# Power series

1. Nov 14, 2005

### overseastar

Hi
I"m having truoble with fnding the power series of the following:

1+(x^3)/3+(x^6)/18+(x^9)/162+...

Can anyone give me a hand?

Also, any hints in finding a power series?

Thanks !

2. Nov 14, 2005

### Hurkyl

Staff Emeritus
That IS a power series...

3. Nov 14, 2005

### quasar987

Hahaha... or did you mean, you have trouble putting it into compact $\sum$ notation?

4. Nov 14, 2005

### Tide

I am guessing that overseastar is looking for a closed form of the sum.

5. Nov 15, 2005

### Joffe

Looks like:

$$\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}$$

6. Nov 15, 2005

### roger

Why do you use the factorial when 3 ! = 6 which is constant ?

7. Nov 15, 2005

### amcavoy

That fails to work when the denominator is 3.

8. Nov 15, 2005

### Moo Of Doom

I think he means $$\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}$$

9. Nov 15, 2005

### Joffe

Thankyou Moo Of Doom, that is what I meant to write.

10. Nov 16, 2005

### HallsofIvy

And $$\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}$$
is equal to
$$\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n$$
Anyone recognize THAT??

11. Nov 16, 2005

### Joffe

No I don't, please enlighten me.

12. Nov 16, 2005

### amcavoy

$$\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n=\exp{\left(\frac{x^3}{3}\right)}$$

13. Nov 19, 2005

### overseastar

oh wow, yes, that's what i meant, sorry

14. Nov 19, 2005

### overseastar

And thank you for all your help !