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Factorial of n from the function f(x)=f^x

Name: Anonymous 2008-04-23 18:11

Sup /sci/,

I was just playing with lists in python,and I noticed something-if you take the differences of the numbers in a list of n to the power x,where both x and n are natural,and do this recursively x times,you get a list of the same number-the factorial of x.
Now as you can obviously see,I can't speak 'maths',so I'll just give you some examples.The question is-Why does this happen?

x^2
[1, 4, 9, 16, 25, 36, 49]
[3, 5, 7, 9, 11, 13]
[2, 2, 2, 2, 2]
x^3
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331]
[7, 19, 37, 61, 91, 127, 169, 217, 271, 331]
[12, 18, 24, 30, 36, 42, 48, 54, 60]
[6, 6, 6, 6, 6, 6, 6, 6]
x^4
[1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625]
[15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209]
[50, 110, 194, 302, 434, 590, 770, 974, 1202, 1454, 1730, 2030, 2354]
[60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324]
[24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24]

Name: Anonymous 2008-04-23 18:12

>>1
shit-'do this recursively N times'

Name: Anonymous 2008-04-23 21:07

>>1
Your "take the differences of the numbers" is in fact a coarse method of numerical differentiation.
Thus, x2'' = k, x3''' = k, etc.
Dunno about the factorial thing.

Name: Anonymous 2008-04-23 21:10

wow this is the most interesting thing on /sci/ since that determinant trick last summer

Name: Anonymous 2008-04-23 21:14

Oh and I guess the factorial thing is like compounding the derivative law for a polynomial.

(x^2)'' = (2x)' = 2
(x^3)''' = 3(x^2)'' = 6(x)' = 6
(x^4)'''' = 4(x^3)''' = well you get the idea

Name: Anonymous 2008-04-24 8:17

I thought everyone knew this.

It's how we used to work out formulas for sequences back when I was like 10.

Name: Anonymous 2008-04-25 14:21

what was the determinant trick

Name: Anonymous 2008-04-25 15:32

NO U!

Name: Anonymous 2008-04-26 19:06

It makes a sequence
a_1(x) = x^n
a_[k+1](x) = a_k(x+1) - a_k(x)

Then the assertion is that a_(n+1)(x) = n!

I tried proving it but binomial theorem? orz

I got up to guessing that

a_[k+1](x) = SUM{j=0 to k, C(k,j) (-1)^j (x+(k-j))^n} and then I dunno lol.

Name: Anonymous 2008-04-26 19:07

C(k,j) is the binomial coefficient

Name: Anonymous 2008-04-27 15:53

>>9
This actually works. Take a_[n+1](x) = n!
Simplifies to
SUM{j=0 to n, (-1)^j * (x+n-j)^n / [ j! (n-j)! ] = 1
No proof as such but try it and it works.

Name: Anonymous 2008-04-27 19:51


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