Sine function - Revision history

Vipul: /* Taylor series equals power series */

2012-07-08T14:46:29Z

Taylor series equals power series

← Older revision		Revision as of 14:46, 8 July 2012
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	===Taylor series equals power series===		===Taylor series equals power series===

	~~There are~~ a ~~number of ways of showing that~~ the Taylor series for the sine function does in fact converge to the function everywhere. ~~One~~ of ~~these follows from the max-estimate on the remainder term from the Taylor polyonmials~~. ~~Since the sine function~~ is a bounded ~~function~~, ~~this max-estimate approaches zero as~~ <math>n \~~to \infty~~</math>~~. For more on this reasoning~~, ~~see [[uniformly bounded derivatives implies analytic]]~~.		The sine function is a [[globally analytic function]]: the Taylor series for the sine function does in fact converge to the function everywhere. This can be proved in a number of ways. One method is to use that [[uniformly bounded derivatives implies analytic]]. Alternatively, we could note that <math>\sin</math> satisfies a certain differential equation <math>y'' = -y</math>, forcing it to be given by a power series.

	Thus we have that:		Thus we have that:

Vipul: /* Taylor series equals power series */

2012-07-08T14:44:27Z

Taylor series equals power series

← Older revision		Revision as of 14:44, 8 July 2012
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	===Taylor series equals power series===		===Taylor series equals power series===

	There are a number of ways of showing that the Taylor series for the sine function does in fact converge to the function everywhere. One of these follows from the max-estimate on the remainder term from the Taylor polyonmials. Since the sine function is a bounded function, this max-estimate approaches zero as <math>n \to \infty</math>.		There are a number of ways of showing that the Taylor series for the sine function does in fact converge to the function everywhere. One of these follows from the max-estimate on the remainder term from the Taylor polyonmials. Since the sine function is a bounded function, this max-estimate approaches zero as <math>n \to \infty</math>. For more on this reasoning, see [[uniformly bounded derivatives implies analytic]].

	Thus we have that:		Thus we have that:

Vipul: /* Higher order limits */

2012-07-08T14:37:57Z

Higher order limits

← Older revision		Revision as of 14:37, 8 July 2012
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	\| Using the [[L'Hopital rule]] \|\| <math>\lim_{x \to 0} \frac{x - \sin x}{x^3} \stackrel{}{=} \lim_{x \to 0} \frac{1 -\cos x}{3x^2} \stackrel{}{=} \lim_{x \to 0} \frac{\sin x}{6x} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x}{6} = 1/6</math>.		\| Using the [[L'Hopital rule]] \|\| <math>\lim_{x \to 0} \frac{x - \sin x}{x^3} \stackrel{}{=} \lim_{x \to 0} \frac{1 -\cos x}{3x^2} \stackrel{}{=} \lim_{x \to 0} \frac{\sin x}{6x} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x}{6} = 1/6</math>.
	\|-		\|-
	\| Using the power series \|\| We have <math>\sin x = x - x^3/3! + O(x^5)</math>, so <math>\frac{x - \~~sinx~~}{x^3} = \frac{1}{3!} - O(x^2)</math>, so the limit as <math>x \to 0</math> is <math>1/3! = 1/6</math>.		\| Using the power series \|\| We have <math>\sin x = x - x^3/3! + O(x^5)</math>, so <math>\frac{x - \sin x}{x^3} = \frac{1}{3!} - O(x^2)</math>, so the limit as <math>x \to 0</math> is <math>1/3! = 1/6</math>.
	\|}		\|}

Vipul: /* Higher antiderivatives */

2012-02-13T00:05:33Z

Higher antiderivatives

← Older revision		Revision as of 00:05, 13 February 2012
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	In general, the <math>k^{th}</math> antiderivative is <math>\pm \cos</math> or <math>\pm \sin</math>, depending on the value of <math>k</math> mod 4. The general expression is the particular antiderivative plus an arbitrary polynomial of degree at most <math>k - 1</math>.		In general, the <math>k^{th}</math> antiderivative is <math>\pm \cos</math> or <math>\pm \sin</math>, depending on the value of <math>k</math> mod 4. The general expression is the particular antiderivative plus an arbitrary polynomial of degree at most <math>k - 1</math>.

			===Integration of product with polynomials===

			Using [[integration by parts]], we know that knowledge of the first <math>k</math> antiderivatives of <math>f</math> is sufficient to determine <math>\int x^{k-1} f(x) \, dx</math> via repeated application of integration by parts. Since the sine function can be antidifferentiated any number of times, this allows us to antidifferentiate any polynomial times the sine function.

			For instance, the function <math>x \mapsto x \sin x</math>, i.e., the [[product of identity function and sine function]], can be integrated via knowledge of how to integrate the sine function twice:

			<math>\int x \sin x \, dx = x \int \sin x \, dx - \int \int \sin x \, dx \, dx = -x\cos x + \sin x + C</math>

	==Differential equations==		==Differential equations==

Vipul: /* Power series and Taylor series */

2012-02-12T23:49:47Z

Power series and Taylor series

← Older revision		Revision as of 23:49, 12 February 2012
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	In general, the <math>k^{th}</math> antiderivative is <math>\pm \cos</math> or <math>\pm \sin</math>, depending on the value of <math>k</math> mod 4. The general expression is the particular antiderivative plus an arbitrary polynomial of degree at most <math>k - 1</math>.		In general, the <math>k^{th}</math> antiderivative is <math>\pm \cos</math> or <math>\pm \sin</math>, depending on the value of <math>k</math> mod 4. The general expression is the particular antiderivative plus an arbitrary polynomial of degree at most <math>k - 1</math>.

			==Differential equations==

			The sine function and its transforms arises as the solution to many differential equations, including [[polynomial differential equation]]s. Some of these are listed below.

			{\| class="sortable" border="1"
			! Equation !! Other forms !! General solution
			\|-
			\| <math>\frac{dx}{dt} = \sqrt{a^2 - x^2}, \qquad a > 0</math> \|\| \|\| <math>x = a \sin (t + \varphi)</math> for <math>-\pi/2 \le t + \varphi \le \pi/2</math>
			\|-
			\| <math>\frac{dx}{dt} = a \sqrt{x(b- x)}, \qquad a,b > 0</math> \|\| <math>\left(\frac{dx}{dt}\right)^2 = a^2x(b- x), \qquad a,b > 0</math> \|\| See [[Verhulst process]]
			\|-
			\| <math>\frac{d^2x}{dt^2} = -x</math> \|\| \|\| <math>x = a \sin(t + \varphi)</math> or <math>x = A \sin t + B \cos t</math>
			\|}

	==Power series and Taylor series==		==Power series and Taylor series==

Vipul: /* Power series and Taylor series */

2012-02-12T22:29:40Z

Power series and Taylor series

← Older revision		Revision as of 22:29, 12 February 2012
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	<math>\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}</math>		<math>\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}</math>

			==Limit computations==

			===Order of zero===

			We have the following limit:

			<math>\lim_{x \to 0} \frac{\sin x}{x} = 1</math>

			Thus, the order of zero at 0 is 1 and the residue is 1.

			===Higher order limits===

			We have the limit:

			<math>\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}</math>

			The limit can be computed in either of two ways:

			{\| class="sortable" border="1"
			! Name of method for computing the limit !! Details
			\|-
			\| Using the [[L'Hopital rule]] \|\| <math>\lim_{x \to 0} \frac{x - \sin x}{x^3} \stackrel{}{=} \lim_{x \to 0} \frac{1 -\cos x}{3x^2} \stackrel{}{=} \lim_{x \to 0} \frac{\sin x}{6x} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x}{6} = 1/6</math>.
			\|-
			\| Using the power series \|\| We have <math>\sin x = x - x^3/3! + O(x^5)</math>, so <math>\frac{x - \sinx}{x^3} = \frac{1}{3!} - O(x^2)</math>, so the limit as <math>x \to 0</math> is <math>1/3! = 1/6</math>.
			\|}

Vipul: /* Composition with other functions */

2012-02-12T22:26:19Z

Composition with other functions

← Older revision		Revision as of 22:26, 12 February 2012
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	\| [[square root function]] \|\| [[square root of sine function]] -- note that this makes sense only on a restricted domain \|\| [[sine of square root function]]		\| [[square root function]] \|\| [[square root of sine function]] -- note that this makes sense only on a restricted domain \|\| [[sine of square root function]]
	\|-		\|-
	\| [[natural logarithm]] of absolute value \|\| [[natural logarithm of absolute value of sine function]] <math>x \mapsto \ln\|\sin x\|</math> \|\| [[sine of natural logarithm of absolute value]] <math>x \mapsto \ln(\|~~\sin~~ x\|)</math>		\| [[natural logarithm]] of absolute value \|\| [[natural logarithm of absolute value of sine function]] <math>x \mapsto \ln\|\sin x\|</math> \|\| [[sine of natural logarithm of absolute value]] <math>x \mapsto \sin(\ln\|x\|)</math>
	\|}		\|}

Vipul: /* Composition with other functions */

2012-02-12T22:25:54Z

Composition with other functions

← Older revision		Revision as of 22:25, 12 February 2012
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	\|-		\|-
	\| [[square root function]] \|\| [[square root of sine function]] -- note that this makes sense only on a restricted domain \|\| [[sine of square root function]]		\| [[square root function]] \|\| [[square root of sine function]] -- note that this makes sense only on a restricted domain \|\| [[sine of square root function]]
			\|-
			\| [[natural logarithm]] of absolute value \|\| [[natural logarithm of absolute value of sine function]] <math>x \mapsto \ln\|\sin x\|</math> \|\| [[sine of natural logarithm of absolute value]] <math>x \mapsto \ln(\|\sin x\|)</math>
	\|}		\|}

Vipul: /* Related functions */

2012-02-12T22:24:23Z

Related functions

@@ Line 102: / Line 102: @@
 |-
 | [[square root function]] || [[square root of sine function]] -- note that this makes sense only on a restricted domain || [[sine of square root function]]
 |}

Vipul: /* Identities */

2012-02-12T22:19:59Z

Identities

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 |-
 | other symmetries || periodicity: <math>\sin(2\pi + x) = \sin x</math><br>anti-periodicity: <math>\sin(\pi + x) = -\sin x</math><br>odd: <math>\sin(-x) = -\sin x</math><br>mirror symmetry about <math>\pi/2</math>: <math>\sin(\pi - x) = \sin x</math>
 |}