Question

Use Cauchy-Hadamard to find the Radius of convergence and the center of the series.

((-1)^n)/((2n)!) * (z- 1/4 pi)^2n

Answer 1

Thing is , my answer is absolutely different from the one I am supposed to get. I get infinity as radius sure enough, but my radius is pi/4 when it should be pi/2

Answer 2

So, if I understand this theorem correctly you need to find the actual c_n? To evaluate the lim sup?

Answer 3

That is the complex valued cosine function expanded about the point z = pi / 4. I am not sure of the problem, exactly.

Answer 4

The radius of convergence is indeed infinite.

Answer 5

@malevolence19 it's the summed up version, just \[R = \lim_{n \rightarrow infinity} \left| an/an+1 \right|\] where the series is \[\sum_{n=0}^{infinity} an (z-z0)^n\] Z0 being the center of the series.
and I meant my center, z0, is supposed to be pi/2, but I get pi/4.

I even tried this with a ratio test and still not the right answer

Answer 6

It is not pi/2. It is pi/4.

Answer 7

And this theorem is the complex equivalent of the ratio test, FYI.

Answer 8

Thanks I know, but the correct answer for z0 is pi/2.

I don't even know how they get it like that

Answer 9

No, I am quite sure it is not. Your source may have a typo.

Answer 10

The textbook? I don't know man

Answer 11

Here is another one, and now the radius is making no sense.

\[\sum_{n=0}^{infinity} (n(n-1))/(2^n) * (z+i)^(2n)\]

Answer 12

If you are talking about the series
\[ \sum_{n=0}^\infty (-1)^n\frac{(z - \frac{\pi}{4})^{2n}}{(2n)!} \]
Then the radius of convergence is infinity, the center of the series is pi / 4, and it converges to
\[\cos(z-\frac{\pi}{4}) \]
I am quite sure of this.

Answer 13

The radius of convergence I get is \[\sqrt{2}\] and the z0 is -i, however the answer is \[\sqrt{3}\] and i

Answer 14

And I'm sure of my answer, but I just don't understand how the book's answer is so different.

Answer 15

Are you sure you're looking at the right answers ... ?

Answer 16

Yeah!

Answer 17

Well, I'm afraid I can't really give any further input. As far as I can tell, the answers that you gave are correct, and the ones you say are from the book are not. You may of course seek other opinions but I am quite confident that I'm not insane.

Answer 18

What book are you using, by the way?

Answer 19

Advanced engineering mathematics by erwin kryszig, international edition

Answer 20

I found your book, so if you're still around, where exactly are these problems?

Answer 21

Problem set 15.2, odd questions starting from 7 since they have their answers in the back

Answer 22

I don't see any of those series, so maybe I have a different edition than you do... What are the page numbers / problem numbers of the two you listed above?

Answer 23

They are on page 684.. Chapter 15, Section 2

Answer 24

Mine is tenth edition

Answer 25

I just looked up the book, and you wrote the series down wrong........

Answer 26

Read the problems again. I'm glad this got settled and I'm not crazy, but please in the future make sure that you read the problem correctly... :)

Answer 27

How?

7. \[\sum_{n=0}^{infinity} ((-1)^n)/(2n!)) (z-1/4 \pi)^(2n)\]

I don't think I wrote it wrong..

Answer 28

If I did please tell me because this is driving me insane

Answer 29

Oh bloody flutterers. The answers are to the questions you've put up. They didn't update the answers to the changed questions

Answer 30

Can I have the link to that book?

Answer 31

You may indeed

http://megaupload.com/?d=1GND4DET

Answer 32

Thank you, now I can actually do the real questions.