On Mon, 25 Jul 2005 23:55:08 +0000 (UTC), Dave Seaman <dseaman@no.such.host> wrote:>On Mon, 25 Jul 2005 19:32:47 -0700, quasi wrote:>> ... by agreement 0^0 is undefined.>If you read the history of sci.math and check the FAQ, I think you will >find that there is no such agreement.Maybe not in sci.math. (but I admit, I haven't read the FAQ)>> we ... want the function >> >> (x,y) --> x^y >> >> to preserve limits, but any value assigned to 0^0 would make x^y >> discontinuous at (0,0).>We do we want that function to preserve limits? Why should we ask the >impossible?It's not impossible. So long as 0^0 is undefined, x^y is continuous everywhere else in the first quadrant, including the positive axes.>> To avoid the discontinuity, we declare 0^0 undefined.All fixed. No more dangerous drops.>Why is it necessary to avoid the discontinuity? Is there something wrong >with discontinuous functions?Discontinuities can cause serious injuries if you step on one. But seriously, yes, there is something wrong with discontinuity in this context. Since all the other basic arithmetic operations are continuous (and for division we force it by disallowing division by 0), we intuitively expect exponentiation to be similarly well behaved. To amplify this point: If (a,b) is sufficiently close to (c,d) in R^2 then a+b is close to c+d a-b is close to c-d a*b is close to c*d a/b is close to c/d (provided both fractions are defined) so ... if a^b and c^d are both defined, then we expect a^b to be close to c^d which means we want x^y to be continuous on it's domain. To get continuity of x^y on it's domain, we have to sacrifice (0,0), a small price to pay. quasi
collection of mathematically elegant tricks
Started by ●July 24, 2005
Reply by ●July 25, 20052005-07-25
Reply by ●July 25, 20052005-07-25
On Mon, 25 Jul 2005 22:00:06 -0700, quasi wrote:> On Mon, 25 Jul 2005 23:55:08 +0000 (UTC), Dave Seaman ><dseaman@no.such.host> wrote:>>> we ... want the function>>> (x,y) --> x^y>>> to preserve limits, but any value assigned to 0^0 would make x^y >>> discontinuous at (0,0).>>We do we want that function to preserve limits? Why should we ask the >>impossible?> It's not impossible. So long as 0^0 is undefined, x^y is continuous > everywhere else in the first quadrant, including the positive axes.It's impossible to make x^y continuous at (0,0). However, that is not a reason to leave 0^0 undefined.>>Why is it necessary to avoid the discontinuity? Is there something wrong >>with discontinuous functions?> Discontinuities can cause serious injuries if you step on one.So you admit you have no actual reason?> But seriously, yes, there is something wrong with discontinuity in > this context. Since all the other basic arithmetic operations are > continuous (and for division we force it by disallowing division by > 0), we intuitively expect exponentiation to be similarly well behaved.Intuition is often wrong. I suggest you read the FAQ. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Reply by ●July 26, 20052005-07-26
On Tue, 26 Jul 2005 02:52:35 +0000 (UTC), Dave Seaman <dseaman@no.such.host> wrote:>On Mon, 25 Jul 2005 22:00:06 -0700, quasi wrote: >> On Mon, 25 Jul 2005 23:55:08 +0000 (UTC), Dave Seaman >><dseaman@no.such.host> wrote:>>>Why is it necessary to avoid the discontinuity? Is there something wrong >>>with discontinuous functions? > >> Discontinuities can cause serious injuries if you step on one. > >So you admit you have no actual reason?I admit no such thing -- I was just being silly, of course, and I followed by giving an actual reason:>> But seriously, yes, there is something wrong with discontinuity in >> this context. Since all the other basic arithmetic operations are >> continuous (and for division we force it by disallowing division by >> 0), we intuitively expect exponentiation to be similarly well behaved.>Intuition is often wrong.This is not a matter of right or wrong, it's a definition -- it's arbitrary. If we value the intuition, based on experience, that the arithmetic operations are continuous, then it's natural to disallow 0^0 in the same way that we disallow x/0. In other words, we have a choice -- we can either agree to assign a value to 0^0 (although then there would be a fight as to which value, presumably 0 or 1), or we can declare 0^0 undefined. Either way, it's a tradeoff. If we assign a value to 0^0 we lose continuity of x^y at (0,0) so exponentiation becomes the only arithmetic operation which is not continuous on it's domain. For that loss, what is the gain? On the other hand, if 0^0 is declared undefined, then all the arithmetic operations are continuous on their domains, and that supports the intuition that we all have from working with numbers. The benefits of continuity are easily seen: If a is sufficiently close to 2 and b is sufficiently close to 3 you can be confident that a+b is close to 5 a-b is close to -1 a*b is close to 6 a/b is close to 2/3 and ... a^b is close to 8>I suggest you read the FAQ.Maybe I will at some point, although if I do, I now sense I will read it with a skeptical bias. quasi
Reply by ●July 26, 20052005-07-26
On 25 Jul 2005 23:22:08 GMT, rusin@vesuvius.math.niu.edu (Dave Rusin) wrote:>The intuition with limits may >be important in some contexts but there is a priori nothing wrong >with a definition like > > 1. x^0 =1 if x > 0. > 2. 0^x = 0 if x > 0. > 3. 0^0 = 17Sure the definition of 0^0 is arbitrary but the only natural choices are "undefined", 0, or 1. As for 0 or 1, even disregarding the continuity argument, why should 0^x have priority over x^0 or vice versa. Ah, I have an idea. How about making it a quantum variable, either 0 or 1 but at any given instant it could be either, no way to determine which, except by a probability distrinution. Oh, but I bet that's already been proposed, maybe it's even in the FAQ.>It's not like Moses descended from the mountains with the Official >Definition of exponentiation written on stone tablets; we define the >operation as we see fit.If Moses came down and proclaimed "0^0=17", they wouldn't have followed him anywhere. quasi
Reply by ●July 26, 20052005-07-26
quasi <quasi@null.set> writes:> On Tue, 26 Jul 2005 02:52:35 +0000 (UTC), Dave Seaman > <dseaman@no.such.host> wrote: > >>On Mon, 25 Jul 2005 22:00:06 -0700, quasi wrote: >>> On Mon, 25 Jul 2005 23:55:08 +0000 (UTC), Dave Seaman >>><dseaman@no.such.host> wrote: > >>>>Why is it necessary to avoid the discontinuity? Is there something wrong >>>>with discontinuous functions? >> >>> Discontinuities can cause serious injuries if you step on one. >> >>So you admit you have no actual reason? > > I admit no such thing -- I was just being silly, of course, and I > followed by giving an actual reason: > >>> But seriously, yes, there is something wrong with discontinuity in >>> this context. Since all the other basic arithmetic operations are >>> continuous (and for division we force it by disallowing division by >>> 0), we intuitively expect exponentiation to be similarly well behaved. > >>Intuition is often wrong. > > This is not a matter of right or wrong, it's a definition -- it's > arbitrary. If we value the intuition, based on experience, that the > arithmetic operations are continuous, then it's natural to disallow > 0^0 in the same way that we disallow x/0.So you think that the polynomial 3 ---- \ k f(x) = > a x / k ---- k = 0 should be undefined for x=0? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
Reply by ●July 26, 20052005-07-26
quasi <quasi@null.set> writes:> On 25 Jul 2005 23:22:08 GMT, rusin@vesuvius.math.niu.edu (Dave Rusin) > wrote: > >>The intuition with limits may >>be important in some contexts but there is a priori nothing wrong >>with a definition like >> >> 1. x^0 =1 if x > 0. >> 2. 0^x = 0 if x > 0. >> 3. 0^0 = 17 > > Sure the definition of 0^0 is arbitrary but the only natural choices > are "undefined", 0, or 1. As for 0 or 1, even disregarding the > continuity argument, why should 0^x have priority over x^0 or vice > versa.Because 0^x is discontinous at 0 even if you set 0^x=0? Setting it at 0 only for one-sided continuity seems to be less useful than setting it to 1 for total continuity of x^0 everywhere on C. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
Reply by ●July 26, 20052005-07-26
On Tue, 26 Jul 2005 00:16:51 -0700, quasi wrote:> Sure the definition of 0^0 is arbitrary but the only natural choices > are "undefined", 0, or 1. As for 0 or 1, even disregarding the > continuity argument, why should 0^x have priority over x^0 or vice > versa.Because x^0 = 1 for all x. That's a theorem of ZF according to Suppes, _Axiomatic_Set_Theory_. The x in that case is an arbitrary cardinal number (finite or infinite), but it is easily extended to other values of x. The theorem in Suppes is based on the definition of cardinal exponentiation, which is that if A and B are cardinals, then A^B is the cardinality of the set of mappings from B into A. In the case A = B = 0 = {}, there is exactly one such mapping, hence 0^0 = 1. According to Lang's _Algebra_, if M is a multiplicative monoid with identity e, then m^n makes sense for each m in M and each natural number n, and furthermore m^0 denotes an empty product, which evaluates to e (the identity in M). Notice that nothing I have said above has anything at all to do with limits. In fact, there isn't even a topology involved. There are some who prefer to say that 0^x = 1 whenever the exponent is viewed as an integer or a rational number, but that 0^0.0 remains undefined if the exponent is regarded as real. <http://db.uwaterloo.ca/~alopez-o/math-faq/node40.html#SECTION00530000000000000000> -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Reply by ●July 26, 20052005-07-26
Dave Seaman <dseaman@no.such.host> writes:> There are some who prefer to say that 0^x = 1 whenever the exponent > is viewed as an integer or a rational number,Cough, cough. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
Reply by ●July 26, 20052005-07-26
In <1h09nx3.19tuzby1lbqkqoN%mh@michael-hoppe.de>, on 07/25/2005 at 07:18 PM, mh@michael-hoppe.de (Michael Hoppe) said:>The proof of> sin(a + b) = (sin a) * (cos b) + (sin b) * (cos a)>according to E. Schmidt runs as follows:> f(x) := sin(a + b - x)*cos(x) + cos(a + b - x)*sin(x)>is constant on R (as well on C), because f' = 0. From> f(0) = f(b)>follows the desired theorem.There's an easier trick. cos(a+b) + i*sin(a+b) = e^i*(a+b) = e^i*a * e^i*b = (cos(a) + i*sin(a)) * (cos(b) + i*sin(b)) = (cos(a)*cos(b) - sin(a)*sin(b)) + i*(cos(a)sin(b) + sin(a)*cos(b)) It's fast and easy to work out just about any common trig identity with this tool. -- Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel> Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org
Reply by ●July 26, 20052005-07-26
On Tue, 26 Jul 2005 14:45:18 +0200, David Kastrup wrote:> Dave Seaman <dseaman@no.such.host> writes:>> There are some who prefer to say that 0^x = 1 whenever the exponent >> is viewed as an integer or a rational number,> Cough, cough.Make that x^0 = 1. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
