Math 118 :: Winter 2009 :: Homework 1 Solution∗
January 18, 2009
1
(a)
Z
+∞
ĝ(k) =
−∞
+∞
e−ikx g(x)dx
Z
=
e−ikx f (x/a)dx
−∞
Z +∞
e−ikay f (y)dy
=a
−∞
= afˆ(ak)
(b)
Z
+∞
ĝ(k) =
−∞
Z +∞
=
e−ikx g(x)dx
e−ikx f (x)dx
−∞
Z
+∞
eikx f (x)dx
=
−∞
= fˆ(−k)
∗ contact
Kaiyuan Zhang through kzhang@math.stanford.edu for any questions.
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(c)
Z
+∞
fˆ(k) =
e−ikx f (x)dx
−∞
+∞
Z
=
Z
−ikx
e
0
Z
+∞
=
e−ikx f (x)dx +
eikx f (−x)dx
0
+∞
=
Z
+∞
e−ikx f (x)dx +
0
Z
e−ikx f (x)dx
−∞
Z +∞
0
Z
0
f (x)dx +
eikx f (x)dx
0
+∞
=
(e−ikx + eikx )f (x)dx
0
Z
+∞
=
2 cos(kx)f (x)dx
0
is real and it is clear that
Z +∞
Z
ˆ
f (−k) =
2 cos(−kx)f (x)dx =
0
+∞
2 cos(kx)f (x)dx = fˆ(k)
(1)
0
(d)
Z
fˆ(k) =
+∞
−∞
Z +∞
=
e−ikx f (x)dx
Z
−ikx
e
0
Z
+∞
=
e−ikx f (x)dx +
0
Z
e−ikx f (x)dx
−∞
Z +∞
eikx f (−x)dx
0
+∞
=
Z
e−ikx f (x)dx +
0
Z
0
f (x)dx +
+∞
eikx (−f (x))dx
0
+∞
=
(e−ikx − eikx )f (x)dx
0
Z
=
+∞
−2i sin(kx)f (x)dx
0
is imaginary and it is clear that
Z +∞
Z
fˆ(−k) =
−2i sin(−kx)f (x)dx =
0
0
2
+∞
2i sin(kx)f (x)dx = −fˆ(k)
(2)
2
(a)
If (2k + 16 )π ≤ x ≤ (2k + 56 )π for some integer k, then sin x ≥ 12 . Therefore,
Z
+∞
|sinc(x)|dx ≥
+∞ Z
X
(2k+ 56 )π
(2k+ 61 )π
−∞
=
k=0
+∞ Z (2k+ 5 )π
X
6
1
2
1
≥
2
=
1/2
dx
x
1
6
1
=
6
k=0
+∞
X
(2k+ 16 )π
ÃZ
1
3
k=0
Z
+∞
X
(2k+ 56 )π
1
dx +
x
(2k+ 61 )π
(2k+ 13
6 )π
(2k+ 16 )π
k=0
Z +∞
1
6π
1
dx
x
Z
(2k+ 96 )π
(2k+ 65 )π
1
dx +
x
Z
1
dx
x
1
dx = +∞
x
(b)
Let
½
rect(λ) =
π
0
−1 < λ < 1
otherwise
and consider its Fourier inverse transform
Z +∞
1
rect(λ)eiλx dλ
2π −∞
Z
1 1 iλx
=
e dλ
2 −1
¯1
1 eiλx ¯¯
=
2 ix ¯−1
1 eix − e−ix
2
ix
sin x
=
= sinc(x)
x
=
Therefore, in the sense of Fourier theory,
Z +∞
Z +∞
sinc(x) =
e−i·0·x sinc(x) = rect(0) = π
−∞
−∞
3
(2k+ 13
6 )π
(2k+ 69 )π
!
1
dx
x
3
Take Fourier transform, we have
û(k) + K̂(k)û(k) = fˆ(k)
Hence
û(k) = fˆ(k)/(1 + K̂(k))
and therefore
Z
+∞
u(x) =
eikx
−∞
fˆ(k)
1 + K̂(k)
dk
(3)
The condition is
K̂(k) 6= −1
4
(a)
Z
fˆk =
π
e−ikx f (x)dx
−π
Z π
e−ikx u0 (x)dx
Z π
¯π
−ikx
¯
=e
u(x) −π −
u(x)(−ik)e−ikx dx
−π
Z π
= 0 + ik
u(x)e−ikx dx
=
−π
−π
= ikûk
(b)
fˆ0 must be zero by the above equation. If the condition holds, the solution is
not unique since u(x) + c would also be a solution if u is.
(c)
h0 =
Rπ
−π
1dx = 2π and for k 6= 0,
Z
π
ĥk =
e−ikx dx
−π
1 −ikx π
e
|−π
−ik
=0
=
4
(d)
In the Fourier coefficients world, the solution is given by
ûk =
fˆk
ik
for k 6= 0 and û0 is arbitrary. In the world of the explicit expression, the solution
is up to a constant. The conclusion in (c) is consistent with these two results.
Since adding a constant to u is equivalent to adding an arbitrary multiple of
h(x), which has the Fourier coefficients given above, and therefore it results in
the arbitrary selection of û0 .
(e)
Since we require that u(−π) = u(π), then we have
0 = u(π) − u(−π)
Z π
=
f (x)dx
−π
= fˆ(0)
5
For distinct k, l ∈ Z,
Z
π
< v k , vl > =
vk (x)vl (x)dx
−π
π
Z
ce−ikx · c̄eilx dx
Z π
2
= |c|
ei(l−k)x dx
=
−π
−π
¯π
¯
1
i(l−k)x ¯
e
= |c| ·
¯
i(l − k)
−π
2
=0
and we also have
Z
π
< v k , vk > =
vk (x)vk (x)dx
−π
Z π
ce−ikx · c̄eikx dx
Z π
1dx
= |c|2
=
−π
−π
2
= |c| · 2π
5
To normalize it, we need
c=
√
2π
(4)
6
Let N = 3M where M is a positive integer.
N
1 X
fˆk =
fn e−ink/N
N n=1
=
M
´
1 X³
f3m−2 e−(3m−2)·ik/N + f3m−1 e−(3m−1)·ik/N + f3m e−(3m)·ik/N
N m=1
=
M
M
M
e2ik/N X
eik/N X
1 X
f3m−2 e−(3m−2)·ik/N +
f3m−1 e−(3m−1)·ik/N +
f3m e−3m·ik/N
N m=1
N m=1
N m=1
=
M
M
eik/N 1 X
e2ik/N 1 X
·
f3m−2 e−imk/M +
·
f3m−1 e−imk/M
3
M m=1
3
M m=1
+
M
1 1 X
·
f3m e−imk/M
3 M m=1
The discrete Fourier transform of {f1 , f2 , · · · , fN } of length N is decomposed
into three discrete Fourier transform of length M = N/3. The three sequences
are {f1 , f4 , · · · , fN −2 }, {f2 , f5 , · · · , fN −1 } and {f3 , f6 , · · · , fN }, respectively.
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