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Topic: Some derivative/continuity questions
Replies: 15   Last Post: Aug 5, 2013 9:47 AM

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 David Bernier Posts: 3,892 Registered: 12/13/04
Re: Some derivative/continuity questions
Posted: Aug 4, 2013 1:58 PM

On 08/04/2013 11:22 AM, G Patel wrote:
> On Saturday, August 3, 2013 6:51:47 PM UTC-4, phol...@gmail.com wrote:
>> On Saturday, August 3, 2013 1:37:09 PM UTC-7, G Patel wrote:
>>

>>> 1) If f is differentiable at x, is it continuous on an interval about x?
>>
>>>
>>
>>> It seems intuitively like it should be.
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>>>
>>
>>>
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>>>
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>>> 2) Do you know a function that is differentiable at an isolated point?
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>>>
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>>>
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>>>
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>>> Thanks
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>>
>>
>> Well, what is normally taught in Calc I is the relationship between a limit, continuity and differentiability.
>>
>> A function is differentiable at a point if it is .....
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>> 1/ Defined at that point in the domain of the function.
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>> 2/ Continuous at that point (both sides, left and right). This precludes the end points of a line segment.
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>> 3/ The left determined derivative equals the right derivative (using the limit definition of a derivative). This precludes a corner point or cusp.
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>>
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>> So if you apply those requirements to an isolated point.......?
>>
>> Phil H

>
> Ooops. I now realize my problem in stating my question.
> I meant, the function is differentiable at a point, but not anywhere near it.
> That is, it causes f' to be defined at an isolated point.
>
> Does such a function exit?
>

Suppose we want the function to be differentiable at 0, but not near 0.
Then, we can fix the interval K of definition to be [a, b],
with a = -1/2 and b = 1/2.

Suppose W: R-> R is a Weierstrass-type continuous, nowhere
differentiable function.

Then, the proposed function will be
f(x) := h(x) W(x), for x in [a, b],
with h(0) = 0 and h(x)> 0 if a <= x < 0 or 0 < x <= 1/2 .

Further, h should be smooth (infinitely differentiable) on [a, b].

There are things called "bump functions", cf.
http://en.wikipedia.org/wiki/Bump_function

The Examples section there gives a standard 1-D bump function,
Psi: R -> R.

Psi(-1) = Psi(1) = 0, and Psi is supported on [-1, 1].

Psi is positive in [-1, 1].

Suppose
h(x) := Psi(1-x^2), for -1/2 <= x <= 1/2 .

Then h(.) is very smooth and very small near x = 0, but is still
strictly positive if x =/= 0 and -1/2 <= x <= 1/2.

So the proposed function is
f: [-1/2, 1/2] -> R,

f(x) := Psi(1-x^2) W(x) .

I expect that f will be differentiable at x=0.

If x ~= 0, but x =/= 0, x in [-1/2, 1/2], then
Psi(1-x^2) > 0 and it is "well-behaved" near x. So
possibly in this case with x in [-1/2, 1/2] and x =/= 0 we have that
f(x) = Psi(1-x^2) W(x) is non-differentiable at x.

david

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