Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Xeda112358 on December 07, 2010, 06:49:07 pm
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Okee, so it seems that people here like math, oddly enough :D . 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 :D 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 :D
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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.
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I dunno... I have tested some very large coefficients... But yeah, imaginary... Oh noes! I am trying for the sum of x^(-a), now :D It is so much more fun right now!
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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
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Nope, it is right :D
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What you sent me said: X, (X^2+X)/2, (2x^3+3x^2+x)/6,...
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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.
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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
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math * mat * ma * m... m4+a3+t2+h. clearly. this looks interesting.. just looking at it
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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!
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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...
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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:
24,143,145
24,7,25
24,10,26
To test for 24, you do:
24:prgmPYTHAG:L1
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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.
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I can't read it all, but Xeda, it's unfair, you're half a maths teacher XD
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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.
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I can't read it all, but Xeda, it's unfair, you're half a maths teacher XD
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.
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Okee, here is some food for thought.
(7x-2) mod 9=0 when x=9K-1
K is an integer
This means that all number that can be expressed as 7(9k-1)-2 can always be divided evenly by 9. This isn't really complex, but it is fun if you like prime numbers
So, when:
x=9k-1 (7x-2) mod 9=0 (7x+7) mod 9=0
x=9k-2 (7x-4) mod 9=0 (7x+5) mod 9=0
x=9k-3 (7x-6) mod 9=0 (7x+3) mod 9=0
x=9k-4 (7x-8) mod 9=0 (7x+1) mod 9=0
x=9k-5 (7x-1) mod 9=0 (7x+8) mod 9=0
x=9k-6 (7x-3) mod 9=0 (7x+6) mod 9=0
x=9k-7 (7x-5) mod 9=0 (7x+4) mod 9=0
x=9k-8 (7x-7) mod 9=0 (7x+2) mod 9=0
When I did this, I learned some things about how numbers work. If you wanna have fun, try using values other than 7 and 9. Something this simple can reveal quite a bit about numbers.
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Okee, here is some food for thought.
(7x-2) mod 9=0 when x=9K-1
K is an integer
This means that all number that can be expressed as 7(9k-1)-2 can always be divided evenly by 9. This isn't really complex, but it is fun if you like prime numbers
So, when:
x=9k-1 (7x-2) mod 9=0 (7x+7) mod 9=0
x=9k-2 (7x-4) mod 9=0 (7x+5) mod 9=0
x=9k-3 (7x-6) mod 9=0 (7x+3) mod 9=0
x=9k-4 (7x-8) mod 9=0 (7x+1) mod 9=0
x=9k-5 (7x-1) mod 9=0 (7x+8) mod 9=0
x=9k-6 (7x-3) mod 9=0 (7x+6) mod 9=0
x=9k-7 (7x-5) mod 9=0 (7x+4) mod 9=0
x=9k-8 (7x-7) mod 9=0 (7x+2) mod 9=0
When I did this, I learned some things about how numbers work. If you wanna have fun, try using values other than 7 and 9. Something this simple can reveal quite a bit about numbers.
Modular arithmetic, it was thought to me and I came up with two independent results:
- you can use modular arithmetic to find divisible criteria like you did, here is a simple example:
let n be a integer number in decimal form
let d(i) be the sequence of its digits in decimal form, i. e., n=sum from i=0 to floor(log(n))+1 of d(i)*10^i or n can be expressed as d(0)+d(1)*10+d(2)*10^2+...+d(n)*10^n
then you have
3 divides n if
3 divides sum d(i)*10^i, that is equivalent to
sum d(i)*10^i = 0 mod 3 <=>
sum d(i)*1^i = 0 mod 3 <=>
sum d(i) = 0 mod 3
so,
3 divides n if the sum of digits of n is divisible by 3
Exercises: do the same for 2, 10, 7 or any number you like.
- I noted what happened to (x+1)^n mod n when n is prime
tips: Pascal triangle, divisibility
Note I don't know how to write mathematics in english conventions, I did the best to translate... Also writing in plain text does not help.
Sorry if you don't understand but I wrote this quickly.
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That is okay, I do not know how to properly "write math" yet. But yeah, my goal was to use that to test for primes. I'll look into this when my brain is functioning again, but I think I understand :D
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That is okay, I do not know how to properly "write math" yet. But yeah, my goal was to use that to test for primes. I'll look into this when my brain is functioning again, but I think I understand :D
Incredible how some people come up results already known, independently on math after learning the start of things.
And you are on a possible right track, although it gets complicated and you need results like Little Fermat Theorem and Euler's Theorem, many prime testing methods use congruences ideas. :)
I am curious. Are you thinking of following Mathematics to your superior course and career, or keep just as a hobby and possibly work on something related? Remember you are not obligated to answer or reply fast.
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I am a math major in college and I do plan to go as far as to get a doctorate. I am not 100% sure, but I am fairly certain I will choose a math oriented career such as being a professor.
*By the way, Galandros, I'm Thunderbolt from UTI :D
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I am a math major in college and I do plan to go as far as to get a doctorate. I am not 100% sure, but I am fairly certain I will choose a math oriented career such as being a professor.
*By the way, Galandros, I'm Thunderbolt from UTI :D
Good luck then. :)
I know because the projects you show off here match those of ThunderBolt in UTI. :P
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I am a math major in college and I do plan to go as far as to get a doctorate. I am not 100% sure, but I am fairly certain I will choose a math oriented career such as being a professor.
*By the way, Galandros, I'm Thunderbolt from UTI :D
Professor like University Professor? (In my country you don't choose if you want to be a professor, you're invited)
Professor like Teacher? That sounds nice.
Also, I think that a Professor (in all ways) is better than becoming a teacher.
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I would love to be a teacher for other things, but I would prefer to be a professor for math.
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Does anybody know where I can find a list of all numbers that are part of three pythagorean triples? My poor calculator found 24 and 840 but it didn't find any others (even at around 10000). I will probably end up trying to create an equation for that, but I want to make sure it isn't just a simple rounding error that is causing me to miss other such numbers.
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(m^2-n^2, 2mn, m^2+n^2) will generate all the pythagorean triples. For instance, m=2 n=1 gives (3,4,5).
Also, the set of all numbers that are part of a pythagorean triple isn't terribly interesting; all numbers larger than 2 are part of a pythagorean triple :P (3,4,5); (5,12,13); (7,24,25); etc. for odd numbers, even numbers just double these triples.
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I beleive Zeda wants to know how many numbers are in 3 different Pythagorean triplets
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Oh... misinterpreted, sorry.
Well, I suppose you could do individual tests on m^2-n^2=N, 2mn=N, m^2+n^2=N...
In particular, the first two are factorings of N.
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i would just right a C program with an overflow at like 5 million run it and see if there are any more
See is fast so it shouldn't take too long if you use a few optimizations
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I'm still wondering though... even given the new conditions, we should still expect to see a LOT more than just 2 numbers in the first 10,000.
For example, 60 should work with (36, 48, 60); (45, 60, 75); (60, 80, 100).
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It might also depend on whether the number must be in 3 triples, or exactly 3 triples, since then 60 wouldn't work (60,63,87 and 60,448,452, for instance would make it at least 5).
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24: (24,32,40), (18,24,30), (24,70,74), (7, 24, 25)...
840: see 24 example but multiply by 35
So it's not exactly 3.
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Ah, sorry I forgot a few things... While I was taking the Putnam exam I decided to prove that all numbers greater than two were part of a Pythagorean triple. To prove it I came up with an equation that given the length of one leg would return a length of another leg and then I made another that returned a possible length for the hypotenuse.
So what I meant to ask was if there was a list of all numbers that are the leg lengths of three Pythagorean. Here are the equations I came up with:
8b=(3a2-6)-(-1)a(a2+2)
8c=(3a2+6)-(-1)a(a2-2)
EDIT: Here are the ones I came up with on the Putnam:
Even values of 'a'
b=a2/4-1
c=a2/4+1
Odd values of 'a'
b=(a2-1)/2
c=(a2+1)/2
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Oh, you're limiting yourself to primitive pythagorean triples only? (so [6,8,10] is invalid because it is simply 2x[3,4,5] ?)
Also, is it three exactly, or at least three?
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Three exactly. I do not believe there are more than 3 with those strictures.
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Hm... are you counting certain triples like [20,21,29]? Not sure if they're fitting in the current description.
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Yes, that should count... is there something special about it that I am not seeing or is it part of two other triples?
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Currently, you're using the following:
Even values of 'a'
b=a^2/4-1
c=a^2/4+1
Odd values of 'a'
b=(a^2-1)/2
c=(a^2+1)/2
which means you're assuming that for odd a, b and c are 1 apart; for even a, b and c are 2 apart. This works usually, but sometimes fails. (e.g. a=20, c-b=8, not 2)
As mentioned a bit earlier in this thread, the generalized Pythagorean triple can be written in the form (m^2-n^2,2mn,m^2+n^2). If you're looking at only primitive triples, then m,n should be coprime.
Note that the first leg is (m+n)(m-n), so factoring should make this task easy.
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Yes, I know about the equations, I was just wondering if there was a proven list I could compare against for what I am working on.
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Using your definition: http://oeis.org/A111766
EDIT: Notice that there's a closed form!
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Aha! Thank you♥! I knew there had to be one between 24 and 840 ♥ Thank you ♥
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Necro update!
So, if you look at the image attached, that is a formula I was working on for the past few years. I made a rough estimate of how many pages of notes I put into this problem and it was over a hundred :D Well, guess what? In the time that I have worked on it, I was able to prove the formula worked and that it extended to any function that you can find derivatives and integrals for (and since you can use Fourier analysis to find a function for just about anything, you are good for just about anything). However, I was still having doubts that it was original. Well, last week I was looking up something only slightly related when I noticed a particular sequence:
1
1/2
1/6
-1/30
1/42
-1/30
At this point, I got excited. I know these numbers! Then the next term: 5/66. I was ecstatic! These were the coefficients to my function and they are known as the Bernoulli numbers. See, the issue is that when I was writing down the numbers, I computed the coefficients, but when you look them up on the internet and whatnot, most people simply refer to them as Bn. Not very helpful for me. On top of that, my function is a generalisation of the Euler-Macluarin formula which cannot be google searched very easily (unless you know the name) and most people don't know of it or have never worked with it! So what have I done with it in the past few months?
I have made some pretty cool identities including one dealing with the coefficients and "e" (so now I have gone back and rewritten e in terms of the Bernoulli numbers). I now have two representations for sum(sin(x),x,1,n) which can be compared and I can derive more relationships (one of them uses imaginary numbers, the other uses sine, cosine, and tangent. We know where this is going :P Another proof of Euler's identity). But more importantly to me, I can apply this to analysing the Riemann Zeta function which I did a bunch of research on over the summer (and I've even presented about it a few times).
I also use this quite a bit in my statistics class and it has been very useful. For the curious, any polynomial with positive integer powers will result in a finite closed form expression. Any polynomial using exponentials, trig functions, or non-integer powers will have an series with infinitely many terms. Luckily, it converges pretty rapidly to the desired sum.
If you want to have fun, look at "identity 0" below. If you replace a with e, all of those ln(a) terms get cancelled to 1 :D
Also, you can use this to derive that the "first Bernoulli numbers" divided by the "second Bernoulli numbers" (the only difference is that B1 is negative in the first, positive in the second) will equal e :)
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:crazy:
Uhh. Must've been a ton of work to make up that formula.
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More randomness here :) Basically, I was trying to evaluate pi when I was in 8th grade, and I found some notes on it. After updating the syntax a little (I had not been exposed to limits at that point), I realised a neat little identity and I tried to prove it :) This reveals an interesting thing about the limit of sin(x) as x approaches zero at an exponential rate compared to the limit of 2^x as x approaches zero. Anyways, another neat fact that this revealed to me is that if you divide pi by the nested radical (see the attached image, at the bottom), it will converge to a power of 2. For example:
pi/sqrt(2) is about 2.22144 (21=2)
pi/sqrt(2-sqrt(2)) is about 4.104688612 (22=4)
pi/sqrt(2-sqrt(2+sqrt(2))) is about 8.051636 (23=8)
pi/sqrt(2-sqrt(2+sqrt(2+sqrt(2)))) is about 16.025731 (24=16)
pi/sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(2))))) is about 32.01285466 (25=32)
Anyways, take a look at the proof :) I devised it when I was trying to show that it did in fact converge :)
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I don't know if I'm understanding right or wrong, but in the last part, shouldn't it be 2^(x+1) instead of 2^(x-1) ?
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No, I took out the /2. 2x/2=2x*2-1=2x-1
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oh. what was I thinking? XP
I was thinking of multiplying 2 by "both" sides :x/me hides
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2^15 /= 32. 2^5 = 32 xD
Little typo....
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Hehe, thanks, I just fixed it XD
EDIT: Oh, and I will at some point have to make a post about this and how it works:
(http://img.removedfromgame.com/imgs/sine%20program.gif)
I also extended it to other denominators recently, not just powers of two (though those are a bit easier). With other denominators, it has the interesting ability to compute all the smaller integers relatively prime to the numerator.
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As a little update on my most recent proof, I spent the last half hour of so on my statistics exam generalising the proof further. Instead of using 2x, you can use any function that diverges to infinity as x goes to infinity. And to further generalise, given that lim x→infinity of g(x) goes to infinity, then lim x→infinity of g(x)sin(k/g(x))=k. However, using g(x)=2x reveals the neat identity with the nested radicals :)
Also, g(x)tan(k/g(x))=k as x→infinity. It converges to k from above, whereas the sine version converges from below. The reason is simple : g(x)tan(k/g(x))=g(x)sin(k/g(x))/cos(k/g(x)). As x→infinty, this converges to g(x)sin(k/g(x))/1=g(x)sin(k/g(x)) XD
I also created a really complicated looking problem that evaluated to the partial sum of k2, too, just for fun. I used:
lim a→inf sum(e2k+1sin2(k*e-2k),k,a,n+a)
Hehe ^_^
EDIT: In the last equation, I had to replace the 'x' with 'k' :P Also, a prettier image is attached.
EDIT2: In the picture, pretend I put the 2 symbol for sin2() >.>