Extended mean value theorem: Difference between revisions

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


Line 18: Line 18:
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(b)).
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  
Line 29: Line 29:
</math>
</math>
and apply Rolle's theorem.
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>.


<jsxgraph width="600" height="400" box="box">
<jsxgraph width="600" height="400" box="box">
Line 34: Line 42:
var p = [];
var p = [];


p[0] = board.create('point', [0, -2], {size:2});
p[0] = board.create('point', [0, -2], {size:2, name: 'C(a)'});
p[1] = board.create('point', [-1.5, 5], {size:2});
p[1] = board.create('point', [-1.5, 5], {size:2, name: ''});
p[2] = board.create('point', [1, 4], {size:2});
p[2] = board.create('point', [1, 4], {size:2, name: ''});
p[3] = board.create('point', [3, 3], {size:2});
p[3] = board.create('point', [3, 3], {size:2, name: 'C(b)'});


// Curve
// Curve
Line 51: Line 59:
// Usually, the extended mean value theorem is formulated as
// 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())
// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y())
// We can avoid division by zero with that 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());
Line 59: Line 67:
             function() { return fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
             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)); },
             function() { return fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
             graph], {name: '&xi;', size: 4, fixed:true, color: 'blue'});
             graph], {name: 'C(&xi;)', size: 4, fixed:true, color: 'blue'});


board.create('tangent', [r], {strokeColor:'#ff0000'});
board.create('tangent', [r], {strokeColor:'#ff0000'});
Line 70: Line 78:
var p = [];
var p = [];


p[0] = board.create('point', [0, -2], {size:2});
p[0] = board.create('point', [0, -2], {size:2, name: 'C(a)'});
p[1] = board.create('point', [-1.5, 5], {size:2});
p[1] = board.create('point', [-1.5, 5], {size:2, name: ''});
p[2] = board.create('point', [1, 4], {size:2});
p[2] = board.create('point', [1, 4], {size:2, name: ''});
p[3] = board.create('point', [3, 3], {size:2});
p[3] = board.create('point', [3, 3], {size:2, name: 'C(b)'});


// Curve
// Curve
Line 87: Line 95:
// Usually, the extended mean value theorem is formulated as
// 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())
// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y())
// We can avoid division by zero with that 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());
Line 95: Line 103:
             function() { return fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
             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)); },
             function() { return fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
             graph], {name: '', size: 4, fixed:true, color: 'blue'});
             graph], {name: 'C(&xi;)', size: 4, fixed:true, color: 'blue'});


board.create('tangent', [r], {strokeColor:'#ff0000'});
board.create('tangent', [r], {strokeColor:'#ff0000'});

Latest revision as of 11:38, 4 February 2019

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

Let

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

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

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

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

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

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

The proof is similar, just use the function

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

with the above prerequisites for [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math], there exists a [math]\displaystyle{ \xi }[/math] such that the tangent to the curve in the point [math]\displaystyle{ C(\xi) }[/math] is parallel to the secant through [math]\displaystyle{ C(a) }[/math] and [math]\displaystyle{ 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(&xi;)', size: 4, fixed:true, color: 'blue'});

board.create('tangent', [r], {strokeColor:'#ff0000'});