Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

Example 23.2.4 Consider a series expansion about the origin of the equation
w

+
1 − z
z
w

−
1
z
2
w = 0.
The indicial equation is
α
2
− 1 = 0
α = ±1.
Substituting a Frobenius series into the differential equation,
z
2
∞

n=0
(n + α)(n + α − 1)a
n
z
n−2
+ (z −z
2

)
∞

n=0
(n + α)a
n
z
n−1
−
∞

n=0
a
n
z
n
= 0
∞

n=0
(n + α)(n + α − 1)a
n
z
n
+
∞

n=0
(n + α)a
n

z
n
−
∞

n=1
(n + α − 1)a
n−1
z
n
−
∞

n=0
a
n
z
n
= 0

α(α − 1) + α − 1

a
0
+
∞

n=1

n + α)(n + α − 1)a

n
+ (n + α − 1)a
n
− (n + α − 1)a
n−1

z
n
= 0.
Equating powers of z to zero,
a
n
(α) =
a
n−1
(α)
n + α + 1
.
We know that the first solution has the form
w
1
= z
∞

n=0
a
n
z
n
.

Setting α = 1 in the reccurence formula,
a
n
=
a
n−1
n + 2
=
2a
0
(n + 2)!
.
1214
Thus the first solution is
w
1
= z
∞

n=0
2a
0
(n + 2)!
z
n
= 2a
0
1
z
∞


n=0
z
n+2
(n + 2)!
=
2a
0
z

∞

n=0
z
n
n!
− 1 − z

=
2a
0
z
(
e
z
−1 − z).
Now to fin d the sec ond solution. Setting α = −1 in the reccurence formula,
a
n
=

a
n−1
n
=
a
0
n!
.
We see that in this case there is no trouble i n defining a
2
(α
2
). The second solution is
w
2
=
a
0
z
∞

n=0
z
n
n!
=
a
0
z
e

z
.
Thus we see that the general solution is
w =
c
1
z
(
e
z
−1 − z) +
c
2
z
e
z
w =
d
1
z
e
z
+d
2

1 +
1
z

.

1215
23.3 Irregular Singular Points
If a point z
0
of a differential equation is not ordinary or regular singular, then it is an irregular singular point. At least
one of the solutions at an irregular singular point will not be of the Frobenius form. We will examine how to obtain
series expansions about an irregular singular point in the chapter on asymptotic expansions.
23.4 The Point at Infinity
If we want to determine the behavior of a function f(z) at infinity, we can make the transformation ζ = 1/z and
examine the point ζ = 0.
Example 23.4.1 Consider the behavior of f(z) = sin z at infinity. This is the same as considering the point ζ = 0 of
sin(1/ζ), which has the series expansion
sin

1
ζ

=
∞

n=0
(−1)
n
(2n + 1)!ζ
2n+1
.
Thus we see that the point ζ = 0 is an esse ntial singularity of sin(1/ζ). Hence sin z has an essential singularity at
z = ∞.
Example 23.4.2 Consider the behavior at infinity of z
e

1/z
. We make the transformation ζ = 1/z.
1
ζ
e
ζ
=
1
ζ
∞

n=0
ζ
n
n!
Thus z
e
1/z
has a pole of order 1 at infi nity.
1216
In order to classify the point at infinity of a differential equation in w(z), we apply the transformation ζ = 1/z,
u(ζ) = w(z). We write the derivatives with respect to z in terms of ζ.
z =
1
ζ
dz = −
1
ζ
2
dζ

d
dz
= −ζ
2
d
dζ
d
2
dz
2
= −ζ
2
d
dζ

−ζ
2
d
dζ

= ζ
4
d
2
dζ
2
+ 2ζ
3
d
dζ

Now we apply the transformation to the differential equation.
w

+ p(z)w

+ q(z)w = 0
ζ
4
u

+ 2ζ
3
u

+ p(1/ζ)(−ζ
2
)u

+ q(1/ζ)u = 0
u

+

2
ζ
−
p(1/ζ)
ζ
2


u

+
q(1/ζ)
ζ
4
u = 0
Example 23.4.3 Classify the singular points of the differential equation
w

+
1
z
w

+ 2w = 0.
There is a regular singular point at z = 0. To examine the point at infinity we make the transformation ζ = 1/z,
u(ζ) = w(z).
u

+

2
ζ
−
1
ζ

u


+
2
ζ
4
u = 0
u

+
1
ζ
u

+
2
ζ
4
u = 0
1217
Thus we see that the differential equation for w(z) has an irregular singular point at infinity.
1218
23.5 Exercises
Exercise 23.1 (mathematica/ode/series/series.nb)
f(x) satisfies the Hermite eq uation
d
2
f
dx
2
− 2x
df

dx
+ 2λf = 0.
Construct two linearly independent solutions of the equation as Taylor series about x = 0. For what values of x do the
series converge?
Show that for certain values of λ, called eigenvalues, one of the solutions is a polynomial, called an eigenfunction.
Calculate the first four eigenfunctions H
0
(x), H
1
(x), H
2
(x), H
3
(x), ordered by degree.
Hint, Solution
Exercise 23.2
Consider the Legendre equation
(1 − x
2
)y

− 2xy

+ α(α + 1)y = 0.
1. Find two linearly independent solutions in the form of power series about x = 0.
2. Compute the radius of convergence of the series. Explain why it is possible to predict the radius of convergence
without actually deriving the series.
3. Show that if α = 2n, with n an integer and n ≥ 0, the series for one of the solutions reduces to an even
polynomial of degree 2n.
4. Show that if α = 2n + 1, with n an integer and n ≥ 0, the series for one of the solutions reduces to an odd

polynomial of degree 2n + 1.
5. Show that the first 4 polynomial solutions P
n
(x) (known as Legendre polynomials) ordered by their degree and
normalized so that P
n
(1) = 1 are
P
0
= 1 P
1
= x
P
2
=
1
2
(3x
2
− 1) P
4
=
1
2
(5x
3
− 3x)
1219
6. Show that the Legendre equation can also be written as
((1 − x

2
)y

)

= −α(α + 1)y.
Note that two Legendre polynomials P
n
(x) and P
m
(x) must satisfy this relation for α = n and α = m respectively.
By multiplying the first relation by P
m
(x) and the second by P
n
(x) and integrating by parts show that Legendre
polynomials satisfy the orthogonality relation

1
−1
P
n
(x)P
m
(x) dx = 0 if n = m.
If n = m, it can be shown that the value of the integral is 2/(2n + 1). Verify this for the first three polynomials
(but you needn’t prove it in general).
Hint, Solution
Exercise 23.3
Find the forms of two linearly independent series expansions about the point z = 0 for the differential equation

w

+
1
sin z
w

+
1 − z
z
2
w = 0,
such that the series are real-valued on the positive real axis. Do not calculate the coefficients in the expansions.
Hint, Solution
Exercise 23.4
Classify the singular points of the equation
w

+
w

z −1
+ 2w = 0.
Hint, Solution
1220
Exercise 23.5
Find the series expansions about z = 0 for
w

+

5
4z
w

+
z −1
8z
2
w = 0.
Hint, Solution
Exercise 23.6
Find the series expansions about z = 0 of the fundamental solutions of
w

+ zw

+ w = 0.
Hint, Solution
Exercise 23.7
Find the series expansions about z = 0 of the two linearly independent solutions of
w

+
1
2z
w

+
1
z

w = 0.
Hint, Solution
Exercise 23.8
Classify the singularity at in finity of the differential equation
w

+

2
z
+
3
z
2

w

+
1
z
2
w = 0.
Find the forms of the series solutions of the differential equation about infinity that are real-valued when z is real-valued
and positive. Do not calculate the coefficients in the expansions.
Hint, Solution
1221
Exercise 23.9
Consider the second order differential equation
x
d

2
y
dx
2
+ (b − x)
dy
dx
− ay = 0,
where a, b are real constants.
1. Show that x = 0 is a regular singular point. Determine the location of any additional singular points and classify
them. Incl ude the point at infinity.
2. Compute the indicial equation for the point x = 0.
3. By solving an appropriate recursion relation, show that one solution has the form
y
1
(x) = 1 +
ax
b
+
(a)
2
x
2
(b)
2
2!
+ ··· +
(a)
n
x

n
(b)
n
n!
+ ···
where the notation (a)
n
is defined by
(a)
n
= a(a + 1)(a + 2) ···(a + n −1), (a)
0
= 1.
Assume throughout this problem that b = n where n is a non-negative integer.
4. Show that when a = −m, where m is a non-negative integer, that there are polynomial solutions to this equation.
Compute the radius of convergence of the series above when a = −m. Verify that the result you get is in accord
with the Frobenius theory.
5. Show that if b = n + 1 where n = 0, 1, 2, . . ., then the second solution of this equation has logarithmic terms.
Indicate the form of the second solution in this case. You need not compute any coefficients.
Hint, Solution
1222
Exercise 23.10
Consider the equation
xy

+ 2xy

+ 6
e
x

y = 0.
Find the first three non-zero terms in each of two linearly independent series solutions about x = 0.
Hint, Solution
1223
23.6 Hints
Hint 23.1
Hint 23.2
Hint 23.3
Hint 23.4
Hint 23.5
Hint 23.6
Hint 23.7
Hint 23.8
Hint 23.9
Hint 23.10
1224
23.7 Solutions
Solution 23.1
f(x) is a Taylor series about x = 0.
f(x) =
∞

n=0
a
n
x
n
f

(x) =

∞

n=1
na
n
x
n−1
=
∞

n=0
na
n
x
n−1
f

(x) =
∞

n=2
n(n − 1)a
n
x
n−2
=
∞

n=0
(n + 2)(n + 1)a

n+2
x
n
We substitute the Taylor series into the differential equation.
f

(x) − 2xf

(x) + 2λf = 0
∞

n=0
(n + 2)(n + 1)a
n+2
x
n
− 2
∞

n=0
na
n
x
n
+ 2λ
∞

n=0
a
n

x
n
Equating coefficients gives us a difference equation for a
n
:
(n + 2)(n + 1)a
n+2
− 2na
n
+ 2λa
n
= 0
a
n+2
= 2
n − λ
(n + 1)(n + 2)
a
n
.
1225
The first two coefficients, a
0
and a
1
are arbitrary. The remaining coefficients are determined by the recurrence relation.
We will find the fundamental set of solutions at x = 0. That is, for the first solution we choose a
0
= 1 and a
1

= 0; for
the second solution we choose a
0
= 0, a
1
= 1. The difference equation for y
1
is
a
n+2
= 2
n − λ
(n + 1)(n + 2)
a
n
, a
0
= 1, a
1
= 0,
which has the solution
a
2n
=
2
n

n
k=0
(2(n − k) −λ)

(2n)!
, a
2n+1
= 0.
The difference equation for y
2
is
a
n+2
= 2
n − λ
(n + 1)(n + 2)
a
n
, a
0
= 0, a
1
= 1,
which has the solution
a
2n
= 0, a
2n+1
=
2
n

n−1
k=0

(2(n − k) −1 −λ)
(2n + 1)!
.
A set of linearly independent solutions, (in fact the fundamental set of solutions at x = 0), is
y
1
(x) =
∞

n=0
2
n

n
k=0
(2(n − k) −λ)
(2n)!
x
2n
, y
2
(x) =
∞

n=0
2
n

n−1
k=0

(2(n − k) −1 −λ)
(2n + 1)!
x
2n+1
.
Since the coefficient functions in the differential equation do not have any singularities in the finite complex plane, the
radius of convergence of the series is infinite.
If λ = n is a positive even integer, then the first solution, y
1
, is a polynomial of order n. If λ = n is a positive odd
integer, then the second solution, y
2
, is a polynomial of order n. For λ = 0, 1, 2, 3, we have
H
0
(x) = 1
H
1
(x) = x
H
2
(x) = 1 −2x
2
H
3
(x) = x −
2
3
x
3

1226
Solution 23.2
1. First we write the differential equation in the standard form.

1 − x
2

y

− 2xy

+ α(α + 1)y = 0 (23.2)
y

−
2x
1 − x
2
y

+
α(α + 1)
1 − x
2
y = 0. (23.3)
Since the coefficients of y

and y are analytic in a neighborhood of x = 0, We can find two Taylor series solutions
about that point. We find the Taylor series for y and its derivatives.
y =

∞

n=0
a
n
x
n
y

=
∞

n=1
na
n
x
n−1
y

=
∞

n=2
(n − 1)na
n
x
n−2
=
∞


n=0
(n + 1)(n + 2)a
n+2
x
n
Here we used index shifting to explicitly write the two forms that we will need for y

. Note that we can take the
lower bound of summation to be n = 0 for all above sums. The terms added by this operation are zero. We
substitute the Taylor s eries into Eq uation 23.2.
∞

n=0
(n + 1)(n + 2)a
n+2
x
n
−
∞

n=0
(n − 1)na
n
x
n
− 2
∞

n=0
na

n
x
n
+ α(α + 1)
∞

n=0
a
n
x
n
= 0
∞

n=0

(n + 1)(n + 2)a
n+2
−

(n − 1)n + 2n − α(α + 1)

a
n

x
n
= 0
1227
We equate coefficients of x

n
to obtain a recurrence relation.
(n + 1)(n + 2)a
n+2
= (n(n + 1) −α(α + 1))a
n
a
n+2
=
n(n + 1) − α(α + 1)
(n + 1)(n + 2)
a
n
, n ≥ 0
We can solve this difference equation to determine the a
n
’s. (a
0
and a
1
are arbitrary.)
a
n
=
















a
0
n!
n−2

k=0
even k

k(k + 1) − α(α + 1)

, even n,
a
1
n!
n−2

k=1
odd k

k(k + 1) − α(α + 1)


, odd n
We will find the fundamental set of solutions at x = 0, that is the set {y
1
, y
2
} that satisfies
y
1
(0) = 1 y

1
(0) = 0
y
2
(0) = 0 y

2
(0) = 1.
For y
1
we take a
0
= 1 and a
1
= 0; for y
2
we take a
0
= 0 and a
1

= 1. The rest of the coefficients are determined
from the recurrence relation.
y
1
=
∞

n=0
even n



1
n!
n−2

k=0
even k

k(k + 1) − α(α + 1)




x
n
y
2
=
∞


n=1
odd n



1
n!
n−2

k=1
odd k

k(k + 1) − α(α + 1)




x
n
1228
2. We determine the radius of convergence of the series solutions with the ratio test.
lim
n→∞




a
n+2

x
n+2
a
n
x
n




< 1
lim
n→∞





n(n+1)−α(α+1)
(n+1)(n+2)
a
n
x
n+2
a
n
x
n






< 1
lim
n→∞




n(n + 1) − α(α + 1)
(n + 1)(n + 2)






x
2


< 1


x
2


< 1

Thus we see that the radius of convergence of the series is 1. We knew that the radius of convergence would be
at least one, because the n earest si ngularities of the coefficients of (23.3) occur at x = ±1, a distance of 1 from
the origin. This implies that the solutions of the equation are analytic in the unit circle about x = 0. The radius
of convergence of the Taylor series expansion of an analytic function is the di stance to the nearest singularity.
3. If α = 2n then a
2n+2
= 0 in our first solution. From the recurrence relation, we see that all subsequ ent coefficients
are also zero. The solution becomes an even polynomial.
y
1
=
2n

m=0
even m



1
m!
m−2

k=0
even k

k(k + 1) − α(α + 1)





x
m
4. If α = 2n + 1 then a
2n+3
= 0 in our second solution. From the recurrence relation, we see that all subsequent
coefficients are also zero. The solution becomes an odd polynomial.
y
2
=
2n+1

m=1
odd m



1
m!
m−2

k=1
odd k

k(k + 1) − α(α + 1)




x
m

1229
Figure 23.4: The First Four Legendre Polynomials
5. From our solutions above, the first four polynomials are
1
x
1 − 3x
2
x −
5
3
x
3
To obtain the Legendre polynomials we normalize these to have value unity at x = 1
P
0
= 1
P
1
= x
P
2
=
1
2

3x
2
− 1

P

3
=
1
2

5x
3
− 3x

These four Legendre polynomials are plotted in Figure 23.4.
1230
6. We note that the first two terms in the Legendre equation form an exact derivative. Thu s the Legendre equation
can also be written as

(1 − x
2
)y



= −α(α + 1)y.
P
n
and P
m
are solutions of the Legendre equation.

(1 − x
2
)P


n


= −n(n + 1)P
n
,

(1 − x
2
)P

m


= −m(m + 1)P
m
(23.4)
We multiply the first relation of Equation 23.4 by P
m
and integrate by parts.

(1 − x
2
)P

n


P

m
= −n(n + 1)P
n
P
m

1
−1

(1 − x
2
)P

n


P
m
dx = −n(n + 1)

1
−1
P
n
P
m
dx

(1 − x
2

)P

n

P
m

1
−1
−

1
−1
(1 − x
2
)P

n
P

m
dx = −n(n + 1)

1
−1
P
n
P
m
dx


1
−1
(1 − x
2
)P

n
P

m
dx = n(n + 1)

1
−1
P
n
P
m
dx
We multiply the secord relation of Equation 23.4 by P
n
and integrate by p arts. To obtain a different expression
for

1
−1
(1 − x
2
)P


m
P

n
dx.

1
−1
(1 − x
2
)P

m
P

n
dx = m(m + 1)

1
−1
P
m
P
n
dx
We equate the two expressions for

1
−1

(1 − x
2
)P

m
P

n
dx. to obtain an orthogonality relation.
(n(n + 1) − m(m + 1))

1
−1
P
n
P
m
dx = 0

1
−1
P
n
(x)P
m
(x) dx = 0 if n = m.
1231
We verify that for the first four polynomials the value of the integral is 2/(2n + 1) for n = m.

1

−1
P
0
(x)P
0
(x) dx =

1
−1
1 dx = 2

1
−1
P
1
(x)P
1
(x) dx =

1
−1
x
2
dx =

x
3
3

1

−1
=
2
3

1
−1
P
2
(x)P
2
(x) dx =

1
−1
1
4

9x
4
− 6x
2
+ 1

dx =

1
4

9x

5
5
− 2x
3
+ x

1
−1
=
2
5

1
−1
P
3
(x)P
3
(x) dx =

1
−1
1
4

25x
6
− 30x
4
+ 9x

2

dx =

1
4

25x
7
7
− 6x
5
+ 3x
3

1
−1
=
2
7
Solution 23.3
The indicial equation for this problem is
α
2
+ 1 = 0.
Since the two roots α
1
= i and α
2
= −i are distinct and do not differ by an integer, there are two solutions in the

Frobenius form.
w
1
= z
i
∞

n=0
a
n
z
n
, w
1
= z
−i
∞

n=0
b
n
z
n
However, these series are not real-valued on the positive real axis. Recalling that
z
i
=
e
i log z
= cos(log z) + i sin(log z), and z

−i
=
e
−i log z
= cos(log z) − i sin(log z),
we can write a new set of solutions that are real-valued on the positive real axis as linear combinations of w
1
and w
2
.
u
1
=
1
2
(w
1
+ w
2
), u
2
=
1
2i
(w
1
− w
2
)
u

1
= cos(log z)
∞

n=0
c
n
z
n
, u
1
= sin(log z)
∞

n=0
d
n
z
n
1232
Solution 23.4
Consider the equation w

+ w

/(z −1) + 2w = 0.
We see that there is a regular singular point at z = 1. All other finite values of z are ordinary points of the equation.
To examine the point at infinity we introduce the transformation z = 1/t, w(z) = u(t). Writing the derivatives with
respect to z in terms of t yields
d

dz
= −t
2
d
dt
,
d
2
dz
2
= t
4
d
2
dt
2
+ 2t
3
d
dt
.
Substituting into the differential equation gives us
t
4
u

+ 2t
3
u


−
t
2
u

1/t − 1
+ 2u = 0
u

+

2
t
−
1
t(1 − t)

u

+
2
t
4
u = 0.
Since t = 0 is an irregular singular point in the equation for u(t), z = ∞ is an irregular singular point in the equation
for w(z).
Solution 23.5
Find the series expansions about z = 0 for
w


+
5
4z
w

+
z −1
8z
2
w = 0.
We see that z = 0 is a regular singular point of the equation. The indicial equation is
α
2
+
1
4
α −
1
8
= 0

α +
1
2

α −
1
4

= 0.

Since the roots are distinct and do not differ by an integer, there will be two solutions in the Frobenius form.
w
1
= z
1/4
∞

n=0
a
n
(α
1
)z
n
, w
2
= z
−1/2
∞

n=0
a
n
(α
2
)z
n
1233
We multiply the differential equation by 8z
2

to put it in a better form. Substituting a Frobenius series into the
differential equation,
8z
2
∞

n=0
(n + α)(n + α − 1)a
n
z
n+α−2
+ 10z
∞

n=0
(n + α)a
n
z
n+α−1
+ (z −1)
∞

n=0
a
n
z
n+α
8
∞


n=0
(n + α)(n + α − 1)a
n
z
n
+ 10
∞

n=0
(n + α)a
n
z
n
+
∞

n=1
a
n−1
z
n
−
∞

n=0
a
n
z
n
.

Equating coefficients of powers of z,
[8(n + α)(n + α − 1) + 10(n + α) − 1] a
n
= −a
n−1
a
n
= −
a
n−1
8(n + α)
2
+ 2(n + α) − 1
.
The First Solution. Setting α = 1/4 in the recurrence formula,
a
n
(α
1
) = −
a
n−1
8(n + 1/4)
2
+ 2(n + 1/4) − 1
a
n
(α
1
) = −

a
n−1
2n(4n + 3)
.
Thus the first solution is
w
1
= z
1/4
∞

n=0
a
n
(α
1
)z
n
= a
0
z
1/4

1 −
1
14
z +
1
616
z

2
+ ···

.
The Second Solution. Setting α = −1/2 in the recurrence formula,
a
n
= −
a
n−1
8(n − 1/2)
2
+ 2(n − 1/2) − 1
a
n
= −
a
n−1
2n(4n − 3)
1234
Thus the second linearly independent solution is
w
2
= z
−1/2
∞

n=0
a
n

(α
2
)z
n
= a
0
z
−1/2

1 −
1
2
z +
1
40
z
2
+ ···

.
Solution 23.6
We consider the series solutions of,
w

+ zw

+ w = 0.
We would like to find the expansions of the fundamental set of solutions about z = 0. Since z = 0 is a regular
point, (the coefficient functions are analytic there), we expand the solutions in Taylor series. Differentiating the series
expansions for w(z),

w =
∞

n=0
a
n
z
n
w

=
∞

n=1
na
n
z
n−1
w

=
∞

n=2
n(n − 1)a
n
z
n−2
=
∞


n=0
(n + 2)(n + 1)a
n+2
z
n
We may take the l ower limit of summation to be zero without changing the sums. Substituting these expressions into
the differential equation,
∞

n=0
(n + 2)(n + 1)a
n+2
z
n
+
∞

n=0
na
n
z
n
+
∞

n=0
a
n
z

n
= 0
∞

n=0

(n + 2)(n + 1)a
n+2
+ (n + 1)a
n

z
n
= 0.
1235
Equating the coefficient of the z
n
term gives us
(n + 2)(n + 1)a
n+2
+ (n + 1)a
n
= 0, n ≥ 0
a
n+2
= −
a
n
n + 2
, n ≥ 0.

a
0
and a
1
are arbitrary. We determine the rest of the coefficients from the recurrence relation. We consider the cases
for even and odd n separately.
a
2n
= −
a
2n−2
2n
=
a
2n−4
(2n)(2n − 2)
= (−1)
n
a
0
(2n)(2n − 2) ···4 · 2
= (−1)
n
a
0

n
m=1
2m
, n ≥ 0

a
2n+1
= −
a
2n−1
2n + 1
=
a
2n−3
(2n + 1)(2n − 1)
= (−1)
n
a
1
(2n + 1)(2n − 1) ···5 · 3
= (−1)
n
a
1

n
m=1
(2m + 1)
, n ≥ 0
If {w
1
, w
2
} is the fundamental set of solutions, then the initial conditions demand that w
1

= 1 + 0 · z + ··· and
w
2
= 0 + z + ···. We see that w
1
will have only even powers of z and w
2
will have only odd powers of z.
w
1
=
∞

n=0
(−1)
n

n
m=1
2m
z
2n
, w
2
=
∞

n=0
(−1)
n


n
m=1
(2m + 1)
z
2n+1
1236
Since the coefficient functions in the differential equation are entire, (analytic in the finite comp lex plane), the radius
of convergence of these series solutions is infinite.
Solution 23.7
w

+
1
2z
w

+
1
z
w = 0.
We can find the indicial equation by su bstituting w = z
α
+ O(z
α+1
) into the differential equation.
α(α − 1)z
α−2
+
1

2
αz
α−2
+ z
α−1
= O(z
α−1
)
Equating the coefficient of the z
α−2
term,
α(α − 1) +
1
2
α = 0
α = 0,
1
2
.
Since the roots are distinct and do not differ by an integer, the solutions are of the form
w
1
=
∞

n=0
a
n
z
n

, w
2
= z
1/2
∞

n=0
b
n
z
n
.
1237