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Messages - Victor Ivrii

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31
Quiz-6 / Q6 TUT 0101
« on: November 17, 2018, 03:49:22 PM »
The coefficient matrix contains a parameter $\alpha$.

(a)  Determine the eigenvalues in terms of $\alpha$.
(b)  Find the critical value or values of  $\alpha$  where the qualitative nature of the phase portrait for the system changes.
(c) Draw a phase portrait for a value of  $\alpha$ slightly below, and for another value slightly above, each critical value.
$$\mathbf{x}' =\begin{pmatrix}
\frac{5}{4} &\frac{3}{4}\\
\alpha & \frac{5}{4}
\end{pmatrix}\mathbf{x}.$$

32
MAT244--Misc / Re: How to post the typed solutions properly
« on: November 16, 2018, 05:28:28 PM »
There are no mind readers or clairvoyants around, AFAIK. You need to be far more specific.

33
MAT334--Lectures & Home Assignments / Re: 2.5 Q 28
« on: November 15, 2018, 06:14:47 AM »
Quote
point because left side limit and right side limit are different
There is no "left / right side limit" in complex variables. Statement like this will bring you 0 mark for the whole problem in the exam paper.

34
If $\xi$ is an eigenvector, corresponding to eigenvalue $k$, so is $\alpha \xi$ ($\alpha$, $\beta$  are scalars).

If $\xi^{(1)}$ and $\xi^{(2)}$ are eigenvectors, corresponding to the same eigenvalues $k$, so is $\alpha \xi^{(1)} +\beta \xi^{(2)}$

35
MAT334--Lectures & Home Assignments / Re: 2.5 Q20
« on: November 13, 2018, 04:37:12 PM »
Aleena It is not the answer because you rely upon unsustained claim that if $f(z)=0$ then it's Laurent's coefficients are $0$ which is an equivalent form of the same question.
Hint: Consider $\int_\gamma (z-z_0)^m  f(z)\,dz $ where $\gamma$ is a counter-clockwise circle around $z_0$ and prove that it is equal to $2\pi i a_{m-1}$.

36
MAT244--Lectures & Home Assignments / Re: Complex eigenvalues
« on: November 13, 2018, 12:42:46 PM »
If the characteristic polynomial $\det (A-kI)$ has a multiple root (say root $k$ of multiplicity $m$) then it can have any number $l=1,\ldots, m$ of linearly independent eigenvectors, associated with this eigenvalue (we do not distinguish here real or complex eigenvalues). Then one needs to find Jordan (echelon?) form.

In particular, let $m=2$. Then either there are two l.i. eigenvectors $\mathbf{e}_1$, $\mathbf{e}_2$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \mathbf{e}_1+C_2\mathbf{e}_2\Bigr)
$$
or there exists a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}=(A-k)\mathbf{f}\ne 0$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}+\mathbf{f}\bigr]+C_2\mathbf{e}\Bigr)
$$
If $m=3$ then there are three cases: either there are three l.i. $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, in which case solution is similar to the former case of $m=2$, or there are a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}_1=(A-k)\mathbf{f}\ne 0$, and $\mathbf{e}_2$, also eigenvector but not proportional $\mathbf{e}_1$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}_1+\mathbf{f}\bigr]+C_2\mathbf{e}_1+C_3\mathbf{e}_2\Bigr);
$$
or there exists a vector $\mathbf{g}$ s.t. $(A-k)^3\mathbf{g}=0$, but $\mathbf{e} =(A-k)^2\mathbf{g}\ne 0$; let $\mathbf{f} =(A-k)\mathbf{g}$ and
the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[\frac{t^2}{2}\mathbf{e}_1+t\mathbf{f}+\mathbf{h}\bigr]+C_2\bigl[t\mathbf{e} +\mathbf{f}\bigr]+C_3\mathbf{e}\Bigr).
$$
and so on...

37
MAT244--Misc / Re: Question about quiz4
« on: November 13, 2018, 12:43:26 AM »
If you wrote Quiz with your section, it should travel between TA of your section, and TA who graded it (David graded Q4) and back. Usually TAs organize it rather well but sometimes Quiz is lost. Did you get your mark?


Crowdmarking would help reduce if not eliminate such cases but there are some problems, rather technical.

38
MAT244--Misc / Re: concern about the quiz6 grade
« on: November 12, 2018, 04:01:45 PM »
Do you have a time machine? Quiz 6 will be written Tue Nov 13 the earliest

39
Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4
« on: November 12, 2018, 03:21:48 PM »
Estimate over large semi-circle contains fixable errors (I would show on the typed solution) and is over-complicated: The integrand does not exceed $(R-1)^{-2}(|\ln R +5)^2$ (rough estimate) and the  integral does not exceed this multiplied by $\pi R$, so it tends to $0$ as $R\to \infty$

Estimate over large semi-circle also contains errors and is over-complicated: Integrand does not exceed $2|\ln \varepsilon|^2$ nd the  integral does not exceed this multiplied by $\pi
\varepsilon$, , so it tends to $0$ as $\varepsilon\to 0$.

40
Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4
« on: November 12, 2018, 04:14:18 AM »
Difficult to read. Completely insufficient space between lines.

On test and exam grader could just miss correct elements of the solution.

41
MAT334--Lectures & Home Assignments / Re: Question 2 from 2.5
« on: November 12, 2018, 03:15:47 AM »
Post scans, not crappy snapshots

42
MAT244--Lectures & Home Assignments / Re: integral problems
« on: November 11, 2018, 01:13:57 AM »
$*$ IS N OT A SIGN OF MULTIPLICATION

43
Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4M
« on: November 10, 2018, 11:42:07 PM »
Jeff
What you are doing is correct but does not make sense. More precisely $\int_{-\infty}^\infty e^{ibz}\cos(z)\,dz$ diverges in the "normal" sense but converges in the sense of distributions, which are more advanced ..

In fact, you can calculate $\int_0^R e^{ix}\cos (x)\, dx$ without taking the limit first. However one needs to justify the differentiation.

Since this problem is more difficult than the corresponding problems in Test 2 or posted alternative, I sketch a correct solution. First, we consider $\int_C \frac{e^{iz}}{z}\,dz=0$ due to Cauchy theorem. Next, integral over large semicircle indeed tends to $0$ because
$$
|\int_C \frac{e^{iz}}{z}\,dz|\le \int_0^{\pi} e^{-\Im z(\theta)}\,d\theta=
2  \int_0^{\pi/2} e^{-R\sin (\theta)}\,d\theta\le
2  \int_0^{\pi/2} e^{-R\theta/2}\,d\theta\le \frac{4}{R}
$$
where we used $\sin (\theta)\ge \theta /2$ as $0<\theta <\pi/2$.

Next, integral over small semicircle equals
$$
\int _{C_4} \frac{e^{iz}-1}{z}\,dz +\int _{C_4} \frac{1}{z}\,dz
$$
where
$$
|\int _{C_4} \frac{e^{iz}-1}{z}\,dz |\le \int _{0}^{\pi} \frac{|e^{iz(\theta)}-1|} d\theta\le  \int_0^\pi \varepsilon\,d\theta \le \pi \varepsilon
$$
tends to $0$ as $\varepsilon \to 0^+$ and $\int _{C_4} \frac{1}{z}\,dz=\int_\pi^0 i d\theta =-\pi i$.

Therefore
$$
\Bigl[\int_{-R}^{-\varepsilon} \frac{e^{ix}}{x}\,dx + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]\to \pi i\tag{*}\label{eq-X}
$$
as $R\to +\infty$, $\varepsilon to 0^+$. However it is true for the sum only, limits of each part simply do not exist! $\int_0^* \frac{e^{ix}}{x}\,dx $ diverges at $0$.

Observe that the LHE in (\ref{eq-X}) equals
$$
\Bigl[\int_{R}^{\varepsilon} \frac{e^{-ix}}{-x}\,d(-x) + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]=
\Bigl[-\int_{\varepsilon}^{R} \frac{e^{-ix}}{x}\,dx + \int_{\varepsilon }^{R} \frac{e^{ix}}{x}\,dx\Bigr]=
\int_{\varepsilon}^R \frac{e^{ix}-e^{-ix}}{x}\,dx= 2i\int_\varepsilon^R \frac{\sin(x)}{x}\,dx
$$
which tends to $2i J $ as $R\to \infty$, $\varepsilon\to 0^+$ and therefore $J= \pi i/2i =\pi/2$.

44
MAT334--Lectures & Home Assignments / Re: Poles and several singularities
« on: November 10, 2018, 10:52:24 PM »
Indeed, decomposition is valid only for $0<|z-z_1|<r_1$ but polynomial $P_1((z-z_1)^{-1})$ is defined for all $z\in \mathbb{C}$ and therefore $g(z)=f(z)-P_1(z)$ is defined for all $z\in D$, and analytic there except $z_2,\ldots z_N$.

45
MAT334--Lectures & Home Assignments / Re: Poles and several singularities
« on: November 10, 2018, 02:51:29 PM »
1. If $z_0$ is a pole, then $|f(z)|\to \infty$ as $z\to z_0$ and therefore $\forall M \exists r>0: |z-z_0|<r\implies |f(z)|\ge M$.

2. If $z_0$ is a pole of order $m$ them $f(z)=\sum_{n=-m}^\infty a_n(z-z_0)^n$. Then $f(z)=P((z-z0)^{-1}+g(z)$ with $P((z-z_0)^{-1})=\sum_{n=-m}^{-1}a_n(z-z_0)^n$.

If $f(z)$ is analytic in $D$ except poles $z_1,\ldots, z_N$, then $f(z)=P_1((z-z_1)^{-1} +g(z)$ where $g(z)$ is analytic in $D$ except poles $z_2,\ldots, z_N$. "Repeat" then $g(z)=P_1((z-z_2)^{-1} +h(z)$ where $g(z)$ is analytic in $D$ except poles $z_3,\ldots, z_N$ and so on...


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