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**APM346--Misc / Scope of TT2?**

« **on:**March 11, 2019, 05:17:23 PM »

As titled, what chapters will be on TT2?

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As titled, what chapters will be on TT2?

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Find the general solution and then plug to initial conditionsHello professor, could u give me a hint as to why we have to impose the restriction for absolute value of x to be smaller or equal to 1?

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There is NO root. You need to parametrize before integration

but, when we parametrize Y in terms of X, don't we have to use $$y^2+x^2=C$$

and thus $$y= +/- \sqrt{C-x^2}$$?

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I have attached my attempt. However I'm having trouble solving the problem fully with the initial condition provided. Anyone knows how to do it? Or did I make any mistakes along the way?

Thanks.

Thanks.

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Heller professor

By parametrizing Y in terms of X, do we need to put a plus/minus sign in front of the root? If yes, does that mean when we put down the final solution, we need to include plus and minus as well?

By parametrizing Y in terms of X, do we need to put a plus/minus sign in front of the root? If yes, does that mean when we put down the final solution, we need to include plus and minus as well?

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Junjing, you are right but I am not sure if anyone but me would be able to read your solutionOkay. Next time I will make it 'skinnier'. Sorry.

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Not sure if this will work

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I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.what i do is since it is a double-pole, i find the coefficient for the degree 2 term of the sine squared function, then i find the coefficient of the degree 1 term for the cosine function, then divide it out.

Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$

while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $

thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.

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So we need to know how to do TT1 + TT2 + Q7 ***AND*** the Sample Final correct?same question here

Also, the Sample Final contains questions about the stretch and rotation angle of a mobius transformation. Isn't that part of Chapter 3.4? I don't see anything in 3.3 that discusses rotation and stretch.

but i guess it never hurts to know how to do it.

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Can you explain how you figured it out?so refer to textbook page 50, the book provides a method. For x just follow the book, for y follow it until the end to prove that y1+y2=0, however the range of y is not limited to be positive in this question. In this case, we will have to expand:

cos(x+iy) = cosx coshy - i sinx sinhy

and realize that sinx is never 0, meaning sinhy will always be there, and sinhy is injective for a positive or negative y of same magnitude.

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There are only answers of part a and part b and it seems that we cannot reply to the question anymore.never mind, it is one-to-one. I figured it out.

Can somebody please post the answer of part d e f?

Thanks

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There are only answers of part a and part b and it seems that we cannot reply to the question anymore.also, for the graphing of the domains, textbook section 1.5 question 26 refers to figure 1.26 on page 53 as the domain of the mapping function. In the question it states that the mapping is one-to-one, but from what I can see it is not one-to-one.

Can somebody please post the answer of part d e f?

Thanks

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There are only answers of part a and part b and it seems that we cannot reply to the question anymore.same, I am stuck at proving the function to be 1 to 1.

Can somebody please post the answer of part d e f?

Thanks

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