# Anyone recognize this series expansion?

1. Apr 27, 2008

Anyone recognize this series expansion??

$$1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{51t^4}{4!}+.....$$

I looks kind of like $e^t$ but i am not sure how to deal with it.

Can I factor something... I kind of suck at these. Someone give me a hint.

2. Apr 27, 2008

### nicksauce

I don't quite see the pattern... how do you get from 27 to 51, and what comes next?

3. Apr 27, 2008

Oh...yes that should be an 81

$$1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{81t^4} {4!}+.....$$

4. Apr 27, 2008

### nicksauce

So you have (3t)^0 / 0! + (3t)^1 / 1! + (3t)^2 /2! + (3t)^3 / 3! + ...
Surely you can see what this function is?

5. Apr 28, 2008

I can now! By the way what the Christ is 0! ??? I want to say that for some strange reason it is 1...... but I don't know why??

6. Apr 28, 2008

### cristo

Staff Emeritus
0!=1, by definition.

7. Apr 28, 2008

Yeah. That is what I thought.... though I thought that there was more to it than 'because the math gods said so.'

But I'll take it if that's all there is to it

8. Apr 28, 2008

### Dick

The math gods said so for a good reason. You want the factorial function to satisfy n!=n*(n-1)!. If you put n=1, then you'd better define 0!=1. You can see you're in big trouble trying to define (-1)!. But that's ok. This is also related to the properties of the gamma function. gamma(n+1)=n!. And gamma(1)=1. gamma(0) is undefined, it's a pole of the gamma function. So we'd better leave (-n)! undefined.

9. Apr 28, 2008

### HallsofIvy

Staff Emeritus
0!= 1 because I SAY SO!!