
Re: Is (t^29)/(t3) defined at t=3?
Posted:
Oct 6, 2013 11:51 PM


On Sunday, October 6, 2013 8:55:23 PM UTC5, Hetware wrote: > On 9/29/2013 9:46 PM, Arturo Magidin wrote: > > > On Sunday, September 29, 2013 7:30:49 PM UTC5, Hetware wrote: > > >> On 9/29/2013 8:06 PM, quasi wrote: > > >> > > >>> Hetware wrote: > > >> > > >>>> > > >> > > >>>> What I am saying is that if I encountered an expression such > > >> > > >>>> as (t^29)/(t3) in the course of solving a problem in > > >> > > >>>> applied math, I would not hesitate to treat it as t+3 and not > > >> > > >>>> haggle over the case where t = 3. > > >> > > >>> > > >> > > >>> And you would be wrong unless either > > >> > > >>> > > >> > > >>> (1) You know by the context of the application that the value > > >> > > >>> t = 3 is impossible. > > >> > > >>> > > >> > > >>> (2) You know by the context that the underlying function must > > >> > > >>> be continuous, thus providing justification for canceling the > > >> > > >>> common factor of t3, effectively removing the discontinuity. > > >> > > >>> > > >> > > >>> I challenged you to find a book  _any_ book, which agrees > > >> > > >>> with your naive preconception. > > >> > > >>> > > >> > > >>> Math book, applied math book, physics book, chemistry book, > > >> > > >>> economics book  whatever. > > >> > > >>> > > >> > > >>> If all the books and all the teachers say you're wrong, > > >> > > >>> don't you think that maybe it's time to admit that you > > >> > > >>> had a flawed conception about this issue and move on? > > >> > > >>> > > >> > > >>> quasi > > >> > > >>> > > >> > > >> > > >> > > >> I don't answer to the authority of mortals. I answer to the > > >> dictates of > > >> > > >> reason. > > > > > > Would that be the reason that you have learned through the authority > > > of mortals, or through direct contact with an immortal of some sort? > > > Come on, now. Get off that high horse... > > > > It does me no good to simply accept what someone tells me without being > > able to derive it in some sense from first principles.
Then you won't get anywhere.
What "first principles" did you apply to derive the notion of "function"? Of "domain"? Of "ordered pair"?
Again: get off the high horse; because you are being either ignorant, or hypocritical.
You aren't trying to "derive it in some sense from first principles." Rather, you are trying to justify your initial mistake, instead of simply accepting it and moving on.
> > Here's the thing: what is your definition of a function? How do you > > > determine whether two functions are equal or not? Are you viewing the > > > expression as purely algebraic expression or as a function? > > > > > > In the context of calculus, which is after all the context you find > > > yourself in, a function is a rule that assigns to every valid input > > > one and only one output. > > > > Thomas (1953) did not ascribe to that definition of a function.
Thomas distinguishes between regular functions and "multvalued functions". This, however, does not invalidate the definition of function, because the other notion that you are now bringing to the table extraneously is a different notion, that of "multivalued function". The adjective signals that you are dealing with a different notion.
Really. It's not that hard, once you stop trying to justify your errors.
> Though > > yours is the one that I prefer. I didn't ask if what I did violated > > tradition. I asked if it leads to a contradiction.
And that question is meaningless in the abstract.
A "contradiction" only arises **in context**. You must have some axioms, some rules of inference, some background logic, and some definitions.
Just asking if something or other "leads to a contradiction" without providing full context (or having that context understood) is nonsense. The fact that you not only did it once but now insist on repeating it, well, you can guess what that implies...
> > Two functions are considered to be the same > > > function if **and only if** they have the same domain (the same set > > > of "valid inputs") and the same value at each element of their > > > domain. > > > > > > Strictly speaking, then, in order to discuss a function, we must > > > agree on two things: (i) what is the domain of the function; and (ii) > > > what is the rule that assigns to each element of the domain a value. > > > That means that each and every time we mention a function, we must > > > say what the domain is. > > > > In this case, the domain is all real numbers were f(t) is defined.
That was the point. But you are trying to extend this from "where it is defined" to "wherever I can define it irrespective of the formula given".
I say right after:
> > Because this becomes both onerous and complicated, there is a > > > standard convention that is, I am positive, mentioned in your book. > > > This convention is: > > > > > > If a function is described by giving a formula, and no domain is > > > explicitly specified, then it is agreed that the domain of the > > > function is the natural domain: that is, the domain is the set of all > > > numbers for which the expression, *as given*, makes sense. > > > > > > Now, the function > > > > > > f(t) = (t^29)/(t3) > > > > > > with no domain specified, is therefore assumed to have as domain the > > > real numbers, **and only the real numbers** for which the expression > > > *as given* makes sense. And this collection is exactly the real > > > numbers different from 3. > > > > If I say that f(t) = (t^29)/(t3) is continuous for all reals, then you
You can say whatever you want. If you say that, however, you will be saying something false.
> can protest that the expression (t^29)/(t3) is meaningless when t=3.
I don't have to protest anything. I merely have to point at the definitions and the conventions at work to establish that your statement is false.
> The assumption that f(t) is continuous,
And again you run into trouble, because you insist on being careless.
A real valued function of real variable f(t) is "continuous at x=a" if and only if:
(i) It is defined at a; (ii) The limit of f(t) as t approaches a exists; and (iii) the value of the function at a equals the value of the limit of f(t) as t approaches a.
A real valued function of real variable f(t) is "continuous at all real numbers" or "continuous everywhere" if and only if it is continuous at x=a for all real numbers a.
A real valued function of real variable f(t) is "continuous" if and only if it is continuous at all points in its domain.
So the function f(t) = (t^29)/(t3) is continuous; it is not, however, continuous at all real numbers.
(There are slight variations if you want to consider a function to be continuous at isolated points or at endpoints of an interval, but I won't go into them since you are having enough trouble as it is).
> and the observation that > > (t^29)/(t3) is meaningful for all t!=3 dictates that f(3)=6.
Nope. Because
(i) "f(t) is continuous" does not mean the same thing as "f(t) is continuous everywhere"; and
(ii) Because by asserting that f(t) is continuous you are ASSUMING that f is defined at 3. However, whether or not f(t) is defined at 3 is *precisely* the crux of the matter here. So if you want to invoke an assumption of continuity to justify that f is defined at 3, then your argument is circular.
What do your "first principles" say about deriving something via a circular argument?
Now, a completely different question would be: "Is there a function g that is defined everywhere, is continuous everywhere, and agrees with f where f is defined?"
In **that** case the answer is "yes", and one can indeed show that such a function g would have value 6 at t=3. However, such a function g is not the function f you have in front of you.
> > On the other hand, the function g(t) = t+3 with no domain specified > > > is assumed to have as domain the real numbers, **and only the real > > > numbers** for which the expression, *as given* makes sense. And this > > > collection is exactly the set of all real numbers. > > > > > > That means that the function f(t) and g(t) have different domains, > > > and therefore are different functions. > > > > > > This is simply a matter of definitions and conventions. Those > > > definitions and conventions exist because they are *important* and > > > *necessary*. The fact that you don't see this yet is not a slight on > > > you: it is the consequence of centuries of work by many > > > mathematicians who have actually thought about this issue. You have > > > the benefit of being the inheritor of these many years of thought, > > > given to you distilled as precise, clear definitions that you are > > > expected to abide by. > > > > History is not reason. Mathematics is founded on axioms and constructed > > of definitions conjoined with logic.
History teaches mathematicians what axioms and what definitions to use. Refusing to accept them because you want to reject history makes you a willful ignoramus, not the searcher for knowledge that you try to tell yourself you are being.
> > The "contradiction" in your assertion arises simply because you > > > assert that f(t) is equal to g(t), when it can be demonstrated that > > > they are not equal: as functions, they have different domains and > > > therefore are different, not equal. The fact that you refuse to apply > > > the definition and say "I don't answer to mortals" does not excuse > > > this fact, nor does your claim that your assertions follows from > > > "reason". The plain definitions contradict your assertion, and that's > > > the end of it. > > > > No. The reasons I was given for not accepting Thomas's development were > > not presented as sound and valid. They were predominately appeals to > > authority.
Yes. The reasons you were given were the definitions. You did not realize you were using your own definitions and thus fell into error. Then you decided to go to unreasonable lengths to justify yourself (only in your own mind) that you were right all along, and then justified *this* by lofty appeals to "not answer[ing] to mortals" and "only answering to reason" and "rejecting authority."
Such proclamations make you sound like a pompous idiot. Rejecting definitions because you don't like them or, worse, because they show that you made a mistake to being with, makes you a willful ignoramus.
And there is nothing worse than willful ignorance.
 Arturo Magidin

