<p>So I figured out a formula today and I want to know how original it is, so I’m challenging any math wizzes to try and either figure it out by </p>
<p>a) using my starting hint (actually i saw the hint on collegeconfidential because some kid said it was an easy way to square numbers) and calculating the formula</p>
<p>or</p>
<p>b) find it somewhere on the internet</p>
<p>c) or just say it looks fimiliar and tell me where you’ve seen it crop up before</p>
<p>The problem is: express x^a (x raised to the power of a) using a series within a series.</p>
<p>The starting hint is: </p>
<p>2^2=[sigma(2n+1) from n=0 to n=1]=1+3=4
3^2=[sigma(2n+1) from n=0 to n=2]=1+3+5=9
4^2=[sigma(2n+1) from n=0 to n=3]=1+3+5+7=16</p>
<p>also if possible, let the inner series have use the value “i” and the outer use the value “n”</p>
<p>If no one can figure it out, I might post it more widespread and offer a prize or something, but if people find the answer soon, I’ll feel ashamed for thinking I was so original and brilliant!</p>
<p>Good luck.</p>
<p>send me a private message if you have any questions!</p>
<p>there’s an infinite number of ways to do this. how would you like it done? you require exactly 2 nested summations? you require a linear function within the summations? a finite number of terms?</p>
<p>Im sorry
how about i give you a head start:</p>
<p>if x^2=sigma(2n+1) from n=0 to n=x-1
then x^3=sigma(function of n) from n=0 to n=x-1</p>
<p>we could find that function of n by taking cubes: 1,8,27,64 and then finding a function that describes the differences between them:
so 1=1
1+7=8
1+7+19=27
1+7+19+37=64</p>
<p>so we have the following terms
n=0 f(n)=1,
n=1 f(n)=7
n=2 f(n)=19
…</p>
<p>so take:
1
7
19
…</p>
<p>subtract 1:
0
6
18
…
divide by 3:
0
2
6</p>
<p>add one:
1
3
7
…</p>
<p>add n:
1+0=1
3+1=4
7+2=9
…</p>
<p>take square root:
1
2
3</p>
<p>Go backward to find function of n: square n, subtract n, subtract 1, multiply by 3, add one:</p>
<p>therefore, x^3=sigma(3n^2+3n+1) from n=0 to n=x-1</p>
<p>now try to find x^4 and x^5 and then recognize a pattern here’s where it starts:</p>
<p>x^2=sigma(<strong>2n</strong> +1) from n=0 to n=x-1
x^3=sigma(<strong>3n^2+3n</strong> +1) from n=0 to n=x-1
find x^4
x^5
…
x^a=sigma(sigma((some function of a, i, and n) from i=some value to some value) from n= some value to some value… I hope that helps.</p>
<p>of course I’ve heard of it, and i used it here. but the answer is not “pascal’s triangle” thats just a step, but at least its in the right direction.</p>
<p>Lol it was me who made up a formula. I am going to post it soon… its easier imo than doing the thing like 116^2 = (120-4)^2. And that is not the way I did it. I am still kind of amazed that people havn’t figured it out… actually now that I think about it its just a TAD bit easier than the squaring formula (foil i mean).</p>
<p>All you’ve done is write a polynomial of order a as a summation of polynomials of order one less than a. Good work, but as pebbles notes, that’s pretty trivial
If you somehow have managed to write it as a sum of single un-nested summations of terms that are linear in n, that would be reasonably more impressive</p>
<p>Look, keys88, no matter what your grades are, or what did you invent, they would always say it’s “trivial” or “average”, or “pretty standard”.</p>
<p>But you figured out the formula on your own, without searching the books, and I think this fact deserves some positive feedback. You should definitely mention it on your application</p>
<p>Nobody “made up” this formula. I personally think it is cool that you found it out yourself. The easiest way to demonstrate how the sequence generated by a difference of terms in a sequence of polynomial degree n is to look at:</p>
<p>x^n - (x -1)^n</p>
<p>Clearly the n-degree term cancels, leaving you with an n-1 polynomial as the difference</p>
<p>hey we’re all adults here we don’t need to be patronized for our mediocrity. he wanted to know if anyone could figure it out. clearly, we can. he wanted to know if it was original, clearly, it was not. Is it a well-done proof? There are much more elegant ways to show the relationship, top-down instead of bottom-up. It’s all well and good he figured it out for himself but this is more along the lines of something you’d have to prove on a math problem set here than something that should blow us out of the water. Just tell it like it is. No one is trying to smack anyone down here, but no one needs to have their hands held either. (“AWw, look at wickle math boy, he worked so hard and all these big bad MIT bullies are hurting his emotion critter…give him a gold star for effort!”) You’re just insulting him thinking he can’t handle a straight answer to his questions.</p>
<p>Agreed. I wudnt send something like this to MIT. Especially after they realize what a trivial proof this is. And I agree with the post that said this is something you’d likely find in Psets rather than… a truly original proof. Might as well send them your multivariable calculus hw if you plan on sending this.</p>
<p>Here it is. By the way my feelings aren’t hurt or anything; i never even planned on sending it to MIT anyway!!! (im sending a description of a song I composed about sloths instead!)
I just wanted to see what ideas where out there!! but thanks for the enthusiam guys!</p>
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