smooth "splice" fxn between two lines
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From: kj on 14 May 2010 16:41 I've seen the function I'm looking for described before, but I can't remember where, nor the details of the definition. This function, f:R->R, is such that f(0) = 0, f(2) = 1, and the function g defined as g(x) = x if x < 0 g(x) = f(x) if 0 <= x <= 2 g(x) = 1 if x > 2 is smooth everywhere. Of course, the problem boils down to achieving smoothness at the "splice" points (0, 0) and (2, 1). Does anyone remember a function f with these properties? Thx! ~K
From: Robert Israel on 14 May 2010 17:19 kj <no.email(a)please.post> writes: > I've seen the function I'm looking for described before, but I > can't remember where, nor the details of the definition. This > function, f:R->R, is such that f(0) = 0, f(2) = 1, and the function > g defined as > > g(x) = x if x < 0 > g(x) = f(x) if 0 <= x <= 2 > g(x) = 1 if x > 2 > > is smooth everywhere. Of course, the problem boils down to achieving > smoothness at the "splice" points (0, 0) and (2, 1). > > Does anyone remember a function f with these properties? Try f(x) = (exp(-1/x) + x exp(-1/(2-x)))/(exp(-1/x) + exp(-1/(2-x))) for 0 < x < 2 -- 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 14 May 2010 19:57 On May 14, 2:19 pm, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote: > kj <no.em...(a)please.post> writes: >> I've seen the function I'm looking for described before, but I >> can't remember where, nor the details of the definition. This >> function, f:R->R, is such that f(0) = 0, f(2) = 1, and the >> function g defined as >> >> g(x) = x if x < 0 >> g(x) = f(x) if 0 <= x <= 2 >> g(x) = 1 if x > 2 >> >> is smooth everywhere. Of course, the problem boils down to >> achieving smoothness at the "splice" points (0, 0) and (2, 1). >> >> Does anyone remember a function f with these properties? > > Try > > f(x) = (exp(-1/x) + x exp(-1/(2-x)))/(exp(-1/x) + exp(-1/(2-x))) > for 0 < x < 2 f(x) = (exp(-1/x) - x exp(-1/(2-x)))/(exp(-1/x) - exp(-1/(2-x))) also works.
From: Ray Koopman on 15 May 2010 03:37 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.
From: kj on 15 May 2010 13:48
Thank you both! ~K |