Question 7 Of Ekman's Compilation (1 Viewer)

frog1944 · Nov 8, 2016

Hi,

I eventually gave up and looked for the answer on the net to question 7 complex numbers in ekman's compilation of questions asking "Simplify $\sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$ ".
When I looked at how to derive it, it all involved $e^{ix}$ and $t+t^2 +...+t^n = \frac {t(1-t^n)}{1-t}$ . Both of these I have not either seen before, or is in the syllabus.

So how would you tackle it?

Thanks

jathu123 · Nov 8, 2016

you can write that infinite sum as Im(cisx+cis2x+cis3x+...+cisnx) = Im(cisx+cis^2(x)+cis^3 (x) + ...+ cis^n (x) [DMT]. This is a geometric series, and so you use the formula for the sum of geometric series; Sn=(a(r^n-1)/(r-1)). They have basically written that in terms of t. So both the common ratio and 'a' is cisx. So they let t=cisx and they got that. I think you can realize and simplify that, and find the imaginary part to get the answer (but very tedious I guess), so they used the alternative version of cisx which is e^(ix) which i think makes it easier, but not entirely sure.

InteGrand · Nov 8, 2016

Just replace any e^{ix} you see with cis(x) (they are the same thing) and it becomes HSC-friendly.

InteGrand · Nov 8, 2016

$\noindent Another way to derive the result is as follows. Let $S = \sum _{k=1}^{n}\sin (kx)$. Multiply both sides by $\sin \frac{x}{2}$: $\left(\sin \frac{x}{2}\right)S = \sum_{k=1}^{n} \sin (kx)\sin \frac{x}{2}$. Now, from products to sums, we have $\sin (kx)\sin \frac{x}{2} = \frac{1}{2}\left(\cos \left(kx - \frac{x}{2}\right) - \cos\left(kx + \frac{x}{2}\right)\right) = \frac{1}{2}\left(\cos \left(\frac{(2k-1)x}{2}\right) - \cos \left(\frac{(2k+1)x}{2}\right)\right) =\frac{1}{2}\left(a_{k-1} - a_{k}\right)$, where $a_{k}:=\cos \left(\frac{(2k+1)x}{2}\right)$. Thus$

$\begin{align*}\left(\sin \frac{x}{2}\right)S &= -\frac{1}{2}\sum _{k=1}^{n} \left(a_{k}-a_{k-1}\right) \\ &= -\frac{1}{2} \left(a_{n} - a_{0}\right) \quad (\text{telescoping}) \\ &= \frac{\cos \frac{x}{2} - \cos \left(\frac{(2n+1)x}{2}\right)}{2} \\ \Rightarrow S &= \frac{\cos \frac{x}{2} - \cos \left(\frac{(2n+1)x}{2}\right)}{2\sin \frac{x}{2}}, \quad \sin \frac{x}{2} \neq 0.\end{align*}$

(If sin(x/2) is 0, then x/2 is a multiple of pi, so kx is a multiple of pi for each positive integer k, whence S = 0.)

jathu123 · Nov 8, 2016

InteGrand said:
$\noindent Another way to derive the result is as follows. Let $S = \sum _{k=1}^{n}\sin (kx)$. Multiply both sides by $\sin \frac{x}{2}$: $\left(\sin \frac{x}{2}\right)S = \sum_{k=1}^{n} \sin (kx)\sin \frac{x}{2}$. Now, from products to sums, we have $\sin (kx)\sin \frac{x}{2} = \frac{1}{2}\left(\cos \left(kx - \frac{x}{2}\right) - \cos\left(kx + \frac{x}{2}\right)\right) = \frac{1}{2}\left(\cos \left(\frac{(2k-1)x}{2}\right) - \cos \left(\frac{(2k+1)x}{2}\right)\right) =\frac{1}{2}\left(a_{k-1} - a_{k}\right)$, where $a_{k}:=\cos \left(\frac{(2k+1)x}{2}\right)$. Thus$

$\begin{align*}\left(\sin \frac{x}{2}\right)S &= -\frac{1}{2}\sum _{k=1}^{n} \left(a_{k}-a_{k-1}\right) \\ &= -\frac{1}{2} \left(a_{n} - a_{0}\right) \quad (\text{telescoping}) \\ &= \frac{\cos \frac{x}{2} - \cos \left(\frac{(2n+1)x}{2}\right)}{2} \\ \Rightarrow S &= \frac{\cos \frac{x}{2} - \cos \left(\frac{(2n+1)x}{2}\right)}{2\sin \frac{x}{2}}, \quad \sin \frac{x}{2} \neq 0.\end{align*}$

(If sin(x/2) is 0, then x/2 is a multiple of pi, so kx is a multiple of pi for each positive integer k, whence S = 0.)

oh man thats genius

Question 7 Of Ekman's Compilation (1 Viewer)

frog1944

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InteGrand

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InteGrand

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