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anonymous
 4 years ago
Two interesting relationships.\[\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty\frac1{k^k}\\\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3\]
anonymous
 4 years ago
Two interesting relationships.\[\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty\frac1{k^k}\\\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm working on proving them... it'll happen, one day.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm posting it here so you guys can prove it alongside me. Right now, don't answer it thoughI want to figure it out on my own.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and there is a similar integral \[\int_{0}^{1} x^x \text{d}x=\sum_{k=1}^{\infty } \frac{(1)^{k+1}}{k^k}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh lawd, I'm still behind on the first proof. There was no need to give me another funny identity.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3use the definition of the integral of a function f(x) over an interval [a,b] \[\int_a^bf(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f(a+i\Delta x)\Delta x~~~\text{ where }~~~\Delta x=\frac{ba}n\]and the first one sort of answers itself for the other use induction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0second one is a really interesting identity... and the short and neat answer to this question : Show that for any given positive integer \(n\) there are \(n\) distinct positive integers such that product of them is a complete cube and sum of them is a complete square.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3I am looking for a really awesome visual proof of that identity I saw once, I hope I find it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Exper gave me a link about triangles i cant remember...santosh what was it?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0lol ... i forgot ... what was that related to?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i cant remember \[1^2+2^2+3^2+...+n^2=?\]or\[1^3+2^3+3^3+...+n^3=?\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0oh ... that was just visual proof of ... http://mathoverflow.net/questions/8846/proofswithoutwords

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3http://mathoverflow.net/questions/8846/proofswithoutwords about halfway down this page, the image with the colored square is the proof I was referring to

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3that page is not as pretty

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1345404870788:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0nice geometry ... i never thought they would add up to make a single square again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got the second one before the first... I guess I'm moving onto the identity @mukushla posted.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I got pretty stuck up evaluating this one .. dw:1345406507509:dw I don't understand why Mathematica or Maple doesn't give it's value.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it 0 to \(\infty\)?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0ah yes ... i tried few days back.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well what was the result then?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0no result ... not even approximation. it's pretty obvious where it converges http://www.wolframalpha.com/input/?i=integrate+1%2Fx^x+from+0+to+infinity
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