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!
 

1. What is a series expansion?

A series expansion is a mathematical representation of a function as an infinite sum of terms, typically expressed in powers of a variable.

2. How is a series expansion useful?

A series expansion can be used to approximate complex functions, making them easier to manipulate and analyze. It is also a powerful tool for solving differential equations and computing integrals.

3. What does it mean to recognize a series expansion?

Recognizing a series expansion means being able to identify the pattern or formula that describes the infinite sum of terms. This can help in simplifying the expression and making it easier to work with.

4. How do you determine the convergence of a series expansion?

The convergence of a series expansion can be determined by evaluating its terms and checking if they approach a finite value. If the terms approach a finite value, the series is said to converge. Otherwise, it diverges.

5. Can a function have more than one series expansion?

Yes, a function can have multiple series expansions depending on the point of expansion. For example, the function e^x has different series expansions at x=0 and x=1. However, these expansions may not necessarily converge for all values of x.

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