# Thread: Trig series.

1. ## Trig series.

I need to show that the triconometric series $\sum_{n=0}^{\infty} 3^{-n}sin(nx)$ converges uniformt on R. That the sumfunction f is an odd, 2pi-periodic continious function, also that the sumfunction is given by $f(x)= \frac{3sin(x)}{10-6cos(x)}$ by the use of eulers formula $e^{ix}=cosx + isinx$

2. Originally Posted by Zaph
I need to show that the triconometric series $\sum_{n=0}^{\infty} 3^{-n}sin(nx)$ converges uniformt on R. That the sumfunction f is an odd, 2pi-periodic continious function, also that the sumfunction is given by $f(x)= \frac{3sin(x)}{10-6cos(x)}$ by the use of eulers formula $e^{ix}=cosx + isinx$
I would consider:

$\sum_{n=0}^{\infty} 3^{-n}\sin(nx)=\mathfrak{Im} \sum_{n=0}^{\infty} 3^{-n} e^{{\bf{i}}.n.x}$

and the series on the right is a convergent geometric series with common ratio:

$
\frac{e^{{\bf{i}}.x}}{3}
$

CB

3. Originally Posted by CaptainBlack
I would consider:

$\sum_{n=0}^{\infty} 3^{-n}\sin(nx)= \sum_{n=0}^{\infty} 3^{-n} e^{{\bf{i}}.n.x}$

and the series on the right is a convergent geometric series with common ratio:

$
\frac{e^{{\bf{i}}.x}}{3}
$

CB
Thanx for the reply, im with you that

$\sum_{n=0}^{\infty} 3^{-n}\sin(nx)= \sum_{n=0}^{\infty} 3^{-n} e^{{\bf{i}}.n.x}$

But what is my next step?

4. Originally Posted by Zaph
Thanx for the reply, im with you that

$\sum_{n=0}^{\infty} 3^{-n}\sin(nx)= \sum_{n=0}^{\infty} 3^{-n} e^{{\bf{i}}.n.x}$

But what is my next step?
1. Your quote of what I wrote seems to have lost the taking of the imaginary part of the right hand side.

2. Use the sum of the geometric series formulae to write down the partial sum to $n$ terms and to $\infty$ of the series and with these show that the series converges uniformly on $\mathbb{R}$.

CB