# Infintite product true or not

1. Sep 19, 2005

### Werg22

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: Sep 19, 2005
2. Sep 19, 2005

### TD

It doesn't seem right to me, try x = 2pi.

3. Sep 19, 2005

### Werg22

Sorry I did a mistake. I meant sin(x).

4. Sep 19, 2005

### TD

Try x = pi.

5. Sep 19, 2005

### Werg22

I'm affraid I do not see what you mean... the serie is infinite...

6. Sep 19, 2005

### shmoe

No, your infintite product does not even converge.

Infinite product form for sine:

$$\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)$$

Last edited: Sep 19, 2005
7. Sep 19, 2005

### TD

I was trying to point out that sin(pi) = 0 but your serie would never go to 0.

8. Sep 19, 2005

### Werg22

Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.

9. Sep 19, 2005

### shmoe

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