nyuuzyou/wikis huggingface dataset:
Convergent Trigonometric Series is Periodic

== Theorem ==
Let $\map S x$ be a trigonometric series:
:$\map S x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
Let $S$ be convergent.
Then $S$ is periodic:
:$\forall r \in \Z: \map S {x + 2 r \pi} = \map S x$
== Proof ==
Let $\map S x$ converge to some $L \in \R$.
Let $r \in \Z$ be arbitrary.
Then:
== Sources ==
* : Chapter One: $\S 1$. Trigonometrical Series
Category:Trigonometric Series
Category:Periodic Functions
Convergents are Best Approximations

== Theorem ==
Let $x$ be an irrational number.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of its continued fraction expansion.
Let $\dfrac {p_n} {q_n}$ be the $n$th convergent.
Let $\dfrac a b$ be any rational number such that $0 < b < q_{n + 1}$.
Then:
:$\forall n > 1: \size {q_n x - p_n} \le \size {b x - a}$
The equality holds only if $a = p_n$ and $b = q_n$.
=== Corollary ===
== Proof ==
Let $\dfrac a b$ be a rational number in canonical form such that $b < q_{n + 1}$.
Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds.
Consider the system of equations:
Multiplying the first by $q_n$, and the second by $p_n$, then subtracting, we get:
:$a q_n - b p_n = s \paren {p_{n + 1} q_n - p_n q_{n + 1} }$
After applying Difference between Adjacent Convergents of Simple Continued Fraction we get:
So $r$ and $s$ are integers.
Neither of them is $0$ because:
:if $r = 0$ then $a q_{n + 1} = b p_{n + 1}$, and Euclid's Lemma means $q_{n + 1} \divides b$ as $p_{n + 1} \perp q_{n + 1}$, which contradicts $0 < b < q_{n + 1}$
:if $s = 0$ we have $\dfrac a b = \dfrac {p_n} {q_n}$ and this we have already excluded as a special case.
From Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent, the convergents are alternately greater than and less than $x$.
Hence since $0 < b = r q_n + s q_{n + 1} < q_{n + 1}$, the integers $r$ and $s$ must have opposite sign.
It follows that $r \paren {q_n x - p_n}$ and $s \paren {q_{n + 1} x - p_{n + 1} }$ have the same sign.
This is necessary for the Triangle Inequality to hold.
So:
as we wanted to prove.
== Sources ==
* 
Category:Continued Fractions
Convergents are Best Approximations/Corollary

== Corollary to Convergents are Best Approximations ==
Let $x$ be an irrational number.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of its continued fraction expansion.
Each convergent $\dfrac {p_n} {q_n}$ is a best rational approximation to $x$.
That is, for any rational number $\dfrac a b$ such that $1 \le b \le q_n$:
:$\size {x - \dfrac {p_n} {q_n} } \le \size {x - \dfrac a b}$
The equality holds only if $a = p_n$ and $b = q_n$.
== Proof ==
Assume otherwise:
:$\exists \dfrac a b$ such that $1 \le b \le q_n$
and:
:$\size {x - \dfrac {p_n} {q_n} } > \size {x - \dfrac a b}$
Then:
which contradicts the result of Convergents are Best Approximations.
Category:Continued Fractions
Convergents of Simple Continued Fraction are Rationals in Canonical Form

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
For all $k \ge 1$, $\dfrac {p_k} {q_k}$ is in canonical form:
:$p_k$ and $q_k$ are coprime
:$q_k > 0$.
== Proof ==
Let $k \ge 1$.
Let $d = \gcd \set {p_k, q_k}$.
From Common Divisor Divides Integer Combination:
:$p_k q_{k - 1} - p_{k - 1} q_k$ is a multiple of $d$.
From Difference between Adjacent Convergents of Simple Continued Fraction:
:$d \divides \paren {-1}^{k + 1}$
where $\divides$ denotes divisibility.
It follows that:
:$d = 1$
By:
:$q_0 = 1$
:Denominators of Simple Continued Fraction are Strictly Increasing
It follows that $q_k > 0$ for all $k \ge 0$.
Category:Simple Continued Fractions
Converse Hinge Theorem

== Theorem ==
If two triangles have two pairs of sides which are the same length, the triangle in which the third side is longer also has the larger angle contained by the first two sides.
== Proof ==
:
Let $\triangle ABC$ and $\triangle DEF$ be two triangles in which:
:$AB = DF$
:$AC = DE$
:$BC > EF$
 that $\angle BAC \not > \angle EDF$.
Then either:
:$\angle BAC = \angle EDF$
or:
:$\angle BAC < \angle EDF$
Let $\angle BAC = \angle EDF$.
Then by Triangle Side-Angle-Side Congruence:
:$BC = EF$
But we know this is not the case, so by Proof by Contradiction:
:$\angle BAC \ne \angle EDF$
Suppose $\angle BAC < \angle EDF$.
Then by Greater Angle of Triangle Subtended by Greater Side:
:$EF > BC$
But we know this is not the case, so by Proof by Contradiction:
:$\angle BAC \not < \angle EDF$
Thus:
:$\angle BAC > \angle EDF$
== Also known as ==
This theorem is also known as the side-side-side inequality theorem, or SSS inequality theorem.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.20$
Category:Triangles
Category:Named Theorems
Converse and Contrapositive Statements/Examples/Triangles of 2 and 3 Equal Sides

== Examples of Converse Statements ==
Let $P$ be the statement:
:If a triangle has $3$ equal sides, then it has $2$ equal sides
which is trivially true.
$P$ can also be written:
:A triangle has $3$ equal sides only if it has $2$ equal sides
:A sufficient condition for a triangle to have $2$ equal sides is for it to have $3$ equal sides
:A necessary condition for a triangle to have $3$ equal sides is for it to have $2$ equal sides
The converse of $P$ is:
:If a triangle has $2$ equal sides, then it has $3$ equal sides
which is false.
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs
Category:Examples of Converse Statements
Converse and Contrapositive Statements/Examples/x less than 5 and x less than or equal to 5

== Examples of Converse Statements and Contrapositive Statements ==
Let:
: $P$ be the statement $x < 5 \implies x \le 5$
: $Q$ be the statement $x \le 5 \implies x < 5$
: $R$ be the statement $x > 5 \implies x \ge 5$
: $S$ be the statement $x \ge 5 \implies x > 5$
for $x \in \R$.
Then:
:$P$ and $Q$ are converse statements
:$R$ and $S$ are converse statements
:$P$ and $R$ are contrapositive statements
:$Q$ and $S$ are contrapositive statements
:$P$ and $R$ are true
:$Q$ and $S$ are false.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $2$
Category:Examples of Contrapositive Statements
Category:Examples of Converse Statements
Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio

== Theorem ==
=== Lemma ===
== Proof ==
:
Let the square on $AB$ be five times the square on $AC$.
Let $CD = 2 \cdot AC$.
It is to be demonstrated that when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.
Let the squares $AF$ and $CG$ be described on $AB$ and $CF$ respectively.
Let the figure $AF$ be drawn.
Let $BE$ be produced from $FB$.
We have that:
:$BA^2 = 5 \cdot AC^2$
That is:
:$AF = 5 \cdot AH$
Therefore the gnomon $MNO$ is $4$ times $AH$.
We have that:
:$DC = 2 \cdot CA$
Therefore:
:$DC^2 = 4 \cdot CA^2$
that is:
:$CG = 4 \cdot AH$
But:
:$MNO = 4 \cdot AH$
Therefore:
:$MNO = CG$
We have that:
:$DC = CK$
and:
:$AC = CH$
Therefore from :
:$KB = 2 \cdot BH$
But we also have:
:$LH + HB = 2 \cdot BH$
Therefore:
:$KB = LH + HB$
But:
:$MNO = CG$
Therefore:
:$HF = BG$
and as $CD = DG$:
:$BG = CD \cdot DB$
Also:
:$HF = CB^2$
Therefore:
:$CD \cdot DB = CB^2$
Therefore:
:$DC : CB = CB : BD$
But:
:$DC > CB$
therefore:
:$CB > BD$
Therefore, when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio/Lemma

== Lemma to Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio ==
== Proof ==
:
Suppose that $BC = 2 \cdot CA$.
Therefore:
:$BC^2 = 4 \cdot CA^2$
Therefore:
:$BC^2 + CA^2 = 5 CA^2$
But by hypothesis:
:$BA^2 = 5 CA^2$
Therefore by :
:$BA^2 = BC^2 + CA^2$
which is impossible.
Therefore:
:$BC \ne 2 \cdot CA$
Similarly it can be shown that it is not the case that:
:$BC < 2 \cdot CA$
as it leads to a much greater absurdity.
Therefore:
:$BC > 2 \cdot CA$
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Converse of Conditional is Contrapositive of Inverse

== Theorem ==
Let $p \implies q$ be a conditional.
Then the converse of $p \implies q$ is the contrapositive of its inverse.
== Proof ==
The converse of $p \implies q$ is:
:$q \implies p$
The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The contrapositive of $\neg p \implies \neg q$ is:
:$\neg \neg q \implies \neg \neg p$
By Double Negation, the two are seen to be equal.
Category:Conditional
Converse of Conditional is Inverse of Contrapositive

== Theorem ==
Let $p \implies q$ be a conditional.
Then the converse of $p \implies q$ is the inverse of its contrapositive.
== Proof ==
The converse of $p \implies q$ is:
:$q \implies p$
The contrapositive of $p \implies q$ is:
:$\neg q \implies \neg p$
The inverse of $\neg q \implies \neg p$ is:
:$\neg \neg q \implies \neg \neg p$
By Double Negation, the two are seen to be equal.
Category:Conditional
Converse of Euclid's Lemma does not Hold

== Lemma ==
Let $n \in Z_{>0}$ be a positive composite number
Let $a b \equiv 0 \pmod n$.
Then it is not necessarily the case that either $a \equiv 0 \pmod n$ or $b \equiv 0 \pmod n$.
== Proof ==
Let $n = 6$.
We have that:
:$6 \equiv 0 \pmod 6$
but:
:$2 \equiv 2 \pmod 6$
:$3 \equiv 3 \pmod 6$
Hence the result by Proof by Counterexample.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $7$
Category:Euclid's Lemma
Converse of Fermat's Little Theorem does not hold

== Theorem ==
Let $n$ be a natural number.
Let:
:$n \divides a^n - a$
where:
:$a$ is a natural number such that $n$ is not a divisor of $a$.
:$\divides$ denotes divisibility.
Then it is not necessarily the case that $n$ is prime
== Proof ==
;Proof by Counterexample
We have that:
:$2^{341} \equiv 2 \pmod {341}$
despite the fact that $341$ is not prime:
:$341 = 11 \times 31$
== Sources ==
* 
* 
Category:Fermat's Little Theorem
Category:Fermat Pseudoprimes
Converse of Tangent Secant Theorem

== Theorem ==
Let $D$ be a point outside a circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Let $DB$ intersect the circle at $B$ such that $DB^2 = AD \cdot DC$.
Then $DB$ is tangent to the circle $ABC$.
== Proof ==
:
Let $DE$ be drawn tangent to the circle $ABC$.
Let $F$ be the center of $ABC$ and join $FB, FD, FE$.
From Radius at Right Angle to Tangent, $\angle FED$ is a right angle.
We have that $DE$ is tangent to the circle $ABC$ and $DA$ cuts it.
So from the Tangent Secant Theorem $AD \cdot DC = DE^2$.
But we also have that $AD \cdot DC = DB^2$.
So $DE^2 = DB^2$ and so $DE = DB$.
Also $FE = FB$ and so $DE, EF = DB, DF$ and $DF$ is common.
So from Triangle Side-Side-Side Congruence $\angle DEF = \angle DBF$.
But $\angle DEF$ is a right angle and so also is $\angle DBF$.
So from Line at Right Angles to Diameter of Circle $DB$ is tangent to the circle $ABC$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Tangent Secant Theorem
Conversion between Cartesian and Cylindrical Coordinates

== Theorem ==
Let $S$ be a $3$-dimensional space.
Let a Cartesian $3$-space $\CC$ be applied to $S$.
Let a cylindrical coordinate system $\PP$ be superimposed upon $\CC$ such that:
:$(1): \quad$ The origin of $\CC$ and $P$ coincide with the pole of the polar coordinate plane of $\PP$ corresponding to $z = 0$ 
:$(2): \quad$ The $z$-axis of $\CC$ coincides with the $z$-axis of $\PP$.
:$(3): \quad$ The $x$-axis of $\CC$ coincides with the polar axis the polar coordinate plane of $\PP$ corresponding to $z = 0$.
Let $p$ be a point in $S$.
Let $p$ be specified as $p = \tuple {r, \theta, z}$ expressed in the cylindrical coordinates of $\PP$.
Then $p$ is expressed as $\tuple {r \cos \theta, r \sin \theta, z}$ in $\CC$.
Contrariwise, let $p$ be expressed as $\tuple {x, y, z}$ in the cartesian coordinates of $\CC$.
Then $p$ is expressed in $\PP$ as $\tuple {r, \theta, z}$, where:
where:
:$\sqbrk {\, \cdot \,}$ is Iverson's convention.
:$\arctan$ denotes the arctangent function.
== Proof ==
== Sources ==
* 
* 
Category:Cartesian Coordinate Systems
Category:Cylindrical Coordinates
Conversion between Cartesian and Polar Coordinates in Plane

== Theorem ==
Let $S$ be the plane.
Let a Cartesian plane $\CC$ be applied to $S$.
Let a polar coordinate plane $\PP$ be superimposed upon $\CC$ such that:
:$(1): \quad$ The origin of $\CC$ coincides with the pole of $\PP$.
:$(2): \quad$ The $x$-axis of $\CC$ coincides with the polar axis of $\PP$.
Let $p$ be a point in $S$.
Let $p$ be specified as $p = \polar {r, \theta}$ expressed in the polar coordinates of $\PP$.
Then $p$ is expressed as $\tuple {r \cos \theta, r \sin \theta}$ in $\CC$.
Contrariwise, let $p$ be expressed as $\tuple {x, y}$ in the cartesian coordinates of $\CC$.
Then $p$ is expressed in $\PP$ as:
:$p = \polar {\sqrt {x^2 + y^2}, \arctan \dfrac y x + \pi \sqbrk {x < 0 \text{ or } y < 0} + \pi \sqbrk {x > 0 \text{ and } y < 0} }$
where:
:$\sqbrk {\, \cdot \,}$ is Iverson's convention.
:$\arctan$ denotes the arctangent function.
== Proof ==
Let $P$ be a point in the plane expressed:
:in Cartesian coordinates as $\tuple {x, y}$
:in polar coordinates as $\polar {r, \theta}$.
:
As specified, we identify:
:the origins of both coordinate systems with a distinguished point $O$
:the $x$-axis of $C$ with the polar axis of $P$.
Let a perpendicular $PM$ be dropped from $P$ to the $x$-axis.
The triangle $OMP$ is a right triangle:
:whose hypotenuse is $OP$, whose length is $r$
:whose legs are $OM$ and $MP$
:whose angle $POM$ is $\theta$.
By definition of sine and cosine
:$x = r \cos \theta$
:$y = r \sin \theta$
The result follows.
== Examples ==
== Sources ==
* : Chapter $\text I$. Coordinates: $3$. Relation between cartesian and polar coordinates
* : $\S 2$. Geometrical Representations: $(2.7)$
* : $\S 10$: Formulas from Plane Analytic Geometry: $10.14$: Polar Coordinates $\tuple {r, \theta}$
* 
* 
* 
Category:Conversion between Cartesian and Polar Coordinates in Plane
Category:Cartesian Coordinate Systems
Category:Polar Coordinates
Conversion between Cartesian and Polar Coordinates in Plane/Examples

== Examples of Use of Conversion between Cartesian and Polar Coordinates in Plane ==
=== Example: $\polar {4, \dfrac \pi 3}$ ===
=== Example: $\polar {-2, -\dfrac \pi 4}$ ===
=== Example: $\polar {3, \dfrac \pi 2}$ ===
=== Example: $\tuple {-1, -1}$ ===
Category:Conversion between Cartesian and Polar Coordinates in Plane
Conversion between Cartesian and Polar Coordinates in Plane/Examples/(-1, -1)

== Example of Use of Conversion between Cartesian and Polar Coordinates in Plane ==
The point $P$ defined in Cartesian coordinates as:
:$P = \tuple {-1, -1}$
can be expressed in the corresponding polar coordinates as:
:$P = \polar {\sqrt 2, \dfrac {5 \pi} 4} = \polar {\sqrt 2, 225 \degrees}$
== Proof ==
Let $P$ be expressed in polar coordinates as:
:$P = \polar {r, \theta}$
From Conversion between Cartesian and Polar Coordinates in Plane:
Now we are to find $\theta$ where
where in this case:
Hence:
and:
Thus:
and:
Hence the result.
== Sources ==
* 
* 
Category:Conversion between Cartesian and Polar Coordinates in Plane
Conversion between Cartesian and Polar Coordinates in Plane/Examples/(-2, -pi over 4)

== Example of Use of Conversion between Cartesian and Polar Coordinates in Plane ==
The point $P$ defined in polar coordinates as:
:$P = \polar {-2, -\dfrac \pi 4}$
can be expressed in the corresponding Cartesian coordinates as:
:$P = \tuple {\sqrt 2, -\sqrt 2}$
== Proof ==
Let $P$ be expressed in Cartesian coordinates as:
:$P = \tuple {x, y}$
From Conversion between Cartesian and Polar Coordinates in Plane:
where in this case:
Hence:
and:
Hence the result.
== Sources ==
* : Chapter $\text I$. Coordinates: $5$. Distance between two points in polar coordinates: Examples: $1 \ \text {(ii)}$
Category:Conversion between Cartesian and Polar Coordinates in Plane
Conversion between Cartesian and Polar Coordinates in Plane/Examples/(3, pi over 2)

== Example of Use of Conversion between Cartesian and Polar Coordinates in Plane ==
The point $P$ defined in polar coordinates as:
:$P = \polar {3, \dfrac \pi 2} = \polar {3, 90 \degrees}$
can be expressed in the corresponding Cartesian coordinates as:
:$P = \tuple {0, 3}$
== Proof ==
Let $P$ be expressed in Cartesian coordinates as:
:$P = \tuple {x, y}$
From Conversion between Cartesian and Polar Coordinates in Plane:
where in this case:
Hence:
and:
Hence the result.
== Sources ==
* 
* 
Category:Conversion between Cartesian and Polar Coordinates in Plane
Conversion between Cartesian and Polar Coordinates in Plane/Examples/(4, pi over 3)

== Example of Use of Conversion between Cartesian and Polar Coordinates in Plane ==
The point $P$ defined in polar coordinates as:
:$P = \polar {4, \dfrac \pi 3}$
can be expressed in the corresponding Cartesian coordinates as:
:$P = \tuple {2, 2 \sqrt 3}$
== Proof ==
Let $P$ be expressed in Cartesian coordinates as:
:$P = \tuple {x, y}$
From Conversion between Cartesian and Polar Coordinates in Plane:
where in this case:
Hence:
and:
Hence the result.
== Sources ==
* : Chapter $\text I$. Coordinates: $5$. Distance between two points in polar coordinates: Examples: $1 \ \text {(i)}$
Category:Conversion between Cartesian and Polar Coordinates in Plane
Conversion from Hexadecimal to Binary

== Theorem ==
Let $n$ be a (positive) integer expressed in hexadecimal notation as:
:$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H$
Then $n$ can be expressed in binary notation as:
:$n = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2$
where $\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2$ is the expression of the hexadecimal digit $a_j$ in binary notation.
That is, you take the binary expression of each hexadecimal digit, padding them out with zeroes to make them $4$ bits long, and simply concatenate them.
== Proof ==
We have:
We have that:
:$0 \le a_j < 16$
and so:
and so:
Hence the result.
== Examples ==
== Sources ==
* : $16$
Category:Hexadecimal Notation
Category:Binary Notation
Category:Conversion from Hexadecimal to Binary
Conversion from Hexadecimal to Binary/Examples/6C5

== Example of Conversion from Hexadecimal to Binary ==
The integer expressed in hexadecimal as $6 \mathrm C 5_H$ can be expressed in binary as:
:$6 \mathrm C 5_H = 110 \, 1100 \, 0101_2$
== Proof ==
Each of the hexadecimal digits $6$, $\mathrm C$ and $5$ can be expressed in binary as:
Hence from Conversion from Hexadecimal to Binary, the result follows by concatenating these strings:
:$6 \mathrm C 5_H = 110 \, 1100 \, 0101_2$
== Sources ==
* : $16$
Category:Conversion from Hexadecimal to Binary
Conversion from Spherical Coordinates to Cartesian Coordinates

== Theorem ==
Let $S$ be ordinary $3$-space.
Let a Cartesian $3$-space $\CC$ be applied to $S$.
Let a spherical coordinate system $\PP$ be superimposed upon $\CC$ such that:
:$(1): \quad$ The origin of $\CC$ coincides with the pole of $\PP$.
:$(2): \quad$ The $x$-axis of $\CC$ coincides with the horizontal axis of $\PP$.
:$(3): \quad$ The $z$-axis of $\CC$ coincides with the polar axis of $\PP$.
Let $p$ be a point in $S$.
Let $p$ be specified as $p = \polar {r, \theta, \phi}$ expressed in the spherical coordinates of $\PP$.
Then $p$ is expressed as $\tuple {r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta}$ in $\CC$.
Contrariwise, let $p$ be expressed as $\tuple {x, y, z}$ in the cartesian coordinates of $\CC$.
Then $p$ is expressed in $\PP$ as:
:$p = \polar {\sqrt {x^2 + y^2 + z^2}, \arctan \dfrac {\sqrt x^2 + y^2} z, \arctan \dfrac y x}$
where:
:$\theta$ is such that $0 \le \theta < \pi$
:$\phi$ is such that $x : y : r \sin \theta = \cos \phi : \sin \phi : 1$
:$\arctan$ denotes the arctangent function.
== Proof ==
== Sources ==
* 
* 
Category:Cartesian Coordinate Systems
Category:Spherical Coordinates
Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE

== Theorem ==
Consider the Cauchy-Euler equation:
:$(1): \quad x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y = 0$
By making the substitution:
:$x = e^t$
it is possible to convert $(1)$ into a constant coefficient homogeneous linear second order ODE:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$
=== General Result ===
== Proof ==
We have:
Then:
and:
Substituting back into $(1)$:
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.9$: Euler or Cauchy Equation
* : $\S 3.17$: Problem $4$
Category:Cauchy-Euler Equation
Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE/General Result

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
The ordinary differential equation:
:$a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} c + a_0 \, \map f x = 0$
can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$.
== Proof ==
Let $y = \map f x$.
First the following are established:
The proof now proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$
can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus:
:$a_1 x \, \map {f'} x + a_0 \, \map f x = 0$
has been transformed into:
:$a_1 \dfrac \d {\d t} \map f {e^t} + a_0 \, \map f {e^t} = 0$
which is a linear first Order ODE with constant coefficients.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$a_k x^k \, \map {f^{\paren k} } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$
can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$ thus:
Then we need to show that:
:$a_{k + 1} x^{k + 1} \map {f^{\paren {k + 1} } } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$
can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$ thus:
=== Induction Step ===
This is our induction step:
Hence the result by the Principle of Mathematical Induction.
Category:Cauchy-Euler Equation
Conversion of Number in Golden Mean Number System to Simplest Form

== Theorem ==
Let $x \in \R_{\ge 0}$ have a representation $S$ in the golden mean number system.
Then $S$ can be converted to its simplest form as follows:
:$(1): \quad$ Replace any infinite string on the right hand end of $S$ of the form $010101 \ldots$ with $100$
:$(2): \quad$ Repeatedly replace the leftmost instance of $011$ with $100$.
== Proof ==
Note that step $(2)$ is an instance of a simplification of $S$.
From 100 in Golden Mean Number System is Equivalent to 011, it has been established that $011$ is equivalent to $100$.
The following need to be established:
:$010101 \ldots$ is equivalent to $100$
:Replacing the leftmost $011$ with $100$ reduces the overall number of instances of $011$.
Let the first $1$ of $010101 \ldots$ represent the instance of $\phi^n$ for some $n \in \Z$.
It follows that the first $0$ of $010101 \ldots$ represents the instance of $\phi^{n + 1}$ for some $n \in \Z$.
Thus $010101 \ldots$ represents the real number $y$ where:
:$y = \phi^n + \phi^{n - 2} + \phi^{n - 4} + \cdots$
and so:
This can represented in the golden mean number system by $100$, where the $1$ corresponds to the first instance of $\phi^{n + 1}$.
Hence $010101 \ldots$ is equivalent to $100$.
== Sources ==
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Category:Golden Mean Number System