0
$\begingroup$

I tried to evaluate the limit of the following sequence:

$$ \frac {1}{n} \left( \sin \frac{\phi}{n} + \sin \frac{2\phi}{n} + \, … \, + \sin \frac{(n-1)\phi}{n} \right)$$

but i can’t seem to come up with anything. Asymptotic estimates don’t help, i tried substituting phi with a random x and the graph looks like a sinusoidal odd function with something like an exponential multiplied. So the limit should be finite. That’s about as far as i got.

$\endgroup$
4
  • 2
    $\begingroup$ This looks like a Riemann sum. So there should (could) be an integral related to it. $\endgroup$
    – MasB
    Dec 28, 2021 at 13:33
  • $\begingroup$ I thought about Riemann sum but one knows that $\varphi\left(a + \frac{b-a}{n}i\right) \cdot \frac{b-a}{n}$ is summand. I don't see easily such an expression. But it is possible to find partial sum by using complex exponential function representation of sine. Or he can multiply and divide by an opted cosine function to procure differences that cancel out each other $\endgroup$ Dec 28, 2021 at 13:40
  • $\begingroup$ @sergeiivanov use $b=\phi$ and $a=0$ $\endgroup$
    – Andrei
    Dec 28, 2021 at 13:43
  • $\begingroup$ Knowing that $\varphi (a + hi) = \varphi (b-hi)$ given the segment of integration $[a,b]$ with partition step $h$. Yes, it might work for $\sin x\psi$ on some segment where the difference is 1. So, $$\frac{1}{\psi}\sum_{k = 1}^{n-1} \sin \left(0+ \frac{\psi}{n}k\right) \cdot \frac{\psi}{n}$$ $\endgroup$ Dec 28, 2021 at 13:48

1 Answer 1

6
$\begingroup$

You could use the following $$ \sin(x)+\sin(2x)+\dots+\sin(kx)=\dfrac{\sin\left(\frac{k+1}{2}x\right)\cdot \sin\left(\frac{kx}{2}\right)}{\sin\left(\frac{x}{2}\right)} $$ Then we taking $x=\frac{\phi}{n}$ and $k=n-1$ we have \begin{align} \frac {1}{n} \left( \sin \frac{\phi}{n} + \sin \frac{2\phi}{n} + \dots + \sin \frac{(n-1)\phi}{n} \right)&=\frac{1}{n}\dfrac{\sin\left(\frac{n}{2}\frac{\phi}{n}\right)\cdot \sin\left(\frac{(n-1)}{2}\frac{\phi}{n}\right)}{\sin\left(\frac{\phi}{2n}\right)}\\ &=\frac{\sin(\frac{\phi}{2})\cdot \sin\Big((1-\frac{1}{n})\frac{\phi}{2}\Big)}{\frac{\sin(\phi/2n)}{1/n}} \end{align} Doing $n\to+\infty$ we take $$\displaystyle\lim_{n\to +\infty}\frac {1}{n} \left( \sin \frac{\phi}{n} + \sin \frac{2\phi}{n} + \dots + \sin \frac{(n-1)\phi}{n} \right)=\frac{2}{\phi}\sin^2(\phi/2).$$

You can also use Riemann sums, for example in this case \begin{align} \displaystyle\lim_{n\to +\infty}\frac {1}{n} \left( \sin \frac{\phi}{n} + \sin \frac{2\phi}{n} + \dots + \sin \frac{(n-1)\phi}{n} \right)&=\lim_{n\to +\infty}\sum_{k=1}^{n-1}\sin(\frac{k\phi}{n})\frac{1}{n}\\ &=\int_0^1 \sin(\phi x)dx\\ &=\frac{1}{\phi}(1-\cos(\phi))\\ &=\frac{2}{\phi}\sin^2(\phi/2) \end{align}

$\endgroup$
1
  • $\begingroup$ Thanks, i hadn’t even thought to think it as a Riemann sum $\endgroup$
    – gent96
    Dec 28, 2021 at 21:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .