mark_pfannenstiel@yahoo.com writes:> well, I learned a new thing today. > I had always believed that 0^0 was an undefined quantity due to the 2 > statements > > 1. x^0 =1 > 2. 0^x = 0 x > 0.It is an indeterminate _form_ in the context of limits: without a specified dependence of x and y, lim (x,y)->(0,0) x^y can't be determined. But that is just the limit. The expression can have a value at the point even though it does not have a limit there. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
collection of mathematically elegant tricks
Started by ●July 24, 2005
Reply by ●July 25, 20052005-07-25
Reply by ●July 25, 20052005-07-25
"David Kastrup" <dak@gnu.org> wrote in message news:854qaiqyn1.fsf@lola.goethe.zz...> mark_pfannenstiel@yahoo.com writes: > >> well, I learned a new thing today. >> I had always believed that 0^0 was an undefined quantity >> due to the 2 >> statements >> >> 1. x^0 =1 >> 2. 0^x = 0 x > 0. > > It is an indeterminate _form_ in the context of limits: > without a > specified dependence of x and y, lim (x,y)->(0,0) x^y > can't be > determined. But that is just the limit. The expression > can have a > value at the point even though it does not have a limit > there. >I admit that my math might be a little rusty but to me it's "intuitively obvious and trivial" that lim (t^0) = 1 t->0 Where's the beef?
Reply by ●July 25, 20052005-07-25
In article <854qaiqyn1.fsf@lola.goethe.zz>, David Kastrup <dak@gnu.org> wrote:>mark_pfannenstiel@yahoo.com writes: > >> I had always believed that 0^0 was an undefined quantity due to the 2 >> statements >> >> 1. x^0 =1 >> 2. 0^x = 0 x > 0.>The expression can have a >value at the point even though it does not have a limit there.Just to underscore Kastrup's point, let me observe that statement "1." ("correctly") defines 0^0, and statement "2." does not define 0^0. Had the qualifier "x>0" been omitted, we would have a contradictory set of definitions; had an additional qualifier "x>0" been added to statement "1." then 0^0 would have been undefined. 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 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. dave PS -- Why do people like to delete the word "if", which turns the statements into grammatically correct sentences that can be read aloud and which explains the connection between the two clauses e.g. "x^0=1" and "x>0" ?
Reply by ●July 25, 20052005-07-25
On 25 Jul 2005 14:52:59 -0700, mark_pfannenstiel@yahoo.com wrote:>well, I learned a new thing today. >I had always believed that 0^0 was an undefined quantity due to the 2 >statements > >1. x^0 =1 >2. 0^x = 0 x > 0. > > >After an afternoon of considering such things as binomial theorem: >(1 + 0)^1 really ought to be equal to 1 >as well as some of the other arguements in the FAQ has convinced me 0^0 >= 1.No, by agreement 0^0 is undefined. One explanation is the reason you stated at the beginning, the conflict between the 2 laws. x^0=1 if x is not 0 0^x=0 if x>0 Another way to explain it, but also based on the conflict between the 2 laws, focuses on continuity: All the other arithmetic operations are continuous on their domains, that is, as functions of 2 variables, the following functions are continuous whenever they are defined: (x,y) --> x+y (x,y) --> x-y (x,y) --> x*y (x,y) --> x/y So we also 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). To see this, suppose 0^0 exists and x^y is continuous at (0,0). First approach (0,0) along the x axis (y=0): As x gets closer and closer to 0, x^0 must approach 0^0, so by this criterion, 0^0 should be 1. Next approach (0,0) along the positive y axis (x=0, y>0): As y gets closer and closer to 0 while staying positive, 0^y should approach 0^0, so by this criterion, 0^0 should be 1. We can't have both, so continuity must fail at (0,0). To avoid the discontinuity, we declare 0^0 undefined. quasi Mark, now that I've replied, I just noticed ... why did you not start a new thread? Your post seems to have nothing to do with the subject line "collection of mathematically elegant tricks".
Reply by ●July 25, 20052005-07-25
On Mon, 25 Jul 2005 19:32:47 -0700, quasi <quasi@null.set> wrote:>... >To see this, suppose 0^0 exists and x^y is continuous at (0,0). > >First approach (0,0) along the x axis (y=0): As x gets closer and >closer to 0, x^0 must approach 0^0, so by this criterion, 0^0 should >be 1. > >Next approach (0,0) along the positive y axis (x=0, y>0): As y gets >closer and closer to 0 while staying positive, 0^y should approach >0^0, so by this criterion, 0^0 should be 1.Correction -- the above line is a typo -- it should have been: Next approach (0,0) along the positive y axis (x=0, y>0): As y gets closer and closer to 0 while staying positive, 0^y should approach 0^0, so by this criterion, 0^0 should be 0.>We can't have both, so continuity must fail at (0,0). > >To avoid the discontinuity, we declare 0^0 undefined.quasi
Reply by ●July 25, 20052005-07-25
"John E. Hadstate" <jh113355@hotmail.com> writes:> "David Kastrup" <dak@gnu.org> wrote in message > news:854qaiqyn1.fsf@lola.goethe.zz... >> mark_pfannenstiel@yahoo.com writes: >> >>> well, I learned a new thing today. >>> I had always believed that 0^0 was an undefined quantity >>> due to the 2 >>> statements >>> >>> 1. x^0 =1 >>> 2. 0^x = 0 x > 0. >> >> It is an indeterminate _form_ in the context of limits: >> without a >> specified dependence of x and y, lim (x,y)->(0,0) x^y >> can't be >> determined. But that is just the limit. The expression >> can have a >> value at the point even though it does not have a limit >> there. > > I admit that my math might be a little rusty but to me it's > "intuitively obvious and trivial" that > > lim (t^0) = 1 > t->0 > > Where's the beef?That you did not read what I actually wrote above? That _is_ a specified dependence of x and y in the limiting process, namely forcing y=0 when doing the limit for x. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
Reply by ●July 25, 20052005-07-25
On Mon, 25 Jul 2005 19:32:47 -0700, quasi wrote:> On 25 Jul 2005 14:52:59 -0700, mark_pfannenstiel@yahoo.com wrote:> No, by agreement 0^0 is undefined. One explanation is the reason you > stated at the beginning, the conflict between the 2 laws.If you read the history of sci.math and check the FAQ, I think you will find that there is no such agreement.> So we also 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?> To see this, suppose 0^0 exists and x^y is continuous at (0,0).> First approach (0,0) along the x axis (y=0): As x gets closer and > closer to 0, x^0 must approach 0^0, so by this criterion, 0^0 should > be 1.> Next approach (0,0) along the positive y axis (x=0, y>0): As y gets > closer and closer to 0 while staying positive, 0^y should approach > 0^0, so by this criterion, 0^0 should be 1.> We can't have both, so continuity must fail at (0,0).> To avoid the discontinuity, we declare 0^0 undefined.Why is it necessary to avoid the discontinuity? Is there something wrong with discontinuous functions? -- 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 25, 20052005-07-25
"David Kastrup" <dak@gnu.org> wrote in message news:85u0iipi0h.fsf@lola.goethe.zz...> "John E. Hadstate" <jh113355@hotmail.com> writes: > >> "David Kastrup" <dak@gnu.org> wrote in message >> news:854qaiqyn1.fsf@lola.goethe.zz... >>> mark_pfannenstiel@yahoo.com writes: >>> >>>> well, I learned a new thing today. >>>> I had always believed that 0^0 was an undefined >>>> quantity >>>> due to the 2 >>>> statements >>>> >>>> 1. x^0 =1 >>>> 2. 0^x = 0 x > 0. >>> >>> It is an indeterminate _form_ in the context of limits: >>> without a >>> specified dependence of x and y, lim (x,y)->(0,0) x^y >>> can't be >>> determined. But that is just the limit. The expression >>> can have a >>> value at the point even though it does not have a limit >>> there. >> >> I admit that my math might be a little rusty but to me >> it's >> "intuitively obvious and trivial" that >> >> lim (t^0) = 1 >> t->0 >> >> Where's the beef? > > That you did not read what I actually wrote above? That > _is_ a > specified dependence of x and y in the limiting process, > namely > forcing y=0 when doing the limit for x. >I did read what you wrote. My question is, "why is what you wrote relevant to explaining why 0^0 is defined to be 1"?
Reply by ●July 25, 20052005-07-25
"John E. Hadstate" <jh113355@hotmail.com> writes:> "David Kastrup" <dak@gnu.org> wrote in message > news:85u0iipi0h.fsf@lola.goethe.zz... >> "John E. Hadstate" <jh113355@hotmail.com> writes: >> >>> "David Kastrup" <dak@gnu.org> wrote in message >>> news:854qaiqyn1.fsf@lola.goethe.zz... >>>> mark_pfannenstiel@yahoo.com writes: >>>> >>>>> well, I learned a new thing today. >>>>> I had always believed that 0^0 was an undefined >>>>> quantity >>>>> due to the 2 >>>>> statements >>>>> >>>>> 1. x^0 =1 >>>>> 2. 0^x = 0 x > 0. >>>> >>>> It is an indeterminate _form_ in the context of limits: >>>> without a >>>> specified dependence of x and y, lim (x,y)->(0,0) x^y >>>> can't be >>>> determined. But that is just the limit. The expression >>>> can have a >>>> value at the point even though it does not have a limit >>>> there. >>> >>> I admit that my math might be a little rusty but to me >>> it's >>> "intuitively obvious and trivial" that >>> >>> lim (t^0) = 1 >>> t->0 >>> >>> Where's the beef? >> >> That you did not read what I actually wrote above? That >> _is_ a >> specified dependence of x and y in the limiting process, >> namely >> forcing y=0 when doing the limit for x. >> > > I did read what you wrote. My question is, "why is what you > wrote relevant to explaining why 0^0 is defined to be 1"?Who claimed that it was? I was explaining that the limit is undefined without a specific path, and that some people take that to imply something for the value itself, which is not a necessary conclusion. That's all. I merely pointed out and dissected one particular common reasoning for wanting to keep 0^0 undefined, and said why I don't think this particular argument holds water. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
Reply by ●July 25, 20052005-07-25
