Real Analysis Counter-Example?
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From: junoexpress on 27 Jan 2010 18:37 Hi, I was evaluating the value of the second derivative of the Dirichlet kernel at zero, and this problem got me wondering about the following question: Let f and g be two smooth real-valued functions whose domain is some interval in R, [a,b] and let us define h = f/g. Suppose c is some point in (a,b) such that the limit of h exists, but the limits of f and g both go to zero as x->c. Is it always true that L'Hospital's rule will yield the true limit of h (given that we work out enough derivatives of f and g)? Intuitively, I want to say that this must be the case probably due to some type of argument using a Taylor series expansion, but it is not obvious to me exactly how to argue this precisely. OTOH, if it is not true, is there a counter-example which shows this? Thanks, Matt
From: Robert Israel on 27 Jan 2010 19:06 junoexpress <mtbrenneman(a)gmail.com> writes: > Hi, > > I was evaluating the value of the second derivative of the Dirichlet > kernel at zero, and this problem got me wondering about the following > question: > > Let f and g be two smooth real-valued functions whose domain is some > interval in R, [a,b] and let us define h = f/g. > > Suppose c is some point in (a,b) such that the limit of h exists, but > the limits of f and g both go to zero as x->c. > > Is it always true that L'Hospital's rule will yield the true limit of > h (given that we work out enough derivatives of f and g)? No. For example, f(x) = exp(-1/x^2) for x <> 0, f(0) = 0. Note that f is smooth, and all derivatives of f are 0 at 0. Let g be another such function, say f(x) + f(x^2). > Intuitively, I want to say that this must be the case probably due to > some type of argument using a Taylor series expansion, but it is not > obvious to me exactly how to argue this precisely. OTOH, if it is not > true, is there a counter-example which shows this? It would be true if f and g were analytic at 0. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Omega Cubed on 27 Jan 2010 22:59 On 2010-01-28, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: > junoexpress <mtbrenneman(a)gmail.com> writes: >> Let f and g be two smooth real-valued functions whose domain is some >> interval in R, [a,b] and let us define h = f/g. >> >> Suppose c is some point in (a,b) such that the limit of h exists, but >> the limits of f and g both go to zero as x->c. >> >> Is it always true that L'Hospital's rule will yield the true limit of >> h (given that we work out enough derivatives of f and g)? > > No. For example, f(x) = exp(-1/x^2) for x <> 0, f(0) = 0. Note that > f is smooth, and all derivatives of f are 0 at 0. > Let g be another such function, say f(x) + f(x^2). > >> Intuitively, I want to say that this must be the case probably due to >> some type of argument using a Taylor series expansion, but it is not >> obvious to me exactly how to argue this precisely. OTOH, if it is not >> true, is there a counter-example which shows this? > > It would be true if f and g were analytic at 0. Correct me if I am wrong, but isn't analyticity a bit strong here? (Sufficient but not necessary.) Take two polynomial perturbations of your f(x) above: g(x) = f(x) + p(x), h(x) = f(x) + q(x), with p(x) = q(x) = 0. Then L'Hospital's rule will still recover the correct limit for g/h. L'Hospital's rule for C^\infty functions is just a re-expression of the product rule for (finitely) differentiable function, no? If f, g are smooth and h = f/g is C^k, then D^i (hg) = D^i f where D is the derivaive operator, as long as i <= k. Now take j to be the largeset positive integer such that D^jg = D^jf = 0 (which we assume to be < k), then the only non-vanishing terms by the product rule for i = j+1 is h D^i g = D^i f So conditioning on that the limit of h exists (which is how I read the original query) and that h is sufficiently differentiable, the only counterexample I can see is precisely when the case that f and g are non-analytic at a point with derivatives vanishing to all order. Cheers, W
From: W^3 on 5 Feb 2010 13:41
In article <rbisrael.20100127235719$7602(a)news.acm.uiuc.edu>, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: > junoexpress <mtbrenneman(a)gmail.com> writes: > > > Hi, > > > > I was evaluating the value of the second derivative of the Dirichlet > > kernel at zero, and this problem got me wondering about the following > > question: > > > > Let f and g be two smooth real-valued functions whose domain is some > > interval in R, [a,b] and let us define h = f/g. > > > > Suppose c is some point in (a,b) such that the limit of h exists, but > > the limits of f and g both go to zero as x->c. > > > > Is it always true that L'Hospital's rule will yield the true limit of > > h (given that we work out enough derivatives of f and g)? > > No. For example, f(x) = exp(-1/x^2) for x <> 0, f(0) = 0. Note that > f is smooth, and all derivatives of f are 0 at 0. > Let g be another such function, say f(x) + f(x^2). There are two questions here: The one actually asked, and the one meant. The one asked is: If f/g -> L, does LHR always yield L for the answer? The one meant is: If f/g -> L, will L = D^nf(c)/D^ng(c) for an appropriate n? Your example shows the answer to the second question is no, but doesn't address the first question (because f'/g' -> 1 in your example, which is the limit of f/g.) Here's an example to show the answer to the first question is no: Let f(x) = exp(-1/x^2)*(1 + xsin(1/x^3)), g(x) = exp(-1/x^2). Then f/g -> 1, but f'/g' fails to have a limit. > > Intuitively, I want to say that this must be the case probably due to > > some type of argument using a Taylor series expansion, but it is not > > obvious to me exactly how to argue this precisely. OTOH, if it is not > > true, is there a counter-example which shows this? > > It would be true if f and g were analytic at 0. |