I’ve discovered a sequence based on differences, I’m trying to convert this to a series formula. It converges to zero so this should be possible.
Here is the sequence using $x^n$ (where for simplicity I use $x^2$):
$$ x = 1,2,3,4,5,\dots $$ $$ x^2 = 1,4,9,16,25,\dots $$
1st difference $D_1$ (using $25-16, 16-9, 9-4, \dots$): $$ D_1 = 9,7,5,\dots $$
2nd difference $D_2$ (using $9-7, 7-5, \dots$): $$ D_2 = 2! $$
3rd difference $D_3$: $$ D_3 = 2-2 = 0 $$
I used factorial 2 here as an example because whatever values of $x^n$ is used the factorial ($n!$) appears. This leads to similarities with derivatives: $$ f^n x^n = n! \quad \text{where my } D\text{'s give } D^n x^n = n! $$
I have shown a few professors this and they basically say “wonderful spot” etc but that’s all.
Now I have converted this to algebraic form:
$$ D_1 = (x+1)^n - x^n \quad \text{for } x^1, x^2, x^3, \dots $$ $$ D_2 = \bigl((x+2)^n - (x+1)^n\bigr) - D_1 $$ $$ D_3 = \bigl((x+3)^n - (x+2)^n\bigr) - D_2 \quad (\text{here } D_2 \text{ would include } D_1) $$ $$ D_4 = \bigl((x+4)^n - (x+3)^n\bigr) - D_3 \quad (\text{as above}) $$
$$ \vdots $$
$$ D_{n-1} = \bigl((x+(n-1))^n - (x+(n-2))^n\bigr) - D_{n-2} \quad \text{for } x^{n-1} $$ $$ D_n = \bigl((x+n)^n - (x+(n-1))^n\bigr) - D_{n-1} $$
$$ D_{n+1} = 0 $$
I hope I haven’t made any fundamental mistakes in my sequence.
There are 3 take-aways from this (for all n’s): a. Dnx^n=n! or (Dnx^n)/n! = 1 for (-∞>n>∞) and -∞…-2,-1,0,1,2,…∞ b. D(n+1)=0 which shows series converges to 0 c.fn^n= n! and Dn^n= n! but Dn ≠ fn (to be shown)
I realise now that this is very similar to classical finite difference calculus and I need to study the subject fully, can anyone point me in the right direction to start off.
