The More You Know

[i- !LoVeEKpnh2 No.2474779]

In case you were wondering,
<span class="math">\sum_{i=1}^x i^{23} = \frac{1}{24}x^{24}+\frac{1}{2}x^{23}+\frac{23}{12}x^{22}\frac{-1771}{120}x^{20}+\frac{4807}{36}x^{18
}\frac{-81719}{80}x^{16}+\frac{37145}{6}x^{14}\frac{-5133439}{180}x^{12}+\frac{572033}{6}x^{10}\frac
{-17386919}{80}x^{8}+\frac{11098351}{36}x^{6}\frac{-28112371}{120}x^{4}+\frac{854513}{12}x^{2}[/spoiler]

You're welcome.

[i- !LoVeEKpnh2 No.2474805]

Fucking fractions and their stupid signs. Let's try...

<span class="math">\frac{1}{24}x^{24}+\frac{1}{2}x^{23}+\frac{23}{12}x^{22}-\frac{1771}{120}x^{20}+\frac{4807}{36}x^{18
}-\frac{81719}{80}x^{16}+\frac{37145}{6}x^{14}-\frac{5133439}{180}x^{12}+\frac{572033}{6}x^{10}-\fra
c{17386919}{80}x^{8}+\frac{11098351}{36}x^{6}-\frac{28112371}{120}x^{4}+\frac{854513}{12}x^{2}[/spoiler]

[Anonymous No.2474807]

Ntripfagfilter++;

[Belisaurius No.2474818]

OH GOD MY BRAIN IS BLEDING

[i- !LoVeEKpnh2 No.2474819]

<span class="math">\frac{1}{24}x^{24} + \frac{1}{2}x^{23} + \frac{23}{12}x^{22} - \frac{1771}{120}x^{20} + \frac{4807}{36}x^{18} - \frac{81719}{80}x^{16} + \frac{37145}{6}x^{14} - \frac{5133439}{180}x^{12} + \frac{572033}{6}x^{10} - \frac{17386919}{80}x^{8} + \frac{11098351}{36}x^{6} - \frac{28112371}{120}x^{4} + \frac{854513}{12}x^{2}[/spoiler]

[Anonymous No.2474836]

horribly inacurate calculation of spacetime
what a waste of time

[i- !LoVeEKpnh2 No.2474864]

>>2474807
>>>/lit/

[Anonymous No.2474874]

Now to something which actually has implications: <div class="math">\frac{x^n-y^n}{x-y}=\sum _{k=1} ^n x^{n-k} y^{k-1}.</div>

[Anonymous No.2474904]

>>2474874
See, this is what I don't understand about math. What the fuck does this mean? Do the letters stand for something or do you plug in numbers where they are, and if you do that, how do you know what numbers to plug in? Is it just a random guessing game or something?

[i- !LoVeEKpnh2 No.2474908]

>>2474874
Actually this is just a quick offshoot of code I wrote to generate interpolating polynomials, given a set of discrete data. It's fairly trivial to use it to generate closed-form sums of powers. Thought someone might find it intredasting, even though it is pretty worthless overall.

[Anonymous No.2474938]

>>2474904
I forgot to add <span class="math">x,y \in \mathbb{C}, x \neq y[/spoiler], i.e. x and y can be any two complex numbers which are not equal.

>>2474908
Yea, I just felt like sharing and took whatever opportunity presented itself. Did you obtain the result by iteration? Getting the solution for <span class="math">i^{n-1}[/spoiler] and then plugging it into <span class="math">i^{n}[/spoiler]?

[i- !LoVeEKpnh2 No.2474984]

>>2474938
Used some recent math I picked up, discrete calc / finite differences. Really cool. I was already working on doing polynomial math for other reasons and it just fell into my lap.

[Anonymous No.2474992]

>>2474779
OP has to prove it or they are banished from this board for ever.

[Anonymous No.2475003]

>>2474908

That relation is also used to show that holomorphic functions have holomorphic derivatives. :D

[Anonymous No.2475029]

>>2474874
This looks... Fibonacious

[i- !LoVeEKpnh2 No.2475034]

>>2474992
If this thread is still alive later tonight when I have time (at work now) I could post the methodology.

[Anonymous No.2475058]

>>2475003
Indeed, that's where I picked it up.
<div class="math">\frac{x^n-y^n}{x-y}=\sum _{k=1} ^n x^{n-k} y^{k-1} - nw^{n-1} = \sum _{k=1} ^{n-1} k x^{k-1} y^{n-k-1}.</div>

[Anonymous No.2475068]

>>2475058
Wow, I messed that up
<div class="math">\frac{x^n-y^n}{x-y} - ny^{n-1}=- ny^{n-1} + \sum _{k=1} ^n x^{n-k} y^{k-1} = (x-y)\sum _{k=1} ^{n-1} k x^{k-1} y^{n-k-1}.</div>

[i- !LoVeEKpnh2 No.2477106]

A little drunk now, forgive me for stupidity.

Finite derivatives:
<span class="math">\delta_x f(x) = f(x+1) - f(x)[/spoiler]
Falling powers: (I don't know how to do underlines so...)
<span class="math">x^{\downarrow m} = x(x-1)\cdots(x-m+1)[/spoiler] eg. <span class="math">4^{\downarrow 3} = 4(4-1)(4-2)[/spoiler].

Using algebra and the derivative definition you can then show that, similar to the power rule of analysis,
<span class="math">\delta_x x^{\downarrow n} = nx^{\downarrow n-1}[/spoiler].

So, suppose you have a series of values, <span class="math">f(0), f(1), \ldots[/spoiler]. Then you can use these to calculate the discrete derivative until you are left with just a constant, and then reintegrate.

(more coming)

[i- !LoVeEKpnh2 No.2477149]

So, suppose we have <span class="math">f(0) = 4, f(1) = 7, f(2)=12[/spoiler] and we wish to interpolate this, and we forgot the stupid lagrangian interpolation shit. Well, we know we want to fit this with a second order, so, calculate the discrete derivative until we're left with just a constant, then reintegrate.
4 7 12
goes to
3 5 (notice it is f(x+1)-f(x) as mentioned)
goes to
2.

Now we reintegrate
<span class="math">2 \rightarrow 2x^{\downarrow 1} + 3[/spoiler]
<span class="math">2x^{\downarrow 1} + 3 \rightarrow x^{\downarrow 2} + 3x^{\downarrow 1} + 4[/spoiler]
<span class="math">x^{\downarrow 2} + 3x^{\downarrow 1} + 4 = x(x-1) + 3x + 4 = x^2 + 2x + 4[/spoiler] which, if you check, indeed interpolates the three points given. (2 pts determine a line, 3 a parabola, etc)

[Anonymous No.2477180]

What the fuck is going on in here.

[Anonymous No.2477189]

>>2477180
OP posted a probably-made-up formula without a full, detailed proof and will be b& shortly if he doesn't come up with one.

[Anonymous No.2477193]

>>2477180
Mathematics. Now go play with your toys, men are talking.

[i- !LoVeEKpnh2 No.2477228]

As the discrete integral is just a kind of summation, we can use this interpolation by integrating one more time to find an explicit formula for our summation.

So, for instance, suppose we wanted to find the sum of n cubes. We calculate out our preliminary values and whip out our difference table
1 9 36 100 225 441 784
8 27 64 125 216 343
19 37 61 92 127
18 24 30 36
6 6 6 ...

Suddenly we see that this derivative is a constant. So, let's integrate, taking the constant of integration, as before, as the leftmost value.
<span class="math">6 \rightarrow 6x+18[/spoiler]
<span class="math">6x+18 \rightarrow 3x^{\downarrow 2} + 18x + 19[/spoiler]
etc. When you have reached the top column, you integrate just one more time to serve as the sum, with a constant of integration of zero, and you have
<div class="math">\sum_{i=1}^n i^3 = \frac{1}{4}x^{4} + \frac{1}{2}x^{3} + \frac{1}{4}x^{2}</div>

This can be easily put into code to come up with the closed form sum of the OP, which would be way too tedious by hand.

[i- !LoVeEKpnh2 No.2477261]

For much better details, please see
"The Finite Difference Calculus and Applications to the Interpolation of Sequences" by Kunin, a short 10-page piece that uses finite difference ideas to develop the closed-form formula for the nth Fibonacci number. (Google it, pdf)

[i- !LoVeEKpnh2 No.2477291]

>>2477228
sorry, I wrote this little routine to print a polynomial with latex and it defaults to x, clearly the variable should be the limit of the summation.

[国防语言学院的学生 !sRg2SqGvH. No.2478117]

>>2477291
>with latex
>latex