/sci/ - Science & Math

[Anonymous No.5184887]

I was playing around with the serie <span class="math">\sum_{k=0}^n k^2[/spoiler] and came up with the interesting identity <span class="math">\sum_{k=0}^{2n} k^2 = 2 n^2 \left ( n+\frac{1}{2} \right )+2 \sum_{k=0}^n k^2[/spoiler] (I can show you how I did it but it will take some time to write everything even though it's very easy)... So I wondered if I could make an approximation of <span class="math">\sum_{k=0}^{2n} k^2[/spoiler] from this by substituting <span class="math">2*\sum_{k=0}^{n} k^2[/spoiler] every time and of course you can.. So I did it and there are two approximations I came up with <span class="math">\sum_{k=0}^{2n} k^2=\sum_{k=0}^{\infty}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right ) AND \sum_{k=0}^{2n} k^2=\sum_{k=0}^{n}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )+1[/spoiler] and as I expected <span class="math">\sum_{k=0}^{2n} k^2=\sum_{k=0}^{n}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )+1[/spoiler] was much more close to the real answer by the other equation seemed to have more in common with the equation resulting from <span class="math">\sum_{k=0}^{2n}\left (k^{2}\right )[/spoiler] which is <span class="math">\sum_{k=0}^{2n}\left (k^{2}\right )= \frac{1}{3} n \left (2 n+1\right ) \left (4 n+1\right )[/spoiler]

/Continuation in the next post because field is too big to post

[Anonymous No.5184889]

/Continuation is below

And so I examined them and with a little bit of algebra, I got <span class="math">\sum_{k=0}^{2n}\left (k^{2}\right )-\sum_{k=0}^{\infty}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )=\frac{n}{3}[/spoiler].... Now my question is... Why is that? I mean how could you get <span class="math"> ]\sum_{k=0}^{2n}\left (k^{2}\right )=\sum_{k=0}^{\infty}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )+\frac{n}{3}[/spoiler] from the first time without having to expand both equations and subtract them from each other to get because my primary goal was to solve <span class="math">\sum_{k=0}^{2n}[/spoiler].. So if I knew where the <span class="math">frac{n}{3}[/spoiler], it would be tremendously helpful.
Note: I think it goes without saying that this isn’t homework and it’s simply for the math. Besides, nobody gives such homework where ever you go in the world.

[Anonymous No.5184890]

>>5184887
Why the hell is everything on the right of the page as if I was writing in Japanese or Arabic?

[Anonymous No.5184902]

>>5184890
It looks fine to me.

[Anonymous No.5184904]

>>5184887
Right version of OP:

I was playing around with the serie <span class="math">\sum_{k=0}^n k^2[/spoiler] and came up with the interesting identity <span class="math">\sum_{k=0}^{2n} k^2 = 2 n^2 \left ( n+\frac{1}{2} \right )+2 \sum_{k=0}^n k^2[/spoiler] (I can show you how I did it but it will take some time to write everything even though it's very easy)... So I wondered if I could make an approximation of <span class="math">\sum_{k=0}^{2n} k^2[/spoiler] from this by substituting <span class="math">2*\sum_{k=0}^{n} k^2[/spoiler] every time and of course you can.. So I did it and there are two approximations I came up with <span class="math">\sum_{k=0}^{2n} k^2=\sum_{k=0}^{\infty}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right ) AND \sum_{k=0}^{2n} k^2=\sum_{k=0}^{n}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )+1[/spoiler] and as I expected <span class="math">\sum_{k=0}^{2n} k^2=\sum_{k=0}^{n}\left ( 2^{k+1}\left ( \frac{n}{2^k} \right )\left ( \frac{n}{2^k}+\frac{1}{2} \right ) \right )+1[/spoiler] was much more close to the real answer by the other equation seemed to have more in common with the equation resulting from <span class="math">\sum_{k=0}^{2n}\left (k^{2}\right )[/spoiler] which is <span class="math">\sum_{k=0}^{2n}\left (k^{2}\right )= \frac{1}{3} n \left (2 n+1\right ) \left (4 n+1\right )[/spoiler]
/Continuation in the next post because field is too big to post

[Anonymous No.5184906]

>>5184890
what the fuck are you smoking and where can I get some?

captcha: WORSHIP nagsru

[Anonymous No.5184910]

>>5184902
It's all pushed to the right on my screen. If every body sees it okay, I will delete >>5184904.
But thanks for telling me that it was okay on your screen so I could see what's possibly wrong with my browser.

[Anonymous No.5184915]

>So I wondered if I could make an approximation of <span class="math">\sum_{k=0}^{2n}k^2[/spoiler] from this by substituting <span class="math">2*\sum{k=0}^{n}k^2[/spoiler] every time and of course you can.

I don't understand, what does this mean?

[Anonymous No.5184918]

>>5184915
*<span class="math">\sum_{k=0}^n k^2[/spoiler], obviously.

[Anonymous No.5184925]

>>5184915
>2*\sum{k=0}^{n}k^2
I apologize. I wrote 2*\sum{k=0}^{n}k^2 instead of 2*\sum_{k=0}^{n}k^2. Here is what I meant <span class="math">2*\sum_{k=0}^{n}k^2[/spoiler]

[Anonymous No.5184943]

>>5184925
>Here is what I meant
And that's the tex that's displayed. The weird sum in >>5184915 was just a typo.

I'm confused by what you're trying to communicate.

[Anonymous No.5184978]

>>5184943
What I'm trying to say is that you can substitute <span class="math">2*\sum_{k=0}^{n}k^2[/spoiler] by <span class="math">2*\left (2\left (\frac{n}{2} \right)^2\left ( \frac{n}{2}+\frac{1}{2} \right )+2*\sum_{k=0}^{\frac{n}{2}}k^2 \right)[/spoiler] and then you can keep substuting it for infinity. That's my reasoning. I can explain more if I didn't explain it right.

[Anonymous No.5185000]

>>5184978
Anybody can help me and explain to me where the <span class="math">\frac{n}{3}[/spoiler]

[Anonymous No.5185115]

Any body good enough in math to help, please?
>>5184943
And to you. Here is a wolframalpha test for the equations : bit(doot)ly/UyDHAz
It proves it's true. Think about it a little bit if you still don't understand it.

[Anonymous No.5185144]

Somebody please explain the <span class="math">frac{n}{3}[/spoiler]. You don't even need any special math knowledge for this, just logic.

[Anonymous No.5185147]

Somebody please explain the <span class="math">\frac{n}{3}[/spoiler]. You don't even need any special math knowledge for this, just logic.

[Anonymous No.5185477]

Bump.