nyuuzyou/wikis huggingface dataset:
Rising Factorial in terms of Falling Factorial of Negative

== Theorem ==
:$x^{\overline k} = \paren {-1}^k \paren {-x}^{\underline k}$
where:
:$x^{\underline k}$ is the $k$ to the $x$ falling
:$x^{\overline k}$ is the $k$ to the $x$ rising.
== Proof ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(20)$
Category:Falling Factorials
Category:Rising Factorials
Rising Sum of Binomial Coefficients

== Theorem ==
Let $m, n \in \Z_{\ge 0}$ be positive integers.
Then:
:$\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1} = \binom {n + m + 1} m$
where $\dbinom n k$ denotes a binomial coefficient.
That is:
:$\dbinom n n + \dbinom {n + 1} n + \dbinom {n + 2} n + \cdots + \dbinom {n + m} n = \dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$
== Proof by Induction ==
== Direct Proof ==
== Proof 3 ==
== Marginal cases ==
Just to make sure, it is worth checking the marginal cases:
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.9$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.33$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.9.$
Category:Rising Sum of Binomial Coefficients
Category:Binomial Coefficients
Rising Sum of Binomial Coefficients/Direct Proof

== Theorem ==
== Direct Proof ==
This proof adds up all the terms of the summation to obtain the desired result.
Since the first term equals $1$, it may be replaced with $\dbinom {n + 1} {n + 1}$.
So:
:$\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + 1} {n + 1} + \sum_{j \mathop = 1}^m \binom {n + j} n$
The sum will be computed in $m$ steps, combining the first two terms with Pascal's Rule in each step. 
Step 1:
:$\ds \binom {n + 1} {n + 1} + \binom {n + 1} n + \sum_{j \mathop = 2}^m \binom {n + j} n = \binom {n + 2} {n + 1} + \sum_{j \mathop = 2}^m \binom {n + j} n$
Step 2:
:$\ds \binom {n + 2} {n + 1} + \binom {n + 2} n + \sum_{j \mathop = 3}^m \binom {n + j} n = \binom {n + 3} {n + 1} + \sum_{j \mathop = 3}^m \binom {n + j} n$ 
After $m - 1$ steps, we obtain:
:$\dbinom {n + m} {n + 1} + \dbinom {n + m} n$
Step $m$:
:$\dbinom {n + m} {n + 1} + \dbinom {n + m} n = \dbinom {n + m + 1} {n + 1}$
Hence the result.
Category:Rising Sum of Binomial Coefficients
Rising Sum of Binomial Coefficients/Marginal Cases

== Theorem ==
=== $n = 0$ ===
When $n = 0$ we have:
So the theorem holds for $n = 0$.
=== $n = 1$ ===
When $n = 1$ we have:
So the theorem holds for $n = 1$.
From Symmetry Rule for Binomial Coefficients:
:$\dbinom {m + n} n = \dbinom {m + n} m$
thus:
:$\dbinom {m + 1} 1 = \dbinom {m + 1} m$
and:
:$\dbinom {m + 2} 2 = \dbinom {m + 2} m$
Category:Rising Sum of Binomial Coefficients
Rising Sum of Binomial Coefficients/Proof by Induction

== Theorem ==
== Proof ==
Proof by induction:
Let $n \in \Z$.
For all $m \in \N$, let $\map P m$ be the proposition:
:$\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$
$\map P 0$ is true, as this just says:
:$\dbinom n n = \dbinom {n + 1} {n + 1}$
But $\dbinom n n = \dbinom {n + 1} {n + 1} = 1$ from the .
=== Basis for the Induction ===
$\map P 1$ is the case:
So:
:$\ds \sum_{j \mathop = 0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$
and $\map P 1$ is seen to hold.
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:
:$\ds \sum_{j \mathop = 0}^k \binom {n + j} n = \binom {n + k + 1} {n + 1}$
Then we need to show:
:$\ds \sum_{j \mathop = 0}^{k+1} \binom {n + j} n = \binom {n + k + 2} {n + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$
Finally, from Symmetry Rule for Binomial Coefficients:
:$\dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$
== Proof ==
* : $\text {3-1}$ Permutations and Combinations: Exercise $5$
Category:Rising Sum of Binomial Coefficients
Ritz Method implies Not Worse Approximation with Increased Number of Functions

== Theorem ==
Consider the Ritz method.
Let $\eta_n = \boldsymbol \alpha \boldsymbol \phi$.
Let $J \sqbrk {\eta_n} = \mu_n$.
Then:
:$\mu_n \ge \mu_{n + 1}$
== Proof ==
Denote $\eta_{n + 1} = \eta_n + \alpha_{n + 1} \phi_{n + 1}$.
For $\alpha_{n + 1} = 0$ we have:
:$\eta_n = \eta_{n + 1}$
Suppose that $J \sqbrk {\eta_n}$ has been minimised $\boldsymbol \alpha$.
If:
:$\exists \boldsymbol \alpha \in \R^n: \nexists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$
then $J \sqbrk {\eta_{n + 1} }$ is minimised for previously determined $\boldsymbol \alpha$ and $\alpha_{n + 1} = 0$:
:$J \sqbrk {\eta_{n + 1} } = J \sqbrk {\eta_n}$
where $\eta_{n + 1} = \eta_n$.
Suppose:
:$\exists \boldsymbol \alpha \in \R^n: \exists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$
Then $\eta_{n + 1} \ne \eta_n$, because their respective $\boldsymbol \alpha$ differ by at least one value: $\alpha_{n + 1}$.
Hence, for this supposition:
:$J \sqbrk {\eta_{n + 1} } > J \sqbrk {\eta_n}$
Finally, both cases together imply that:
:$J \sqbrk {\eta_{n + 1} } \ge J \sqbrk {\eta_n}$
or equivalently:
:$\mu_n \ge \mu_{n + 1}$
== Sources ==
* : $\S 8.40 $: The Ritz Method and the Method of Finite Differences
Category:Calculus of Variations
Robustness/Examples/T-Test

== Example of Robustness ==
The $t$-test is an example of a statistical test which is robust departures from normality if the observations are from nearly symmetrical distributions with aopproximately the same variance.
However, it is less robust in the face of distributions which are skew, or if the variances are very different.
== Sources ==
* 
* 
Category:Examples of Robustness
Rodrigues' Formula for Legendre Polynomials

== Theorem == 
:$\map {P_n} x = \dfrac 1 {2^n n!} \dfrac {\d^n} {\d x^n} \paren {x^2 - 1}^n$
where:
:$n \in \N$ is a natural number
:$P_n$ is the $n$th Legendre polynomial.
== Proof == 
== Sources == 
* : $15.11$: Rodrigues' formula for the Legendre polynomials.
* : $\S 25$: Legendre Functions: Legendre Polynomials: $25.2$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials: $28.2.$
* 
Category:Legendre Polynomials
Rodrigues' Formula for Legendre Polynomials/Mistake

== Source Work ==
:Chapter $25$: Legendre Functions
::Legendre Polynomials: $25.2$
== Mistake ==
:If $n = 0, 1, 2, \ldots$, solutions of $25.1$ are Legendre polynomials $\map {P_n} x$ given by Rodrigue's formula 
== Correction ==
The grammar of Rodrigue's formula is incorrect.
It should say Rodrigues' formula.
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials: $25.2$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Rokhlin's Theorem (Intersection Forms)

== Theorem ==
Let $M$ be a smooth 4-manifold.
Then:
:$\map {\omega_2} {\map T M} = 0 \implies \operatorname {sign} Q_M = 0 \pmod {16}$
where:
:$Q_M$ is the intersection form
:$\map T M$ is the tangent bundle
:$\omega_2$ is the second Stiefel-Whitney class.
== Proof ==
Category:Manifolds
Rokhlin's Theorem on Bounded Manifolds and Induced Spin Structures

== Theorem ==
=== Part 1 ===
Let $M$ be a smooth oriented $4$-manifold.
Let $\operatorname {sign} Q_M = 0$.
Then there exists a smooth oriented $5$-manifold $W$ such that:
:$\partial W = M$
=== Part 2 ===
Let $M$ be a smooth oriented $4$-manifold.
Let $\operatorname {sign} Q_M = 0$.
Let $M$ be endowed with a spin structure.
Then there exists a smooth oriented $5$-manifold $W$ such that:
:$\partial W = M$
and the spin structure of $W$ induces the spin structure of $M$.
== Proof ==
=== Part 1 ===
By the Whitney Immersion Theorem, there exists an immersion of $M$ into $\R^7$. 
Suppose $\exists M'$ such that $M'$ embeds in $\R^6$ and that $M'$ and $M$ are cobordant.
By Images of Smooth Embeddings bound Oriented Manifolds, $M'$ must bound a $5$-manifold $W'$.
The union of the cobordism and $W'$ are necessarily a $5$-manifold $W$ which satisfy the theorem.
Hence it suffices to show that for any smooth, orientable $4$-manifold, there exists a similar manifold which is cobordant to the original and embeds in $\R^6$.
=== Part 2 ===
Category:Manifolds
Rolle's Theorem

== Theorem ==
Let $f$ be a real function which is:
:continuous on the closed interval $\closedint a b$
and:
:differentiable on the open interval $\openint a b$.
Let $\map f a = \map f b$.
Then:
:$\exists \xi \in \openint a b: \map {f'} \xi = 0$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Extended Rolle's Theorem
* Mean-Value Theorem
== Sources ==
* 
* 
* 
Category:Rolle's Theorem
Category:Differentiable Real Functions
Category:Continuous Real Functions
Rolle's Theorem/Proof 1

== Theorem ==
== Proof ==
We have that $f$ is continuous on $\closedint a b$.
It follows from Continuous Image of Closed Interval is Closed Interval that $f$ attains:
:a maximum $M$ at some $\xi_1 \in \closedint a b$
and:
:a minimum $m$ at some $\xi_2 \in \closedint a b$.
Suppose $\xi_1$ and $\xi_2$ are both end points of $\closedint a b$.
Because $\map f a = \map f b$ it follows that $m = M$ and so $f$ is constant on $\closedint a b$.
Then, by Derivative of Constant, $\map {f'} \xi = 0$ for all $\xi \in \openint a b$.
Suppose $\xi_1$ is not an end point of $\closedint a b$.
Then $\xi_1 \in \openint a b$ and $f$ has a local maximum at $\xi_1$.
Hence the result follows from Derivative at Maximum or Minimum‎.
Similarly, suppose $\xi_2$ is not an end point of $\closedint a b$.
Then $\xi_2 \in \openint a b$ and $f$ has a local minimum at $\xi_2$.
Hence the result follows from Derivative at Maximum or Minimum‎.
== Sources ==
* : $\S 11.4$
Category:Rolle's Theorem
Rolle's Theorem/Proof 2

== Theorem ==
== Proof ==
First take the case where:
:$\forall x \in \openint a b: \map f x = 0$
Then:
:$\forall x \in \openint a b: \map {f'} x = 0$
Otherwise:
:$\exists c \in \openint a b: \map f c \ne 0$
Let $\map f c > 0$.
Then there exists an absolute maximum at a point $\xi \in \openint a b$.
Hence:
As $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.
That is:
:$\map {f'} \xi = 0$
Similarly, let $\map f c < 0$.
Then there exists an absolute minimum at a point $\xi \in \openint a b$.
Hence:
Again, as $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.
That is:
:$\map {f'} \xi = 0$
Hence the result.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.3$ Rolle's Theorem
* : $\S 1.12$: Valid Arguments: Proposition $1.12.3$
Category:Rolle's Theorem
Roman Number Wordplay/9 from 6, 10 from 9, 50 from 40

== Puzzle ==
:If from six ye take nine, and from nine ye take ten
:(Ye youths, now the mystery explain),
:And if fifty from forty be taken, there then,
:Shall just half a dozen remain.
== Solution ==
Putting them together gives:
:$\mathtt{SIX}$
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $147$
Category:Roman Numerals
Category:Trick Puzzles
Root is Commutative

== Theorem ==
Let $x \in \R_{> 0}$ be a (strictly) positive real number.
Let $a$ and $b$ be nonzero integers.
Then:
:$\sqrt [a] {\sqrt [b] x} = \sqrt [b] {\sqrt [a] x}$
== Proof ==
Let $y = \sqrt [a] {\sqrt [b] x}$.
Then
Category:Roots of Numbers
Root is Strictly Increasing

== Theorem ==
Let $x \in \R_{> 0}$.
Let $n \in \N$.
Let $f: \R_{> 0} \to \R$ be the real function defined as:
:$\map f x = \sqrt [n] x$
where $\sqrt [n] x$ denotes the $n$th root of $x$.
Then $f$ is strictly increasing.
== Proof ==
Let $x, y \in \R$ such that $0 < x < y$.
 that:
:$\sqrt [n] x \ge \sqrt [n] y$
We have:
This contradicts the hypothesis that $x < y$.
Therefore, by Proof by Contradiction:
:$0 < x < y \implies \sqrt [n] x \le \sqrt [n] y$
Hence the result by definition of strictly increasing real function.
Category:Examples of Strictly Increasing Real Functions
Category:Roots of Numbers
Root of Area contained by Rational Straight Line and Fifth Binomial

== Theorem ==
== Proof ==
:
Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the fifth binomial $AD$.
Let $E$ divide $AD$ into its terms such that $AE > ED$.
From this is possible at only one place.
By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.
Thus $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$.
Let $ED$ be bisected at $F$.
Using :
:Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.
By :
:$AG$ is commensurable in length with $EG$.
Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.
Using :
:let the square $SN$ be constructed equal to the parallelogram $AH$
and:
:let the square $NQ$ be constructed equal to the parallelogram $GK$.
Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$.
Therefore $RN$ is also in a straight line with $NP$.
Let the parallelogram $SQ$ be completed.
From :
:$SQ$ is a square.
We have that $AD$ is a third binomial straight line.
Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only.
We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.
Then by :
:$AG : EF = EF : EG$
and so by :
:$AH : EL = EL : KG$
Therefore $EL$ is a mean proportional between $AH$ and $GK$.
But:
:$AH = SN$
and:
:$GK = NQ$
therefore $EL$ is a mean proportional between $SN$ and $NQ$.
But from :
:$MR$ is a mean proportional between $SN$ and $NQ$.
Therefore:
:$EL = MR$
and so:
:$EL = PO$
But:
:$AH + GK = SN + NQ$
Therefore:
:$AC = SQ$
That is:
:$AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$.
From :
:$AE$ is incommensurable in length with $AD$.
By:
:
and:
:
it follows that:
:$AH$ is commensurable with $HE$.
That is:
:$MN^2$ is incommensurable with $NO^2$.
Therefore $MN$ and $NO$ are incommensurable in square.
We have that $AD$ is a fifth binomial with $ED$ the lesser segment.
Therefore by :
:$ED$ is commensurable in length with $AB$.
But $AE$ is incommensurable in length with $ED$.
Therefore by 
:$AB$ is incommensurable in length with $AE$.
Therefore, by definition, $AK = MN^2 + NO^2$ is medial.
We have that:
:$DE$ is commensurable in length with $AB = EK$
and:
:$DE$ is commensurable in length with $EF$.
So by :
:$EF$ is commensurable in length with $EK$.
We have that $EK$ is a rational straight line.
Therefore by :
:$EL = MR = MN \cdot NO$ is a rational area.
Therefore $MN$ and $NO$ are straight lines which are incommensurable in square such that:
:$MN^2 + NO^2$ is medial
:$MN \cdot NO$ is rational.
Therefore by definition, $MO$ is the side of a rational plus a medial area.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Root of Area contained by Rational Straight Line and First Binomial/Lemma

== Lemma to Root of Area contained by Rational Straight Line and First Binomial ==
== Proof ==
:
We have that $DB = BF$ and $BE = BG$.
Thus $DE = FG$.
But from :
:$DE = AH = KC$
and:
:$FG = AK = HC$.
Therefore:
:$AH = AK = HC = KC$
Therefore the parallelogram $AKCH$ is equilateral.
The parallelogram $AKCH$ is also rectangular.
Therefore by definition $AKCH$ is a square.
We have that:
:$FB : BG = DB : BE$
and:
:$FB : BG = AB : DG$
and from :
:$DB : BE = DG : BC$
it follows from :
:$AB : DG = DG : BC$
Therefore $DG$ is a mean proportional between $AB$ and $BC$.
We also have that:
:$AD : DK = KG : GC$
and so from :
:$AK : KD = KC : CG$
while:
:$AK : KD = AC : CD$
and from :
:$KC : CG = DC : CB$
it follows from :
:$AC : DC = DC : BC$
Therefore $DC$ is a mean proportional between $AC$ and $BC$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Root of Area contained by Rational Straight Line and Fourth Binomial

== Theorem ==
== Proof ==
:
Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the fourth binomial $AD$.
Let $E$ divide $AD$ into its terms such that $AE > ED$.
From this is possible at only one place.
By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.
Thus by definition of fourth binomial: 
:$AE^2$ is greater than $ED^2$ by the square on a straight line which is incommensurable in length with $AE$.
By , the greater term $AE$ is commensurable in length with $AB$.
Let $ED$ be bisected at $F$.
Using :
:Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.
By :
:$AG$ is incommensurable in length with $EG$.
Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.
Using :
:let the square $SN$ be constructed equal to the parallelogram $AH$
and:
:let the square $NQ$ be constructed equal to the parallelogram $GK$.
Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$.
Therefore $RN$ is also in a straight line with $NP$.
Let the parallelogram $SQ$ be completed.
From :
:$SQ$ is a square.
We have that $AD$ is a fourth binomial straight line.
Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only.
We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.
Then by :
:$AG : EF = EF : EG$
and so by :
:$AH : EL = EL : KG$
Therefore $EL$ is a mean proportional between $AH$ and $GK$.
But:
:$AH = SN$
and:
:$GK = NQ$
therefore $EL$ is a mean proportional between $SN$ and $NQ$.
But from :
:$MR$ is a mean proportional between $SN$ and $NQ$.
Therefore:
:$EL = MR$
and so:
:$EL = PO$
But:
:$AH + GK = SN + NQ$
Therefore:
:$AC = SQ$
That is:
:$AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$.
We have that:
:$AG$ is incommensurable in length with $EG$
and so by:
:
and:
::
it follows that:
:$AH$ is incommensurable with $GK$
that is:
:$SN$ is incommensurable with $NQ$.
Therefore $MN$ and $NO$ are incommensurable in square.
We have that:
:$AE$ is commensurable in length with $AB$
and therefore from :
:$AK$ is rational.
But $AK = MN^2 + NO^2$.
Therefore $MN^2 + NO^2$ is rational.
Since:
:$DE$ is incommensurable in length with $AB = EK$
while:
:$DE$ is commensurable in length with $EF$
from :
:$EF$ is incommensurable in length with $EK$.
Therefore $EF$ and $EK$ are rational straight lines which are commensurable in square only.
Therefore $LE = MR$ is medial.
But $MR$ is a rectangle contained by $MN$ and $NO$.
Therefore the rectangle contained by $MN$ and $NO$ is medial.
We also have that:
:$MN^2 + NO^2$ is rational
:$MN$ and $NO$ are incommensurable in square.
Thus by definition $MN + NO$ is major.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Root of Area contained by Rational Straight Line and Sixth Binomial

== Theorem ==
== Proof ==
:
Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the sixth binomial $AD$.
Let $E$ divide $AD$ into its terms such that $AE > ED$.
Let $E$ divide $AD$ into its terms such that $AE > ED$.
From this is possible at only one place.
By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.
Thus $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$.
Let $ED$ be bisected at $F$.
Using :
:Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.
By :
:$AG$ is commensurable in length with $EG$.
Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.
Using :
:let the square $SN$ be constructed equal to the parallelogram $AH$
and:
:let the square $NQ$ be constructed equal to the parallelogram $GK$.
Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$.
Therefore $RN$ is also in a straight line with $NP$.
Let the parallelogram $SQ$ be completed.
From :
:$SQ$ is a square.
We have that $AD$ is a third binomial straight line.
Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only.
We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.
Then by :
:$AG : EF = EF : EG$
and so by :
:$AH : EL = EL : KG$
Therefore $EL$ is a mean proportional between $AH$ and $GK$.
But:
:$AH = SN$
and:
:$GK = NQ$
therefore $EL$ is a mean proportional between $SN$ and $NQ$.
But from :
:$MR$ is a mean proportional between $SN$ and $NQ$.
Therefore:
:$EL = MR$
and so:
:$EL = PO$
But:
:$AH + GK = SN + NQ$
Therefore:
:$AC = SQ$
That is:
:$AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$.
We have that $EA$ is incommensurable in length with $AB$.
Therefore $EA$ and $AB$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $AK = MN^2 + NO^2$ is medial.
We have that:
:$ED$ is incommensurable in length with $AB = EK$
So by :
:$FE$ is incommensurable in length with $EK$.
Therefore $FE$ and $EK$ are rational straight lines which are commensurable in square only.
Therefore by definition:
:$EL = MR = MN \cdot NO$ is a medial area.
Since:
:$AE$ is incommensurable in length with $EF$
by:
: :
and:
::
:$AK$ is incommensurable with $EL$.
But:
:$AK = MN^2 + NO^2$
and:
:$EL = MN \cdot NO$
Therefore:
:$MN^2 + NO^2$ is incommensurable with $MN \cdot NO$.
So by definition, $MO$ is the side of the sum of two medial areas.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Root of Equation/Examples/Arbitrary Quadratic

== Examples of Roots of Equation ==
Consider the equation:
:$\map E x := x^2 - x - 6 = 0$
The roots of $\map E x$ are:
== Proof ==
$\map E x$ can be expressed as:
:$\paren {x - 3} \paren {x + 2} = 0$
The result follows.
== Sources ==
* 
* 
Category:Examples of Roots of Equations
Root of Equation e^x (x - 1) = e^-x (x + 1)

== Theorem ==
The equation:
:$e^x \paren {x - 1} = e^{-x} \paren {x + 1}$
has a root:
:$x = 1 \cdotp 19966 \, 78640 \, 25773 \, 4 \ldots$
== Proof ==
Let $\map f x = e^x \paren {x - 1} - e^{-x} \paren {x + 1}$.
Then if $\map f c = 0$, $c$ is a root of $e^x \paren {x - 1} = e^{-x} \paren {x + 1}$.
Notice that:
:$\map f 1 = e^1 \paren {1 - 1} - e^{-1} \paren {1 + 1} = -\dfrac 2 e < 0$
:$\map f 2 = e^2 \paren {2 - 1} - e^{-2} \paren {2 + 1} = e^2 - \dfrac 3 {e^2} > 0$
By Intermediate Value Theorem:
:$\exists c \in \openint 1 2: \map f c = 0$.
This shows that our equation has a root between $1$ and $2$.
The exact value of this root can be found using any numerical method, e.g. Newton's Method.
== Also see ==
* Definition:Kepler's Equation
== Sources ==
* 
* : $1,19966 78640 25773 4 \ldots$
Category:Analysis