Author Topic: Q5 TUT 0101  (Read 154 times)

Victor Ivrii

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Q5 TUT 0101
« on: November 02, 2018, 03:28:51 PM »
Give the order of each of the zeros of the given function:
$$
z^2(1-\cos(z)).
$$

Tianfangtong Zhang

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Re: Q5 TUT 0101
« Reply #1 on: November 02, 2018, 03:29:12 PM »
Let $z^2(1-\cos(z)) = 0$

Then $z = 0$ or $\cos(z) = 1$

thus $z = 2k\pi$

 

case $1$: $z = 0$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(0)=0$

$f^{'}(0) = 2z|_{z=0} = 0$

$f^{''}(0)=2\neq 0$

thus order = 2

Then $h(0) = 0$

$h^{'}(0) = 2\sin(z)|_{z=0} = 0$

$f^{''}(0)=\cos(z)\neq 0$

thus order = 2

Therefore order(0) = 4



case $2$: $z = 2k\pi$ ($k\neq 0)$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(z) = z^2 = (2k\pi)^2 \neq 0$

thus order = 0

$h(z) = 1- \cos(z) = 0 $

$h^{'}(z) = \sin(z) = 0$

$h^{''} = \cos(z) = 0$

thus order = 2

Therefore order($2k\pi$) = 2, $k\neq0$
« Last Edit: November 03, 2018, 01:34:37 AM by Tianfangtong Zhang »

Yuechen Huang

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Re: Q5 TUT 0101
« Reply #2 on: November 02, 2018, 04:59:54 PM »
z² (1 − cos(z))=0

z = 0 or cos (z) = 1

z = 0 or 𝑧 = 2𝑘𝜋
When z = 0

let 𝑓(𝑧) = z², ℎ(𝑧) = 1− cos (𝑧)

then 𝑓(0) = 0, 𝑓′(0) = 2𝑧 = 0, 𝑓″(0) = 2 ≠ 0

thus order = 2

ℎ(0)=0, ℎ′(0)= 2 sin(0)=0, h″(0)= cos(𝑧)≠0

thus order = 2

Therefore, order (0) = 4


When 𝑧 = 2𝑘𝜋 (𝑘≠0)

let 𝑓(𝑧) = z² and ℎ(𝑧) = 1−cos(𝑧)

𝑓(𝑧) = z² = (2𝑘𝜋)² ≠ 0
 
thus order = 0 

ℎ(𝑧)= 1− cos(𝑧)=0, ℎ′(𝑧)= sin(𝑧)=0, ℎ″= cos(𝑧)=0

thus order = 2
Therefore order (2𝑘𝜋) = 2, 𝑘 ≠ 0

Victor Ivrii

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Re: Q5 TUT 0101
« Reply #3 on: November 04, 2018, 09:39:43 PM »
Yuechen
no need to post the same solution