Download Solutions to Problem Paper 1

Solutions to Problem Paper 1
Due September 13th , 2010
Problem 1 (Lec. I)
Since f (n) (x) = ex we have f (n) (0) = 1 for all b ∈ N. Applying the definition
of Taylor series we have
∞
∞
X
f (n) (0) n X xn
x
x =
.
e =
n!
n!
n=0
n=0
The ratio test gives
n+1
x
n! n!
n!
lim · n = |x| lim
= |x| lim
=
n→∞ (n + 1)! x
n→∞ (n + 1)!
n→∞ (n + 1)n!
1
|x| lim
= 0 · |x| < 1 ∀x ∈ R.
n→∞ n + 1
Equivalently, we can observe that L = 0 and hence the radius of convergence
is ρ = 1/L = ∞.
Problem 2 (c) (Lec. I)
Taking into account that
∞
X
1
=
xn
1 − x n=0
whenever |x| < 1
and observing that
1
d 1
=
dx 1 − x
(1 − x2 )
we have
∞
∞
1
d 1
d X n X n−1
=
=
x =
nx .
(1 − x2 )
dx 1 − x
dx n=0
n=0
Noticing that the first term in the sum is zero and shifting the index according
to m = n − 1 we have
∞
∞
∞
∞
X
X
X
X
n−1
n−1
m
nx
=
nx
=
(m + 1)x =
(n + 1)xn .
n=0
n=1
m=0
n=0
Hence, we conclude that
∞
X
1
=
(n + 1)xn
(1 − x2 ) n=0
1
whenever |x| < 1.
Problem 3 (Lec. I)
Remember we want to compute the first three nonzero terms in the Taylor
expansion and that the last term in such expansion will contain the fifth
power of x. Observe that
f (x) = 2 sin x cos x,
x2 x4
x3 x5
+
− ···
1−
+
− ··· ,
= 2 x−
3!
5!
2!
4!
x3 x5
x3
x5
x5
= 2 x−
+
− ··· −
+
− ··· +
− ··· ,
2!
4!
3!
2!3!
5!
1 1
1
1
1
3
= 2x + 2 − −
x +2
+
+
x5 + · · · ,
2 6
24 12 120
4 5
4 3
= 2x − x + x − · · · .
3
15
Compare the above result with the Taylor series of sin 2x given by
sin 2x =
∞
X
n=0
(−)n
1
1
(2x)2n+1 = 2x − (2x)3 + (2x)5 − · · · ,
(2n + 1)!
6
5!
4
4
= 2x − x3 + x5 − · · · .
3
15
The two series coincide in virtue of the trigonometric indentity
sin 2x = 2 sin x cos x.
Problem 1 (d) (Lec. II)
The standard form of the given differential equation is obtained by dividing
the original equation by x(1 − x) and one has
d2 y
dy
2
−
+
y = 0.
2
dx
dx 1 − x
There is only one singular point at x = 1 and we conclude that the ordinary
points of the differential equations are the set of all x ∈ R without x = 1,
that is R\{1}.
Problem 2 (Lec. II)
Concerning (a) we make the shift m = n − 2. Then
∞
X
n=2
n(n − 1)an x
n−2
=
∞
X
m=0
2
(m + 2)(m + 1)am+2 xm
and m0 = 0. Concerning (b) the shift we need is m = n + 1. Therefore
x
2
∞
X
an x
n−1
=
∞
X
n=0
an x
n+1
=
n=0
∞
X
am−1 xm
m=1
and in this case m0 = 1. Finally, in the case (c) the shift reads m = n + 2
and we obtain
∞
∞
X
X
n+2
an+2 x
=
am xm
n=1
m=3
and m0 = 3.
Problem 3 (c) (Lec. II)
Consider the differential equation
d2 y
dy
+ (x − 1) + y = 0.
2
dx
dx
2
Let the dependent variable y be expressed in terms of a power series
y(x) =
∞
X
an x n
n=0
around the point x = 0. We want to find a recurrence relation for the coefficients an . To this purpose, we first compute the first and second derivative
of y. We obtain
∞
∞
d2 y X
n(n − 1)xn−2 ,
=
2
dx
n=0
dy X n−1
nx .
=
dx n=0
Substituting them into the original differential equation we find
2
∞
X
n(n − 1)an x
n−2
+ (x − 1)
n=0
∞
X
nan x
n−1
n=0
+
∞
X
an xn = 0,
n=0
which can be rewritten as follows
∞
∞
∞
∞
X
X
X
X
n−2
n
n−1
2
n(n − 1)an x
+
nan x −
nan x
+
an xn = 0.
n=0
n=0
n=0
(1)
n=0
From the expression above, we see that only the indices in the first and third
series have to be shifted by m = n−2 and m = n−1, respectively. Therefore,
∞
X
n=0
n(n − 1)an x
n−2
=
∞
X
n(n − 1)an x
n=2
n−2
=
∞
X
n=0
3
(n + 2)(n + 1)an+2 xn
and
∞
X
nan x
n−1
=
n=0
∞
X
nan x
n−1
n=1
=
∞
X
(n + 1)an+1 xn .
n=0
Substituting these last two results into (1) we get
2
∞
X
n
(n + 2)(n + 1)an+2 x +
n=0
∞
X
n
nan x −
n=0
∞
X
n
(n + 1)an+1 x +
n=0
∞
X
an xn = 0,
n=0
that is
∞
X
[2(n + 2)(n + 1)an+2 + nan − (n + 1)an+1 + an ] xn = 0.
n=0
The above equation will be satisfied for every x if
2(n + 2)(n + 1)an+2 + nan − (n + 1)an+1 + an = 0
for all n ∈ N. The above expression can be further simplified by noticing
that we can factorize an . This gives
2(n + 2)(n + 1)an+2 + (n + 1)an − (n + 1)an+1 = 0.
Since n + 1 can never be zero we can divide the above equation by n + 1 and
obtain the recurrence relation
2(n + 2)an+2 + an − an+1 = 0
whch can also be written as
an+2 = −
an − an+1
.
2(n + 2)
4