Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Apr 20, 2017 16:08:38 GMT
Gary I think all we can say when $R$ is finite is that the complete trip round the closed loop $\Gamma$ results in $F(s)$ going round $0$ $m$ times or a rotation of $2m\pi$ where $m$ is the number of roots inside $\Gamma$. Vasco Vasco, I'm not sure of your point, but I think that $R$ is never finite in this problem because, by the hint, it is allowed to tend to infinity. I think I agree with the rest, but the complete trip around is almost beside the point, because the focus is on how much winding occurs in one traversal of just the imaginary axis. To know that $\mathcal{R}$ is $n\pi$ one only has to know that the roots have negative real parts. Then by construction the imaginary axis is a segment of a circle that goes round them and traversal on it generates windings. Gary |
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Post by Admin on Apr 25, 2017 13:30:09 GMT
Gary
I have published a new updated version of this exercise. From what you have posted previously I expect you will consider my solution a bit over the top, but I wanted to use a finite value for the radius $R$ of the circle and then let it tend to infinity as suggested in the hint.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 25, 2017 14:52:55 GMT
Vasco,
You wrote
I like your more explicit answer better in all respects except the summing of the windings that include roots with Re(z) > 0. To me, including a loop on the right is not wrong, and has the virtue of presenting another view and a larger view of the context, but I think it is unnecessary and a little bit distracting, and Needham explicitly excludes it by his construction of the loop. Doesn't this create two loops, both including the imaginary axis$-$one to the right and one to the left?
I see that I need to add a phrase to my own to make it clear that the angle $\pi$ arises from traversal only on the segment -iR to iR as R tends to infinity.
Gary
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Post by Admin on Apr 25, 2017 16:14:12 GMT
Vasco, You wrote I like your more explicit answer better in all respects except the summing of the windings that include roots with Re(z) > 0. To me, including a loop on the right is not wrong, and has the virtue of presenting another view and a larger view of the context, but I think it is unnecessary and a little bit distracting, and Needham explicitly excludes it by his construction of the loop. Doesn't this create two loops, both including the imaginary axis$-$one to the right and one to the left? I see that I need to add a phrase to my own to make it clear that the angle $\pi$ arises from traversal only on the segment -iR to iR as R tends to infinity. Gary Gary I am not using two loops, only the one specified by Needham. When I am travelling round the semicircle I have to include the positive roots when calculating the net rotation of $F(s)$ about $0$ as $s$ travels round the semicircle. In [8] on page 345 roots outside $\Gamma$ make no net contribution (eg the point c), but when the loop is not closed (as in my case when I'm just going round the semicircle, I have to include the rotation of $(s-s_j)$ about $0$, where $s_j$ is a root with positive real part, because its contribution is only zero if we go around the whole loop $\Gamma$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 25, 2017 19:58:38 GMT
Vasco, You wrote I like your more explicit answer better in all respects except the summing of the windings that include roots with Re(z) > 0. To me, including a loop on the right is not wrong, and has the virtue of presenting another view and a larger view of the context, but I think it is unnecessary and a little bit distracting, and Needham explicitly excludes it by his construction of the loop. Doesn't this create two loops, both including the imaginary axis$-$one to the right and one to the left? I see that I need to add a phrase to my own to make it clear that the angle $\pi$ arises from traversal only on the segment -iR to iR as R tends to infinity. Gary Gary I am not using two loops, only the one specified by Needham. When I am travelling round the semicircle I have to include the positive roots when calculating the net rotation of $F(s)$ about $0$ as $s$ travels round the semicircle. In [8] on page 345 roots outside $\Gamma$ make no net contribution (eg the point c), but when the loop is not closed (as in my case when I'm just going round the semicircle, I have to include the rotation of $(s-s_j)$ about $0$, where $s_j$ is a root with positive real part, because its contribution is only zero if we go around the whole loop $\Gamma$. Vasco Vasco, OK, I understand now and it all looks valid. Looking at the continuation paragraph at the top of p. 345, it appears that one can either add in the external p-points, or one can take license to disregard p-points that lie outside $\Gamma$ because "the direction of C merely oscillates." Gary |
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Post by Admin on Apr 25, 2017 20:04:12 GMT
Gary
When calculating the rotation of $F(s)$ due to movement of $s$ along a curve which is not closed, the rotation of all the $(s-s_j)$ factors must be taken into account. It's only when the curve is closed that these factors only oscillate and make no net contribution to the rotation.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 25, 2017 22:29:25 GMT
Gary When calculating the rotation of $F(s)$ due to movement of $s$ along a curve which is not closed, the rotation of all the $(s-s_j)$ factors must be taken into account. It's only when the curve is closed that these factors only oscillate and make no net contribution to the rotation. Vasco Vasco, I see your point. As z traverses the imaginary axis from $-i \infty$ to $i \infty$, there is positive rotation around negative roots and cancelling negative rotation around positive roots. The net rotation is the sum of all the rotation about all the roots. If the net (summed) rotation $\mathcal{R}$ on the axis is not the same as the number of roots counted with their multiplicities times $\pi$, then the equation won't decay, because some roots are positive and maintain positive terms in F(s). Gary |
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