Re: Analysis Differentiation question

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/12/04 

Date: Mon, 12 Jul 2004 17:48:11 GMT

On Mon, 12 Jul 2004 13:16:20 -0400, "mr0x" <mr00xx@nospam.hotmail.com>
wrote:

>
>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>news:hqr4f0dumhmqmca391qfe1i9o65t95paf8@4ax.com...
>> On Sun, 11 Jul 2004 17:26:07 -0400, "mr0x" <mr00xx@nospam.hotmail.com>
>> wrote:
>>
>> >Let f be differentiable on [a,b] (some closed interval).
>> >
>> >How would I go about constructing a function so that the range of f is
>> >
>> >a). an open interval
>> >b). open on one side and closed on the other side interval.
>>
>> This is clearly impossible, since f is continuous, so its
>> range is compact (and connected, hence a closed interval).
>>
>> William says something about a "correction", but your
>> two posts look identical to me...
>>
>
>Sorry I did mean the range of f' not f.
>
>Of course, it wouldn't make any sense for f to map some closed interval to
>an open interval since it's continuous.
>
>My idea was to somehow make f' discontinuous at the end-points. But, the
>Darboux property says I can't start off with f' with a removable or jump
>discontinuity. However, the only discontinuity of the derivative I've learnt
>about is the form of sin(1/x) kind.
>
>I was having a hard time trying to construct a function where f' =
>sin(1/(x-a)) * sin (1/(b-x)) and thought maybe it was a more simpler problem
>and I wasn't seeing the answer.

Well, I don't see how f' = sin(1/(x-a)) * sin (1/(b-x)) is going to
do it, and you should note that Wade has already given a big hint.

But for future reference, you seem to be missing something.
If some time you _do_ want to construct a function f with
f' = sin(1/(x-a)) * sin (1/(b-x)), you do so like so:

Let f(x) = int_a^x sin(1/(t-a)) * sin (1/(b-t)) dt.

("int_a^x" means "integral from a to x".)

>Thanks.
>
>

David C. Ullrich