# Difference between revisions of "Extended mean value theorem"

A WASSERMANN (talk | contribs) |
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then there exists a value <math>\xi \in (a,b)</math> such that | then there exists a value <math>\xi \in (a,b)</math> such that | ||

:<math> | :<math> | ||

− | \frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g( | + | \frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}. |

</math> | </math> | ||

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Then there exists a value <math>\xi \in (a,b)</math> such that | Then there exists a value <math>\xi \in (a,b)</math> such that | ||

:<math> | :<math> | ||

− | f'(\xi)\cdot (g(b)-g(a)) = g'(\xi) \cdot (f(b)-f( | + | f'(\xi)\cdot (g(b)-g(a)) = g'(\xi) \cdot (f(b)-f(a)). |

</math> | </math> | ||

This second formulation avoids the need that | This second formulation avoids the need that | ||

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and apply Rolle's theorem. | and apply Rolle's theorem. | ||

− | '''Visualization''' | + | '''Visualization:''' |

The extended mean value theorem says that given the curve | The extended mean value theorem says that given the curve | ||

:<math> C: [a,b]\to\mathbb{R}, \quad t \mapsto (f(t), g(t)) </math> | :<math> C: [a,b]\to\mathbb{R}, \quad t \mapsto (f(t), g(t)) </math> | ||

Line 96: | Line 96: | ||

// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y()) | // 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: | // We can avoid division by zero with the following formulation: | ||

− | |||

var quot = function(t) { | var quot = function(t) { | ||

return df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X()); | return df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X()); |

## Latest revision as of 12:38, 4 February 2019

The *extended mean value theorem* (also called *Cauchy's mean value theorem*) is usually formulated as:

Let

- [math] f, g: [a,b] \to \mathbb{R}[/math]

be continuous functions that are differentiable on the open interval [math](a,b)[/math]. If [math]g'(x)\neq 0[/math] for all [math]x\in(a,b)[/math], then there exists a value [math]\xi \in (a,b)[/math] such that

- [math] \frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}. [/math]

**Remark:**
It seems to be easier to state the extended mean value theorem in the following form:

Let

- [math] f, g: [a,b] \to \mathbb{R}[/math]

be continuous functions that are differentiable on the open interval [math](a,b)[/math]. Then there exists a value [math]\xi \in (a,b)[/math] such that

- [math] f'(\xi)\cdot (g(b)-g(a)) = g'(\xi) \cdot (f(b)-f(a)). [/math]

This second formulation avoids the need that [math]g'(x)\neq 0[/math] for all [math]x\in(a,b)[/math] and is therefore much easier to handle numerically.

The proof is similar, just use the function

- [math] h(x) = f(x)\cdot(g(b)-g(a)) - (g(x)-g(a))\cdot(f(b)-f(a)) [/math]

and apply Rolle's theorem.

**Visualization:**
The extended mean value theorem says that given the curve

- [math] C: [a,b]\to\mathbb{R}, \quad t \mapsto (f(t), g(t)) [/math]

with the above prerequisites for [math]f[/math] and [math]g[/math], there exists a [math]\xi[/math] such that the tangent to the curve in the point [math]C(\xi)[/math] is parallel to the secant through [math]C(a)[/math] and [math]C(b)[/math].

### The underlying JavaScript code

```
var board = JXG.JSXGraph.initBoard('box', {boundingbox: [-5, 10, 7, -6], axis:true});
var p = [];
p[0] = board.create('point', [0, -2], {size:2, name: 'C(a)'});
p[1] = board.create('point', [-1.5, 5], {size:2, name: ''});
p[2] = board.create('point', [1, 4], {size:2, name: ''});
p[3] = board.create('point', [3, 3], {size:2, name: 'C(b)'});
// Curve
var fg = JXG.Math.Numerics.Neville(p);
var graph = board.create('curve', fg, {strokeWidth:3, strokeOpacity:0.5});
// Secant
line = board.create('line', [p[0], p[3]], {strokeColor:'#ff0000', dash:1});
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 = function(t) {
return df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X());
};
var r = board.create('glider', [
function() { return fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
function() { return fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
graph], {name: 'C(ξ)', size: 4, fixed:true, color: 'blue'});
board.create('tangent', [r], {strokeColor:'#ff0000'});
```