Comments on: Why Do Taylor Series Converge? (part 1) http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/ Playing on the Sea-Shore, Rough Pebbles Welcome Fri, 25 Dec 2009 08:01:26 +0000 http://wordpress.com/ hourly 1 By: Mark Eichenlaub http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-238 Mark Eichenlaub Fri, 17 Jul 2009 12:06:36 +0000 http://arcsecond.wordpress.com/?p=889#comment-238 Thanks, Kiwi. Good points. I will keep thinking about this. Thanks, Kiwi. Good points. I will keep thinking about this.

]]> By: kiwi http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-232 kiwi Sun, 12 Jul 2009 23:12:41 +0000 http://arcsecond.wordpress.com/?p=889#comment-232 Ahhh I cannot seem to get the function to appear - I'm not sure why it's getting dropped (I guess I need to escape the inequality symbol because html is interpreting it as an embedding command). The function is exp(-1/x) for x greater than zero, and zero otherwise. Ahhh I cannot seem to get the function to appear – I’m not sure why it’s getting dropped (I guess I need to escape the inequality symbol because html is interpreting it as an embedding command). The function is exp(-1/x) for x greater than zero, and zero otherwise.

]]> By: kiwi http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-231 kiwi Sun, 12 Jul 2009 23:09:19 +0000 http://arcsecond.wordpress.com/?p=889#comment-231 Some symbols did get dropped. The example, and it's not that pathological, is $latex f(x) = 0$ for $latex x 0$. This agrees with the zero function for $latex x <= 0$ but not otherwise. Also, the interval over which continuous functions agree is necessarily closed. Incidentally, your example a couple of posts back gives a function which has a Taylor series which converges to a different function - this is case (c) in my list. Some symbols did get dropped. The example, and it’s not that pathological, is f(x) = 0 for x 0. This agrees with the zero function for x <= 0 but not otherwise.

Also, the interval over which continuous functions agree is necessarily closed.

Incidentally, your example a couple of posts back gives a function which has a Taylor series which converges to a different function – this is case (c) in my list.

]]> By: Mark Eichenlaub http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-230 Mark Eichenlaub Sun, 12 Jul 2009 22:00:34 +0000 http://arcsecond.wordpress.com/?p=889#comment-230 Hi Kiwi, I'm not sure I understand your example completely because there is some sort of symbol missing, but I think I get your point. I assumed that my $latex c$ was not actually in the interval over which the functions agree, but that is necessarily correct. So the proof only works is the interval over which the functions agree is open. Gah, Thanks for pointing out that example. I started out talking about Taylor series in this post, but I got around to thinking about analytic functions. I think the pathological example does not actually agree with the function f(x) = 1 except at x=0. Hi Kiwi,

I’m not sure I understand your example completely because there is some sort of symbol missing, but I think I get your point. I assumed that my c was not actually in the interval over which the functions agree, but that is necessarily correct. So the proof only works is the interval over which the functions agree is open.

Gah,

Thanks for pointing out that example. I started out talking about Taylor series in this post, but I got around to thinking about analytic functions. I think the pathological example does not actually agree with the function f(x) = 1 except at x=0.

]]> By: kiwi http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-229 kiwi Sun, 12 Jul 2009 17:03:30 +0000 http://arcsecond.wordpress.com/?p=889#comment-229 The problem is ... smooth is not enough. Basically, a function f having derivatives to all orders at a point a has a Taylor series centered at a which may (a) converge to f everywhere; (b) converge to f in a finite interval centered at a and diverge outside that interval; (c) converge but to a completely different function in some interval centered at a; (d) diverge everywhere except at a. It is quite possible for two different smooth functions to agree on an interval. Example: the zero function and the function define as 0 for x 0. Your proof fails because h(c) = 0 and only for x > c will h become nonzero (c = 0 in the example here). Analytic functions (i.e. functions which equal their Taylor series) are very strange in that knowing them in a neighborhood of a point, no matter how small, fixes them everywhere else. I liken it to standing at a rail station and determining exactly where the tracks go from the limited portion you can see at the station. The problem is … smooth is not enough. Basically, a function f having derivatives to all orders at a point a has a Taylor series centered at a which may (a) converge to f everywhere; (b) converge to f in a finite interval centered at a and diverge outside that interval; (c) converge but to a completely different function in some interval centered at a; (d) diverge everywhere except at a.

It is quite possible for two different smooth functions to agree on an interval. Example: the zero function and the function define as 0 for x 0. Your proof fails because h(c) = 0 and only for x > c will h become nonzero (c = 0 in the example here).

Analytic functions (i.e. functions which equal their Taylor series) are very strange in that knowing them in a neighborhood of a point, no matter how small, fixes them everywhere else. I liken it to standing at a rail station and determining exactly where the tracks go from the limited portion you can see at the station.

]]> By: gah http://arcsecond.wordpress.com/2009/07/11/why-do-taylor-series-converge-part-1/#comment-228 gah Sun, 12 Jul 2009 03:43:12 +0000 http://arcsecond.wordpress.com/?p=889#comment-228 Functions that agree with their Taylor series are called analytic functions. Analytic functions are very like polynomials. Specify the roots of a polynomial and the polynomial is fixed (except for a scale factor). The moment you bend a polynomial a little bit you no longer have a polynomial. One difference between a polynomial and an analytic function is that an analytic function is that an analytic function can have a countable infinity of roots (think sin(x) ). The function Exp[-1/(x^2)] has all derivatives at zero but is not analytic there because the derivatives are all zero (really flat). For a graph of this function and more type Exp[-1/(x^2)] into Wolfram|Alpha. Functions that agree with their Taylor series are called analytic functions. Analytic functions are very like polynomials. Specify the roots of a polynomial and the polynomial is fixed (except for a scale factor). The moment you bend a polynomial a little bit you no longer have a polynomial. One difference between a polynomial and an analytic function is that an analytic function is that an analytic function can have a countable infinity of roots (think sin(x) ).

The function Exp[-1/(x^2)] has all derivatives at zero but is not analytic there because the derivatives are all zero (really flat). For a graph of this function and more type Exp[-1/(x^2)] into Wolfram|Alpha.

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