Anyone recognize this series expansion??

Saladsamurai

[tex]1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{51t^4}{4!}+.....[/tex]

I looks kind of like [itex]e^t[/itex] 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.

nicksauce

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

Saladsamurai

Oh...yes that should be an 81

[tex]
1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{81t^4} {4!}+.....[/tex]

nicksauce

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

Saladsamurai

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

0!=1, by definition.

Saladsamurai

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

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

0!= 1 because I SAY SO!!