Question

What's the closed form for this power series? ;)

Answer 1

\[\Large \sum_{n=0}^\infty \binom{n}{k}x^n\] Those are binomial terms as coefficients for this power series.

Answer 2

k doesnt change?

Answer 3

k is an arbitrary positive integer.

Answer 4

I guess I should say that since this is sort of uncommon to come across, but the extension of the factorial to negative values is complex infinity. What that means to us is specifically if you try to evaluate this:

\[\large \binom{3}{10} = \frac{3!}{10!(3-10)!} = \frac{3!}{10!} \frac{1}{(-7)!}=\frac{3!}{10!} \frac{1}{\infty } = 0\] So all factorials with a larger value on bottom than top go to zero.

Answer 5

pascals triangle can be expanded such that the usual triangle is a subsection of a larger structure.

Answer 6

\[\binom{-2}{3},~or~\binom{3}{7}\]
have value

Answer 7

Sure, but here n is only summing over positive values and here I'm explicitly saying that the inverse of a negative factorial is 0 since that's how the answer I found works out, but maybe there's some magical stuff you can do.

Answer 8

all ive got is switch or switches

Answer 9

I don't even know what that means hahaha

Answer 10

I guess I should just give away the answer since there's not much attention haha

Answer 11

a switch is something that is 1 or 0 for a given interval of a domain.

Answer 12

its basically a sgn function, raised by 1, and split in half.

\[\frac12(1+\frac{f(x)}{|f(x)|})\]
the roots of f(x) define where the switch turns to 1 or 0

Answer 13

It follows from these two facts that

$$ \frac{d^k}{dx^k} \left( x^n \right) = \frac{n!}{(n-k)!} x^{n-k} $$

Then we look at the geometric series and see it has this general form

$$ \frac{d^k}{dx^k} \left( \frac{1}{1-x} \right) = \frac{k!}{(1-x)^{k+1}} $$

Plug this into the kth derivative of the geometric series we have the general form for the kth derivative of the geometric series

$$ \sum_{n=0}^{\infty} \frac{n!}{(n-k)!} x^{n-k} = \frac{k!}{(1-x)^{k+1}} $$


Now we can move \(x^{-k}\) and \(k!\) to the other sides to finish

$$ \sum_{n=0}^{\infty} \binom{n}{k}x^n = \frac{x^k}{(1-x)^{k+1}} $$

Answer 14

oh, so this was a known solution. good show then.

Answer 15

lol oh well