/sci/ - Science & Math

>>15013174
>The geometric series is not a large concept
It's actually a cool topic with many directions

https://en.wikipedia.org/wiki/Neumann_series

In functional analysis.

For just numbers, some angles on the series and the sum which come in handy are

[math]\dfrac{z^s}{(1-z)^{s+1}} = \sum_{k=s}^\infty {k\choose s} z^{k}[/math]

[math]\dfrac{1}{(1-z)^s} = \sum_{k,m \geq 0} \left[\begin{matrix} k \\ m \end{matrix} \right] \dfrac{1}{k!} s^m z^k[/math]

[math] \dfrac{1}{1-(z+w)} = \sum_{k,m \geq 0} \binom{k+m}{k}\, w^m\, z^i [/math]

[math]\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z}\cdot\int_0^1 f(s)\,{\mathrm d}_zs[/math]
(see q-integral)

[math]\prod_{k=0}^{n-1} \frac{1}{(1-q^kt)}=\sum_{k=0}^\infty {n+k-1\choose{}k}_q t^k[/math]

There's more on the "neighboring" product formulas

[math]\dfrac{1}{1-x}=\exp(-\log(1-x))=\exp(\sum_n\frac{x^n}{n})=\prod_n\exp(\frac{x^n}{n})[/math]

which is the crucial relation in some Weierstrass factorizations of more general functions.
You got the factoring out of (a-b) in

[math]a^{n+1}-b^{n+1} = (a-b)\sum_{k=0}^n a^k\,b^{n-k}[/math]

(and similarly in)

[math]a^n-b^n = (a-b)\prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}}) [/math]

The interplay of sum and product comes back in integral equation theory, where for an operator A on a Banach space you can define the resolvent, basically

R(z, A) = -1/(1-A/z)

and the difference of two such geometric series for A and B still has a factor (A-B).
See pic related.

The map from a to 1/(1-a) with a in a ring is generalized to Star semirings, and that's a whole other can of nutritious worms

https://en.wikipedia.org/wiki/Semiring#Star_semirings