Solve the Tricky Series Question: Express \sum n^2x^n as a Function of x

[Zythyr]

My professor was going over some powers relating to series, and power series. He came across the problem below which he didn't even know how to solve. I am trying to figure out how to solve it, but can't get anywhere. Does anyone know how to approach this problem? Express this series as a function of x.
\sum n^2x^n from n=1 to \infty

So, f(x) = ?

 

[mathwonk]

come on. who's your professor, george w bush?

 

[Haelfix]

I've blanked once or twice when teaching a class before. Its embarrasing, particularly when its easy, but I've seen it happen to field medal winners

 

[Zythyr]

Woops sorry guys. I forgot to post the actually question. The guestion is to express that series as a function of x.

 

[Gib Z]

Quite simple really, f(x) = \sum_{n=1}^{\infty} n^2 x^n.

Looking for a nicer (ie closed form) answer? There is no such closed form answer, unless you are looking for values of x where |x| <1, in which case there may be.

 

[HallsofIvy]

It's not just that there is no "closed form answer"- that series only converges for |x|< 1.

 

[Zythyr]

I am not looking for the radius of convergence. I am looking how to epxress that series as a function.

 

[Gib Z]

It's not just that there is no "closed form answer"- that series only converges for |x|< 1.

I was alluding to that point =]

I am not looking for the radius of convergence. I am looking how to express that series as a function.

Don't you like my answer?

 

[atqamar]

Just to see if the summation converges to a single function, I Maple-ed it, and Maple gives a solution to f(x)=\sum_{n=1}^{\infty} n^2 x^n. HINT: It is the quotient of a quadratic polynomial and a cubic polynomial.

Perhaps this identity can be of some help (from Wikipedia):
\sum_{i=0}^n i^2 x^i = \frac{x}{(1-x)^3} (1+x-(n+1)^2x^n+(2n^2+2n-1)x^{n+1}-n^2x^{n+2}).

This simpler identity helps too (from here):
\sum_{i=0}^\infty i^2 x^i = \frac{x^2+x}{(1-x)^3}

EDIT: Do a little manipulation to the above identity, and you'll end up with your solution.

 

[yasiru89]

My word,
1 + x + x^2 + x^3 + .. . + x^n + ... = (1-x)^(-1)

differentiating and multiplying both sides by x we have,

1+ 2x^2 + 3x^3 + ... + nx^n + ... = x (1-x)^(-2)
do the same again and you'll have a result valid inside the unit disc with the origin as a centre.

 

[sennyk]

Can one go backwards from the series to the quotient of polynomials? Or must one have to know that each are equivalent from experience?

 

[arildno]

Quite simple really, f(x) = \sum_{n=1}^{\infty} n^2 x^n.

Looking for a nicer (ie closed form) answer? There is no such closed form answer, unless you are looking for values of x where |x| &lt;1, in which case there may be.

Here's how to derive it:
f(x)=x\sum_{i=1}^{\infty}n^{2}x^{n-1}=x\frac{d}{dx}\sum_{i=1}^{\infty}nx^{n}=x\frac{d}{dx}x\frac{d}{dx}\sum_{i=1}^{\infty}x^{n}=x\frac{d}{dx}x\frac{d}{dx}\frac{x}{1-x}

Now, do the differentiations, and get the rational representation of f(x).

 

[yasiru89]

Can one go backwards from the series to the quotient of polynomials? Or must one have to know that each are equivalent from experience?

Well we need an infinite series, S satisfying the following relation,

S = 1 + xS

Not all that far off, it can be more 'traditionally' derived by considering the finite geometric progression and evaluating the limit. I've provided a simple derivation that can be easily followed.