Question

Say I wanted to take the k-th difference of the sequence of numbers 1^k, 2^k, 3^k, etc, what does that mean? Would the first difference be 2^k-1^k, 3^k-2^k, etc.. and then would the second difference be 3^k-2^k-(2^k-1^k), 4^k-3^k-(3^k-2^k), etc ?

I've tried to find what is meant by this terminology, but can't find it... Thanks...

Answer 1

i would try with actual values of k before i tried to do this in general

Answer 2

say \(k=2\) then you have a sequence that looks like
\[1^2,2^2,3^2,...\]
first difference is
\[2^2-1^2=4-1=3,3^2-2^2=9-4=5,4^2-3^2=16-9=7...\] and second differences are
\[5-3=2,7-5=2,... \] looks like it is the constant 2

Answer 3

try it with \(k=3\) then you get
\[1^3,2^3,3^3,4^3,...\] first difference is
\[7,19,37,61...\]
second difference is
\[12,18,24...\] third is
\[6,6,6,...\] hmm maybe a pattern

Answer 4

Thanks for your help. I've found that it seems to be true that the kth difference is equal to k! or something... Not quite sure how to say it precisely...

For example, just letting k stay as k, I find the first element of the sequence of taking the first difference is 2^k-1^k.

The first difference of the sequence from taking the difference, and then the difference again (maybe call it 2nd difference sequence?) is 3^k-2*2^k+1^k...

Then, taking the difference, then, again, again, and again (four times) you get 5^k-4*4^k+6*3^k-4*2^k+1^k. Since I've taken the difference four times, if I let k=4, then it gives me 4!... The same with the ones above... I think this is true in general, but I can't prove it myself yet.

I'm doing this because here http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares in Euler's proof almost at the end of part 5, it says something to this effect and I wasn't really sure what they meant... I think this is what they mean... but I lack the knowledge of maths in this area to know why it's true... It's to do with difference operators or something... I reckon this is what the article means. I think this is quite a strange identity for factorial!

Answer 5

it sure does say that, doesn't it. well we got it for k = 2 and k = 3, i wonder what a general proof would look like, or is it just supposed to be obvious

Answer 6

I think that part of the proof on wikipedia could do with some extra information. I noticed in the discussion on it that another person was as confused as I am. I will keep going on it...

Answer 7

ok i think i see it, but i am not sure i can write it convincingly,
we want the kth difference, so start with say \(d_1=n^k-(n-1)^k\) as the first difference.
expand and the \(n^k\) term is gone, leaving by the binomial theorem \(k(n-1)^{k-1}\) plus lower terms repeat the process k times and you will end up with \(k!\)

Answer 8

Hmn. I'll need to think about it some more. My knowledge of the binomial theorem is neglected.