Taylor series operator commutes with differentiation - Revision history

Vipul: /* Taylor series of the derivative */

2012-07-04T16:25:28Z

Taylor series of the derivative

← Older revision		Revision as of 16:25, 4 July 2012
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	By <math>(\dagger)</math>, we can rewrite <math>(f')^{(k)}(x_0)</math> on the right side as <math>f^{(k+1)}(x_0)</math>, and get:		By <math>(\dagger)</math>, we can rewrite <math>(f')^{(k)}(x_0)</math> on the right side as <math>f^{(k+1)}(x_0)</math>, and get:

	<math>\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k+1)}}(x_0)}{k!}(x - x_0)^k</math>		<math>\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k+1)}(x_0)}{k!}(x - x_0)^k</math>

	===Checking equality===		===Checking equality===

	It is now easy to verify that we have precisely the same expressions for the derivative of the Taylor series of <math>f</math> and the Taylor series of <math>f'</math>.		It is now easy to verify that we have precisely the same expressions for the derivative of the Taylor series of <math>f</math> and the Taylor series of <math>f'</math>.

Vipul: /* Derivative of the Taylor series */

2012-07-04T16:17:39Z

Derivative of the Taylor series

← Older revision		Revision as of 16:17, 4 July 2012
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	The expression <math>\frac{f^{(k)}(x_0)}{k!}</math> is constant, so we use the [[differentiation rule for power functions]] on <math>(x - x_0)^k</math> (with the chain rule) and obtain:		The expression <math>\frac{f^{(k)}(x_0)}{k!}</math> is constant, so we use the [[differentiation rule for power functions]] on <math>(x - x_0)^k</math> (with the chain rule) and obtain:

	<math>>\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right] = \frac{kf^{(k)}(x_0)}{k!}(x - x_0)^{k-1}</math>		<math>\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right] = \frac{kf^{(k)}(x_0)}{k!}(x - x_0)^{k-1}</math>

	For <math>k = 0</math>, this becomes 0. For <math>k \ge 1</math>, set <math>l= k - 1</math> (note that <math>l \ge 0</math> now) and this becomes:		For <math>k = 0</math>, this becomes 0. For <math>k \ge 1</math>, set <math>l= k - 1</math> (note that <math>l \ge 0</math> now) and this becomes:

Vipul at 15:35, 30 June 2012

2012-06-30T15:35:12Z

← Older revision		Revision as of 15:35, 30 June 2012
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	* [[Taylor series operator is linear]]		* [[Taylor series operator is linear]]
	* [[Taylor series operator is multiplicative]]		* [[Taylor series operator is multiplicative]]

			==Proof==

			'''Given''': <math>f</math> is a function defined on a subset of the reals and it is infinitely differentiable at a point <math>x_0</math> in its domain.

			'''To prove''': The derivative <math>f'</math> is defined and infinitely differentiable at <math>x_0</math> and the Taylor series for <math>f'</math> is the derivative of the Taylor series for <math>f</math>.

			'''Proof''': We have the following relation between the derivatives of <math>f</math> and <math>f'</math>:

			<math>(f')^{(k)}(x_0) = f^{(k+1)}(x_0) \ \forall \ k \in \{ 0,1,2,\dots,\} \qquad (\dagger)</math>

			In particular, this means that, since <math>f</math> is infinitely differentiable at <math>x_0</math>, so is <math>f'</math>, with the derivatives given by the above relationship. Thus, it makes sense to take the Taylor series of <math>f'</math>.

			===Derivative of the Taylor series===

			We note that:

			<math>\mbox{Taylor series for } f \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k</math>

			Let's try differentiating this. Differentiation of power series is term-wise, so we compute the derivative for each term, i.e., we try to calculate:

			<math>\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right]</math>

			The expression <math>\frac{f^{(k)}(x_0)}{k!}</math> is constant, so we use the [[differentiation rule for power functions]] on <math>(x - x_0)^k</math> (with the chain rule) and obtain:

			<math>>\frac{d}{dx}\left[\frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\right] = \frac{kf^{(k)}(x_0)}{k!}(x - x_0)^{k-1}</math>

			For <math>k = 0</math>, this becomes 0. For <math>k \ge 1</math>, set <math>l= k - 1</math> (note that <math>l \ge 0</math> now) and this becomes:

			<math>\frac{f^{(l + 1)}(x_0)}{l!}(x - x_0)^l</math>

			Now adding together all the terms, we get:

			<math>\mbox{Derivative of Taylor series for } f \mbox{ at } x_0 = \sum_{l=0}^\infty \frac{f^{(l + 1)}(x_0)}{l!}(x - x_0)^l</math>

			Since <math>l</math> is a dummy variable, we can use the dummy variable <math>k</math> instead, and get:

			<math>\mbox{Derivative of Taylor series for } f \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k + 1)}(x_0)}{k!}(x - x_0)^k</math>

			===Taylor series of the derivative===

			We note that:

			<math>\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{(f')^{(k)}(x_0)}{k!}(x - x_0)^k</math>

			By <math>(\dagger)</math>, we can rewrite <math>(f')^{(k)}(x_0)</math> on the right side as <math>f^{(k+1)}(x_0)</math>, and get:

			<math>\mbox{Taylor series for } f' \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k+1)}}(x_0)}{k!}(x - x_0)^k</math>

			===Checking equality===

			It is now easy to verify that we have precisely the same expressions for the derivative of the Taylor series of <math>f</math> and the Taylor series of <math>f'</math>.

Vipul: /* Statement */

2011-09-07T11:43:55Z

Statement

← Older revision		Revision as of 11:43, 7 September 2011
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	==Statement==		==Statement==

	Suppose <math>f</math> if a function defined on a subset of the reals that is [[infinitely differentiable function\|infinitely differentiable]] at a point <math>x_0/math> in its [[domain]]. Then, the [[derivative]] <math>f'</math> is also defined and infinitely differentiable at <math>x_0</math>, and the [[fact about::Taylor series]] for <math>f'</math> is the [[fact about::derivative of power series\|derivative]] (in the sense of derivative of power series) of the Taylor series for <math>f</math>.		Suppose <math>f</math> if a function defined on a subset of the reals that is [[infinitely differentiable function\|infinitely differentiable]] at a point <math>x_0</math> in its [[domain]]. Then, the [[derivative]] <math>f'</math> is also defined and infinitely differentiable at <math>x_0</math>, and the [[fact about::Taylor series]] for <math>f'</math> is the [[fact about::derivative of power series\|derivative]] (in the sense of derivative of power series) of the Taylor series for <math>f</math>.

	==Related facts==		==Related facts==

Vipul: Created page with "==Statement== Suppose $f$ if a function defined on a subset of the reals that is infinitely differentiable at a point $x_0..."$

2011-09-07T11:43:21Z

Created page with "==Statement== Suppose <math>f</math> if a function defined on a subset of the reals that is infinitely differentiable at a point <math>x_0..."

New page

==Statement==

Suppose <math>f</math> if a function defined on a subset of the reals that is [[infinitely differentiable function|infinitely differentiable]] at a point <math>x_0/math> in its [[domain]]. Then, the [[derivative]] <math>f'</math> is also defined and infinitely differentiable at <math>x_0</math>, and the [[fact about::Taylor series]] for <math>f'</math> is the [[fact about::derivative of power series|derivative]] (in the sense of derivative of power series) of the Taylor series for <math>f</math>.

==Related facts==

* [[Differentiation theorem for power series]]
* [[Taylor series operator is linear]]
* [[Taylor series operator is multiplicative]]

Taylor series operator commutes with differentiation - Revision history

Vipul: /* Taylor series of the derivative */

Vipul: /* Derivative of the Taylor series */

Vipul at 15:35, 30 June 2012

Vipul: /* Statement */

Vipul: Created page with "==Statement== Suppose f if a function defined on a subset of the reals that is infinitely differentiable at a point x_0..."

Vipul: Created page with "==Statement== Suppose $f$ if a function defined on a subset of the reals that is infinitely differentiable at a point $x_0..."$