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'''NEWS FLASH:''' The due date for HWK 8 has been extended to Monday, Nov. 9 | '''NEWS FLASH:''' The due date for HWK 8 has been extended to Monday, Nov. 9 | ||
| − | ''Hint for V.16. | + | ''Hint for V.16.2:'' We know that |
<math>f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}</math> | <math>f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}</math> | ||
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What are the power series for <math>zf'(z)</math> and <math>z^2f''(z)</math>? How can you combine these to get the series in the question? --[[User:Bell|Steve Bell]] | What are the power series for <math>zf'(z)</math> and <math>z^2f''(z)</math>? How can you combine these to get the series in the question? --[[User:Bell|Steve Bell]] | ||
| − | + | ------------------------------------------------------------- | |
| − | Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be <math>(z- | + | |
| + | Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be <math>[(z-z_0)+z_0]^{k}</math>. Did anybody do it another way? | ||
--Adrian Delancy | --Adrian Delancy | ||
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| + | For 10.2 compare <math>\sum_{n=0}^\infty n^nz^{n}</math> to <math>\sum_{n=0}^\infty c^nz^{n}</math> where <math>c</math> is a constant to establish the inequality in the RoC's. Then let <math>c \rightarrow \infty</math> to squeeze the RoC of the first power series to zero.--[[User:Rgilhamw|Robert Gilham Westerman]] 14:21, 8 November 2009 (UTC) | ||
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| + | Or you could use a similar idea to show that, if <math>z\ne0</math>, | ||
| + | then the terms <math>n^n z^n</math> in the series do not tend to zero as n goes to infinity. Therefore, the series does not converge if <math>z\ne0</math>, and we conclude that the radius of convergence must be zero. --[[User:Bell|Steve Bell]] | ||
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| + | Also for problem 10.2 instead of using the comparison test you can just use the Hadamard Theorem since your <math>a_n=n^n</math>. Then <math>\frac{1}{lim sup|a_n|^{1/n}} = \lim_{n\to\infty}\frac{1}{n}</math> which equals zero. | ||
| + | --[[User:Kfernan|Kevin Fernandes]] 15:47, 8 November 2009 (UTC) | ||
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| + | Hey if any one can give me a hint to get somewhere with the last 4 problems which are 16.2 16.3 17.1 and 18.1 that would be awesome. | ||
| + | --[[User:Kfernan|Kevin Fernandes]] 16:37, 8 November 2009 (UTC) | ||
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| + | I think 16.2 was started above by Prof. Bell. For 16.3, I used the substitution <math>w=\sqrt z</math>, which expresses sum in the form of a power series. For 17.1, there is a trick that appears in section 17 involving the manipulation of 2 sums that appear from the product <math>exp(z_1)exp(z_2)</math>. It seems that 18.1 depends on the result of 8.1, which is giving me troubles as well. For this problem, with the above hints, all the coefficients were 1, but this does not seem right. Does anybody else have other ideas? --[[User:Phebda|Phil Hebda]] 20:46, 8 November 2009 (UTC)Phil Hebda | ||
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| + | For problem 8.1, I set up the general formula: | ||
| + | |||
| + | <math>z^k \equiv f(z) = \displaystyle\sum_{n=0}^{\infty}a_n(z-z_0)^n</math> | ||
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| + | First off, the coefficients for n>k will be zero, as we do not need higher powers in z. | ||
| + | You can differentiate both sides to obtain: | ||
| + | |||
| + | <math> \frac{d^m f(z)}{dz^m} = \displaystyle\sum_{n=m}^{\infty} \frac{n!}{(n-m)!}a_n(z-z_0)^{n-m} </math> | ||
| + | |||
| + | From this, you should be able to set z = z_0 to get the required coefficients as factorials and derivatives of f(z_0) | ||
| + | --[[User:adbohn|Andy Bohn]] 22:23, 8 November 2009 | ||
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| + | I'm still having some difficulty piecing together 16.1 even after Professor Bell's hint. | ||
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| + | Just use a Taylor series for 16.1. This homework is much harder than any of the others. | ||
| + | Also, for 16.2 I try to do something similar to the book but keep failing. Has anyone actually done it? --[[User:Aata|Adam Ata]] | ||
| + | |||
| + | Yes I did 16.2 successfully... I think. The hint by Professor Bell above is for 16.2. Try using that as a starting point. --[[User:Ysuo|Yu Suo]] 23:14, 8 November 2009 (UTC) | ||
Latest revision as of 09:00, 9 November 2009
Homework 8
NEWS FLASH: The due date for HWK 8 has been extended to Monday, Nov. 9
Hint for V.16.2: We know that
$ f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z} $
if $ |z|<1 $. Notice that
$ f'(z)=\sum_{n=1}^\infty nz^{n-1} $,
and
$ f''(z)=\sum_{n=2}^\infty n(n-1)z^{n-2} $.
What are the power series for $ zf'(z) $ and $ z^2f''(z) $? How can you combine these to get the series in the question? --Steve Bell
Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be $ [(z-z_0)+z_0]^{k} $. Did anybody do it another way?
--Adrian Delancy
For 10.2 compare $ \sum_{n=0}^\infty n^nz^{n} $ to $ \sum_{n=0}^\infty c^nz^{n} $ where $ c $ is a constant to establish the inequality in the RoC's. Then let $ c \rightarrow \infty $ to squeeze the RoC of the first power series to zero.--Robert Gilham Westerman 14:21, 8 November 2009 (UTC)
Or you could use a similar idea to show that, if $ z\ne0 $, then the terms $ n^n z^n $ in the series do not tend to zero as n goes to infinity. Therefore, the series does not converge if $ z\ne0 $, and we conclude that the radius of convergence must be zero. --Steve Bell
Also for problem 10.2 instead of using the comparison test you can just use the Hadamard Theorem since your $ a_n=n^n $. Then $ \frac{1}{lim sup|a_n|^{1/n}} = \lim_{n\to\infty}\frac{1}{n} $ which equals zero. --Kevin Fernandes 15:47, 8 November 2009 (UTC)
Hey if any one can give me a hint to get somewhere with the last 4 problems which are 16.2 16.3 17.1 and 18.1 that would be awesome. --Kevin Fernandes 16:37, 8 November 2009 (UTC)
I think 16.2 was started above by Prof. Bell. For 16.3, I used the substitution $ w=\sqrt z $, which expresses sum in the form of a power series. For 17.1, there is a trick that appears in section 17 involving the manipulation of 2 sums that appear from the product $ exp(z_1)exp(z_2) $. It seems that 18.1 depends on the result of 8.1, which is giving me troubles as well. For this problem, with the above hints, all the coefficients were 1, but this does not seem right. Does anybody else have other ideas? --Phil Hebda 20:46, 8 November 2009 (UTC)Phil Hebda
For problem 8.1, I set up the general formula:
$ z^k \equiv f(z) = \displaystyle\sum_{n=0}^{\infty}a_n(z-z_0)^n $
First off, the coefficients for n>k will be zero, as we do not need higher powers in z. You can differentiate both sides to obtain:
$ \frac{d^m f(z)}{dz^m} = \displaystyle\sum_{n=m}^{\infty} \frac{n!}{(n-m)!}a_n(z-z_0)^{n-m} $
From this, you should be able to set z = z_0 to get the required coefficients as factorials and derivatives of f(z_0) --Andy Bohn 22:23, 8 November 2009
I'm still having some difficulty piecing together 16.1 even after Professor Bell's hint.
Just use a Taylor series for 16.1. This homework is much harder than any of the others. Also, for 16.2 I try to do something similar to the book but keep failing. Has anyone actually done it? --Adam Ata
Yes I did 16.2 successfully... I think. The hint by Professor Bell above is for 16.2. Try using that as a starting point. --Yu Suo 23:14, 8 November 2009 (UTC)
