|
|
Post by Admin on Jul 29, 2017 8:31:56 GMT
|
|
Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on Aug 12, 2017 18:43:52 GMT
Vasco, I just finished mine. I'll have a look now at what you did. nh.ch8.ex22.pdf (179.34 KB) Gary |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 13, 2017 0:28:00 GMT
Vasco,
I have been studying your answers to Ex. 22. I have a few comments and questions regarding (iii).
(a) It is not clear that the answer deduces the vanishing from the Fundamental Theorem, unless it is just an argument from the result $\frac{1}{24}f’’(a_j)\epsilon^3_j $.
(b) In the note half way down, I wonder if you meant “arguments of $f$” rather than “values of f which lie on the loop $C$”. If values of $f$ lie on the loop $C$, that would mean that parts of the image lie on parts of the contour --- a partial mapping of the loop to itself.
(c) It appears that you are using the midpoint of $f$ on the endpoints of a chord $j$ of $C$ as a proxy for $f(a)_j$. I’ll call that midpoint $f(b)_j$. You show that $f(b)_j$ approaches $f(a)_j$ with shrinking $\epsilon$. Then you show that the sum of all $f(b)_j\hspace{.2em}\epsilon$ goes to zero as $\epsilon \rightarrow 0$. I don't question the correctness of this, but wouldn’t it be easier to argue that $f(a)_j\hspace{.2em}\epsilon$ goes to zero as $\epsilon \rightarrow 0$. Why is the proxy midpoint a better guarantee that the integral vanishes?
Gary
|
|
|
|
Post by Admin on Aug 16, 2017 14:24:35 GMT
Vasco, I have been studying your answers to Ex. 22. I have a few comments and questions regarding (iii). (a) It is not clear that the answer deduces the vanishing from the Fundamental Theorem, unless it is just an argument from the result $\frac{1}{24}f’’(a_j)\epsilon^3_j $. (b) In the note half way down, I wonder if you meant “ arguments of $f$” rather than “ values of f which lie on the loop $C$”. If values of $f$ lie on the loop $C$, that would mean that parts of the image lie on parts of the contour --- a partial mapping of the loop to itself. (c) It appears that you are using the midpoint of $f$ on the endpoints of a chord $j$ of $C$ as a proxy for $f(a)_j$. I’ll call that midpoint $f(b)_j$. You show that $f(b)_j$ approaches $f(a)_j$ with shrinking $\epsilon$. Then you show that the sum of all $f(b)_j\hspace{.2em}\epsilon$ goes to zero as $\epsilon \rightarrow 0$. I don't question the correctness of this, but wouldn’t it be easier to argue that $f(a)_j\hspace{.2em}\epsilon$ goes to zero as $\epsilon \rightarrow 0$. Why is the proxy midpoint a better guarantee that the integral vanishes? Gary Gary Regarding point (a): In (iii) I am expressing the integral round the loop as the sum of the integrals along each chord. In part (i) I have used the FT to obtain an expression for each of these integrals, and so I am automatically using the FT also in part (iii). However I take your point and I will make this more explicit in my re-write. Regarding point (b): I agree that my sloppy use of language here causes confusion and I will change this to read: '... with an expression which only contains values of $f(z)$ at points which lie on the loop $C$.' Regarding point (c): We have to find an approximation that converges to the integral round $C$ and by using the result of part (ii) we are using the average of two different Riemann sums or, if you like, the trapezoidal rule. In the limit both of these converge to the value of the integral. If we use the value of $f$ at the midpoint of the chord then we need to show that the sum of the $f_{mid}\epsilon_j$ converges to the value of the integral. By substituting for $f_{mid}$ we obtain a sum which we already know (from the book pages 378-380) converges to the value of the integral. Again I will make this clearer in my rewrite I hope. Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 16, 2017 14:52:55 GMT
Vasco,
It's an interesting problem. I look forward to seeing the revision. Then I will reevaluate my answer.
Gary
|
|
|
|
Post by Admin on Aug 17, 2017 14:42:11 GMT
|
|
|
|
Post by Admin on Aug 17, 2017 18:05:13 GMT
Gary
I have looked at your solution to exercise 22. It seems to me from your answer to part (iii) that you are treating the the disc centred at $a$ with radius $h$ as the disc of convergence of the Taylor series. This does not seem right to me. We can expand the Taylor series about any point $a$ inside its disc of convergence. The disc of radius $h$ is about approximation, not convergence.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 17, 2017 22:00:06 GMT
Gary I have looked at your solution to exercise 22. It seems to me from your answer to part (iii) that you are treating the disc centred at $a$ with radius $h$ as the disc of convergence of the Taylor series. This does not seem right to me. We can expand the Taylor series about any point $a$ inside its disc of convergence. The disc of radius $h$ is about approximation, not convergence. Vasco Vasco, Yes, you are right. I didn't know quite what to do with it. I thought I had changed that part. Gary |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 22, 2017 4:35:09 GMT
Vasco,
I reread my answer to 22:(iii) and decided it was just wrong, so with some inspiration from your answer I started over. I have posted my new answer in the original location. It still departs from yours significantly.
I have studied your answer of Aug 17, stamped 15:33 carefully and I have a few comments ranging from trivial to more fundamental.
p.3, last paragraph, line 4 “dominant terms of this series”. Does “this series” refer to the RHS of the equation immediately above the paragraph? I think it must, but Needham used “this series” in (iii) to refer to the series given in (i), so the phrase for me has a kind of protected status.
In the same paragraph, you assert “with the series in its present form, we cannot demonstrate (a)”. My comment is, that given the construction of the contour with chords, the discussion on pp. 410-411 should demonstrate (a). It seems obvious by inspection that the difference between a chord and the segment it marks off must shrink as the chord shrinks, so the chord-contour approaches the loop with shrinking $\epsilon$.
p. 4, sentence 1. I don’t see why it is necessary or even useful to replace the value of $f$ at the midpoint $a_j$ of $K_j$ with an expression which only contains values of $f$ for points which lie on the loop $C$. By the construction of chords, $a_j$ is still within the loop and therefore within the disc of convergence.
p. 4, dagger footnote: “The sums above are two different Riemann sums over $C$.” They do not appear to me to be Riemann sums because $a_j$ is not a midpoint for $\frac{e_j}{2}$. They are half-chords times $\epsilon_j$. But it does appear as you say that they both equal the integral of $f$ on $C$.
It appears to me that your answer is generally correct. I wonder if you agree that my answer is correct or if you see some deficiency that I have overlooked.
Gary
|
|
|
|
Post by Admin on Aug 22, 2017 10:36:42 GMT
Vasco, I reread my answer to 22:(iii) and decided it was just wrong, so with some inspiration from your answer I started over. I have posted my new answer in the original location. It still departs from yours significantly. I have studied your answer of Aug 17, stamped 15:33 carefully and I have a few comments ranging from trivial to more fundamental. p.3, last paragraph, line 4 “dominant terms of this series”. Does “this series” refer to the RHS of the equation immediately above the paragraph? I think it must, but Needham used “this series” in (iii) to refer to the series given in (i), so the phrase for me has a kind of protected status. yes What you say I agree with.However it is not the chord or arc on its own that concerns us here, but the limit of the products of the chord and $f$ at the midpoint. We need to show that the sum of these products tends to the value of the integral and so we need to express the product in terms of Riemann sums, whose behaviour we understand in the limit. See above. They are Riemann sums, just not midpoint Riemann sums.In the limit all Riemann sums are equal to the integral. Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 22, 2017 15:32:50 GMT
Vasco,
If $K$ approaches $C$ in the limit, the integral of $f$ on $K$ must approach the integral of $f$ on $C$. If further proof is needed, $F$ is an analytic function. Since $K$ and $C$ become equal in the limit, $F=\int\hspace{.2em} f\hspace{.2em} dz$ amplitwists $f$ equally on both contours at the limit. Then the contour integrals must be equal on $K$ and $C$.
Gary
|
|
|
|
Post by Admin on Aug 27, 2017 15:03:40 GMT
Gary
I have read your solution to exercise 22. In part (iii) three lines from the bottom of page 3 you write "... and their midpoints $(a_j)$ fall in the interior of $C$, ..."
But this is not so because any part of $C$ which looks concave from outside $C$ will have chords which lie outside $C$.
Also, on page 4 at the end of the first paragraph you write "Every summand ... vanishes on $K$ for all $k_j$ in $K$."
Again this does not seem right to me since we integrate along each chord $k_j$, which is not a loop, and so the integral of powers of $z$ do not vanish along the $k_j$.
In your post above you write
Your first sentence above seems to me to require proof and I don't understand your proof in the two sentences which follow in the above quote.
Needham skips over some of the details of Riemann integration in the book. In the theory of Riemann Integration (RI) the integral is defined as the limit of any Riemann sum (RS), as the interval tends to zero. One way to be sure that a given expression is a suitable approximation to an integral is to show that the expression can be written as a RS. Then we know that in the limit its value tends to that of the integral.
For me the fact that in part (ii) Needham asks us to show something that seems to bear no relation to the rest of the exercise unless you use it to show that the approximation to the integral on $K$ from part (i) can be written in terms of a value of $f$ on $C$ and then the sum of all the integrals on the $K$s becomes a Riemann sum, is a very strong hint that this is the way to prove part (iii).
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Aug 27, 2017 21:19:10 GMT
Vasco,
Thank you for the comments. I have some responses.
True. I should have written, “and their midpoints fall inside the disc of convergence.” Because the loop lies within the disc of convergence, any chord on the loop must also lie within the disc. And that is what I was trying to establish at that point.
The phrase “for all $k_j$ in $K$” was unfortunate, because it implies that the integral vanishes on every $k_j$. What I meant to say is for any term in the series, the sum of the integrals on all the $k_j$ (each with its own interval of integration) vanishes. One reason for this is that when the integrals for a single term are summed, the result is the integral on a closed loop for a power function other than 1/z. So with the change of phrasing, I think it is correct.
The proof I suggested above was based on subsection 4 The Integral as Antiderivative, particularly paragraph 3, p. 406, which contains (16). We have established the equivalence or near equivalence of $K$ and $C$ in the limit. Then the integral on $K$ sums all the $F_j$ corresponding to chords. The amplitwist of $F$ at any point is f(P) (or $f(a_j)$ on $k_j$). It is the same on $K$ and $C$ at the limit. So the integral of $f$ on $K$ must approach the integral of $f$ on $C$ at the limit. The amplitwist and the $\Delta$'s are equal in the limit. As the chords shrink, they approximate infinitesimal segments of $C$. This seems to me to prove the deformation theorem, which also supports the conclusion that $\oint_C\hspace{.3em}f(z)\hspace{.3em}dz$ = $\oint_K\hspace{.3em}f(z)\hspace{.3em}dz$.
That is useful to know. I saw it in your answer, but I didn’t grasp it.
I agree entirely. I think your answer is a better response to the question as it was constructed. But I also hate to miss the opportunity to take a more direct approach using general principles rather than calculations if the opportunity presents itself, so long as it is not wrong. I can summarize my argument as follows:
$\hspace{1em}$The sum of the chords of $K$ approaches $C$ in the limit.
$\hspace{1em}$For any single term of $f(a+h)$, the sum of the integrals on the chords of $K$ vanishes by Cauchy's Theorem and the Fundamental Theorem.
$\hspace{1em}$When we sum the sums of the integrals of all the chords of all the terms, the integral of $f$ on $K$ vanishes.
$\hspace{1em}$Since $K$ approaches $C$ in the limit, the integral of $f$ on $C$ must also vanish.
Gary
|
|
|
|
Post by telemeter on Mar 14, 2020 18:04:29 GMT
Hi both
Apologies if I'm duplicating anything above that I've missed.
On looking at part iii, I went down the same track as the published answer but realised that it does not work for two good reasons...and one of aesthetics
a) The published answer seems to work for any line in the disc of convergence not just a loop (ie if you substitute and integral over line J for the Contour integral you get the same result)
That is to say it seems irrelevant that the integral is a contour...you could be approximating chords to any line.
b) The published answer seems to rely on saying as epsilon tends to zero a product of anything * epsilon will also tend to zero. This is incorrect....consider the general sum of f(z)deltaz...as deltaz tends to zero the sum tends to the integral f(z)dz, not zero. It is similar here as epsilon tends to zero the number of terms approximating the contour tends to infinity.
c) As Vasco noted, the FT theorem hasn't really been used much
After mulling this I think we both made it far more complex than it is and assumed we had to use part ii. I don't think we do. Consider this approach...
By the FT, the integral between a and b of f(z) dz = F(b)-F(a). This only works if f(z) is F(z)' ie only if F(z) exists. The existence of the Taylor series guarantees that there will be such an F(z) for any f(z)....where there is convergence.
Thus any integral of a function with a Taylor series between a and b will be path independent and just depend on F(b) and F(a).
In particular, when we have a loop and b=a then the integral is F(a)-F(a) = 0.
This argument is I think a better way of fulfilling the question...happy to be shown wrong though.
I think Needham's pedagogical aim is showing how the Taylor series expansion in the question is essentially intertwined with Cauchy...a fact that he intimated earlier in the chapter.
telemeter
|
|
|
|
Post by Admin on Mar 15, 2020 11:30:18 GMT
telemeter
I take your point and will take another look at this as soon as I can.
Vasco
|
|
