Taylor's Formula
in [Math]
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From: W^3 on 28 Jul 2010 14:32 In article <1701515789.15370.1280310473387.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > The World Wide Wade wrot: > > > In article > > <1944347450.9590.1280243069459.JavaMail.root(a)gallium.m > > athforum.org>, > > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > > > The World Wide Wade wrote: > > > > > > > In article > > > > > > <1690076286.3865.1280165656552.JavaMail.root(a)gallium.m > > > > athforum.org>, > > > > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > > > > > > > Hello, > > > > > let R be the field of real numbers, E an open > > set > > > > > of R^(n+1), and f:E -> R a C^m function, with m > > >= > > > > 1. > > > > > Let us use the notation (t,x) for a point of > > > > R^(n+1), > > > > > with t in R and x in R^n. If (a,b) is a point > > of E, > > > > > we define in a neighborhood of (0,b) the > > function > > > > > (remainder of Taylor's formula with respect to > > t) > > > > > > > > > > r(h,x) = f(a+h,x) - > > > > > > > > > > sum_{k= 0 to m} [1/(k!)]*[(D_1)^k f](a,x) * > > h^k, > > > > > > > > > > where (D_1)^k f denotes the k-th partial > > derivative > > > > of > > > > > f with respect to t. In the same neighborhood, > > > > define > > > > > > > > > > g(h,x) = r(h,x)/(h^m) if h =/= 0, > > > > > > > > > > g(0,x) = 0. > > > > > > > > > > Is g continuous in (0,b)? > > > > > > > > I'll use D in place of D_1. By Taylor's theorem > > in > > > > one variable, for > > > > each (h,x) > > > > > > > > f(a+h,x) - sum{k=0,m} 1/k! * D^kf(a,x) * h^k > > > > > > > > = 1/m! * [D^mf(c,x) - D^mf(a,x)] * h^m, > > > > > > > > for some c = c(h,x) between a and a+h. The last > > term > > > > in brackets -> 0 > > > > as (h,x) -> (0,b) by the continuity of D^mf, so > > the > > > > answer is yes. > > > > > > Yes, ok. Do you think the same holds in the weaker > > > hypothesis that f is m times differentiable in E? > > > > No, the idea is to look at something like > > (x^2+y^2)sin(1/(x^2+y^2)). I > > couldn't quite get that to work, so I did the below > > (which is actually > > easier in a sense). > > > > Let h be any smooth bounded function on (0,oo) such > > that > > h(n)-h(n^2/(n+1)) = 1 for n = 1, 2, ... (Define > > f(0,0) = 0, f(x,y) = > > (x^2+y^2)h(1/(x^2+y^2)) otherwise. Then f is > > differentiable everywhere > > on R^2 (thinking m = 1 here), and D_yf(x,0) = 0 for > > all x. So f(x,y) - > > [f(x,0) + D_yf(x,0)y] = (x^2+y^2)h(1/(x^2+y^2)) - > > x^2h(1/(x^2). Now > > along the curve (sqrt(y),y), the last expression = > > (y+y^2)h(1/(y+y^2)) > > - yh(1/y). Divide by y, and you'll have an expression > > that does not > > tend to 0 as y -> 0; to see this, look along the > > values y = 1/n. This > > violates the continuity of your expression at (0,0). > > That's right. Your original example > > f(x,y) = (x^2+y^2)sin(1/(x^2+y^2)) if (x,y)=/=(0,0), > > f(0,0) = (0,0) > > works too. You have > > f(x,y) - f(x,0) - [D_yf(x,0)]y = > > = (x^2+y^2)sin(1/(x^2+y^2)) - (x^2)sin(1/(x^2)), > > so along the curve (sqrt(y),y) you obtain, dividing > by y, > > (I) (1+y)sin(1/(y+y^2)) - sin(1/y). > > Now set 1/y = 2n*pi. Since > > 1/(y+y^2) = 1/y - 1/(y+1), > > the limit of (I) as n tends to infinity is sin(-1)=/=0. Right! |