Compute the sum $\sum_{i=1}^n 2\cos{ix}$ by using the fact that $\cos{kx} = \real\left({(\cos{x}+i\sin{x})^k}\right)$
#Complex Numbers Question
sour mirage
wet otterBOT
martin(ping-me-when-reply)
sour mirage
If someone could please assist me with maybe the first step to solving this that'd be nice as I am super stuck
quaint marten
$2\cos nx = (\cos nx + i\sin nx) + (\cos (-nx) + i\sin (-nx)) = (\cos x+i\sin x)^n + (\cos x + i\sin x)^{-n}$
wet otterBOT
k12byda5h
quaint marten
Compute the sum $\sum_{i=1}^n 2\cos{ix}$ by using the fact that $\cos{kx} = \rea...
btw, you use the same variables in the sigma. It shouldn't be like that
it can be change to $\sum_{i=1}^n 2\cos ix$
wet otterBOT
k12byda5h
rigid tulip
To compute the sum $\sum_{n=1}^\infty 2\cos{nx}$, we can use the fact that $\cos{kx} = \real\left({(\cos{x}+i\sin{x})^k}\right)$ as follows:
\begin{align*}
\sum_{n=1}^\infty 2\cos{nx} &= 2\sum_{n=1}^\infty \cos{nx}
&= 2\sum_{n=1}^\infty \real\left({(\cos{x}+i\sin{x})^n}\right)
&= 2\real\left(\sum_{n=1}^\infty (\cos{x}+i\sin{x})^n\right)
&= 2\real\left(\frac{1}{1-(\cos{x}+i\sin{x})}\right)
&= 2\real\left(\frac{1-\cos{x}+i\sin{x}}{1-2\cos{x}+1}\right)
&= 2\real\left(\frac{1-\cos{x}+i\sin{x}}{2\sin^2{x}}\right)
&= \frac{2(1-\cos{x})}{2\sin^2{x}}
&= \frac{1-\cos{x}}{\sin^2{x}}.
\end{align*}
This is the final result for the sum $\sum_{n=1}^\infty 2\cos{nx}$. Note that the sum is infinite, so it doesn't make sense to specify an upper limit of $n$
wet otterBOT
gicu
sour mirage
wait
the sum isn't meant to be infinite
I am meant to compute it such that the upperbound of the limit is n
I shouldn't have used n=1 as the variable of the sum
I should have just used i
Compute the sum $\sum_{i=1}^n 2\cos{ix}$ by using the fact that $\cos{kx} = \real\left({(\cos{x}+i\sin{x})^k}\right)$
wet otterBOT
martin(ping-me-when-reply)
rigid tulip
Sure, we can use the fact that $\cos kx = \real((\cos x + i\sin x)^k)$ to compute the sum $\sum_{i=1}^n 2\cos ix$.
First, we can rewrite the sum as follows:
$$\sum_{i=1}^n 2\cos ix = 2\sum_{i=1}^n \cos ix$$
Now, we can use the formula for $\cos kx$ to rewrite each term in the sum:
$$\begin{aligned} 2\sum_{i=1}^n \cos ix &= 2\sum_{i=1}^n \real((\cos x + i\sin x)^i) \ &= 2\sum_{i=1}^n \real((\cos x + i\sin x)^{i-1}(\cos x + i\sin x)) \ &= 2\sum_{i=1}^n \real((\cos x + i\sin x)^{i-1})(\cos x + i\sin x) \ &= 2\cos x\sum_{i=1}^n \real((\cos x + i\sin x)^{i-1}) + 2i\sin x\sum_{i=1}^n \real((\cos x + i\sin x)^{i-1}) \ &= 2\cos x\sum_{i=0}^{n-1} \real((\cos x + i\sin x)^{i}) + 2i\sin x\sum_{i=0}^{n-1} \real((\cos x + i\sin x)^{i}) \ \end{aligned}$$
Now, we just need to compute the two sums on the right-hand side. To do this, we can use the formula for a finite geometric series:
$$\sum_{i=0}^{n-1} a^i = \frac{a^n-1}{a-1}$$
Substituting the expression for the sum into our expression for the original sum, we get:
$$\begin{aligned} 2\cos x\sum_{i=0}^{n-1} \real((\cos x + i\sin x)^{i}) + 2i\sin x\sum_{i=0}^{n-1} \real((\cos x + i\sin x)^{i}) &= 2\cos x\frac{(\cos x + (n-1)\sin x)^{n}-1}{\cos x + (n-1)\sin x-1} \ &\quad + 2i\sin x\frac{(\cos x + (n-1)\sin x)^{n}-1}{\cos x + (n-1)\sin x-1} \ \end{aligned}$$
This is the final answer for the sum.
wet otterBOT
gicu
