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Messages - Victor Ivrii

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31
APM346--Misc / Welcome to APM346
« on: January 07, 2019, 06:30:22 AM »
Welcome to this forum!

The main purpose of this forum is not the communication with instructors but peer-to-peer. I noticed that some of you answer the questions of your peers and I really appreciate, but I would like to see more -- much more -- of this!

It runs as aid to the class website. More permanent and systematic information is placed on website, while forum is for discussion. Please read before registering!

Registration
• You can select any username (login) you want (take a rather short one). Username is what only you (and Admin) can see.
• However  change your (screen) Name  (the one which everybody sees) to one easily identifiable (Admin should be able to identify you with a student enrolled to this course--and matching BlackBoard name; otherwise your posting privileges could be reduced or revoked):  Open Profile > Account Settings and look for "Name".
• It would be preferable to use your University of Toronto email address.
• BTW you can have a cool avatar too. Please do not display any confidential info.
• You cannot change the loginname (used for login).  But there is no need.  Nobody except you knows your password - even Admins - but Admins don't need  to know it.

Some forum rules
• Please use different boards according to their description (I will create new ones if needed).
• Do not hijack topics: if a topic is devoted to some question, do not post anything which is not related. Start a new topic.
• On the other hand, do not start new topic answering to the existing post; use the same topic.
• Do not put in the same post several not related things - make several posts instead (in different topics).
• Do not post solution which coincides with the solution already posted by someone else; exception: typed solution beats the scanned one.
• Do not spam (post irrelevant messages, especially to promote something), flood (post many messages with very little or no sense, like "me too", "I like this post", ...) or flame (messages, containing personal attacks and heated arguments).
• Do not thank instructors.
• Do not post snapshots of paper documents, scan them (with the scanner or smartphone).
• Do not use red color (it belongs to Admins and Moderators), large fonts
• Read and search before postings.
• Use common sense.

Karma
• This semester karma does not translates into Bonus mark points (there are plenty reasons for this)
• On the other hand, your posts and active participation in the class will be a decisive factor if you ask me for a recommendation letter for a Graduate School. To write such letter I need to know something about you beyond your marks (and your final mark is in the transcript anyway). Participation in the forum automatically leaves a trace.

Have questions?
• If you have a question which you think is of general interest--ask it on the Forum rather than via email.

MathJax
• Both this forum and website use MathJax to display mathematics.

32
MAT244--Misc / Re: Quiz 5 explanation (was Unfair quiz)
« on: December 21, 2018, 05:30:52 PM »
For those who had < 4 for this Quiz Quiz mark was based on the best 4.5 quizzes (which means that the 5-th best had a half weight).

33
MAT244--Misc / Re: How do we calculate the WB bonus mark?
« on: December 21, 2018, 02:40:43 PM »
Karma * 0.4. I extracted karma from database but because I did it in rush to submit emarks before Dean's office closes , I could miss several students, especially those whose Screen name on forum differed from their Quercus name. Please email me, if it is the case: I am planning to amend marks Dec 24. My changes will not affect ACORN marks until Jan 8 for sure, and may be a bit later

34
Technical Questions / emoji
« on: December 20, 2018, 08:37:39 PM »
😈 👿 👹 👺 💀 👻 👽 🤖 💩 😺 😸 😹 😻 😼 😽 🙀 😿 😾

35
MAT244--Misc / Re: When should we expect the marks?
« on: December 20, 2018, 01:58:07 PM »
for  Quizzes and Bonuses Fri
Final -- I hope Fri (actually, I am required to do it in 7 business days)  but it depends on circumstances beyond my control. First, other graders. I graded the most difficult to grade P5 in 4 days, then I graded MAT334, and some of graders still have not finished. If they finish tonight, I will do my overhead overnight and submit Friday morning.

Second, marks need to be approved. If they are not approved by our undergraduate chair and then Dean's office before 5 pm, they are delayed until the first day of the next semester, and I am not allowed to release Crowdmark links until their approval.

36
Final Exam / FE-P4 official
« on: December 20, 2018, 05:15:13 AM »
(a) We consider inverse transform $z=g(w)$, since we know that it maps $\frac{1}{2}$ into $0$. It must be
\begin{gather*}
g(w)=\lambda \frac{w-a}{1- w\bar{a}}\\
\text{with $|\lambda|=1$ and $a=\frac{1}{2}$:}\\
g(w)=\lambda \frac{2w-1}{2- w}.
\end{gather*}
Since $g(-1)=1$ we have $1=\lambda \frac{-2-1}{2+1}\implies \lambda=-1$ and therefore
$$g(w)= - \frac{2w-1}{2- w}=z \implies (2w-1)=-z(2-w)\implies w= \frac{2z-1}{z-2}=: f(z).$$

(b)
$$\frac{2z-1}{z-2}=z\implies 2z-1=z^2-2z\implies z^2-4z +1=0\implies z=2\pm \sqrt{3}.$$
These are "black" points on the picture.

(c) Then $f' = \frac{2(z-2)-(2z-1)}{(z-2)^2}=-3 (z-2)^{-2}$; $f'(0)= -3/4$ and therefore stretch is $|-3/4|= 3/4$ and rotation is $\arg(-3/4)= \pi$.

37
Final Exam / Re: FE-P5
« on: December 20, 2018, 05:09:56 AM »
(a) As $|z|=2$ consider $Q(z)=z^3$; then  $|Q(z)|=8$,
$$|P(z)-Q(z)|=|2iz -3-i|\le 4+|3+i|=4+\sqrt{10}<8;$$
therefore $P$ has as many roots in $\{z\colon |z|<2\}$ as $Q(z)$ has, which is $3$ (we count orders).

(c) Then there are no roots in $\{z\colon |z|>2\}$.

(b) Consider $z=w+1$ with $|w|=1$; then $P(w+1)=w^3+3w^2 +5w -i$ and we set $Q(w)=5w$; then $|Q(w)|=5$,
$$|P(w)-Q(w)|=|w^3+3w^2 -i| < 5.$$
Indeed, $|P(w)-Q(w)|\le |w+3|+1< 5$ except $w=1$, and for $w=1$ we have $|w^3+3w^2 -i|=|4-i|=\sqrt{17}<5$. Therefore $P(w+1)$ has as many roots in $\{w\colon |w|<1\}$ as $Q(w)$ has, which is $1$ (we count orders).

Finally, in the domain  $\{z\colon |z-1|>1, |z|<2\}$, there are $3-1=2$ roots.

38
Final Exam / FE-P3 official
« on: December 20, 2018, 05:04:59 AM »
$\newcommand{\Res}{\operatorname{Res}}$
(a) Singular points are of $g(z)=\tan(z)$ and $h(z)=z\cot ^2(z)$, that is $z_n= (n+\frac{1}{2})\pi$ and $w_n = \pi n$.

(b) $z_n$ are simple poles and $\Res (f,z_n)= \Res (\tan(z), z_n)= \frac{\sin(z)}{(\cos (z))'}\bigr|_{z=z_n} =-1$.

$w_0$ is a simple pole
$$\Res (f,w_0)= \Res (z\cot^2(z), w_0)= \Res (z\cot (z) \times \cot(z), 0)= \Res (\cot(z),0)= 1$$
because $\lim _{z\to 0} z\cot(z)=1$.

$w_n$ with $n\ne 0$ are double poles and
\begin{align*}
\Res (f,w_n)=&\Res (z\cot^2 (z), w_n)= \Res ((\pi n +w)\cot^2 (w) , 0) =\\
&\pi n\Res (\cot^2 (w) , 0) +\Res (w\cot^2(w),0) =1
\end{align*}
because $\Res (\cot^2 (w) , 0)=0$ (since $\cot^2(w)$ is an even function) and $\Res (w\cot^2(w),0) =1$ we already calculated.

$\infty$ is a not isolated singularity and therefore residue here is not defined.

39
Final Exam / FE-P2 official
« on: December 20, 2018, 04:49:22 AM »
(a)--(b) If $|z|=r$ and $\arg{w}=t$, then $z=r e^{it}$ and
\begin{align*}
w= \frac{1}{2}\bigl(re^{it} +r^{-1}e^{-it}\bigr)&= \cosh(s)\cos(t)+i\sinh(s)\sin (t)
\end{align*}
with $s=\ln (r)$ (and $r=e^s$).

Then $w=u+iv$ with $\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$, $a=\cos(s)$, $b=\sinh(s)$ and also
$\frac{u^2}{A^2}-\frac{v^2}{B^2}=1$,
with  $A=\cos (\theta)$, $B=\sin(\theta)$, $A^2+B^2=1$.

(c) Consider, what the circle $\{z\colon |z|=1\}$ is mapped to: $z=e^{it}$, then
$w =\frac{1}{2}(e^{it}+e^{-it})=\cos(t)$ and it runs a line segment $[-1,1]\subset\mathbb{R}$. So $f$ maps the unit disk onto compliment of it $\mathbb{C}^*\setminus [-1,1]$ (and $0$ is mapped to $\infty$).

(d) See attachment

(e) $w=\frac{1}{2}(z+z^{-1}) \implies z^2 -2wz +1=0$; it has two roots $z_1$ and $z_2$ s.t.
$z_1z_2=1$ and $z_1+z_2=2w$; exactly one of them for $w\notin [-1,1]$ is in unit disk, and the second one is in $\{z\colon |z|>1\}$.

40
Final Exam / FE-P1 official
« on: December 20, 2018, 04:41:00 AM »
Observe that
$$f(z)=\frac{1}{(z+1+i)(z+1-i)}= \frac{1}{2i}\Bigl(\frac{1}{z-z_2}-\frac{1}{z-z_1}\Bigr)$$
has two singular points $z_{1,2} =-1\pm i=\sqrt{2}e^{\pm 5\pi i/4}$.

(a)   So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z_2-z}-\frac{1}{z_1-z}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^n z_2^{-n-1} - \sum_{n=0}^\infty z^n z_1^{-n-1}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^n \Bigr(z_2^{-n-1} - z_1^{-n-1}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^n 2^{-(n+1)/2}\Bigl(e^{5(n+1)\pi i/4} - e^{-5(n+1)\pi i/4}\Bigr)=\\
& \sum_{n=0}^\infty 2^{-(n+1)/2}\sin (5(n+1))\pi /4)z^n
\end{align*}
As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.

(b) So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z-z_1}-\frac{1}{z-z_2}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^{-n-1} z_1^{n} - \sum_{n=0}^\infty z^{-n-1} z_2^{n}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} \Bigr(z_1^{n} - z_2^{n}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} 2^{n/2}\Bigr(e^{5n\pi i/4} - e^{-5n\pi i/4}\Big)=\\
&\sum_{n=0}^\infty 2^{n/2}\sin (5n\pi /4)z^{-n-1} = \sum_{-\infty}^{-2} 2^{-(n+1)/2}\sin (5z(-n-1)\pi /4)z^{n}
\end{align*}
where we plugged $n=m-1$ with $m=1,2,\ldots$, observed that the term with $m=1$ is $0$ and finally replaced $m$ by $n$.

As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.

41
Final Exam / FE-P6 Official
« on: December 20, 2018, 04:37:41 AM »
There is a single pole of $f(z)=\frac{1}{1+z^n}$ inside $\Gamma$, namely $z=e^{i\pi/n}$, which is a simple pole and the residue is $\frac{1}{(1+z^n)'}\bigr|_{z=e^{i\pi/n}}= \frac{1}{nz^{n-1}}\bigr|_{z=e^{i\pi/n}}=-\frac{1}{n}e^{i\pi/n}$.

Therefore due to the residue theorem $I_R+J_R+K_R= -\frac{2}{n}\pi i \times e^{i\alpha/2}$, where $K_R$ is an integral over an arc and $J_R$ is an integral over the second straight segment.

Then
$$J_R=\int _{R}^0 \frac{e^{i\alpha}\,dt }{1+e^{in\alpha}t^n}=-e^{i\alpha}I_R$$
with $I_R=\int_0^R\frac{dx}{1+x^n}$.

On the other hand,
$$|K_R|=|\int_0^\alpha \frac{iRe^{it}\,dt}{1+e^{itn}R^n }|\le \frac{R}{R^n-1}\int_0^\alpha\, dt=\frac{\alpha R}{R^n-1}\to 0 \qquad\text{as }\ \ R\to \infty.$$

Then $(1-e^{i\alpha}) I =-\frac{2}{n}\pi i \times e^{i\alpha/2}$ and
\begin{align*}
I =& -\frac{2}{n}\pi i \times  \frac{e^{i\alpha/2}}{1-e^{i\pi \alpha}}=
&&-\frac{2}{n}\pi i \times  \frac{1}{e^{-i\alpha/2}-e^{i\pi \alpha/2}}=\\
&-\frac{2}{n}\pi i \times  \frac{1}{-2i\sin(\alpha/2)}=
&&\frac{\pi}{n\sin(\pi /n)}.
\end{align*}

42
Final Exam / FE-P4 official
« on: December 20, 2018, 02:01:42 AM »
$\renewcommand{\Re}{\operatorname{Re}}$
Characteristic equation:
$$\left|\begin{matrix} 1-k & -2\\ 1 & -1-k \end{matrix}\right|= k^2+1=0\implies k_{1,2}=\pm i.$$
Finding eigenvector corresponding to $k_1=i$
$$\begin{pmatrix} 1-i & -2\\ 1 & -1-i \end{pmatrix}\begin{pmatrix} \alpha\\ \beta \end{pmatrix}=0\implies \alpha =(1+i)\beta\implies \mathbf{e}_1 =\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$$
and $\mathbf{e}_2$ is complex conjugate. Then the general solution to the homogeneous equation is
\begin{align}
\begin{pmatrix}x\\y\end{pmatrix}=&\Re \Bigl[(C_1+iC_2) (\cos (t)+i\sin(t))\begin{pmatrix} 1+i\\ 1 \end{pmatrix}\Bigr]\notag\implies \\
&x=  C_1 (\cos(t)-\sin(t))+ C_2( -\cos(t)-\sin(t) )\\
&y=  C_1\cos(t)    -C_2\sin(t)  \,.
\label{eq-4-1}
\end{align}
We are looking for solution to inhomogeneous equation in the same form albeit with variable $C_1,C_2$. Then
\begin{align*}
&\left\{\begin{aligned}
&C'_1 (\cos(t)-\sin(t))+ C'_2( -\cos(t)-\sin(t) )=\sec(t),\\
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
&C'_1\sin(t)+ C'_2\cos(t)=-\sec(t),\\
&C'_1\cos(t)      -\ C'_2\sin(t) =0
\end{aligned}\right.\implies\\
&\left\{\begin{aligned}
&C'_1= -\sec(t)\sin(t)=-\tan(t)\implies C_1=\ln (\cos(t))+c_1\\
&C'_2=-\sec(t)\cos(t)=-1\implies C_2=-t +c_2\,.
\end{aligned}\right.
\end{align*}
Finally
\begin{align*}
&x=  [\ln (\cos(t))+c_1] (\cos(t)-\sin(t))+ [t -c_2]( \cos(t)+\sin(t) )\,\\
&y=[\ln (\cos(t))+c_1]\cos(t)   +[t-c_2]\sin(t) \,.
\end{align*}

43
Final Exam / FE-P3 official
« on: December 20, 2018, 01:59:33 AM »
Writing characteristic equation: $L(k):= k^3-2k^2-k+2=0$, with $L(k)=(k-2)k^2 -(k-2)=(k-2)(k^2-1)$; then $k_1=1, k_2=-1, k_3=2$. Then

y^*= C_1e^{t} + C_2 e^{-t}  + C_3 e^{2t}
\label{eq-3-1}

is a general solution to the homogeneous equation.

We are looking for solution to the inhomogeneous equation as (\ref{eq-3-1}) with unknown functions $C_1,C_2,C_3$ s.t.
\begin{align*}
&\left\{\begin{aligned}
&C_1' e^{t} + C_2' e^{-t}+ C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
&C_1' e^{t} + C_2' e^{-t}+4 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
2&C_1' e^{t} + 3C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
2&C_1' e^{t} +  6 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&C_1=-\int \frac{6e^{t}\,dt}{e^t+1}= -6 \ln (e^t+1) +c_1,\\[4pt]
&C_2= \int \frac{2e^{3t}\,dt}{e^t+1}= \int \Bigl[2e^{2t}- 2 e^{t}\Bigr] \,dt +\int  \frac{2e^{t}\,dt}{e^t+1}
=e^{2t} -2e^{t} +2\ln (e^t+1)+c_2,\\[4pt]
&C_3=\int \frac{4\,dt}{e^t+1}=\int \frac{4e^{-t}\,dt}{1+e^{-t}}= -4\ln (1+e^{-t})+c_3= -4\ln (e^t+1)+4t +c_3.
\end{align*}
Then
\begin{align*}
y= &\bigl[-6 \ln (e^t+1) +c_1\bigr]e^{t}+\bigl[e^{2t} -2e^{t} +2\ln (e^t+1)+c_2\bigr]e^{-t}+
\bigl[-4\ln (e^t+1)+4t +c_3\bigr]e^{2t}=\\
&-6  e^{t}\ln (e^t+1)+ 2e^{-t}\ln (e^t+1)
-4e^{2t}\ln (e^t+1)+4te^{2t}-2 +
c_1e^{t} + c_2 e^{-t}  + c_3 e^{2t}
\end{align*}
with $c_1:= c_1+1$ in the last transition.

44
Final Exam / FE-P2 official
« on: December 20, 2018, 01:53:44 AM »
Writing characteristic equation: $L(k):= k^3-3k^2+4k-2=0$. Obviously, one root $k_1=1$; then $k_2+k_3= 2$, $k_1k_2=2$ and they satisfy $k^2-2k+2=0\implies k_{1,2}= 1\pm i$. Then

y^*= C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)
\label{eq-2-1}

is a general solution to the homogeneous equation.

Solving inhomogeneous equations with RHE $f_1=10e^{t}$, $f_2=10 e^{-t}$, $f_3=\cos(t)$:
\begin{align*}
&y_{p1}= At e^t,\\
&y_{p2}=Be^{-t},\\
&y_{p3}= C\cos(t)+D\sin(t).
\end{align*}
Here $A L'(k)|_{k=1} = A(3k^2-6k+4)|_{k=1}=10\implies A=10$,
$BL(-1) =-10 B=10\implies B=-1$ and
$$(C+iD)L(i)= (A+iB) (1+3i)=20\implies C+iD= \frac{20}{1+3i}=\frac{20(1-3i)}{10}= 2-6i\implies C=2, D=6.$$
Then
\begin{align*}
&y_{p1}= 10t e^t,\\
&y_{p2}=-e^{-t},\\
&y_{p3}= 2\cos(t)+6\sin(t).
\end{align*}
Finally
$$y= \underbracket{10t e^{t}}_{y_{p1}}\underbracket{-e^{-t}}_{y_{p2}}+ \underbracket{2\cos(t)+6\sin(t)}_{y_{p3}} + \underbracket{C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)}_{y^*}.$$

45
Final Exam / FE-P1 official
« on: December 20, 2018, 01:49:48 AM »
As $M=2x\sin(y) +1$, $N= 4x^2\cos(y) + 3x\cot(y)+10 \sin(y)\cos(y)$ we have
$$M_y-N_x= 2x\cos(y)-[8x\cos(y)+3\cot(y)]=-6x \cos(y) -3\frac{\cos(y)}{\sin(y)} \implies \frac{N_x-M_y}{M}= 3\cot(y).$$
Therefore we are looking for an integrating factor $\mu(y)$, satisfying
$$\frac{\mu'}{\mu}= 3\cot(y)\implies \ln (\mu)=3\int \cot(y)\,dy=3\ln (\sin(y))$$
(we take constant equal $0$). Then $\mu =\sin^3(y)$.

After multiplication we get
$$\bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx + \bigl[4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\bigr]\,dy=0\,.$$
Looking for $H(x,y)$ satisfying
\begin{align}
&H_x = 2x\sin^4(y) +\sin^3(y)\,,\label{eq-1-1}\\
&H_y=4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\,.
\label{eq-1-2}

\end{align}
(\ref{eq-1-1}) implies
$$H(x,y)= \int \bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx= x^2\sin^4(y)+ x\sin^3(y) +h(y).$$
Then
$$H_y= 4x^2\sin^3(y)\cos(y)+ 3x\sin^2\cos(y) +h'(y)$$
and comparing with (\ref{eq-1-2}) we see that
$$h'(y)=10\sin^4(y)\cos(y)\implies h(y)=\int 10\sin^4(y)\cos(y)\,dy= 2\sin^5(y);$$
we pick-up constant $0$.

Finally
$$H(x,y)= x^2\sin^4(y)+ x\sin^3(y) +2\sin^5(y) =C$$
is a solution to the original problem.

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