# Anyone recognize this series expansion?

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.

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

I don't quite see the pattern... how do you get from 27 to 51, and what comes next?
Oh...yes that should be an 81

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

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

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

cristo
Staff Emeritus
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??
0!=1, by definition.

0!=1, by definition.
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 Dick
Homework Helper
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 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.

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