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Mean value theorem

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Theorem in mathematics

For the theorem in harmonic function theory, seeHarmonic function § The mean value property.

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\int _{a}^{b}f'(t)\,dt=f(b)-f(a)

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Inmathematics, themean value theorem (orLagrange's mean value theorem) states, roughly, that for a given planararc between two endpoints, there is at least one point at which thetangent to the arc is parallel to thesecant through its endpoints. It is one of the most important results inreal analysis. This theorem is used to prove statements about a function on aninterval starting from local hypotheses about derivatives at points of the interval.

History

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A special case of thistheorem for inverse interpolation of the sine was first described byParameshvara (1380–1460), from theKerala School of Astronomy and Mathematics inIndia, in his commentaries onGovindasvāmi andBhāskara II.^[1] A restricted form of the theorem was proved byMichel Rolle in 1691; the result was what is now known asRolle's theorem, and was proved only for polynomials, without the techniques of calculus. The mean value theorem in its modern form was stated and proved byAugustin Louis Cauchy in 1823.^[2] Many variations of this theorem have been proved since then.^[3]^[4]

Statement

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The function $f {\displaystyle f}$ attains the slope of the secant between $a {\displaystyle a}$ and $b {\displaystyle b}$ as the derivative at the point $c\in (a,b)$ .

The function $f {\displaystyle f}$ attains the slope of the secant between $a {\displaystyle a}$ and $b {\displaystyle b}$ as the derivative at the point $c\in (a,b)$ .

It is also possible that there are multiple tangents parallel to the secant.

Let $f:[a,b]\to \mathbb {R}$ be acontinuous function on theclosed interval $[a,b]$ , anddifferentiable on theopen interval $(a,b)$ , where $a<b$ . Then there exists some $c {\displaystyle c}$ in $(a,b)$ such that:^[5]

f'(c)={\frac {f(b)-f(a)}{b-a}}.

The mean value theorem is a generalization ofRolle's theorem, which assumes $f(a)=f(b)$ , so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that $f:[a,b]\to \mathbb {R}$ iscontinuous on $[a,b]$ , and that for every $x {\displaystyle x}$ in $(a,b)$ thelimit

\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}

exists as a finite number or equals $\infty$ or $-\infty$ . If finite, that limit equals $f'(x)$ . An example where this version of the theorem applies is given by the real-valuedcube root function mapping $x\mapsto x^{1/3}$ , whosederivative tends to infinity at the origin.

Proof

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The expression ${\textstyle {\frac {f(b)-f(a)}{b-a}}}$ gives theslope of the line joining the points $(a,f(a))$ and $(b,f(b))$ , which is achord of the graph of $f {\displaystyle f}$ , while $f'(x)$ gives the slope of the tangent to the curve at the point $(x,f(x))$ . Thus the mean value theorem says that given any chord of a smooth curve, we can find a point on the curve lying between the end-points of the chord such that the tangent of the curve at that point is parallel to the chord. The following proof illustrates this idea.

Define $g(x)=f(x)-rx$ , where $r {\displaystyle r}$ is a constant. Since $f {\displaystyle f}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , the same is true for $g {\displaystyle g}$ . We now want to choose $r {\displaystyle r}$ so that $g {\displaystyle g}$ satisfies the conditions ofRolle's theorem. Namely

{\begin{aligned}g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\&\iff r(b-a)=f(b)-f(a)\\&\iff r={\frac {f(b)-f(a)}{b-a}}.\end{aligned}}

ByRolle's theorem, since $g {\displaystyle g}$ is differentiable and $g(a)=g(b)$ , there is some $c {\displaystyle c}$ in $(a,b)$ for which $g'(c)=0$ , and it follows from the equality $g(x)=f(x)-rx$ that,

{\begin{aligned}&g'(x)=f'(x)-r\\&g'(c)=0\\&g'(c)=f'(c)-r=0\\&\Rightarrow f'(c)=r={\frac {f(b)-f(a)}{b-a}}\end{aligned}}

Implications

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Theorem 1: Assume that $f {\displaystyle f}$ is a continuous, real-valued function, defined on an arbitrary interval $I {\displaystyle I}$ of the real line. If the derivative of $f {\displaystyle f}$ at everyinterior point of the interval $I {\displaystyle I}$ exists and is zero, then $f {\displaystyle f}$ isconstant on $I {\displaystyle I}$ .

Proof: Assume the derivative of $f {\displaystyle f}$ at everyinterior point of the interval $I {\displaystyle I}$ exists and is zero. Let $(a,b)$ be an arbitrary open interval in $I {\displaystyle I}$ . By the mean value theorem, there exists a point $c {\displaystyle c}$ in $(a,b)$ such that

0=f'(c)={\frac {f(b)-f(a)}{b-a}}.

This implies that $f(a)=f(b)$ . Thus, $f {\displaystyle f}$ is constant on the interior of $I {\displaystyle I}$ and thus is constant on $I {\displaystyle I}$ by continuity. (See below for a multivariable version of this result.)

Remarks:

Only continuity of $f {\displaystyle f}$ , not differentiability, is needed at the endpoints of the interval $I {\displaystyle I}$ . No hypothesis of continuity needs to be stated if $I {\displaystyle I}$ is anopen interval, since the existence of a derivative at a point implies the continuity at this point. (See the sectioncontinuity and differentiability of the articlederivative.)
The differentiability of $f {\displaystyle f}$ can be relaxed toone-sided differentiability, a proof is given in the article onsemi-differentiability.

Theorem 2: If $f'(x)=g'(x)$ for all $x {\displaystyle x}$ in an interval $(a,b)$ of the domain of these functions, then $f-g$ is constant, i.e. $f=g+c$ where $c {\displaystyle c}$ is a constant on $(a,b)$ .

Proof: Let $F(x)=f(x)-g(x)$ , then $F'(x)=f'(x)-g'(x)=0$ on the interval $(a,b)$ , so the above theorem 1 tells that $F(x)=f(x)-g(x)$ is a constant $c {\displaystyle c}$ or $f=g+c$ .

Theorem 3: If $F {\displaystyle F}$ is an antiderivative of $f {\displaystyle f}$ on an interval $I {\displaystyle I}$ , then the most general antiderivative of $f {\displaystyle f}$ on $I {\displaystyle I}$ is $F(x)+c$ where $c {\displaystyle c}$ is a constant.

Proof: It directly follows from the theorem 2 above.

Cauchy's mean value theorem

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Cauchy's mean value theorem, also known as theextended mean value theorem, is a generalization of the mean value theorem.^[6]^[7] It states: if the functions $f {\displaystyle f}$ and $g {\displaystyle g}$ are both continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ , then there exists some $c\in (a,b)$ , such that

(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).

Of course, if $g(a)\neq g(b)$ and $g'(c)\neq 0$ , this is equivalent to:

{\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}.

Geometrically, this means that there is sometangent to the graph of thecurve^[8]

{\begin{cases}[a,b]\to \mathbb {R} ^{2}\\t\mapsto (f(t),g(t))\end{cases}}

which isparallel to the line defined by the points $(f(a),g(a))$ and $(f(b),g(b))$ . However, Cauchy's theorem does not claim the existence of such a tangent in all cases where $(f(a),g(a))$ and $(f(b),g(b))$ are distinct points, since it might be satisfied only for some value $c {\displaystyle c}$ with $f'(c)=g'(c)=0$ , in other words a value for which the mentioned curve isstationary; in such points no tangent to the curve is likely to be defined at all. An example of this situation is the curve given by

t\mapsto \left(t^{3},1-t^{2}\right),

which on the interval $[-1,1]$ goes from the point $(-1,0)$ to $(1,0)$ , yet never has a horizontal tangent; however it has a stationary point (in fact acusp) at $t=0$ .

Cauchy's mean value theorem can be used to proveL'Hôpital's rule. The mean value theorem is the special case of Cauchy's mean value theorem when $g(t)=t$ .

Proof

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The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.

Define $h(x)=(g(b)-g(a))f(x)-(f(b)-f(a))g(x)$ , then we easily see $h(a)=h(b)=f(a)g(b)-f(b)g(a)$ .

Since $f {\displaystyle f}$ and $g {\displaystyle g}$ are continuous on $[a,b]$ and differentiable on $(a,b)$ , the same is true for $h {\displaystyle h}$ . All in all, $h {\displaystyle h}$ satisfies the conditions ofRolle's theorem. Consequently, there is some $c {\displaystyle c}$ in $(a,b)$ for which $h'(c)=0$ . Now using the definition of $h {\displaystyle h}$ we have:

0=h'(c)=(g(b)-g(a))f'(c)-(f(b)-f(a))g'(c)

The result easily follows.

Mean value theorem in several variables

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The mean value theorem generalizes to real functions of multiple variables. The trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.

Let $G {\displaystyle G}$ be an open subset of $\mathbb {R} ^{n}$ , and let $f:G\to \mathbb {R}$ be a differentiable function. Fix points $x,y\in G$ such that the line segment between $x, y {\displaystyle x,y}$ lies in $G {\displaystyle G}$ , and define $g(t)=f{\big (}(1-t)x+ty{\big )}$ . Since $g {\displaystyle g}$ is a differentiable function in one variable, the mean value theorem gives:

g(1)-g(0)=g'(c)

for some $c {\displaystyle c}$ between 0 and 1. But since $g(1)=f(y)$ and $g(0)=f(x)$ , computing $g'(c)$ explicitly we have:

f(y)-f(x)=\nabla f{\big (}(1-c)x+cy{\big )}\cdot (y-x)

where $\nabla$ denotes agradient and $\cdot$ adot product. This is an exact analog of the theorem in one variable (in the case $n=1$ thisis the theorem in one variable). By theCauchy–Schwarz inequality, the equation gives the estimate:

{\Bigl |}f(y)-f(x){\Bigr |}\leq {\Bigl |}\nabla f{\big (}(1-c)x+cy{\big )}{\Bigr |}\ {\Bigl |}y-x{\Bigr |}.

In particular, when $G {\displaystyle G}$ is convex and the partial derivatives of $f {\displaystyle f}$ are bounded, $f {\displaystyle f}$ isLipschitz continuous (and thereforeuniformly continuous).

As an application of the above, we prove that $f {\displaystyle f}$ is constant if the open subset $G {\displaystyle G}$ is connected and every partial derivative of $f {\displaystyle f}$ is 0. Pick some point $x_{0}\in G$ , and let $g(x)=f(x)-f(x_{0})$ . We want to show $g(x)=0$ for every $x\in G$ . For that, let $E=\{x\in G:g(x)=0\}$ . Then $E {\displaystyle E}$ is closed in $G {\displaystyle G}$ and nonempty. It is open too: for every $x\in E$ ,

{\Big |}g(y){\Big |}={\Big |}g(y)-g(x){\Big |}\leq (0){\Big |}y-x{\Big |}=0

for every $y {\displaystyle y}$ in open ball centered at $x {\displaystyle x}$ and contained in $G {\displaystyle G}$ . Since $G {\displaystyle G}$ is connected, we conclude $E=G$ .

The above arguments are made in a coordinate-free manner; hence, they generalize to the case when $G {\displaystyle G}$ is a subset of a Banach space.

Mean value theorem for vector-valued functions

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There is no exact analog of the mean value theorem for vector-valued functions (see below). However, there is an inequality which can be applied to many of the same situations to which the mean value theorem is applicable in the one dimensional case:^[9]

Theorem—For a continuous vector-valued function $\mathbf {f} :[a,b]\to \mathbb {R} ^{k}$ differentiable on $(a,b)$ , there exists a number $c\in (a,b)$ such that

|\mathbf {f} (b)-\mathbf {f} (a)|\leq (b-a)\left|\mathbf {f} '(c)\right|

Proof

Take $\varphi (t)=({\textbf {f}}(b)-{\textbf {f}}(a))\cdot {\textbf {f}}(t)$ . Then $\varphi$ is real-valued and thus, by the mean value theorem,

\varphi (b)-\varphi (a)=\varphi '(c)(b-a)

for some $c\in (a,b)$ . Now, $\varphi (b)-\varphi (a)=|{\textbf {f}}(b)-{\textbf {f}}(a)|^{2}$ and $\varphi '(c)=({\textbf {f}}(b)-{\textbf {f}}(a))\cdot {\textbf {f}}'(c).$ Hence, using theCauchy–Schwarz inequality, from the above equation, we get:

|{\textbf {f}}(b)-{\textbf {f}}(a)|^{2}\leq |{\textbf {f}}(b)-{\textbf {f}}(a)||{\textbf {f}}'(c)|(b-a).

If ${\textbf {f}}(b)={\textbf {f}}(a)$ , the theorem holds trivially. Otherwise, dividing both sides by $|{\textbf {f}}(b)-{\textbf {f}}(a)|$ yields the theorem.

Mean value inequality

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Jean Dieudonné in his classic treatiseFoundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and one cannot find the mean value and in applications one only needs mean inequality.Serge Lang inAnalysis Iuses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative. If one uses theHenstock–Kurzweil integral one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock–Kurzweil integrable.

The reason why there is no analog of mean value equality is the following: Iff :U →R^m is a differentiable function (whereU ⊂Rⁿ is open) and ifx +th,x,h ∈Rⁿ,t ∈ [0, 1] is the line segment in question (lying insideU), then one can apply the above parametrization procedure to each of the component functionsf_i (i = 1, …,m) off (in the above notation sety =x +h). In doing so one finds pointsx +t_ih on the line segment satisfying

f_{i}(x+h)-f_{i}(x)=\nabla f_{i}(x+t_{i}h)\cdot h.

But generally there will not be asingle pointx +t*h on the line segment satisfying

f_{i}(x+h)-f_{i}(x)=\nabla f_{i}(x+t^{*}h)\cdot h.

for allisimultaneously. For example, define:

{\begin{cases}f:[0,2\pi ]\to \mathbb {R} ^{2}\\f(x)=(\cos(x),\sin(x))\end{cases}}

Then $f(2\pi )-f(0)=\mathbf {0} \in \mathbb {R} ^{2}$ , but $f_{1}'(x)=-\sin(x)$ and $f_{2}'(x)=\cos(x)$ are never simultaneously zero as $x {\displaystyle x}$ ranges over $\left[0,2\pi \right]$ .

The above theorem implies the following:

Mean value inequality^[10]—For a continuous function ${\textbf {f}}:[a,b]\to \mathbb {R} ^{k}$ , if ${\textbf {f}}$ is differentiable on $(a,b)$ , then

|{\textbf {f}}(b)-{\textbf {f}}(a)|\leq (b-a)\sup _{(a,b)}|{\textbf {f}}'|

In fact, the above statement suffices for many applications and can be proved directly as follows. (We shall write $f {\displaystyle f}$ for ${\textbf {f}}$ for readability.)

Proof

First assume $f {\displaystyle f}$ is differentiable at $a {\displaystyle a}$ too. If $f^{'} {\displaystyle f'}$ is unbounded on $(a,b)$ , there is nothing to prove. Thus, assume $\sup _{(a,b)}|f'|<\infty$ . Let $M>\sup _{(a,b)}|f'|$ be some real number. Let $E=\{0\leq t\leq 1\mid |f(a+t(b-a))-f(a)|\leq Mt(b-a)\}.$ We want to show $1\in E$ . By continuity of $f {\displaystyle f}$ , the set $E {\displaystyle E}$ is closed. It is also nonempty as $0 {\displaystyle 0}$ is in it. Hence, the set $E {\displaystyle E}$ has the largest element $s {\displaystyle s}$ . If $s=1$ , then $1\in E$ and we are done. Thus suppose otherwise. For $1>t>s$ ,

{\begin{aligned}&|f(a+t(b-a))-f(a)|\\&\leq |f(a+t(b-a))-f(a+s(b-a))-f'(a+s(b-a))(t-s)(b-a)|+|f'(a+s(b-a))|(t-s)(b-a)\\&+|f(a+s(b-a))-f(a)|.\end{aligned}}

Let $\epsilon >0$ be such that $M-\epsilon >\sup _{(a,b)}|f'|$ . By the differentiability of $f {\displaystyle f}$ at $a+s(b-a)$ (note $s {\displaystyle s}$ may be 0), if $t {\displaystyle t}$ is sufficiently close to $s {\displaystyle s}$ , the first term is $\leq \epsilon (t-s)(b-a)$ . The second term is $\leq (M-\epsilon )(t-s)(b-a)$ . The third term is $\leq Ms(b-a)$ . Hence, summing the estimates up, we get: $|f(a+t(b-a))-f(a)|\leq tM|b-a|$ , a contradiction to the maximality of $s {\displaystyle s}$ . Hence, $1=s\in M$ and that means:

|f(b)-f(a)|\leq M(b-a).

Since $M {\displaystyle M}$ is arbitrary, this then implies the assertion. Finally, if $f {\displaystyle f}$ is not differentiable at $a {\displaystyle a}$ , let $a'\in (a,b)$ and apply the first case to $f {\displaystyle f}$ restricted on $[a',b]$ , giving us:

|f(b)-f(a')|\leq (b-a')\sup _{(a,b)}|f'|

since $(a',b)\subset (a,b)$ . Letting $a'\to a$ finishes the proof.

Cases where the theorem cannot be applied

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All conditions for the mean value theorem are necessary:

${\boldsymbol {f(x)}}$ is differentiable on ${\boldsymbol {(a,b)}}$
${\boldsymbol {f(x)}}$ is continuous on ${\boldsymbol {[a,b]}}$
${\boldsymbol {f(x)}}$ is real-valued

When one of the above conditions is not satisfied, the mean value theorem is not valid in general, and so it cannot be applied.

The necessity of the first condition can be seen by the counterexample where the function $f(x)=|x|$ on [-1,1] is not differentiable.

The necessity of the second condition can be seen by the counterexample where the function $f(x)={\begin{cases}1,&{\text{at }}x=0\\0,&{\text{if }}x\in (0,1]\end{cases}}$ satisfies criteria 1 since $f'(x)=0$ on $(0,1)$ but not criteria 2 since ${\frac {f(1)-f(0)}{1-0}}=-1$ and $-1\neq 0=f'(x)$ for all $x\in (0,1)$ so no such $c {\displaystyle c}$ exists.

The theorem is false if a differentiable function is complex-valued instead of real-valued. For example, if $f(x)=e^{xi}$ for all real $x {\displaystyle x}$ , then $f(2\pi )-f(0)=0=0(2\pi -0)$ while $f'(x)\neq 0$ for any real $x {\displaystyle x}$ .

Mean value theorems for definite integrals

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First mean value theorem for definite integrals

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Geometrically: interpretingf(c) as the height of a rectangle andb –a as the width, this rectangle has the same area as the region below the curve froma tob^[11]

Letf : [a,b] →R be a continuous function. Then there existsc in (a,b) such that

\int _{a}^{b}f(x)\,dx=f(c)(b-a).

This follows at once from thefundamental theorem of calculus, together with the mean value theorem for derivatives. Since the mean value off on [a,b] is defined as

{\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx,

we can interpret the conclusion asf achieves its mean value at somec in (a,b).^[12]

In general, iff : [a,b] →R is continuous andg is an integrable function that does not change sign on [a,b], then there existsc in (a,b) such that

\int _{a}^{b}f(x)g(x)\,dx=f(c)\int _{a}^{b}g(x)\,dx.

Second mean value theorem for definite integrals

[edit]

There are various slightly different theorems called thesecond mean value theorem for definite integrals. A commonly found version is as follows:

G:[a,b]\to \mathbb {R}

is a positivemonotonically decreasing function and

\varphi :[a,b]\to \mathbb {R}

is an integrable function, then there exists a numberx in (a,b] such that

\int _{a}^{b}G(t)\varphi (t)\,dt=G(a^{+})\int _{a}^{x}\varphi (t)\,dt.

Here $G(a^{+})$ stands for ${\textstyle {\lim _{x\to a^{+}}G(x)}}$ , the existence of which follows from the conditions. Note that it is essential that the interval (a,b] containsb. A variant not having this requirement is:^[13]

G:[a,b]\to \mathbb {R}

is amonotonic (not necessarily decreasing and positive) function and

\varphi :[a,b]\to \mathbb {R}

is an integrable function, then there exists a numberx in (a,b) such that

\int _{a}^{b}G(t)\varphi (t)\,dt=G(a^{+})\int _{a}^{x}\varphi (t)\,dt+G(b^{-})\int _{x}^{b}\varphi (t)\,dt.

If the function $G {\displaystyle G}$ returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of $G {\displaystyle G}$ is also multi-dimensional.

For example, consider the following 2-dimensional function defined on an $n {\displaystyle n}$ -dimensional cube:

{\begin{cases}G:[0,2\pi ]^{n}\to \mathbb {R} ^{2}\\G(x_{1},\dots ,x_{n})=\left(\sin(x_{1}+\cdots +x_{n}),\cos(x_{1}+\cdots +x_{n})\right)\end{cases}}

Then, by symmetry it is easy to see that the mean value of $G {\displaystyle G}$ over its domain is (0,0):

\int _{[0,2\pi ]^{n}}G(x_{1},\dots ,x_{n})dx_{1}\cdots dx_{n}=(0,0)

However, there is no point in which $G=(0,0)$ , because $|G|=1$ everywhere.

Generalizations

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Linear algebra

[edit]

Assume that $f, g, {\displaystyle f,g,}$ and $h {\displaystyle h}$ are differentiable functions on $(a,b)$ that are continuous on $[a,b]$ . Define

D(x)={\begin{vmatrix}f(x)&g(x)&h(x)\\f(a)&g(a)&h(a)\\f(b)&g(b)&h(b)\end{vmatrix}}

There exists $c\in (a,b)$ such that $D'(c)=0$ .

Notice that

D'(x)={\begin{vmatrix}f'(x)&g'(x)&h'(x)\\f(a)&g(a)&h(a)\\f(b)&g(b)&h(b)\end{vmatrix}}

and if we place $h(x)=1$ , we getCauchy's mean value theorem. If we place $h(x)=1$ and $g(x)=x$ we getLagrange's mean value theorem.

The proof of the generalization is quite simple: each of $D(a)$ and $D(b)$ aredeterminants with two identical rows, hence $D(a)=D(b)=0$ . The Rolle's theorem implies that there exists $c\in (a,b)$ such that $D'(c)=0$ .

Probability theory

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LetX andY be non-negativerandom variables such that E[X] < E[Y] < ∞ and $X\leq _{st}Y$ (i.e.X is smaller thanY in theusual stochastic order). Then there exists an absolutely continuous non-negative random variableZ havingprobability density function

f_{Z}(x)={\Pr(Y>x)-\Pr(X>x) \over {\rm {E}}[Y]-{\rm {E}}[X]}\,,\qquad x\geqslant 0.

Letg be ameasurable anddifferentiable function such that E[g(X)], E[g(Y)] < ∞, and let its derivativeg′ be measurable andRiemann-integrable on the interval [x,y] for ally ≥x ≥ 0. Then, E[g′(Z)] is finite and^[14]

{\rm {E}}[g(Y)]-{\rm {E}}[g(X)]={\rm {E}}[g'(Z)]\,[{\rm {E}}(Y)-{\rm {E}}(X)].

Complex analysis

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As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:^[15]

Letf : Ω →C be aholomorphic function on the open convex set Ω, and leta andb be distinct points in Ω. Then there exist pointsu,v on the interior of the line segment froma tob such that

\operatorname {Re} (f'(u))=\operatorname {Re} \left({\frac {f(b)-f(a)}{b-a}}\right),

\operatorname {Im} (f'(v))=\operatorname {Im} \left({\frac {f(b)-f(a)}{b-a}}\right).

Where Re() is the real part and Im() is the imaginary part of a complex-valued function.

Notes

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^J. J. O'Connor and E. F. Robertson (2000).Paramesvara,MacTutor History of Mathematics archive.
^Ádám Besenyei."Historical development of the mean value theorem"(PDF).
^Lozada-Cruz, German (2020-10-02)."Some variants of Cauchy's mean value theorem".International Journal of Mathematical Education in Science and Technology.51 (7):1155–1163.Bibcode:2020IJMES..51.1155L.doi:10.1080/0020739X.2019.1703150.ISSN 0020-739X.S2CID 213335491.
^Sahoo, Prasanna. (1998).Mean value theorems and functional equations. Riedel, T. (Thomas), 1962-. Singapore: World Scientific.ISBN 981-02-3544-5.OCLC 40951137.
^Rudin 1976, p. 108.
^Weisstein, Eric W."Extended Mean-Value Theorem".MathWorld.
^Rudin 1976, pp. 107–108.
^"Cauchy's Mean Value Theorem".Math24. Retrieved2018-10-08.
^Rudin 1976, p. 113.
^Hörmander 2015, Theorem 1.1.1. and remark following it.
^"Mathwords: Mean Value Theorem for Integrals".www.mathwords.com.
^Michael Comenetz (2002).Calculus: The Elements. World Scientific. p. 159.ISBN 978-981-02-4904-5.
^Hobson, E. W. (1909)."On the Second Mean-Value Theorem of the Integral Calculus".Proc. London Math. Soc.S2–7 (1):14–23.Bibcode:1909PLMS...27...14H.doi:10.1112/plms/s2-7.1.14.MR 1575669.
^Di Crescenzo, A. (1999). "A Probabilistic Analogue of the Mean Value Theorem and Its Applications to Reliability Theory".J. Appl. Probab.36 (3):706–719.doi:10.1239/jap/1032374628.JSTOR 3215435.S2CID 250351233.
^1 J.-Cl. Evard, F. Jafari, A Complex Rolle’s Theorem, American Mathematical Monthly, Vol. 99, Issue 9, (Nov. 1992), pp. 858-861.

References

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Rudin, Walter (1976).Principles of Mathematical Analysis. Auckland: McGraw-Hill Publishing Company.ISBN 978-0-07-085613-4.
Hörmander, Lars (2015),The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, Classics in Mathematics (2nd ed.), Springer,ISBN 9783642614972

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Steinmetz solid

v t e Joseph-Louis Lagrange
Lagrange multiplier Lagrange polynomial Lagrange's four-square theorem Lagrange's theorem (group theory) Lagrange's identity Lagrange's identity (boundary value problem) Lagrange's trigonometric identities Lagrangian mechanics Lagrange's mean value theorem Lagrange stability Lagrange point