Author Topic: Section 1.4 Example 8  (Read 1409 times)

Min Gyu Woo

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Section 1.4 Example 8
« on: September 20, 2018, 08:15:41 PM »
Can someone prove this using the definition of limits for sequences:

$\lim_{n\rightarrow\infty} (1/n)(\cos{(n\pi/4)}+i\sin{(n\pi/4)}) = 0$

Alexander Elzenaar

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Re: Section 1.4 Example 8
« Reply #1 on: September 20, 2018, 10:27:24 PM »
We simply need to show that the absolute value of $(1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr) $ tends to zero as $ n \to \infty $. Note first that $ \lvert (1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr)\rvert = 1/n $ (the trig functions vanish due to the Pythagorean identity).

I claim that for every $\varepsilon > 0 $ there exists some $ N \in \mathbb{N} $ such that for all $ n > N $ we have $ \lvert 1/n \rvert < \varepsilon $. Let such a $ \varepsilon $ be given; then by the Archimedian property of the real numbers, there exists a natural number $ N $ such that $ 1/\varepsilon < N $. If we pick any $ n > N $, then $ 1/\varepsilon < n $, and $ 1/n < \varepsilon $; we are done. (Note: I am justified in dropping the absolute value bars because everything is positive.)
« Last Edit: September 20, 2018, 11:41:25 PM by Victor Ivrii »

Min Gyu Woo

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Re: Section 1.4 Example 8
« Reply #2 on: September 21, 2018, 07:55:57 AM »
Could you explain more about the trig function vanishing?

Victor Ivrii

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Re: Section 1.4 Example 8
« Reply #3 on: September 21, 2018, 09:17:12 AM »
Could you explain more about the trig function vanishing?
" trig function vanishing" is the product of your imagination. Essentially you are told that
$$
| \frac{1}{n}\bigl(\cos (\theta)+i\sin(\theta)\bigr| = \frac{1}{n}  |\cos (\theta)+i\sin(\theta|= \frac{1}{n}.
$$
« Last Edit: September 23, 2018, 09:12:20 PM by Victor Ivrii »

Vedant Shah

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Re: Section 1.4 Example 8
« Reply #4 on: September 23, 2018, 05:25:09 PM »
Can we use squeeze theorem on $\cos{\frac{n \pi}{4}}$ and $ i \sin{\frac{n \pi}{4}}$?
Then the sum of the limits is the limit of the sums (provided they both exist)

Victor Ivrii

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Re: Section 1.4 Example 8
« Reply #5 on: September 23, 2018, 09:13:50 PM »
Explain, how you use squeeze theorem.

Vedant Shah

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Re: Section 1.4 Example 8
« Reply #6 on: September 24, 2018, 11:22:43 AM »
Start with:
$-1 \le \sin(\frac{n \pi}{4}) \le 1$
$\frac{-1}{n} \le \frac{\sin(\frac{n \pi}{4})}{n} \le \frac{1}{n} $
$\lim_{n \to \infty} \frac{-1}{n} \le \lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le \lim_{n \to \infty} \frac{1}{n}$
$ 0 \le lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le 0$

Thus by squeeze theorem,
$\lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} = 0$

After a similar argument, we get that:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0$

We also have:
$lim_{n \to \infty} i = i$

Since all three limits exist, we have:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + i \frac{sin(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + lim_{n \to \infty} i *lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0+i0 = 0$


« Last Edit: September 24, 2018, 12:06:29 PM by Victor Ivrii »

Victor Ivrii

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Re: Section 1.4 Example 8
« Reply #7 on: September 24, 2018, 12:05:02 PM »
Yes, it is an overcomplicated proof.

"mathoperators" like lim, sin, cos , ... should be typed as \lim, \sin, \cos ... to produce a proper output
« Last Edit: September 24, 2018, 12:07:51 PM by Victor Ivrii »