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 thou
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
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
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
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 1
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
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 Ca
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
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 h
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. Howe
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
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
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 exp
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, (
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
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
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 det
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].nyu.edu
Homework set 1: Due Friday September 19 at
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 1
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
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
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
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
b
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)
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 soluti
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 ca
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 ( ) +
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. L
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 p
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 co