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Messages - Victor Ivrii

Pages: [1] 2 3 ... 170
1
Chapter 4 / Re: 4.2 question 28
« on: October 25, 2020, 01:13:50 PM »
you need to write it, if you hope for any answer

2
Test 1 / Re: 2020TT1 Deferred Sitting #1
« on: October 15, 2020, 08:17:16 AM »
Indeed, instead of $\log(\pm w)$ with $+2\pi mi$ we write $\log(w)$  with $+\pi mi$ since $\log (-w)=\log(w)+\i i$

3
Test 1 / Re: Abel's Theorem
« on: October 15, 2020, 08:11:23 AM »
Of some fundamental set (remember a constant factor!)

4
Chapter 2 / Re: Section 2.2 "closed form" Qs
« on: October 13, 2020, 09:35:57 AM »
Yes, some of them are geometric series, and some of $e^{z}$, $\sin(z)$, $\sinh(z)$ and so on. However some can be derived from those, ether by substitution (f.e. $z^2$ instead of $z$), some by integration, differentiation, multiplication by $z^m$ or combination of both. F.e. consider geometric $\dfrac{1}{1-z}$. Integratinfg we can get power series for $-\Log (1-z)$, diffeerentiating for $\frac{1}{(1-z)^m}$ ,...

5
Chapter 1 / Re: Section 1.4: Question 29 Proof Check
« on: October 08, 2020, 09:27:59 AM »
Indeed

6
Chapter 2 / Re: Problem of Section 2.6 Q28
« on: October 03, 2020, 07:54:48 AM »
Hi there, I think there might be some problem with Q28. After I found the integrating factor and then solve for the solution, it's impossible to find the solution of my h(y) in this case (see the last line of my writing).
If you found an integrating factor correctly there should be no problem to $h(y)$. There is a problem with your solution, not with the problem.
In the textbook, it is e^(-2y). Maybe this one is correct.
Both are correct

7
Chapter 2 / Re: Question on 2.1 Example 10
« on: October 03, 2020, 04:50:32 AM »
is du/dx+i dv/dx? Why can't we take the derivative with respect to y?
You can take also derivative by $y$ but you need to multiply it by $i$ (think why)

8
Try to solve it by yourself (it follows right from the definition)

9
Chapter 1 / Re: why complex plane closed
« on: October 03, 2020, 04:46:26 AM »
why the complex plane is both open and closed? why did it close? Because by definition, a set is called closed if it contains its boundary. And I don't see the complex plane contains its boundary.
And what is the boundary of $\mathbb{C}$?

10
Chapter 1 / Re: Section 1.4: Question 29 Proof Check
« on: October 03, 2020, 04:45:02 AM »
You should remember that each segment of the polygonal curve may be served not by two discs, but several discs (think why)

11
Chapter 1 / Re: What is the meaning of complex number
« on: October 03, 2020, 04:42:37 AM »
This "meaning" has no sense. You should worry about math, not about this crasp

12
Chapter 2 / Re: Help for section 2.4, questions 14-15
« on: September 29, 2020, 04:54:13 PM »
Almost done, but you need also define interval, on which solution exists; it must include $0$

13
Chapter 1 / Re: Definition for limit is infinity?
« on: September 29, 2020, 04:35:15 AM »
That's right but it is a custom to denote by $\varepsilon,\delta$ something small. In this case capital letters, say $R,M$ would be better, for arbitrarily large

14
It was explained in the lectures that solutions in the parametric form are also admissible. You could ask it on the tutorial.

15
Chapter 2 / Re: W3L3 Exact solutions to inexact equations
« on: September 29, 2020, 04:31:19 AM »
In week 3 lecture 3, we get the example $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$.

My question is why is this good enough? We didn't answer the original question. We answered a modified version which we chose specifically because it seems easier to us. Shouldn't we still find a solution to the original equation $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$? Is there some way we can "divide out" $y^2$ from $y^3\cos(x)+y^5\sin(x)=C$ to get it?
We have not modified equation, but simply multiplied it by an integrating factor. These two equations are equivalent, except as $y=0$, that means $C=0$. But $y=0$ os also a solution to the original equation. Checking this would give you a 100% correct solution, otherwise it is almost perfect

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