Stochastic Integrals.
In dening the integral
T
I=
f (s)dx(s)
0
the Lebesgue theory assumes that f () is a bounded continuous function and x(s) is a
function of bounded variation so that dx can be thought of as a measure on [0, T ]. With
that we get the bo
MATH-GA.2450-001
Complex Variables I Fall 2014
November 6, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 9: Due Friday November 14 at 12 noon.
Homework should be given to me (or my possible r
MATH-GA.2450-001
Complex Variables I Fall 2014
November 13, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 10: Due Friday November 21 at 12 noon.
Homework should be given to me (or my possible
MATH-GA.2450-001
Complex Variables I Fall 2014
October 30, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 8: Due Friday November 7 at 12 noon.
Homework should be given to me (or my possible re
MATH-GA.2450-001
Complex Variables I Fall 2014
December 4, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 11: Due Friday December 12 at 12 noon.
Homework should be given to me (or my possible
MATH-GA.2450-001
Complex Variables I Fall 2014
October 22, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 7: Due Friday October 31 at 12 noon.
Homework should be given to me (or my possible re
Complex Homework 2
Jiawei Sun
1
Let f (z) = u + iv, since |f (z)| is a constant, u2 + v 2 = C, then we have
u ux + v vx = 0
(1)
u uy + v vy = 0
Further more, since f (z) is analytic, so we have the Cauchy-Riemann
Equation: ux = vy and uy = vx , thus (1) t
Complex Variable Homework1
Jiawei Sun
September 17, 2014
2i
1 i 3
1 + i 3
1 + i 3 3
2i
= e 3 , so (
) = e = 1; Similarly,
=
1. Since
2
2
2
4i
1 i 3 3
e 3 , so (
) = e4i = 1.
2
1
2
2
3
= +
i, so Arg (z)=
. 3 i = 2e 6 i , ( 3 i)6 =
2.
2
2
3
1+i 3
26 ei =
Complex Homework3
Jiawei Sun
1
We can directly show that
L(f g) =
(f g)
f g + fg
f
g
=
=
+ = L(f ) + L(g)
fg
fg
f
g
2
1 u = x2 + 2axy + by 2 . If it is the real part of an analytic function, then
we have u = 0. However, uxx = 2, uyy = 2b, so only on condi
Complex Homework 4
Jiawei Sun
1
Let z = x + iy, then we have |e2z | < 1 |e2(x+iy) | < 1 e2x
|e2iy | < 1 e2x < 1 x = Re(z) > 0
2
We know that Logz = log |z| + iArg(z). So, Log(i3 ) = Log(i) =
i. However, Log(i) = i. Clearly, Log(i3 ) = 3Log(i). Gener2
2
Jiawei Sun
Complex Homework 5
1
On the curve y = x3 , we can denote z = t + t3 i. So that dz =
(1 + 3t2 i)dt. Let be the directed curve from 1 i to 1 + i, and we
have
1
0
4t3 (1 + 3t2 i)dt
(1 + 3t2 i)dt +
f (z)dz =
1
0
= (t + t3 i)
0
1
+ (t4 + 2t6 i)
1
0
6
Complex Homework6
Jiawei Sun
1
Note that
f (z)(z z0 ) f (z)
f (z)
=
2
(z z0 )
z z0
=: g (z)
We denote R be the area such that R = C, and D be the circle area
included in R and centered with z0 with radius r, then since f (z) is
analytic, so g(z) (and so
Complex Homework 7
Jianbo Sun
1
In Problem 6 of Homework 4 we have shown that
ez
z
1
1
1 4
= 1 z + z2
z +
1
2
12
720
Thus, we immediately have
1
1 1
1
1 3
= + z
z +
ez 1
z 2 12
720
In Laurent expansion, we see
1 = a1 =
So,
C
1
2i
f (z)dz
C
f (z)dz =
Complex Homework 8
Jiawei Sun
1
No, it is not a contradiction.
1
1
1
1
=
=
zn
z1
z (1 1/z) n=
(1)
See, (1) is valid only when |z| > 1. Moreover,
1
=
1z
zn
(2)
n=0
(2) is valid only when |z| < 1. So, (1) and (2) can not be valid
simultaneously.
2
We consid
Complex Homework 9
Jiawei Sun
1
Clearly,we have
Im =
x1
x
1
2m+1
We consider the function
z1
z
1
f (z) =
2m+1
We calculate the integral as shown below
I=
f (z)dz
C
C is the cantour in Figure 1.
Figure 1: Cantour
C1 is a upper circle with radius R, when R
Complex Homework 10
Jiawei Sun
1
Note that we have the following
n=1
(1)n
(2n + 1)3
=
n=1
=
n=1
=
1
(4n 3)3
1
(4n 3)3
n=1
n=0
1
(4n 1)3
1
(4n + 3)3
1
(4n 3)3
n=
3
cot(z)
, then f (z) has a pole with order 3 at z = .
(4z 3)3
4
Recall the stu in the class,
MATH-GA.2450-001
Complex Variables I Fall 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Sample questions for nal. For full credit always show all details.
1. Let a, b, and c be complex numbers which are t
MATH-GA.2450-001
Complex Variables I Fall 2013
September 7, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 1: Due Friday September 19 at 12 noon.
Homework should be given to me (or my possible
MATH-GA.2450-001
Complex Variables I Fall 2014
September 26, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 4: Due Friday October 10 at 12 noon.
Homework should be given to me (or my possible
MATH-GA.2450-001
Complex Variables I Fall 2013
September 17, 2014
Professor Olof Widlund
Oce: CIWW 612, Phone: 998-3110
Electronic mail: [email protected]
Homework set 2: Due Friday September 26 at 12 noon.
Homework should be given to me (or my possibl
7. Brownian Motion as a Markov process.
As a process with independent increments given Fs , x(t) x(s) is independent and has a
normal distribution with mean 0 and variance t s. Therefore
P [x(t) A|Fs ] =
p(t s, x(s), y )dyy
A
where
p(t, xy ) =
(y x)2
1
e
6. Brownian Motion.
A stochastic process can be thought of in one of many equivalent ways. We can begin
with an underlying probability space (, , P ) and a real valued stochastic process can
be dened as a collection of random variables cfw_x(t, ) indexed
9. Diusion proceses.
A diusion process is a Markov process with continuous paths with values in some Rd .
Given the past history up to time s the conditional distribution at a future time t is given
by the transition probability p(s, x, t, dy ). The cfw_p
Assignment 4.
1. Consider the region t 0, |x| 1. Let = inf cfw_t : |x(t)| = 1. Show that for any
staring point |x| < 1.
P [ < |x(0) = x] = 1
If u(t, x) is a bounded solution of
1
ut + uxx = 0, u(t, 1) = 0 for t > 0
2
then show that u 0. Can you construct
11. Dynamic Programming.
If x(t) =
t
0
c(s, )d (s) and |c(s, )| C , then it is not hard to see that
E [x(t)]2] CT
But actually if (x) is any convex function
E [(x(t)]
1
2Ct
y2
(y )e 2Ct dy
The solution u(s, x) of
us +
C
; u(t, x) = (x)
2 xx
is given by t
8. Stochastic Dierential Equations.
Brownian motion has the property that the distribution of x(t + h) x(t) given the eld Ft of information up to time t, is Gaussian with mean 0 and variance h. One can
visualize a Markov process x(t) for which the corresp
2. Independent random variables.
The Law of large Numbers. If cfw_Xi : i 1 are a sequence of independent identically
distributed random variables (on some (, F , P ) with E [Xi] = m, then
1
[X1 ( ) + + Xn ( )] = m = 1
n n
P : lim
This is the strong law of
Section 12. Markov Chain Approximations
It is often necessary to approximate models in continuous time by discrete versions. The
simplest example is approximation of Brownian motion by random walks. Let us consider
1
independent random variables Xi = 1 wi
5. Markov Processes.
A stochastic process in discrete time is just a sequence cfw_Xj : j 0 of random variables with values in some (X , F ) dened on a probability (, , P ).
It can also be specied by prescribing, in a self consistent manner, the joint
dist
3. Martingales I.
Let us start with a sequence cfw_Xi of independent random variables with E [Xi ] = 0 and
2
E [Xi ] = 1. We saw earlier that for a sequence cfw_aj of constants
S=
ai X i
i=1
will converge with probability 1 and in mean square provided
a