10.4: Power Series and Taylor’s Theorem

A power series is like an infinite polynomial. It has the form

∞∑n=0

an(x− c)n = a0 + a1(x− c)+ a2(x− c)2 + ...+ an(x− c)n + ...

here c is any real number and a series of this form is called a powerseries centered at c. Note that c = 0 is ok and then the powerseries will look like a0 + a1x + a2x

2 + a3x3 + .... Note that a power

series is just like any other series except that it depends on x.

Radius of Convergence of a power series

Let

f(x) =∞∑

n=0

an(x− c)n

be the function defined by this power series. Note that f(x) is onlydefined if the power series converges so we will consider thedomain of the function f to be the set of x values for which theseries converges. There are three possible cases:

1 The power series converges at x = c (note f(c) = a0)

2 The power series converges for all x, i.e. (−∞,∞)3 There is a number R called the Radius of convergence such

that the series converges for all c−R < x < c + R and theseries diverges outside this interval.

Find the Radius of Convergence of∞∑

n=0

xn

n!:

First note that this is a power series centered at c = 0, and thecoefficients an = 1

n! . We will use the ratio test to find the radius ofconvergence. We look at the terms of the series

limn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = limn→∞

∣∣∣∣xn+1/(n + 1)!xn/n!

∣∣∣∣= lim

n→∞

∣∣∣∣ x

n + 1

∣∣∣∣= 0.

Since the ratio test implies that this converges and final answer i.e.0 does not depend on x we see that this series will converge for allx meaning that the Radius of Convergence is infinite.

Find the Radius of Convergence of∞∑

n=0

(−1)n(x− 2)n

3n:

First note that this is a power series centered at c = 2 and that thecoefficients an = (−1)n

3n . We use the ratio test and have

limn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = limn→∞

∣∣∣∣(−1)n+1(x− 2)n+1/3n+1

(−1)n(x− 2)n/3n

∣∣∣∣= lim

n→∞

∣∣∣∣(−1)(x− 2)3

∣∣∣∣= lim

n→∞

∣∣∣∣x− 23

∣∣∣∣ .Using the ratio test we know that this will converge if |x− 2|/3 < 1or if |x− 2| < 3. This gives the Radius of Convergence is R = 3and the interval is (−1, 5) since the center is c = 2.

Taylor and Maclaurin series

If you start with a function f(x) and want to find the power seriesrepresentation for it, there is a nice formula, called the TaylorSeries. (If c = 0 it is called the Maclaurin Series). If f(x) isrepresented by a power series centered at c, then

f(x) =∞∑

n=0

f (n)(c)n!

(x− c)n

This can be written out the long way as

f(x) = f(c) + f ′(c)(x− c) +f ′′(c)

2(x− c)2 +

f ′′′(c)3!

(x− c)3 + ...

Finding a Maclaurin Series

Find the power series for f(x) = ex centered at x = 0. This is anice and easy one since all the derivatives of ex are also ex. Ingeneral you start by writing the derivatives out at x = c andfinding a pattern.

f(x) = ex f(0) = 1f ′(x) = ex f ′(0) = 1f ′′(x) = ex f ′′(0) = 1

...

So in general f (n)(0) = 1 so

ex = 1 + x +x2

2!+

x3

3!+ ... =

∞∑n=0

xn

n!.

We saw from earlier that this series converges for all values of x,because the radius of convergence was infinite.

Finding a Taylor Series

Find the power series for f(x) = lnx centered at x = 1.

f(x) = ln x f(1) = ln 1 = 0

f ′(x) =1x

f ′(1) = 1/1 = 1

f ′′(x) = −x−2 f ′′(1) = −1

f (3)(x) = 2x−3 f (3)(1) = 2

f (4)(x) = 2(−3)x−4 f (4)(1) = −6

f (5)(x) = 2(−3)(−4)x−5 f (4)(1) = 4 · 3 · 2...

Continued

In general we see that

f (n)(1)=(−1)nn!

so the Taylor Series

ln x = f(1) + f ′(1)(x− 1) +f ′′(1)

2(x− 1)2 +

f ′′′(1)3!

(x− 1)3 + ...

= 0 + (x− 1)− 12(x− 1)2 +

13(x− 1)3 − 1

4(x− 1)4 + ..

We can write this out as

ln x =∞∑

n=1

(−1)n+1

n(x− 1)n.

Using known power series to help find new ones...

Find the Maclaurin series for e−x, and ex2.

We know that

ex =∞∑

n=0

xn

n!= 1 + x +

x2

2+

x3

3!+ ...

so we can plug in −x for x to get the series for e−x which gives

e−x = 1 + (−x) +(−x)2

2+

(−x)3

3!+ ...

so the negative sign will cancel on even terms leaving

e−x = 1− x +x2

2− x3

3!+

x4

4!+ ... =

∞∑n=0

(−1)n

n!xn.

Using known power series to help find new ones...

Find the Maclaurin series for e−x, and ex2.

We know that

ex =∞∑

n=0

xn

n!= 1 + x +

x2

2+

x3

3!+ ...

so we can plug in x2 for x to get the series for ex2which gives

ex2= 1 + (x2) +

(x2)2

2+

(x2)3

3!+ ...

so the square will return only even terms giving

ex = 1 + x2 +x4

2+

x6

3!+

x8

4!+ ... =

∞∑n=0

1n!

x2n.

Basic List of Power Series

(I will give you this or something like it on the test)

ex = 1 + x + x2

2 + x3

3! + ... + xn

n! + ..., −∞ < x <∞

ln x = (x− 1)− (x−1)2

2 + ... + (−1)n(x−1)n

n + ... 0 < x ≤ 2

1x = 1− (x− 1) + (x− 1)2 + ... + (−1)n(x− 1)n + ... 0 < x < 2

1x+1 = 1− x + x2 − x3 + ...(−1)nxn... −1 < x < 1

(1 + x)k = 1 + kx + k(k−1)x2

2! + k(k−1)(k−2)x3

3! + ... −1 < x < 1

Using the basic list:

Find the power series for f(x) =√

1 + x centered at c = 0. Wecan use the binomial series (1 + x)k with k = 1/2 :

(1 + x)k = 1 + kx +k(k − 1)x2

2!+

k(k − 1)(k − 2)x3

3!+ ...

and so for k = 1/2 we have

(1+x)1/2 = 1+x

2+(1/2)(1/2−1)

x2

2!+(1/2)(1/2−1)(1/2−2)

x3

3!+...

Which gives

√1 + x = 1 +

x

2− x2

8+

3x3

48− ...

Using the basic list:

Find the power series for 2xx+1 We can use the basic list to tell us

the power series for 11+x and multiply each term by 2x :

1x + 1

= 1− x + x2 − x3 + ...(−1)nxn + ...

So2x

x + 1= 2x

[1− x + x2 − x3 + ...(−1)nxn + ...

]2x

x + 1= 2x− 2x2 + 2x3 − 2x4 + ...(−1)n2xn+1 + ...

or2x

x + 1=∞∑

n=0

(−1)n2xn+1