Power series

  • #1
Hi
I"m having truoble with fnding the power series of the following::frown:

1+(x^3)/3+(x^6)/18+(x^9)/162+...

Can anyone give me a hand?

Also, any hints in finding a power series?

Thanks !
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,950
19
That IS a power series...
 
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
4,784
18
Hahaha... or did you mean, you have trouble putting it into compact [itex]\sum[/itex] notation?
 
  • #4
Tide
Science Advisor
Homework Helper
3,089
0
I am guessing that overseastar is looking for a closed form of the sum.
 
  • #5
36
0
Looks like:

[tex]\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}[/tex]
 
  • #6
319
0
Joffe said:
Looks like:

[tex]\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}[/tex]


Why do you use the factorial when 3 ! = 6 which is constant ?
 
  • #7
665
0
Joffe said:
Looks like:

[tex]\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}[/tex]
That fails to work when the denominator is 3.
 
  • #8
367
1
I think he means [tex]\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}[/tex]
 
  • #9
36
0
Thankyou Moo Of Doom, that is what I meant to write.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,847
966
And [tex]\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}[/tex]
is equal to
[tex]\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n[/tex]
Anyone recognize THAT??
 
  • #11
36
0
No I don't, please enlighten me.
 
  • #12
665
0
[tex]\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n=\exp{\left(\frac{x^3}{3}\right)}[/tex]
 
  • #13
oh wow, yes, that's what i meant, sorry
 
  • #14
And thank you for all your help !
 

Related Threads on Power series

  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
921
Replies
3
Views
2K
Top