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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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Roland Franzius

Posts: 444
Registered: 12/7/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Sep 30, 2013 3:01 AM
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Am 30.09.2013 00:47, schrieb Hetware:
> On 9/29/2013 8:45 AM, Michael F. Stemper wrote:
>> On 09/28/2013 04:20 PM, Hetware wrote:
>>> On 9/28/2013 4:24 PM, Richard Tobin wrote:
>>>> In article <9cSdnYGhH7CMpNrPnZ2dnUVZ_sCdnZ2d@megapath.net>,
>>>> Hetware <hattons@speakyeasy.net> wrote:

>>
>>>>> So the answer is consensus among mathematicians holds that F(t) =
>>>>> (t^2 -
>>>>> 9)/(t - 3) is undefined at t=3?

>>>>
>>>> Yes.
>>>>

>>>>> Perhaps what I should have said at the
>>>>> outset is something along the lines of: on any given day, if I'm
>>>>> setting
>>>>> up an equation in physics, and produce an expression such as F(t) =
>>>>> (t^2
>>>>> - 9)/(t - 3), I treat it as t+3, and do not expect any adverse
>>>>> consequence from doing so.

>>>>
>>>> Your simplification is not valid for t=3.

>>
>>> I believe most mathematicians solving for x as a function of t given
>>>
>>> t^2 - 9 = x (t - 3)
>>>
>>> would not hesitate to factor the left hand side and divide both sides by
>>> t - 3 without treating t = 3 as a special case.

>>
>> This is high-school stuff. Literally.
>>
>> I just pulled my text[1] from Algebra II (tenth grade) off the shelf.
>>
>> Turning to Section 3.6, "Rational Expressions: Reduction to Simplest
>> Form", I see:
>>
>> It should be observed that since division by zero is excluded
>> as an operation, a fraction whose denominator is zero has no
>> meaning. For example, the fraction 9/(y-7) has no meaning
>> when y=7. Also, the fraction (x+5)/(x^2-9) has no meaning
>> when x=3 or x=-3.
>>
>> We sometimes encounter fractions such as (x-3)/(3-x)
>> (x!=3). [...]
>>
>> The authors then proceed to work through several examples of fractions
>> with polynomials in the numerator and denominator. In every case, when
>> the fraction is specified, they include:
>>
>> where (d(x)!=0)
>>
>> (I'm using d(x) here to represent whatever the denominator polynomial
>> happens to be.)
>>
>> If your high school math classes neglected to point out that this
>> exclusion is necessary, they were faulty. However, the errors and
>> omissions of your past teachers do not change the fact that it is.
>>
>>
>> [1] _Algebra and Trigonometry: A Modern Approach_, Peters & Schaaf,
>> van Nostrand, (c) 1965

>
> What I am saying is that if I encountered an expression such as
> (t^2-9)/(t-3) 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.
>
> If I wanted to express x as a function of t given (t^2-9) = x(t-3), I
> would have no compunction about "dividing out" t-3. Since I can arrive
> at the same result through a different method, I do not see why treating
> (t-3)/(t-3) = 1 in this context is an error.
>
> When I do something like
>
> (t^2-9) = x(t-3) <=> (t^2-9)/(t-3) = x(t-3)/(t-3) <=> (t^2-9)/(t-3) = x
>
> I don't feel a twinge of unease. I have no experience of it leading to
> an incorrect final result. If I encounter something of the form
> f(t)/(3-t) and the problem domain includes t = 3, I anticipate a problem
> at t = 3. If I can fiddle with f(t) and factor out 3-t, I simply cancel
> them out, and keep going.



In the context of meromorphic complex functions, functions that have as
singularities only multiple zeros and poles, its always enough to
multiply, divide and concatenate functions and to determine the
singularities of the outcome. Here automatic cancellation of common
zeroes or poles in fractions are done by the implied process of so
called "removing removable singularities.

The situation changes dramatically if inverse functions of powers,
rational functions or of transcendental function are included in the
function algebra.

This problem is not tractable in the trival setting of a complex plane
because each root function introduces cuts of discontinuity between
zeros and poles. The cut lines are artificially and deliberately
choosen by the user. The general solution of the problem is the
representation of complex functions by their Riemann surfaces, a theme
not taught in undergraduate courses generally .

But there is a quite different setting originating from Fourier
transforms and normed function spaces. Here on considers very general
classes of functions that are considered to differ only if their
absolute difference integrated yields zero.

For these setting in functional anylysis in any class of equivalent
functions with respect to their integrated difference there exists a
representative that has a maximally extended domain of continuity, all
holes closed.

So eg in Hilbert spaces, used in most areas of physics and engineering,
a double Fourier transform smoothes out all point holes in the direct
definition. That happens automatically for all discontinuities with
equal limits from all complex directions.

For normed function spaces, the principle in general is to take the
double dual.

The dual space is the space of all linear functionals with respect to
the given norm. The double dual then is a subspace of the original space
with all artifical point discontiniuties removed, that do not contribute
to the norm integral which is choosen to be the Lebesgue integral for
that purpose.

--

Roland Franzius


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9/28/13
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