MA22S3, sample for my four questions on the
schol exams: outline solutions
Eight questions do six in a three hour exam.
1. Prove that for two periodic functions f (t) and g(t) with the same period
L then
Z
Z ∞
1 L/2
∗
f (t)g (t)dt =
cn d∗n
L −L/2
n=−∞
where cn and dn are the coefficients in the complex Fourier series for
f (t) and g(t) respectively. Deduce Parseval’s theorem from this.
Solution: Just substitute in
1
l
Z
c+l
c
dtf (t)g(t) =
1
l
Z
c+l
c
dt
X
cn eintl/2π
X
dm eimtl/2π
(1)
and then do a change of index to send m to −m and use d−m = d∗m to
get
XX
1 Z c+l
1 Z c+l
dtf (t)g(t) =
dteintl/2π e−imtl/2π
(2)
cn d∗m
l c
l
c
n m
and then the integral gives the required Kronecker δ-function. Setting
f = g gives Parseval’s theorem.
2. Write the triangular pulse



At/T + A
−T < t < 0
f (t) = −At/T + A 0 < t < T


0
|t| > T
as a Fourier integral.
Solution: So, substituting in
1 ∞
f (t)e−ikt dt
2π Z∞ 1 0 At
=
+ A e−ikt dt
2π −T
T
Z At
1 T
−
+ A e−ikt dt
+
2π 0
T
˜
f (k)
=
Z
1
(3)
Now, doing a change of variable from t to −t gives
0
Z
−T
Z T
At −ikt
At ikt
e dt = −
e dt
T
T
0
(4)
so we can put everything together
A
2π
˜ =
f (k)
Z
T
e−ikt dt −
−T
A
πT
Z
T
t cos ktdt
(5)
0
The second integral has to be done by parts
Z
T
T
1
1 T
t sin kt −
sin ktdt
k
k 0
0
T
1
1
=
sin kT + 2 cos kT − 2
k
k
k
t cos ktdt =
0
Z
(6)
The first integral is
Z
T
e
−ikt
−T
1 −ikt
=
e
−ik
T
=
−T
2
sin kT
k
(7)
Thus, we have
A
A T
1
1
sin kT −
sin kT + 2 cos kT − 2
πk
πT k
k
k
A
(1 − cos kT )
=
πT k 2
˜
f (k)
=
(8)
3. Solve the Euler-Cauchy equation
t2
d2 y
dy
+
3t
+ y = 0.
dt2
dt
Solution: By the usual change of variable, see the notes,
t = ez
(9)
this become
dy
d2 y
+2 +z =0
2
dz
dz
This equation has auxiliary
λ2 + 2λ + 1 = 0
2
(10)
(11)
which has a repeated root λ = −1; hence the solution is
y = C1 e−z + C2 ze−z
or
y=
C1 + C2 ln t
t
(12)
(13)
4. Using the method of Fröbenius, or otherwise, find the general solution
to
dy
d2 y
t 2 + 2 − αty = 0
dt
dt
where α > 0.
Solution: So, to use the method of Froebenius, we assume there is a
solution of the form
∞
y=
X
an tn+r
(14)
n=0
This gives
tÿ =
2ẏ =
tẏ =
∞
X
n=0
∞
X
n=0
∞
X
an (n + r)(n + r − 1)tn+r−1
2an (n + r)tn+r−1
an tn+r+1
(15)
n=0
which means we need to shift the indices for the first two terms forward
two steps by setting n = m+2, after taking terms out of the sum to get
the range of the index correct, and renaming m back to n, this gives
tÿ = r(r − 1)a0 tr−1 + (r + 1)ra1 tr
+
∞
X
an+2 (n + r + 2)(n + r + 1)tn+r+1
n=0
2ẏ = 2ra0 tr−1 + 2(r + 1)a1 tr +
∞
X
n=0
Thus, the equation becomes
r(r + 1)a0 tr−1 + (r + 1)(r + 2)a1 tr
3
2an+2 (n + r + 2)tn+r+1 (16)
+
∞
X
[an+2 (n + r + 2)(n + r + 3) − αan ]tn+r+1
n=0
= 0
(17)
The indicial equation, which is a consequence of the assumption that
a0 6= 0 is, therefore
r(r + 1) = 0
(18)
so r = 0 and r = −1 are the possilities, hence r = 0 or r = −1, giving
general solution
y = C1 y 1 + C2 y 2
(19)
where
∞
X
an tn
(20)
α
(n + 2)(n + 1)
(21)
y1 =
n=0
where
an+2 =
with a0 = 1 and a1 = 0 and, where where
y2 =
∞
1X
an tn
t n=0
where
an+2 =
with a0 = 1 and a1 = 0.
4
α
(n + 1)n
(22)
(23)