User talk:StatisticsMan

Sorry for not responding sooner -- I've been busy with other non-Wikipedia related tasks this week. Unfortunately, I don't have convenient access to box scores from the 1970s. If I do come across a five-by-five performance from that time period, however, I'll be sure to add it to the article you've created. Myasuda 00:08, 18 April 2007 (UTC)Reply

Regarding your last question, I recommend that you consult Wikipedia:Redirect and Wikipedia:Criteria for speedy deletion. If you like, you can create the new page, gut the old one, remove all links you've made to the old one, and then request speedy deletion for the old page. The other, somewhat less satisfactory, option is to redirect. Redirection is more often used when there are multiple natural ways of referencing an article. Myasuda 01:02, 19 April 2007 (UTC)Reply

Hello. I would like to respectfully ask that you consider using edit summaries to go along with your edits to help other editors better understand your edits (such as to Greatest_common_divisor_of_two_polynomials). Thanks. Doctormatt 06:17, 26 May 2007 (UTC)Reply

Where does this series converge?

${\displaystyle \sum _{k=0}^{\infty }p(k)\cdot z^{k}}$ where p is a polynomial.

My thought is this converges for all |z| < 1 and that's it. But, I'm not exactly sure how the root test would work. I guess with a one term polynomial, ${\displaystyle k^{n}}$, that's easy:

${\displaystyle \varlimsup |k^{n}|^{1/k}=\varlimsup |k^{1/k}|^{n}=1^{n}=1}$

And, on D(0, 1), ${\displaystyle z=e^{i\theta }}$ so the absolute value of the terms is

${\displaystyle |k^{n}||e^{i\theta }|^{k}=|k^{n}|\,}$

which goes to infinity, so there is no convergence on the boundary.

But, say I'm dealing with p(x) = x^2 + 7x + 5.

If p(k) = k² + 7k + 5, then why not just do this:

${\displaystyle \sum _{k=0}^{\infty }(k^{2}+7k+5)z^{k}=\sum _{k=0}^{\infty }k^{2}z^{k}+7\sum _{k=0}^{\infty }kz^{k}+5\sum _{k=0}^{\infty }z^{k}.}$

Then you can apply your result for k = n to the three cases n = 0, 1, 2. Michael Hardy (talk) 01:40, 6 April 2009 (UTC)Reply

Hi StatisticsMan. I know it doesn't really matter that much, but in adding back the "i", you let the "x" fall in "(e^ix)^3" -- I suppose that you need "(e^{ix})^3". -- 124.157.254.146 (talk) 04:06, 4 October 2010 (UTC)Reply

Since you've not been on, I went ahead and fixed it myself before it archives. Hope you don't mind. -- ToE^T 12:33, 6 October 2010 (UTC)Reply

Funny thing is that I had forgotten de Moivre's formula by name, so I first confused it with De Morgan's theorem. When I finally figured out what was really being discussed, I tried to recall De Morgan's theorem 's real name, first looking at Inclusion–exclusion principle, feeling that there was a vague connection. There I did find a link to De Morgan's theorem, but also mention that it was Abraham de Moivre who was responsible for the general inclusion–exclusion formula. Small world inhabited by to de M* mathematicians. -- 124.157.254.146 (talk) 04:15, 4 October 2010 (UTC)Reply