Hetware
Posts:
148
Registered:
4/13/13


Re: Is (t^29)/(t3) defined at t=3?
Posted:
Oct 13, 2013 8:49 PM


On 10/13/2013 4:54 PM, Peter Percival wrote: > Hetware wrote: >> On 10/9/2013 4:49 AM, Peter Percival wrote: >>> Hetware wrote: >>>> On 10/7/2013 8:39 PM, Peter Percival wrote: >>>>> Hetware wrote: >>>>>> On 10/7/2013 4:56 AM, David Bernier wrote: >>>>>>> On 10/07/2013 03:21 AM, Robin Chapman wrote: >>>>>>>> On 07/10/2013 04:34, Hetware wrote: >>>>>>>>> On 9/30/2013 4:03 AM, Robin Chapman wrote: >>>>>>>> >>>>>>>>>> Hetware: 0/0 = 3 >>>>>>>>>> >>>>>>>>>> Ciekaw: 0/0 = 1 >>>>>>>>>> >>>>>>>>>> Any more entrants? >>>>>>>>>> >>>>>>>>> >>>>>>>>> To be correct; Hetware: 0/0 = 1 (under certain circumstances). >>>>>>>> >>>>>>>> Not according to your original posting in this thread :( >>>>>>> >>>>>>> Heware, if 0/0 = 1, then 0/0 = (100*0)/0 = 100*(0/0) = 100*1 = 100. >>>>>>> >>>>>>> So, assuming 0/0 = 1, we find that 1 = 100 :( >>>>>>> >>>>>>> David >>>>>>> >>>>>> >>>>>> That statement came with a qualification. That is, given a function >>>>>> defined by f(t) = (t^29)/(t3), I could assume (t3)/(t3) = 1, even >>>>>> where t=3. >>>>> >>>>> "given a function defined by" is irrelevant. At t=3 (t3)/(t3) is >>>>> undefined. >>>>> >>>>>> I've already shown that a modified version of that >>>>>> proposition does make sense. >>>>> >>>>> No you haven't. >>>>> >>>>>> Given a function f(t) continuous for all real numbers t and >>>>>> defined by >>>>>> (t^29)/(t3) everywhere the expression is meaningful, that >>>>>> function is >>>>>> identical to g(t) = (t+3). My original mistake was to assume >>>>>> continuity >>>>>> after using (t^29)/(t3) to define the entire function. >>>>> >>>>> What you wish to say is that the function g:R>R defined by >>>>> >>>>> g(t) = f(t) if t=/=3 >>>>> = 6 otherwise >>>>> >>>>> has these properties: >>>>> i) g = f where f is defined, >>>>> ii) g is defined on the whole of R, >>>>> iii) g is continuous. >>>>> >>>>> >>>>> >>>> >>>> The value of 6 at t=3 follows from the stipulation of continuity. >>> >>> The value of what? f or g? f has no value at 3. >>> >> >> Given a function f(t) continuous for all real numbers t and defined by >> (t^29)/(t3) everywhere the expression is meaningful, that function is >> identical to g(t) = (t+3). >> >>>> It is >>>> meaningful to say that f(t) is continuous over the domain of real >>>> numbers. >>> >>> So what? Is it true? Yes, by proof, not by stipulation. > > Whoops. I didn't mean to claim that f is continuous at every real. > >> You cannot prove a definition. > > To prove that f is continuous throughout its domain (which is what I > meant) is not proving a definition. > > Is this the situation: "Here is a function f not defined at every real. > Can I find a function g with these properties: (i) it agrees with f > where f is defined, and (ii) it continuous at every real?" The answer > is yes. > >>>> It is also meaningful to say that f(t) = (t^29)/(t3) >>>> everywhere that the rhs is meaningful. >>> >>> I would say "f is defined everywhere (in R) that (t^29)/(t3) is >>> defined". >> >> And you would not be giving the same definition as I am giving. >> >>>> That is sufficient information >>>> to determine that f(3) = 6. >>> >>> f(3) doesn't equal anything. One may "fill the gap" by defining g as I >>> have done above, *then* g(3) = 6. But g isn't f. >>> >> >> "Definition 21.1. A function f is a set of ordered pairs, no two of >> which have the same first element. The set of first elements of the >> pairs is called the domain of the function, whereas the set of second >> elements of the pairs is called the range. The domain and range >> elements are related by a given rule" >> >> The definition of continuity provides the rule for determining the value >> of the function f(3) which is *DEFINED* as continuous, and is given by >> f(t) = (t^29)/(t3) where t!=3. By definition f(3) has a definite, >> finite value at f(3) and that value is f(3) = limit[(t^29)/(t3), t>3]. >> >> Where do you get the idea that one cannot define a function as >> continuous? How would you prove, for example, the mean value theorem, >> if you could not define an arbitrary function to be continuous? > > I can define an arbitrary function to be continuous? What if my chosen > function is not continuous? >
I appreciate your objection. The same thing has been in my mind. It is a question of the chicken and the egg.
Let f(t) = (t^29)/(t3) where t!=3, and f(3) = limit[(t^29)/(t3), t>3]. Is, by this definition, f(t) a continuous function over the set of real numbers t? For the sake of further exposition, I will answer for you in the affirmative.
I believe we have agreed that (t^29)/(t3) is continuous everywhere except for t=3. When f(3) is defined as
f(3) = limit[(t^29)/(t3), t>3],
We have arrived at a function which is defined for all real numbers. So now we have a function f(t) which is defined for all real numbers t, and is globally continuous.
If we assert that f(t) is continuous, we are stipulating that
f(t) = limit[(f(s), s>t]
for all real numbers t. The function defined as
g(t) = limit[(s^29)/(s3), s>t]
is identical to f(t) as defined above. The difference in connotation is that the definition g(t) is not a piecewise definition.
So. To answer your question regarding defining a function to be continuous and then being able to show that it is not continuous. That results in a contradiction. You get your hiney spanked, and you are expelled from Plato's Heaven, and sent to the purgatory of teaching calculus to business majors.

