nyuuzyou/wikis huggingface dataset:
Euler-Binet Formula/Proof 2

== Theorem ==
== Proof ==
Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.
First by the lemma to Cassini's Identity: 
:$(1): \quad \forall n \in \Z_{>1}: A^n = \begin {bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix}$
Recall from Eigenvalues of Real Square Matrix are Roots of Characteristic Equation, we can find the eigenvalues of $\mathbf A$ by solving the equation $\map \det {\mathbf A - \lambda \mathbf I} = 0$
Therefore, taking the determinant of the square matrix below:
:$\mathbf A - \lambda \mathbf I = \begin {bmatrix} 1 - \lambda & 1 \\ 1 & -\lambda \end{bmatrix}$
We obtain the equation:
Therefore, $A$ has the eigenvalues $\phi$ and $\paren {1 - \phi}$. 
We will now determine the eigenvectors of $A$
For $\lambda = \phi$, we have:
We see from the above, when $v_2 = 1$ then $v_1 = \phi$ and we have:
For $\lambda = \paren {1 - \phi} $, we have:
We see from the above, when $v_2 = 1$ then $v_1 = \paren {1 - \phi}$ and we have:
We have now demonstrated that:
:$\begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$
and:
:$\begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\paren {1 - \phi}$.
We now observe that:
Then we notice that:
From Eigenvalue of Matrix Powers for a positive integer $n$, we have:
 
Putting all of the pieces together, we obtain:
Hence the result.
== Also known as ==
Category:Euler-Binet Formula
Category:Eigenvectors
Category:Eigenvalues
Euler-Binet Formula/Proof 3

== Theorem ==
== Proof ==
This follows as a direct application of the first Binet form:
:$U_n = m U_{n - 1} + U_{n - 2}$
where:
has the closed-form solution:
:$U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$
where:
where $m = 1$.
== Also known as ==
Category:Euler-Binet Formula
Euler-Binet Formula/Proof 4

== Theorem ==
== Proof ==
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
:$\map G z = \dfrac z {1 - z - z^2}$
Hence:
where:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
By Sum of Infinite Geometric Sequence:
:$\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$
and so:
:$\map G z = \dfrac 1 {\sqrt 5} \paren {1 + \phi z + \phi^2 z^2 + \cdots - 1 - \hat \phi z - \hat \phi^2 z^2 - \cdots}$
By definition, the coefficient of $z^n$ in $\map G z$ is exactly the $n$th Fibonacci number.
That is:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
== Also known as ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: $(14)$
Category:Euler-Binet Formula
Euler-Lagrange Equation for Conservative System

== Theorem ==
Let $P$ be a conservative system composed of $n \in \N$ particles.
The Euler-Lagrange equation for $P$ is given by:
:$\map {\dfrac \d {\d t} } {\dfrac {\partial L} {\partial \dot q_j} } - \dfrac {\partial L} {\partial q_j} = 0$
where:
:$L$ denotes the Lagrangian of $P$
:$q_j$ denotes the positions of the $n$ particles composing $P$
:$t$ denotes time.
== Proof ==
== Sources ==
* 
* 
Category:Euler-Lagrange Equation
Euler-Maclaurin Summation Formula

== Theorem ==
Let $f$ be a real function which is appropriately differentiable and integrable.
Then:
where:
:$f^{\paren k}$ denotes the $k$th derivative of $f$
:$B_n$ denotes the $n$th Bernoulli number.
== Proof ==
== Also known as ==
== Also see ==
* Cauchy Integral Test
== Sources ==
* : $3$: Elementary Analytic Methods: $3.6$ Infinite Series: Euler-Maclaurin Summation Formula: $3.6.28$
* : $\S 19$: The Euler-Maclaurin Summation Formula: $19.45$
* 
* 
* 
* 
* : $\S 21$: Series of Constants: The Euler-Maclaurin Summation Formula: $21.45.$
Category:Euler-Maclaurin Summation Formula
Category:Integral Calculus
Category:Series
Euler-Mascheroni Constant as Difference of Integrals involving Cosine

== Theorem ==
:$\ds \int_0^1 \frac {1 - \cos x} x \rd x - \int_1^\infty \frac {\cos x} x \rd x = \gamma$
 
where $\gamma$ is the Euler-Mascheroni Constant.
== Proof ==
From Characterization of Cosine Integral Function, we have: 
:$\ds \int_t^\infty \frac {\cos x} x \rd x = -\gamma - \ln t + \int_0^t \frac {1 - \cos x} x \rd x$
for all real $t > 0$.
Rearranging: 
:$\ds \int_0^t \frac {1 - \cos x} x \rd x - \int_t^\infty \frac {\cos x} x \rd x = \gamma + \ln t$
Setting $t = 1$ gives the result, from Natural Logarithm of 1 is 0.
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.65$
Category:Definite Integrals involving Cosine Function
Category:Euler-Mascheroni Constant
Euler-Poincaré Characteristic on Homeomorphic Simplicial Complexes

== Theorem ==
Let $K$ and $L$ be simplicial complexes.
Let $K$ and $L$ be homeomorphic.
Then:
:$\map \chi K = \map \chi L$
where $\map \chi K$ denotes the Euler-Poincaré characteristic of $K$.
== Proof ==
== Sources ==
* 
* 
Category:Euler-Poincaré Characteristic
Category:Simplicial Complexes
Euler-Poincaré Characteristic on Homotopy-Equivalent Simplicial Complexes

== Theorem ==
Let $K$ and $L$ be simplicial complexes.
Let $K$ and $L$ be homotopy-equivalent.
Then:
:$\map \chi K = \map \chi L$
where $\map \chi K$ denotes the Euler-Poincaré characteristic of $K$.
== Proof ==
== Sources ==
* 
* 
Category:Euler-Poincaré Characteristic
Category:Simplicial Complexes
Euler Buckling Formula

== Theorem ==
:$F = \dfrac {\pi^2 E I} {\paren {K L}^2}$
where:
:$F$ = maximum or critical force (vertical load on column)
:$E$ = modulus of elasticity
:$I$ = area moment of inertia of the cross section of the rod
:$L$ = unsupported length of column
:$K$ = column effective length factor, whose value depends on the conditions of end support of the column, as follows:
::For both ends pinned (hinged, free to rotate), $K = 1.0$
::For both ends fixed, $K = 0.50$
::For one end fixed and the other end pinned, $K \approx 0.699$
::For one end fixed and the other end free to move laterally, $K = 2.0$
:$K L$ is the effective length of the column
== Proof ==
== Also known as ==
The Euler buckling formula is also known as the Euler column formula.
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : Chapter $\text {A}.21$: Euler ()
Category:Mechanics
Euler Characteristic is not Dependent upon Triangulation

== Theorem ==
Let $S$ be a surface.
Let $T_1$ and $T_2$ be triangulations of $S$.
Let:
:$\map {\chi_1} S$ be the Euler characteristic of $S$ as calculated using $T_1$
:$\map {\chi_2} S$ be the Euler characteristic of $S$ as calculated using $T_2$.
Then:
:$\map {\chi_1} S = \map {\chi_2} S$
== Proof ==
Recall:
where:
:$\map v {T_1}$ denotes the order of $T_1$ (that is, the number of its vertices)
:$\map e {T_1}$ denotes the size of $T_1$ (that is, the number of its edges)
:$\map f {T_1}$ denotes the number of faces of $T_1$.
== Sources ==
* 
Category:Euler Characteristic
Category:Triangulation (Topology)
Euler Characteristic of General Surface

== Theorem ==
Let $S$ be a surface.
Let $\map \chi S$ denote the Euler characteristic of $S$.
Then:
:$\map \chi S = 2 - 2 h - c$
where:
:$h$ is the number of handles
:$c$ is the number of cross-caps
that make up $S$.
== Proof ==
== Sources ==
* 
Category:Euler Characteristic of Surface
Euler Diagram/Examples

== Examples of Euler Diagrams ==
The following examples show:
:$(\text a)$ The subset relation $T \subseteq S$
:$(\text b)$ Examples of disjoint sets $S \cap T = \O$
:$(\text c)$ The most general case: $S \cap T \ne \O$, $T \nsubseteq S$, $S \nsubseteq T$
:
The shape of the areas is irrelevant, but usually circles are used.
== Sources ==
* : Chapter $1$: Set Theory: $1.2$: Sets and subsets
:::(where this is referred to as a Venn diagram)
* : $\S 1.2$.Subsets: Example $10$
:::(where this is referred to as a Venn diagram)
* 
Category:Examples of Euler Diagrams
Euler Form of Gamma Function at Positive Integers

== Theorem ==
The Euler form of the Gamma function:
:$\ds \map \Gamma z := \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \cdots \paren {z + m} }$
converges to the factorial function at positive integers:
:$\ds \lim_{m \mathop \to \infty} \frac {m^n m!} {\paren {n + 1} \paren {n + 2} \cdots \paren {n + m} } = n!$
== Proof ==
Now we have from Factorial of Integer plus Reciprocal of Integer that:
:$\ds \lim_{m \mathop \to \infty} \dfrac {\paren {m + n}!} {m! m^n} = 1$
Now:
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $8$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $22$
Category:Gamma Function
Euler Formula for Cosine Function

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Euler Formula for Sine Function
* Euler Formula for Hyperbolic Sine Function
* Euler Formula for Hyperbolic Cosine Function
== Sources ==
* : $\S 38$: Infinite Products: $38.2$
Category:Euler Formula for Cosine Function
Category:Cosine Function
Category:Examples of Infinite Products
Euler Formula for Cosine Function/Proof 1

== Theorem == 
== Proof ==
We have that $\cos x$ has a power series representation:
:$\cos x = 1 - \dfrac {x^2} {2!} + \dfrac {x^4} {4!} - \dfrac {x^6} {6!} + \dotsb$
The roots of cosine are the numbers $\paren {k + \dfrac 1 2} \pi$, where $k$ is any integer.
From the Polynomial Factor Theorem, the following (after simplification) might be true:
:$\ds \cos x = A \prod \paren {1 - \frac {2 x} {\paren {2 k + 1} \pi} }$
where the product is taken over all $k \in \Z$, and $A$ is some constant.
The intuition is as follows.
Letting $x$ tend to $0$ in the above equation implies that $A = 1$.
It remains to formalize the above claims.
Category:Euler Formula for Cosine Function
Euler Formula for Hyperbolic Cosine Function

== Theorem ==
== Proof ==
== Also see ==
* Euler Formula for Sine Function
* Euler Formula for Cosine Function
* Euler Formula for Hyperbolic Sine Function
== Sources ==
* : $\S 38$: Infinite Products: $38.4$
Category:Euler Formula for Hyperbolic Cosine Function
Category:Hyperbolic Cosine Function
Category:Examples of Infinite Products
Euler Formula for Hyperbolic Sine Function

== Theorem ==
== Proof ==
== Also see ==
* Euler Formula for Sine Function
* Euler Formula for Cosine Function
* Euler Formula for Hyperbolic Cosine Function
== Sources ==
* : $\S 38$: Infinite Products: $38.3$
Category:Euler Formula for Hyperbolic Sine Function
Category:Hyperbolic Sine Function
Category:Examples of Infinite Products
Euler Formula for Sine Function

== Theorem ==
=== Real Numbers ===
=== Complex Numbers ===
== Also see ==
* Euler Formula for Cosine Function
* Euler Formula for Hyperbolic Sine Function
* Euler Formula for Hyperbolic Cosine Function
== Historical Note ==
== Sources ==
* : $\S 38$: Infinite Products: $38.1$
Category:Euler Formula for Sine Function
Category:Sine Function
Category:Examples of Infinite Products
Euler Formula for Sine Function/Complex Numbers/Proof 1

== Theorem ==
== Proof ==
For $z \in \C$ and $n \in \N$, let:
:$\ds \map {I_n} z = \int_0^{\pi / 2} \cos {z t} \cos^n t \rd t $
Observe that $\map {I_0} 0 = \dfrac {\pi} 2$ and:
which yields:
:$(1): \quad \map \sin {\dfrac {\pi z} 2} = \dfrac {\pi z} 2 \dfrac {\map {I_0} z} {\map {I_0} 0}$
Integrating by parts twice with $n \ge 2$, we have:
which yields the reduction formula:
:$n \paren {n - 1} \map {I_{n - 2} } z = \paren {n^2 - z^2} \map {I_n} z$
Substituting $z = 0$ we obtain:
:$n \paren {n - 1} \map {I_{n - 2} } 0 = n^2 \map {I_n} 0$
From Shape of Cosine Function, it is clear that $\map {I_n} 0 > 0$ for $n \ge 0 $.
Therefore we can divide the two equations to get:
:$(2): \quad \dfrac {\map {I_{n - 2} } z} {\map {I_{n - 2} } 0} = \paren {1 - \dfrac {z^2} {n^2} } \dfrac {\map {I_n} z} {\map {I_n} 0}$
We have:
From Sine Inequality we have that $2 \map {\sin^2} {\dfrac {x t} 2} \le \dfrac 1 2 x^2 t^2$.
By Lemma 1, $\dfrac {\sinh x} x$ is an increasing function for $x \ge 0$, so for $t \in \closedint 0 {\dfrac {\pi} 2}$:
So we deduce:
By Relative Sizes of Definite Integrals we have:
which yields the inequality:
:$\cmod {1 - \dfrac {\map {I_n} z} {\map {I_n} 0} } \le \dfrac {\map C {x, y} } n$
It follows from Squeeze Theorem that:
:$(3): \quad \ds \lim_{n \mathop \to \infty} \frac {\map {I_n} z} {\map {I_n} 0} = 1$
Consider the equation, for even $n$:
:$\ds \map \sin {\dfrac {\pi z} 2} = \dfrac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \dfrac {z^2} {\paren {2 i}^2} } \dfrac {\map {I_n} z} {\map {I_n} 0}$
This is true for $n = 0$ by $(1)$.
Suppose it is true for some $n = k$.
Then:
So it is true for all even $n$ by induction.
Taking the limit as $n \to \infty$ we have:
or equivalently, letting $\dfrac {\pi z} 2 \mapsto z$:
:$\ds \sin z = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2 \pi^2} }$
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Complex Numbers/Proof 1/Lemma 1

== Theorem ==
The function:
:$\dfrac {\sinh x} x$ 
is increasing for positive real $x$.
== Proof ==
Let $\map f x = \dfrac {\sinh x} x$.
By Quotient Rule for Derivatives and Derivative of Hyperbolic Sine:
:$\map {f'} x = \dfrac {x \cosh x - \sinh x} {x^2}$
From Hyperbolic Tangent Less than X, we have $\tanh x \le x$ for $x \ge 0$.
Since $\cosh x \ge 0$, we can rearrange to get $x \cosh x - \sinh x \ge 0$.
Since $x^2 \ge 0$, we have $\map {f'} x \ge 0$ for $x \ge 0$.
So by Derivative of Monotone Function it follows that $\map f x$ is increasing for $x \ge 0$.
Category:Hyperbolic Sine Function
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Complex Numbers/Proof 2

== Theorem ==
== Proof ==
For $z \in \C$ and $n \in \N_{> 0}$, let:
:$\map {f_n} z = \dfrac 1 2 \paren {\paren {1 + \dfrac z n}^n - \paren {1 - \dfrac z n}^n}$
Then $\map {f_n} z = 0$ :
Let $n = 2 m + 1$.
Then the roots of $\map {f_{2 m + 1} } z$ are:
:$\paren {2 m + 1} i \map \tan {\dfrac {k \pi} {2 m + 1} }$ 
for $- m \le k \le m$.
Observe that $\map {f_{2 m + 1} } z$ is a polynomial of degree $2 m + 1$. 
Then for some constant $C$, we have:
It can be seen from Binomial Theorem for Integral Index that the coefficient of $z$ in $\map {f_{2 m + 1} } z$ is $1$.
Hence $C = 1$, and we obtain:
:$\ds \map {f_{2 m + 1} } z = z \prod_{k \mathop = 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$
First we consider $z = x$ where $x$ is a non-negative real number.
Let $l < m$.
Then:
:$\ds x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$
Taking the limit as $m \to \infty$ we have:
By Tangent Inequality, we have:
:$\map \tan {\dfrac {k \pi} {2 m + 1} } \ge \dfrac {k \pi} {2 m + 1}$
hence:
:$\ds \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$
Taking the limit as $l \to \infty$ we have by Squeeze Theorem:
:$(1): \quad \ds x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$
Now let $1 < l < m$.
By Complex Modulus of Product of Complex Numbers and the Triangle Inequality, we can deduce:
Taking the limit as $m \to \infty$ we have:
:$\ds \cmod {\sinh z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {k^2 \pi^2} } } \le \sinh \cmod z - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {k^2 \pi^2} }$
Now take the limit as $l \to \infty$.
By $(1)$ and Squeeze Theorem, we have:
:$\ds \sinh z = z \prod_{k \mathop = 1}^\infty \paren {1 + \frac {z^2} {k^2 \pi^2} }$
Finally, substituting $z \mapsto i z$, we obtain:
:$\ds \sin z = z \prod_{k \mathop = 1}^\infty \paren {1 - \frac {z^2} {k^2 \pi^2} }$
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Real Numbers/Proof 1

== Theorem ==
== Proof ==
For $x \in \R$ and $n \in \N$, let:
:$\ds \map {I_n} x = \int_0^{\pi / 2} \cos {x t} \cos^n t \rd t $
Observe that:
:$\map {I_0} 0 = \dfrac {\pi} 2$
and:
which yields:
:$(1): \quad \map \sin {\dfrac {\pi x} 2} = \dfrac {\pi x} 2 \dfrac {\map {I_0} x} {\map {I_0} 0}$
Integrating by parts twice with $n \ge 2$, we have:
which yields the reduction formula:
:$n \paren {n - 1} \map {I_{n - 2} } x = \paren {n^2 - x^2} \map {I_n} x$
Substituting $x = 0$ we obtain:
:$n \paren {n - 1} \map {I_{n - 2} } 0 = n^2 \map {I_n} 0$
From Shape of Cosine Function, it is clear that $\map {I_n} 0 > 0$ for $n \ge 0 $.
Therefore we can divide the two equations to get:
:$(2): \quad \dfrac {\map {I_{n - 2} } x} {\map {I_{n - 2} } 0} = \paren {1 - \dfrac {x^2} {n^2} } \dfrac {\map {I_n} x} {\map {I_n} 0}$
By Relative Sizes of Definite Integrals we have:
which yields the inequality:
:$\size {1 - \dfrac {\map {I_n} x} {\map {I_n} 0} } \le \dfrac {x^2} {2 n}$
It follows from Squeeze Theorem that:
:$\ds (3): \quad \lim_{n \mathop \to \infty} \frac {\map {I_n} x} {\map {I_n} 0} = 1$
Consider the equation, for even $n$:
:$\ds \map \sin {\frac {\pi x} 2} = \frac {\pi x} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_n} x} {\map {I_n} 0}$
This is true for $n = 0$ by $(1)$.
Suppose it is true for some $n = k$.
Then:
So it is true for all even $n$ by induction.
Taking the limit as $n \to \infty$ we have:
or equivalently, letting $\dfrac {\pi x} 2 \mapsto x$:
:$\ds \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Real Numbers/Proof 2

== Theorem ==
== Proof ==
Using De Moivre's Formula:
:$\sin x = \dfrac {\left({\cos \dfrac x n + i \sin \dfrac x n}\right)^n - \left({\cos \dfrac x n - i \sin \dfrac x n}\right)^n} {2i}$
The difference between two $n$th powers can be extracted into linear factors using $n$th roots of unity.
For large $n$, we can replace:
: $\cos \dfrac x n$ by $1$
: $\sin \dfrac x n$ by $\dfrac x n$
He proved it in vol. 1 of his 1748 work Introductio in analysin infinitorum using De Moivre's Formula as above.
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Real Numbers/Proof 3

== Theorem ==
== Proof ==
We have that $\sin x$ has a power series representation:
:$\sin x = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots$
The roots of sine are the numbers $k \pi$, where $k$ is any integer.
From the Polynomial Factor Theorem, the following might be true:
:$\ds \sin x = A x \prod \paren {1 - \frac x {k \pi} }$
where the product is taken over all $n \in \Z \setminus \set 0$, and $A$ is some constant.
The intuition is as follows.
From Limit of $\dfrac {\sin x} x$ at Zero:
:$\dfrac {\sin x} x \to 1$ as $x \to 0$
Letting $x$ tend to $0$ in the above equation implies that $A = 1$.
We now formalize the above claims.
== Sources ==
* : Chapter $\text {B}.14$: Euler's Discovery of the Formula $\ds \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
Category:Euler Formula for Sine Function
Euler Formula for Sine Function/Real Numbers/Proof 4

== Theorem ==
== Proof ==
For $x \in \R$ and $n \in \N_{> 0}$, let:
:$\map {f_n} x = \dfrac 1 2 \paren {\paren {1 + \dfrac x n}^n - \paren {1 - \dfrac x n}^n }$
Then $\map {f_n} x = 0$ :
Hence the roots of $\map {f_{2 m + 1} } x$ are:
:$\paren {2 m + 1} i \, \map \tan {\dfrac {k \pi} {2 m + 1} }$ 
for $-m \le k \le m$.
Observe that $\map {f_{2 m + 1} } x$ is a polynomial of degree $2 m + 1$. 
Then for some constant $C$, we have:
It can be seen from Binomial Theorem for Integral Index that the coefficient of $x$ in $\map {f_{2 m + 1} } x$ is $1$.
Hence $C = 1$, and we obtain:
:$\ds \map {f_{2 m + 1} } x = x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$
Let $l < m$.
Then:
:$\ds x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$
Taking the limit as $m \to \infty$ we have:
By Tangent Inequality, we have:
:$\map \tan {\dfrac {k \pi} {2 m + 1} } \ge \dfrac {k \pi} {2 m + 1}$
and hence:
:$\ds \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$
Taking the limit as $l \to \infty$, we have by the Squeeze Theorem:
:$\ds x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$
Substituting $x \mapsto i x$, we obtain:
:$\ds \sin x = x \prod_{k \mathop = 1}^\infty \paren {1 - \frac {x^2} {k^2 \pi^2} }$
Category:Euler Formula for Sine Function
Euler Lucky Number/Examples

== Examples of Euler Lucky Numbers ==
=== $n^2 + n + 2$ ===
=== $n^2 + n + 3$ ===
=== $n^2 + n + 5$ ===
=== $n^2 + n + 11$ ===
=== $n^2 + n + 17$ ===
=== $n^2 + n + 41$ ===
Category:Euler Lucky Numbers
Euler Lucky Number/Examples/2

== Example of Euler Lucky Number ==
The expression:
:$n^2 + n + 2$
yields primes for $n = 0$ and for no other $n \in \Z_{\ge 0}$.
This demonstrates that $2$ (trivially) is a Euler lucky number.
== Proof ==
Let $n > 0$.
Then:
and so is divisible by $2$.
Category:Euler Lucky Numbers
Category:2
Euler Lucky Number/Examples/41

== Example of Euler Lucky Number ==
The expression:
:$n^2 + n + 41$
yields primes for $n = 0$ to $n = 39$.
It also generates the same set of primes for $n = -1 \to n = -40$.
These are not the only primes generated by this formula.
No other quadratic function of the form $x^2 + a x + b$, where $a, b \in \Z_{>0}$ and $a, b < 10000$ generates a longer sequence of primes.
== Proof ==
Then we have:
and so replacing $0$ to $39$ with $-1$ to $-40$ yields exactly the same sequence of primes.
We note in addition the example:
:$581^2 + 581 + 41 = 338 \, 183$
which is prime.
== Sources ==
* : $41$
* : $41$
Category:Euler Lucky Numbers
Category:41
Euler Lucky Number/Examples/41/Mistake

== Source Work ==
:
:The Dictionary
::$41$
== Mistake ==
:[ $x^2 + x + 41$ ] has many further prime values, including $581$, among its first $1000$ values.
This is misworded. $581$ is not itself prime -- it is $581^2 + 581 + 41 = 338 \, 183$ which is prime.
== Sources ==
* : $41$
Category:Mistakes/Curious and Interesting Numbers
Euler Lucky Number Function for n equals p

== Definition ==
Let $p$ be a prime number.
Let $f_p: \Z \to \Z$ be the mapping defined as:
:$\forall n \in \Z: f_p \left({n}\right) = n^2 - n + p$
Then $f_p \left({p}\right)$ is not prime.
== Proof ==
Let $n = p$.
Then:
which is not prime.
Category:Euler Lucky Numbers
Euler Method

== Proof Technique ==
Consider the first order ODE:
:$(1): \quad y' = \map f {x, y}$ subject to the initial condition $\map y {x_0} = y_0$
where $\map f {x, y}$ is a continuous real function.
Let $\map y x$ be the particular solution of $(1)$.
For all $n \in \N_{>0}$, we define:
:$x_n = x_{n - 1} + h$
where $h \in \R_{>0}$.
Then for all $n \in \N_{>0}$ such that $x_n$ is in the domain of $y$:
:$y_{n + 1} = y_n + h \map f {x_n, y_n}$
is an approximation to $\map y {x_{n + 1} }$.
== Proof ==
Let $(1)$ be integrated $x$ from $x_0$ to $x_1$.
Because $f$ is continuous, the assumption holds.
By making $h$ small, the difference:
:$y_0 + h \, \map f {x_0, y_0} - \map f {x_1, y_1}$
can be made arbitrarily small.
$y_{n + 1}$ can be defined recursively:
:
The errors accumulate; with increasing $n$ the values of $y_{n + 1}$ are based on increasingly inaccurate values of $y_n$.
These can be reduced by making $h$ smaller, so the inaccuracies can be reduced by increasing the computation needed.
== Also known as ==
== Examples ==
== Also see ==
* Improved Euler Method
== Sources ==
* : Chapter $\text I$: Existence of Solutions: $2$. Simple ideas about solutions
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
* 
* 
* 
Category:Euler Method
Category:Proof Techniques
Category:Differential Equations
Category:Numerical Analysis
Euler Method/Examples/Arbitrary Example 1

== Example of Use of Euler Method ==
Consider the differential equation:
:$y' = \paren {1 - x} y + \cos x$
with the initial condition:
:$\map y 0 = 1$
Then the Euler Method generates:
:$y_1 = 1 + 2 h$
as an approximatiion to $\map y h$.
== Proof ==
By definition of the Euler Method:
:$y_{n + 1} = y_n + h \map f {x_n, y_n}$
where in this case:
:$\map f {x_n, y_n} = \paren {1 - x_n} y_n + \cos x_n$
and:
:$x_0 = 0$
:$y_0 = \map y {x_0} = \map y 0 = 1$
Hence:
== Sources ==
* 
* 
Category:Examples of Use of Euler Method