nyuuzyou/wikis huggingface dataset:
Position-Ratio of Point/Examples/R Approaching Q

== Example of Position-Ratio of Point ==
Let $P$ and $Q$ be points in space.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
As $R$ approaches $Q$ where $R$ is on the line segment $PQ$, $k \to +\infty$.
As $R$ approaches $Q$ where $R$ is not on the line segment $PQ$, $k \to -\infty$.
:
== Sources ==
* : Chapter $\text I$. Coordinates: $11$. Position-ratio of a point
Category:Position-Ratios
Position-Ratio of Point/Examples/R Midway between P and Q

== Example of Position-Ratio of Point ==
Let $P$ and $Q$ be points.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
When $R$ is on the midpoint of the line segment $PQ$, $k = +1$.
This is the centroid of $P$ and $Q$.
:
== Sources ==
* : Chapter $\text I$. Coordinates: $12$. Mean points and centres of gravity
Category:Position-Ratios
Position Vector of Midpoint of Line

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be the position vectors of points $A$ and $B$.
The position vector $\mathbf r$ of the midpoint of the line segment $AB$ is given by:
:$\mathbf r = \dfrac {\mathbf a + \mathbf b} 2$
== Proof ==
From Point dividing Line Segment between Two Points in Given Ratio:
:the position vector $\mathbf r$ of a point $R$ on $AB$ which divides $AB$ in the ratio $m : n$ is given by:
:$\mathbf r = \dfrac {n \mathbf a + m \mathbf b} {m + n}$
In this case the ratio $m : n$ is $1 : 1$.
Hence when $\mathbf r$ is the position vector of the midpoint of $AB$:
== Sources ==
* 
* 
Category:Bisection
Category:Vector Algebra
Category:Straight Lines
Position of Card after n Modified Perfect Faro Shuffles

== Theorem ==
Let $D$ be a deck of cards $D$ of size $2 r$.
Let $C$ be a card in position $x$ of $D$.
Let $n$ modified perfect faro shuffles be performed on $C$.
Then $C$ will be in position $w$, where:
:$w \equiv 2^n x \pmod {2 r + 1}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:After $n$ modified perfect faro shuffles, $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod {2 r + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:After $1$ modified perfect faro shuffle, $C$ will be in position $a_1$ where $a_1 \equiv 2 x \pmod {2 r + 1}$
which is the definition of a modified perfect faro shuffle.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:After $k$ modified perfect faro shuffles, $C$ will be in position $a_k$ where $a_k \equiv 2^k x \pmod {2 r + 1}$
from which it is to be shown that:
:After $k + 1$ modified perfect faro shuffles, $C$ will be in position $a_{k + 1}$ where $a_{k + 1} \equiv 2^{k + 1} x \pmod {2 r + 1}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}$: after $n$ modified perfect faro shuffles, $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod {2 r + 1}$
== Sources ==
* : $\text {4-3}$ Riffling
Category:Cards
Position of Cart attached to Wall by Spring

== Theorem ==
=== Problem Definition ===
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = C_1 \cos \alpha t + C_2 \sin \alpha t$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m}$
=== $x = x_0$ at time $t = 0$ ===
== Proof ==
From Motion of Cart attached to Wall by Spring, the horizontal position of $C$ is given as:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac k m \mathbf x = 0$
As both $k$ and $m$ are positive, $\dfrac k m$ can be expressed as:
:$\dfrac k m = \alpha^2$
for some $\alpha \in \R_{>0}$.
Hence:
:$\dfrac {\d^2 x} {\d t^2} + \alpha^2 x = 0$
and so from Linear Second Order ODE: $y + k^2 y = 0$:
:$x = C_1 \cos \alpha t + C_2 \sin \alpha t$
Hence the result.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(4)$
Category:Mechanics
Position of Cart attached to Wall by Spring/x = x0 at t = 0

== Theorem ==
=== Problem Definition ===
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = x_0 \cos \alpha t$
where $\alpha = \sqrt {\dfrac k m}$.
== Proof ==
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as:
:$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m}$
Differentiating $(1)$ $x$ gives:
:$(2): \quad x' = -C_1 \alpha \sin \alpha t + C_2 \alpha \cos \alpha t$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
Hence:
:$x = x_0 \cos \alpha t$
where $\alpha = \sqrt {\dfrac k m}$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(6)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = \begin{cases}
C_1 e^{m_1 t} + C_2 e^{m_1 t} & : b > a \\
& \\
C_1 e^{-a t} + C_2 t e^{-a t} & : b = a \\
& \\
e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) & : b < a
\end{cases}$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
::$m_1 = -b + \sqrt {b^2 - a^2}$
::$m_2 = -b - \sqrt {b^2 - a^2}$
:$\alpha = \sqrt {a^2 - b^2}$
== Proof ==
=== Overdamped ===
=== Critically Damped ===
=== Underdamped ===
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Critically Damped

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b = a$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = C_1 e^{-a t} + C_2 t e^{-a t}$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
Such a system is defined as being critically damped.
=== Initial Conditions: $x = x_0$ at $t = 0$ ===
== Proof ==
When $b = a$, we have $b^2 - a^2 = 0$ and so:
:$m_1 = m_2 = -b = -a$
So from Solution of Constant Coefficient Homogeneous LSOODE: Equal Real Roots of Auxiliary Equation:
:$C_1 e^{-a x} + C_2 x e^{-a x}$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(16)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Critically Damped/x = x0 at t = 0

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b = a$.
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = x_0 e^{-a t} \paren {1 + a t}$
Such a system is defined as being critically damped.
== Proof ==
From Position of Cart attached to Wall by Spring under Damping: Critically Damped:
:$x = C_1 e^{-a t} + C_2 t e^{-a t}$
where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$.
It remains to evaluate $C_1$ and $C_2$ under the given conditions.
Differentiating $(1)$ $t$ gives:
:$(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
Hence:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(17)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Overdamped

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b > a$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
::$m_1 = -b + \sqrt {b^2 - a^2}$
::$m_2 = -b - \sqrt {b^2 - a^2}$
Such a system is defined as being overdamped.
=== Initial Conditions: $x = x_0$ at $t = 0$ ===
== Proof ==
When $b > a$, we have $b^2 - a^2 > 0$ and so $m_1$ and $m_2$ are real and distinct.
So from Solution of Constant Coefficient Homogeneous LSOODE: Real Roots of Auxiliary Equation:
:$\mathbf x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$
where
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(14)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Overdamped/x = x0 at t = 0

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b > a$.
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = \dfrac {x_0} {m_1 - m_2} \paren {m_1 e^{m_2 t} - m_2 e^{m_1 t} }$
where $m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
::$m_1 = -b + \sqrt {b^2 - a^2}$
::$m_2 = -b - \sqrt {b^2 - a^2}$
Such a system is defined as being overdamped.
== Proof ==
From Position of Cart attached to Wall by Spring under Damping: Overdamped:
:$(1): \quad x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$
where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$.
It remains to evaluate $C_1$ and $C_2$ under the given conditions.
Differentiating $(1)$ $t$ gives:
:$(2): \quad x' = C_1 m_1 e^{m_1 t} + C_2 m_2 e^{m_2 t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
Then we have:
Hence:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(15)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Preliminary

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Then the horizontal position of $C$ at time $t$ is determined by the roots of the auxiliary equation:
:$m^2 + 2 b + a^2 = 0$
according to Solution of Constant Coefficient Homogeneous LSOODE.
== Proof ==
From Motion of Cart attached to Wall by Spring under Damping, the horizontal position of $C$ is given as:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$
With the given substitutions $a$ and $b$, this resolves to:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + 2 b \dfrac {\d \mathbf x} {\d t} + a^2 \mathbf x = 0$
This is a homogeneous linear second order ODE with constant coefficients.
Recall that $m_1$ and $m_2$ are the roots of the auxiliary equation:
:$m^2 + 2 b + a^2 = 0$
By Solution to Quadratic Equation with Real Coefficients:
From the initial problem definition, we have that $k, m, c \in \R_{>0}$.
Hence $a, b \in \R_{>0}$.
Hence the nature of $m_1$ and $m_2$ is dependent upon whether $b > a$, $b = a$ or $b < a$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(13)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Underdamped

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b < a$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {a^2 - b^2}$
Such a system is defined as being underdamped.
:
=== Initial Conditions: $x = x_0$ at $t = 0$ ===
== Proof ==
When $b < a$, we have $b^2 - a^2 < 0$ and so:
So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation:
:$\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$
where:
:$\alpha = \sqrt {a^2 - b^2}$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(18)$
Category:Mechanics
Position of Cart attached to Wall by Spring under Damping/Underdamped/x = x0 at t = 0

== Theorem ==
=== Problem Definition ===
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
Let $b < a$.
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$
where $\alpha = \sqrt {a^2 - b^2}$.
Such a system is defined as being underdamped.
:
== Proof ==
From Position of Cart attached to Wall by Spring under Damping: Underdamped:
:$(2): \quad x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$.
It remains to evaluate $C_1$ and $C_2$ under the given conditions.
Differentiating $(1)$ $t$ gives:
:$(2): \quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
Hence:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(19)$
Category:Mechanics
Position of Centroid of Triangle on Median

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$ meeting at the centroid $G$ of $\triangle ABC$.
Then $G$ is $\dfrac 1 3$ of the way along $AL$ from $L$, and similarly for the other medians.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* 
* 
Category:Position of Centroid of Triangle on Median
Category:Centroids of Triangles
Position of Centroid of Triangle on Median/Proof 1

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be embedded in a Cartesian plane such that $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$.
The coordinates of $L$ are $\tuple {\dfrac {x_2 + x_3} 2, \dfrac {y_2 + y_3} 2}$.
Let $G$ be the point dividing $AL$ in the ratio $2 : 1$.
The coordinates of $G$ are $\tuple {\dfrac {x_1 + \paren {x_2 + x_3} } {1 + 2}, \dfrac {y_1 + \paren {y_2 + y_3} } {1 + 2} }$.
By similarly calculating the coordinates of $M$ and $N$, we get:
Similarly:
:calculating the position of the point $G'$ dividing $BM$ in the ratio $2 : 1$
:calculating the position of the point $G$ dividing $CN$ in the ratio $2 : 1$
we find that:
:$G = G' = G = \tuple {\dfrac {x_1 + x_2 + x_3} 3, \dfrac {y_1 + y_2 + y_3} 3}$
and the result follows.
== Sources ==
* : Chapter $\text I$. Coordinates: $12$. Mean points and centres of gravity
* : $\text {III}$. Analytical Geometry: The Straight Line: Centre of gravity of the triangle $ABC$
Category:Position of Centroid of Triangle on Median
Position of Centroid of Triangle on Median/Proof 2

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$.
Let the medians be concurrent at the centroid, $G$.
By the definition of median, the sides of $\triangle ABC$ are bisected.
Hence:
:the distance from the vertex to the centroid is $2/3$ of the whole length of the median.
and so:
:the distance from the centroid to the side is $1/3$ of the whole length of the median.
The result follows.
Category:Position of Centroid of Triangle on Median
Position of Centroid of Triangle on Median/Proof 3

== Theorem ==
== Proof ==
:
By Medians of Triangle Form Six Triangles of Equal Area‎:
:these six triangles formed by the medians of $\triangle ABC$ have equal area:
* $\triangle AGN$
* $\triangle BGN$
* $\triangle BGL$
* $\triangle CGL$
* $\triangle CGM$
* $\triangle AGM$
 consider one of the three medians of $\triangle ABC$, $AGL$.
The following triangles both have their base on $AGL$ and share vertex $C$:
* $\triangle CGA$
* $\triangle CGL$
Since $\triangle CGA$ contains two of the small triangles:
Since 
:$\AA \paren { \triangle AGM } = \AA \paren { \triangle CGM } = \AA \paren { \triangle CGL }$
:
:$BG = 2 \cdot GM$
:$CG = 2 \cdot GN$
Hence:
:the distance from the vertex to the centroid is $2/3$ of the whole length of the median.
and so:
:the distance from the centroid to the side is $1/3$ of the whole length of the median.
The result follows.
Category:Position of Centroid of Triangle on Median
Position of Centroid on Euler Line

== Theorem ==
Let $\triangle ABC$ be a triangle which is not equilateral.
Let $O$ be the circumcenter of $\triangle ABC$.
Let $G$ be the centroid of $\triangle ABC$.
Let $H$ be the orthocenter of $\triangle ABC$.
Then $G$ lies on the straight line connecting $O$ and $H$ such that:
:$OG : GH = 1 : 2$
The line $OGH$ is the Euler line of $\triangle ABC$.
== Proof ==
First it is necessary to dispose of the case where $\triangle ABC$ is equilateral.
From Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral, in that case $O$, $G$ and $H$ are the same point.
For all other triangles, $O$, $G$ and $H$ are distinct.
:
Let $A'$ be the midpoint of $BC$.
Let $B'$ be the midpoint of $AC$.
By definition of centroid, $G$ is the point at which $AA'$ and $BB'$ intersect.
By Circumscribing Circle about Triangle, $O$ is the point at which the perpendicular bisectors of $AC$ and $BC$ intersect.
By construction, the perpendicular bisectors of $BC$ and $AC$ pass through $A'$ and $B'$ respectively.
Let $OG$ be produced to $H$ such that $OG : HG = 1 : 2$.
It will be demonstrated that $H$ is the orthocenter of $\triangle ABC$.
Draw $AH$ and produce it to intersect $BC$ at $P$.
From Medians of Triangle Meet at Centroid:
:$A'G : AG = 1 : 2$
Consider $\triangle A'OG$ and $\triangle AHG$.
From above:
:$OG : HG = A'G : AG$
From Two Straight Lines make Equal Opposite Angles:
:$\angle OGA' = \angle HGA$
From Triangles with One Equal Angle and Two Sides Proportional are Similar:
:$\triangle A'OG$ and $\triangle AHG$ are similar.
Thus:
:$\angle GHA = \angle GOA'$
From Equal Alternate Angles implies Parallel Lines:
:$AH \parallel A'O$
From Parallelism implies Equal Corresponding Angles:
:$\angle APC = \angle OA'C$
As $OA'$ is perpendicular to $BC$, it follows that $AP$ is also perpendicular to $BC$.
Thus $AP$ is a straight line through $A$ perpendicular to $BC$.
So by definition $AP$ is an altitude of $\triangle ABC$ which passes through $H$.
Similarly, draw $BH$ and produce it to intersect $AC$ at $Q$.
By a similar analysis of the triangles $\triangle B'OG$ and $\triangle BHG$:
:$BQ$ is an altitude of $\triangle ABC$ which passes through $H$.
From Altitudes of Triangle Meet at Point, the altitude of $\triangle ABC$ from $C$ to $AB$ likewise will pass through $H$.
Thus $H$ is by definition the orthocenter of $\triangle ABC$.
By construction, $OGH$ is a straight line such that:
:$OG : GH = 1 : 2$
Hence the result.
== Historical Note ==
== Sources ==
* : $3$
* : $3$
* 
* 
* 
* 
Category:Euler Line
Category:Orthocenters of Triangles
Category:Circumcircles of Triangles
Category:Centroids of Triangles
Position of Centroid on Euler Line/Historical Note

== Historical Note on Position of Centroid on Euler Line ==
It was who published the result that the circumcenter, centroid and orthocenter of a triangle all lie on the same straight line.
He also determined the ratio of the distances between these points.
Hence this line is now known as the triangle's Euler line.
== Sources ==
* : $3$
* : $3$
Category:Historical Notes
Position of Interpolated Point under Linear Interpolation

== Theorem ==
Let $f$ be a real function.
Let $y_1, y_2, \ldots, y_n$ be known values of $f$ corresponding to $x_1, x_2, \ldots, x_n$ respectively.
Let $x'$ be in the domain of $f$ such that $x_i < x' < x_{i + 1}$.
Let $y' = \map f {x'}$ be determined according to linear interpolation.
Then:
:$y' = y_i + \dfrac {\paren {x' - x_i} \paren {y_{i + 1} - x_i} } {x_{i + 1} - x_i}$
== Proof ==
By definition of linear interpolation, the three points $\tuple {x_i, y_i}$, $\tuple {x', y'}$ and $\tuple {x_{i + 1}, y_{i + 1} }$ are all collinear.
Then:
== Sources ==
* 
* 
Category:Linear Interpolation
Position of Underdamped Cart attached to Wall by Spring under Forced Vibration

== Theorem ==
=== Problem Definition ===
Then the horizontal position of $C$ at time $t$ can be expressed as:
:$\mathbf x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
:$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$
== Proof ==
From Forced Vibration of Cart attached to Wall by Spring, the motion of $C$ is described by the second order ODE:
:$(1): \quad m \dfrac {\d^2 \mathbf x} {\d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$
Setting:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
this can be written as:
:$(2): \quad \dfrac {\d^2 \mathbf x} {\d t^2} + 2 b \dfrac {\d \mathbf x} {\d t} + a^2 \mathbf x = \dfrac {\mathbf F_0} m \cos \omega t$
We are given that $C$ is underdamped, and so $b < a$.
From Linear Second Order ODE: $y + 2 b y' + a^2 y = K \cos \omega x$ for $b < a$, $(2)$ has the general solution:
:$\mathbf x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega t - \phi}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
Substituting back:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
:$K = \dfrac {\mathbf F_0} m$
we obtain:
and:
and:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(28)$
Category:Mechanics
Positions of Instances of a in Fibonacci String

== Theorem ==
Let $S_n$ denote the $n$th Fibonacci string.
Let $F_n$ denote the $n$th Fibonacci number.
Let $m \in \Z$ such that $m \le F_n$.
Let $m - 1$ be expressed in Zeckendorf representation as $Z_{m - 1}$.
Then the $m$th letter of $S_n$ is $\text a$ :
: $k_r = 2$
where $k_r$ denotes the final digit of $Z_{m - 1}$.
== Proof ==
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Positions of Instances of b in Fibonacci String

== Theorem ==
Let $S_n$ denote the $n$th Fibonacci string.
Let $F_n$ denote the $n$th Fibonacci number.
Let $k \in \Z$ such that $k \le F_n$.
Then the $k$th letter of $S_n$ is $\text b$ :
: $\left\lfloor{\left({k + 1}\right) \phi^{-1} }\right\rfloor - \left\lfloor{k \phi^{-1} }\right\rfloor = 1$
where:
: $\left\lfloor{\, \cdot \,}\right\rfloor$ denotes the floor function
: $\phi$ denotes the golden mean.
== Proof ==
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Positive-Term Generalized Sum Converges iff Supremum

== Theorem ==
Let $\struct {G, \circ, \le}$ be an abelian totally ordered group, considered under the order topology.
Let $\set {x_i: i \in I}$ be an indexed set of positive elements of $G$.
Then:
:the generalized sum $\ds \sum \set {x_i: i \in I}$ converges to a point $x \in G$
:$x$ is the supremum of:
::$P := \ds \set {\sum_{i \mathop \in F} x_i: \text{$F \subseteq I$ and $F$ is finite} }$
== Proof ==
=== Sufficient Condition ===
Let $\ds \sum \set {x_i: i \in I}$ converge to $x$.
We first show that $x$ is an upper bound of $P$.
 that for some finite subset $F$ of $I$:
:$\ds \sum_{i \mathop \in F} x_i > x$
Then the net of finite sums is eventually less than $\ds \sum_{i \mathop \in F} x_i$.
Let $F \subseteq F' \subseteq I$.
Let $F'$ be finite.
Since finite sums are monotone:
:$\ds \sum_{i \mathop \in F'} x_i \ge \sum_{i \mathop \in F} x_i$
which is a contradiction.
Thus we conclude that $x$ is an upper bound of $P$.
Let $m \in G$ such that $m < x$.
Then the net of finite sums is eventually greater than $m$.
Thus $m$ is certainly not an upper bound of $P$.
So we have shown that $x$ is the supremum of $P$.
=== Necessary Condition ===
Let $x$ be the supremum of $P$.
Then:
:$\forall b > x: \forall y \in P: b > y$
so the net of finite sums is always (hence eventually) less than $b$.
For all $a < x$, $a$ is not an upper bound of $P$.
Therefore there exists a finite subset $F$ of $I$ such that:
:$\ds \sum_{i \mathop \in F} x_i > a$
Since finite sums are monotone increasing, the net of finite sums is eventually greater than $a$.
Thus $\ds \sum \set {x_i: i \in I}$ converges to $x$.
Category:Topology
Category:Group Theory