mm-4309 === Subject: Re: Another Inconvenient Truth > The axiom of infinity could be expressed as: There exists a > set that is equinumerous with a proper subset of itself The statement 'there exists a Dedekind infinite set' entails the axiom of infinity in ZF-I, or rather is some form of choice needed? infinity) states: In many theories it is convenient to use the so-called Dedekind axiom of infinity: There exists a set which can be mapped one-to-one into one of its proper subsets. With the aid of the axiom of choice it is easily shown that Dedekind's axiom of infinity is equivalent to the other above-mentioned forms of the axiom of infinity. It is known, however, that this equivalence cannot be proved by usual set-theoretic means without the use of the axiom of choice. MoeBlee === Subject: Re: PARADISE LOST: Debunking Cantor's theory > If I say When there are no mathematicians in Heidelberg, there will > be no mathematicians in the whole of Germany, the whole emphasises > that Germany is different from (and bigger than) Heidelberg. > So when Cantor says that if the set of finitely definable reals > is countable, then the *whole* set of reals is countable, he is > giving a big hint that he thinks the *whole* set is different > and larger. > But this whole set is countable (if Koenig is right). So what should remain? That a set is countable does not mean automatically that any superset is > also countable. The countability of supersets must be separately > established. And it has not been. > This issue is not under discussion here. A totally irrelevant remark --- off topic. Only in the tiny mind of an anti-mathematician like WM. The issue is whether the countability of the set of finitely definable > reals necessarily implies that of the set of all reals, which it does > not. So my point is right on target. > No. All I can do is to shorten his original sentence > W?re K?nigs Satz, da? alle endlich definirbaren reellen Zahlen einen Inbegriff von der M?chtigkeit 0 ausmachen, /richtig/, so hie?e dies, das ganze Zahlencontinuum sei abz?hlbar, was doch sicherlich / falsch/ ist. > to an easily understandable version: > If the definable numbers had cardinality aleph_0, then the whole continuum was countable. Deliberate mistranslation. Even those whose German is non-existent can > tell the difference between endlich definirbaren and definirbaren- > This was not a translation but a brief rephrasing. Infinite has been omitted because, according to Cantor, only finitely definable is definable. Deliberate misrepresentation, then, as we have no evidence here that He regarded non-finitely definable items as nonsense. But as in the view of modern mathematicians nonsense seems to be an important ingredient of mathematics, you cannot accept my identifiying nonsense with nothing. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Turing > gave precise definitions of computable number and the result (that is, > there are numbers that are not computable). > *Where* are those numbers that are not computable? Not in computers. > That is a tautology. The question was *where* are those numbers that are not computable? In the axiom systems on which they are based My dear mathematician: Axiom systems in general do not contain numbers - except for enumerating the axioms. === Subject: Re: TeX for sci.math? Cc: cronholm144@gmail.com > Is their any way to getTeXto render within these groups, it would > make the reading much easier. --Cronholm http://thewe.net/tex === Subject: Re: Help with a trigonometric function > I'm trying to solve the following inequation for d: arctan(a/d) * arctan (b/d) < c a, b and c are knowns; d is the unknown I need an expression of the kind d > .... but I could not found how > to get rid of the atan The problem Is not easy because tan [arctan (a) * arctan (b)] is not a * b I need a form to convert from two arctans to one, ie, from arctan(a)*arctan(b) to arctan(....) or any other way to simplify the arctan Nothing of that sort exists. Assuming a, b, c > 0 and c < (pi/2)^2, about the best you can say is d > b/t where t > 0 and c = arctan(t) arctan(a t/b). Let's let a/b = r, so we have one fewer parameter. For specific values of c and r, you could use numerical methods (e.g. Newton's method). Or you might try series expansions. For example, the solution has the following power series in c, according to Maple: t = 1/r^(1/2)*c^(1/2)+1/6*(r^2+1)/r^(3/2)*c^(3/2) -1/360*(r^4-50*r^2+1)/r^(5/2)*c^(5/2) -1/5040*(r^2+1)*(11*r^4-158*r^2+11)/r^(7/2)*c^(7/2) +1/1814400*(1357*r^8-4260*r^6+45486*r^4-4260*r^2+1357)/r^(9/2)*c^(9/2) -1/119750400*(r^2+1)*(12965*r^8-2196*r^6-552226*r^4-2196*r^2+12965)/r^(11/2) *c^(11/2) + ... -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: Re: Calculus I: What to expect? >Re Mathematica, IMO it's certainly not a must have, but it is darn >handy to have a subset of the full product, for example to check your >work. In that regard, here are some web sites you should be aware of: I know this sounds like a weird question..... but I do not have a computer right now of any kind and in market for one. Given that I may need to run math software and other engineering type apps at some point..... would you advise AGINST a laptop? Or are laptops so powerful now days its no problem? If yes to a laptop....any size you would AVOID? Example.... is a 14 laptop too small for such software? Need bigger display such as 15 laptop? === Subject: Re: Calculus I: What to expect? > excellent links!! === Subject: Re: Calculus I: What to expect? >excellent links!! > Foo! I forgot another good one. Add this one to the list: The math stuff at Wikipedia is useful resource. The capabilities of the first three sites above are pretty clear. The fourth site (the Wolfram Functions site) merits some exploration because there's more there than might be apparent at first. As one example ... o Click on Elementary Functions o Click on sin(z) Click on o Plotting o Evaluation o Visualizations o any of the other long list things to get an idea of what's what === Subject: Re: Calculus I: What to expect? >I'll second what Arturo said. Re Mathematica, IMO it's certainly not a must have, but it is darn >handy to have a subset of the full product, for example to check your >work. In that regard, here are some web sites you should be aware of: don't see any problem with that. For me, chemistry was hard! Putting >a little milk in the coffee is about all the chemistry I know how to >do. 8-) Well I was going to start Chem I, but the teacher died unexpectedly!!! he was also the Physics teacher so neither one is offered this semester. So I'm trying to think of other classes to take that will not waste my time.... such as music appreciation!! === Subject: Re: Blaschke product convergence <10535141.1187088436435.JavaMail.jakarta@nitrogen.mathforum.org > On Aug 13, 10:28 pm, jane sum (1 - |a_n|) = oo, B_n (z) = prod {z-a_n}/(1-bar{a_n}z), then lim_{n -> oo} B_n(z) = 0, for |z| < 1 ? Yes .See Burckel,Robert B ,An introduction to classical complex analysis ,Burkhauser Verlag Basil (1979) Exercise 6.9 i (liberal hints given in all his exercises) ,this one says for all a ,z in the open unit disk that |z-a/1-bar(a)z| <= exp(1-|a|)(|z|-1)/|z|+1). Are you sure about that ? Cause then, if sum(1-|a_n|) = oo the product of exp(1-|a_n|) tends to oo ? No it goes to 0 because |z|-1 Is NEGATIVE! FOR z IN THE The proof of this uses exercise 6.3 iv ,a corollary of - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Do any integers occur in both sequences? > I submitted these two interdependent sequences to the > Encyclopedia Of > Integer Sequences. > welcome to the club :-) > (Those familiar with the EIS will be familiar with > the formatting.) > %S A131937 1,4,8,14,21,29,38,49,61 %N A131937 a(1)=1; a(2)=4. a(n) = a(n-1) + (nth > positive integer which does not occur in sequence A131938). %e A131937 A131938: 2,5,10,16,23,32,42,53,... Positive integers not in A131938: > 1,3,4,6,7,8,9,11,... So A131937(8) = A131937(7) + 11 = 49. %Y A131937 A131938 %O A131937 1 %K A131937 ,more,nonn, > %S A131938 2,5,10,16,23,32,42,53,65,78,93,109 %N A131938 a(1)=2; a(2)=5. a(n) = a(n-1) + (nth > positive integer which does not occur in sequence A131937). %e A131938 A131937: 1,4,8,14,21,29,... Positive integers not in A131937: > 2,3,5,6,7,9,10,11,... So A131938(8) = A131938(7) + 11 = 53. %Y A131938 A131937 %O A131938 1 %K A131938 ,more,nonn, > Do any positive integers occur in both A131937 and > A131938? In other words: Does A131937(k) = > A131938(j) for any j and k {j and k are >= 1}, where > j need not equal > k? I have not thought about this too hard; so for all I > know, the proof > is quite easy. I conjecture that no particular positive integer > occurs in both > sequences. > Here is a smaller result related to these sequences, > which I doubt > will > help (dis)prove the main conjecture: Let a(n) = A131937(n), b(n) = A131938(n), a(0) = > b(0) = 0. Let n = any positive integer. Then n occurs (a(n) - a(n-1) - 1) times in sequence > {b(n) - b(n-1) - n + 1}. And n occurs (b(n) - b(n-1) - 1) times in sequence > {a(n) - a(n-1) - n + 1}. Not too earth-shattering -- but while we're on the > subject... > Also, I wonder if anyone can come up with a closed > form for > {A131937(n)} > and {A131938(n)}. > They seem like they might be related to Beautty > sequences somehow. plz give link to beautty sequence Leroy Quet > tommy1729 === Subject: Re: (mathematical) proof that there is no God <5ic0veF3mq1viU2@mid.individual.net>