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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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Peter Percival

Posts: 1,215
Registered: 10/25/10
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 13, 2013 9:41 AM
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Hetware wrote:
> On 10/13/2013 1:18 AM, Arturo Magidin wrote:

>>
>> In particular, functions are not "declared" to be continuous, they
>> are not "assumed" to be continuous, and they are not "asserted" to be
>> continuous.

>
> That's nonsense. In variational calculus we assert the existence of an
> infinite number of continuous, twice differentiable functions which
> coincide at the boundaries of the parameter domain, but are otherwise
> arbitrary.
>
> A statement such as "let M be a smooth manifold", asserts the existence
> of a continuous function.


That is quite different. You have a function f defined by a formula,
and more than once you have asserted that it is continuous. If f's
continuity matters to you, you _prove_ its continuity. f is indeed
continuous at all points in its domain, but it isn't continuous at 3;
it's not even defined there.

--
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9/28/13
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Hetware
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9/29/13
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Richard Tobin
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10/16/13
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10/19/13
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10/20/13
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Peter Percival
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10/19/13
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Peter Percival
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Virgil
10/18/13
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Hetware
10/19/13
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Peter Percival
10/19/13
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Peter Percival
10/19/13
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Hetware
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Peter Percival
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Hetware
10/19/13
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fom
10/19/13
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magidin@math.berkeley.edu
10/19/13
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Hetware
10/19/13
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10/20/13
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quasi
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10/20/13
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10/20/13
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