/sci/ - Science & Math » Thread #11110553

23KiB, 500x375, 1514227542453.jpg

View Same Google iqdb SauceNAO

Need arXiv endorsement.

Anonymous Fri 01 Nov 20:14:47 2019 No.11110553 View Reply Original Report

Quoted By: >>11110565 >>11110568 >>11110806

Does anyone want to endorse me on arXiv? I found an alternative way to derive the formula for the sum of sequence of cubes, but I don't want to share it because someone might steal it.

I just need an endorsement because surprisingly all my previous profs don't publish.

Anonymous

Anonymous Fri 01 Nov 2019 20:21:53 No.11110565 Report

Quoted By: >>11110603 >>11110612

>>11110553
> I found an alternative way to derive the formula for the sum of sequence of cubes
That's not worth publishing. There are hundreds of ways to sum a sequence of cubes, and nth powers in general. Consider publishing when you've discovered something original.
>but I don't want to share it because someone might steal it.
Fucking kek. Way to out yourself as a moron

Anonymous

Anonymous Fri 01 Nov 2019 20:23:19 No.11110568 Report

Quoted By: >>11110603

>>11110553
vixra it is for you lmao

Anonymous

Anonymous Fri 01 Nov 2019 20:42:01 No.11110603 Report

Quoted By: >>11110614 >>11110620

>>11110568
>vixra
but I don't have a proof for the Riemann Hypothesis that is wrong

>>11110565
There's an infinite number of ways to sum a sequence of cubes, but this is a derivation of the formula rather than using the formula itself, but you're probably right.

Anonymous

Anonymous Fri 01 Nov 2019 20:49:13 No.11110612 Report

Quoted By:

>>11110565
>That's not worth publishing.
Sure it is. It's not a super noteworthy result that belongs in a big journal or anything, but if it's legitimately new and not already known or a rearrangement of something already known (which it probably is) there are plenty of journals devoted to fun little elementary things that would be happy to publish a one or two page note on something like this.

Anonymous

Anonymous Fri 01 Nov 2019 20:50:37 No.11110614 Report

Quoted By:

>>11110603
You know that the sum of cubes is a polynomial, call it
y_x = ax^4 + bx^3 + cx^2 + dx. Then the discrete derivative is just
y_(x+1) - y_(x) = a(4x^3 + 6x^2 + 4x^1 + 1)+ b(3x^2 + 3x + 1) + c(2x + 1) + d
= (4a)x^3 + (6a + 3b)x^2 + (4a +3b + 2c)x + (a+b+c+d)
we want this to be equal to x^3. Comparing the coefficients, we get four equations, so
a=1/4
b= -1/2
c= -1/2 ( 1 - 3/2 ) = -1/2 (-1/2) = 1/4
d= - (a+b+c)= - (1/4 - 1/2 + 1/4 ) = 0
so the sum of cubes from 0 to n-1 is (1/4)n^4 - 1/2 n^3 + 1/4 n^2

Anonymous

Anonymous Fri 01 Nov 2019 20:54:47 No.11110620 Report

Quoted By:

>>11110603
>but this is a derivation of the formula rather than using the formula itself
Share your derivation here.

Anonymous

Anonymous Fri 01 Nov 2019 22:48:16 No.11110806 Report

Quoted By:

>>11110553
write it up and send it to American Math Monthly
don't fuck around with arXiv if you care about the result and want a paper trail that establishes you as the source

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches