RE: [math-fun] Re: another calculus question
<< Jim asked about the integrability of 2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero. The function is Riemann integrable on [0,1]. Consider the intervals from 0 to \eps, and from \eps to 1. By uniform continuity, there is some \delta such that on the larger interval, any partitions of mesh \delta, on a subinterval of \eps to 1, with any associated Riemann sums, will yield values within \eps of one another. . . . . . .
Huh? Uniform continuuity of what? Because of the cos(1/x) term, f is not continuous on [0,1]. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Feb 8, 2008 2:38 PM, Dan Asimov <dasimov@earthlink.net> wrote:
<< Jim asked about the integrability of
2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero.
The function is Riemann integrable on [0,1]. Consider the intervals from 0 to \eps, and from \eps to 1. By uniform continuity, there is some \delta such that on the larger interval, any partitions of mesh \delta, on a subinterval of \eps to 1, with any associated Riemann sums, will yield values within \eps of one another. . . . . . .
Huh? Uniform continuuity of what? Because of the cos(1/x) term, f is not continuous on [0,1].
f is uniformly continuous on [\eps, 1], which I think is all that is used in the above argument. -- Andy.Latto@pobox.com
I certainly did mean uniform continuity only on the interval [\eps, 1]; that's why there are separate estimates elsewhere. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Andy Latto Sent: Friday, February 08, 2008 3:29 PM To: Dan Asimov; math-fun Subject: Re: [math-fun] Re: another calculus question On Feb 8, 2008 2:38 PM, Dan Asimov <dasimov@earthlink.net> wrote:
<< Jim asked about the integrability of
2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero.
The function is Riemann integrable on [0,1]. Consider the intervals from 0 to \eps, and from \eps to 1. By uniform continuity, there is some \delta such that on the larger interval, any partitions of mesh \delta, on a subinterval of \eps to 1, with any associated Riemann sums, will yield values within \eps of one another. . . . . . .
Huh? Uniform continuuity of what? Because of the cos(1/x) term, f is not continuous on [0,1].
f is uniformly continuous on [\eps, 1], which I think is all that is used in the above argument. -- Andy.Latto@pobox.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
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Andy Latto -
Dan Asimov -
Fred Kochman