So I recently learned about Sequence Calculus from a Mathologer video on the subject, and have been pondering about a number of different sequences of which I can apply the Gregory-Newton formula to. One that intrigued me was to, in a sense, shift the binomial expansion backwards, by having $\Delta^nf(0) = n$ . This led to a sequence of the following:
$$0,1,4,12,32,80...$$
If you lay out the terms and differences, then $\Delta^nf(0) = n$. Then, I used the Gregory-Newton Method and plugged in 6, to continue the pattern.
$$0\frac{6^\underline{0}}{0!}+1\frac{6^\underline{1}}{1!}+2\frac{6^\underline{2}}{2!}...+6\frac{6^\underline{6}}{6!}=0\,+\,6\,+\,30+\,60\,+\,60\,+\,30\,+\,6 = 192$$
One thing I noticed was the inclusion of $\Delta^nf(0) = n$ seemed to offset the Binomial Expansion's factorial denominator, $n\frac{q}{n!} = \frac{q}{(n-1)!}$. Notably, this is obviously present with the sequence constructed, and I would be lying if i said i didn't consider this beforehand, but what has happened is instead of the regular binomial expansion apprearing, the factorial has decreased by one.
Now, instead of the regular binomial sum being:
$$\sum_{m=0}^\infty \frac{n^\underline{m}}{m!} $$
It is now:
$$\sum_{m=1}^\infty \frac{n^\underline{m}}{(m-1)!} $$
And just like this, for $n \geq 1 $ the expansion gives values of $f(n)$ in the sequence.
If I try the original binomial theorem with a negative you get the iconic $1-1+1-1+1-1+...$ that everyone knows. If i try it with the new theorem i get:
$$\frac{-1}{1}+\frac{2}{1}+\frac{-6}{2}+\frac{24}{6} ... = -1+2-3+4-5+6- . . .$$
Then I tried $\frac{1}{2}$ and got a converging series. What I want to find is:
$$\sum_{m=1}^\infty \frac{\frac{1}{2}^\underline{m}}{m!}$$
What does this approach? Thank you.
-Adri
$n^{\underline m}$ appears to be notation for the "falling factorial", $n^{\underline m}=n(n-1)(n-2)\cdots(n-m+1)$. Then $${n^{\underline m}\over m!}={n\choose m}$$ for all $n$, not just positive integer values of $n$. So, $$\sum_1^{\infty}{(1/2)^{\underline m}\over m!}=\sum_1^{\infty}{1/2\choose m}$$ By the Binomial Theorem, $$\sum_0^{\infty}{1/2\choose m}x^m=(1+x)^{1/2}$$ Letting $x=1$, we get $$\sum_1^{\infty}{(1/2)^{\underline m}\over m!}=\sqrt2-1$$