## Junior Member

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Showing Visitor Messages 1 to 5 of 5
1. praise_Jesus wrote
at 8:26 am Jun 15, '09
Ah no problem. It's not as hard as it sounds. If you have any questions feel free to ask me. I'll be able to help.
2. praise_Jesus wrote
at 8:42 am Jun 12, '09
It doesn't involve overly impossible combinatorics (which was good) but you'll want to be familiar with n choose k, Pascal's triangle, permutations, combinations, and of course n factorial. And then the principle of inclusion exclusion. Thanks for the find though... it creates a sort of alternating pascal's triangle in terms of the leading coefficients to the powers I may add an additional proof that is a simple generalization of the first part.
3. praise_Jesus wrote
at 10:48 pm Jun 11, '09
The proof is finally up. I started a forum thread on the group for discussion about the proof if you have any questions. It's not extremely pretty looking so hopefully you can read through it, but do ask questions I'd be happy to explain. Oh and be sure to check the captions because the pages appear out of order but I made sure the captions reflected the proper page.
4. praise_Jesus wrote
at 8:23 am Jun 10, '09
I have a proof relating to your find with the differences of consecutive powers. I will post it in the next day or so once I get it scanned and I'll put it in the forums.

Are you familiar with any combinatorics? The proof relies on combinatorics. And uses Pascal's Identity, and the Principle of Inclusion and Exclusion. Anyways expect to see it up there soon.
5. praise_Jesus wrote
at 10:22 pm Apr 26, '09
I posted an answer to your 0.9999... = 1 question on the wall in mathematics

Biography
Math Nerd :D and proud of it.
Religion
Catholic
Interests
MATH! What else? Reading, I guess...
• Signature
LIFE begins at CONCEPTION! (when the zygote has its own unique DNA pattern...)
PRO-LIFE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!:

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