If you have a series N^x, where N may be any integer from 0 to ∞ or from 0 to -∞, the xth absolute arithmetic difference in that series will be equal to the factorial x. OK, so I just made up the term "absolute arithmetic difference," but I'll explain! Let's take the series N^2. I'll plug some values in it. 0 results in 0 1... 1 2... 4 3... 9 4... 16 5... 25 6... 36 7... 49 8... 64 9... 81 10... 100 OK! So, now, I'll find the differences between each of 'em. 1 3 5 7 9 11 13 15 17 19 The odd numbers. These were what really sparked my interest! So, I'll find the difference a second time! 2 2 2 2 2 2 2 2 2 So I found the difference TWO times! The same as x! Once I did that, I found a constant difference of 2. Of course... 2 = 2! I hope you guys get me now. This will work with any whole number x. It can also work with negative N values, if you have an initial value N = 0. I'm calling the "finding the difference" process the "absolute arithmetic difference", since it seems mroe practical with negative N values if one uses the absolute difference and not the difference. I call it arithmetic because... it is. Please Register or Log in to view the hidden image! :bugeye: I'll define it lΔlp, p being a subscript to indicate the pth "finding the difference"... In my use of lΔlp, p = x I'll try to express myself in math... ∞, -∞ ∑ N^x i = 0 x = [1, ∞] lΔlx = x! Of course, the pattern goes like... For N^1, lΔl1 = 1! = 1 For N^2, lΔl2 = 2! = 2 For N^3, lΔl3 = 3! = 6 For N^4, lΔl4 = 4! = 24 For N^5, lΔl5 = 5! = 120 For N^6, lΔl6 = 6! = 720 ... and so on into infinity... ... though I can't prove that, I suppose... What I really want to know people, is what this thing is called, seeing as I can't be the first to find it. Is there a specific name for it? ... and, if at all possible, could someone take a wack at explaining why it happens... 'cause I sure don't know why! Edit: Sorry... where I defined the domain of x, I accidentally wrote negative infinity instead of positive infinity (as the upper bound). I changed it though.
Here's the proof for the x=2 case. The 1st-order difference between the nth and (n+1)th terms is: (n+1)<sup>2</sup> - n<sup>2</sup> = 2n + 1 The second-order difference is: 2(n+1) + 1 - (2n + 1) = 2 = 2! For the x=3 case, we have: 1st-order: (n+1)<sup>3</sup> - n<sup>3</sup> = 3n<sup>2</sup> + 3n + 1 2nd-order: 3(n+1)<sup>2</sup> + 3(n+1) + 1 - (3n<sup>2</sup> + 3n + 1) = 6n+6 3rd-order: 6(n+1) + 6 - (6n + 6) = 6 = 3! Now, can we construct a general proof (perhaps by induction)?
Yeah! I think you're on the right track! There must be some sort of way to express the "absolute arithmetic difference" operation. There's probably a formula to be had somewhere out in the nether. Now... to get on to that induction you mentioned... :bugeye: This may be difficult... The equalities you provided are smooth and intuitive, very nice, and it's not too difficult to understand what's going on. But I'm not sure how to express it.
The case for general x is the equation (follow what JamesR did, only don't expand and simplify): x ∑ (-1)^i * xCi *(n+x-i)^x = x! i=0 Where xCi denotes "x choose i", aka x!/(i!*(x-i)!) This can be proven directly using a counting argument. Let S be the set of ordered x-tuples with entries from {1,2,..,x+n}. Let S(i) be the n-tuples of S that don't have any i's in them. Let P=S-{intersection of S(1) to S(x)}. Now an x-tuple in P must have at least one i in it, since it lies outside of S(i). So the elements of P have x entries, and must contain each of {1,..,x} at least one time, so it contains each exactly once. Thus, P is the set of all permutations on x-numbers. We count this 2 ways. First it has x! elements. Secondly, we use inclusion/exclusion on S and find |P|= |S| -|S(1)|-..-|S(x)| +|S(1) intersect S(2)| +all other double intersections -all possible triple intersections +etc.....(-1)^x * |intersection of all S(i)| This is very symmetric and can be reduced to: |P|=xC0*|S|-xC1*|S(1)|+xC2*|intersection of S(1), S(2)| -...(-1)^x * |intersection of all S(i)| |S|=(n+x)^x |S(1)|=number of x-tuples with entries from {2,..,n+x}=(n+x-1)^x |intersection S(1), S(2)|=x-tuples from {3,..n+x}=(n+x-2)^x and so on, down to |intersection of all S(i)|=x-tuples from {x+1,..x+n}=n^x This gives the sum we are looking for. You could also consider the initial sum to be an x-th degree polynomial in n (no requirement n is an integer here!). Use the binomial theorem on the terms ((n)+(x-i))^x. Use this to determine the coefficients of the n^k terms. You'll find that if k=0, you get x!, otherwise, 0 (this will require a similar inclusion/exclusion argument). I'm not sure how to get an induction argument to work.
