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== Homework 1 collaboration area  ==
 
== Homework 1 collaboration area  ==
  
Feel free to toss around ideas here. Feel free to form teams to toss around ideas. Feel free to create your own workspace for your own team. --[[User:Bell|Steve Bell]] 12:11, 20 August 2010 (UTC)
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Feel free to toss around ideas here.--[[User:Bell|Steve Bell]]
  
 
Here is my favorite formula:  
 
Here is my favorite formula:  
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<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math>
 
<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math>
  
[[2011 MA 530 Bell|Back to the MA 530 Rhea start page]]  
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Here is Stirling's Formula:
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<math>\lim_{n\to\infty}\frac{n!}{\sqrt{n}\left(\frac{n}{e}\right)^n} =\sqrt{2\pi}.</math>--[[User:Bell|Steve Bell]] 10:28, 19 January 2011 (UTC)
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== Problem 1 ==
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== Problem 2 ==
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== Problem 3 ==
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Difference quotient should include a special case when <math>f(z)=f(z_0)</math>.
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== Problem 4 ==
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== Problem 5 ==
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Use Problem 4.
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== Problem 6 ==
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Hint:  Notice that if
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<math>|a_n r^n|< M,</math>
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then
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<math>|a_n z^n|=|a_n r^n|\left(\frac{|z|}{r}\right)^n<
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M\left(\frac{|z|}{r}\right)^n,</math>
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and you can compare the series to a convergent geometric series
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if
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<math>|z|<r.</math>
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About the trick in the Problem 6, one direction is easy;
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The other direction can be proved using a trick by considering <math>r-\epsilon</math> where <math>\epsilon>0</math> is some arbitrarily small quantity.  This yields a convergent geometric series, which serves as an upper-bound of the original absolute series.  Finally, let <math>\epsilon</math> go to zero.  Result from Problem 5 is involved.
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[[2011 Spring MA 530 Bell|Back to the MA 530 Rhea start page]]  
  
 
[[Course List|To Rhea Course List]]
 
[[Course List|To Rhea Course List]]
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[[Category:MA530Spring2011Bell]]

Latest revision as of 10:25, 19 January 2011

Homework 1 collaboration area

Feel free to toss around ideas here.--Steve Bell

Here is my favorite formula:

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $

Here is Stirling's Formula:

$ \lim_{n\to\infty}\frac{n!}{\sqrt{n}\left(\frac{n}{e}\right)^n} =\sqrt{2\pi}. $--Steve Bell 10:28, 19 January 2011 (UTC)

Problem 1

Problem 2

Problem 3

Difference quotient should include a special case when $ f(z)=f(z_0) $.

Problem 4

Problem 5

Use Problem 4.

Problem 6

Hint: Notice that if

$ |a_n r^n|< M, $

then

$ |a_n z^n|=|a_n r^n|\left(\frac{|z|}{r}\right)^n< M\left(\frac{|z|}{r}\right)^n, $

and you can compare the series to a convergent geometric series if

$ |z|<r. $

About the trick in the Problem 6, one direction is easy;

The other direction can be proved using a trick by considering $ r-\epsilon $ where $ \epsilon>0 $ is some arbitrarily small quantity. This yields a convergent geometric series, which serves as an upper-bound of the original absolute series. Finally, let $ \epsilon $ go to zero. Result from Problem 5 is involved.

Back to the MA 530 Rhea start page

To Rhea Course List

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