Exercise 8 chapter 7

Vasco,

Here is my attempt at 8.

nh.ch7.ex8.pdf (133.71 KB)     (rewritten)

Gary

Vasco,

I noticed the difference, but in my answer they are actually the same quantity in that $iR$ is an exponent of e, and so $R = y$ induces the rotation $\mathcal{R}$ as a multiple of y, so I wasn't sure that Needham intended them to be different. I will fix the the $R$'s. I still need to read your answer.

Gary

Vasco,

That looks like a problem that I should have allowed to sit on the shelf for a day or two before posting. I will revisit it and repost as soon as possible.

Gary

Gary

While reading your solution to exercise 8 and re-reading my own, I realised that there were a couple of typos on page 2 where I talked about the winding number of $F(\Gamma)$ about $c_0$ when it should really be about $0$.

I have corrected this and republished.

I am still reading your solution, but I noticed that in your diagram the black dots denoting the roots of $F(s)$ which are complex do not have a complex conjugate root indicated on the diagram.

Vasco

Gary

The way I interpret the hint is that Needham is suggesting we apply the Argument principle to $\Gamma$ when $R$ the radius of the semicircle is finite, and then let $R$ tend to infinity.

Vasco

Vasco,

It was just a figure of speech. It was actually only one line of code. I also added a bit more explanation of why the DE dies away.

Gary

Vasco,

Here is my view of it:

The total winding angle for each p-point should depend in part on where a wind begins. I read the hint to mean that one first traverses the segment $-iR$ to $iR$, which winds by an angle of $\pi$ around every p-point and winds $n\pi$ around 0 in the image, but only if Re(s) < 0 for every p-point. Even the first traversal produces this winding. A non- or null-traversal would give 0 winding, but an initial angle of $-\frac{\pi}{2}$. If one adopts an initial starting point for z of 1, the initial angle is different for every p-point. If one adopts an initial starting angle of 0, as in [8], then the full traversal of the segment $-iR$ to $iR$ is split into two segments and is not completed until $z$ has completed a traversal of $\Gamma$. But no matter where one begins the traversal, the winding $nu[F(s), 0]$ resulting from one traversal of the segment $-iR$ to $iR$ is just $n\pi$, but only if Re(s) < 0 for all s. Only the degree of decay depends on the total number of traversals, but I don't think we have to take the number of traversals into account.

Rereading my comment, I see that it probably didn't respond very well to your posting, but I don't see the need to count external and internal points, since by construction all the internal points have Re(s) < 0, only internal points are p-points, traversal of the segment on the imaginary axis produces winds in the image only for the p-points, and Q(t) decays only if all Re(s) < 0. Given that Q(t) dies away, then Re(s) < 0 for all roots, all roots are p-points, and it follows that one traversal of the segment $-iR$ to $iR$ produces one winding $\mathcal{R}$ equaling the number of p-points times $\pi$.

Gary