nyuuzyou/wikis huggingface dataset:
McEliece's Theorem (Integer Functions)

== Theorem ==
Let $f: \R \to \R$ be a continuous, strictly increasing real function defined on an interval $A$.
Let:
:$\forall x \in A: \floor x \in A \text { and } \ceiling x \in A$
where:
:$\floor x$ denotes the floor of $x$
:$\ceiling x$ denotes the ceiling of $x$
Then:
:$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z} \iff \floor {\map f x} = \floor {\map f {\floor x} }$
:$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z} \iff \ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$
== Proof ==
Let $x \in A$.
Hence we have that both $\floor x \in A$ and $\ceiling x \in A$.
=== Necessary Condition ===
Let $\map f x \in \Z$.
Let:
:$\floor {\map f x} = \floor {\map f {\floor x} }$
Then:
 $x \notin \Z$.
Thus by Proof by Contradiction:
:$x \in \Z$
Similarly, let:
:$\ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$
Then:
 $x \notin \Z$.
Thus by Proof by Contradiction:
:$x \in \Z$
Thus:
:$\forall x \in A: \floor {\map f x} = \floor {\map f {\floor x} } \implies \paren {\map f x \in \Z \implies x \in \Z}$
and:
:$\forall x \in A: \ceiling {\map f x} = \ceiling {\map f {\ceiling x} } \implies \paren {\map f x \in \Z \implies x \in \Z}$
=== Sufficient Condition ===
Let $f$ be such that:
:$\map f x \in \Z \implies x \in \Z$
 there exists $x \in A$ such that:
:$\floor {\map f x} \ne \floor {\map f {\floor x} }$
We have by definition of the floor function that $x \ge \floor x$.
As $f$ is strictly increasing, it cannot be the case that $\floor {\map f x} < \floor {\map f {\floor x} }$.
So it must be that:
:$\floor {\map f {\floor x} } < \floor {\map f x}$
Because $f$ is continuous:
:$\exists y \in A: \floor x < y \le x$
such that $\map f y \in \Z$
But by definition of the floor function, $y$ cannot be an integer.
Thus by Proof by Contradiction:
:$\paren {\map f x \in \Z \implies x \in \Z} \implies \floor {\map f x} = \floor {\map f {\floor x} }$
, similarly, that there exists $x \in A$ such that:
:$\ceiling {\map f x} \ne \ceiling {\map f {\ceiling x} }$
We have by definition of the ceiling function that $x \le \ceiling x$.
As $f$ is strictly increasing, it cannot be the case that $\ceiling {\map f x} > \ceiling {\map f {\ceiling x} }$.
So it must be that:
:$\ceiling {\map f {\ceiling x} } > \ceiling {\map f x}$
Because $f$ is continuous:
:$\exists y \in A: \ceiling x > y \ge x$
such that $\map f y \in \Z$
But by definition of the ceiling function, $y$ cannot be an integer.
Thus by Proof by Contradiction:
:$\paren {\map f x \in \Z \implies x \in \Z} \implies \ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$
Thus we have:
:$\paren {\map f x \in \Z \implies x \in \Z} \iff \floor {\map f x} = \floor {\map f {\floor x} }$
and
:$\paren {\map f x \in \Z \implies x \in \Z} \iff \ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$
Hence the result.
== Historical Note ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $34$ (Answers to Exercises)
Category:Floor Function
Category:Ceiling Function
McEliece's Theorem (Integer Functions)/Historical Note

== Historical Note on McEliece's Theorem (Integer Functions) ==
 reports in that this generalisation of Conditions for $\floor {\log_b x}$ to equal $\floor {\log_b \floor x}$ was established by .
 refers to it (in passing) as McEliece's Theorem.
Whiile this name for it is not backed up in the literature, it is convenient for , because of its unwieldy nature, to refer to it thus.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $34$ (Answers to Exercises)
Category:Historical Notes
McNemar's Test/Examples/Political Allegiance

== Example of Use of McNemar's Test ==
A sample of voters in an electorate are given the chance of voting for party $\text A$ or party $\text B$.
Their party allegiance is determined both before and after they have been shown a party political broadcast on behalf of each party.
After the broadcast, their allegiance is once more determined.
McNemar's test can then be used to indicate whether the proportion changing from $\text A$ to party $\text B$ is significantly different from that changing from $\text B$ to party $\text A$.
== Sources ==
* 
* 
Category:Examples of McNemar's Test
Meager Sets in Arens-Fort Space

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $A \subseteq S$.
Then $A$ is meager $A = \set {\tuple {0, 0} }$.
== Proof ==
First let $A = \set {\tuple {0, 0} }$.
From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is closed because $S \setminus \set {\tuple {0, 0} }$ is open.
From Closed Set Equals its Closure:
:$\set {\tuple {0, 0} }^- = \set {\tuple {0, 0} }$
where $\set {\tuple {0, 0} }^-$ denotes the closure of $\set {\tuple {0, 0} }$.
From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is not open.
Therefore the smallest open set contained in $\set {\tuple {0, 0} }$ is $\O$.
Hence:
:$\set {\tuple {0, 0} }^\circ = \O$ 
where $\set {\tuple {0, 0} }^\circ$ denotes the interior of $\set {\tuple {0, 0} }$.
Thus we have that:
:$\paren {\set {\tuple {0, 0} }^-}^\circ = \O$
and so by definition $\set {\tuple {0, 0} }$ is nowhere dense in $T$.
From Union of Singleton it trivially follows that $\set {\tuple {0, 0} }$ is the union of a countable set of subsets of $S$ (that is: just the one) which are nowhere dense in $S$.
Hence by definition $\set {\tuple {0, 0} }$ is meager.
Now assume $A \ne \set {\tuple {0, 0} }$.
Then $\exists x \in A: x \ne \tuple {0, 0}$. 
By definition of the Arens-Fort space, $\set x$ is open in $T$. 
Thus, from Interior of Open Set $\set x^\circ = \set x$.
Hence $\set x \subseteq \paren {\set x^-}^\circ$ and by definition is not nowhere dense in $T$.
So $A$ is not the union of a countable set of subsets of $S$ which are nowhere dense in $S$.
So, by definition, $A$ is not meager.
Category:Meager Spaces
Category:Arens-Fort Space
Mean Deviation about Mean equals Zero

== Theorem ==
Let $S$ be an object upon which an arithmetic mean and a mean deviation $\map D S$ are both defined.
Let $\map D S$ be the mean deviation about the arithmetic mean $\bar x$ of $S$.
Then:
:$\map D S = 0$
== Proof ==
== Sources ==
* 
* 
Category:Mean Deviation
Category:Arithmetic Mean
Mean Deviation about Median equals Zero iff Median equals Mean

== Theorem ==
Let $S$ be an object upon which an arithmetic mean, a median, and a mean deviation $\map D S$ are both defined.
Let $\map D S$ be the mean deviation about the median $m$ of $S$.
Then:
:$\map D S = 0$
 the median of $S$ equals the arithmetic mean of $S$.
== Proof ==
== Sources ==
* 
* 
Category:Mean Deviation
Category:Arithmetic Mean
Category:Medians
Mean Distance between Two Random Points in Cuboid

== Theorem ==
Let $B$ be a cuboid in the Cartesian $3$-space $\R^3$ as:
:$\size x \le a$, $\size y \le b$, $\size z \le c$
Let $E$ denote the mean distance $D$ between $2$ points chosen at random from the interior of $B$.
Then:
where:
== Proof ==
Let $X_1$, $X_2$; $Y_1$, $Y_2$; $Z_1$, $Z_2$ be pairs of independent random variables with a continuous uniform distribution in $\closedint {-a} a$, $\closedint {-b} b$ and $\closedint {-c} c$.
The random variables $U = \size {X_1 - X_2}$, $V = \size {Y_1 - Y_2}$, $W = \size {Z_1 - Z_2}$ are likewise independent.
Thus the probability density functions $\map f U$, $\map g V$, $\map h W$ can be shown to be:
Hence the sought expectation is given by:
:$E = \ds \dfrac 1 {8 a^2 b^2 c^2} \int_0^{2 c} \int_0^{2 b} \int_0^{2 a} \sqrt {u^2 + v^2 + v^2} \paren {2 a - u} \paren {2 b - v} \paren {2 c - w} \rd u \rd v \rd w$
Let $\map P {a, b, c}$ be the pyramid defined by the planes $u = 2 a$, $v = 0$, $w = 0$, $a v = b u$ and $a w = c u$.
Set:
:$\map F {a, b, c} = \ds \dfrac 1 {8 a^2 b^2 c^2} \iiint_{\map P {a, b, c} } \sqrt {u^2 + v^2 + v^2} \paren {2 a - u} \paren {2 b - v} \paren {2 c - w} \rd u \rd v \rd w$
By symmetry:
:$\map F {a, b, c} = \map F {a, c, b}$
and so:
:$(1): \quad E = \dfrac 1 {8 a^2 b^2 c^2} \paren {\map F {a, b, c} + \map F {c, a, b} + \map F {b, c, a} }$
All that remains is to calculate $\map F {a, b, c}$.
Using spherical coordinates we obtain:
$\map F {a, b, c} = \ds \int_0^{\arctan \frac b a} \int_{\arccot \frac c a \cos \theta}^{\frac \pi 2} \int_0^{2 a \cosec \phi \sec \theta} \Phi \rd \rho \rd \phi \rd \theta$
where:
:$\Phi = \rho^3 \paren {2 a - \rho \sin \phi \cos \theta} \paren {2 b - \rho \sin \phi \sin \theta} \paren {2 c - \rho \cos \phi} \sin \phi$
After integration:
where:
Substituting in $(1)$ we obtain:
== Sources ==
* 
* 
Category:Solid Geometry
Mean Distance between Two Random Points in Unit Cube

== Theorem ==
The mean distance $R$ between $2$ points chosen at random from the interior of a unit cube is given by:
The value $R$ is known as the Robbins constant.
== Proof ==
From Mean Distance between Two Random Points in Cuboid:
The result follows by setting $a = b = c = \dfrac 1 2$.
Hence we have:
and:
So:
:$r - r_1 = r - r_2 = r - r_3 = \dfrac {\sqrt 3 - \sqrt 2} 2$
Thus:
== Sources ==
* : $0,66170 7 \ldots$
* 
Category:Solid Geometry
Mean Ergodic Theorem

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Let $\map {L^2_\C} \mu$ be the complex-valued $L^2$ space of $\mu$.
Let $U_T : \map {L^2_\C} \mu \to \map {L^2_\C} \mu$ of $T$ be the Koopman operator.
Let $I := \set {f \in \map {L^2_\C} \mu : \map {U_T} f = f}$.
Then for each $f \in \map {L^2_\C} \mu$:
:$\ds \lim_{N \mathop \to \infty} \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} \map {U_T^n} f = \map {P_T} f$ converges in the $L^2$-norm
where:
:$U_T^n$ denotes the $n$ times composition of $U_T$
:$P_T: \map {L^2_\C} \mu \to \map {L^2_\C} \mu$ denotes the orthogonal projection on the closed linear subspace $I$.
== Proof ==
Recall $L^2$ space forms Hilbert space.
That is, $\map {L^2_\C} \mu$ is a Hilbert space with the $L^2$ inner product:
:$\ds \innerprod f g := \int f \; \overline g \rd \mu$
Let $\norm \cdot$ be the norm of $\map {L^2_\C} \mu$.
By Koopman Operator is Isometry:
:$\forall f \in \map {L^2_\C} \mu : \norm {\map U f} = \norm f$
Therefore, the result follows from Mean Ergodic Theorem (Hilbert Space).
== Sources ==
* $2.5$ The Mean Ergodic Theorem
Category:Mean Ergodic Theorem
Category:Ergodic Theory
Category:Operator Theory
Mean Ergodic Theorem (Hilbert Space)

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F$.
Let $U : \HH \to \HH$ be a bounded linear operator such that:
:$\forall f \in \HH : \norm {\map U f} \le \norm f$
Then for each $f \in \HH$:
:$\ds \lim_{N \mathop \to \infty} \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} \map {U^n} f = \map P f$
where:
:$U^n$ denotes the $n$ times composition of $U$
:$I := \set {f \in \HH : \map U f = f}$
:$P : \HH \to I$ denotes the orthogonal projection on $I$
== Proof ==
Note that $I$ is a closed linear subspace of $\HH$, since $U$ is bounded.
Especially, $P : \HH \to I$ is well-defined.
Moreover, by Direct Sum of Subspace and Orthocomplement:
:$\HH = I \oplus I^\perp$
Let $f \in \HH$.
We can write:
:$ f = \map P f + f^\perp$
where $f^\perp \in I^\perp$.
Then we have:
Thus we need to show:
:$\ds \lim_{N \mathop \to \infty} \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} \map {U^n} {f^\perp} = 0$
=== Lemma ===
Let $\epsilon > 0$ be arbitrary.
Since $f^\perp \in \overline B$, there is a $g \in B$ such that:
:$\norm {f^\perp - g} < \epsilon$
where $g = \map U h - h$ for an $h \in \HH$.
Thus for all $N \ge 2 \norm h \epsilon^{-1}$:
Category:Mean Ergodic Theorem
Mean Ergodic Theorem (Hilbert Space)/Lemma

== Lemma ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F$.
Let $U : \HH \to \HH$ be a bounded linear operator such that:
:$\forall f \in \HH : \norm {\map U f} \le \norm f$
Let $B \subseteq \HH$ be the linear subspace defined as:
:$B := \set {\map U h - h : h \in \HH }$
Then:
:$I^\perp \subseteq \overline B$
== Proof ==
It suffice to show $B^\perp \subseteq I$.
Indeed, then it follows:
Recall that the adjoint $U^\ast$ of $U$ exists by Existence and Uniqueness of Adjoint.
Let $f \in B^\perp$.
Then $\map U f = f$, as:
Therefore $\map U f = f$, as:
Category:Mean Ergodic Theorem
Mean Number of Elements Fixed by Self-Map

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be a finite set of cardinality $n$.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\map \mu n$ denote the arithmetic mean of the number of fixed points of all the mappings in $S^S$.
Then:
:$\map \mu n = 1$
== Proof ==
Let $f \in S^S$ be an arbitrary mapping from $S$ to itself.
Let $s \in S$ be an arbitrary element of $S$.
$s$ has an equal probability of being mapped to any element of $S$.
Hence the probability that $\map f s = s$ is equal to $\dfrac 1 n$.
There are $n$ elements of $S$.
By the above argument, each one has a probability of $\dfrac 1 n$ that it is a fixed point.
Thus the expectation of the number of fixed points is $n \times \dfrac 1 n = 1$.
== Examples ==
== Sources ==
* : Expectations: $104$
* : The misaddressed letters: $131$
Category:Combinatorics
Category:Probability Theory
Category:Mean Number of Elements Fixed by Self-Map
Mean Number of Elements Fixed by Self-Map/Examples/7

== Example of Mean Number of Elements Fixed by Self-Map ==
$7$ letters are placed at random into $7$ addressed envelopes, not necessarily making sure that one envelope contains only one letter.
How many letters, on average, can be expected to be in its correct envelope?
== Proof ==
This is an instance of Mean Number of Elements Fixed by Self-Map.
Hence we can expect, on average, $1$ letter to be in its correct envelope.
== Sources ==
* : The misaddressed letters: $131$
Category:Mean Number of Elements Fixed by Self-Map
Mean Squared Error equals Variance for Unbiased Estimator

== Theorem ==
Let $T$ be an unbiased estimator for a population parameter of a population $P$.
Then the mean squared error for $T$ equals the variance of $T$.
== Proof ==
Let $T$ be a general estimator.
From Mean Squared Error for Biased Estimator:
:$M = \var T + \paren {\map B T}^2$
where:
:$\var T$ denotes the variance of $T$
:$\map B T$ denotes the bias on $T$.
By definition, if $T$ is unbiased:
:$\map B T = 0$
The result follows.
== Sources ==
* 
* 
Category:Mean Squared Error
Category:Variance
Mean Squared Error for Biased Estimator

== Theorem ==
Let $T$ be an estimator for a population parameter of a population $P$.
Let $M$ be the mean squared error for $T$.
Then:
:$M = \var T + \paren {\map B T}^2$
where:
:$\var T$ denotes the variance of $T$
:$\map B T$ denotes the bias on $T$.
== Proof ==
== Sources ==
* 
* 
Category:Mean Squared Error
Category:Variance
Mean Value Theorem

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Then:
:$\exists \xi \in \openint a b: \map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$
== Proof 1 ==
:
 
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
== Also known as ==
== Examples ==
== Also see ==
* Mean Value Theorem for Integrals
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean
* 
* 
* 
Category:Mean Value Theorem
Category:Differential Calculus
Category:Named Theorems
Mean Value Theorem/Examples/x^3/Formulation 1

== Example of Use of Mean Value Theorem ==
Let $f$ be the real function defined as:
:$\map f x = x^3$
Let:
:$a = 1$, $b = 2$
Then when $\xi = \sqrt {\dfrac 7 3}$:
:$\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$
== Proof ==
We have:
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean: Example $\text G$
Category:Mean Value Theorem
Mean Value Theorem/Examples/x^3/Formulation 2

== Example of Use of Mean Value Theorem ==
Let $f$ be the real function defined as:
:$\map f x = x^3$
Let:
:$c = 2$, $h = -1$
Then when $\theta = 2 - \sqrt {\dfrac 7 3}$:
:$\map {f'} {c + \theta h} = \dfrac {\map f {c + h} - \map f c} h$
== Proof ==
We have:
and we see that:
:$1 < 2 - \sqrt {\dfrac 7 3} < 2$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean: Example $\text G$
Category:Mean Value Theorem
Mean Value Theorem/Proof 1

== Theorem ==
== Proof ==
For any constant $h \in \R$ we may construct the real function defined on $\closedint a b$ by:
:$\map F x = \map f x + h x$
We have that $h x$ is continuous on $\closedint a b$ from Linear Function is Continuous.
From the Sum Rule for Continuous Real Functions, $F$ is continuous on $\closedint a b$ and differentiable on $\openint a b$.
Let us calculate what the constant $h$ has to be such that $\map F a = \map F b$:
Since $F$ satisfies the conditions for the application of Rolle's Theorem:
:$\exists \xi \in \openint a b: \map {F'} \xi = 0$
But then:
:$\map {F'} \xi = \map {f'} \xi + h = 0$
The result follows.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean
* : $\S 11.6$
Category:Mean Value Theorem
Mean Value Theorem/Proof 2

== Theorem ==
== Proof ==
Let $g : \closedint a b \to \R$ be a real function with: 
:$\map g x = x$
for all $x \in \closedint a b$. 
By Power Rule for Derivatives, we have: 
:$g$ is differentiable with $\map {g'} x = 1$ for all $x \in \closedint a b$.
Note that in particular: 
:$\map {g'} x \ne 0$ for all $x \in \openint a b$.
Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, we can apply the Cauchy Mean Value Theorem.
We therefore have that there exists $\xi \in \openint a b$ such that: 
:$\dfrac {\map {f'} \xi} {\map {g'} \xi } = \dfrac {\map f b - \map f a} {\map g b - \map g a}$
Note that: 
:$\map {g'} \xi = 1$
and: 
:$\map g b - \map g a = b - a$
so this can be rewritten: 
:$\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$
Category:Mean Value Theorem
Mean Value Theorem for Holomorphic Functions

== Theorem ==
Let $D$ be a region.
Let $f : D \to \C$ be a holomorphic function.
Let $z \in D$. 
Let $r$ be such that $\map {B_r} z \subseteq D$. 
Then:
:$\ds \map f z = \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$
== Proof ==
By Cauchy's Integral Formula, we have: 
:$\ds \map f z = \frac 1 {2 \pi i} \oint_{\partial \map {B_r} z} \frac {\map f t} {t - z} \rd t$
where $\partial \map {B_r} z$ is the boundary of $\map {B_r} z$. 
That is, $\partial \map {B_r} z$ is the circle of radius $r$, centred at $z$.
Note that we can parameterise $\partial \map {B_r} z$ by the function $\gamma : \hointr 0 {2 \pi} \to \partial \map {B_r} z$ defined by: 
:$\map \gamma \theta = z + r e^{i \theta}$
for each $\theta \in \hointr 0 {2 \pi}$.
Then, by the definition of a contour integral, we have: 
Category:Mean Value Theorem
Category:Holomorphic Functions
Mean Value Theorem for Integrals

== Theorem ==
Let $f$ be a continuous real function on the closed interval $\closedint a b$.
Then there exists a real number $k \in \closedint a b$ such that:
:$\ds \int_a^b \map f x \rd x = \map f k \paren {b - a}$
=== Generalization ===
== Proof 1 ==
 
== Proof 2 ==
== Also known as ==
== Also see ==
* Definition:Mean Value of Function
* Darboux's Theorem
* Mean Value Theorem
== Sources ==
* : $\S 15$: General Formulas involving Definite Integrals: $15.12$
* 
* : $\S 1.4, \ \S 4.4$
* 
* : $\S 18$: Definite Integrals: General Formulas involving Definite Integrals: $18.12$
Category:Mean Value Theorem for Integrals
Category:Definite Integrals
Category:Named Theorems
Mean Value Theorem for Integrals/Generalization

== Theorem ==
Let $f$ and $g$ be continuous real functions on the closed interval $\closedint a b$ such that:
:$\forall x \in \closedint a b: \map g x \ge 0$
Then there exists a real number $k \in \closedint a b$ such that:
:$\ds \int_a^b \map f x \map g x \rd x = \map f k \int_a^b \map g x \rd x$
== Proof ==
Let:
:$\ds \int_a^b \map g x \rd x = 0$
We are given that:
:$\forall x \in \closedint a b: \map g x \ge 0$
Hence by Continuous Non-Negative Real Function with Zero Integral is Zero Function:
:$\forall x \in \closedint a b: \map g x = 0$
Hence:
:$\ds \int_a^b \map f x \cdot 0 \rd x = \map f k \cdot 0$
and so the result holds for any choice of $k$.
Let:
:$\ds \int_a^b \map g x \rd x \ne 0$
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that:
:$\ds \map f m = \min_{x \mathop \in \closedint a b} \map f x$
:$\ds \map f M = \max_{x \mathop \in \closedint a b} \map f x$
Then the following inequality holds for all $x$ in $\closedint a b$:
:$\map f m \le \map f x \le \map f M$
Multiplying by $\map g x$, and using that:
:$\forall x \in \closedint a b: \map g x \ge 0$
we get:
:$\map f m \map g x \le \map f x \map g x \le \map f M \map g x$
Integrating from $a$ to $b$ gives:
:$\ds \int_a^b \map f m \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \int_a^b \map f M \map g x \rd x$
By Linear Combination of Definite Integrals:
:$\ds \map f m \int_a^b \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \map f M \int_a^b \map g x \rd x$
Dividing by $\ds \int_a^b \map g x \rd x$ gives:
:$\ds \map f m \le \dfrac {\int_a^b \map f x \map g x \rd x} {\int_a^b \map g x \rd x} \le \map f M $
By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that:
== Sources ==
* : $\S 15$: General Formulas involving Definite Integrals: $15.13$
* : $\S 18$: Definite Integrals: General Formulas involving Definite Integrals: $18.13$
Category:Definite Integrals
Category:Named Theorems
Mean Value Theorem for Integrals/Proof 1

== Theorem ==
== Proof ==
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that:
:$\ds \map f m = \min_{x \mathop \in \closedint a b} \map f x$
:$\ds \map f M = \max_{x \mathop \in \closedint a b} \map f x$
Then, from Darboux's Theorem:
:$\ds \map f m \paren {b - a} \le \int_a^b \map f x \rd x \le \map f M \paren {b - a}$ 
Dividing all terms by $\paren {b - a}$ gives:
:$\ds \map f m \le \frac 1 {b - a}\int_a^b \map f x \rd x \le \map f M$
By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that:
Category:Mean Value Theorem for Integrals
Mean Value Theorem for Integrals/Proof 2

== Theorem ==
== Proof ==
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
Let $F : \closedint a b \to \R$ be a real function defined by: 
:$\ds \map F x = \int_a^x \map f x \rd x$
We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$.
From Fundamental Theorem of Calculus: First Part, we have: 
: $F$ is continuous on $\closedint a b$
: $F$ is differentiable on $\openint a b$ with derivative $f$
By the Mean Value Theorem, there therefore exists $k \in \openint a b$ such that: 
:$\map {F'} k = \dfrac {\map F b - \map F a} {b - a}$
As $F$ is differentiable on $\openint a b$ with derivative $f$: 
:$\map {F'} k = \map f k$
We therefore have: 
giving: 
:$\ds \int_a^b \map f x \rd x = \paren {b - a} \map f k$
as required.
Category:Mean Value Theorem for Integrals
Mean Value of Concave Real Function

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Let $f$ be concave on $\openint a b$.
Then:
:$\forall \xi \in \openint a b: \map f x - \map f \xi \le \map {f'} \xi \paren {x - \xi}$
== Proof ==
By the Mean Value Theorem:
:$\exists \eta \in \openint x \xi: \map {f'} \eta = \dfrac {\map f x - \map f \xi} {x - \xi}$
From Real Function is Concave iff Derivative is Decreasing, the derivative of $f$ is decreasing.
Thus:
:$x > \xi \implies \map {f'} \eta \le \map {f'} \xi$
:$x < \xi \implies \map {f'} \eta \ge \map {f'} \xi$
Hence:
:$\map f x - \map f \xi \le \map {f'} \xi \paren {x - \xi}$
== Also see ==
* Mean Value of Convex Real Function
Category:Concave Real Functions
Category:Differential Calculus