nyuuzyou/wikis huggingface dataset:
Radical of Power of Prime Ideal

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak p \subseteq A$ be a prime ideal.
Let $n > 0$ be a natural number.
Then the radical of the $n$th power of $\mathfrak p$ equals $\mathfrak p$:
:$\map {\operatorname{Rad} } {\mathfrak p^n} = \mathfrak p$
== Proof ==
$(\subseteq):$
Let $a \in \map {\operatorname{Rad} } {\mathfrak p^n}$.
Then by definition, $a^k \in \mathfrak p^n$ for some integer $k$.
By Power of Ideal is Subset, $\mathfrak p^n \subseteq \mathfrak p$.
Hence, $a^k \in \mathfrak p$.
By Power in Prime Ideal, $a \in \mathfrak p$.
$(\supseteq):$
Let $b \in \mathfrak p$.
Then by definition $b^n \in \mathfrak p^n$.
Hence $b \in \map {\operatorname{Rad} } {\mathfrak p^n}$.
Category:Radical of Ideals
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
Radical of Primary Ideal is Smallest Larger Prime Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak q$ be a primary ideal of $R$.
Let $\map \Rad {\mathfrak q}$ be the radical of $\mathfrak q$.
Then $\map \Rad {\mathfrak q}$ is the smallest prime ideal containing $\mathfrak q$.
== Proof ==
First, we show that $\map \Rad {\mathfrak q}$ is a prime ideal.
Let $x y \in \map \Rad {\mathfrak q}$.
Then by definition of radical of ideal:
:$\exists n \in \N_{>0} : \paren {xy}^n \in \mathfrak q$
By Commutativity $(\text M 2)$:
:$x^n y^n \in \mathfrak q$
By definition of primary ideal:
:$x^n \in \mathfrak q \lor \exists m \in \N_{>0} : \paren {y^n}^m = y^{n m} \in \mathfrak q$
Thus, by definition of radical of ideal:
:$x \in \map \Rad {\mathfrak q} \lor y \in \map \Rad {\mathfrak q}$
So, $\map \Rad {\mathfrak q}$ is a prime ideal.
Finally, by Prime Ideal Including Ideal Includes Radical, every prime ideal $\mathfrak p$ satisfies:
:$\mathfrak q \subseteq \mathfrak p \implies \map \Rad {\mathfrak q} \subseteq \mathfrak p$
That is, $\map \Rad {\mathfrak q}$ is the smallest prime ideal including $\mathfrak q$.
Category:Radical of Ideals
Category:Primary Ideals
Category:Prime Ideals of Rings
Radical of Prime Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Let $\map \Rad {\mathfrak p}$ be the radical of $\mathfrak p$.
Then:
:$\map \Rad {\mathfrak p} = \mathfrak p$
== Proof ==
=== $\supseteq$ ===
Let $x \in \mathfrak p$.
Since $x = x^1$, by :
:$x \in \map \Rad {\mathfrak p}$
=== $\subseteq$ ===
Let $x \in \map \Rad {\mathfrak p}$.
Then:
:$\exists n \in \N_{>0} : x^n \in \mathfrak p$
 $x \notin p$.
Then by :
:$x^n \notin \mathfrak p$
which contradicts the assertion that $x^n \in \mathfrak p$.
Thus:
:$x \in \mathfrak p$
Category:Radical of Ideals
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
Radical of Sum of Ideals

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be ideals.
Then for the radical of their sum we have:
:$\map \Rad {\mathfrak a + \mathfrak b} = \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$
== Proof ==
From Ideal of Ring is Contained in Radical:
:$\mathfrak a \subseteq \map \Rad {\mathfrak a}$
From Sum of Larger Ideals is Larger:
:$\mathfrak a + \mathfrak b \subseteq \map \Rad {\mathfrak a} + \map \Rad {\mathfrak b}$
From Radical of Ideal Preserves Inclusion:
:$\map \Rad {\mathfrak a + \mathfrak b} \subseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$
Let $x \in \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$.
Then there exists $n \in \N$ such that the power $x^n \in \map \Rad {\mathfrak a} + \map \Rad {\mathfrak b}$.
By definition of sum of ideals, there exist $a \in \map \Rad {\mathfrak a}$ and $b \in \map \Rad {\mathfrak b}$ with $x^n = a + b$.
Let $p, q \in \N$ with $a^p \in \mathfrak a$ and $b^q \in \mathfrak b$.
By the Binomial Theorem:
:$\paren {a + b}^{p + q - 1} \in \mathfrak a + \mathfrak b$
Thus:
:$x^{n \cdot \paren {p + q - 1} } \in \mathfrak a + \mathfrak b$
That is:
:$x \in \map \Rad {\mathfrak a + \mathfrak b}$
Hence by definition of subset:
:$\map \Rad {\mathfrak a + \mathfrak b} \supseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$
The result follows by definition of set equality.
Category:Radical of Ideals
Radical of Unit Ideal

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\ideal 1$ be its unit ideal.
Then its radical equals $\ideal 1$:
:$\map \Rad {\ideal 1} = \ideal 1$.
== Proof ==
By definition of ideal:
:$\map \Rad {\ideal 1} \subseteq A$
By Ideal of Ring is Contained in Radical:
:$\ideal 1 = A \subseteq \Rad {\ideal 1}$.
By definition of set equality:
:$\map \Rad {\ideal 1} = \ideal 1$
Category:Radical of Ideals
Radiocarbon Dating/Example/7000 BCE

== Example of Radiocarbon Dating ==
It was reported in the newspaper that a prehistoric village had been dated using radiocarbon dating.
The carbon-$14$ test had been used to measure the amount of radioactivity still present in the organic material found in the ruins.
It was determined that a settlement existed in that place as long ago as .
Given that:
:the estimated half-life of carbon-$14$ used was $5600$ years
:the newspaper report dated from about $1960$
the carbon-$14$ test must have shown that approximately $32 \%$ or $33 \%$ of carbon-$14$ was still present in the organic material at the time of discovery.
== Proof ==
From First-Order Reaction, we have:
:$x = x_0 e^{-k t}$
where:
:$x$ is the quantity of carbon-$14$ at time $t$
:$x_0$ is the quantity of carbon-$14$ at time $t = 0$
:$k$ is a positive number.
By definition of half-life, when $x = \dfrac {x_0} 2$, we have $t = 5600$.
So:
:$e^{-5600 k} = \dfrac 1 2$
So:
:$k = \dfrac {\ln 0.5} {-5600} = \dfrac {\ln 2} {5600}$
After $7000 + 1960$ years:
So there is between $32 \%$ and $33 \%$ remaining.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate: Exercise $1.3$
Category:Examples of Radiometric Dating
Radiocarbon Dating/Examples/Lascaux Caves

== Example of Radiocarbon Dating ==
Charcoal from the remains of the fire found in the was analysed.
It was determined that $85.5 \%$ of the amount of carbon-14 had decomposed since the wood was alive.
The Formula for Radiocarbon Dating is used:
:$t = -8060 \ln r$
where:
:$t$ denotes the age in years of the sample of wood which is to be determined
:$r$ denotes ratio of the quantity of carbon-14 remaining in the sample after time $t$ to the quantity of carbon-14 in the sample at the time of its death.
Thus:
Hence it is determined that the fire in the cave, and hence the dwellers therein, dates from approximately $15 \, 500$ years ago.
== Historical Note ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions
Category:Examples of Radiometric Dating
Radiocarbon Dating/Examples/Lascaux Caves/Historical Note

== Historical Note on Lascaux Caves ==
The were discovered on $12$ September $1940$.
The walls of the caves were found to be covered in artwork from thousands of years ago.
Within the cave were found the remains of a fire.
By using radiocarbon dating on the charcoal remains of that fire, it was possible to obtain an approximate date of when the caves were occupied.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate
Category:Historical Notes
Radiometric Dating/Example/Radium in Lead

== Example of Radiometric Dating ==
The radium in a piece of lead decays at a rate governed by a first-order reaction.
=== Example: Percentage after $1000$ Years ===
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate: Exercise $1.1$
Category:Examples of Radiometric Dating
Radiometric Dating/Example/Radium in Lead/1000 Years

== Example of Radiometric Dating ==
Let $Q$ be a sample of lead.
Let it be established that $10 \%$ of the radium decays in $200$ years.
After $1000$ years, there will be approximately $59.05 \%$ of the original amount of radium in $Q$.
== Proof ==
From First-Order Reaction, we have:
:$x = x_0 e^{-k t}$
where:
:$x$ is the quantity of radium at time $t$
:$x_0$ is the quantity of radium at time $t = 0$
:$k$ is a positive number.
We are given that when $t = 200$, $x = 0.9 \times x_0$.
Hence:
So after $1000$ years, we have:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate: Exercise $1.1$
Category:Examples of Radiometric Dating
Radiometric Dating/Example/Radium in Lead/100 Years

== Example of Radiometric Dating ==
Let $Q$ be a sample of lead.
It is established that the half-life of radium-226 is $1600$ years.
So, after $100$ years, the amount of radium-226 that has decayed will be $4.2 \%$.
== Proof ==
From First-Order Reaction, we have:
:$x = x_0 e^{-k t}$
where:
:$x$ is the quantity of radium at time $t$
:$x_0$ is the quantity of radium at time $t = 0$
:$k$ is a positive number.
By definition of half-life, when $x = \dfrac {x_0} 2$, we have $t = 1600$.
So:
:$e^{-1600 k} = \dfrac 1 2$
So:
:$k = \dfrac {\ln 0.5} {-1600} = \dfrac {\ln 2} {1600}$
After $100$ years:
So there is $95.76\%$ remaining, and so $4.2\%$ will be lost.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate: Exercise $1.2$
Category:Examples of Radiometric Dating
Radius at Right Angle to Tangent

== Theorem ==
== Proof ==
:
Let $DE$ be tangent to the circle $ABC$ at $C$.
Let $F$ be the center of $ABC$.
Let $FC$ be the radius in question.
 $FC$ were not perpendicular to $DE$.
Instead, suppose $FG$ were drawn perpendicular to $DE$.
Since $\angle FGC$ is a right angle, then from Two Angles of Triangle are Less than Two Right Angles it follows that $\angle FCG$ is acute.
From Greater Angle of Triangle Subtended by Greater Side it follows that $FC > FG$.
But $FC = FB$ and so $FB > FG$, which is impossible.
So $FG$ can not be perpendicular to $DE$.
Similarly it can be proved that no other straight line except $FC$ can be perpendicular to $DE$.
Therefore $FC$ is perpendicular to $DE$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Tangents to Circles
Radius of Convergence from Limit of Sequence/Complex Case

== Theorem ==
Let $\xi \in \C$ be a complex number.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.
Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converge.
Then $R$ is given by:
:$\ds \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$
If:
:$\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$
then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.
== Proof ==
Let the sequence $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ converge.
Let $\epsilon \in \R_{>0}$, and let $z \in \C$.
Let $\cmod {z - \xi} = R - \epsilon$.
By definition of radius of convergence, it follows that $\map S z$ is absolutely convergent.
Then from the Ratio Test:
:$\lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \le 1$
By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:
Let $\cmod {z - \xi} = R + \epsilon$.
Then $\map S z$ is divergent, so the Ratio Test shows that:
:$\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \ge 1$
Similarly, this inequality can be rearranged as:
:$\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } \ge \dfrac 1 {R + \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
:$\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = \dfrac 1 R$
Hence the result.
== Also presented as ==
This result can also be seen presented as:
:$\ds R = \lim_{n \mathop \to \infty} \cmod {\frac {a_{n - 1} } {a_n} }$
but in this case the condition under which the radius of convergence is infinite is less conveniently stated.
Category:Radius of Convergence from Limit of Sequence
Category:Complex Power Series
Radius of Convergence from Limit of Sequence/Real Case

== Theorem ==
Let $\xi \in \R$ be a real number.
Let $\ds S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.
Then the radius of convergence $R$ of $S \paren x$ is given by:
:$\ds \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$
if this limit exists and is nonzero.
If
:$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$
then the radius of convergence is infinite and therefore the interval of convergence is $\R$.
== Proof ==
From the ratio test, $S \paren x$ is convergent if:
:$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$
Thus:
The result follows from the definition of radius of convergence.
== Also presented as ==
This result can also be seen presented as:
:$\ds R = \lim_{n \mathop \to \infty} \size {\frac {a_{n - 1} } {a_n} }$
but in this case the condition under which the radius of convergence is infinite is less conveniently stated.
== Sources ==
* : $\S 15.2$
Category:Radius of Convergence from Limit of Sequence
Category:Real Power Series
Radius of Convergence of Derivative of Complex Power Series

== Theorem ==
Let $\xi \in \C$.
For all $z \in \C$, define the power series:
:$\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$
and:
:$\ds \map {S'} z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Let $R$ be the radius of convergence of $\map S z$, and let $R'$ be the radius of convergence of $\map {S'} z$.
Then $R = R'$.
== Proof ==
Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$.
Then $\map {S'} z$ converges absolutely by Existence of Radius of Convergence of Complex Power Series, so:
From the $n$th Root Test, it follows that $S \paren z$ converges absolutely.
Hence, $R \ge R'$.
Suppose that $z \in \C$ with $\cmod {z - \xi} < R$.
Find $z_o \in \C$ such that $\cmod {z - \xi} < \cmod {z_0 - \xi} < R$.
By Existence of Radius of Convergence of Complex Power Series, $\map S {z_0}$ converges absolutely.
From the $n$th Root Test, it follows that $\cmod {a_n \paren {z_0 - \xi}^n}^{1/n} < 1$ for all $n \ge N$ for some $N \in \N$.
Then:
From the $n$th Root Test, it follows that $\map {S'} z$ converges absolutely.
Hence, $R' \ge R$, so $R' = R$.
== Sources ==
* : $\S 1.4$
Category:Complex Power Series
Category:Complex Analysis
Radius of Convergence of Power Series Expansion for Cosine Function

== Theorem ==
The cosine function has the complex power series expansion:
which is the power series expansion of the cosine function.
This is valid for all $z \in \C$.
== Proof ==
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that:
Hence the result.
== Also see ==
* Power Series Expansion for Cosine Function
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(iii)}$
Category:Radius of Convergence
Radius of Convergence of Power Series Expansion for Cosine Function/Mistake

== Source Work ==
:
: Chapter $4$: Elementary Functions of a Complex Variable: 
:: Section $4$: Power Series:
::: Example $\text{(iii)}$
This mistake can be seen in the $1960$ edition as published by Routledge & Kegan Paul.
== Mistake ==
:The series $C \paren z = 1 - \dfrac {z^2} {2!} + \dfrac {z^4} {4!} - z \dfrac 6 {6!} + \cdots$ and $S \paren z = z - \dfrac {z^3} {3!} + \dfrac {z^5} {5!} - \dfrac {z^7} {7!} + \cdots$ also converge for all $z$, ...
== Correction ==
The expression for $C \paren z$ should read $1 - \dfrac {z^2} {2!} + \dfrac {z^4} {4!} - \dfrac {z^6} {6!} + \cdots$
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(iii)}$
Category:Mistakes/Complex Numbers/Walter Ledermann
Radius of Convergence of Power Series Expansion for Sine Function

== Theorem ==
The sine function has the complex power series expansion:
which is the power series expansion of the sine function.
This is valid for all $z \in \C$.
== Proof ==
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that:
Hence the result.
== Also see ==
* Power Series Expansion for Sine Function
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(iii)}$
Category:Radius of Convergence
Radius of Convergence of Power Series in Complex Plane

== Theorem ==
Consider the complex power series:
:$S = \ds \sum_{k \mathop = 0}^\infty z^n$
The radius of convergence $S$ is $1$.
== Proof ==
By the Ratio Test, it follows that:
:$S$ is convergent for $\cmod z < 1$
:$S$ is divergent for $\cmod z > 1$.
Hence the result by definition of radius of convergence.
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(i)}$
Category:Complex Power Series
Category:Radius of Convergence
Radius of Convergence of Power Series over Factorial/Complex Case

== Theorem ==
Let $\xi \in \C$ be a complex number.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$.
Then $\map f z$ converges absolutely for all $z \in \C$.
That is, the radius of convergence of the power series $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.
== Proof ==
This is a power series in the form $\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$.
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that:
Hence the result.
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(ii)}$
Category:Radius of Convergence of Power Series over Factorial
Category:Complex Power Series
Radius of Convergence of Power Series over Factorial/Real Case

== Theorem ==
Let $\xi \in \R$ be a real number.
Let $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!}$.
Then $\map f x$ converges for all $x \in \R$.
That is, the interval of convergence of the power series $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!}$ is $\R$.
== Proof ==
This is a power series in the form $\ds \sum_{n \mathop= 0}^\infty a_n \paren {x - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$.
Applying Radius of Convergence from Limit of Sequence, we find that:
Hence the result.
== Sources ==
* : $\S 15.3$
Category:Radius of Convergence of Power Series over Factorial
Category:Real Power Series
Radius of Curvature in Cartesian Form

== Theorem ==
Let $C$ be a curve defined by a real function which is twice differentiable.
Let $C$ be embedded in a cartesian plane.
The radius of curvature $\rho$ of $C$ at a point $P = \tuple {x, y}$ is given by:
:$\rho = \dfrac {\paren {1 + y'^2}^{3/2} } {\size {y' '} }$
where:
:$y' = \dfrac {\d y} {\d x}$ is the derivative of $y$ $x$ at $P$
:$y' ' = \dfrac {\d^2 y} {\d x^2}$ is the second derivative of $y$ $x$ at $P$.
== Proof ==
By definition, the radius of curvature $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the curvature, given in Cartesian form as:
:$\kappa = \dfrac {y' '} {\paren {1 + y'^2}^{3/2} }$
As $\paren {1 + y'^2}^{3/2}$ is positive, it follows that:
:$\size {\dfrac {y' '} {\paren {1 + y'^2}^{3/2} } } = \dfrac {\size {y' '} } {\paren {1 + y'^2}^{3/2} }$
Hence the result.
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(i)}$ Radius of curvature
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Radius of Curvature
Radius of Curvature in Parametric Cartesian Form

== Definition ==
Let $C$ be a curve defined by a real function which is twice differentiable.
Let $C$ be embedded in a cartesian plane and defined by the parametric equations:
:$\begin{cases} x = \map x t \\ y = \map y t \end{cases}$
The radius of curvature $\rho$ of $C$ at a point $P = \tuple {x, y}$ is given by:
:$\rho = \size {\dfrac {\tuple {x'^2 + y'^2}^{3/2} } {x' y - y' x} }$
where:
:$x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ $t$ at $P$
:$y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ $t$ at $P$
:$x$ and $y$ are the second derivatives of $x$ and $y$ $t$ at $P$.
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
:$\kappa = \dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} }$
where:
:$x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ $t$ at $P$
:$y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ $t$ at $P$
:$x$ and $y$ are the second derivatives of $x$ and $y$ $t$ at $P$.
By definition, the radius of curvature $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the curvature, given in parametric Cartesian form as:
:$\kappa = \dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} }$
As $\tuple {x'^2 + y'^2}^{3/2}$ is positive, it follows that:
:$\size {\dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} } } = \dfrac {\size {x' y - y' x} } {\tuple {x'^2 + y'^2}^{3/2} }$
Hence the result.
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(i)}$ Radius of curvature
Category:Radius of Curvature
Radius of Curvature in Whewell Form

== Theorem ==
Let $C$ be a curve defined by a real function which is twice differentiable.
The radius of curvature $\kappa$ of $C$ at a point $P$ can be expressed in the form of a Whewell equation as:
:$\rho = \size {\dfrac {\d s} {\d \psi} }$
where:
:$s$ is the arc length of $C$
:$\psi$ is the turning angle of $C$
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
By definition, the radius of curvature $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the curvature, given in Whewell form as:
:$\kappa = \dfrac {\d \psi} {\d s}$
Hence the result.
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(i)}$ Radius of curvature
Category:Radius of Curvature
Radius of Gyration of Rotating Body

== Theorem ==
Let $B$ be a rigid body which is rotating in space about some axis $\LL$.
The radius of gyration of $B$ about $\LL$ is:
:$k = \sqrt {\dfrac I m}$
where:
:$I$ is the moment of inertia of $B$ about $\LL$
:$m$ is the mass of $B$.
== Sources ==
* 
* 
Category:Radius of Gyration
Category:Moments of Inertia
Radon's Theorem

== Theorem ==
Let $S$ be a set of $n + 2$ points in a real Euclidean space $\R^n$ of $n$ dimensions.
Then $S$ can be partitioned into $2$ subsets whose convex hulls have at least one point in their intersection.
Such a point of intersection is known as a Radon point of $S$.
== Proof ==
== Also see ==
* Definition:Radon Point
== Historical Note ==
== Sources ==
* 
Category:Radon's Theorem
Category:Convex Hulls