What is this summation?

[Colic]

No, it's not a homework problem, it's a random thing I was working on when playing around with algorithms to convert between music and systems of math.

I have these terms,

 

[MATH](1/1)x+(1/3)x^3+(1/15)x^5+(1/105)x^7+(1/945)x^9[/MATH]...

 

And I'm trying to come up with a formula to describe the coefficients in terms of a summation as part of a larger formula that I've already broken down. I've spent more than two hours testing different formulas and none of them work, but I know the concise pattern. If you want the nth coefficient, you'd do 1 divided by the derivative of the nth term times the coefficient (n-1)th term times the coefficient of the (n-2)nd term times the coefficient of the (n-3)rd term and so on. In a sense, it's like a factorial based off of derivatives, but I can't figure out a formula for it in terms of Sum(n=1) n->infinity. if n starts at 0, it's x^(2n-1), I got that down, I just can't figure out the coefficients, it's some kind of alternative factorial.

Edited February 3, 2014 by Colic

[John]

You'll want to use a double factorial.

 

Using that, I have [math]\sum_{k=0}^{\infty} \frac{1}{(2k+1)!!}x^{2k+1}[/math].

 

Summing from 1 instead of from 0 (if you prefer that for whatever reason) simply requires changing each instance of 2k+1 to 2k-1.

Edited February 3, 2014 by John

[mathematic]

[latex]\frac{k!2^kx^{2k+1}}{(2k+1)!}[/latex]

 

Start sum at k = 0.

[Someguy1]

No, it's not a homework problem, it's a random thing I was working on when playing around with algorithms to convert between music and systems of math.

I have these terms,

 

[MATH](1/1)x+(1/3)x^3+(1/15)x^5+(1/105)x^7+(1/945)x^9[/MATH]...

 

And I'm trying to come up with a formula to describe the coefficients in terms of a summation as part of a larger formula that I've already broken down. I've spent more than two hours testing different formulas and none of them work, but I know the concise pattern. If you want the nth coefficient, you'd do 1 divided by the derivative of the nth term times the coefficient (n-1)th term times the coefficient of the (n-2)nd term times the coefficient of the (n-3)rd term and so on. In a sense, it's like a factorial based off of derivatives, but I can't figure out a formula for it in terms of Sum(n=1) n->infinity. if n starts at 0, it's x^(2n-1), I got that down, I just can't figure out the coefficients, it's some kind of alternative factorial.

 

Denominators are double factorials of odd numbers. Next few are 10395,135135, 2027025, 34459425, 654729075, 13749310575, ...

 

http://oeis.org/search?q=1%2C3%2C15%2C105%2C945&language=english&go=Search

Edited February 4, 2014 by Someguy1

[SamBridge]

Well thanks for the formula, it actually worked (btw colic wasn't a sockpuppet I just forgot my old username and email associated with (making it my main account) it since I haven't been here for a long time and I received absolutely no help from the staff member for looking them up and so spent a lot of time looking through my 12 emails for any info after the other acc was banned, so sorry for the delay too).

I had one other question too:

Apparently with the sequences I was dealing with, they ended up turning into something that resembles a power series for a function, but I can't quite nail down one of the properties for manipulating summation it seems. The series I have converges, but I'm trying to divide out a whole other power series so I can just have some irrational number as a remainder, and my basic attempts seem to yield a remainder that doesn't converge somehow.

Just as an example, if I have

 

[math]1=sum[((x+1)^n)(2^n)/((2n+1)!)][/math]

 

and I divide the whole summation by a power series where say,

 

[math]f(x)=sum[((x+1)^n)(n!)][/math],

 

would I or would I not get

 

[math]1/f(x)=sum[((2^n)/(((2n+1)!)(n!))][/math]?

 

And why or why not? Because I searched all over the internet and for whatever reason there doesn't seem to be anything describing sum(a)/sum(b) or factoring out a function from a power series.

Edited February 5, 2014 by SamBridge

[mathematic]

You need to detail the derivation of the series for 1/f(x).

[SamBridge]

You need to detail the derivation of the series for 1/f(x).

i don't mean for that specific series, just in general, how would you generally do series(a)/series(b)? Because I know I didn't do it right, and I don't find anything when I google "dividing one powers series by another".

[mathematic]

i don't mean for that specific series, just in general, how would you generally do series(a)/series(b)? Because I know I didn't do it right, and I don't find anything when I google "dividing one powers series by another".

Short answer: there is no general method. To understand the difficulty see what happens in a simpler case, dividing a polynomial by a polynomial.

[SamBridge]

Wow, that sucks.