Find a way to obtain an explicit formula, in terms of r such that -1 < r < 1, for , for any positive integral choice of p.

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- Oct 4th 2008, 05:59 PMicemanfanPseudo-Geometric Series
Find a way to obtain an explicit formula, in terms of r such that -1 < r < 1, for , for any positive integral choice of p.

- Oct 4th 2008, 07:03 PMSoroban
Hello, icemanfan!

I haven't solved this yet, but I have a primitive start on it.

Maybe someone will be inspired to continue . . .

Quote:

Find a way to obtain an explicit formula, in terms of , such that

, for , for any positive integral choice of

For

Subtract : .

We have: .

Therefore: . ... for

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For

Subtract: .

. .

Therefore: . ... for

. . . Anyone getting excited yet?

- Oct 5th 2008, 03:53 AMLaurent
Note that (by usual properties of power series), for :

,

and, in general,

.

...so that a general method to get a simple expression for consists in writing as a linear combination of .

For instance, , . And for each of the terms, .

There is no "extremely simple" formula valid for all , since you need the coefficients in the decomposition . These coefficients are called Stirling numbers of the second kind, denoted by .

One has . A closed formula would then be (up to possible computation errors):

.