# Toronto Math Forum

## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Min Gyu Woo on September 20, 2018, 08:15:41 PM

Title: Section 1.4 Example 8
Post by: Min Gyu Woo on September 20, 2018, 08:15:41 PM
Can someone prove this using the definition of limits for sequences:

$\lim_{n\rightarrow\infty} (1/n)(\cos{(n\pi/4)}+i\sin{(n\pi/4)}) = 0$
Title: Re: Section 1.4 Example 8
Post by: Alexander Elzenaar 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.)
Title: Re: Section 1.4 Example 8
Post by: Min Gyu Woo on September 21, 2018, 07:55:57 AM
Could you explain more about the trig function vanishing?
Title: Re: Section 1.4 Example 8
Post by: Victor Ivrii 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
$$| \frac{1}{n}\bigl(\cos (\theta)+i\sin(\theta)\bigr| = \frac{1}{n} |\cos (\theta)+i\sin(\theta|= \frac{1}{n}.$$
Title: Re: Section 1.4 Example 8
Post by: Vedant Shah 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)
Title: Re: Section 1.4 Example 8
Post by: Victor Ivrii on September 23, 2018, 09:13:50 PM
Explain, how you use squeeze theorem.
Title: Re: Section 1.4 Example 8
Post by: Vedant Shah on September 24, 2018, 11:22:43 AM
$-1 \le \sin(\frac{n \pi}{4}) \le 1$
$\frac{-1}{n} \le \frac{\sin(\frac{n \pi}{4})}{n} \le \frac{1}{n}$
$\lim_{n \to \infty} \frac{-1}{n} \le \lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le \lim_{n \to \infty} \frac{1}{n}$
$0 \le lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le 0$

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:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + i \frac{sin(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + lim_{n \to \infty} i *lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0+i0 = 0$

Title: Re: Section 1.4 Example 8
Post by: Victor Ivrii 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