CMS: Euler's Formula for sin(x)
Han de Bruijn
http://www.mathlinks.ro/Forum/viewtopic.php?t=91010 : message #4
I'm completely lost after the sentence "After algebraic manipulation
and for large N, we get". (IMHO this is not a proper argument at all)
Whatever .. Can this line of reasoning be improved: such that it works?
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>Found the following reference on the Internet:
>
>http://www.mathlinks.ro/Forum/viewtopic.php?t=91010 : message #4
>
>I'm completely lost after the sentence "After algebraic manipulation
>and for large N, we get". (IMHO this is not a proper argument at all)
It's evidently supposed to leave part of the argument to the reader.
>Whatever .. Can this line of reasoning be improved: such that it works?
I _suspect_ that if you say
(1-x^2/N^2) cos() =
(1 - x^2/N^2) + (1 - x^2/N^2)(cos() - 1),
see what cancels when you do that, and then use the fact that
cos(t) - 1 is approximately -t^2/2 for small t you may see what
the author has in mind.
But actually making this a _proof_ will require justifying
the passage to the limit - why is the limit of the product
equal to the product of the limits?
There are two proofs with all the details given in Chapter
6 of CMS, btw; one of them is the same as the second
proof on the web page, with details added.
>Han de Bruijn
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Han de Bruijn
> On Tue, 26 May 2009 10:02:43 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>
>>Found the following reference on the Internet:
>>
>>http://www.mathlinks.ro/Forum/viewtopic.php?t=91010 : message #4
>>
>>I'm completely lost after the sentence "After algebraic manipulation
>>and for large N, we get". (IMHO this is not a proper argument at all)
>
> It's evidently supposed to leave part of the argument to the reader.
>
>>Whatever .. Can this line of reasoning be improved: such that it works?
>
> I _suspect_ that if you say
>
> (1-x^2/N^2) cos() =
>
> (1 - x^2/N^2) + (1 - x^2/N^2)(cos() - 1),
>
> see what cancels when you do that, and then use the fact that
> cos(t) - 1 is approximately -t^2/2 for small t you may see what
> the author has in mind.
Sure. But he doesn't give those excruciating details a la David Ullrich.
So I'm still at lost. Almost for two days now ..
> But actually making this a _proof_ will require justifying
> the passage to the limit - why is the limit of the product
> equal to the product of the limits?
>
> There are two proofs with all the details given in Chapter
> 6 of CMS, btw; one of them is the same as the second
> proof on the web page, with details added.
I'm developing some "feeling" for the subject first, if you don't mind.
Especially wonder how Euler _himself_ did it, with the tools available
at that time ..
Han de Bruijn
Toni...@yahoo.com
Read the following about what the author calles "Euler Nonstandard
Nonsense": I think it is a very nice way to show how some things
developed:
http://cornellmath.wordpress.com/2007/07/13/eulers-nonstandard-nonsense/
There are several formal proofs of the product formula for sinx, and
of the nicest ones, imo, is the proof using the gamma function (the
reflexion formula).
Regards
Tonio
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
It's hard to see what you want here. You find a sketch of a proof with
many missing details and you ask how to fill in the details - you get
a big hint (that may or may not actually work, I haven't tried it) and
you're not happy with that either. Then you get a pointer to two
complete proofs in a book you own and you're not happy with that
either. Make up your mind! If you want to do it yourself then do it.
If you want a pre-chewed proof then read one... I mean really. In
just one post you complain that that sketch is insufficiently
CMS-ish and also explain why you don't want to just read the
proof in CMS.
People who are fans of CMS tend to say that one of its virtues is
the attempts to explain where things come from - see the first
few pages of Chapter 6, before the start of the formal proofs.
>Especially wonder how Euler _himself_ did it, with the tools available
>at that time ..
_Again_ - make up your mind what you want!
The tools available at that time were not sufficient to give an actual
proof of _anything_ in analysis, in the sense that we use the word
"proof" at present. In fact it's clear that proofs were impossible
since the relevant _definitions_ didn't even exist. In many (most?)
cases filling in the missing details in Euler's proofs about infinite
sums and so on is easy given what's now basic real and complex
analysis.
Which of course is not to deny that he was one of the great
mathematicians of all time. There's a fascinating little book
"Euler: The Master of Us All"
that includes a discussion of how he "proved" that product
formula. (If I recall correctly the author claims he started
with the idea that sin should have a factorization as a
"polynomial of infinite degree" just like an actual polynomial.
Which of course is literally nonsense, since the same argument
implies exp(z) = 1. But it's also where, for example, Chapter
6 of CMS starts.)
Han de Bruijn
http://www.mathlinks.ro/Forum/viewtopic.php?t=91010 : message #4
That reference makes good start, but when it becomes sort of interesting
then it says the rest is trivial while I can't possibly see why it would
indeed be so. Don't make a person hungry when there isn't a free lunch.
> If you want a pre-chewed proof then read one... I mean really. In
> just one post you complain that that sketch is insufficiently
> CMS-ish and also explain why you don't want to just read the
> proof in CMS.
Chewing on a proof is not the same as becoming the owner of an idea. Let
me point you to another example: Rouche's Theorem. I have no doubt that
the treatment in your book is quite correct, but nevertheless I got NO
good "feeling" for the theorem, until I encountered _this_ reference:
http://en.wikipedia.org/wiki/Rouche's_theorem
Especially the "Geometric explanation" at that page. I can also see now,
with an alike geometric interpretation, what the virtue is of the CMS
version, with |f(z) - g(z)| < |f(z)| + |g(z)| . After absorbing this
stuff with the visual part of my cortex, the actual proof becomes more
of a live experience. (And hopefully "my own" at last)
> People who are fans of CMS tend to say that one of its virtues is
> the attempts to explain where things come from - see the first
> few pages of Chapter 6, before the start of the formal proofs.
>
>>Especially wonder how Euler _himself_ did it, with the tools available
>>at that time ..
>
> _Again_ - make up your mind what you want!
> The tools available at that time were not sufficient to give an actual
> proof of _anything_ in analysis, in the sense that we use the word
> "proof" at present. In fact it's clear that proofs were impossible
> since the relevant _definitions_ didn't even exist. In many (most?)
> cases filling in the missing details in Euler's proofs about infinite
> sums and so on is easy given what's now basic real and complex
> analysis.
>
> Which of course is not to deny that he was one of the great
> mathematicians of all time. There's a fascinating little book
> "Euler: The Master of Us All"
>
> http://www.amazon.com/Euler-Master-Dolciani-Mathematical-Expositions/dp/0883853280/ref=sr_1_1?ie=UTF8&s=books&qid=1243424219&sr=8-1
>
> that includes a discussion of how he "proved" that product
> formula. (If I recall correctly the author claims he started
> with the idea that sin should have a factorization as a
> "polynomial of infinite degree" just like an actual polynomial.
> Which of course is literally nonsense, since the same argument
> implies exp(z) = 1. But it's also where, for example, Chapter
> 6 of CMS starts.)
BTW. Calculations indicate that convergence of this infinite product for
the sine is terribly slow. Guess that's well known ..
Han de Bruijn
Han de Bruijn
> http://www.mathlinks.ro/Forum/viewtopic.php?t=91010 : message #4
>
> That reference makes good start, but when it becomes sort of interesting
> then it says the rest is trivial while I can't possibly see why it would
> indeed be so. Don't make a person hungry when there isn't a free lunch.
My end-result so far is (for n = odd):
sin(pi*x) = sin(pi*x/n) *
Product(k=1,(n-1)/2) [ -4*(sin(pi*(x+k)/n)*sin(pi*(x-k)/n)) ]
Guess it's still a long way to go from here to the Euler-Wallis formula.
Han de Bruijn
Martin Musatov
CERN," by Martin Musatov:
Han de Bruijn
> Has to close:=======<><>{�NP_Complete:�
"Always be prepared for the impossible", I wrote. Shouldn't have written
that. You deserve no goodwill.
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
Well it's good you feel you got that. In fact the argument on that
Wikipedia page is right there in CMS: Look at Proposition 5.3. Both
5.3 and the W page say "if two curves are 'close' then they have
the same winding number about the origin." The book actually
gives a _proof_ of this geometric fact is the main difference (which
is not a criticism of W, that's not supposed to be a text book).
Anyway, forget what I said about using a Taylor expansion for
cos. I believe I got that proof of Euler's formula. We got this
far: If n is odd then
(1+z/n)^n - (1-z/n)^n = 2z/n prod_{k=1}^{(n-1)/2}
[2+2z^2/n^2 - 2(1-z^2/n^2)cos(2pi k/n)]
Take that factor and write it in the form
2+2z^2/n^2 - 2(1-z^2/n^2)cos(2pi k/n)
= A(1 + Bz^2)
where A and B have no z's in them and you can perhaps begin to
see what's going on. That worked for me, anyway...
I'll probably post details tomorrow morning - right now I'm stuck
typing on a very inconvenient tiny keyboard on a 'netbook' (also
I haen't yet got the knack of keeping my hands off the trackpad
so the cursor keeps jumping around...)
Should be more or less the same as the convergence of sum 1/k^2
(which I guess counts as slow, on the order of 10^d terms needed
to get d decimal places right.)
>Han de Bruijn
rancid moth
> I'm developing some "feeling" for the subject first, if you don't
> mind. Especially wonder how Euler _himself_ did it, with the tools
> available at that time ..
I think it was really quite simple. I think Euler came up with the formulae
for the bassel problem - or as part of it - a pre cursor perhaps. He was
big on polynomials and treated infinite series _as_ polynomials. So his
argument was loosley
p(x) = 1 -x^2/3! + x^4/5! - x^6/7! + ....
which was an "infinite polynomial". he said p(0) = 1 and noted that for
p(x)=0 for x!=0 then
p(x) = ( (x - x^3/3! + x^5/5! - x^7/7! +....) /x ) = sin(x)/x
but he knew that sin(x) = 0 --> x= +/- n*pi for n = 1,2,3.... he doesnt
include x=0 becuase p(0)=1.
so _because_ p(x) = 0 has the same roots as sin(x) he simply writes
p(x) = (1-x/pi)*(1 - x/(-pi) )*(1 - x/(2*pi))*(1-x/(-2*pi))*....
in fact he actually writes (translated - you can find it online here
http://math.dartmouth.edu/~euler/)
"the fundamnetal equation becomes 0 = x - x^3/(1.2.3) +
x^5/(1.2.3.4.5) -x^7/(1.2.3.4.5.6.7) + etc., and the roots of this equation
give all the arcs whose sine is = 0. There is one minimal root x=0, so that
when the equation is devided by x, it exhibits all the remaining arcs whose
sine is =0, so that these arcs are all roots of the equation 1 - x^2/(1.2.3)
+ x^4/(1.2.3.4.5) -x^6/(1.2.3.4.5.6.7) + etc. of course these arcs whose
sine is =0 are pi, -pi, +2pi,-2pi.3pi,-3pi, etc., of which the second of
each pair is negative, which the equation itself also tells us, because the
dimensions of x are even. Hence the divisors of this equation will be
1-x/pi, 1+x/pi, 1-x/2pi, 1+x/2pi, etc., and by combining each of these pairs
of divisors together it will be
1 - x^2/1.2.3 + x^4/1.2.3.4.5 - x^6/1.2.3.4.5.6.7 +etc. =
(1-x^2/pi^2)*(1-x^2/(4*pi^2))*(1-x^2/(9*pi^2))*(1-x^2/(16*pi^2)) etc."
Note: i have paraphrased from the translated copy, using "x" rather than
their "s" and "pi" rather than their "p".
Han de Bruijn
See? _Real_ mathematicians (like you) simply can't resist :-) Thanx !
Han de Bruijn
> in fact he actually writes (translated - you can find it online here
> http://math.dartmouth.edu/~euler/)
Great! Thanks! But .. in _which one_ of those, say, thousand documents?
Han de Bruijn
Han de Bruijn
Okay. Think I've found it: E101capitel9.11.pdf . Right?
( Wow! Can't believe my eyes .. )
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
Right.
It really does follow from just "some algebraic manipulations"
from what we saw on that web page (at least if we're willing to
ignore a technical point that I'm certain _you_'s be happy to
ignore) - the proof on that page really is much more complete
than you've been thinking.
Anyway. Writing just cos for cos(2 pi k/n) to save space and
typing, just factoring the factors above shows that
(1+z/n)^n - (1-z/n)^n = 2z/n prod_{k=1}^{(n-1)/2}
[2+2z^2/n^2 - 2(1-z^2/n^2)cos]
= 2z/n prod 2(1 - cos) [1 + (z^2/n^2)(1+cos)/(1-cos)] .
The first amazing but simple thing is we can find prod 2(1 - cos)
exactly. This all started with the fact that if n is odd and t <> 1
then
(1 - t^n)/(1-t) = prod [1 + t^2 - 2t cos(2 pi k/n)].
If you let t -> 1 there it follows that prod 2(1 - cos) = n.
So we have
(1+z/n)^n - (1-z/n)^n
= 2z prod [1 + (z^2/n^2)(1+cos)/(1-cos)] .
Now it's just a calculus exercise to show that for
fixed k
lim_n (1/n^2)(1+cos)/(1-cos) = 1/(pi^2 k^2).
You can do that by (shudder) L'Hopital's rule, or much
better use the fact that for every t in R we have
cos(t) = 1 - t^2/2 + O(t^4).
If we're willing to simply believe that the limit of the
profuct is the product of the limits then it follows that
e^z - e^(-z) = 2z prod_{k=1}^infinity (1 + z^2/(pi^2 k^2)),
QED.
And there's really nothing here that you couldn't have done
yourself...
As I conjectured at the start of this, actually proving that the
limit of the product is the product of the limits is harder than
the formal manipulations above. It's still straightforward,
if we know some basic facts about infinite products.
In particular we're going to use
Lemma 1. If sum |a_k| < 1/2 then
|1 - prod(1 + a_k)| <= 2 sum |a_k|.
That's Lemma 6.1.3 in CMS. It's easy to show from there that
Lemma 2. If |b_k| < 1/2 and sum |a_k - b_k| < 1/4 then
|1 - prod(1 + a_k)/(1 + b_k)| <= 4 sum |a_k - b_k|.
Now if we've fixed a real number z then there exists N such
that |z/(pi^2 k^2)| < 1/4 for all k > N. Since N is finite
(and fixed, independent of n) the limit of the product
of the first N factors is certainly the product of the
limits; we apply Lemma 2 to the product for k = N+1
to (n-1)/2.
As far as I can see the fact that cos(t) = 1 - t^2/2 + O(t^4)
is not enough - we need to keep track of the implicit constant
there in order to show that the "error term" in the denominator
is actually smaller than the "main term". Taylor's Theorem
shows that
cos(t) = 1 - t^2/2 + E(t)
for real t, where
|E(t)| <= t^4/24.
If I use that for cos then a little algebra, if I did it
right, shows that
|1/(pi^2 k^2) - (1/n^2)(1+cos)/(1-cos)| <= c/n^2.
(If anyone wants to check that, note that
k^2/n^2 < 1/4
is used once or twice, and then at the end the fact that
pi^2/12 < 1
is important.)
So the sum from k=1 to (n-1)/2 is less than c/n,
which tends to 0, QED.
David C. Ullrich
Han de Bruijn
Believe it or not: I've been _stuck_ exactly at that "technical point",
among other things (see below. Don't forget I'm in learning mode these
days, so I'm [ schizophrenically perhaps ] trying to work according to
the rules of mainstream mathematics [ which is not so bad after all ] )
> Anyway. Writing just cos for cos(2 pi k/n) to save space and
> typing, just factoring the factors above shows that
>
> (1+z/n)^n - (1-z/n)^n = 2z/n prod_{k=1}^{(n-1)/2}
> [2+2z^2/n^2 - 2(1-z^2/n^2)cos]
>
> = 2z/n prod 2(1 - cos) [1 + (z^2/n^2)(1+cos)/(1-cos)] .
So far so good. But now comes one thing I couldn't come up with:
> The first amazing but simple thing is we can find prod 2(1 - cos)
> exactly. This all started with the fact that if n is odd and t <> 1
> then
>
> (1 - t^n)/(1-t) = prod [1 + t^2 - 2t cos(2 pi k/n)].
>
> If you let t -> 1 there it follows that prod 2(1 - cos) = n.
> So we have
>
> (1+z/n)^n - (1-z/n)^n
>
> = 2z prod [1 + (z^2/n^2)(1+cos)/(1-cos)] .
Yeah!
> Now it's just a calculus exercise to show that for
> fixed k
>
> lim_n (1/n^2)(1+cos)/(1-cos) = 1/(pi^2 k^2).
>
> You can do that by (shudder) L'Hopital's rule, or much
> better use the fact that for every t in R we have
>
> cos(t) = 1 - t^2/2 + O(t^4).
Right!
> If we're willing to simply believe that the limit of the
> profuct is the product of the limits then it follows that
>
> e^z - e^(-z) = 2z prod_{k=1}^infinity (1 + z^2/(pi^2 k^2)),
>
> QED.
>
> And there's really nothing here that you couldn't have done
> yourself...
Ough ! How I *was* stupid ..
> As I conjectured at the start of this, actually proving that the
> limit of the product is the product of the limits is harder than
> the formal manipulations above. It's still straightforward,
> if we know some basic facts about infinite products.
>
> In particular we're going to use
>
> Lemma 1. If sum |a_k| < 1/2 then
>
> |1 - prod(1 + a_k)| <= 2 sum |a_k|.
Yes, I recognize that:
Whew!
My gratitude is infinite, actually.
Han de Bruijn
Han de Bruijn
Now that I have read the treatment by Euler himself, I think it can be
characterized as heuristics at best, not really a proof. (Unexpectedly,
perhaps, David C. Ullrich nevertheless has succeeded in completing this
line of thought. See one of the postings elsewhere in this thread.)
It remains astonishing, though, how geniouses at that time could come
up with true results. And throw around quite a few numerical examples,
without even having a decent calculator at their disposal.
Han de Bruijn
Jesse F. Hughes
> My gratitude is infinite, actually.
Er, given your oft-repeated opinions, this suggests that your
gratitude doesn't exist.
--
I thought the wreck was over. I thought the fire was out.
I thought the storm had passed and I was safe at last.
I thought the wreck was over, but here she comes again.
--The Flatlanders
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>My gratitude is infinite, actually.
>
> Er, given your oft-repeated opinions, this suggests that your
> gratitude doesn't exist.
Pun intended, yes. (But as a matter of fact, I *am* grateful to Ullrich)
Han de Bruijn
Han de Bruijn
Here is a "neat" version of his explanation (and PM's):
http://hdebruijn.soo.dto.tudelft.nl/jaar2009/wallis.pdf
Han de Bruijn
Rob Johnson
I used a very similar argument, but it was made a bit shorter using
the identity
2 1 + cos(x)
cot (x/2) = ----------
1 - cos(x)
We can factor
n-1
z^n - 1 ---
------- = | | ( z - cis(2pi k/n) ) [1]
z - 1 k=1
Taking the limit of [1] as z -> 1, we get
n-1
---
n = | | ( 1 - cis(2pi k/n) ) [2]
k=1
Multiplying [2] by its conjugate, we get
n-1
2 ---
n = | | 2 ( 1 - cos(2pi k/n) ) [3]
k=1
Since cos(2pi - x) = cos(x), if n is odd, we can use half the terms
of the product in [3] to get
(n-1)/2
---
n = | | 2 ( 1 - cos(2pi k/n) ) [4]
k=1
Assume n is odd, then
(n-1)/2
--- x^2 x^2
2x/n | | 2 (( 1 + --- ) - ( 1 - --- ) cos(2pi k/n))
k=1 n^2 n^2
(n-1)/2
--- x^2
= 2x/n | | 2 ( (1 - cos(2pi k/n)) + (1 + cos(2pi k/n)) --- )
k=1 n^2
(n-1)/2
--- 1 + cos(2pi k/n) x^2
= 2x | | ( 1 + ---------------- --- )
k=1 1 - cos(2pi k/n) n^2
(n-1)/2
--- x^2
= 2x | | ( 1 + cot^2(pi k/n) --- ) [5]
k=1 n^2
Since cot(pi k/n)/n -> 1/(pi k) monotonically, [5] tends to
oo
--- x^2
2x | | ( 1 + -------- )
k=1 pi^2 k^2
as n tends to infinity.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
If that was the only thing you couldn't do you can still finish
the proof without evaluating that bit exactly, like so:
>> So we have
>>
>> (1+z/n)^n - (1-z/n)^n
>>
>> = 2z prod [1 + (z^2/n^2)(1+cos)/(1-cos)] .
Instead, if you haven't got that one bit, you have
(1+z/n)^n - (1-z/n)^n
= c_n 2z prod [1 + (z^2/n^2)(1+cos)/(1-cos)] .
Now since the left-hand side converges to something as
n -> infinity and the other factor on the right converges
to something it follows that c_n must converge to
something, so you get
(e^z - e^(-z))/z = c 2 prod_{k=1}^infinity (1 + z^2/(pi^2 k^2)),
and then letting z -> 0 shows that c = 1.
David C. Ullrich
David C. Ullrich
wrote:
Ah. The "monotonically" simplifies the argument considerably.
>[5] tends to
>
> oo
> --- x^2
> 2x | | ( 1 + -------- )
> k=1 pi^2 k^2
>
>as n tends to infinity.
>
>Rob Johnson <r...@trash.whim.org>
>take out the trash before replying
>to view any ASCII art, display article in a monospaced font
David C. Ullrich
umu...@gmail.com
> Ah. The "monotonically" simplifies the argument considerably.
Even stronger. Think I can't really accept the argument without it.
So thank you, Rob Johnson ! I'll make a new "neat" summary of this
debate, with proper credits for everyone who made it possible to
finish it in such a satisfactory manner.
Han de Bruijn
David C. Ullrich
>On 29 mei, 11:52, David C. Ullrich <dullr...@sprynet.com> wrote:
>
>> Ah. The "monotonically" simplifies the argument considerably.
>
>Even stronger. Think I can't really accept the argument without it.
Guffaw. Exactly what step in the proof I gave was incorrect?
>So thank you, Rob Johnson ! I'll make a new "neat" summary of this
>debate,
Debate?
> with proper credits for everyone who made it possible to
>finish it in such a satisfactory manner.
>
>Han de Bruijn
David C. Ullrich
umu...@gmail.com
> On Fri, 29 May 2009 13:19:06 -0700 (PDT), umum...@gmail.com wrote:
> >On 29 mei, 11:52, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> >> Ah. The "monotonically" simplifies the argument considerably.
>
> >Even stronger. Think I can't really accept the argument without it.
>
> Guffaw. Exactly what step in the proof I gave was incorrect?
Not so much incorrect, but not considerably simplified either.
> >So thank you, Rob Johnson ! I'll make a new "neat" summary of this
> >debate,
>
> Debate?
Allright. What shall we say then? A free lunch? Or simply "thread"?
David C. Ullrich
>On 30 mei, 12:38, David C. Ullrich <dullr...@sprynet.com> wrote:
>
>> On Fri, 29 May 2009 13:19:06 -0700 (PDT), umum...@gmail.com wrote:
>> >On 29 mei, 11:52, David C. Ullrich <dullr...@sprynet.com> wrote:
>>
>> >> Ah. The "monotonically" simplifies the argument considerably.
>>
>> >Even stronger. Think I can't really accept the argument without it.
>>
>> Guffaw. Exactly what step in the proof I gave was incorrect?
>
>Not so much incorrect, but not considerably simplified either.
Yes, a key step (that the limit of the product is the product of the
limit) is considerably simpler if you note the convergence is
monotone, as Rob did.
But saying you "accept" the simplified argument but not the
more complicated one makes no sense unless you simply
don't _understand_ the original one. If so your original
enthusiastic "acceptance" of the original was a little
disingenuous.
And in any case this is all silly, since you don't actually
understand either version.
If you want to claim you _do_ understand the simplified
version you need to answer the following question
(no help from the peanut gallery please). Note that
Ron's simplified version seems to use the following
result:
Theorem: Suppose that a[[n,j] > 0 for n = 1, 2. ... and
j = 1, 2, ... . Suppose that a[n,j] increases to a_j as n
tends to infinity. Then prod_j a[n,j] tends to prod_j a_j
as n tends to infinity.
Two questions, both trivial for someone who understands
all this:
Q1. Give a simple example showing that the theorem as
stated is false.
Q2. Give one or two additional hypotheses that make the
theorem true, and give a proof of the amended version
of the theorem. (Of course the added hypotheses should
he such that the amended version of the theorem is still
sufficient to prove the result about the factorization of
the sine function.)
No fair asking for help elsewhere or trying to find the
answer on the internet. Both questions are very easy,
shouldn't take more than a day or two. Or at least both
questions _should_ be very easy for someone who feels
that whether or not _he_ "accepts" the argument has
any relevance to anything.
>> >So thank you, Rob Johnson ! I'll make a new "neat" summary of this
>> >debate,
>>
>> Debate?
>
>Allright. What shall we say then? A free lunch? Or simply "thread"?
What you should say depends on what you mean to say. "Debate"
is silly since there's no debate here (or there hasn't been until just
now...)
>> > with proper credits for everyone who made it possible to
>> >finish it in such a satisfactory manner.
>
>Han de Bruijn
David C. Ullrich
David C. Ullrich
<dull...@sprynet.com> wrote:
>On Sat, 30 May 2009 12:48:00 -0700 (PDT), umu...@gmail.com wrote:
>
>>On 30 mei, 12:38, David C. Ullrich <dullr...@sprynet.com> wrote:
>>
>>> On Fri, 29 May 2009 13:19:06 -0700 (PDT), umum...@gmail.com wrote:
>>> >On 29 mei, 11:52, David C. Ullrich <dullr...@sprynet.com> wrote:
>>>
>>> >> Ah. The "monotonically" simplifies the argument considerably.
>>>
>>> >Even stronger. Think I can't really accept the argument without it.
>>>
>>> Guffaw. Exactly what step in the proof I gave was incorrect?
>>
>>Not so much incorrect, but not considerably simplified either.
>
>Yes, a key step (that the limit of the product is the product of the
>limit) is considerably simpler if you note the convergence is
>monotone, as Rob did.
>
>But saying you "accept" the simplified argument but not the
>more complicated one makes no sense unless you simply
>don't _understand_ the original one. If so your original
>enthusiastic "acceptance" of the original was a little
>disingenuous.
>
>And in any case this is all silly, since you don't actually
>understand either version.
>
>If you want to claim you _do_ understand the simplified
>version you need to answer the following question
>(no help from the peanut gallery please). Note that
>Ron's
The "b" key is next to the "n" key, sorry.
>simplified version seems to use the following
>result:
>
>Theorem: Suppose that a[[n,j] > 0 for n = 1, 2. ... and
>j = 1, 2, ... . Suppose that a[n,j] increases to a_j as n
>tends to infinity. Then prod_j a[n,j] tends to prod_j a_j
>as n tends to infinity.
>
>Two questions, both trivial for someone who understands
>all this:
Since it's been two days I'll give the solutions, just to
demonstrate that they are indeed trivial.
>Q1. Give a simple example showing that the theorem as
>stated is false.
Define a[n,j] = 1 if j < n, a[n,j] = 1/2 if j >= n.
Then prod_j a[n,j] = 0 for all n. On the other hand,
for every j, a[n,j] increases to a_j = 1 as n tends to
infinity, and prod a_j = 1.
(Of course prod_j a[n,j] "diverges to 0"; one could
also give an example with convergent products.)
>Q2. Give one or two additional hypotheses that make the
>theorem true, and give a proof of the amended version
>of the theorem. (Of course the added hypotheses should
>he such that the amended version of the theorem is still
>sufficient to prove the result about the factorization of
>the sine function.)
The theorem is true if we assume a[n,j] >= 1 (which
holds in the application to that Euler thing).
In fact we're going to allow infinity as a value for
an infinite product; now any product of factors >= 1
exists, just as any sum of non-begative terms exists.
Say p_n = prod_j a[n,j], and let p = prod a_j.
Since a[n+1,j] >= a[n, j] it follows that
p_{n+1} >= p_n, and hence the limit of the
p_n exists. Say q = lim p_n.
Since a[n,j] <= a_j it is clear that p_n <= p
and hence q <= p.
To show q >= p: Define
P_N = prod_{j=1}^N a_j.
Since P_N -> p by definition, it's enough to show that
q >= P_N for all N. But since a[n,j] >= 1 and N is finite
it's clear that
q >= lim_n prod_{j=1}^N a[n, j]
= prod_{j=1}^N a_j
= P_N.
Han de Bruijn
> On Sat, 30 May 2009 12:48:00 -0700 (PDT), umu...@gmail.com wrote:
>
>>On 30 mei, 12:38, David C. Ullrich <dullr...@sprynet.com> wrote:
>>
>>>On Fri, 29 May 2009 13:19:06 -0700 (PDT), umum...@gmail.com wrote:
>>>
>>>>On 29 mei, 11:52, David C. Ullrich <dullr...@sprynet.com> wrote:
>>>
>>>>>Ah. The "monotonically" simplifies the argument considerably.
>>>
>>>>Even stronger. Think I can't really accept the argument without it.
>>>
>>>Guffaw. Exactly what step in the proof I gave was incorrect?
>>
>>Not so much incorrect, but not considerably simplified either.
>
> Yes, a key step (that the limit of the product is the product of the
> limit) is considerably simpler if you note the convergence is
> monotone, as Rob did.
http://hdebruijn.soo.dto.tudelft.nl/jaar2009/wallis.pdf
The end result, just before "taking the limit" on the right hand side:
(exp(z)-exp(-z))/2 =
z lim(n->oo) prod(k=1,(n-1)/2)(1 + (z/(n.tan(k.pi/n)))^2)
> But saying you "accept" the simplified argument but not the
> more complicated one makes no sense unless you simply
> don't _understand_ the original one. If so your original
> enthusiastic "acceptance" of the original was a little
> disingenuous.
Yes it was. The original was dropped as soon as I saw the better idea.
Some people call that opportunism.
> And in any case this is all silly, since you don't actually
> understand either version.
It doesn't really matter what my little brain understands or not.
What matters is how we can arrive in a natural way at the desired
result, how to turn the heuristics so far into a proof.
Challenge dismissed (for the moment being. Later). Reason: Off Topic.
What's ON topic is the following.
Define: F(N,n) = prod(k=1,N)(1 + (z/(n.tan(k.pi/n)))^2)
Both you and Rob and the rest of the world claim that:
lim(N->oo) [ lim(n->oo) F(N,n) ] = lim(n->oo) F((n-1)/2,n)
As simple as that. There is _nothing_ wrong with my understanding of the
right _question_. Now let's try for the right answer.
I've finally absorbed the Lemmas 6.1.3 and 6.1.4 in CMS. Lemma 6.1.4 is
employed (which is exactly what you derived with Lemma 1,2 in a previous
posting but with Rob's contribution it's simpler). Because we must only
show now that sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| is finite:
sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| < sum(k=1,oo) |(z/(k.pi)^2)| =
|z/pi|^2 . sum(k=1,oo) 1/k^2 = |z|^2/6 < oo . QED (at least I hope so).
Han de Bruijn
Han de Bruijn
Hey! Make that two _*working*_ days! It's been Pentecost! For heavens
sake, do you never take any rest? How about getting a nice sun tan in
the first place? (It's been great weather in the Netherlands)
>>Q1. Give a simple example showing that the theorem as
>>stated is false.
>
> Define a[n,j] = 1 if j < n, a[n,j] = 1/2 if j >= n.
> Then prod_j a[n,j] = 0 for all n. On the other hand,
> for every j, a[n,j] increases to a_j = 1 as n tends to
> infinity, and prod a_j = 1.
>
> (Of course prod_j a[n,j] "diverges to 0"; one could
> also give an example with convergent products.)
Yep. It's almost a copy from your CMS book ..
[ .. some old stuff snipped .. ]
Han de Bruijn
David C. Ullrich
"matter" to what? The question of whether you actually understand
either proof certainly matters in regard to the question of whether
your "accepting" one proof but not the other is silly.
Depends on what the topic _is_. When you say you accept one
proof but not the other you change the topic to include what
_you_ "accept" and why. The challenge is certainly relevant
to _that_.
>What's ON topic is the following.
>
>Define: F(N,n) = prod(k=1,N)(1 + (z/(n.tan(k.pi/n)))^2)
>
>Both you and Rob and the rest of the world claim that:
[*]
>lim(N->oo) [ lim(n->oo) F(N,n) ] = lim(n->oo) F((n-1)/2,n)
>
>As simple as that. There is _nothing_ wrong with my understanding of the
>right _question_.
I didn't claim that there was. I made a claim about your understanding
of the _proofs_ that (*) holds.
>Now let's try for the right answer.
>
>I've finally absorbed the Lemmas 6.1.3 and 6.1.4 in CMS.
That sounds like a concession that you hadn't done so previously.
I said "don't understand", not "will never understand".
>Lemma 6.1.4 is
>employed (which is exactly what you derived with Lemma 1,2 in a previous
>posting but with Rob's contribution it's simpler).
Simpler _because_ it uses that fact about monotonicity. So it's fair
for _me_ to say it's simpler (which I did), since the reason that
that fact about increasing products was true was immediately
clear. But if _you_ don't know a proof of the result about
increasing products then _you_ have no basis for thinking
it's simpler! Because for all you know the result about
increasing products could be very complicated in itself.
Really. Look back at that web site where you found the
sketch that started all this. There's an Eulerian sort of
comment about "for infinite n this becomes..." or
something. That's simpler yet, right? No, that's not
simpler, because it simply pretends that the
technical detail about the limit of the product being
the product of the limits doesn't exist. It _looks_
simpler just because a large part of the argument
is left out. Before you can concluse that Rob's
argument is simpler you need to know that the
same thing's not happening there.
(Ok, if the thing about increasing products was
something you've known for years that would give
you another possible reason for saying Ron's
version was simpler, even if the proof of the
thing about increasing products was very
complicated - using difficult but universally
well-known facts counts as a simplification.
But it's not so that you've known that fact
for years. Or at least I don't believe that it's
so = although it's "obvious" (see the proof I
gave above) it's not something that _I've_
known for years...)
>Because we must only
>show now that sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| is finite:
Unimportant quibble: We need to show that that sum is
_bounded_, not just finite. That is, we need to show that
sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| <= c < infinity,
where c does not depend on n; if, say, we showed that the
sum was <= n that would show that it was _finite_ but
that would not be enough.
Note that I'm calling that an unimportant quibble because
it's not a comment on your understanding of the math,
just on your understanding of the correct terminology.
I suspect that "bounded" is what you meant - below
you show that it's bounded.
But the next question is _not_ an unimportant quobble,
if you're going to claim that there are no gaps in your
understanding of that argument:
Question: Exactly _why_ is it enough to show that that
sum is bounded?
>sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| < sum(k=1,oo) |(z/(k.pi)^2)| =
>|z/pi|^2 . sum(k=1,oo) 1/k^2 = |z|^2/6 < oo . QED (at least I hope so).
That's a QED for the fact that the sum is bounded. Whether or not it's
a QED for the proof of the main result depends on whether you can
answer the question above correctly.
Han de Bruijn
It is not. Remember what Victor Meldrew said when he judged about using
Rouches Theorem - suggested by YOU - for the purpose of "CMS: where are
the zeroes ?". Quote:
> (Also using Rouche is a sledgehammer to crak a nut).
From: http://groups.google.nl/group/sci.math/msg/f8ed0aa55ce38fe3
Assuming that Victor Meldrew is possibly right, you're free to _explain_
to us _why_ Rouche is a sledgehammer.
And if your whole chapter 6 in CMS can be summarized in a page or three,
then you're almost certainly putting everybody on the wrong leg. Read:
http://hdebruijn.soo.dto.tudelft.nl/jaar2009/wallis.pdf
Yes, David C. Ullrich, you're a true MASTER in Making Simple mathematics
unneccessarily Complex.
Our topic is how to have a simple heuristics = proof of the Euler-Wallis
formula for sin(x). Can't you read the header of a thread? And can't you
just do your own thing: Complex Made Simple, not the other way around.
>>What's ON topic is the following.
>>
>>Define: F(N,n) = prod(k=1,N)(1 + (z/(n.tan(k.pi/n)))^2)
>>
>>Both you and Rob and the rest of the world claim that:
>
> [*]
>
>>lim(N->oo) [ lim(n->oo) F(N,n) ] = lim(n->oo) F((n-1)/2,n)
>>
>>As simple as that. There is _nothing_ wrong with my understanding of the
>>right _question_.
>
> I didn't claim that there was. I made a claim about your understanding
> of the _proofs_ that (*) holds.
Mainly _your_ proofs. I know of at least one other mathematician who has
given me another proof. And I am able to understand all steps of it. How
can that happen?
>>Now let's try for the right answer.
>>
>>I've finally absorbed the Lemmas 6.1.3 and 6.1.4 in CMS.
>
> That sounds like a concession that you hadn't done so previously.
> I said "don't understand", not "will never understand".
It _is_ a concession. But, in retrospect, these lemmas are not at all
relevant for a decent proof of the Euler-Wallis formula. You're again
using a sledgehammer to crack a nut.
Meanwhile, I've abandoned that QED. Because it's again a sledgehammer to
crack a peanut. You're a true MASTER ..
Han de Bruijn
Michael Press
David C. Ullrich <dull...@sprynet.com> wrote:
[...]
I kept verifying the formulae by checking that the
polynomials had the same roots and eventually recast the
formal manipulations slightly differently.
Take n odd. Define u = pi/n, and
p(x) = (1 + x)^n - (1 - x)^n.
The roots of p are i.tan(k.u), k = 0, 1, ..., n-1, because
|1 + i.tan(k.u)| = |1 - i.tan(k.u)|
and
arg([1 + i.tan(k.u)]^n) = n.arctan(tan(k.pi/n)
= k.pi
== -k.pi (mod 2.pi)
= arg([1 + i.tan(k.u)]^n)
Therefore
n-1
---
p(x) = 2.x | | [ x - i.tan(k.u)]
k=1
Comparing coefficients of x:
n-1
---
| | i.tan(pi.k/n) = n
k=1
so
n-1 n-1
--- ---
p(x) = 2.n.x | | [-x.i.cot(k.u) - 1] = 2.n.x | | [ x.i.cot(k.u) + 1]
k=1 k=1
As cot((n-k)u) = cot(pi - k.u) = -cot(k.u)
---
p(x) = 2.n.x | | [ 1 + (x.cot(k.u))^2].
1<=k<n/2
Setting x = z/n,
---
p(z/n) = 2.z | | [ 1 + (z/n.cot(k.u))^2].
1<=k<n/2
--- ( z^2 1 + cos(k.u) )
= 2.z | | ( 1 + ---- -------------- )
1<=k<n/2 ( n^2 1 - cos(k.u) )
--
Michael Press
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
Erm, asking him would make more sense.
>And if your whole chapter 6 in CMS can be summarized in a page or three,
>then you're almost certainly putting everybody on the wrong leg. Read:
>
>http://hdebruijn.soo.dto.tudelft.nl/jaar2009/wallis.pdf
Ok, I read that. It doesn't summarize Chapter 6 in the book - almost
none of the material in that chapter is mentioned in that pdf. The
point to the chapter is to teach the techniques used there. (Did
you notice that there are _two_ proofs given in that chapter?
Have you thought about why that might be when one is enough
to prove the theorem?)
At most the pdf gives a proof of the formula. Except it doesn't.
You simply _state_ without any justification that the limit of
the product is the product of the limits. That needs to be proved -
I gave you an example right here in this thread where it's not so,
the point being to demonstrate that it _does_ need to be proved.
And the complications you complain about are all in the proof that
the limit of the product _is_ the product of the limits.
So you're "simplifying" the proof by simply omitting the hard part.
You can do that if you want. Seriously, there's nothing wrong
with that, if you want to just concentrate on the formalisms and
not worry about the technical details. But that's not the point
to the book - it's an actual math text, the point being to learn
techniques for actual _proofs_.
Good question. Hard to answer unless you show us the proof you're
referring to (for example, there's _no_ proof in that pdf - if someone
showed you a proof you understood why didn't you type it up instead
of what you did put there?
Or maybe you're referring to the proof that Rob posted. You didn't
understand all the steps in that proof - if you did you would have
been able to eplain why that montonicity showed the limit of the
product was the product of the limits, and you certainly wouldn't
have said my question about that was "off topic".)
umu...@gmail.com
> On Wed, 03 Jun 2009 14:41:21 +0200,Han de Bruijn
> <Han.deBru...@DTO.TUDelft.NL> wrote:
>
> >http://hdebruijn.soo.dto.tudelft.nl/jaar2009/wallis.pdf
>
> Ok, I read that. It doesn't summarize Chapter 6 in the book - almost
> none of the material in that chapter is mentioned in that pdf. The
> point to the chapter is to teach the techniques used there. (Did
> you notice that there are _two_ proofs given in that chapter?
> Have you thought about why that might be when one is enough
> to prove the theorem?)
>
> At most the pdf gives a proof of the formula. Except it doesn't.
> You simply _state_ without any justification that the limit of
> the product is the product of the limits. That needs to be proved -
> I gave you an example right here in this thread where it's not so,
> the point being to demonstrate that it _does_ need to be proved.
>
> And the complications you complain about are all in the proof that
> the limit of the product _is_ the product of the limits.
>
> So you're "simplifying" the proof by simply omitting the hard part.
> You can do that if you want. Seriously, there's nothing wrong
> with that, if you want to just concentrate on the formalisms and
> not worry about the technical details. But that's not the point
> to the book - it's an actual math text, the point being to learn
> techniques for actual _proofs_.
Allright. You've made your point and I capitulate. Considering the
rest as (my) homework, I've finished the abovementioned article in
style. (Its final version will appear on the web next monday.)
Han de Bruijn
umu...@gmail.com
> And wrote:
>
> >If you want to claim you _do_ understand the simplified
> >version you need to answer the following question
> >(no help from the peanut gallery please). Note that
> >Ron's
>
> The "b" key is next to the "n" key, sorry.
>
> >simplified version seems to use the following
> >result:
>
> >Theorem: Suppose that a[[n,j] > 0 for n = 1, 2. ... and
> >j = 1, 2, ... . Suppose that a[n,j] increases to a_j as n
> >tends to infinity. Then prod_j a[n,j] tends to prod_j a_j
> >as n tends to infinity.
>
>
> The theorem is true if we assume a[n,j] >= 1 (which
> holds in the application to that Euler thing).
I would rather like to call this: "Complex made Real". Because:
a[n,k] = 1 + [z/(k.pi).T(k.pi/n)]^2 in that Euler thing: k,n e N
and T(x) = x/tan(x) for x < pi/2 and T(x) = 0 for x >= pi/2: reals,
but z is COMPLEX and by that I mean really complex. Though the rest
of your argument could make sense if it is assumed that by 'a' you
mean the absolute value of a . So let's give it a try and proceed.
> In fact we're going to allow infinity as a value for
> an infinite product; now any product of factors >= 1
> exists, just as any sum of non-begative terms exists.
>
> Say p_n = prod_j a[n,j], and let p = prod a_j.
> Since a[n+1,j] >= a[n, j] it follows that
> p_{n+1} >= p_n, and hence the limit of the
> p_n exists. Say q = lim p_n.
All as absolute values. Is that what you mean?
> Since a[n,j] <= a_j it is clear that p_n <= p
> and hence q <= p.
>
> To show q >= p: Define
>
> P_N = prod_{j=1}^N a_j.
>
> Since P_N -> p by definition, it's enough to show that
> q >= P_N for all N. But since a[n,j] >= 1 and N is finite
> it's clear that
>
> q >= lim_n prod_{j=1}^N a[n, j]
>
> = prod_{j=1}^N a_j
>
> = P_N.
Sure, but there you go! Indeed it can be shown for complex z that
|a[n+1,j]| < |a[n,j]| but it can NOT be shown that |a[n,j]| >= 1 .
Argument refuted. What else?
Han de Bruijn
umu...@gmail.com
Oh well, in order for the infinite product not to disappear, it's
enough to assume that a[n,j] >= 1 for sufficient _large_ n . It's
clear that our function T(x) takes care of this. With such a slight
modification, we can continue your argument and finally find that:
| Limit of Product | = | Product of Limits | : absolute values.
So the two are indeed equal, but still apart from a "phase factor"
(i.e. complex number with absolute value = 1). The bad news is that
this phase factor may be dependent on z ..
Han de Bruijn
umu...@gmail.com
Elsewhere in the thread, I've "reproduced" your proof that the limit
of the product is the product of the limits (but apart from a phase
factor). That proof does not exclude the possibility that the product
may be infinite (i.e. does not exist). The above does exclude this.
Han de Bruijn
umu...@gmail.com
Oops! Exchanged the indices n<->j, so this is a non-argument. Alas,
only if one can think of a better reasoning it could be valid that:
David C. Ullrich
>On 3 jun, 12:43, David C. Ullrich <dullr...@sprynet.com> wrote:
>> On Tue, 02 Jun 2009 12:02:04 +0200, Han de Bruijn
>> <Han.deBru...@DTO.TUDelft.NL> wrote:
>>
>> But the next question is _not_ an unimportant quobble,
>> if you're going to claim that there are no gaps in your
>> understanding of that argument:
>>
>> Question: Exactly _why_ is it enough to show that that
>> sum is bounded?
>>
>> >sum(k=1,oo) |(z/(n.tan(k.pi/n)))^2| < sum(k=1,oo) |(z/(k.pi)^2)| =
>> >|z/pi|^2 . sum(k=1,oo) 1/k^2 �= |z|^2/6 < oo . QED (at least I hope so).
>>
>> That's a QED for the fact that the sum is bounded. Whether or not it's
>> a QED for the proof of the main result depends on whether you can
>> answer the question above correctly.
>
>Elsewhere in the thread, I've "reproduced" your proof that the limit
>of the product is the product of the limits
No, you've mangled it.
> (but apart from a phase
>factor). That proof does not exclude the possibility that the product
>may be infinite (i.e. does not exist).
I'm not sure which argument you're referring to, the one I
gave originally or my expansion of the details needed in
the argument Rob gave. But regardless of what follows from
your mangled version, yes, both those arguments do show
that the product is finite.
>The above does exclude this.
>
>Han de Bruijn
David C. Ullrich
David C. Ullrich
>On 2 jun, 11:33, David C. Ullrich <dullr...@sprynet.com> wrote:
>> And wrote:
>>
>> >If you want to claim you _do_ understand the simplified
>> >version you need to answer the following question
>> >(no help from the peanut gallery please). Note that
>> >Ron's
>>
>> The "b" key is next to the "n" key, sorry.
>>
>> >simplified version seems to use the following
>> >result:
>>
>> >Theorem: Suppose that a[[n,j] > 0 for n = 1, 2. ... and
>> >j = 1, 2, ... . Suppose that a[n,j] increases to a_j as n
>> >tends to infinity. Then prod_j a[n,j] tends to prod_j a_j
>> >as n tends to infinity.
>>
>[ Questions 1 and 2 snipped ]
>>
>> The theorem is true if we assume a[n,j] >= 1 (which
>> holds in the application to that Euler thing).
>
>I would rather like to call this: "Complex made Real". Because:
>
>a[n,k] = 1 + [z/(k.pi).T(k.pi/n)]^2 in that Euler thing: k,n e N
>and T(x) = x/tan(x) for x < pi/2 and T(x) = 0 for x >= pi/2: reals,
>but z is COMPLEX and by that I mean really complex. Though the rest
>of your argument could make sense if it is assumed that by 'a' you
>mean the absolute value of a .
No, that's not what I meant. I meant exactly what I said.
Right now we're not proving the Euler thing, quite.
We're procoing that if x is real then
(*) exp(x) - exp(-x) = a certain product.
Everything in sight is real.
_Then_ after we prove (*) we note that both sides
of (*) define entire functions. Two entire functions
that agree on the line must be the same, and the
Euler factorization for sin(z) follows on replacing
x by ix in (*).
>So let's give it a try and proceed.
>
>> In fact we're going to allow infinity as a value for
>> an infinite product; now any product of factors >= 1
>> exists, just as any sum of non-begative terms exists.
>>
>> Say p_n = prod_j a[n,j], and let p = prod a_j.
>> Since a[n+1,j] >= a[n, j] it follows that
>> p_{n+1} >= p_n, and hence the limit of the
>> p_n exists. Say q = lim p_n.
>
>All as absolute values. Is that what you mean?
No, I meant what I wrote.
>> Since a[n,j] <= a_j it is clear that p_n <= p
>> and hence q <= p.
>>
>> To show q >= p: Define
>>
>> � � �P_N = prod_{j=1}^N a_j.
>>
>> Since P_N -> p by definition, it's enough to show that
>> q >= P_N for all N. But since a[n,j] >= 1 and N is finite
>> it's clear that
>>
>> � �q >= lim_n prod_{j=1}^N a[n, j]
>>
>> � � � � �= prod_{j=1}^N a_j
>>
>> � � � � �= P_N.
>
>Sure, but there you go! Indeed it can be shown for complex z that
>|a[n+1,j]| < |a[n,j]| but it can NOT be shown that |a[n,j]| >= 1 .
>
>Argument refuted.
No, your mangled version of the argument is refuted, not
the argument I gave.
Amusing note: In the original argument I posted, the one
you didn't "accept" because it was too complicated, this
problem doesn't come up - one could make that argument
work for complex x from the start (although the proofs
of some of the basic inequalities might be a little different).
It's only in the simplified argument that Rob gave that
we need to restrict to real x to begin with - that's because
the argument depends on monotonicity. The things I wrote
above that you believe you've refuted are my filling in
the details of the argument Rob gave.
This is amusing, given that you complained so bitterly
about how complicated my argument was, while you
claimed to understand Rob's. It's Rob's argument that
you erroneously think you've refuted, not mine.
>What else?
>
>Han de Bruijn
David C. Ullrich
David C. Ullrich
I'm not sure what you're talking about. The (valid) argument
for real x, which involves a[j,n] >= 1, does _not_ prove
what you say about complex x by taking absolute values,
because for x complex there's simply no reason why the
absolute value of the relevant factors should be monotone.
The absolute valus of 1 + z is not 1 + |z|.
>So the two are indeed equal, but still apart from a "phase factor"
>(i.e. complex number with absolute value = 1). The bad news is that
>this phase factor may be dependent on z ..
Not that anything you're saying here is correct, but if you _were_
right about this that would be incredibly easy to fix: Suppose we
know that sin(z) = that product times a "phase" dependent on z.
Then f = sin(z)/product would be an entire function with
|f(z)| = 1 for all z. Basic complex analysis (have you _read_
any of the book?) would then show that f was constant.
Han de Bruijn
Real ! Ah, ah, indeed I couldn't find anything _complex_ in relationship
to monotone convergence. But herewith you're adding a entire new element
to the argument.
> _Then_ after we prove (*) we note that both sides
> of (*) define entire functions.
Excuse me? How can you be so sure that the _right_ hand side is entire?
Yes I mangled it, because you're a master in putting people on the wrong
leg. That argument of yours with reals instead of complex numbers hasn't
been present in a previous posting. And I think you make it up _ad hoc_.
Han de Bruijn
This is indeed nonsense (as I've found last night myself):
> Not that anything you're saying here is correct, but if you _were_
> right about this that would be incredibly easy to fix: Suppose we
> know that sin(z) = that product times a "phase" dependent on z.
> Then f = sin(z)/product would be an entire function with
> |f(z)| = 1 for all z. Basic complex analysis (have you _read_
> any of the book?) would then show that f was constant.
Reading a bunch of Propositions, Lemmas and Theorems doesn't reveal any
clue per se in what circumstances to apply them.
Han de Bruijn
David C. Ullrich
These comments on the book are getting to be amusing.
See, in the opinion of everyone but you this aspect of things
is exactly what makes the book a _good_ book - it includes
the sort of motivation and explanation that other books
tend to leave out.
But it doesn't work if you don't _read_ it. No reason you
_should_ read it, but blaming your ignorance on the book
is funny.
("Everyone", you ask? Yes, every single person who's
expressed an opinion. That's people here at OSU, people
on sci.math, one user review on amazon, and the independent
review on Zentralblatt:
http://www.zentralblatt-math.org/zmath/en/search/?q=an:05360956&type=pdf&format=complete
says, among other things,
"...motivation, on exhaustive detailedness, and on teaching the
features of complex-analytic thinking throughout the entire text"
"...illustrate the actual significance of
various results, to show how to tackle a complex-analytic problem in
possibly different
ways"
"In general, the entire exposition stands out by its ... expository
mastery, and by its lucid style helping students grasp both the
matter and the beauty of complex function theory profoundly."
No, I didn't pay him to write that.)
But you have to actually _read_ it (and also do the exercises).
No, merely skimming through it the way you've been doing
is not going to help.
David C. Ullrich
That element has been there all along. When Rob says something about
monotone convergence, I say something about monotone convergence
and also about things like a >= 1 and you think somehow we're talking
about complex numbers that's not _our_ fault.
>> _Then_ after we prove (*) we note that both sides
>> of (*) define entire functions.
>
>Excuse me? How can you be so sure that the _right_ hand side is entire?
That's obvious if you know the most basic things about infinite
products. See for example the bottom of page 94 in the book
(and note that P(z) is entire if and only if P(iz) is entire).
If you _don't_ know the most basic things about infinite products,
which you have just _shown_ you don't, one might think you'd
expcet to have to _learn_ some basic things about infinite products
before starting to prove things about them.
At least that's what _I'd_ think. If you prefer to continue
floundering about in ignorance that's up to you.
Giggle.
>> Amusing note: In the original argument I posted, the one
>> you didn't "accept" because it was too complicated, this
>> problem doesn't come up - one could make that argument
>> work for complex x from the start (although the proofs
>> of some of the basic inequalities might be a little different).
>> It's only in the simplified argument that Rob gave that
>> we need to restrict to real x to begin with - that's because
>> the argument depends on monotonicity. The things I wrote
>> above that you believe you've refuted are my filling in
>> the details of the argument Rob gave.
>>
>> This is amusing, given that you complained so bitterly
>> about how complicated my argument was, while you
>> claimed to understand Rob's. It's Rob's argument that
>> you erroneously think you've refuted, not mine.
>>
>>>What else?
>
>Han de Bruijn
David C. Ullrich
Han de Bruijn
> On Mon, 08 Jun 2009 16:31:43 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>>
>>Reading a bunch of Propositions, Lemmas and Theorems doesn't reveal any
>>clue per se in what circumstances to apply them.
>
> These comments on the book are getting to be amusing.
>
> See, in the opinion of everyone but you this aspect of things
> is exactly what makes the book a _good_ book - it includes
> the sort of motivation and explanation that other books
> tend to leave out.
If your book is supposed to be a good one, guess how bad the others are.
Serious. Because I'm not going to treat you the way you're treating me.
I have another book on Real and Complex Analysis (only the last chapter)
in my bookshelves and find it little insightful. I've lost almost all of
my old college notes on the subject. But still remember I was very much
impressed by the complex analysis we had to learn at that time (35 years
ago) especially by applications of the residue theorem on real integrals
and conformal mappings to describe ideal flow. As soon as you announced
your book, I decided to buy it for a couple of reasons:
1. You're a well known personality in this group and I respect you as a
knowledgable mainstream mathematician. (Don't mis-use that respect!)
2. I was curious how complex analysis is done by mainstream mathematics.
3. Was triggered by "Complex Made Simple"; especially needed the Simple.
4. Was curious how a much more abstract subject works out for Infinity.
Look at "Where are the zeroes" and you will see that I have not given
up my old interests, only want to look upon them differently.
5. "bol.com" in the Netherlands offered it for a reasonable price (which
is nevertheless lowering. Why? If it is such a vaulable book ..)
> But it doesn't work if you don't _read_ it. No reason you
> _should_ read it, but blaming your ignorance on the book
> is funny.
It does _not_ read like a novel, but nobody should expect that. I had to
reread the first few chapters several times before some of the structure
became apparent to me (I can only speak for myself). And even then it's
been like learning French. I can more or less understand something that
has been cooked up by others, but being creative in this field (such as
making the excercises) is another matter. At this stage, I can hear, but
not speak. And yes, your book _could_ have been helpful with this, if it
were only with "answers to odd numbered excercises" or Summary/Overview
sections; such captions haven't been invented by coincidence.
> ("Everyone", you ask? Yes, every single person who's
> expressed an opinion. That's people here at OSU, people
> on sci.math, one user review on amazon, and the independent
> review on Zentralblatt:
>
> http://www.zentralblatt-math.org/zmath/en/search/?q=an:05360956&type=pdf&format=complete
>
> says, among other things,
>
> "...motivation, on exhaustive detailedness, and on teaching the
> features of complex-analytic thinking throughout the entire text"
>
> "...illustrate the actual significance of
> various results, to show how to tackle a complex-analytic problem in
> possibly different
> ways"
>
> "In general, the entire exposition stands out by its ... expository
> mastery, and by its lucid style helping students grasp both the
> matter and the beauty of complex function theory profoundly."
>
> No, I didn't pay him to write that.)
> But you have to actually _read_ it (and also do the exercises).
> No, merely skimming through it the way you've been doing
> is not going to help.
I'll find my way through it, as soon as YOU stop being de-motivating and
personal, and stick to the issues at hand. I didn't ask much more than a
lending hand. You could have omitted all of your "wise words" and simply
say, for example: hey Han, first prove that the infinite products exist,
by looking at Lemma x.x at page yy. Can you do it? Okay, now we're there
you have two iterated limits:
lim_{n->oo} { lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
lim_{N->oo} { lim_{n->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
And you have already proved that:
lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi)}^2] = ? =
lim_{n->oo} prod_{k=1,oo} [1 + {z/(k.pi).T(k.pi/n)}^2]
In order to proceed, take a look at .. Etcetera. See? Without calling a
person ignorant (of course I am ignorant in this field, what else do you
expect). And without saying you're so good. You don't have to say that;
it's much more satisfactory to let other people say you're good. And you
are only good as a mathematician. Other people are good in other things.
Glad that we are different. No? Are you a decent music composer? I guess
you're not:
http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm
http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm#karaoke
You will find the name "Edwin van Veldhoven" with the last URL. It shows
that I have no trouble with finding someone superior in comparison with
myself. On the contrary, _recognizing_ that fact has resulted in a most
enjoyable collaboration where he wrote the music and I wrote the lyrics.
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
Of course you did. So does anyone. To learn a subject like this you
have to _study_ it - a person would be very remarkable if he were
able to really understand everything just reading through it once.
You also have to do the exercises.
>(I can only speak for myself). And even then it's
>been like learning French. I can more or less understand something that
>has been cooked up by others, but being creative in this field (such as
>making the excercises) is another matter. At this stage, I can hear, but
>not speak. And yes, your book _could_ have been helpful with this, if it
>were only with "answers to odd numbered excercises" or Summary/Overview
>sections; such captions haven't been invented by coincidence.
I'm not aware of a graduate text in math that includes answers
to the even-numbered exercises. Nor one with sections labelled
as summaries or overviews, for that matter.
The intended audience (as stated clearly in the introduction, and
note that you could have read the introduction before buying the
book by following a link I posted) is beginning mathematics
graduate students. It's not supposed to be a book for people
who want a quick overview of the field, it's a book for people
who want to become mathematicians. Such people need to learn
much more than an overview.
>> ("Everyone", you ask? Yes, every single person who's
>> expressed an opinion. That's people here at OSU, people
>> on sci.math, one user review on amazon, and the independent
>> review on Zentralblatt:
>>
>> http://www.zentralblatt-math.org/zmath/en/search/?q=an:05360956&type=pdf&format=complete
>>
>> says, among other things,
>>
>> "...motivation, on exhaustive detailedness, and on teaching the
>> features of complex-analytic thinking throughout the entire text"
>>
>> "...illustrate the actual significance of
>> various results, to show how to tackle a complex-analytic problem in
>> possibly different
>> ways"
>>
>> "In general, the entire exposition stands out by its ... expository
>> mastery, and by its lucid style helping students grasp both the
>> matter and the beauty of complex function theory profoundly."
>>
>> No, I didn't pay him to write that.)
>
>> But you have to actually _read_ it (and also do the exercises).
>> No, merely skimming through it the way you've been doing
>> is not going to help.
>
>I'll find my way through it, as soon as YOU stop being de-motivating and
>personal, and stick to the issues at hand. I didn't ask much more than a
>lending hand. You could have omitted all of your "wise words" and simply
>say, for example: hey Han, first prove that the infinite products exist,
>by looking at Lemma x.x at page yy. Can you do it?
My first reply said this:
"There are two proofs with all the details given in Chapter
6 of CMS, btw; one of them is the same as the second
proof on the web page, with details added."
This was your reply:
"I'm developing some "feeling" for the subject first, if you don't
mind."
I somehow got the impression that you wanted to know how
to prove the formula. You complained about missing details
in the proofs you'd seen on that web site. So I told you
exactly where to find two proofs with all the details filled
in.
That wasn't good enough. So for _your_ benefit I worked
out and posted a detailed version of the argument on that
site. When Rob posted a simpler version you said that
his was the first one you "accepted".
Which was a little irritating, given that (as seemed likely
at the time, and as you've repeatedly made very clear since
then) you obviously didn't actually understand _either_ version.
>Okay, now we're there
>you have two iterated limits:
>
>lim_{n->oo} { lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
>lim_{N->oo} { lim_{n->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
>
>And you have already proved that:
>
>lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi)}^2] = ? =
>lim_{n->oo} prod_{k=1,oo} [1 + {z/(k.pi).T(k.pi/n)}^2]
>
>In order to proceed, take a look at .. Etcetera. See? Without calling a
>person ignorant (of course I am ignorant in this field, what else do you
>expect).
If you know you're ignorant about something then it's surprising
you're not more careful about suggesting that I meant something
other than what I wrote, then announcing a "refutation" what
you think I meant. That sort of thing certainly doesn't sound
like a person who's aware of his ignorance.
>And without saying you're so good.
I didn't say _I_ was good. I pointed out that others feel the book
iis good in precisely the sort of ways that you find it's lacking.
That could be useful information for you - it _could_ tell you
where the source of your difficulties lies. (In fact you simply
don't have the background expected of a reader of CMS.
And again you could have found that out before buying it,
by reading the introduction.)
>You don't have to say that;
>it's much more satisfactory to let other people say you're good. And you
>are only good as a mathematician. Other people are good in other things.
>
>Glad that we are different. No? Are you a decent music composer? I guess
>you're not:
Giggle. You've determined that how, exactly?
>http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm
>http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm#karaoke
>
>You will find the name "Edwin van Veldhoven" with the last URL. It shows
>that I have no trouble with finding someone superior in comparison with
>myself. On the contrary, _recognizing_ that fact has resulted in a most
>enjoyable collaboration where he wrote the music and I wrote the lyrics.
>
>Han de Bruijn
David C. Ullrich
Han de Bruijn
That's why so many mathematics textbooks are virtually _unreadable_.
But don't think there is a reason to be proud of bad writing manners.
> The intended audience (as stated clearly in the introduction, and
> note that you could have read the introduction before buying the
> book by following a link I posted) is beginning mathematics
> graduate students. It's not supposed to be a book for people
> who want a quick overview of the field, it's a book for people
> who want to become mathematicians. Such people need to learn
> much more than an overview.
Why don't you simply say that you don't want an _amateur_ to invade into
your territory ? Ah, how I have experience with this attitude! Once upon
a time, another professor said to me: "Look, mr. de Bruijn, it is _us_
who do the research here". And professional music makers .. Oh well, I'm
not even going to try to penetrate into those circles anymore.
Yes, because it is more elegant even at first sight. While your muddling
with the (1+cos)/(1-cos) instead of (1/tan)^2 seems clumsy even at first
sight. (Have enough common calculus experience to recognize such things)
> Which was a little irritating, given that (as seemed likely
> at the time, and as you've repeatedly made very clear since
> then) you obviously didn't actually understand _either_ version.
That's true, but I _suspected_ the second version would bring us closer
to a nice solution. In fact, I _accepted_ the solution as formulated at
the end of my article as the final one, until you (trouble maker) threw
doubt on "the limit of the product is the product of the limits". I must
admit that you're quite right in the end but it took me a while to have
peace with it.
>>Okay, now we're there
>>you have two iterated limits:
>>
>>lim_{n->oo} { lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
>>lim_{N->oo} { lim_{n->oo} prod_{k=1,N} [1 + {z/(k.pi).T(k.pi/n)}^2] }
>>
>>And you have already proved that:
>>
>>lim_{N->oo} prod_{k=1,N} [1 + {z/(k.pi)}^2] = ? =
>>lim_{n->oo} prod_{k=1,oo} [1 + {z/(k.pi).T(k.pi/n)}^2]
>>
>>In order to proceed, take a look at .. Etcetera. See? Without calling a
>>person ignorant (of course I am ignorant in this field, what else do you
>>expect).
>
> If you know you're ignorant about something then it's surprising
> you're not more careful about suggesting that I meant something
> other than what I wrote, then announcing a "refutation" what
> you think I meant. That sort of thing certainly doesn't sound
> like a person who's aware of his ignorance.
>
>>And without saying you're so good.
>
> I didn't say _I_ was good. I pointed out that others feel the book
> iis good in precisely the sort of ways that you find it's lacking.
> That could be useful information for you - it _could_ tell you
> where the source of your difficulties lies. (In fact you simply
> don't have the background expected of a reader of CMS.
> And again you could have found that out before buying it,
> by reading the introduction.)
I have not the background to be on this earth anyway, and yet here I am.
>>You don't have to say that;
>>it's much more satisfactory to let other people say you're good. And you
>>are only good as a mathematician. Other people are good in other things.
>>
>>Glad that we are different. No? Are you a decent music composer? I guess
>>you're not:
>
> Giggle. You've determined that how, exactly?
Because there's no music on your web page? But, uh, I see that there is
no mathematics there either. In fact, there's almost nothing at all:
http://www.math.okstate.edu/~ullrich/
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
Unreadable by people who don't know the prereqisites and don't
have the required mathematical "maturity", you mean.
>But don't think there is a reason to be proud of bad writing manners.
Giggle. I'll tell you what the very first comment on a very early
version of the book was. It was from a student in the class I was
teaching. She said "At last, a math book you can actually _read_!"
You really need to grow up. It's not reasonable to expect every
book in every technical field to be accessible to people who
know nothing about the field. That would be theoretically
possible, but it would mean that every book would have to
contain the contents of what's normally included in ten other
books that are instead taken as prerequsites. And _that_
_would_ make the book unreadable by the intended
audience.
I use the quadratic formula for solving quadratic equations
at one or two points in the book. Does this mean I should
include a proof of the quadratic formula, to make the book
readable by someone who doesn't know it already? What
about readers who never studied any high-school algebra?
Should I explain what it means to say that x = 2 is a solution
to the equation x + 2 = 4?
Of course not. In _any_ book there _has_ to be _some_
place where the line is drawn - there _must_ be _some_
set of prerequisites that the author simply assumes of
the reader, or every book in every technical field would
be more than 10,000 pages long. Any book on even
moderately advanced mathematics _is_ going to be
unreadable for 99% of the people on the planet. But
you seem to feel that in spite of this every book _should_
be accessible to _you_.
You need to grow up a little. There are plenty of math
books that I would have no chance of reading - in fact
that probably applies to _most_ "advanced" math books.
That doesn't make them bad books, or unreadable books.
>> The intended audience (as stated clearly in the introduction, and
>> note that you could have read the introduction before buying the
>> book by following a link I posted) is beginning mathematics
>> graduate students. It's not supposed to be a book for people
>> who want a quick overview of the field, it's a book for people
>> who want to become mathematicians. Such people need to learn
>> much more than an overview.
>
>Why don't you simply say that you don't want an _amateur_ to invade into
>your territory ?
Because that's not true. Deducing this from the fact that there are
prerequisites for CMS is just silly. And thinking this in spite of
the amount of time you see me spending here on sci.math trying
to explain math to people indicates you haven't been paying
attention.
Giggle. Yes, let's concentrate on the trivial calculations involved.
In _fact_ the reason Rob's version was much better is because he
noticed the convergence was _monotone_.
>> Which was a little irritating, given that (as seemed likely
>> at the time, and as you've repeatedly made very clear since
>> then) you obviously didn't actually understand _either_ version.
>
>That's true, but I _suspected_ the second version would bring us closer
>to a nice solution.
For no good reason, since you were simply ignoring the part that
might _not_ be just a trivial calculation. You had _no_ evidence
that that part was simpler, since you didn't follow either version.
What I'd posted _looked_ more complicated since I included
the details that Rob just wrapped into the word "monotone".
>In fact, I _accepted_ the solution as formulated at
>the end of my article as the final one, until you (trouble maker) threw
>doubt on "the limit of the product is the product of the limits".
Not "until". I'd pointed out that this needs to be proved several
times _before_ you showed us that pdf. That's what was so
funny about your calling the pdf a simpler version - "simplifying"
the proof by simply leaving out a major detail.
Guggle again. So you _realize_ that the evidence that led you to
conclusions about my lack of musical abilities could be used
equally well to deduce that I know nothing about math. It's
not clear to me whether you realize what this says about the
_validity_ of that deduction...
giggle.
>http://www.math.okstate.edu/~ullrich/
>
>>>http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm
>>>http://hdebruijn.soo.dto.tudelft.nl/www/muziek/download.htm#karaoke
>>>
>>>You will find the name "Edwin van Veldhoven" with the last URL. It shows
>>>that I have no trouble with finding someone superior in comparison with
>>>myself. On the contrary, _recognizing_ that fact has resulted in a most
>>>enjoyable collaboration where he wrote the music and I wrote the lyrics.
>
>Han de Bruijn
David C. Ullrich
Han de Bruijn
That student is not even wrong. Your book may be the best of its _kind_.
But the question remains if that _kind_ could be improved. I find it can
be done with relatively little effort. Just adopting writing manners as
with common books on quite different topics than mathematics. See below.
> You really need to grow up. It's not reasonable to expect every
> book in every technical field to be accessible to people who
> know nothing about the field. That would be theoretically
> possible, but it would mean that every book would have to
> contain the contents of what's normally included in ten other
> books that are instead taken as prerequsites. And _that_
> _would_ make the book unreadable by the intended
> audience.
I'm not talking about content here, just talking about giving _form_ to
content. Headers like "Summary", "Introduction", "Overview" haven't been
invented for nothing. I can also think of "decision tree", found in some
books. More pictures maybe. Some numerical work (eg Euler was not shy of
the latter, despite of the fact that he had no electronic calculators at
his disposal). But on the other hand I can imagine that your purpose has
not been the publication of a glossy magazine (i.e. the other extreme as
with some literally heavy textbooks like Calculus Early Transcendentals)
> I use the quadratic formula for solving quadratic equations
> at one or two points in the book. Does this mean I should
> include a proof of the quadratic formula, to make the book
> readable by someone who doesn't know it already? What
> about readers who never studied any high-school algebra?
> Should I explain what it means to say that x = 2 is a solution
> to the equation x + 2 = 4?
>
> Of course not. In _any_ book there _has_ to be _some_
> place where the line is drawn - there _must_ be _some_
> set of prerequisites that the author simply assumes of
> the reader, or every book in every technical field would
> be more than 10,000 pages long. Any book on even
> moderately advanced mathematics _is_ going to be
> unreadable for 99% of the people on the planet. But
> you seem to feel that in spite of this every book _should_
> be accessible to _you_.
Of course ! Did I deny anything ? Wouldn't you be frustrated if you buy
something beyond your comprehension ? Giggle ..
> You need to grow up a little. There are plenty of math
> books that I would have no chance of reading - in fact
> that probably applies to _most_ "advanced" math books.
> That doesn't make them bad books, or unreadable books.
Not per se. But it wouldn't be the first time people use sledgehammers
instead of a simple screw driver. I've seen more than one example of it
in my would-be career as a would-be scientist.
>>>The intended audience (as stated clearly in the introduction, and
>>>note that you could have read the introduction before buying the
>>>book by following a link I posted) is beginning mathematics
>>>graduate students. It's not supposed to be a book for people
>>>who want a quick overview of the field, it's a book for people
>>>who want to become mathematicians. Such people need to learn
>>>much more than an overview.
>>
>>Why don't you simply say that you don't want an _amateur_ to invade into
>>your territory ?
>
> Because that's not true. Deducing this from the fact that there are
> prerequisites for CMS is just silly. And thinking this in spite of
> the amount of time you see me spending here on sci.math trying
> to explain math to people indicates you haven't been paying
> attention.
Okay. Therefore I should withdraw that accusation. Sorry.
No comment anymore.
Nothing. Just curious ..
Chip Eastham
> These comments on the book are getting to be amusing.
[snip]
> But you have to actually _read_ it (and also do the exercises).
> No, merely skimming through it the way you've been doing
> is not going to help.
A book sale is a book sale! Perhaps Han should
read yours side by side with the one James self-
published.
regards, chip
Han de Bruijn
James Stewart Jr. ?
Han de Bruijn
David C. Ullrich
No, James Harris.
Han de Bruijn
> On Wed, 10 Jun 2009 15:50:04 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>
>>Chip Eastham wrote:
>>
>>>On Jun 8, 10:56 am, David C. Ullrich <dullr...@sprynet.com> wrote:
>>>
>>>>These comments on the book are getting to be amusing.
>>>
>>>[snip]
>>>
>>>>But you have to actually _read_ it (and also do the exercises).
>>>>No, merely skimming through it the way you've been doing
>>>>is not going to help.
>>>
>>>A book sale is a book sale! Perhaps Han should
>>>read yours side by side with the one James self-
>>>published.
>>
>>James Stewart Jr. ?
>
> No, James Harris.
Should have added that smiley.
Han de Bruijn