Instructor: Frank Liou
Exam Time: 8:10 am to 10:00 am Total Score: 100 Points
Name:
Student ID:
TA’s Name:
(1) Please do not turn this page until told to do so.
(2) You can use either Chinese or English to write this exam.
(3) You can use pen or pencils. You have to write everything clearly.
(4) You cannot use your calculator, iPhone, iPad or any other electronic devices during this exam.
(5) Please place your personal belongings under your seat to give free way to the ex- aminers.
(6) No notes, books, or classmates may be used as resources for this exam. It is a violation of the University honor code too, in any way, assist another person in the completion of this exam. Please place your own work covered up as much as possible during the exam that the others will not be tempted or distracted. Thank you for your cooperation.
(7) Read directions to each problem carefully. Show all work for full credit. In most cases, with no supporting work will not receive full credit. The best way to get maximum partial credit is to write neatly and be organized.
(8) If you use any result from the homework, you need to prove it.
(9) Your score will be min{x,100}, where x is the number of points you get in this exam.
(10) Make sure you have pages, including the cover page.
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Score ( = min{Total,100} Points)
Formula which might be useful.
(1) Product to sum formula:
(a) 2 sinAcosB = sin(A+B) + sin(A−B), (b) 2 cosAsinA= sin(A+B)−sin(A−B), (c) 2 cosAcosB = cos(A+B) + cos(A−B), (d) 2 sinAsinB =−(cos(A+B)−cos(A−B)).
(2) Gamma function Γ(x) = Z ∞
0
e−ttx−1dt.
(3) Beta functionB(x, y) = Γ(x)Γ(y) Γ(x+y) and (a) B(x, y) =
Z ∞
0
tx−1 (1 +t)x+ydt, (b) B(x, y) = 2
Z π
2
0
cos2x−1θsin2y−1θdθ, (c) B(x, y) =
Z 1
0
tx−1(1−t)y−1dt.
(4) Let f(x) be a continuous real-valued periodic function on R of period 2π. The Fourier expansion ofy =f(x) is the infinite series
a0
2 +
∞
X
n=1
{ancosnx+bnsinnx},
wherean= 1 π
Z 2π 0
f(x) cosnxdxforn≥0 andbn= 1 π
Z 2π 0
f(x) sinnxdx,forn≥1.
(5) The Laplace transform of a continuous functiony =f(x) on [0,∞) is L(f)(s) =
Z ∞
0
e−stf(t)dt.
Suppose lim
N→∞e−sNf(N) = 0 and lim
N→∞e−sNf0(N) = 0.Then (a) L(f0)(s) =sL(f)(s)−f(0).
(b) L(f00)(s) =s2L(f)(s)−sf(0)−f0(0).
(1) (12 Points) Letf(x) = sinµxfor 0≤x≤2π. Find the Fourier expansion forf(x).
(Finda0 andan forn≥1 andbn forn≥1.)
Whenµis an integer,a0 =an= 0 for alln≥1 andbm= 0 form6=µandbµ= 1.
When µis not an integer, a0 = 1
π Z 2π
0
sinµxdx=−1 π
cosµx µ
2π 0
= 1−cos 2πµ
πµ .
an= 1 π
Z 2π 0
sinµxcosnxdx
= 1 2π
Z 2π 0
{sin(µ+n)x+ sin(µ−n)x}dx
= −1 2π
cos(µ+n)x
µ+n + cos(µ−n)x µ−n
2π 0
.
bn= 1 π
Z 2π 0
sinµxsinnxdx
= −1 2π
Z 2π 0
{cos(µ+n)x−cos(µ−n)x}dx
= −1 2π
sin(µ+n)x
µ+n −sin(µ−n)x µ−n
2π 0
.
(2) (14 Points) Use Gamma and Beta function to compute the following integrals. You will receive no credits if you use other methods.
(a) (7 Points) Z ∞
0
e−t2t10dt.
(b) (7 Points) Z π/2
0
cos11xsin13xdx.
Let u=t2.Then the integral becomes 1
2 Z ∞
0
e−uu92du= 1 2Γ
11 2
. The second integral is equal to 12B(6,7)
(3) (14 Points) Test the Convergence of the following improper integrals.
(a) (7 Points) Z ∞
0
e−x2sin π 2 +x2dx.
(b) (7 Points) Z ∞
1
2x+ 1
√3
x5+x4+x+ 1dx.
One can see that
0≤e−x2sin π
2 +x2 ≤e−x2 ≤e−x, x≥0.
Since R∞
0 e−xdxis convergent, by comparison test, the improper integral is conver- gent.
Letf(x) = (2x+ 1)/(x5+x4+x+ 1)1/3 and g(x) = 1/x2/3.Then
x→∞lim f(x) g(x) = 2.
Since R∞
1 1/x2/3dx is divergent by p-test, by limit comparison test, the improper integral is divergent.
(4) (10 Points) The integral
Z ∞
0
√ dx
x(1 +x)dx
is improper for two reasons: The interval [0,∞) is infinite and the integrand has an infinite discontinuity at 0. Evaluate it by expression it as a sum of improper integrals of Type 2 and Type 1 as follows:
Z ∞
0
√ dx
x(1 +x)dx= Z 1
0
√ dx
x(1 +x)dx+ Z ∞
1
√ dx
x(1 +x)dx.
Letu=√
x. Thendu=dx/2√
x.The indefinite integral of Z dx
√x(1 +x) =
Z 2du 1 +u2. We know
Z 1 a
√ dx
x(1 +x) = 2 Z 1
√a
du
1 +u2 = 2·(tan−11−tan−1√ a) = π
2 −2 tan−1√ a.
Hence Z 1
0
√ dx
x(1 +x) = lim
a→0
Z 1 a
√ dx
x(1 +x) = lim
a→0
π
2 −2 tan−1√ a
= π 2. We also know
Z b
1
√ dx
x(1 +x) = 2 Z
√ b 1
du
1 +u2 = 2(tan−1
√ b− π
4) = 2 tan−1
√ b−π
2. Hence
Z ∞
1
√ dx
x(1 +x) = lim
b→∞
Z b
1
√ dx
x(1 +x) = lim
b→∞(2 tan−1√ b−π
2) =π−π 2 = π
2. Hence the integral is π/2 +π/2 =π.
(5) (10 Points) Solve the initial value problem d2y
dx2 + 6dy
dx + 13y= 0, y(0) = 2, y0(0) = 3
using Laplace transform. (You can use the Table of Laplace Transform) You will receive no credit if you use other methods.
LetF(s) =L(y)(s).Using Formula 5 (a) and 5 (b), we see (s2F(s)−2s−3) + 6(sF(s)−2) + 13F(s) = 0.
We obtain that
F(s) = 2s+ 15 s2+ 6s+ 13
We know thats2+ 6s+ 13 =s2+ 6s+ 9 + 4 = (s+ 3)2+ 22.This would allow us to use Table of Laplace transform formula 19 and 20, i.e. a=−3 andb= 2.We know
L(e−3tcos 2t)(s) = s+ 3
(s+ 3)2+ 22, L(e−3tsin 2t)(s) = 2 (s+ 3)2+ 22. We know 2s+ 15 = 2(s+ 3) + 9.Hence
2s+ 15
s2+ 6s+ 13 = 2(s+ 3) s2+ 6s+ 13+9
2
2 s2+ 6s+ 13. This implies that
y(x) = 2e−3xcos 2x+9
2e−3xsin 2x=e−3x
2 cos 2x+9 2sin 2x
.
(6) (15 Points) Letf(x) = tan−11−x2
1 +x2,x∈R. (a) (6 Points) Find f0(x).
(b) (9 Points) Find the Taylor expansion of f atx= 0. Notice thatf(0) = π 4. By Calculation,
f0(x) = 1 1 +
1−x2 1+x2
2 ·−2x(1 +x2)−2x(1−x2) (1 +x2)2
= −4x−2x3+ 2x3 (1 +x2)2+ (1−x2)2
= −4x
(1 + 2x2+x4) + (1−2x2+x4)
= −4x
2 + 2x4
= −2x 1 +x4.
Using geometric series, for|x|<1,
−2x 1 +x4 =
∞
X
n=0
(−2x)(−x4)n=
∞
X
n=0
2(−1)n+1x4n+1. Fundamental theorem of calculus implies that
f(x)−f(0) = Z x
0
f0(t)dt=
∞
X
n=0
2(−1)n+1 Z x
0
t4n+1dt=
∞
X
n=0
2(−1)n x4n+2 4n+ 2. By f(0) =π/4,we obtain
f(x) = π 4 +
∞
X
n=0
2(−1)nx4n+2
4n+ 2, |x|<1.
(7) (15 Points) Letf(x) =
∞
X
n=0
(−1)nxn+1 n+ 1.
(a) (9 Points) Find the interval of convergence.
(b) (6 Points) Evaluate the following series:
1 2 −1
2 1
2 2
+1 3
1 2
3
+· · ·+ (−1)n 1 (n+ 1)
1 2
n+1
+· · ·.
(Hint: You need to find f(x). In order to do this, considerf0(x),or you can use the result proved in class.)
The radius of convergence is R= 1 by ratio test. Forx= 1,the infinite series
∞
X
n=0
(−1)n n+ 1
is convergent by the Leibnitz test and for x=−1,the infinite series
∞
X
n=0
1 n+ 1
is divergent by p-test. The interval of convergence is (−1,1]. The power series is convergent to f(x) = ln(1 +x) as discussed in class. Hence the infinite series is equal tof(1/2) = ln(3/2).
(8) (10 Points) Use integral test to study the convergence/divergence of the infinite series:
∞
X
n=10
1 n(lnn)p.
Let f(x) = 1/x(lnx)p.Forp >0,the function f(x) is positive and decreasing and an=f(n) forn≥10. Hence the infinite series and the improper integralR∞
10 f(x)dx both diverge or converge. Notice that (usingu= lnx,)
Z ∞
10
dx x(lnx)p =
Z ∞
ln 10
du up
is convergent for p >1 and divergent for 0< p <1 byp-test.
(9) (Bonus 15 Points) Choose one of the following problems as your bonus:
(a) Compute Z 1
0
lnxln(1−x)dx.
(b) For each n≥1,definean=
n
X
k=1
1
k −lnn. Show that {an} is convergent.
Proof. The solution to the second problem can be found in the book; this is an exercise.
About (a), I will write in another note.
