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#1 Re: This is Cool » Explanation? » 2013-10-02 19:55:24
I couldn't imagine it would be. Oh well, it was still fun to derive!
#2 Re: This is Cool » Explanation? » 2013-10-02 19:51:09
Well that is cool 
Could it have any practical uses?
#3 Re: This is Cool » Explanation? » 2013-10-02 19:42:37
Yeah! That's what I was looking for. Is this not a known formula?
#4 Re: This is Cool » Explanation? » 2013-10-02 19:33:44
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
#5 Re: This is Cool » Explanation? » 2013-10-02 19:12:29
Somewhat. Is that summation useful at all? I thought it was pretty cool how it followed coefficients from Pascal's triangle.
#6 Re: This is Cool » Explanation? » 2013-10-02 14:38:14
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)
#7 Re: This is Cool » Explanation? » 2013-10-02 13:06:29
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
#8 Re: This is Cool » Explanation? » 2013-09-30 12:04:04
That's really cool! So how can you approximate derivatives with that?
#9 Re: This is Cool » Explanation? » 2013-09-28 07:37:09
Could you explain both the approximation of derivatives and the multiplication? That sounds cool
#10 This is Cool » Explanation? » 2013-09-27 18:07:22
- 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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