Author Topic: Math! (and that is not a factorial) (Read 17927 times)

Xeda112358 · « **on:** December 07, 2010, 06:49:07 pm »

Okee, so it seems that people here like math, oddly enough

. So... I have my.. erm... "notes" if you will. Some of this is so out there and abstract that it makes minds explode (that has only happened 2 in 7 times, so the odds are in your favor!) Some of this also is a way to attack the Riemann Hypothesis, so besides the fact that I believe in the development of knowledge and understanding, I should probably not be releasing some of my notes. However, it is pretty well known that I am indeed attacking these areas of math, so I am willing to give you all some food for thought. Also, if you guys have any equations or anything that you have come up with, please feel free to post. I would prefer if this topic was limited to original ideas (of course, those ideas will need non-original ideas to explain, I am sure), but whatever, math is math is fun is fun. So, here are my notes and y'all better ask questions fast because when I leave Friday, I will be home where internet access is very severely limited. I will be gone until around January 16 or 17 of 2011. Graphmastur so far has a pretty good grasp of my main algorithm, but there is sooooo much more to it that I need to be reminded to explain (like the fact that the ratio of the coefficients seem to converge to 19/750) Anywho, here goes (and sorry for my hand writing

I don't exactly meet the requirements for certain stereotypes). By the way, I have led a fairly sheltered life, so there are many notations that I have come up with, not knowing that notation already existed. I have been using proper notation as I learn it, but I cannot guarantee that this all will exactly fit your styles or tastes in "writing" math

jnesselr · « **Reply #1 on:** December 07, 2010, 07:01:04 pm »

Looks interesting. What all haven't you told me of this algorithm? Answer in pm if you'd like. Also, it's likely not 19/750, but something else, and it just appears that way. We shall see.

I'm curious about if you applied it to complex numbers.

Xeda112358 · « **Reply #2 on:** December 07, 2010, 07:03:20 pm »

I dunno... I have tested some very large coefficients... But yeah, imaginary... Oh noes! I am trying for the sum of x^(-a), now

It is so much more fun right now!

AngelFish · « **Reply #3 on:** December 07, 2010, 07:05:48 pm »

Xeda, I think you made a mistake in p₃. It should read (2x^3+3x^2+x)/6, not (x^3+3x^2+x)/6

Xeda112358 · « **Reply #4 on:** December 07, 2010, 07:08:13 pm »

Nope, it is right

AngelFish · « **Reply #5 on:** December 07, 2010, 07:09:20 pm »

What you sent me said: X, (X^2+X)/2, (2x^3+3x^2+x)/6,...

jnesselr · « **Reply #6 on:** December 07, 2010, 07:11:04 pm »

Quote from: Qwerty.55 on December 07, 2010, 07:09:20 pm

What you sent me said: X, (X^2+X)/2, (2x^3+3x^2+x)/6,...

Let's see, going by the coeffs, (2x^3+3x^2+x)/6 is correct. Unless this is a different algorithm. In which case, I don't know.

Also, xeda, I can make your algorithm work for U=x^-a. I think.

Xeda112358 · « **Reply #7 on:** December 07, 2010, 07:13:04 pm »

It is the sum of the sum of 1, so the equation is (x^3+3x^2+2x)/6
sum of 1=x
sum of x=(x^2+x)/2
sum of (x^2+x)/2=(x^3+3x^2+2x)/6

nemo · « **Reply #8 on:** December 07, 2010, 07:23:40 pm »

math * mat * ma * m... m⁴+a³+t²+h. clearly. this looks interesting.. just looking at it

Xeda112358 · « **Reply #9 on:** December 07, 2010, 08:45:24 pm »

oooh, what is the use of taking a number in some base "x" like 1.33.42.1.36 and doing:
(1.33.42.1.36)*(33.42.1.36)*(42.1.36)*(1.36)*(36)
oooooooh
that looks very fun indeed, especially when you do a.b.c.d.e and you get:
(a.b.c.d.e)*(b.c.d.e)*(c.d.e)*(d.e)*(e)

That looks like it could use a little Pascals Triangle!
...
Which means that if we were very extremely bored, we could extend that to the summations algorithm!

I bet this has a use, like in statistics or number theory...

Xeda112358 · « **Reply #10 on:** December 09, 2010, 11:34:05 pm »

Oooh, I was leafing through my notes and I found something else I did two months ago that might be of interest... The sum of (x^a)-((x-1)^a) from 0 to n is (n+1)^a.
What people do not notice off the bat is that it is taking the sum of a degree n-1 polynomial and results in a degree n polynomial, just like my previous sum.
Also, I have proven this through induction, so I can express x^a as a power series.

On a completely different, yet slightly related, note, during the Putnam math competition, I came up with a way to a) prove that any integer >2 was part of at least one Pythagorean triple and b) I found a way to find which Pythagorean Triple(s) it belonged to. I uploaded this here program which I haven't even tried to optimize. All it does is, given the length of one leg, it outputs a list of the possible lengths of the other leg for a Pythagorean Triple. Just do (A^2+L1^2) for the C^2 values. For example, trying 24 will result in a list {143,7,10}.. So the Pythagorean Triples including 24 are:

Code: [Select]

24,143,145
24,7,25
24,10,26

To test for 24, you do:

Code: [Select]

24:prgmPYTHAG:L1

Xeda112358 · « **Reply #11 on:** January 17, 2011, 12:43:19 pm »

Okee, as a mini trick say you have X^2+2X+1 and you want to find the square root of it. Replace the X's with 10's and you get 121. Find the square root of that and you get 11. 11 is 10+1, so replace the 10 with X and you get x+1.

Here is what is really going on:
121=1*10^2+2*10^1+1
X^2+2x+1 is the same thing if X=10.
So really you are only changing the base of the number

So if you want to find the square root of x^2+8X+16, you get 100+80+16=196. sqrt(196)=14=(X+4). Now what if you wanted to find the square root of 169? If you are bored, you can do:
169=10^2+6*10+9=x^2+6x+9

0=x^2+6x+9
-9=x^2+6x
9-9=(x+3)^2
0=(X+3)^2
0=X+3
10+3=13 is the square root

So what happens if I do 187...?
0=x^2+8x+7
-7=X^2+8x
16-7=(X+4)^2
9=(X+4)^2
STOP
x+4=14
187 is 9 away from 14^2

As a note, the second digit has to be even, so if you have 196, take 1 from the 9 and add 10 to the 6

Also, if I had a copy of my notes, I would show y'all what I am calling the "optimal base" for numbers. For numbers like 225, you need to do 100+100+25, not 100+120+5. The optimal base is a special number base where every number is a three digit number with the first being 1. There is an equation for it ( have it in an excel spreadsheet) and 196 is the only number that fails. Interestingly, if you check the number 196 on Wolfram.com, you will find that 196 is a very interesting number and has some unique properties that might tied into this.

Munchor · « **Reply #12 on:** January 17, 2011, 03:48:56 pm »

I can't read it all, but Xeda, it's unfair, you're half a maths teacher

Xeda112358 · « **Reply #13 on:** January 17, 2011, 04:24:05 pm »

Yeah, sorry. I wish I had a blackboard to work with and I could give all the info in real time. People would be able to actually read and understand some stuff.

jnesselr · « **Reply #14 on:** January 17, 2011, 05:05:46 pm »

Quote from: Scout on January 17, 2011, 03:48:56 pm

I can't read it all, but Xeda, it's unfair, you're half a maths teacher

See, in actuality, it's a bunch of math work to make fun of the non-math people. j/k

Impressive work as always Xeda.

Author Topic: Math! (and that is not a factorial) (Read 17927 times)

Xeda112358

Math! (and that is not a factorial)

jnesselr

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

AngelFish

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

AngelFish

Re: Math! (and that is not a factorial)

jnesselr

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

nemo

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

Munchor

Re: Math! (and that is not a factorial)

Xeda112358

Re: Math! (and that is not a factorial)

jnesselr

Re: Math! (and that is not a factorial)