Taylor series - Revision history

Vipul: /* Computation of Taylor series */

2012-12-22T22:06:33Z

Computation of Taylor series

← Older revision		Revision as of 22:06, 22 December 2012
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	* Evaluate all these at <math>x_0</math>.		* Evaluate all these at <math>x_0</math>.
	* Plug the result into the Taylor series formula.		* Plug the result into the Taylor series formula.

			<center>{{#widget:YouTube\|id=0X_j36q6vsg}}</center>

	===Example of exponential function===		===Example of exponential function===

Vipul at 22:05, 22 December 2012

2012-12-22T22:05:35Z

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	<math>f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \dots +</math>		<math>f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \dots +</math>

			<center>{{#widget:YouTube\|id=wwq3emepmYI}}</center>

	==Well defined on germs of a functions==		==Well defined on germs of a functions==

Vipul: /* Example of sine function */

2012-07-08T14:38:59Z

Example of sine function

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	===Example of sine function===		===Example of sine function===

	{{further\|[[Sine function#Taylor series]]}}		{{further\|[[Sine function#Computation of Taylor series]]}}

	Consider the [[sine function]]:		Consider the [[sine function]]:

Vipul: /* Facts */

2012-07-08T14:26:49Z

Facts

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	* [[Taylor series operator commutes with composition]]		* [[Taylor series operator commutes with composition]]

	===~~Relation with polynomials and power series summation~~===		===Convergence to the original function===

	* [[Taylor series ~~of polynomial~~ is the ~~same polynomial~~]]		* [[Composite of Taylor series operator and power series summation operator is identity map]]: What this essentially says is that ''if'' a power series centered at <math>x_0</math> converges to <math>f</math> on an open interval centered at <math>x_0</math>, then the power series ''must'' equal the [[Taylor series]] of <math>f</math>.
	* [[Composite of ~~Taylor~~ series operator and ~~power~~ series ~~summation~~ operator is identity map]]		* [[Composite of power series summation operator and Taylor series operator is not identity map]]: It is possible to have an everywhere infinitely differentiable function <math>f</math> and a point <math>x_0</math> in the domain such that the sum of the [[Taylor series]] of <math>f</math> at <math>x_0</math> is not equal to <math>f</math> on any interval of positive radius centered at <math>x_0</math>. In fact, we can arrange our example so that the power series sum agrees with <math>f</math> ''only'' at <math>x_0</math>. Note that if this happens, then there cannot be any other power series centered at <math>x_0</math> that converges to <math>f</math> on a positive radius of convergence.

			A function whose Taylor series at a point converges to the function in an open interval centered at the point is termed a [[locally analytic function]] at the point. If the Taylor series converges to the function everywhere, the function is termed a [[globally analytic function]]. We have that [[locally analytic not implies globally analytic]].

Vipul: /* Example of cosine function */

2012-07-08T14:21:09Z

Example of cosine function

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	Note that it is ''also'' true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.		Note that it is ''also'' true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.

			===Example of sine function===

			{{further\|[[Sine function#Taylor series]]}}

			Consider the [[sine function]]:

			<math>f = \sin</math>

			We want to compute the Taylor series of this function at 0.

			Applying the procedure above, we get:

			* Formal expressions for <math>f,f',f'',\dots</math>: These are <math>\sin, \cos, -\sin, -\cos, \sin, \cos, -\sin, -\cos, \dots</math>. The sequence is periodic with period 4.
			* Evaluate at 0: Since <math>\sin 0 = 0, \cos 0 = 1</math>, the values are <math>0,1,0,-1,0,1,0,-1,\dots</math>. The sequence is periodic with period 4.
			* Plug the result into the Taylor series formula: We get:

			Taylor series for <math>\sin x</math> is <math>x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k + 1}}{(2k + 1)!}</math>. Note that the <math>k</math> here is ''roughly half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.

			Note that it is ''also'' true that the Taylor series for the sine function converges to the sine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.

	==Facts==		==Facts==

Vipul: /* Example of cosine function */

2012-07-08T14:18:47Z

Example of cosine function

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	* Plug the result into the Taylor series formula: We get:		* Plug the result into the Taylor series formula: We get:

	Taylor series for <math>\cos x</math> is <math>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k!)}</math>. Note that the <math>k</math> here is ''half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.		Taylor series for <math>\cos x</math> is <math>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!}</math>. Note that the <math>k</math> here is ''half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.

	Note that it is ''also'' true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.		Note that it is ''also'' true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.

Vipul: /* Computation of Taylor series */

2012-07-08T14:17:33Z

Computation of Taylor series

← Older revision		Revision as of 14:17, 8 July 2012
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	Taylor series for <math>\exp(x)</math> is <math>\sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</math>		Taylor series for <math>\exp(x)</math> is <math>\sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</math>

			Note that it is ''also'' true that the Taylor series for the exponential function converges to the exponential function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.

	===Example of cosine function===		===Example of cosine function===
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	Taylor series for <math>\cos x</math> is <math>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k!)}</math>. Note that the <math>k</math> here is ''half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.		Taylor series for <math>\cos x</math> is <math>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k!)}</math>. Note that the <math>k</math> here is ''half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.

			Note that it is ''also'' true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not ''a priori'' obvious, and we are ''not'' asserting it as part of the computation of the Taylor series.

	==Facts==		==Facts==

Vipul at 14:16, 8 July 2012

2012-07-08T14:16:10Z

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	Whether or not the Taylor series of a function converges to the function is determined by whether or not the sequence of Taylor polynomials of the function converges to the function.		Whether or not the Taylor series of a function converges to the function is determined by whether or not the sequence of Taylor polynomials of the function converges to the function.

			==Computation of Taylor series==

			To calculate the Taylor series of a function <math>f</math> at a point <math>x_0</math>, we use the following procedure:

			* Compute formal expressions for <math>f,f',f'',\dots</math>, i.e., <math>f</matH> and all its derivatives, at a generic point.
			* Evaluate all these at <math>x_0</math>.
			* Plug the result into the Taylor series formula.

			===Example of exponential function===

			{{further\|[[Exponential function#Taylor series]]}}

			Consider the [[exponential function]]:

			<math>f = \exp</math>

			i.e., <math>f(x) = e^x</math>.

			We want to compute the Taylor series of this function at 0.

			Applying the procedure above, we get:

			* Formal expressions for <math>f,f',f'',\dots</math>: These are <math>\exp,\exp,\exp,\dots</math>. The sequence is a constant sequence of functions with all its members equal to <math>\exp</math>.
			* Evaluate at 0: Since <math>\exp(0) = 1</math>, we get <matH>1,1,1,\dots</math>. The sequence is a constant sequence with value 1 in all places.
			* Plug the result into the Taylor series formula: We get:

			Taylor series for <math>\exp(x)</math> is <math>\sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</math>

			===Example of cosine function===

			{{further\|[[Cosine function#Taylor series]]}}

			Consider the [[cosine function]]:

			<math>f = \cos</math>

			We want to compute the Taylor series of this function at 0.

			Applying the procedure above, we get:

			* Formal expressions for <math>f,f',f'',\dots</math>: These are <math>\cos, -\sin, -\cos, \sin, \cos, -\sin, -\cos, \sin, \dots</math>. The sequence is periodic with period 4.
			* Evaluate at 0: Since <math>\cos 0 = 1, \sin 0 = 0</math>, the values are <math>1,0,-1,0,1,0,-1,0,\dots</math>. The sequence is periodic with period 4.
			* Plug the result into the Taylor series formula: We get:

			Taylor series for <math>\cos x</math> is <math>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</math> which can be rewritten compactly as <math>\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k!)}</math>. Note that the <math>k</math> here is ''half'' the exponent on <math>x</math>, so this is a little different from the usual way of writing Taylor series.


	==Facts==		==Facts==

Vipul at 14:33, 7 July 2012

2012-07-07T14:33:10Z

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	Here, two functions <math>f</math> and <math>g</matH> are said to have the same ''germ'' about a point <math>x_0</math> if there is an open interval <math>U</math> containing <math>x_0</math> such that <math>f(x) = g(x) \ \forall x \in U</math>.		Here, two functions <math>f</math> and <math>g</matH> are said to have the same ''germ'' about a point <math>x_0</math> if there is an open interval <math>U</math> containing <math>x_0</math> such that <math>f(x) = g(x) \ \forall x \in U</math>.

			==Relation with Taylor polynomials==

			The Taylor series can be viewed as a limit of [[Taylor polynomial]]s. The <math>n^{th}</math> Taylor polynomial for a function <math>f</math> at a point <math>x_0</math> in the domain is the truncation of the Taylor series to powers up to the <math>n^{th}</math> power. If we denote the polynomial by <math>P_n(f;x_0)</math>, it is given as:

			<math>P_n(f;x_0) = x \mapsto \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k</math>

			Note that this is a polynomial of degree ''at most'' <math>n</math>. The degree is exactly <math>n</math> if and only if <math>f^{(n)}(x_0) \ne 0</math>.

			Whether or not the Taylor series of a function converges to the function is determined by whether or not the sequence of Taylor polynomials of the function converges to the function.
	==Facts==		==Facts==

Vipul: /* Preservation of structure */

2012-06-30T16:11:18Z

Preservation of structure

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	===Preservation of structure===		===Preservation of structure===

	Together, ~~these~~ three facts show that the Taylor series operator is a homomorphism of <math>\R</math>-algebras that commutes with the differential structure:		Together, the first three facts show that the Taylor series operator is a homomorphism of <math>\R</math>-algebras that commutes with the differential structure. The fourth fact show that it preserves an additional structure:

	* [[Taylor series operator is linear]]		* [[Taylor series operator is linear]]
	* [[Taylor series operator commutes with differentiation]]		* [[Taylor series operator commutes with differentiation]]
	* [[Taylor series operator is multiplicative]]		* [[Taylor series operator is multiplicative]]
			* [[Taylor series operator commutes with composition]]

	===Relation with polynomials and power series summation===		===Relation with polynomials and power series summation===