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From: Robert Israel on 16 May 2010 13:52
Ray Koopman <koopman(a)sfu.ca> writes:

> On May 14, 4:57 pm, Ray Koopman <koop...(a)sfu.ca> wrote:
> > f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x)))
> >
> > also works.
>
> f(x) = 1 - exp(2x/(x-2)) is simpler, and the plot is nicer.

But it doesn't join smoothly to x at x=0: f'''(0) = -1/2.
Well, maybe kj would be satisfied with a C^2 join, but I interpreted
"smooth" as C^infty.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Ray Koopman on 16 May 2010 16:11
On May 16, 10:52 am, Robert Israel
<isr...(a)math.MyUniversitysInitials.ca> wrote:
> Ray Koopman <koop...(a)sfu.ca> writes:
>> On May 14, 4:57 pm, Ray Koopman <koop...(a)sfu.ca> wrote:
>>> f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x)))
>>>
>>> also works.
>>
>> f(x) = 1 - exp(2x/(x-2))  is simpler, and the plot is nicer.
>
> But it doesn't join smoothly to x at x=0: f'''(0) = -1/2.
> Well, maybe kj would be satisfied with a C^2 join, but I interpreted
> "smooth" as C^infty.

Yes, I wondered about that.
From: Axel Vogt on 16 May 2010 16:29
Robert Israel wrote:
> Ray Koopman <koopman(a)sfu.ca> writes:
>
>> On May 14, 4:57 pm, Ray Koopman <koop...(a)sfu.ca> wrote:
>>> f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x)))
>>>
>>> also works.
>> f(x) = 1 - exp(2x/(x-2)) is simpler, and the plot is nicer.
>
> But it doesn't join smoothly to x at x=0: f'''(0) = -1/2.
> Well, maybe kj would be satisfied with a C^2 join, but I interpreted
> "smooth" as C^infty.

Is that something similar like "partition of 1" in Analysis?
From: kj on 17 May 2010 16:10
In <rbisrael.20100516174609$2241(a)news.acm.uiuc.edu> Robert Israel <israel(a)math.MyUniversitysInitials.ca> writes:

>Ray Koopman <koopman(a)sfu.ca> writes:

>> On May 14, 4:57 pm, Ray Koopman <koop...(a)sfu.ca> wrote:
>> > f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x)))
>> >
>> > also works.
>>
>> f(x) = 1 - exp(2x/(x-2)) is simpler, and the plot is nicer.

>But it doesn't join smoothly to x at x=0: f'''(0) = -1/2.
>Well, maybe kj would be satisfied with a C^2 join, but I interpreted
>"smooth" as C^infty.

Actually, I did have C^infty in mind.

I also had in mind, but neglected to specify, that the second
derivative of the resulting curve would be <= 0.

I thought that there was a pretty standard recipe for constructing
a "connector" having these properties, but maybe my memory is
tricking me.
From: Ray Koopman on 17 May 2010 18:42
On May 17, 1:10 pm, kj <no.em...(a)please.post> wrote:
> In <rbisrael.20100516174609$2...(a)news.acm.uiuc.edu> Robert Israel
> <isr...(a)math.MyUniversitysInitials.ca> writes:
>> Ray Koopman <koop...(a)sfu.ca> writes:
>>> On May 14, 4:57 pm, Ray Koopman <koop...(a)sfu.ca> wrote:
>>>
>>>> f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x)))
>>>> also works.
>>>
>>> f(x) = 1 - exp(2x/(x-2)) is simpler, and the plot is nicer.
>>
>> But it doesn't join smoothly to x at x=0: f'''(0) = -1/2.
>> Well, maybe kj would be satisfied with a C^2 join,
>> but I interpreted "smooth" as C^infty.
>
> Actually, I did have C^infty in mind.
>
> I also had in mind, but neglected to specify, that the second
> derivative of the resulting curve would be <= 0.

If we call Robert's function f1 and my first function f2 then f(x) =
(r*f1(x) + f2(x))/(r + 1) will have C^infty splices at both 0 & 2.
Taking r = .8812 will approximately maximize the minimum f",
subject to f" <= 0.
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