Taylor's Formula
in [Math]
|
From: Maury Barbato on 26 Jul 2010 09:33 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)? Thank you very much for your attention. My Best Regards, Maury Barbato
From: W^3 on 26 Jul 2010 23:44 In article <1690076286.3865.1280165656552.JavaMail.root(a)gallium.mathforum.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.
From: Maury Barbato on 27 Jul 2010 07:03 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? Thank you very much for your great help. Best Regards, Maury Barbato
From: W^3 on 27 Jul 2010 23:33 In article <1944347450.9590.1280243069459.JavaMail.root(a)gallium.mathforum.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).
From: Maury Barbato on 28 Jul 2010 01:47
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. Thank you very very ... much for your invaluable help. Friendly Regards, Maury Barbato Deep into that darkness peering, long I stood there, wondering, fearing, doubting, dreaming dreams no mortal ever dared to dream before. (E.A. Poe) |