Anyone recognize this series expansion?

In summary, the conversation is about the series expansion 1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{51t^4}{4!}+... and how it resembles the function e^t. The participants discuss the pattern and how to get from 27 to 51, with one person realizing the function is (3t)^n / n! and another questioning the value of 0!. The conversation ends with the explanation that 0!=1 by definition for the factorial function to work properly.
  • #1
Saladsamurai
3,020
7
Anyone recognize this series expansion??

[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.
 
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  • #2
I don't quite see the pattern... how do you get from 27 to 51, and what comes next?
 
  • #3
nicksauce said:
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

[tex]
1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{81t^4} {4!}+...[/tex]
 
  • #4
So you have (3t)^0 / 0! + (3t)^1 / 1! + (3t)^2 /2! + (3t)^3 / 3! + ...
Surely you can see what this function is?
 
  • #5
nicksauce said:
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??
 
  • #6
Saladsamurai said:
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.
 
  • #7
cristo said:
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:smile:
 
  • #8
Saladsamurai said:
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:smile:

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
Saladsamurai said:
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:smile:
0!= 1 because I SAY SO!
 

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