Difference between revisions of "HWK8MA425Fall2009" - Rhea

Revision as of 05:48, 9 November 2009

Homework 8

HWK 8 problems

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-zo)^{k} $Did anybody do it another way? --Adrian Delancy

^^^On 8.1, I think you just need to use what you already know about geometric series, as Taylor's Theorem isn't mentioned until a couple of sections after. I just wrote $ z^k = z^k * (1-z)/(1-z) $ , and used the fact that the geometric series (with coefficient 1, center 0) converges to $ 1/(1-z) $

--Dgoodin 10:46, 8 November 2009 (UTC)

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.--Rgilhamw 14:21, 8 November 2009 (UTC)

did any one get this as the answer for 8.1?

$ \sum_{n=0}^\infty (1-z)(z-z0)^{2k} $

--Kfernan 15:41, 8 November 2009 (UTC)

I got something slightly different (though I'm somewhat confused on exactly how to make it centered on z0 so I could be wrong on that part). Most notably, I didn't get 2k, I just got k... where did the 2 come from? $ \sum_{n=0}^\infty (1-(z-z_0))(z-z_0)^n $ (or as I found to be more attractive an equation: $ \sum_{n=0}^\infty ((z-z_0)^{n+1}-(z-z_0)^n) $ ) --Whoskins

Also for problem 10.2 instead of using the comparison test you can just use the Hadamard Theorem since your $ An=n^n $ Then $ \frac{1}{lim sup(An)^{1/n}} = \frac{1}{n} $ which equals zero as n tends to infinity --Kfernan 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. --Kfernan 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? --Phebda 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) --adbohn 22:23, 8 November 2009 (UTC)Andy Bohn

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? --Aata

Yes I did 16.2 successfully... I think. The hint by Professor Bell above is for 16.2, not 16.1 it seems. Try using that as a starting point. --Ysuo 23:14, 8 November 2009 (UTC)