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Extended mean value theorem
Extended mean value theorem
The*'extended mean value theorem* (also called *Cauchy's mean value theorem*) is usually formulated as:
Let
$$ f, g: [a,b] \to \mathbb{R}$$
be continuous functions that are differentiable in the open interval $(a,b)$.
If $g'(x)\neq 0$ for all $x\in(a,b)$,
then there exists a value $\xi \in (a,b)$ such that
$$
\frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}.
$$
__Remark:__
It seems to be easier to state the extended mean value theorem in the following form:
Let
$$f, g: [a,b] \to \mathbb{R}$$
be continuous functions that are differentiable in the open interval $(a,b)$.
Then there exists a value $\xi \in (a,b)$ such that
$$
f'(\xi)\cdot (g(b)-g(a)) = g'(\xi) \cdot (f(b)-f(a)).
$$
This second formulation avoids the need that
$g'(x)\neq 0$ for all $x\in(a,b)$ and is therefore much easier to
handle numerically.
The proof is similar, just use the function
$$
h(x) = f(x)\cdot(g(b)-g(a)) - (g(x)-g(a))\cdot(f(b)-f(a))
$$
and apply *Rolle's theorem*.
__Visualization:__
The extended mean value theorem says that given the curve
$$C: [a,b]\to\mathbb{R}, \quad t \mapsto (f(t), g(t))$$
with the above prerequisites for $f$ and $g$,
there exists a $\xi$ such that the tangent to the curve in the point $C(\xi)$
is parallel to the line through $C(a)$ and $C(b)$.
// Define the id of your board in BOARDID
const board = JXG.JSXGraph.initBoard(BOARDID, {
boundingbox: [-5, 10, 7, -6],
axis: true
});
// Some initial points
var p = [];
p.push(board.create('point', [0, -2], {size: 2, name: 'C(a)'}));
p.push(board.create('point', [-1.5, 5], {size: 2, name: ''}));
p.push(board.create('point', [1, 4], {size: 2, name: ''}));
p.push(board.create('point', [3, 3], {size: 2, name: 'C(b)'}));
// Lagrange interpolation through the points
var fg = JXG.Math.Numerics.Neville(p);
var graph = board.create('curve', fg, {strokeWidth: 3, strokeOpacity: 0.5});
// Line
var line = board.create('line', [p[0], p[3]], {strokeColor: '#ff0000', dash: 1});
// Derivatives of the curve
var df = JXG.Math.Numerics.D(fg[0]);
var dg = JXG.Math.Numerics.D(fg[1]);
// Usually, the extended mean value theorem is formulated as
// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y())
// We can avoid division by zero with the following formulation:
var quot = (t) => df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X());
// Construct the point C(ξ)
var r = board.create('glider', [
() => fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)),
() => fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)),
graph
], {name: 'C(ξ)', size: 4, fixed: true, color: 'blue'});
// Tangent to C through C(ξ)
board.create('tangent', [r], {strokeColor: '#ff0000'});
This example is licensed under a
Creative Commons Attribution 4.0 International License.
Please note you have to mention
The Center of Mobile Learning with Digital Technology
in the credits.