(6 intermediate revisions by 2 users not shown)
Line 1: Line 1:
== Homework 4 collaboration area ==
+
== Homework 4 discussion area ==
 +
 
 +
== Problem 3 ==
 +
 
 +
This is what I come up with.  I am not sure if the following is sharp enough:
 +
 
 +
The successive derivatives of an analytic function at a point <math>a</math> can never satisfy <math>|f^{(n)}(a)|>n!h^n(n)</math>, where <math>h(n)</math> is a function of <math>n</math> such that <math>\limsup_{n\to\infty}h(n)=+\infty</math>.  --[[User:Zhug|Guangwei Zhu]] 15:20, 11 February 2011 (UTC)
  
 
Problem 7 hint:
 
Problem 7 hint:
  
 
<math>e^{\pm f(z)}</math>
 
<math>e^{\pm f(z)}</math>
 +
 +
I came up with a proof by looking at <math>\frac{1}{1+f(z)}</math> and use Liouville's Theorem.  --[[User:Zhug|Guangwei Zhu]] 15:21, 11 February 2011 (UTC)
  
 
Problem 10 hint:
 
Problem 10 hint:
Line 9: Line 17:
 
Parametrize the circular part of the boundary via
 
Parametrize the circular part of the boundary via
  
<math>C_r:\quad z(t)=Re^{it}, 0<t<\pi/4.</math>
+
<math>C_R:\quad z(t)=Re^{it}, 0<t<\pi/4.</math>
  
 
You need to show that
 
You need to show that
Line 17: Line 25:
 
as R goes to infinity.  You won't be able to use the standard estimate to do this.  Write out the definition of the integral to find that
 
as R goes to infinity.  You won't be able to use the standard estimate to do this.  Write out the definition of the integral to find that
  
<math>|I_R|\le\int Re^{-R^2\cos(2t)}\,dt</math>
+
<math>|I_R|\le\int_0^{\pi/4} Re^{-R^2\cos(2t)}\,dt</math>
  
and use freshman calculus ideas to show that this integral tends to zero.  (Don't hit it with the big stick, the Lebesgue Dominated Convergence Theorem.)
+
and use freshman calculus ideas to show that this integral tends to zero.  (Don't hit it with the big stick, the Lebesgue Dominated Convergence Theorem.) Hint:  Draw the graph of cos_2t on the interval and realize that the line connecting the endpoints is under the graph.  Compare the integral with what you would get by replacing cos_2t by the simple linear function underneath it.
  
  

Latest revision as of 11:21, 11 February 2011

Homework 4 discussion area

Problem 3

This is what I come up with. I am not sure if the following is sharp enough:

The successive derivatives of an analytic function at a point $ a $ can never satisfy $ |f^{(n)}(a)|>n!h^n(n) $, where $ h(n) $ is a function of $ n $ such that $ \limsup_{n\to\infty}h(n)=+\infty $. --Guangwei Zhu 15:20, 11 February 2011 (UTC)

Problem 7 hint:

$ e^{\pm f(z)} $

I came up with a proof by looking at $ \frac{1}{1+f(z)} $ and use Liouville's Theorem. --Guangwei Zhu 15:21, 11 February 2011 (UTC)

Problem 10 hint:

Parametrize the circular part of the boundary via

$ C_R:\quad z(t)=Re^{it}, 0<t<\pi/4. $

You need to show that

$ I_R := \int_{C_R}e^{-z^2}\ dz\to 0 $

as R goes to infinity. You won't be able to use the standard estimate to do this. Write out the definition of the integral to find that

$ |I_R|\le\int_0^{\pi/4} Re^{-R^2\cos(2t)}\,dt $

and use freshman calculus ideas to show that this integral tends to zero. (Don't hit it with the big stick, the Lebesgue Dominated Convergence Theorem.) Hint: Draw the graph of cos_2t on the interval and realize that the line connecting the endpoints is under the graph. Compare the integral with what you would get by replacing cos_2t by the simple linear function underneath it.


Back to the MA 530 Rhea start page

To Rhea Course List

Alumni Liaison

Have a piece of advice for Purdue students? Share it through Rhea!

Alumni Liaison