- #1

- 1,425

- 1

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)...

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:

- Thread starter Werg22
- Start date

- #1

- 1,425

- 1

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:

- #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]

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

I was trying to point out that sin(pi) = 0 but your serie would never go to 0.Werg22 said:I'm affraid I do not see what you mean... the serie is infinite...

- #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.

Really sorry.

- #9

shmoe

Science Advisor

Homework Helper

- 1,992

- 1

Still doesn't converge, the absolute value of the terms is growing without bound.Werg22 said:Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 9

- Views
- 2K

- Replies
- 11

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Replies
- 5

- Views
- 933

- Last Post

- Replies
- 20

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 341