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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 6, 2013 11:38 AM
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On 9/30/2013 11:53 PM, Arturo Magidin wrote:
> On Monday, September 30, 2013 7:20:26 PM UTC-5, 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, though any precise definition I would use
would be consistent with that description.

A function is a rule for evaluating an argument. The argument is a
value in the domain, the result is a value in the range. That is, for
each value in the domain (assuming functions are single-valued) 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.

> 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 single-valued and multiple-valued
functions. For example, y^2 = x is said to be a double-valued 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". I can live with
that, but the implication is that the demonstration follows from
something more fundamental.

> 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 takes a lot of work to build up
rigorous definitions of even basic mathematical symbols and operations.
Very few texts at the level of Thomas's Calculus go into that level of
detail.

I now understand what I was missing. If I assume f(t) = (t^2-9)/(t-3)
is continuous at t=3, then it is defined there. In just about
everything I do, functions are assumed to be continuous unless there is
some obvious coordinate, or manifold singularity. It took me a while to
get the assumption of continuity out of my head.


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9/28/13
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Hetware
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