Author Topic: hm6 Q1  (Read 1416 times)

Yumeng Wang

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Yumeng Wang

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Re: hm6 Q1
« Reply #1 on: October 26, 2015, 02:58:45 PM »
This is my answer for question1(a), the prove part.
« Last Edit: October 26, 2015, 03:04:27 PM by Yumeng Wang »

Yumeng Wang

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Re: hm6 Q1
« Reply #2 on: October 26, 2015, 03:24:56 PM »
This is my answer for prove part of (b).

Yumeng Wang

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Re: hm6 Q1
« Reply #3 on: October 26, 2015, 03:55:27 PM »
(d)

Bruce Wu

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Re: hm6 Q1
« Reply #4 on: October 28, 2015, 04:06:50 PM »
I don't see why $A_{n}=\omega_{n}$

Emily Deibert

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Re: hm6 Q1
« Reply #5 on: October 28, 2015, 06:07:27 PM »
After speaking to Professor Ivrii about it we found out that you simplify choose the value of $A_n$ for convenience. You do not need to plug in any tangent term.

Rong Wei

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Re: hm6 Q1
« Reply #6 on: October 28, 2015, 07:32:13 PM »
(d)
YU MENG,
I think in the step By Foundamental Theorem of Calculus, it should be d(-Xn'Xm + XnXm')/dx
the signs of your solution is wrong, but anyway, it doesn't influence the right answer.

Emily Deibert

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Re: hm6 Q1
« Reply #7 on: October 29, 2015, 11:27:54 AM »
Could someone clarify what we are supposed to do in part c? It seems like there is no question being asked.

Yeming Wen

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Re: hm6 Q1
« Reply #8 on: October 29, 2015, 12:00:25 PM »
I think the question is asking the case where $\lambda =0$.

Victor Ivrii

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Re: hm6 Q1
« Reply #9 on: October 29, 2015, 12:35:02 PM »
I think the question is asking the case where $\lambda =0$.

It is only part of the story. Consider plane $(\alpha,\beta)$. After you found when $\lambda=0$ is an e.v. you got an equation to $(\alpha,\beta)$ and it describes a hyperbola, which breaks the plane into 3 zones. Since $\lambda_n=\lambda_n(\alpha,\beta)$ depend on $(\alpha,\beta)$ continuously in each of those zones the number of negative eigenvalues is the same. This number changes when you go from one zone to another and some eigenvalue crosses $0$.

One can use hint: $\lambda_n$ monotone with respect to each of arguments. If $alpha>0,\beta>0$ then there are no negative e.v.