10.4: Power Series and Taylor’s Theorem
A power series is like an infinite polynomial. It has the form
∞
X
an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + ... + an (x − c)n + ...
n=0
here c is any real number and a series of this form is called a power
series centered at c. Note that c = 0 is ok and then the power
series will look like a0 + a1 x + a2 x2 + a3 x3 + .... 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) =
∞
X
an (x − c)n
n=0
be the function defined by this power series. Note that f (x) is only
defined if the power series converges so we will consider the
domain of the function f to be the set of x values for which the
series 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 the
series diverges outside this interval.
Find the Radius of Convergence of
∞
X
xn
n=0
n!
:
First note that this is a power series centered at c = 0, and the
1
coefficients an = n!
. We will use the ratio test to find the radius of
convergence. We look at the terms of the series
n+1
an+1 xn+1 x
/(n
+
1)!
= lim lim n→∞
n→∞
an xn
xn /n!
x = lim n→∞ 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 all
x meaning that the Radius of Convergence is infinite.
Find the Radius of Convergence of
∞
X
(−1)n (x − 2)n
n=0
3n
:
First note that this is na power series centered at c = 2 and that the
coefficients an = (−1)
3n . We use the ratio test and have
(−1)n+1 (x − 2)n+1 /3n+1 an+1 xn+1 = lim lim n→∞ an xn n→∞ (−1)n (x − 2)n /3n
(−1)(x − 2) = lim n→∞
3
x − 2
.
= lim n→∞
3 Using the ratio test we know that this will converge if |x − 2|/3 < 1
or if |x − 2| < 3. This gives the Radius of Convergence is R = 3
and 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 series
representation for it, there is a nice formula, called the Taylor
Series. (If c = 0 it is called the Maclaurin Series). If f (x) is
represented by a power series centered at c, then
f (x) =
∞
X
f (n) (c)
n=0
n!
(x − c)n
This can be written out the long way as
f (x) = f (c) + f 0 (c)(x − c) +
f 00 (c)
f 000 (c)
(x − c)2 +
(x − c)3 + ...
2
3!
Finding a Maclaurin Series
Find the power series for f (x) = ex centered at x = 0. This is a
nice and easy one since all the derivatives of ex are also ex . In
general you start by writing the derivatives out at x = c and
finding a pattern.
f (x) = ex
0
f (0) = 1
x
f 0 (0) = 1
f 00 (x) = ex
f 00 (0) = 1
f (x) = e
...
So in general f (n) (0) = 1 so
∞
ex = 1 + x +
X xn
x2 x3
+
+ ... =
.
2!
3!
n!
n=0
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) = ln x centered at x = 1.
f (x) = ln x
f (1) = ln 1 = 0
1
f 0 (x) =
f 0 (1) = 1/1 = 1
x
f 00 (x) = −x−2
f 00 (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
n
f (n)(1)=(−1) n!
so the Taylor Series
f 000 (1)
f 00 (1)
(x − 1)2 +
(x − 1)3 + ...
2
3!
1
1
1
= 0 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 + ..
2
3
4
ln x = f (1) + f 0 (1)(x − 1) +
We can write this out as
ln x =
∞
X
(−1)n+1
n=1
n
(x − 1)n .
Using known power series to help find new ones...
2
Find the Maclaurin series for e−x , and ex .
We know that
x
e =
∞
X
xn
n=0
n!
=1+x+
x2 x3
+
+ ...
2
3!
so we can plug in −x for x to get the series for e−x which gives
e−x = 1 + (−x) +
(−x)2 (−x)3
+
+ ...
2
3!
so the negative sign will cancel on even terms leaving
∞
−x
e
X (−1)n
x2 x3 x4
−
+
+ ... =
xn .
=1−x+
2
3!
4!
n!
n=0
Using known power series to help find new ones...
2
Find the Maclaurin series for e−x , and ex .
We know that
ex =
∞
X
xn
n=0
n!
=1+x+
x2 x3
+
+ ...
2
3!
2
so we can plug in x2 for x to get the series for ex which gives
2
ex = 1 + (x2 ) +
(x2 )2 (x2 )3
+
+ ...
2
3!
so the square will return only even terms giving
∞
X 1
x4 x6 x8
e =1+x +
+
+
+ ... =
x2n .
2
3!
4!
n!
x
2
n=0
Basic List of Power Series
(I will give you this or something like it on the test)
ex = 1 + x +
x2
2
ln x = (x − 1) −
1
x
+
x3
3!
+ ... +
(x−1)2
2
xn
n!
+ ... +
−∞ < x < ∞
+ ...,
(−1)n (x−1)n
n
+ ...
= 1 − (x − 1) + (x − 1)2 + ... + (−1)n (x − 1)n + ...
1
x+1
= 1 − x + x2 − x3 + ...(−1)n xn ...
(1 + x)k = 1 + kx +
k(k−1)x2
2!
+
k(k−1)(k−2)x3
3!
0<x≤2
0<x<2
−1 < x < 1
+ ...
−1 < x < 1
Using the basic list:
√
Find the power series for f (x) = 1 + x centered at c = 0. We
can use the binomial series (1 + x)k with k = 1/2 :
(1 + x)k = 1 + kx +
k(k − 1)x2 k(k − 1)(k − 2)x3
+
+ ...
2!
3!
and so for k = 1/2 we have
x
x2
x3
(1+x)1/2 = 1+ +(1/2)(1/2−1) +(1/2)(1/2−1)(1/2−2) +...
2
2!
3!
Which gives
√
1+x=1+
x x2 3x3
−
+
− ...
2
8
48
Using the basic list:
2x
Find the power series for x+1
We can use the basic list to tell us
1
the power series for 1+x and multiply each term by 2x :
1
= 1 − x + x2 − x3 + ...(−1)n xn + ...
x+1
So
or
2x
= 2x 1 − x + x2 − x3 + ...(−1)n xn + ...
x+1
2x
= 2x − 2x2 + 2x3 − 2x4 + ...(−1)n 2xn+1 + ...
x+1
∞
X
2x
=
(−1)n 2xn+1
x+1
n=0