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I couldn't imagine it would be. Oh well, it was still fun to derive!

Well that is cool

Could it have any practical uses?

Yeah! That's what I was looking for. Is this not a known formula?

Sure, I will find 4! using the summation:

Substuting 4 in for n in sum((-1)^p*nCp*(n-p)^n, p=0..n) gives sum((-1)^p*4Cp*(4-p)^4, p=0..4).

Taking the sum gives (-1)^0*4C0*(4-0)^4 + (-1)^1*4C1*(4-1)^4 + (-1)^2*4C2*(4-2)^4 + (-1)^3*4C3*(4-3)^4 + (-1)^4*4C4*(4-4)^4

= 1*1*256 + (-1)*4*81 + 1*6*16 + (-1)*4*1 + 1*1*0

= 256 - 324 +96 -4 +0

=24

I converted it into sigma notation, but I'm not sure how to write that on the computer.

In maple, it is sum((-1)^p*nCp*(n-p)^n, p=0..n)

Thank you! That is really cool!

Using those differences and the fact that the last difference before 0 is n!, i have found a formula for n!

n! = nC0(x+n)^n - nC1(x+n-1)^n + nC2(x+n-2)^n - nC3(x+n-3)^n + ... +- nCn(x)^n for any x

the +- is at the end because if n is even it will be +, and if n is odd it will be -

Not sure if that's useful at all. Probably not, but it was fun anyway

That's really cool! So how can you approximate derivatives with that?

Could you explain both the approximation of derivatives and the multiplication? That sounds cool

**Ma123**- Replies: 31

Hi. I'm not well educated in math, just a high school student interested in math. I was playing around with numbers and noticed a pattern. If I lined up consecutive perfect squares like {0,1,4,9,16,25,36...} and then took the difference between each pair of consecutive numbers, I got {1,3,5,7,9,11,13...}. Then I took the difference between each consecutive pair of numbers and got {2,2,2,2,2...}. Taking the difference between those gives {0,0,0,0,0...}. Interesting.

I then thought about using other powers, say cubes. I take the sequence of perfect cubes {1,8,27,64,125...} and make a sequence of the differences like before {7,19,37,61,91...}. I continue the process of taking the differences between consecutive numbers (12,18,24,30...} then {6,6,6,6...} then {0,0,0...}.

I then try the fourth power. {1,16,81,256,625,1296...} --> {15,65,175,369,671,1105...} --> {50,110,194,302,434...} --> {60,84,108,132...} --> {24,24,24,24...} --> {0,0,0...}

So that's the process.. I wrote (x+1)^n-x^n where n is the exponent.

Then I noticed that the number of times you have to take the difference in order to reach a difference of 0 is (n+1).

Also, the final sequence before {0,0,0...} will be a sequence of {n!,n!,n!...}

I think this is really cool! But I have no idea what it means. I was hoping one of you brilliant minds would be able to expand on this pattern for me. So do these things have a relationship? Is it just coincidence? Does it have a use? I appreciate your help!

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