = - 1,

i^{3} = (i^{2}) i = - i,

i^{4} = (i^{2})^{2} = 1,

i^{5} = (i^{4}) i = i,

and so on. This gives rise to the following notion.

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Definition 9.4.11. Euler’s Identity.

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For any real number

θ,

\begin{aligned} e^{i θ} & = 1 + \frac{i θ}{1!} + \frac{(i θ)^{2}}{2!} + \frac{(i θ)^{3}}{3!} + \frac{(i θ)^{4}}{4!} + \frac{(i θ)^{5}}{5!} + \frac{(i θ)^{6}}{6!} + \frac{(i θ)^{7}}{7!} + \frac{(i θ)^{8}}{8!} + \dots \\ = 1 + i θ - \frac{θ^{2}}{2!} - \frac{i θ^{3}}{3!} + \frac{θ^{4}}{4!} + \frac{i θ^{5}}{5!} - \frac{θ^{6}}{6!} - \frac{i θ^{7}}{7!} + \frac{θ^{8}}{8!} + \dots \\ = (1 - \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} - \frac{θ^{6}}{6!} + \dots) + i (θ - \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} - \frac{θ^{7}}{7!} + \dots) \\ = \cos (θ) + i \sin (θ) . \end{aligned}

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Activity 9.4.12.

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Use Euler’s identity to evaluate

e^{i π} .

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Subsection 9.4.2 Videos

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Figure 196. Video: Determine a Taylor or Maclaurin series for a function

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Subsection 9.4.3 Exercises

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Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/PS4/

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