Uniformly bounded derivatives implies globally analytic - Revision history

Vipul: /* Global statement */

2012-07-12T20:27:12Z

Global statement

← Older revision		Revision as of 20:27, 12 July 2012
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	===Global statement===		===Global statement===

	Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:		Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a<b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:

	<math>\|f^{(n)}(t)\| \le C \ \forall t \in [a,b]</math>		<math>\|f^{(n)}(t)\| \le C \ \forall t \in [a,b]</math>

Vipul: /* Proof */

2012-07-08T15:02:12Z

Proof

← Older revision		Revision as of 15:02, 8 July 2012
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	where <math>J</math> is the interval joining <math>x_0</math> to <math>x</math>. Let <math>a = \min \{ x,x_0 \}</math> and <math>b = \max \{ x, x_0 \}</math>. The interval <math>J</math> is the interval <math>[a,b]</math>.		where <math>J</math> is the interval joining <math>x_0</math> to <math>x</math>. Let <math>a = \min \{ x,x_0 \}</math> and <math>b = \max \{ x, x_0 \}</math>. The interval <math>J</math> is the interval <math>[a,b]</math>.

	Now, from the given data, there exists <math>C</math>, dependent on <math>x</math> and <math>x_0</math>, such that:		Now, from the given data, there exists <math>C</math>, dependent on <math>x</math> and <math>x_0</math> but ''not'' on <math>n</math>, such that:

	<math>\max_{t \in J} \|f^{(n+1)}(t)\| \le C \ \forall \ n</math>		<math>\max_{t \in J} \|f^{(n+1)}(t)\| \le C \ \forall \ n</math>

Vipul at 15:01, 8 July 2012

2012-07-08T15:01:13Z

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 Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
-<math>|f^{(n)}(x)| \le C \ \forall x \in [a,b]</math>
+<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
 Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.
@@ Line 17: / Line 17: @@
 The functions <math>\exp, \sin, \cos</math> all fit this description.
-If <math>f = \exp</math>, we know that each of the derivatives equals <math>\exp</math>, so <math>f^{(n)}(x) = f(x)</math> for all <math>x \in [a,b]</math>. Since <math>\exp</math> is continuous, it is bounded on the closed interval <math>[a,b]</math>, and the upper bound for <math>\exp</math> thus serves as a uniform bound for all its derivatives.
+If <math>f = \exp</math>, we know that each of the derivatives equals <math>\exp</math>, so <math>f^{(n)}(t) = f(t)</math> for all <math>t \in [a,b]</math>. Since <math>\exp</math> is continuous, it is bounded on the closed interval <math>[a,b]</math>, and the upper bound for <math>\exp</math> thus serves as a uniform bound for all its derivatives. (In fact, since <math>f</math> is increasing, we can explicitly take <math>C = \exp(b)</math>).
 For <math>f = \sin</math> or <math>f = \cos</math>, we know that all the derivatives are <math>\pm \sin</math> or <math>\pm \cos</math>, so their magnitude is at most 1. Thus, we can take <math>C = 1</math>.

Vipul at 14:49, 8 July 2012

2012-07-08T14:49:59Z

← Older revision		Revision as of 14:49, 8 July 2012
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	Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.		Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.

			==Facts used==

			# [[uses::Max-estimate version of Lagrange formula]]

	==Examples==		==Examples==

	The functions <math>\exp, \sin, \cos</math> all fit this description.		The functions <math>\exp, \sin, \cos</math> all fit this description.

			If <math>f = \exp</math>, we know that each of the derivatives equals <math>\exp</math>, so <math>f^{(n)}(x) = f(x)</math> for all <math>x \in [a,b]</math>. Since <math>\exp</math> is continuous, it is bounded on the closed interval <math>[a,b]</math>, and the upper bound for <math>\exp</math> thus serves as a uniform bound for all its derivatives.

			For <math>f = \sin</math> or <math>f = \cos</math>, we know that all the derivatives are <math>\pm \sin</math> or <math>\pm \cos</math>, so their magnitude is at most 1. Thus, we can take <math>C = 1</math>.

Vipul: Vipul moved page Uniformly bounded derivatives implies analytic to Uniformly bounded derivatives implies globally analytic

2012-07-08T14:44:51Z

Vipul moved page Uniformly bounded derivatives implies analytic to Uniformly bounded derivatives implies globally analytic

← Older revision	Revision as of 14:44, 8 July 2012
(No difference)

Vipul at 14:51, 7 July 2012

2012-07-07T14:51:58Z

← Older revision		Revision as of 14:51, 7 July 2012
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	==Statement==		==Statement==

			===Global statement===

	Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:		Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
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	Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.		Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.

			==Examples==

			The functions <math>\exp, \sin, \cos</math> all fit this description.

Vipul at 14:51, 7 July 2012

2012-07-07T14:51:22Z

← Older revision		Revision as of 14:51, 7 July 2012
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	<math>\|f^{(n)}(x)\| \le C \ \forall x \in [a,b]</math>		<math>\|f^{(n)}(x)\| \le C \ \forall x \in [a,b]</math>

	Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to 0.		Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.

Vipul: Created page with "==Statement== Suppose $f$ is an infinitely differentiable function on $\R$ such that, for any fixed $a,b \in \R$ , there is a constant $C..."$

2012-07-07T14:51:11Z

Created page with "==Statement== Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C..."

New page

==Statement==

Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:

<math>|f^{(n)}(x)| \le C \ \forall x \in [a,b]</math>

Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to 0.

Uniformly bounded derivatives implies globally analytic - Revision history

Vipul: /* Global statement */

Vipul: /* Proof */

Vipul at 15:01, 8 July 2012

Vipul at 14:49, 8 July 2012

Vipul: Vipul moved page Uniformly bounded derivatives implies analytic to Uniformly bounded derivatives implies globally analytic

Vipul at 14:51, 7 July 2012

Vipul at 14:51, 7 July 2012

Vipul: Created page with "==Statement== Suppose f is an infinitely differentiable function on \R such that, for any fixed a,b \in \R, there is a constant C..."

Vipul: Created page with "==Statement== Suppose $f$ is an infinitely differentiable function on $\R$ such that, for any fixed $a,b \in \R$ , there is a constant $C..."$