Proving $\sum_{k=1}^{n}e^{ik\theta}$ Formula

[Dustinsfl]

Prove the formula
$$
\sum_{k=1}^{n}e^{ik\theta} = \frac{e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}
$$

I have a hint that says consider the expression $e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\left(n-\frac{1}{2}\right)\theta}$.
How can I get the second exponential in the numerator to have that expression?

 

[Sudharaka]

Prove the formula
$$
\sum_{k=1}^{n}e^{ik\theta} = \frac{e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}
$$

I have a hint that says consider the expression $e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\left(n-\frac{1}{2}\right)\theta}$.
How can I get the second exponential in the numerator to have that expression?

Hi dwsmith, :)

We shall start from,

\[\sum\limits_{k = 1}^ne^{ik\theta} = \frac{e^{i\theta}\left(1 - e^{in\theta}\right)}{1 - e^{i\theta}}\]

With reference to your thread, http://www.mathhelpboards.com/f13/basics-fourier-series-1717/ we have obtained,

\[e^{i\theta} - 1 = 2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}\]

Therefore,

\begin{eqnarray}

\sum\limits_{k = 1}^ne^{ik\theta} &=& \frac{e^{i\theta}\left(e^{in\theta}-1\right)}{2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}}\\

&=& \frac{e^{\frac{i\theta}{2}}\left(e^{in\theta}-1\right)}{2i\sin\frac{\theta}{2}}\\

&=& \frac{e^{i\left(n + \frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}\\

\end{eqnarray}

\[\therefore \sum\limits_{k = 1}^ne^{ik\theta}=\frac{e^{i\left(n + \frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}\]

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Hi dwsmith, :)

We shall start from,

\[\sum\limits_{k = 1}^ne^{ik\theta} = \frac{e^{i\theta}\left(1 - e^{in\theta}\right)}{1 - e^{i\theta}}\]

With reference to your thread, http://www.mathhelpboards.com/f13/basics-fourier-series-1717/ we have obtained,

Shouldn't the sum be
$$
\sum\limits_{k = 1}^ne^{ik\theta} = \frac{\left(1 - e^{i(n+1)\theta}\right)}{1 - e^{i\theta}}
$$
where did the $e^{i\theta}$ in the front come from?

 

[Sudharaka]

Shouldn't the sum be
$$
\sum\limits_{k = 1}^ne^{ik\theta} = \frac{\left(1 - e^{i(n+1)\theta}\right)}{1 - e^{i\theta}}
$$
where did the $e^{i\theta}$ in the front come from?

Note that \(\sum\limits_{k = 1}^ne^{ik\theta}\) is a geometric series with common ratio \(e^{i\theta}\). So do you know the sum of the first \(n\) terms of a geometric series?

 

[Dustinsfl]

Note that \(\sum\limits_{k = 1}^ne^{ik\theta}\) is a geometric series with common ratio \(e^{i\theta}\). So do you know the sum of the first \(n\) terms of a geometric series?

$$
s_N = 1 + e^{i\theta} + e^{2i\theta} + \cdots + e^{in\theta}
$$
Multiple by $z$.
$$
e^{i\theta}s_N = e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \cdots + e^{i(n+1)\theta}
$$
Next make the subtraction $s_N - e^{i\theta}s_N$.
So
$$
s_n(1 - e^{i\theta}) = 1 - e^{i(n+1)\theta}\iff s_n = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}}.
$$

 

[Sudharaka]

$$
s_N = 1 + e^{i\theta} + e^{2i\theta} + \cdots + e^{in\theta}
$$
Multiple by $z$.
$$
e^{i\theta}s_N = e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \cdots + e^{i(n+1)\theta}
$$
Next make the subtraction $s_N - e^{i\theta}s_N$.
So
$$
s_n(1 - e^{i\theta}) = 1 - e^{i(n+1)\theta}\iff s_n = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}}.
$$

Note that we are starting from \(k=1\). So the first term should be, \(e^{i\theta}\).