# Tricky Series Questions

1. Oct 23, 2007

### 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) = ?

Last edited: Oct 24, 2007
2. Oct 23, 2007

### mathwonk

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

3. Oct 24, 2007

### 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

4. Oct 24, 2007

### Zythyr

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

5. Oct 24, 2007

### 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.

6. Oct 24, 2007

### HallsofIvy

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

7. Oct 24, 2007

### Zythyr

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

8. Oct 25, 2007

### Gib Z

I was alluding to that point =]

Don't you like my answer?

9. Oct 27, 2007

### 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.

Last edited: Oct 27, 2007
10. Nov 5, 2007

### 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.

11. Nov 5, 2007

### 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?

12. Nov 5, 2007

### arildno

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).

13. Nov 6, 2007

### yasiru89

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.