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Khanh: Created page with "==The Mean Value Theorem for Differentiable Functions from Rn to Rm== Recall that if $f$ is a continuous function on the closed interval $[x, y]$ and di..."

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Created page with "==The Mean Value Theorem for Differentiable Functions from Rn to Rm== Recall that if <math>f</math> is a continuous function on the closed interval <math>[x, y]</math> and di..."

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==The Mean Value Theorem for Differentiable Functions from Rn to Rm==

Recall that if <math>f</math> is a continuous function on the closed interval <math>[x, y]</math> and differentiable on the open interval <math>(x, y)</math> (where we assume <math>x < y</math>) then there exists a number <math>c \in (x, y)</math> for which:

<div style="text-align: center;"><math> \begin{align} \quad f'(c) = \frac{f(y) - f(x)}{y - x} \quad \Leftrightarrow \quad f(y) - f(x) = f'(c)(y - x) \end{align}</math></div>

We would like to generalize this extremely important result to differentiable functions from <math>\mathbb{R}^n</math> to <math>\mathbb{R}^m</math>. Doing so is actually not that straightforward though. The equation above does not immediately generalize to differentiable functions from <math>\mathbb{R}^n</math> to <math>\mathbb{R}^m</math> and we will need to do some more work in order to make a meaningful generalization.

To emphasize this, consider the function <math>\mathbf{f} : \mathbb{R} \to \mathbb{R}^2</math> defined for all <math>x \in \mathbb{R}</math> by:

<div style="text-align: center;"><math>\begin{align} \quad \mathbf{f}(x) = (\cos x, \sin x) \end{align}</math></div>

Then the total derivative of <math>\mathbf{f}</math> at <math>x</math> evaluated at any <math>h \in \mathbb{R}</math> is:

<div style="text-align: center;"><math>\begin{align} \quad \mathbf{f}'(x)(h) = (-\sin x, \cos x)h \end{align}</math></div>

Therefore we have that:

<div style="text-align: center;"><math>\begin{align} \quad \mathbf{f}(y) - \mathbf{f}(x) = (\cos y, \sin y) - (\cos x, \sin x) = (\cos y - \cos x, \sin y - \sin x) \quad (*) \end{align}</math></div>

And also:

<div style="text-align: center;"><math>\begin{align} \quad \mathbf{f}'(z)(y - x) = (-\sin z, \cos z)(y - x) \quad (**) \end{align}</math></div>

Now set <math>x = 0</math> and <math>y = 2\pi</math>. Then <math>(*)</math> will always equal the zero vector, <math>\mathbf{0}</math>, and <math>(**)</math> will never equal the zero vector for any choice of <math>z</math> between <math>0</math> and <math>2\pi</math>. Therefore we see that <math>\mathbf{f}(y) - \mathbf{f}(x) \neq \mathbf{f}'(z)(y - x)</math> in general.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem 1 (The Mean Value Theorem):''' Let <math>S \subseteq \mathbb{R}^n</math> be open and let <math>\mathbf{f} : S \to \mathbb{R}^m</math> be differentiable on all of <math>S</math>. Let <math>\mathbf{x}, \mathbf{y} \in S</math> be such that the line segment connecting these two points is contained in <math>S</math>, i.e., <math>L(\mathbf{x}, \mathbf{y}) \subset S</math>. Then for every <math>\mathbf{a} \in \mathbb{R}^m</math> there exists a point <math>\mathbf{z} \in L(\mathbf{x}, \mathbf{y})</math> such that <math>\mathbf{a} \cdot [\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})] = \mathbf{a} \cdot [\mathbf{f}'(\mathbf{z})(\mathbf{y} - \mathbf{x})]</math>.
</blockquote>

''In the following Theorem we use the notation "<math>L(\mathbf{x}, \mathbf{y})</math>" to denote the line segment that joints the point <math>\mathbf{x}</math> to <math>\mathbf{y}</math>. This line segment can be parameterized as <math>L(\mathbf{x}, \mathbf{y}) = \{ (1 - t)\mathbf{x} + t \mathbf{y} : t \in [0, 1] \}</math>.''

*'''Proof:''' Let <math>\mathbf{a} \in \mathbb{R}^m</math> and define a new function <math>F : [0, 1] \to \mathbb{R}</math> for all <math>t \in [0, 1]</math> by:

<div style="text-align: center;"><math>\begin{align} \quad F(t) = \mathbf{a} \cdot \mathbf{f}(\mathbf{x} + t(\mathbf{y} - \mathbf{x})) \end{align}</math></div>

*Since <math>\mathbf{f}</math> is differentiable on <math>S</math> we have from the Differentiable Functions from Rn to Rm are Continuous page that <math>\mathbf{f}</math> is continuous on <math>S</math> and so <math>F</math> must continuous on <math>[0, 1]</math>. Furthermore, <math>F</math> is differentiable on <math>(0, 1)</math> by the chain rule:

<div style="text-align: center;"><math>\begin{align} \quad F'(t) = \mathbf{a} \cdot \mathbf{f}'(\mathbf{x} + t(\mathbf{y} - \mathbf{x})) (\mathbf{y} - \mathbf{x}) \end{align}</math></div>

*So by the Mean Value Theorem for single-variable real-valued functions, for <math>x = 0</math> and <math>y = 1</math> there exists a number <math>h \in (0, 1)</math> for which:<div style="text-align: center;"><math>\begin{align} \quad F(1) - F(0) &= F'(h)(1 - 0) \quad (*)\\ \end{align}</math></div>

*The lefthand side of <math>(*)</math> is:
<div style="text-align: center;"><math>\begin{align} \quad F(1) - F(0) = \mathbf{a} \cdot \mathbf{f}(\mathbf{y}) - \mathbf{a} \cdot \mathbf{f}(\mathbf{x}) = \mathbf{a} \cdot (\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})) \end{align}</math></div>

*The righthand side of <math>(*)</math> is:
<div style="text-align: center;"><math>\begin{align} \quad \mathbf{F}'(h)(1 - 0) = \mathbf{F}'(h) = \mathbf{a} \cdot \mathbf{f}'(\mathbf{x} + h(\mathbf{y} - \mathbf{x})) (\mathbf{y} - \mathbf{x}) \end{align}</math></div>

*Set <math>\mathbf{z} = \mathbf{x} + h(\mathbf{y} - \mathbf{x})</math>. Then <math>\mathbf{z} \in L(\mathbf{x}, \mathbf{y})</math> and we have from the equality at <math>(*)</math> that:
<div style="text-align: center;"><math>\begin{align} \quad \mathbf{a} \cdot [\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})] = \mathbf{a} \cdot \mathbf{f}'(\mathbf{z})(\mathbf{y} - \mathbf{x}) \quad \blacksquare \end{align}</math></div>

== Licensing ==
Content obtained and/or adapted from:
* [http://mathonline.wikidot.com/the-mean-value-theorem-for-differentiable-functions-from-rn The Mean Value Theorem for Differentiable Functions from Rn to Rm, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 38: / Line 38: @@
 <div style="text-align: center;"><math>\begin{align} \quad F'(t) = \mathbf{a} \cdot \mathbf{f}'(\mathbf{x} + t(\mathbf{y} - \mathbf{x})) (\mathbf{y} - \mathbf{x}) \end{align}</math></div>
-*So by the Mean Value Theorem for single-variable real-valued functions, for <math>x = 0</math> and <math>y = 1</math> there exists a number <math>h \in (0, 1)</math> for which:<div style="text-align: center;"><math>\begin{align} \quad F(1) - F(0) &amp;= F'(h)(1 - 0) \quad (*)\\ \end{align}</math></div>
+*So by the Mean Value Theorem for single-variable real-valued functions, for <math>x = 0</math> and <math>y = 1</math> there exists a number <math>h \in (0, 1)</math> for which:<div style="text-align: center;"><math>\begin{align} \quad F(1) - F(0) = F'(h)(1 - 0) \quad (*)\\ \end{align}</math></div>
 *The lefthand side of <math>(*)</math> is:

← Older revision		Revision as of 20:58, 12 November 2021
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−	~~==The Mean Value Theorem for Differentiable Functions from Rn to Rm==~~

	Recall that if <math>f</math> is a continuous function on the closed interval <math>[x, y]</math> and differentiable on the open interval <math>(x, y)</math> (where we assume <math>x < y</math>) then there exists a number <math>c \in (x, y)</math> for which:		Recall that if <math>f</math> is a continuous function on the closed interval <math>[x, y]</math> and differentiable on the open interval <math>(x, y)</math> (where we assume <math>x < y</math>) then there exists a number <math>c \in (x, y)</math> for which:

Mean-Value Theorems for Vector Valued Functions - Revision history

Khanh at 21:43, 12 November 2021

Khanh at 20:58, 12 November 2021

Khanh: Created page with "==The Mean Value Theorem for Differentiable Functions from Rn to Rm== Recall that if f is a continuous function on the closed interval [x, y] and di..."

Khanh: Created page with "==The Mean Value Theorem for Differentiable Functions from Rn to Rm== Recall that if $f$ is a continuous function on the closed interval $[x, y]$ and di..."