This article was Featured Proof between 17 July 2009 and 31 July 2009.
Contents
- 1 Theorem
- 2 Proof
- 2.1 Also presented as
- 2.2 Geometrical Interpretation
- 3 Also known as
- 4 Example
- 5 Also see
- 6 Source of Name
- 7 Sources
Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Suppose:
- $\forall x \in \openint a b: \map {g'} x \ne 0$
Then:
- $\exists \xi \in \openint a b: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f b - \map f a} {\map g b - \map g a}$
Proof
First we check $\map g a \ne \map g b$.
Aiming fora contradiction, suppose $\map g a = \map g b$.
From Rolle's Theorem:
- $\exists \xi \in \openint a b: \map {g'} \xi = 0$.
This contradicts $\forall x \in \openint a b: \map {g'} x \ne 0$.
Thus by Proof by Contradiction $\map g a \ne \map g b$.
Let $h = \dfrac {\map f b - \map f a} {\map g b - \map g a}$.
Let $F$ be the real function defined on $\closedint a b$ by:
- $\map F x = \map f x - h \map g x$.
Then:
| \(\ds \map F b - \map F a\) | \(=\) | \(\ds \paren {\map f b - h \map g b} - \paren {\map f a - h \map g a}\) | as $\map F x = \map f x - h \map g x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\map f b - \map f a} - h \paren {\map g b - \map g a}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map F a\) | \(=\) | \(\ds \map F b\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \xi \in \openint a b: \, \) | \(\ds \map {F'} \xi\) | \(=\) | \(\ds \map {f'} \xi - h \map {g'} \xi\) | Sum Rule for Derivatives, Derivative of Constant Multiple | |||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Rolle's Theorem | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \xi \in \openint a b: \, \) | \(\ds \frac {\map {f'} \xi} {\map {g'} \xi}\) | \(=\) | \(\ds h\) | $\forall x \in \openint a b: \map {g'} x \ne 0$ | |||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\map f b - \map f a} {\map g b - \map g a}\) |
$\blacksquare$
Also presented as
The Cauchy Mean Value Theorem can also be found presented as:
- $\exists \xi \in \openint a b: \map {f'} \xi \paren {\map g b - \map g a} = \map {g'} \xi \paren {\map f b - \map f a}$
Geometrical Interpretation
Consider two functions $\map f x$ and $\map g x$:
- continuous on the closed interval $\closedint a b$
- differentiable on $\openint a b$.
For every $x \in \closedint a b$, we consider the point $\tuple {\map f x, \map g x}$.
If we trace out the points $\tuple {\map f x, \map g x}$ over every $x \in \closedint a b$, we get a curve in two dimensions, as shown in the graph:
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In the drawing, the slope of the red line is $\dfrac {\map g b - \map g a} {\map f b - \map f a}$.
This is because:
- $\dfrac {\Delta y} {\Delta x} = \dfrac {\map g b - \map g a} {\map f b - \map f a}$
assuming that the vertical axis, which contains the value of $\map f x$, is the $y$-axis.
The slope of the green line is $\dfrac {\map {g'} c} {\map {f'} c}$.
This is because:
- $\valueat {\dfrac {\d g} {\d f} } {x \mathop = c} = \valueat {\dfrac {\d g / \d x} {\d f / \d x} } {x \mathop = c} = \dfrac {\map {g'} c} {\map {f'} c}$
The graph illustrates that for the value of $c$ chosen in the graph, the slopes of the red line and green line are the same.
That is:
- $\dfrac {\map g b - \map g a} {\map f b - \map f a} = \dfrac {\map {g'} c} {\map {f'} c}$
Also known as
The Cauchy Mean Value Theorem is also known as the generalized mean value theorem.
Some sources include a possessive apostrophe: Cauchy's Mean Value Theorem
Example
In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of $9.63$ seconds.
By definition, his average speed was the total distance travelled divided by the total time it took:
| \(\ds V_a\) | \(=\) | \(\ds \frac {\map d {t_2} - \map d {t_1} } {t_2 - t_1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 10.384 \ \mathrm {m/s}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 37.38 \ \mathrm {km/h}\) |
The Mean Value Theorem gives:
- $\map {f'} c = \dfrac {\map f b - \map f a} {b - a}$
Hence, at some point Bolt was actually running at the average speed of $37.38 \ \mathrm {km/h}$
Asafa Powell was participating in that same race.
He achieved a time of $11.99 \ \mathrm s = 1.245 \times 9.63 \ \mathrm s$.
So Bolt's average speed was $1.245$ times the average speed of Powell.
The Cauchy Mean Value Theorem gives:
| \(\ds \frac {\map {f'} c} {\map {g'} c}\) | \(=\) | \(\ds \frac {\map f b - \map f a} {\map g b - \map g a}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\dfrac {\map f b - \map f a} {b - a} } {\dfrac {\map g b - \map g a} {b - a} }\) |
Hence, at some point, Bolt was actually running at a speed exactly $1.245$ times that of Powell's.
Also see
- Mean Value Theorem
Source of Name
This entry was named for Augustin Louis Cauchy.
Sources
- 1977:K.G. Binmore: Mathematical Analysis: A Straightforward Approach... (previous)... (next)
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