PHZ 3113 Fall 2010
Homework #9, Due Monday, December 6
1. Let the differential operator
d
d
(1 − x2 )
dx
dx
act on functions defined on the interval (−1, 1).
D=
(a) Suppose DPm (x) = −m(m + 1) Pm (x) and DPn (x) = −n(n + 1) Pn (x) Show that for
n 6= m Pm (x) and Pn (x) are orthogonal on (−1, 1).
Take the differential equation DPm = −m(m + 1)Pm , multiply by Pn , and integrate from
−1 to 1:
Z 1 Z 1
d
dPm
2
(1 − x )
Pm Pn dx.
Pn dx = −m(m + 1)
dx
−1 dx
−1
Integrate the left-hand side by parts, casting the derivative onto Pn , and obtain
Z 1
Z 1
2 dPm dPn
(1 − x )
Pm Pn dx
dx = m(m + 1)
dx dx
−1
−1
(because of the factor 1−x2 the surface term vanishes). Then, do the same thing, exchanging
m and n,
Z 1
Z 1
dPn dPm
2
dx = n(n + 1)
(1 − x )
Pn Pm dx,
dx dx
−1
−1
and subtract:
[m(m + 1) − n(n + 1)]
Z 1
−1
Pm Pn dx = 0.
If m 6= n, the factor in brackets is nonzero, and so
Z 1
−1
Pm Pn dx = 0.
(b) Verify that P0 = 1 and P1 = x are solutions to DPm = −m(m + 1) Pm . What are the
corresponding values of m?
d
d
2
DP0 =
(1 − x )
(1) = 0 = −0(0 + 1)P0 .
dx
dx
i
d
d h
2 d
2
(1 − x )
DP1 =
(x) =
(1 − x )(1) = −2x = −1(1 + 1)P1 .
dx
dx
dx
So, P0 and P1 are solutions to DPm = −m(m + 1)Pm , with m = 0 and m = 1, respectively.
(c) Let P2 = c0 + c1 x + c2 x2 be a polynomial of degree 2. Choose c0 and c1 so that P2 is
orthogonal to P0 and P1 . Choose c2 so that P2 (1) = 1. Write your resulting P2 (x). Is your
P2 a solution of DP2 = −m(m + 1) P2 for some m? If so, what is the your value for m?
The inner products are
Z 1
Z 1
−1
P0 P2 dx =
−1
Z 1
−1
(c0 + c1 x + c2 x2 ) dx = 2c0 +
Z 1
P1 P2 dx =
−1
x(c0 + c1 x + c2 x2 ) dx =
2
c .
3 2
2
c .
3 1
For P2 to be orthogonal to P0 and P1 , both of these must vanish: c1 = 0, c0 = − 13 c2 , leaving
P2 = c2 (− 13 + x2 ). Evaluated at x = 1, P2 = 23 c2 = 1 gives c2 = 32 . Thus,
P2 (x) =
3 2 1
x − .
2
2
Acting with D gives
i
d
3 2 1
d
d h
2
DP2 =
(1 − x )
x −
=
(1 − x2 )(3x) = 3 − 9x2 = −2(2 + 1)P2 .
dx
dx
2
2
dx
So, m = 2 (perhaps you anticipated this). This construction will produce any Legendre
polynomial Pm you need, but it is not the most efficient way to accomplish this.
2. Let
f (x) =
sin x 0 < x < π ,
0
π < x < 2π .
(a) Compute the Fourier series coefficients an , bn .
For n 6= 1, the cosine coefficients are

2/π
Z π
1
1 + cos nπ  − 2
n −1
an =
=
sin x cos nx dx = −

π 0
π(n2 − 1)
0
(n even)
(n odd)
(this also holds for a0 ), and the sine coefficients are
Z
1 π
sin nπ
bn =
sin x sin nx dx = 2
= 0.
π 0
n −1
For n = 1,
Thus,
Z
1 π
a1 =
sin x cos x dx = 0,
π 0
Z
1 π 2
1
b1 =
sin x dx = .
π 0
2
∞
1 1
2 X cos 2kx
f (x) = + sin x −
.
π 2
π
(2k)2 − 1
k=1
∞
X
(−1)k+1
(b) Compute the value of the sum
.
(2k)2 − 1
k=1
[Hint: what is f ( π2 )?]
As it says, evaluate f at π2 :
∞
π
1 1 2 X cos kπ
f( ) = 1 = + −
.
2
π 2 π
(2k)2 − 1
k=1
So,
∞
X
π
(−1)k+1
1 1
π 1
=
1− −
= − .
2
2
2 π
4 2
(2k) − 1
k=1
(c) Compute the value of the sum
∞
X
k=1
1
[(2k)2 − 1]2
.
The square of coefficients appears in Parseval’s theorem,
Z 2π
0
and so
f 2 (x) =
Z π
0
sin2 x dx =
∞
X
π
π
= a20 + π
(a2n + b2n ),
2
2
n=1
∞
1 2 2 1 2
1
4 X
1
+
+ 2
=
,
2 π
2
2
π
[(2k)2 − 1]2
k=1
∞
X
k=1
2 π2 1
π2 1 1
=
−
−
− .
=
4 2 4 π2
16 2
[(2k)2 − 1]2
1
3. Let
f (x) =
cos(k0 x) −a < x < a,
0
otherwise.
with k0 a = π2 .
(a) Compute the Fourier transform f̃ (k). What happens at k → k0 ?
Z a
Z a
1 ik0 x
f̃(k) =
(e
cos k0 x e
dx =
+ e−ik0 x ) e−ikx dx
−a
−a 2
"
#a
i(k
−k)x
−i(k
+k)x
0
0
1 e
e
sin[(k0 − k)a] sin[(k0 + k)a]
=
+
+
=
2 i(k0 − k) −i(k0 + k)
k0 − k
k0 + k
−ikx
−a
=
sin( π2 − ka) sin( π2 + ka)
2k cos ka
+
= 02
.
k0 − k
k0 + k
k0 − k 2
Although the denominator vanishes at k = k0 , the numerator does too, and the function
goes as
2k0 cos ka
π
π
≈
−
(k − k0 ) + · · · .
2k0 4k02
k02 − k 2
(b) Compute the integral (there is an easy way and an easier way to do this)
Z ∞
cos y dy
.
π 2
2
−∞ [( 2 ) − y ]
The “easy” way is as a contour integral, with cos y = Re[eiy ] and with two poles (on the
axis) at y = ± π2 ; but the “easier” way is to evaluate Fourier transform back to f (x) at x = 0:
Z ∞
Z
d(ka) 2k0 a cos ka
dk 2k0 cos ka ikx f (0) =
e =
= 1.
2
2
2π (k0 a)2 − (ka)2
x=0
−∞ 2π k0 − k
Now, let ka = y, with k0 a = π2 , and
Z
1=
So the value of the desired integral is 2.
dy π cos y
.
2π ( π2 )2 − y 2
(c) Compute the integral
Z ∞
cos2 y dy
.
π 2
22
−∞ [( 2 ) − y ]
This time, use Parseval’s theorem for the Fourier integrals,
Z a
Z
2
cos2 k0 x dx = a,
|f (x)| dx =
−a
Z
2k0 a cos ka 2 πa ∞ cos2 y dy
d(ka)
=
.
2π
2 −∞ [( π2 )2 − y 2 ]2
(k0 a)2 − (ka)2
By theorem these are the same, and so
Z
dk 2k0 cos ka 2
=a
2π k02 − k 2
Z
Z ∞
cos2 y dy
2
= .
π
2
2
2
π
−∞ [( 2 ) − y ]