FOURIER ANALYSIS: FIRST MIDTERM EXAM ANSWER KEY
A rod 100 cm long is insulated along its length and at both ends1. Suppose that its initial
temperature is u(x, 0) = x (x in cm, u in o C, t in seconds (s), 0 x 100, t 0), and that
its diusivity coecient is
110.443 Fourier Analysis
1
HW 5 Answer Keys
Page 61
10. Integral by parts for k times,
f (n) =
=
f (x)einx dx = f (x)
1
in
einx
1

in in
f (x)einx dx = =
1
ik nk
f (x)einx dx
f k (x)einx dx
C/nk
12. If sn s, that is (sn s) 0, so we could assume sn 0.
110.443 Fourier Analysis
1
HW 2 Answer Keys
Page 27.
9. By the formula,
Am =
=
=
2h
p
=
2
p
0
2
f (x) sin mxdx
0
2
xh
sin mxdx +
p
x cos mx p
0
m
p
h( x)
sin mxdx
p
cos mx/mdx +
0
2
p
h( x)
sin mxdx
p
2h
( p) cos mp sin mp
2h p cos mp sin mp
(
+
)+
(
+
110.443 Fourier Analysis
1
HW 4 Answer Keys
Page 59
2.a. use Eulers identity and combine the terms for f (n) and f (n) to get the
expression.
b. By denition,
1
f (n) =
2
1
2
=
f ()e(n) d
f (y)e(n)y dy (y = )
=
1
2
f (y)eny dy
= f (n)
Use (a) to see it is
110.443 Fourier Analysis
1
HW 6 Answer Keys
Page 87
1. We only prove the case for Rd . C d case is similar.
Suppose sequence Xk = (xk,1 , xk,2 , .xk,d ) is Cauchy.
For any , there exist N , such that for all k, k > N ,
i=d
Xk Xk
(xk,i xk ,i )2 <
=
i=1
So
110.443 Fourier Analysis
1
HW 11 Answer Keys
Page 164
14. use the Poisson summation formula.
From exercise 2, we see that FN (x) = [N,N ] (1 x/N ).
FN e2nx
FN (x + n) =
N
(1 x/N )e2nx = FN (x)
=
N
15.(a) Use the Poisson summation formula. Here let f =
110.443 Fourier Analysis
1
HW 9 Answer Keys
Page 124
12. In (8) let t = 4 2 and = 2x. Direct computation gives that u(, ) =
(f )h is a solution for the boundary value equation. Moreover, letting r = e ,
we see the heat kernel hr is a Poisson kernel.
Page
110.443 Fourier Analysis
1
HW 7 Answer Keys
Page 88
3. Follows the hint, choose Ik be the intervals covering [0, 2].They are
I1 = [0, ], I2 = [, 2],
I3 = [0, 2/3], I4 = [2/3, 4/3], I5 = [4/3, 2],
.
The length goes to zero and for any , it is in innite int
110.443 Fourier Analysis
1
HW 3 Answer Keys
1. When x = 0, fn (x) is continuous at x, so osc(fn , x) = 0 for any n.
When x = 0, if n > 0,
osc(fn , 0, r) = sup fn (x) fn (y) xn  + y n  2rn
osc(fn , 0) = lim osc(fn , 0, r) = 0
r0
if n = 0, osc(f0 , 0,
110.443 Fourier Analysis
1
HW 8 Answer Keys
Page 120
b
2.a. Integral by parts a x(s)y (s)ds = x(s)y(s)s=b y(s)x (s)ds
s=a
Here it is a closed curve, the rst term above is zero. The rest results follow.
3. Let the two curve segment be 1 = (x, f (x),2 = (x
FOURIER ANALYSIS: SECOND MIDTERM EXAM
Justify your answers as fully as possible. MAXIMUM TOTAL SCORE IS 100 POINTS.
(1) (30 points) True or false:
The area of an ellipse with semiaxes of lengths a, b is A = ab and its perimeter is
= 4ab/(a + b).
False. By