
Re: Is (t^29)/(t3) defined at t=3?
Posted:
Oct 6, 2013 4:29 PM


On Sunday, October 6, 2013 10:38:00 AM UTC5, Hetware wrote: > On 9/30/2013 11:53 PM, Arturo Magidin wrote: > > > On Monday, September 30, 2013 7:20:26 PM UTC5, Hetware wrote: > > > > > >> A function is a mapping from elements of a range to elements of a > > >> > > >> domain. Often the image is required to be single valued, but > > >> Thomas > > >> > > >> does not stipulate that requirement. > > > > > > This "definition" does not agree with your prior assertion that a > > > function is a set of ordered pairs; nor does your aside about the > > > possibility of images being more than "single valued". > > > > My description of a function as a set of ordered pairs was not intended > > to be a precise definition,
Well, could've fooled me, since that is in fact one of the precise ways in which one can define a function (see, e.g., Bourbaki, or Halmos's "Naive Set Theory", which is really about axiomatic set theory).
> though any precise definition I would use > > would be consistent with that description.
Which *which* description? You've given at least two or three descriptions of what you believe to be "a function". Which is the one that you think actually *is* a **precise** definition of a function?
Unless and until you provide one, you can't really complain about arguments that are not precise, since you are not being at all precise yourself.
> A function is a rule for evaluating an argument.
Is that a "precise" definition? What is a rule? What constitutes an argument? What does "evaluate" mean?
> The argument is a > > value in the domain, the result is a value in the range.
That's putting the cart before the horse. The domain is usually defined to be the set of all inputs, while the range is the set of all outputs. You can't then turn around and define the outputs as being elements of the range and the inputs as being elements of the domain, because that yields a circular definition.
> That is, for > > each value in the domain (assuming functions are singlevalued) there is > > one value in the range. The pair of values constitutes and ordered > > pair. That set of ordered pairs is also said to be a mapping.
Look; do you want to have a precise definition of function? Fine, let's give one.
Given sets A and B, the set A x B is the set of all ordered pairs (a,b) with a in A and b in B; the ordered pairs can be defined precisely within set theory (e.g., the Kuratowski definition, (a,b) = {{a},{a,b}}) so that they have the property that (a,b) = (c,d) if and only if a=c and b=d.
Now, a function f is a collection of ordered pairs such that
(i) if (a,b) and (a,b') are in f, then b=b'.
Given sets A and B, a function f is a "function from A to B" if and only if (i) f is a subset of A x B; (ii) For every a in A there exists b in B such that (a,b) in f; (iii) For all a in A, b and b' in B, if (a,b) and (a,b') are both in f, then b=b'.
If f is a function from A to B, then A is called "the domain of f", and B the "codomain of f". The "range of f" is defined to be the set of all b in B for which there exists a in A with (a,b) in f.
We write f(a)=b to mean "(a,b) in f."
> > In any case, I'll wager that Thomas *does* specify that functions > > > must yield a unique output given any particular input (or, if he > > > defines functions as you did before, as sets of ordered pairs > > > satisfying special conditions, then one of the conditions required of > > > functions defined this way is that if (a,b) and (a,b') are in f, then > > > b=b'; i.e., that they be single valued). > > > > Surprisingly Thomas distinguishes singlevalued and multiplevalued > > functions. For example, y^2 = x is said to be a doublevalued function > > of x. I would not call a one to many mapping a function. > > > > >> If I can deterministically > > >> > > >> interpret a formal expression as such a mapping, then my > > >> interpretation > > >> > > >> of that formal expression satisfies the definition of a function. > > > > > > But your *interpretation* of the expression is not necessarily the > > > *intended* meaning of the expression. > > > > > > As I said before, if functions are set of ordered pairs (or if > > > functions are rules assigning to every valid input a corresponding > > > output), then an expression is not a function. The expression in > > > question is supposed to determine a function *in a particular, agreed > > > upon manner*. > > > > So we are back to "that's the way is 'cause I said so".
No, we are back to the standard meanings and conventions. These are not so "because I said so", but because it's been generally decided to be a good idea everyone agrees on. These are conventions, to be sure, but conventions born out of experience and usefulness, not mere petty decisions made to make your life harder.
For example, it is **only a convention** that in a standard algebraic expression with no parentheses, exponentiation is to occur first, then multiplication, then addition. It's so "just 'cause I said so". We could just as well say that addition goes first, then exponentiation, then multiplication. There is no problem with saying that.
However, we pick the order "exponentiation first, then multiplication, then addition" because that means that we can write polynomial expressions without parentheses and have the intended meaning; if we decided to go "addition first, then multiplication, then exponentiation", then to write the function that squares the argument, then multiplies by 2, then adds 1 (what we would normally write simply as 2x^2+1) we would have to write as (2(x^2))+1.
Likewise with conventions about functions.
Technically, two functions are equal if and only if they have the same domain, and they take the same value at each point on the domain. That means that in order to actually specify a function and talk about it, I need to tell you two things: both the domain, and the rule that associates to each element its output (equivalently, describe precisely the set of ordered pairs that make up the function; equivalently, tell you the domain A, and tell you how to determine whether a given pair (a,b) is in the function).
But if every time we need to talk about a function we need to specify the domain, then this becomes both cumbersome and tedious. So we establish **conventions**. These conventions are established in such a way that the *most common* situation is always assumed, and then only when we deviate from this most common situation do extra specifications become necessary. Just like the precedence of operations is done the way it's done so that the most common expressions can be written without parentheses, though that means that other expressions do need parentheses.
The most common situation in real analysis and calculus is that we want to consider a function at all points where the expression is defined, and nowhere else. So we establish the convention called "natural domain of a function". The natural domain of a function given by a formula is the set of all real numbers for which the formula, **as given** makes sense; no fewer numbers, and no more numbers than that.
Under that standard convention, we are not allowed to guess what the function "may" or "may not" do at points in which the expression is not defined. We are not allowed to say the value equals the limit when it exists, we are forced to say the function is not defined at those points.
Under that **convention** (and yes, it's only a convention, but is a useful, tested, wellestablished convention arrived at after decades of experience), the function given by the expression f(t) = (t^29)/(t3) is understood, in the absence of explicit information to the contrary, to be:
(i) A real valued function of real variable (domain is a subset of R, codomain is R); (ii) Whose domain consists exactly and precisely of those values of t for which the expression (t^29)/(t3) makes sense; i.e., all t different from 3; (iii) The set of ordered pairs (t,s) where t is a real number different form 3 (see point (ii)), and where s is equal to (t^29)/(t3) (which an be evaluated since t is not 3).
The fact that it is possible to extend this function to one which agrees with f at all values of t=/=3, and takes a value at t=3, and is continuous, is irrelevant as to what f *is*.
> I can live with > > that, but the implication is that the demonstration follows from > > something more fundamental.
Given how much you seem to be reading into the material ex nihilo, I would challenge the claim that there is that implication. It follows from the *standard convention that exists to interpret expressions as defining real valued functions of real variable*.
> > > While you are free to set up your own particular and personal > > > conventions, you are *not* free to impose those on the book you are > > > reading, which no doubt established its own intended meaning before > > > and that you are ignoring. > > > > > > > I guess that's part of the problem.
It's actually most of the problem.
> I now understand what I was missing. If I assume f(t) = (t^29)/(t3) > > is continuous at t=3, then it is defined there.
Yes, but you have *no warrant whatsoever* to assume that it is defined there, much less to assume it must be continuous there. In fact, you have very good reason to *understand* the expression as *telling you*, implicitly, that f is **not** defined at t=3. You are ignoring this and proceeding from that, so no wonder you get strange conclusions.
 Arturo Magidin

