1
MATH 348 - A Review of Series and Homework II
When you come to a fork in the road, take it. Yogi Berra.
Assume α is a constant in the differential equation
y 0 (x) − αy(x) = 0, y(0) = a0 .
(1)
The equation implies that the derivative y 0 (x) has the same form as the function y(x).
Polynomials (“very long polynomials”) have this property so if y(x) =
∞
X
an xn then
n=0
∞
∞
∞
X
X
dy
d X
y (x) =
=
an x n =
nan xn−1 =
(n + 1)an+1 xn .
dx
dx n=0
n=1
n=0
!
0
(2)
Substitution of y(x) and y 0 (x) into (1) gives
∞
∞
∞
X
X
d X
[(n + 1)an+1 − αan ] xn .
an x n =
an x n − α
0=
dx n=0
n=0
n=0
!
A series is equal to zero if all of the coefficients are zero
[(n + 1)an+1 − αan ] = 0 or an+1 =
αan
for n ≥ 0.
(n + 1)
a0 α n
. The solution of the
This recurrence relation can be written non-recursively, an =
n!
differential equation is
y(x) =
∞
X
an x n =
n=0
∞
X
∞
X
a0 α n n
1
x = a0
(αx)n = a0 eαx ,
n!
n!
n=0
n=0
(3)
since y(x) = a0 eαx is a solution of the differential equation (1) and the uniqueness of solutions
(Theorem 6 with n = 1, page 512 of your text). If α = a0 = 1 in (3) then
y(x) =
∞
X
n=0
an x n =
∞
X
xn
= ex
n!
n=0
|x| < ∞.
(4)
The expansion in (4) is called the Maclaurin expansion of the function f (x) = ex . Another
Maclaurin expansion is obtained from the long division (and induction)
1
xn+1
−
, x 6= 1
1−x 1−x
1
=
, |x| < 1.
1−x
1 + x + x2 + · · · x n =
which for |x| < 1, as n → ∞ gives the series
∞
X
n=0
xn
(5)
2
Each of (4) and (5) are examples of Taylor series. Here is a brief review. A function f (x) is
called analytic at x0 if
∞
X
f n (x0 )
f (x) =
(x − x0 )n ,
n!
n=0
|x − x0 | < R.
This is called the Taylor series of f at x0 . Note that if t = x − x0 then
f (x) = F (x − x0 ) = F (t) =
∞
X
F n (0) n
t
n!
n=0
since F n (x − x0 )|x=x0 = F n (t)|t=0 = F n (0). So there is no loss of generality in setting x0 = 0.
The Maclaurin expansion of a general function f is
f (x) =
∞
X
f n (0) n
x ,
n=0 n!
|x| < R.
(6)
The number R is called the radius of convergence and can be found from the ratio test which
asserts that if
n+1 a
an+1 n+1 x
= |x|L < 1
lim =
lim
|x|
n→∞ an xn n→∞ an then the series
∞
X
ak xn converges for all |x| <
n=0
1
≡ R. Apply this criteria to the series in (4)
L
and (5) to find R = ∞ and 1 respectively.
In a certain real sense the series in (4) and (5) give rise to all of the series for the elementary
functions. This sense is provided by the following two theorems. The first is a substitution
theorem.
Theorem 1.1 Assume that f (x) =
∞
X
an xn converges in |x| < R and g(x) = x2 then the
n=0
composition function has the series expansion f (g(x)) =
∞
X
an x2n , |x| <
√
R.
n=0
The second is a Calculus theorem.
Theorem 1.2 Assume that f (x) =
∞
X
an xn converges in |x| < R then
n=0
f 0 (x) =
∞
X
nan xn−1
Z x
and
0
n=1
f (t)dt =
∞
X
an n+1
x
n=0 n + 1
for |x| < R.
2
To find the Maclaurin series for f (x) = ex is cumbersome using (6). The derivative
evaluations f n (0) needed for (6) get complicated quickly (try a few). Instead, combine
Theorem 1.1 with (4) to find
∞
X
x2n
x2
e =
.
n=0 n!
(7)
3
In a similar fashion replace x by ±ix in (4) where i2 = −1 and rearrange Euler’s identity
eix + e−ix
e±ix = cos(x) ± i sin(x) in the form cos(x) =
to find
2
∞
∞
X
X
eix + e−ix
(−1)n x2n
eix − e−ix
(−1)n x2n+1
cos(x) =
=
and sin(x) =
=
2
(2n)!
2
n=0
n=0 (2n + 1)!
(8)
Alternatively, one could find the expansion of sin(x) using Theorem 1.2 and the expansion of
cos(x). Replace x by ix in (7) to find the expansions of cosh(x) and sinh(x). Lastly, use
Theorem 1.2 with (5) to find
∞
X
xj+1
, |x| < 1.
− ln(1 − x) =
j=0 j + 1
(9)
Homework II
1. In most every Calculus book on the planet one finds the numerical identity
∞
X
∞
X
(−1)k+1
1 1
(−1)N +1
(−1)k+1
ln(2) =
= 1 − + + ··· +
· · · = SN +
k
2 3
N
k
k=1
k=N +1
1
. Obtain the displayed equation from (9) (why
N +1
can you set x = −1?) How many terms of the series is required to find ln(2) with an
error bounded by 10−4 (a modest request in this age of double digit accuracy.
and the estimate | ln(2) − SN | <
1+x
2. Find the series expansion for ln
, |x| < 1. Hint: Use (9), replace x with −x, and
1−x
then use an appropriate law of logarithms.
3. For the differential equation y 00 (x) + xy 0 (x) + y(x) = 0
(a) Find the two (independent) series solutions y1 (x) and y2 (x).
(b) Show that one of these solutions, say y1 (x) is y(x) = e−x
2 /2
.
(c) Now use the reduction of order formula (page 525) to find a second independent
solution. Reconcile this result with the other solution found in part (a).
4. Find two Frobenius solutions to 2x2 y 00 (x) + x(2x + 1)y 0 (x) − y(x) = 0.
5. Do #29 on page 552. Which case of Theorem 2 (page 544) does this represent?
6. Bessel’s differential equation of order p reads x2 y 00 (x) + xy 0 (x) + (x2 − p2 )y(x) = 0 .
(a) If p = 0 show that r1 = r2 = 0. Find one Frobenius solution and use reduction of
order to find (three terms) of a second independent solution.
(b) If p = 1/2 show that r1 − r2 = 1 and there are two independent Frobenius solutions.
(c) If p = 1 show that r1 − r2 = 2 and there is only one Frobenius solutions.