ISI2015-MMA-19

Question

The limit $\:\:\:\underset{n \to \infty}{\lim}  \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is

• A. $2$
• B. $2e$
• C. $2 \pi$
• D. $2i$

Comments

How is this related to mathematical logic?

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I think something is wrong in this question..I m getting 2 if I take + instead of - otherwise zero

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Byusing DeMoivre's Thm,

$e^{ix}$= $\cos(x)$ + $\iota\sin(x)$

By using Trigonomteric Sum and Difference Formula
\[ \cos A - \cos B = -2\sin((A+B)/2)\sin((A-B)/2)\]\[ \sin A - \sin B = +2\cos((A+B)/2)\cos((A-B)/2)\]

    $\lim_{n \to \infty} \sum_{k=1}^{k=n} -2\sin((2k-1)\pi/n)sin(\pi/n) + \iota 2\cos((2k-1)\pi/n)cos(\pi/n)$
    
As limit and summation can be interchanged,on taking $\lim_{n \to \infty}$ and 1/n tends to 0
$\sum_{k=1}^{k=\infty} -2\sin((2k-1)\pi/n)sin(\pi/n) + \iota 2\cos((2k-1)\pi/n)cos(\pi/n)$,

simplifies to
=$-2\sin((2k-1)\pi*0)\sin(\pi*0) + \iota 2\cos((2k-1)\pi*0)cos(\pi*0)$
=$ -2\sin(0)\sin(0) + \iota 2\cos(0)cos(0) $  ($\because$ sin(0)=0and cos(0)=1)
=2$\iota$

Answer : D

Answer 1

Answer: $C$

The given expression can be simplified as:

$$=\bigg| e^{\frac{2\pi ik}{n}}- e^{\frac{2\pi i(k-1)}{n}}\bigg| $$

$$= \bigg| e^{\frac{2\pi i(k-1)}{n}}\bigg |\bigg |e^{\frac{2\pi i}{n}}-1\bigg | $$

$$= \bigg|\cos\bigg (\frac{2\pi}{n}\bigg )-1 + i\sin\bigg (\frac{2\pi}{n}\bigg )\bigg |$$

$$= \bigg | -2\sin^2\bigg (\frac{\pi}{n}\bigg) + 2i\sin\bigg(\frac{\pi}{n}\bigg)\cos\bigg (\frac{\pi}{n}\bigg)\bigg |$$

$$= 2\sin\bigg(\frac{\pi}{n}\bigg )$$

Now, the required limit will be:

$$\therefore \underset{n \to \infty }{\lim} \sum^{n}_{k = 1}2\sin \bigg (\frac{\pi}{n}\bigg ) = \underset {n \to \infty}{\lim}2n\sin\bigg (\frac{\pi}{n}\bigg ) = 2\pi$$

$\therefore$ C is the right answer.

Comments

Here's a mistake in the line shown below.

Besides can't you make some side notes like below?

$|x+iy|=\sqrt{x^2+y^2}$
$e^{i\theta}=\cos{\theta}+i\sin{\theta}\Rightarrow |e^{i\theta}|=\sqrt{\cos^2\theta+\sin^2\theta}=1$

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thanks.

Side notes like?

You mean more explanation?

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Yeah. But please correct the mistake first. Side notes are optional although it helps others to know understand thoroughly. 🙂

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done.

Answer 2

Answer: C

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Explanation:

Note that $$e^{\frac{2\pi \iota k}{n}}\enspace ;1\le k\le n$$ can be thought of as a roots of unity.

Since roots of unity form a regular n-gon. Hence, the side length of the side of n-gon is given by $$|e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|$$.

Thus, the sum $$\sum_{k=1}^n |e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|$$ is the perimeter of the regular n-gon.

 

Further, one should note that as $n\to \infty$, the regular n-gon forms a circle.

Thus, $$\lim_{n\to \infty} \sum_{k=1}^n |e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|=2\pi$$