## Welcome!

PlanetMath is a virtual community which aims to help make mathematical knowledge more accessible. PlanetMath's content is created collaboratively: the main feature is the mathematics encyclopedia with entries written and reviewed by members. The entries are contributed under the terms of the Creative Commons By/Share-Alike License in order to preserve the rights of authors, readers and other content creators in a sensible way. We use LaTeX, the lingua franca of the worldwide mathematical community. On February 13th 2013, PlanetMath.org was updated to use the new software system Planetary. Some release notes are here. Please report bugs in the Planetary Bugs Forum or on Github.

## Latest Messages

Oct 25
As you can probably see, the front page of the site is a bit broken right now. I plan to fix it, and improve it a bit in the process, this weekend. There are some other long-standing bugs that need fixing as well, and I hope to get to those as I have time. I have a LOT of ideas about ways to improve the site -- as I'm sure many of you do too. The problem is that we currently have zero budget. Sincere thanks go to everyone making constructive contributions, and thanks as well for your patience. >

Oct 20
The answer should be in the Picard–Lindelöf theorem.

Oct 20
The search engine of PlanetMath is again functioning.>

Oct 11
I've done a few samples using direct summing and it seems that your: 1/(2^2q) should be 1/(2^q) Why don't you try it; I could have the formula feeding the summation wrong. ---------------- binomial(10,5)/(2^10); sum(binomial(2*k,k)*(-1/2)^k*binomial(10,k),k,0,10); (9/2)!/(5!*sqrt(pi)); ---------------------- Answers 63/256 63/256 63/256

Oct 11
Using Maxima's simplify_sum gives for a limit of 2*n ---- $\sum_{p=0}^{2n}\left(\frac{-1}{2}\right)^{p}\binom{2n}{p}\binom{2p}{p}$ $\frac{\left(\frac{2*n-1}{2}\right)!}{\sqrt{(\pi)}\cdot n!}$ Which seems strange until you evaluate -- $\left(\frac{\left(2\cdot4-1\right)}{2}\right)!=\left(\frac{105}{16}\right)\cdot\sqrt{\pi}$ So the sqrt(pi)'s cancel. Remember $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ And for odd n $\sum_{p=0}^{2n+1}\left(\frac{-1}{2}\right)^{p}\binom{2n+1}{p}\binom{2p}{p}$ 0 ---------------------- Maxima also has the Zeilberger algorithm. I will copy the answer here when I understand it. If you feel you need it. -------------------- Incidently simplify_sum doesn't give a proof certificate but Zeilberger does. ---- Ray

Oct 10
Would anybody be interested in discussing a presentation here on using Pascal/Shift matrices to directly generate various Polynomial sequences via. their generating functions expressed in matrices? The underlying idea is that most generating functions are still true when indexing variables (t) are replaced with full rank singular matrices (i.e. n-1). In particular the Pascal or Shift matrices. This leads to the generating function expressed in terms of matrices and most familiar generating functions directly stating the polynomials. Also that the series is truncated automatically and exactly. I do have some theorems rather than just words :) I have a "blog" where I have stuffed some notes and results. Ray>

Oct 4
Attention Dr. Puzio: Kindly see my request to Administration. Would be glad if you would kindly do the needful.

Sep 30
This is to remind administration about my request to either a) enable " copy and paste operation" on the templates reserved for articles and messages or b) open a page on facebook which will automtically enable copying and pasting as well as uploading snapshots of articles and messages.

Sep 28
Hi parag, There is no set theory needed for the proof. I think only the expressions of m and M are strange for you; they simply mean that m is the least and M the greatest of the given fractions! Jussi

Sep 28
I am not able to understand that proof at planetmath.org/summednumeratorandsummeddenominator as I am not aware of the set theory. Please give a simple proof for that.

Sep 28
parag, do you mean such as in http://planetmath.org/summednumeratorandsummeddenominator?

Sep 28
Many of the Gaussian integers indicated in "Fermat's theorem in k(i)" and " pseudoprimes in k(i) " are too big to be copied manually. I, therefore, suggest that the template for articles and messages be modified to enable "copy and paste" operations. Alternately planetmath.org open a page on facebook. This will enable us to upload snapshots of messages/articles.

Sep 23
hi bro, i think this tip will help you : - Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary. EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19. and i found an harshad number digit sum 19: 874

Sep 23
We can construct pseudoprimes in k(i) as follows: Take a couple of primes in k(1) of the same type i.e. both being of type 4m+3 or both being of type 4m+1. Let us take 3*7 = 21 as an example of first type. Choose as base 21 + i. ((21+i)^20-1)/21 yields a quotient which is a Gaussian integer; hence 21 is a psedoprime in k(i). Note a) This is possible only with the aid of software pari or similar. b) 21 + i, 21 - i, 1 - 21i, -1+21i, all yield identical Gaussian integers when we apply Fermat's theorem. b) Second base is chosen as follows: Partition 22 into two parts such that one is divisible by 3 and the other by 7 i.e. 7 + 15i, 7 - 15i, -7 + 15i and -7 - 15i are four different points on the complex plane which yield identical Gaussian integers when Fermat's theorem is applied. Similar remarks apply to composites in k(1) each of which is of type 4m + 1.