+495 Note: this is an off topic post. Read for your own enjoyment.
Let's say we have a simple function f(x) = x. We can very easily figure out the values of the function, but can you design another function, s(x) that "keeps track of" the sum of the values of the function at integer values? In other words, s(0) = f(0), s(1) = f(0)+f(1), s(2)=f(2)+f(1)+f(0), and so on. You could rewrite this rule as s(x)-s(x-1)=f(x), true for all values of x.
With a simple function like f(x) = x, it's pretty easy to figure out the function that "keeps track of" the cumulative sum of the original. If you never tried it before, try it out and see if you can figure it out!
(by the way, if you know this is already established mathematics or linked to it someway, please inform me because I'm very intersted in learning more)
f(0) = 0, s(0) = 0
f(1) = 1, s(1) = 0+1 = 1
f(2) = 2, s(2) = 0+1+2 = 3
f(3) = 3, s(3) = 0+1+2+3 = 6
and so on.
It turns out that s(x) is the triangle function! It happens to be (x(x+1))/2 or (x^2)/2 + (x/2)
So what about f(x) = x^2? (Note: lets now say that the s(x) function that describes a particular f(x) function will be denoted as s(f(x)=<whatever it equals>) What would s(f(x)=x^2) be? Try to figure it out if you like.
It turns out that s(f(x)=x^2) = (x^3)/3 + (x^2)/2 + x^6, which is nice.
Here are the next few powers up according to hours of my hard work:
s(f(x)=x^3) = (x^4)/4 + (x^3)/2 + (x^2)/4
s(f(x)=x^4) = (x^5)/5 + (x^4)/2 + (x^3)/3 - x/30
s(f(x)=x^5) = (x^6)/6 + (x^5)/2 + 5(x^4)/12 - (x^2)/12
s(f(x)=x^6) = (x^7)/7 + (x^6)/2 + (x^5)/2 - (x^3)/6 + x/42
s(f(x)=x^7) = (x^8)/8 + (x^7)/2 + 7(x^6)/12 - 7(x^4)/24 + (x^2)/12
s(f(x)=x^8) = (x^9)/9 + (x^8)/2 + 2(x^7)/3 - 7(x^5)/15 + 2(x^3)/9 - x/30
s(f(x)=x^9) = (x^10)/10 + (x^9)/2 + 3(x^8)/4 - 7(x^6)/10 + (x^4)/2 -3(x^2)/20
(I realize these can be factored but I think they're more beautiful when written term by term)
The real challenge is to figure out the pattern!
The first thing to notice that the first term has its power raised by one and then divided by the new power. (similar to an antiderivative)
The second term is the original function divided by two.
The third term is a bit more complicated however. You take the original exponent and subtract one from it. Then divide by 12. Then multiply by the original exponent.
The fourth term is much more complicated. If the original exponent is "n", then the 4th term is equal to
((n(n^2-3n+2)x^n-3)/720
I have not found the pattern for the 4th, 5th, etc terms.
Basically, if you repeatedly do the derivative of each s(f(x)) function (down to the n factorial each one leads!), you start to see more obvious patterns which help you find the function for the next higher power. Then you can plug in integer values into your incomplete function and find a correction term (by manually figuring out the true values of the function and subtracting the values you get from the incomplete function) which if "n" is even will be a first power, and if "n" is odd will be a second power term.
I think its a pretty neat concept.
What do you think about it?
If you know anything about it, I'm very interested in learning more.
