Infintite product true or not

  • Thread starter Werg22
  • Start date
  • #1
1,425
1

Main Question or Discussion Point

Is this true?

f(x)=sin(x)=x(x-2pi)(x-4pi)(x-6pi)...

edit:

f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...
 
Last edited:

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
It doesn't seem right to me, try x = 2pi.
 
  • #3
1,425
1
Sorry I did a mistake. I meant sin(x).
 
  • #4
TD
Homework Helper
1,022
0
Try x = pi.
 
  • #5
1,425
1
I'm affraid I do not see what you mean... the serie is infinite...
 
  • #6
shmoe
Science Advisor
Homework Helper
1,992
1
No, your infintite product does not even converge.

Infinite product form for sine:

[tex]\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)[/tex]
 
Last edited:
  • #7
TD
Homework Helper
1,022
0
Werg22 said:
I'm affraid I do not see what you mean... the serie is infinite...
I was trying to point out that sin(pi) = 0 but your serie would never go to 0.
 
  • #8
1,425
1
Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.
 
  • #9
shmoe
Science Advisor
Homework Helper
1,992
1
Werg22 said:
Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.
Still doesn't converge, the absolute value of the terms is growing without bound.
 

Related Threads for: Infintite product true or not

  • Last Post
Replies
3
Views
1K
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
20
Views
3K
Replies
5
Views
830
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
6
Views
2K
Top