nyuuzyou/wikis huggingface dataset:
Real Area Hyperbolic Cosine of x over a in Logarithm Form

== Theorem ==
:$\arcosh \dfrac x a = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a$
== Proof ==
== Also see ==
* $\arsinh \dfrac x a$ in Logarithm Form
* $\artanh \dfrac x a$ in Logarithm Form
* $\arcoth \dfrac x a$ in Logarithm Form
* $\arsech \dfrac x a$ in Logarithm Form
* $\arcsch \dfrac x a$ in Logarithm Form
Category:Inverse Hyperbolic Cosine
Real Area Hyperbolic Cotangent of x over a in Logarithm Form

== Theorem ==
:$\arcoth \dfrac x a = \dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} }$
== Proof ==
== Also see ==
* $\arsinh \dfrac x a$ in Logarithm Form
* $\arcosh \dfrac x a$ in Logarithm Form
* $\artanh \dfrac x a$ in Logarithm Form
* $\arsech \dfrac x a$ in Logarithm Form
* $\arcsch \dfrac x a$ in Logarithm Form
Category:Inverse Hyperbolic Cotangent
Real Area Hyperbolic Secant of x over a in Logarithm Form

== Theorem ==
:$\arsech \dfrac x a = \map \ln {\dfrac {a + \sqrt {a^2 - x^2} } x}$
== Proof ==
== Also see ==
* $\arsinh \dfrac x a$ in Logarithm Form
* $\arcosh \dfrac x a$ in Logarithm Form
* $\artanh \dfrac x a$ in Logarithm Form
* $\arcoth \dfrac x a$ in Logarithm Form
* $\arcsch \dfrac x a$ in Logarithm Form
Category:Inverse Hyperbolic Secant
Real Area Hyperbolic Sine of Reciprocal equals Real Area Hyperbolic Cosecant

== Theorem ==
Everywhere that the function is defined:
:$\map \arsinh {\dfrac 1 x} = \arcsch x$
where $\arsinh$ and $\arcsch$ denote real area hyperbolic sine and real area hyperbolic cosecant respectively.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.61$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cosecant
Category:Inverse Hyperbolic Sine
Category:Reciprocals
Real Area Hyperbolic Sine of x over a in Logarithm Form

== Theorem ==
== Proof ==
== Also see ==
* $\arcosh \dfrac x a$ in Logarithm Form
* $\artanh \dfrac x a$ in Logarithm Form
* $\arcoth \dfrac x a$ in Logarithm Form
* $\arsech \dfrac x a$ in Logarithm Form
* $\arcsch \dfrac x a$ in Logarithm Form
Category:Inverse Hyperbolic Sine
Real Area Hyperbolic Tangent of Reciprocal equals Real Area Hyperbolic Cotangent

== Theorem ==
Everywhere that the function is defined:
:$\map \artanh {\dfrac 1 x} = \arcoth x$
where $\artanh$ and $\arcoth$ denote real area hyperbolic tangent and real area hyperbolic cotangent respectively.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.63$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Tangent
Category:Inverse Hyperbolic Cotangent
Category:Reciprocals
Real Area Hyperbolic Tangent of x over a in Logarithm Form

== Theorem ==
:$\artanh \dfrac x a = \dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} }$
== Proof ==
== Also see ==
* $\arsinh \dfrac x a$ in Logarithm Form
* $\arcosh \dfrac x a$ in Logarithm Form
* $\arcoth \dfrac x a$ in Logarithm Form
* $\arsech \dfrac x a$ in Logarithm Form
* $\arcsch \dfrac x a$ in Logarithm Form
Category:Inverse Hyperbolic Tangent
Real Complex Roots of Unity for Even Index

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is even.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R$ are:
:$x = 1$ or $x \in -1$
That is, $1$ and $-1$ are the only complex $n$th roots of unity which are real number.
== Proof ==
From Positive Real Complex Root of Unity, we have that $1$ is the only element of $U_n$ which is a positive real number.
We note that $\paren {-1}^n = 1$ as $n$ is even.
Thus $-1$ is also an element of $U_n$.
Now let $z \in U_n$ such that $\cmod z \ne 1$.
Let $z > 0$.
From Positive Power Function on Non-negative Reals is Strictly Increasing it follows that:
:$z < 1 \implies z^n < 1$
and:
:$z > 1 \implies z^n > 1$
Let $z < 0$.
From Positive Power Function on Negative Reals is Strictly Decreasing it follows that:
:$z < -1 \implies z^n > 1$
and:
:$z > -1 \implies z^n < 1$
That is, in all cases where $\cmod z \ne 1$ we have that $z^n \ne 1$.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
Real Complex Roots of Unity for Odd Index

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is odd.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R$ is:
:$x = 1$
That is, $1$ is the only complex $n$th root of unity which is a real number.
== Proof ==
From Positive Real Complex Root of Unity, we have that $1$ is the only positive real number in $U_n$.
 $z \in \R$ such that $z \in U_n$ and $z < 0$.
From Odd Power of Negative Real Number is Negative, $z^n < 0$.
But this contradicts the fact that $z_n = 1 > 0$.
Hence by Proof by Contradiction it follows there is no negative real number $z$ such that $z^n = 1$.
Thus $1$ is the only real number in $U_n$.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
Real Cosine Function has Zeroes

== Theorem ==
The real cosine function has at least two zeroes that are additive inverses of each other.
That is:
:$\exists \theta \in \R: \map \cos \theta = 0 \text { and } \map \cos {-\theta} = 0$
== Proof ==
 $\cos x$ is positive everywhere on $\R$.
From Derivative of Cosine Function and Derivative of Sine Function:
:$\map {\dfrac {\d^2} {\d x^2} } {\cos x} = \map {\dfrac \d {\d x} } {-\sin x} = -\cos x$
Thus $\map {\dfrac {\d^2} {\d x^2} } {\cos x} = -\cos x$ would always be negative.
Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$.
But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$.
By Differentiable Bounded Concave Real Function is Constant, $\cos x$ would then be a constant function.
This contradicts the fact that $\cos x$ is not a constant function.
Thus, by Proof by Contradiction, $\cos x$ cannot be positive everywhere on $\R$.
Therefore:
:$\exists \phi \in \R: \cos \phi < 0$
From Cosine of Zero is One:
:$\cos 0 = 1 > 0$
Since the Cosine Function is Continuous, by the Intermediate Value Theorem:
:$\exists \theta \in \R: \cos \theta = 0$
Hence from Cosine Function is Even:
:$\map \cos {-\theta} = 0$
Category:Cosine Function
Real Cosine Function is Bounded

== Theorem ==
Let $x \in \R$.
Then:
:$\size {\cos x} \le 1$
== Proof ==
From the algebraic definition of the real cosine function:
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
it follows that $\cos x$ is a real function.
Similarly, $\sin x$ is also a real function.
Thus it follows that:
From Ordering of Squares in Reals and the definition of absolute value, we have that:
:$x^2 \le 1 \iff \size x \le 1$
The result follows.
== Also see ==
* Complex Cosine Function is Unbounded
== Sources ==
* : $\S 16.3 \ (3) \ \text{(ii)}$
Category:Cosine Function
Real Division is not Closed

== Theorem ==
The operation of division on the set of real numbers $\R$ is not closed.
== Proof ==
From Division by Zero, we have that for all $a \in \R$, the operation $\dfrac a 0$ is not defined.
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $1$
Category:Real Division
Category:Algebraic Closure
Real Elementary Functions are Continuous

== Theorem ==
The elementary functions on the real numbers $\R$ are all continuous at every point of their domain.
== Proof ==
In turn:
:Real Polynomial Function is Continuous
:Real Trigonometric Functions are Continuous
:Real Logarithm Function is Continuous
:Real Exponential Function is Continuous
It remains to show that:
:All real functions that are compositions of the above
:All real functions obtained by adding, subtracting, multiplying and dividing any of the above types any finite number of times
are also continuous at every point of their domain.
== Sources ==
* 
* 
Category:Continuous Real Functions
Category:Elementary Functions
Real Elements of Mandelbrot Set

== Theorem ==
Let $M$ be the Mandelbrot set.
Let $\R$ denote the real number line.
Then:
:$M \cap \R = \closedint {-2} {\dfrac 1 4}$
where:
:$\cap$ denotes set intersection
:the $\closedint {-2} {\dfrac 1 4}$ denotes the closed real interval from $-2$ to $\dfrac 1 4$.
== Proof ==
== Sources ==
* 
* 
Category:Mandelbrot Set
Real Function/Examples/Arbitrary Function 1

== Example of Real Function ==
Let $f$ denote the real function defined as follows:
:$\map f x = \begin {cases} 2 & : 0 \le x \le 1 \\ \sqrt x & : 2 \le x \le 3 \end {cases}$
Thus the domain of $f$ is seen to be:
:$\Dom f = \closedint 0 1 \cup \closedint 2 3$
:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable: Example $2.21$
Category:Examples of Real Functions
Real Function/Examples/Arbitrary Function 1/Mistake

== Source Work ==
:
:Chapter $1$: Basic Concepts:
::Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable
== Mistake ==
:The relationship between two variables $x$ and $y$ is the following. If $x$ is between $0$ and $1$, $y$ is to equal $2$. If $x$ is between $2$ and $3$, $y$ is equal to $\sqrt x$. The equations which express the relationship between the two variables are, with the end points of the interval included,
 
:These two equations now define $y$ as a function of $x$. For each value of $x$ in the specified intervals, a value of $y$ is determined uniquely. The graph of this function is shown in Fig. $2.211$. Note that these equations do not define $y$ as a function of $x$ for values of $x$ outside the two stated intervals.
::
:::::::::Figure $2.211$
== Correction ==
The shape of the second part of the graph is incorrect.
It has been depicted as a convex function, in shape more like a square function than a square root.
The square root is a concave function.
The actual shape of the function in question can be found in the depiction of this arbitrary function.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable: Example $2.21$
Category:Mistakes/Ordinary Differential Equations
Real Function/Examples/Arbitrary Temperature

== Example of Real Function ==
The following is an invented graph of a hypothetical output from an automatic temperature sensing device over the period of $12$ hours.
:
Note that while there is no way to express the temperature as a formula, the relation between the time and the temperature is still a function.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable: Example $2.22$
Category:Examples of Real Functions
Real Function/Examples/Square Root

== Example of Real Function ==
The (real) square root function is the real function $f: \R \to \R$ defined on the positive real numbers as:
:$\forall x \in \R_{\ge 0}: \map f x = \sqrt x$
:
== Also see ==
* Domain of Real Square Root Function
* Image of Real Square Root Function
== Sources ==
* : $\S 2.2$: Functions
Category:Examples of Real Functions
Real Function both Convex and Concave is Linear

== Theorem ==
Let $f$ be a real function which is both convex and concave.
Then $f$ is a linear function.
== Proof ==
Let $f$ be both convex and concave on a subset $S \subseteq \R$ of the real numbers $\R$.
Then by definition:
== Sources ==
* 
Category:Convex Real Functions
Category:Concave Real Functions
Category:Linear Real Functions
Real Function is Concave iff Derivative is Decreasing

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Then $f$ is concave on $\openint a b$ its derivative $f'$ is decreasing on $\openint a b$.
Thus the intuitive result that a concave function "gets less steep".
== Proof ==
=== Necessary Condition ===
Let $f$ be concave on $\openint a b$.
Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that:
:$x_1 < x_2 < x_3 < x_4$
By the definition of concave:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \ge \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$
Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.
Thus:
:$\map {f'} {x_1} \ge \map {f'} {x_4}$
Hence $f'$ is decreasing on $\openint a b$.
=== Sufficient Condition ===
Let $f'$ be decreasing on $\openint a b$.
Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
By the Mean Value Theorem:
:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
where $x_1 < \xi < x_2 < \eta < x_3$.
Since $f'$ is decreasing:
:$\map {f'} \xi \ge \map {f'} \eta$
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
Hence $f$ is concave by definition.
== Also see ==
* Real Function is Strictly Concave iff Derivative is Strictly Decreasing
* Real Function is Convex iff Derivative is Increasing
* Real Function is Strictly Convex iff Derivative is Strictly Increasing
== Sources ==
* : $\S 3.3, \S 3.4$
Category:Differential Calculus
Category:Concave Real Functions
Real Function is Concave iff its Negative is Convex

== Theorem ==
Let $f$ be a real function.
Let $I \subseteq \R$ be an interval of $\R$.
Then $f$ is concave on $I$ $-f$ is convex on $\R$.
== Proof ==
=== Necessary Condition ===
Let $f$ be concave on $I$.
Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.
Then:
and so $-f$ is convex by definition.
=== Sufficient Condition ===
Let $-f$ be convex on $I$.
Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.
Then:
and so $f$ is concave by definition.
== Sources ==
* : $\S 12.13$
* 
Category:Convex Real Functions
Category:Concave Real Functions
Real Function is Continuous at Isolated Point

== Theorem ==
Let $A \subseteq \R$ be any subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $x \in A$ be an isolated point of $A$.
Then $f$ is continuous at $x$.
== Proof ==
By definition of isolation point:
:$\exists \delta \in \R_{> 0}: \openint {x - \delta} {x + \delta} \cap A = \set x$
Pick any $\epsilon > 0$.
We have that for any $z \in \openint {x - \delta} {x + \delta} \cap A = \set x$:
:$\size {\map f z - \map f x} = \size {\map f x - \map f x} = 0 < \epsilon$.
This satisfies the condition of continuity at $x$,
and the limit in this case is trivially equal to $\map f x$.
Category:Continuity
Category:Real Functions
Real Function is Continuous at Point iff Oscillation is Zero

== Theorem ==
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of open subset neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$:
:$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} {I \cap D}: I \in N_x}$
where $\map {\omega_f} {I \cap D}$ denotes the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Then $\map {\omega_f} x = 0$ $f$ is continuous at $x$.
== Proof ==
=== Necessary Condition ===
Let $\map {\omega_f} x = 0$.
Let $\epsilon > 0$.
Suppose that $\forall I: I \in N_x: \map {\omega_f} {I \cap D} \ge \epsilon$.
Then by definition of oscillation at point, $\map {\omega_f} x \ge \epsilon$. 
This contradicts $\map {\omega_f} x = 0$.
From this contradiction we deduce that:
:$\exists J: J \in N_x: \map {\omega_f} {J \cap D} < \epsilon$
For this particular $J$, there is an open set $O \subset J$ that contains $x$ by the definition of open subset neighborhood.
Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a subset of $O$.
So, if $y$ satisfies:
:$\size {x - y} < \delta$
Then:
:$y \in J$
and, if $y \in D$:
:$\size {\map f x - \map f y} \le \map {\omega_f} {J \cap D}$
Since $\map {\omega_f} {J \cap D} < \epsilon$ it follows by the definition of continuity that $f$ is continuous at $x$.
=== Sufficient Condition ===
Let $f$ be continuous at $x$.
Then $\forall \epsilon > 0: \exists \delta \in \R_{>0}$ such that, if $y \in D$:
:$\size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$
Let the interval $I_\delta$ be defined as:
:$I_\delta := \openint {x - \delta} {x + \delta}$
$I_\delta$ is an element of $N_x$ as $I_\delta$ is a neighborhood of $x$.
We have:
This gives:
This holds true for any value of $\epsilon$.
Thus $\map {\omega_f} x$ must be $0$.
Hence the result.
Category:Real Analysis
Real Function is Convex iff Derivative is Increasing

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Then:
:$f$ is convex on $\openint a b$
:
:its derivative $f'$ is increasing on $\openint a b$.
Thus the intuitive result that a convex function "gets steeper".
== Proof ==
=== Necessary Condition ===
Let $f$ be convex on $\openint a b$.
Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that:
:$x_1 < x_2 < x_3 < x_4$
By the definition of convex function:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \le \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$
Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.
Thus:
:$\map {f'} {x_1} \le \map {f'} {x_4}$
Hence $f'$ is increasing on $\openint a b$.
=== Sufficient Condition ===
Let $f'$ be increasing on $\openint a b$.
Let $x_1, x_2, x_3 \in \openint a b$ such that $x_1 < x_2 < x_3$.
By the Mean Value Theorem:
:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
where $x_1 < \xi < x_2 < \eta < x_3$.
Since $f'$ is increasing:
:$\map {f'} \xi \le \map {f'} \eta$
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
Hence $f$ is convex by definition.
== Also see ==
* Real Function is Strictly Convex iff Derivative is Strictly Increasing
* Real Function is Concave iff Derivative is Decreasing
* Real Function is Strictly Concave iff Derivative is Strictly Decreasing
== Sources ==
* : $\S 12.18$
* : $\S 3.3, \S 3.4$
Category:Differential Calculus
Category:Convex Real Functions
Real Function is Expressible as Sum of Even Function and Odd Function

== Theorem ==
Let $f: \R \to \R$ be a real function which is neither an even function nor an odd function.
Then $f$ may be expressed as the pointwise sum of an even function and an odd function.
== Proof ==
Let:
We note that:
Thus $g$ is an even function.
Then:
Thus $h$ is an odd function.
Then:
Hence the result.
== Examples ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(h)}$ Even and Odd Functions $(9)$
Category:Real Function is Expressible as Sum of Even Function and Odd Function
Category:Real Functions
Category:Odd Functions
Category:Even Functions
Real Function is Expressible as Sum of Even Function and Odd Function/Examples/Arbitrary Function 1

== Example of Use of Real Function is Expressible as Sum of Even Function and Odd Function ==
Let $f: \R \to \R$ denote the real function:
:$\map f x = e^{2 x} \sin x$
$f$ can be expressed as the pointwise sum of:
:the even function $\map g x = \dfrac {\paren {e^{2 x} - e^{-2 x} } \sin x} 2$
and:
:the odd function $\map h x = \dfrac {\paren {e^{2 x} + e^{-2 x} } \sin x} 2$
== Proof ==
and:
The result follows.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(h)}$ Even and Odd Functions $(10)$
Category:Real Function is Expressible as Sum of Even Function and Odd Function
Real Function is Linearly Dependent with Zero Function

== Theorem ==
Let $\map f x$ be a real function defined on a closed interval $\closedint a b$.
Let $\map g x$ be the constant zero function on $\closedint a b$:
:$\forall x \in \closedint a b: \map g x = 0$
Then $f$ and $g$ are linearly dependent on $\closedint a b$.
== Proof ==
We have that:
:$\forall x \in \closedint a b: \map g x = 0 = 0 \times \map f x$
and $0 \in \R$.
Hence the result by definition of linearly dependent real functions.
== Sources ==
* : $\S 3.15$: The General Solution of the Homogeneous Equation
Category:Real Analysis
Real Function is Strictly Concave iff Derivative is Strictly Decreasing

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Then $f$ is strictly concave on $\openint a b$ its derivative $f'$ is strictly decreasing on $\openint a b$.
== Proof ==
=== Necessary Condition ===
Let $f$ be strictly concave on $\openint a b$.
Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$.
We are to show that:
:$\map {f'} r > \map {f'} s$
Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that:
:$r < x_1 < x_2 < x_3 < s$
By the definition of strictly concave:
:$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} > \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} > \dfrac {\map f s - \map f {x_3} } {s - x_3}$
and:
:$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} > \dfrac {\map f s - \map f {x_2} } {s - x_2}$
Let $x_1 \to r^+$.
Then:
Similarly, let $x_3 \to s^-$.
We have:
Thus we have:
Thus:
:$\map {f'} r > \map {f'} s$
Hence $f'$ is strictly decreasing on $\openint a b$.
=== Sufficient Condition ===
Let $f'$ be strictly decreasing on $\openint a b$.
Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
By the Mean Value Theorem:
:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
where $x_1 < \xi < x_2 < \eta < x_3$.
Since $f'$ is strictly decreasing:
:$\map {f'} \xi > \map {f'} \eta$
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
Hence $f$ is strictly concave by definition.
== Also see ==
* Real Function is Concave iff Derivative is Decreasing
* Real Function is Convex iff Derivative is Increasing
* Real Function is Strictly Convex iff Derivative is Strictly Increasing
== Sources ==
* 
Category:Differential Calculus
Category:Concave Real Functions
Real Function is Strictly Convex iff Derivative is Strictly Increasing

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Then $f$ is strictly convex on $\openint a b$ its derivative $f'$ is strictly increasing on $\openint a b$.
== Proof ==
=== Necessary Condition ===
Let $f$ be strictly convex on $\openint a b$.
Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$.
We are to show that:
:$\map {f'} r < \map {f'} s$
Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that:
:$r < x_1 < x_2 < x_3 < s$
By the definition of strictly convex:
:$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} < \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} < \dfrac {\map f s - \map f {x_3} } {s - x_3}$
and:
:$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} < \dfrac {\map f s - \map f {x_2} } {s - x_2}$
Let $x_1 \to r^+$.
Then:
Similarly, let $x_3 \to s^-$.
We have:
Thus we have:
Thus:
:$\map {f'} r < \map {f'} s$
Hence $f'$ is strictly decreasing on $\openint a b$.
=== Sufficient Condition ===
Let $f'$ be strictly increasing on $\openint a b$.
Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
By the Mean Value Theorem:
:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
where $x_1 < \xi < x_2 < \eta < x_3$.
Since $f'$ is strictly increasing:
:$\map {f'} \xi < \map {f'} \eta$
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
Hence $f$ is strictly convex by definition.
== Also see ==
* Real Function is Convex iff Derivative is Increasing
* Real Function is Concave iff Derivative is Decreasing
* Real Function is Strictly Concave iff Derivative is Strictly Decreasing
Category:Strictly Increasing Real Functions
Category:Differential Calculus
Category:Convex Real Functions