Composite of increasing functions is increasing - Revision history

Vipul at 03:49, 18 December 2011

2011-12-18T03:49:01Z

← Older revision		Revision as of 03:49, 18 December 2011
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	===Statement for two functions===		===Statement for two functions===

	Suppose <math>f</math> and <math>g</math> are both [[function]]s of one variable that are [[increasing function]]s on their respective [[domain]]s. Consider the [[composite of two functions]] <math>f \circ g</math>. This is also an increasing function on its domain.		Suppose <math>f</math> and <math>g</math> are both [[function]]s of one variable that are [[fact about::increasing function]]s on their respective [[domain]]s. Consider the [[fact about::composite of two functions]] <math>f \circ g</math>. This is also an increasing function on its domain.

	Note that the statement makes ''no'' assumptions about the continuity or differentiability of the functions or even the nature of their domains. In fact, we do not even require that the domains and ranges be subsets of the real numbers, but only require that they be totally ordered sets so that the notion of ''increasing'' makes sense.		Note that the statement makes ''no'' assumptions about the continuity or differentiability of the functions or even the nature of their domains. In fact, we do not even require that the domains and ranges be subsets of the real numbers, but only require that they be totally ordered sets so that the notion of ''increasing'' makes sense.

Vipul: /* Proof */

2011-12-18T03:39:06Z

Proof

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	\| 3 \|\| <math>(f \circ g)(x_1) < (f \circ g)(x_2)</math> \|\| \|\| Step (2) \|\| Just rewrite Step (2) in terms of composite function, using the definition of composite function.		\| 3 \|\| <math>(f \circ g)(x_1) < (f \circ g)(x_2)</math> \|\| \|\| Step (2) \|\| Just rewrite Step (2) in terms of composite function, using the definition of composite function.
	\|}		\|}

			===Compatibility with chain rule for differentiation===

			In case both the functions <math>f</math> and <math>g</math> are differentiable, then we can check that the statement is compatible with the [[chain rule for differentiation]]. In fact, the chain rule for differentiation can also be used to furnish an alternative proof in this case, though we have to deal carefully with the case of zero derivative.

			Note that:

			<math>\! (f \circ g)'(x) = f'(g(x))g'(x)</math>

			In particular

			* If both <math>\! f'</math> and <math>\! g'</math> are positive everywhere on their domain, so is <math>\! (f \circ g)'</math>.
			* If both <math>\! f'</math> and <math>\! g'</math> are nonnegative everywhere on their domain, so is <math>\! (f \circ g)'</math>.

			These ''almost'' prove that a composite of increasing differentiable functions is increasing.

Vipul: /* Related facts */

2011-12-18T03:30:32Z

Related facts

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

			===Related facts about composites of functions===
	* [[Composite of two decreasing functions is increasing]]		* [[Composite of two decreasing functions is increasing]]
	* [[Composite of one-one functions is one-one]]		* [[Composite of one-one functions is one-one]]

			===Related facts about increasing functions===

			* [[Inverse of increasing function is increasing]]
	* [[Increasing functions form a cone in a vector space]]		* [[Increasing functions form a cone in a vector space]]
	* [[Product of increasing functions need not be increasing]]		* [[Product of increasing functions need not be increasing]]

Vipul: /* Related facts */

2011-12-18T03:29:47Z

Related facts

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	* [[Composite of one-one functions is one-one]]		* [[Composite of one-one functions is one-one]]
	* [[Increasing functions form a cone in a vector space]]		* [[Increasing functions form a cone in a vector space]]
			* [[Product of increasing functions need not be increasing]]

	==Proof==		==Proof==

Vipul: /* Related facts */

2011-12-18T03:29:19Z

Related facts

← Older revision		Revision as of 03:29, 18 December 2011
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	* [[Composite of two decreasing functions is increasing]]		* [[Composite of two decreasing functions is increasing]]
	* [[Composite of one-one functions is one-one]]		* [[Composite of one-one functions is one-one]]
			* [[Increasing functions form a cone in a vector space]]

	==Proof==		==Proof==

Vipul: /* Statement for multiple functions */

2011-12-18T03:28:56Z

Statement for multiple functions

← Older revision		Revision as of 03:28, 18 December 2011
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	{{fillin}}		{{fillin}}

			==Related facts==

			* [[Composite of two decreasing functions is increasing]]
			* [[Composite of one-one functions is one-one]]

	==Proof==		==Proof==

Vipul: /* Proof */

2011-12-18T03:21:42Z

Proof

← Older revision		Revision as of 03:21, 18 December 2011
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	==Proof==		==Proof==

			===Proof for two functions===

	'''Given''': <math>f</math> and <math>g</math> are increasing functions. <math>x_1 < x_2</math> are both in the domain of the composite function <math>f \circ g</math>.		'''Given''': <math>f</math> and <math>g</math> are increasing functions. <math>x_1 < x_2</math> are both in the domain of the composite function <math>f \circ g</math>.

Vipul: /* Proof */

2011-12-18T03:21:23Z

Proof

← Older revision		Revision as of 03:21, 18 December 2011
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	==Proof==		==Proof==

	{~~{fillin}~~}		'''Given''': <math>f</math> and <math>g</math> are increasing functions. <math>x_1 < x_2</math> are both in the domain of the composite function <math>f \circ g</math>.

			'''To prove''': <math>(f \circ g)(x_1) < (f \circ g)(x_2)</math>.

			'''Proof''':

			{\| class="sortable" border="1"
			! Step no. !! Assertion !! Given data used !! Previous steps used !! Explanation
			\|-
			\| 1 \|\| <math>g(x_1) < g(x_2)</math> \|\| <math>g</math> is increasing<br><math>x_1 < x_2</math> \|\| \|\| apply definition of increasing
			\|-
			\| 2 \|\| <math>f(g(x_1)) < f(g(x_2))</math> \|\| <math>f</math> is increasing \|\| Step (1) \|\| apply definition of increasing to inputs <math>g(x_1), g(x_2)</math>, use Step (1).
			\|-
			\| 3 \|\| <math>(f \circ g)(x_1) < (f \circ g)(x_2)</math> \|\| \|\| Step (2) \|\| Just rewrite Step (2) in terms of composite function, using the definition of composite function.
			\|}

Vipul: Created page with "==Statement== ===Statement for two functions=== Suppose $f$ and $g$ are both functions of one variable that are increasing functions on their resp..."

2011-12-18T03:17:56Z

Created page with "==Statement== ===Statement for two functions=== Suppose <math>f</math> and <math>g</math> are both functions of one variable that are increasing functions on their resp..."

New page

==Statement==

===Statement for two functions===

Suppose <math>f</math> and <math>g</math> are both [[function]]s of one variable that are [[increasing function]]s on their respective [[domain]]s. Consider the [[composite of two functions]] <math>f \circ g</math>. This is also an increasing function on its domain.

Note that the statement makes ''no'' assumptions about the continuity or differentiability of the functions or even the nature of their domains. In fact, we do not even require that the domains and ranges be subsets of the real numbers, but only require that they be totally ordered sets so that the notion of ''increasing'' makes sense.

===Statement for multiple functions===

{{fillin}}

==Proof==

{{fillin}}

Composite of increasing functions is increasing - Revision history

Vipul at 03:49, 18 December 2011

Vipul: /* Proof */

Vipul: /* Related facts */

Vipul: /* Related facts */

Vipul: /* Related facts */

Vipul: /* Statement for multiple functions */

Vipul: /* Proof */

Vipul: /* Proof */

Vipul: Created page with "==Statement== ===Statement for two functions=== Suppose f and g are both functions of one variable that are increasing functions on their resp..."

Vipul: Created page with "==Statement== ===Statement for two functions=== Suppose $f$ and $g$ are both functions of one variable that are increasing functions on their resp..."