nyuuzyou/wikis huggingface dataset:
Second-Countability is not Continuous Invariant

== Theorem ==
Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping.
If $T_A$ is a second-countable space, then it does not necessarily follow that $T_B$ is also second-countable.
== Proof ==
Let $T_S = \struct {S, \tau_S}$ be the Arens-Fort space.
Let $T_D = \struct {S, \tau_D}$ be the discrete space, also on $S$.
As $S$ is countable, from Arens-Fort Space is Expansion of Countable Fort Space, it follows that $T_D = \struct {S, \tau_D}$ is a countable discrete space.
Let $I_S: S \to S$ be the identity mapping on $S$.
From Mapping from Discrete Space is Continuous, we have that $I_S$ is a continuous mapping.
Then we have that a Countable Discrete Space is Second-Countable.
We have that the Arens-Fort Space is not First-Countable.
It follows from Second-Countable Space is First-Countable that the Arens-Fort space is not second-countable either.
Thus we have demonstrated a continuous mapping from a second-countable space to a space which is not second-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $6$
Category:Second-Countable Spaces
Category:Continuous Invariants
Second-Countable Space is Compact iff Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a second-countable space.
Then $T$ is compact $T$ is countably compact.
== Proof ==
=== Necessary Condition ===
We have that a compact space is countably compact (whether the space is second-countable or not).
=== Sufficient Condition ===
Let $T$ be countably compact.
Since $T$ is second-countable, we can choose a countable (analytic) basis $\BB$ for $\tau$.
Let $\CC$ be an open cover for $S$.
Define:
:$\CC' = \set {B \in \BB: \exists C \in \CC: B \subseteq C}$
Then $\CC' \subseteq \BB$, and so $\CC'$ is countable.
By the definition of an analytic basis, we have $\BB \subseteq \tau$; since $\subseteq$ is a transitive relation, it therefore follows that $\CC' \subseteq \tau$.
From Equivalence of Definitions of Analytic Basis:
:$\forall C \in \CC: \forall x \in C: \exists B \in \CC': x \in B$
That is, by the definition of set union:
:$\ds \forall C \in \CC: C \subseteq \bigcup \CC'$
Hence, by Union is Smallest Superset:
:$\ds S \subseteq \bigcup \CC \subseteq \bigcup \CC'$
Since $\subseteq$ is a transitive relation, it follows that $\CC'$ is an open cover for $S$.
Since $T$ is countably compact, there exists a finite subcover $\FF$ of $\CC'$ for $S$.
By the Principle of Finite Choice, there exists an indexed family $\family {C_B}_{B \mathop \in \FF}$ such that:
:$\forall B \in \FF: B \subseteq C_B \in \CC$
Note that $\set {C_B: B \in \FF}$ is a finite subcover of $\CC$ for $X$.
Hence, $T$ is compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Compact Spaces
Category:Countably Compact Spaces
Category:Second-Countable Spaces
Second-Countable Space is First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Then $T$ is also first-countable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Second-Countable Space is First-Countable
Category:First-Countable Spaces
Category:Second-Countable Spaces
Second-Countable Space is First-Countable/Proof 1

== Theorem ==
== Proof ==
By definition $T$ is second-countable its topology has a countable basis.
Consider the entire set $S$ as an open set.
From Set is Open iff Neighborhood of all its Points, $S$ has that property.
As $T$ has a countable basis, then (trivially) every point in $T$ has a countable local basis.
So a second-countable space is trivially first-countable.
Category:Second-Countable Space is First-Countable
Second-Countable Space is First-Countable/Proof 2

== Theorem ==
== Proof ==
By definition of second-countable space, there exists a countable analytic basis $\BB \subseteq \tau$.
Then each $x \in S$ has a countable local basis:
:$\BB_x := \set {U \in \BB: x \in U}$
Category: Second-Countable Space is First-Countable
Second-Countable Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Then $T$ is also a Lindelöf space.
== Proof ==
Let $T$ be second-countable.
Then by definition its topology has a countable basis.
Let $\BB$ be this countable basis.
Let $\CC$ be an open cover of $T$.
Every set in $\CC$ is the union of a subset of $\BB$.
So $\CC$ itself is the union of a subset of $\BB$.
This union of a subset of $\BB$ is therefore a countable subcover of $\CC$.
That is, $T$ is by definition Lindelöf.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Second-Countable Spaces
Category:Lindelöf Spaces
Second-Countable Space is Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a second-countable topological space.
Then $T$ is also a separable space.
== Proof ==
By definition, there exists a countable basis $\BB$ for $\tau$.
Using the , we can obtain a choice function $\phi$ for $\BB \setminus \set \O$.
Define:
:$H = \set {\map \phi B: B \in \BB \setminus \set \O}$
By Image of Countable Set under Mapping is Countable, it follows that $H$ is countable.
It suffices to show that $H$ is everywhere dense in $T$.
Let $x \in U \in \tau$.
By Equivalence of Definitions of Analytic Basis, there exists a $B \in \BB$ such that $x \in B \subseteq U$.
Then $\map \phi B \in U$, and so $H \cap U$ is non-empty.
Hence, $x$ is an adherent point of $H$.
By Equivalence of Definitions of Adherent Point, it follows that $x \in H^-$, where $H^-$ denotes the closure of $H$.
Therefore, $H^- = S$, and so $H$ is everywhere dense in $T$ by definition.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Second-Countable Spaces
Category:Separable Spaces
Second-Countable T3 Space is T5

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ space which is also second-countable.
Then $T$ is a $T_5$ space.
== Proof ==
Let $A, B \subseteq S$ with $A^- \cap B = A \cap B^- = \O$.
For each $x \in A$, since $T$ is $T_3$:
:$\exists P, Q \in \tau: x \in P, B^- \subseteq Q, P \cap Q = \O$
Let $\BB$ be a basis for $T$.
Then:
:$\exists U \in \BB: x \in U \subseteq P$
Notice that:
By Subset of Empty Set, $U^-$ and $B$ are disjoint.
Since $T$ is second-countable, $\BB$ is countable.
Doing the above process for each $x \in A$ yields a subset $\set {U_n}_{n \mathop \in \N}$ of $\BB$.
Doing a similar process for each $y \in B$ yields another subset $\set {V_n}_{n \mathop \in \N}$ of $\BB$.
These sets are open sets by definition.
Define $\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ and $\ds V'_n = V_n \setminus \bigcup_{i \mathop \le n} U_i^-$.
Define $\ds U' = \bigcup_{n \mathop \in \N} U'_n$ and $\ds V' = \bigcup_{n \mathop \in \N} V'_n$.
We show that $U'$ and $V'$ are disjoint open sets containing $A$ and $B$ respectively.
For any $n \in \N$, we have that $U_n$ is open.
From Topological Closure is Closed:
:$V_i^-$ is closed for each $i \le n$.
From Finite Union of Closed Sets is Closed in Topological Space:
:$\ds \bigcup_{i \mathop \le n} V_i^-$ is closed.
By Open Set minus Closed Set is Open:
:$\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ is open.
By :
:$\ds U' = \bigcup_{n \mathop \in \N} U'_n$ is open.
Similarly, $V'$ is open.
Let $y \in B$.
By construction, there is some $k \in \N$ where $y \in V_k$.
From above we see that $U_i^-$ and $B$ are disjoint for all $i \in \N$.
So $y \notin U_i^-$ for every $i \le k$.
Hence $y \in V_k \setminus \bigcup_{i \mathop \le k} U_i^- = V'_k \subseteq V'$.
By , $B \subseteq V'$.
Similarly, $A \subseteq U'$.
Let $i, j \in \N$.
We show that $U'_i, V'_j$ are disjoint.
 suppose $i \le j$.
Then:
Now:
Hence $U'$ and $V'$ are disjoint.
Since $A, B$ are arbitrary, $T$ is a $T_5$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Second-Countable Spaces
Category:T3 Spaces
Category:T5 Spaces
Second Apotome/Example

== Example of Second Apotome ==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a second apotome :
:$(1): \quad b \in \Q$
:$(2): \quad \dfrac {\sqrt {a^2 - b^2}} a \in \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = 2 \sqrt {3}$ and $b = 3$.
Then:
Therefore $2 \sqrt 3 - 3$ is a second apotome.
Category:Examples of Apotome
Second Apotome of Medial is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ be a medial straight line.
Let a medial straight line $CB$ such that:
:$CB$ is commensurable in square only with $AB$
:the rectangle contained by $AB$ and $BC$ is medial
be cut off from $AB$.
Let $DI$ be a rational straight line.
Using :
:Let $DE$ be a parallelogram set out on $DI$ equal to $AB^2 + BC^2$.
Let its breadth be $DG$.
Similarly:
:Let $DH$ be a parallelogram set out on $DI$ equal to $2 \cdot AB \cdot BC$.
From :
:$FE = AC^2$
We have that $AB^2$ and $BC^2$ are medial areas which are commensurable.
Therefore from:
:
and:
::
it follows that:
:$DE$ is medial.
We have that $DE$ has been applied to the rational straight line $DI$ producing $DG$ as breadth.
So from :
:$DG$ is rational and incommensurable in length with $DI$.
We have that $AB \cdot BC$ is medial.
So from:
:$2 \cdot AB \cdot BC$ is medial.
But $2 \cdot AB \cdot BC = DH$.
Therefore $DH$ is medial.
We have that $DH$ has been applied to the rational straight line $DI$ producing $DF$ as breadth.
So from :
:$DF$ is rational and incommensurable in length with $DI$.
We have that $AB$ and $BC$ are commensurable in square only.
Therefore from :
:$AB$ is incommensurable in length with $BC$.
But from :
:$AB^2 + BC^2$ are commensurable with $AB^2$
and from :
:$2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$
therefore from :
:$2 \cdot AB \cdot BC$ is incommensurable with $AB^2 + BC^2$.
But:
:$DE = AB^2 + BC^2$
and:
:$DH = 2 \cdot AB \cdot BC$
and so $DE$ is incommensurable with $DH$.
But from :
:$DE : DH = GD : DF$
Therefore from :
:$GD$ is incommensurable with $DF$.
But both $GD$ and $DF$ are rational.
Therefore $GD$ and $DF$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $FG$ is an apotome.
But $DI$ is rational.
From :
:a rectangle contained by a rational and an irrational straight line is irrational.
Hence its "side" is irrational.
But $AC$ is the "side" of $FE$.
Therefore $AC$ is irrational.
Such a straight line is known as a second apotome of a medial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Second Bimedial Straight Line is Divisible Uniquely

== Theorem ==
== Proof ==
Let $AB$ be a second bimedial straight line.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
:$AC$ and $CB$ are medial straight lines
:$AC$ and $CB$ are commensurable in square only
:$AC$ and $CB$ contain a medial rectangle.
:
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
WLOG let $AC > DB$.
From :
:$AD^2 + DB^2 < AC^2 + CB^2$
From :
:$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
Let $EF$ be a rational straight line.
Using :
:Let $EK$ be a rectangle set out on $EF$ equal to the square on $AB$.
Let $EG = EK - AC^2 - CB^2$.
Then $HK = 2 \cdot AC \cdot CB$.
Similarly, let $EL = AD^2 + DB^2$.
Then $HL = 2 \cdot AD \cdot DB$.
Since the squares on $AC$ and $CB$ are by definition medial, it follows that $EG$ is medial.
But $EG$ is applied to the rational straight line $EF$.
Therefore from :
:$EH$ is a rational straight line which is incommensurable in length with $EF$.
Using the same reasoning:
:$HN$ is a rational straight line which is incommensurable in length with $EF$.
We have that $AC$ and $CB$ are medial straight lines which are commensurable in square only.
Therefore $AC$ and $CB$ are incommensurable in length.
But:
:$AC : CB = AC^2 : AC \cdot CB$
So from :
:$AC^2$ is incommensurable with $AC \cdot CB$.
But we have that $AC$ and $CB$ are commensurable in square.
So from :
:$AC^2 + CB^2$ is commensurable in square with $AC^2$.
From :
:$2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$.
So from :
:$AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
We have:
: $EG = AC^2 + CB^2$
and:
: $HK = 2 \cdot AC \cdot CB$
Therefore $EG$ is incommensurable with $HK$.
So by:
: 
and:
: :
it follows that:
: $EH$ is incommensurable in length with $HN$.
We also have that $EH$ and $HN$ are rational.
Therefore $EH$ and $HN$ are rational straight lines which are commensurable in square only.
But from :
:Two rational straight lines commensurable in square only, when added together, form an irrational straight line which is a binomial.
Therefore $EN$ is a binomial straight line which is divided at $H$.
In the same way $EM$ and $MN$ can be shown to be rational straight lines commensurable in square only.
Thus $EN$ is a binomial straight line which is divided into terms in two places: $H$ and $M$.
We have that:
:$AC^2 + CB^2 > AD^2 + DB^2$
Also:
:$AD^2 + DB^2 > 2 \cdot AD \cdot DB$
Thus $EG = AC^2 + CB^2 > 2 \cdot AD \cdot DB = MK$
Thus $EH > MN$.
But from :
:$EN$ can be divided into its terms in only one way.
This contradicts the conclusion that has been drawn: that $H$ and $M$ divide $EH$ into terms in two different ways.
So the supposition that $AB$ can be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$ must be false.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Second Bimedial is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only.
Let $AB$ and $BC$ contain a medial rectangle.
Let $DE$ be a rational straight line.
Using :
:Let $DF$ be a parallelogram set out on $DE$ equal to the square on $AC$.
Let its breadth be $DG$.
From :
:$AC^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$
Let $EH$ be a rectangle applied to $DE$ whose area equals $AB^2 + BC^2$.
The rectangle $HF$ which remains from $DF$ having had $EH$ removed is therefore equal to $2 \cdot AB \cdot BC$.
Since each of $AB$ and $BC$ are medial, both of $AB^2$ and $BC^2$ are also medial.
By hypothesis, $2 \cdot AB \cdot BC$ is also medial.
As:
:$EH = AB^2 + BC^2$
and:
:$FH = 2 \cdot AB \cdot BC$
each of the rectangles $EH$ and $FH$ is medial.
Each of $EH$ and $FH$ is applied to the rational straight line $DE$.
From :
:Each of $DH$ and $HG$ is rational and incommensurable in length with $DE$.
We have that $AB$ is incommensurable in length with $BC$.
We also have that:
:$AB : BC = AB^2 : AB \cdot BC$
So by :
:$AB^2$ is incommensurable with $AB \cdot BC$.
But from :
:$AB^2 + BC^2$ is commensurable with $AB^2$
and from :
:$2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$.
Therefore from :
:$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
But:
:$EH = AB^2 + BC^2$
and:
:$HF = 2 \cdot AB \cdot BC$
Therefore $EH$ is incommensurable with $HF$.
So from:
:
and:
:
it follows that $DH$ is incommensurable in length with $HG$.
Therefore $DH$ and $HG$ are rational straight lines which are commensurable in square only.
By :
:$DG$ is irrational.
But $DE$ is rational.
From :
:The rectangle contained by an irrational and a rational straight line is irrational.
Therefore $DF$ is irrational.
From :
:The side of a square equal to an irrational area is irrational.
But $AC$ is the side of a square equal to $DF$.
Therefore $AC$ is irrational.
Such a straight line is called second bimedial.
== Sources ==
* : Book $\text{X}$. Propositions
::(in which a mistake apppears)
Category:Euclidean Number Theory
Second Bimedial is Irrational/Mistake

== Source Work ==
The mistake can be seen in the edition as published by Dover: ISBN 0-486-60089-0.
== Mistake ==
:For let a rational straight line $DE$ be set out, and let the parallelogram $DF$ equal to the square on $AC$ be applied to $DE$, producing $DG$ as breadth. [I.44]
The reference is incorrect.
:
is not the result needed here.
The correct reference should be:
:.
Category:Mistakes/Euclid: The Thirteen Books of The Elements: Volume 3
Second Borel-Cantelli Lemma

== Statement ==
Let the events $E_n$ be independent.
Let the sum of the probabilities of the $E_n$ diverges to infinity.
Then the probability that infinitely many of them occur is $1$.
That is:
If $\ds \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$ and the events $\ds \sequence {E_n}^\infty_{n \mathop = 1}$ are independent, then:
:$\ds \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 1$
== Proof ==
Let $\ds \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$.
Let $\sequence {E_n}^\infty_{n \mathop = 1}$ be independent.
It is sufficient to show the event that the $E_n$s did not occur for infinitely many values of $n$ has probability $0$.
This is just to say that it is sufficient to show that:
:$\ds 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 0$
Noting that:
it is enough to show:
:$\ds \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$
Since the $\sequence {E_n}^\infty_{n \mathop = 1}$ are independent:
This completes the proof.
Alternatively, we can see:
:$\ds \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$
by taking negative the logarithm of both sides to get:
Since $-\map \ln {1 - x} \ge x$ for all $x > 0$, the result similarly follows from our assumption that $\ds \sum^\infty_{n \mathop = 1} \map \Pr {E_n} = \infty$.
== Also see ==
* Borel-Cantelli Lemma
Category:Probability Theory
Second Chebyshev Function is Increasing

== Theorem ==
The second Chebyshev function $\psi$ is increasing.
== Proof ==
Let $x \ge y$. 
Then: 
From Logarithm is Strictly Increasing:
:$\ln p \ge \ln 2 > 0$ 
So, we have: 
:$\ds \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p \ge 0$
so:
So if $x \le y$, then:
:$\map \psi x \le \map \psi y$
so:
:$\psi$ is increasing.
Category:Second Chebyshev Function
Second Column and Diagonal of Pascal's Triangle consist of Triangular Numbers

== Theorem ==
The $2$nd column and $2$nd diagonal of Pascal's triangle consists of the set of triangular numbers.
== Proof ==
Recall Pascal's triangle:
By definition, the entry in row $n$ and column $m$ contains the binomial coefficient $\dbinom n m$.
Thus the $2$nd column contains all the elements of the form $\dbinom n 2$.
The $m$th diagonal consists of the elements in column $n - m$.
Thus the $m$th diagonal contains the binomial coefficients $\dbinom n {n - m}$.
By Symmetry Rule for Binomial Coefficients:
:$\dbinom n {n - m} = \dbinom n m$
Thus the $2$nd diagonal also contains the binomial coefficients $\dbinom n 2$.
By Binomial Coefficient with Two: Corollary, the triangular numbers are precisely those numbers of the form $\dbinom n 2$.
Hence the result.
== Sources ==
* : $15$
* : $15$
Category:Pascal's Triangle
Category:Triangular Numbers
Second Derivative at Maximum is Negative/Mistake

== Source Work ==
:
:Chapter $\text I$. The First Variation:
::$1.2$. Ordinary maximum and minimum theory
== Mistake ==
:... it follows that at a maximum $\map {f' '} a$ is negative and ... that at a minimum $\map {f' '} a$ is positive. Alternatively at a maximum $\map {f'} x$ is a decreasing function of $x$ and at a minimum $\map {f'} x$ is an increasing function of $x$. Thus it is possible to discriminate quite easily between maxima and minima.
== Correction ==
Not necessarily.
Let $\map f x = x^4$.
Then we have that $f$ has a local minimum at $x = 0$.
However, by Power Rule for Derivatives:
both of which are definitely zero at $x = 0$.
Hence, while the above statement is true in general, it needs to be pointed out that there are special cases which need to be investigated carefully.
== Sources ==
* : Chapter $\text I$. The First Variation: $1.2$. Ordinary maximum and minimum theory
Category:Mistakes/An Introduction to the Calculus of Variations
Second Derivative at Point of Inflection

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let $f$ have a point of inflection at $\xi \in \openint a b$.
Then:
:$\map {f' '} \xi = 0$
where $\map {f' '} \xi$ denotes the second derivative of $f$ at $\xi$.
== Proof ==
By definition of point of inflection, $f'$ has either a local maximum or a local minimum at $\xi$.
From Derivative at Maximum or Minimum, it follows that the derivative of $f'$ at $\xi$ is zero, that is:
:$\map {f' '} \xi = 0$
== Also see ==
* Point with Zero Second Derivative is not necessarily Point of Inflection
== Historical Note ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* : Chapter $\text {A}.19$: Leibniz ()
* 
* 
* 
* 
Category:Points of Inflection
Category:Differential Calculus
Second Derivative of Concave Real Function is Non-Positive

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Then $f$ is concave on $\openint a b$ its second derivative $f \le 0$ on $\openint a b$.
== Proof ==
From Real Function is Concave iff Derivative is Decreasing, $f$ is concave $f'$ is decreasing.
From Derivative of Monotone Function, $f'$ is decreasing its second derivative $f \le 0$.
== Also see ==
* Second Derivative of Convex Real Function is Non-Negative
* Real Function with Strictly Positive Second Derivative is Strictly Convex
* Real Function with Strictly Negative Second Derivative is Strictly Concave
Category:Differential Calculus
Category:Concave Real Functions
Second Derivative of Convex Real Function is Non-Negative

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Then $f$ is convex on $\openint a b$ its second derivative $f' ' \ge 0$ on $\openint a b$.
== Proof ==
From Real Function is Convex iff Derivative is Increasing, $f$ is convex $f'$ is increasing.
From Derivative of Monotone Function, $f'$ is increasing its second derivative $f' ' \ge 0$.
== Also see ==
* Real Function with Strictly Positive Second Derivative is Strictly Convex
* Second Derivative of Concave Real Function is Non-Positive
* Real Function with Strictly Negative Second Derivative is Strictly Concave
== Sources ==
* : $\S 12.19$
Category:Differential Calculus
Category:Convex Real Functions
Second Derivative of Inverse Function

== Theorem ==
Let $f$ be a real function which is of differentiability class $2$.
Let $f$ have an inverse $f^{-1}$, likewise of differentiability class $2$.
Then:
:$\dfrac {\d^2 x} {\d y^2} = -\dfrac {\d^2 y} {\d x^2} \paren {\dfrac {\d y} {\d x} }^{-3}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Leibniz's Theorem for Differentiation of a Product: $3.3.10$
Category:Derivative of Inverse Function
Second Derivative of Laplace Transform

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, twice differentiable on any closed interval $\closedint 0 a$.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then, everywhere that $\dfrac {\d^2} {\d s^2} \laptrans f$ exists:
:$\dfrac {\d^2} {\d s^2} \laptrans {\map f t} = \laptrans {t^2 \, \map f t}$
== Proof ==
== Also see ==
* Derivative of Laplace Transform
* Higher Order Derivatives of Laplace Transform
== Sources ==
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.11$
Category:Laplace Transforms