Sereis Solutions

1. Power series
A power series is a series of the form
X∞
an xn . (1.1) pow1
n=0
We say that a power series is convergent at the point x = x0 , if
∞
anxn0 is convergent.
P
the number series
n=0
pow1
Theorem 1.1. If the power series of the form (1.1) is convergent
at some point x0 6= 0 then this series converges absolutely for
|x| < r0 = |x0| .
∞
anxn0 is convergent the sequence
P
Proof. Since the number series
n=0
{anxn0 }tends to zero as n → ∞. Thus there exist a number M > 0
such that
|anxn0 | ≤ M0, n = 1, 2, ...
Therefore we have
n x x n
n
n
|anx | = |anx0 | ≤ M0 .
x0 x0
For |x| < |x0| we have
∞
X x n 1
= .
n=0
x 0 x
1 − x0
∞
|anxn|
P
Therefore thanks to the comparison theorem the series
n=0
is convergent. 
1
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pow1
Theorem 1.2. If a power series (1.1) converges for
pow1
some x0 6=
0, then there exists a number R > 0 such that (1.1) converges
absolutely
pow1
on the inerval (−R, R) and diverges for |x| > R. If
(1.1) converges just for x = 0, then r = 0.
Theorem 1.3. If
an + 1
lim =L
n→∞ an
or p
lim n |an| = L
n→∞
then the radius of convergence of the power series
X∞
an xn
n=0

is the number r = L1 , with r = ∞ if L = 0 and r = 0 if L = ∞.

Theorem 1.4. If
∞
X
anxn = 0, ∀x
n=0
then an = 0, n = 0, 1, 2, ..
If
∞
X ∞
X
n
f (x) = anx , g(x) = bn x n
n=1 n=1
then
∞
X ∞
X
f (x) + g(x) = (an + bn)xn, f (x)g(x) = cn xn , (1.2) pws2
n=1 n=0
Pn
where cn pws2
= k=0 ak bn−k . The interval of convergence of both
series in (1.2) is the common interval of convergence of power series
for f (x) and g(x).
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Theorem 1.5. If
∞
X
f (x) = anxn (1.3) pws1
n=1
∞
nanxn−1 obtained by
P
is convergent on (−r, r), then the series
n=1
termwise differentiation of the series has the radius of convergence
r and
X∞
0
f (x) = nanxn−1.
n=1
TheRseries obtained by termwise integration gives the power series
x
for 0 f (x)ds:
Z x ∞
X 1
f (x)ds = anxn+1, |x| < r
0 n=0
n+1
Exercise 1.6. Show that
X∞ ∞
X
n−1
nan−1x + bnxn+1 = 0
n=1 n=2
implies that
a0 = a1 = a2 = 0 and an = −bn−1/(n + 1), n≥3

Definition 1.7. A function f (x) is said to be analytic at a

point x = x0 if f (x) is sum of the power series
∞
X
an(x − x0)n
n=0
that has a positive radius of convergence.
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2. Power series solutions of ODE’s

A point x0 is called an ordinary point for
y 00 + p(x)y 0 + q(x)y = 0, (2.1) ordp1

if p(x) andordp1
q(x) are analytic at x0. Otherwise x0 is called a singular
point for (2.1).

Definition 2.1. A singular point x0 is called a regular singular

point if both (x − x0)p(x) and (x − x0)2q(x) are analytic at x0.
Otherwise it is called an irregular singular point.

Exercise 2.2. Find the power series solution of the equation

y 00 + xy 0 + (x2 + 2)y = 0.
Solution.
∞
X ∞
X
n 0
y(x) = anx , y (x) = nanxn−1,
n=0 n=1

∞
X ∞
X
y 00(x) = n(n − 1)anxn−2 = (n + 2)(n + 1)an+2xn.
n=2 n=0
Inserting the expressions for y, y 0, y into equation we get
00

X∞ X∞ X∞ ∞
X
(n+2)(n+1)an+2xn + nanxn +2 an xn + anxn+2 = 0
n=0 n=1 n=0 n=0
P∞ n+2
P∞ n
Since n=0 an x = n=2 an−2 x we have
∞
X
[(n+2)(n+1)an+2+(n+2)an+an−2]xn+(6a3+3a1)x+2a2+2a0 = 0
n=2
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From the last equality we deduce that

3a1 + 6a3 = 0; 2a2 + 2a0 = 0 ⇒ a2 = −a0,
1
a3 = − a1,
2
(n + 2)(n + 1)an+2 + (n + 2)an + an−2 = 0,
1 1
an+2 = − an − an−2
(n + 1) (n + 1)(n + 2)
1 1 1 1 1
a4 = − a2 − a0 = a0 − a0 = a0
3 3·4 3 12 4
1 1 1 1 3
a5 = − a3 − a1 = a1 − a1 = a1, ...
4 4·5 8 20 40
1 1 3
y(x) = a0[1 − x2 + x4 + ...] + a1[x − x3 + x5 + ...]
4 2 40
Exercise 2.3. Find the power series solution of the Airy equation
y 00 − xy = 0

∞
X
y= an xn ,
n=0
∞
X ∞
X
y 00 = n(n − 1)anxn−2 = (n + 2)(n + 1)an+2xn
n=2 n=0
∞
X ∞
X
y0 = nanxn−1 = ( (n + 1)an+1xn)
n=1 n=0
∞
X ∞
X
[(n + 2)(n + 1)an+2]xn − anxn+1 = 0
n=0 n=0
X∞ ∞
X
(n + 2)(n + 1)an+2xn − an−1xn = 0
n=0 n=1
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⇒ 2a2 = 0 ⇒ a2 = 0
1
an+2 = an−1
(n + 2)(n + 1)
1
a3 = a0
2·3
1
a4 = a1
4·3
a5 = 0 n = 3
1 1
a6 = a3 = a0 , n = 4
6·5 2·3·5·6
1 1
a7 = a4 = a1, ... n = 5
7·6 7·6·4·3
a8 = 0, n = 6
1 1
a9 = a6 = a0 , n = 7
9·8 2·3·5·6·8·9
1 1
a10 = a7 = a1, n = 8,
9 · 10 3 · 4 · 6 · 7 · 9 · 10
a11 = 0,
a3n−1 = 0,
a0
a3n = ,
2 · 3 · 5 · 6...(3n − 1)3n
a1
a3n+1 = , n ≥ 4.
3 · 4 · 6 · 7 · · · (3n)(3n + 1)
Thus
x3 x6 x3n
 
y(x) = a0 1 + + + + ··· +
2 · 3 2 · 3 · 5 · 6 2 · 3 · 5 · 6 · · · (3n − 1)3n
x4 x7 x3n+1
 
a1 1 + + + + ··· .
3 · 4 3 · 4 · 6 · 7 3 · 4 · 6 · 7 · · · (3n)(3n + 1)