nyuuzyou/wikis huggingface dataset:
Polar Form of Complex Number/Examples/3 cis -2 pi 3^-1/Mistake

== Source Work ==
:Chapter $1$: Complex Numbers
::Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(f)}$
== Mistake ==
:Graph each of the following and express in rectangular form.
::... $\text {(f)} \ \ 3 e^{-2 \pi i / 3}$
:Ans. ... $\text {(f)} \ -3 \sqrt 3 / 2 - \paren {3 / 2} i$
== Correction ==
The correct solution is:
:$-\dfrac 3 2 - \dfrac {3 \sqrt 3} 2 i$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(f)}$
Category:Mistakes/Theory and Problems of Complex Variables
Polar Form of Complex Number/Examples/4 cis 315

== Example of Polar Form of Complex Number ==
The complex number $\polar {4, 315 \degrees}$ can be expressed in Cartesian form as:
:$4 \cis 315 \degrees = 2 \sqrt 2 - 2 \sqrt 2 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(c)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/4 cis 3 pi 5^-1

== Example of Polar Form of Complex Number ==
The complex number $4 \paren {\cos \dfrac {3 \pi} 5 + i \sin \dfrac {3 \pi} 5}$ can be expressed as:
:$4 \paren {\cos \dfrac {3 \pi} 5 + i \sin \dfrac {3 \pi} 5} = 4 \paren {\cos 108 \degrees + i \sin 108 \degrees} = 6 \cis 108 \degrees = 6 \cis \dfrac {3 \pi} 5 = 6 e^{3 \pi / 5}$
and depicted in the complex plane as:
:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $17 \ \text {(b)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/5 cis 7 pi 6^-1

== Example of Polar Form of Complex Number ==
The complex number $\polar {5, \dfrac {7 \pi} 6}$ can be expressed in Cartesian form as:
:$5 \cis \dfrac {7 \pi} 6 = -5 \dfrac {\sqrt 3} 2 - \dfrac 5 2 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(e)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/6 cis 135

== Example of Polar Form of Complex Number ==
The complex number $\polar {6, 135 \degrees}$ can be expressed in Cartesian form as:
:$6 \cis 135 \degrees = -3 \sqrt 2 + 3 \sqrt 2 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(a)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/6 cis 240

== Example of Polar Form of Complex Number ==
The complex number $6 \paren {\cos 240 \degrees + i \sin 240 \degrees}$ can be expressed as:
:$6 \paren {\cos 240 \degrees + i \sin 240 \degrees} = 6 \cis 240 \degrees = 6 \cis \dfrac {4 \pi} 3 = 6 e^{4 \pi / 3}$
and depicted in the complex plane as:
:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $17 \ \text {(a)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/i

== Example of Polar Form of Complex Number ==
The imaginary unit $i$ can be expressed in polar form as $\polar {1, \dfrac \pi 2}$.
== Proof ==
Then:
Hence:
:$\map \arg i = \dfrac \pi 2$
and hence the result.
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/root 2 i

== Example of Polar Form of Complex Number ==
The imaginary number $\sqrt 2 i$ can be expressed in polar form as $\polar {\sqrt 2, \dfrac \pi 2}$.
== Proof ==
Then:
Hence:
:$\map \arg {\sqrt 2 i} = \dfrac \pi 2$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(g)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/root 3 over 2 - 3 i over 2

== Example of Polar Form of Complex Number ==
The complex number $\dfrac {\sqrt 3} 2 - \dfrac {3 i} 2$ can be expressed as a complex number in polar form as $\polar {\sqrt 3, \dfrac {5 \pi} 3}$.
== Proof ==
Then:
Hence:
:$\map \arg {\dfrac {\sqrt 3} 2 - \dfrac {3 i} 2} = \dfrac {5 \pi} 3$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(h)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Reciprocal of Complex Number

== Theorem ==
Let $z := r \paren {\cos \theta + i \sin \theta} \in \C$ be a complex number expressed in polar form.
Then:
:$\dfrac 1 z = \dfrac {\cos \theta - i \sin \theta} r$
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.14)$
Category:Polar Form of Complex Number
Polar is Locus of Harmonic Conjugates wrt Ellipse

== Theorem ==
Let $\EE$ be an ellipse embedded in the plane.
Let $P$ be an arbitrary point in the plane.
Let $\LL$ be the polar of $P$ $\EE$.
Then $\LL$ is the locus of harmonic conjugates $\EE$.
== Proof ==
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $3$.
Category:Polars of Points
Category:Ellipses
Polar of Point is Perpendicular to Line through Center

== Theorem ==
Let $\CC$ be a circle.
Let $P$ be a point.
Let $\LL$ be the polar of $P$ with respect to $\CC$.
Then $\LL$ is perpendicular to the straight line through $P$ and the center of $\CC$.
== Proof ==
Let $\CC$ be positioned so as for its center to be at the origin of a Cartesian plane.
Let $P$ be located at $\tuple {x_0, y_0}$.
From Equation of Straight Line in Plane, $P$ can be described as:
:$y = \dfrac {y_0} {x_0} x$
and so has slope $\dfrac {y_0} {x_0}$.
By definition of polar, $\LL$ has the equation:
:$x x_0 + y y_0 = r^2$
which has slope $-\dfrac {x_0} {y_0}$.
Hence the result from Condition for Straight Lines in Plane to be Perpendicular.
== Sources ==
* : Chapter $\text {III}$. The Circle: $6$.
Category:Polars of Points
Category:Perpendiculars
Polarization Identity/Complex Vector Space

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\C$. 
Let $\norm \cdot$ be the inner product norm on $V$. 
Then, we have: 
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$
for each $x, y \in V$.
== Proof ==
We write: 
We then compute: 
We then have: 
and:
Then we have: 
== Sources ==
* $8.3$: Properties of the Induced Norms 
Category:Polarization Identity
Polarization Identity/Real Vector Space

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\R$. 
Let $\norm \cdot$ be the inner product norm for $V$. 
Then we have: 
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2$
for all $x, y \in V$.
== Proof ==
We have: 
== Sources ==
* : 1. Functions on Euclidean Space: Norm and Inner Product
* : $\S 2$: Riemannian Metrics. Definitions
* $8.3$: Properties of the Induced Norms 
Category:Polarization Identity
Pole (Complex Analysis)/Examples

== Examples of Poles in the context of Complex Analysis ==
=== Example: $\frac 1 {\paren {z - 3}^2 \paren {z + 1} }$ ===
=== Example: $\frac {z^2 + 1} z$ ===
=== Example: $\frac {\sin z} {\paren {z - \pi} \paren {z - 2}^4}$ ===
Category:Poles (Complex Analysis)
Pole (Complex Analysis)/Examples/Reciprocal of (z-3)^2 (z+1)

== Examples of Poles in the context of Complex Analysis ==
Let $f$ be the complex function:
:$\forall z \in \C \setminus \set {-1, 3}: \map f z = \dfrac 1 {\paren {z - 3}^2 \paren {z + 1} }$
Then $f$ has:
:a pole of order $2$ at $z = 3$
:a simple pole at $z = -1$.
== Sources ==
* 
* 
Category:Poles (Complex Analysis)
Pole (Complex Analysis)/Examples/sin z over (z-pi) (z-2)^4

== Examples of Poles in the context of Complex Analysis ==
Let $f$ be the complex function:
:$\forall z \in \C \setminus \set 2: \map f z = \dfrac {\sin z} {\paren {z - \pi} \paren {z - 2}^4}$
Then $f$ has a pole of order $4$ at $z = 2$
Note that at $z = \pi$ there is no pole as the numerator is $0$ at that point.
== Sources ==
* 
Category:Poles (Complex Analysis)
Pole (Complex Analysis)/Examples/z^2 + 1 over z

== Examples of Poles in the context of Complex Analysis ==
Let $f$ be the complex function:
:$\forall z \in \C \setminus \set 0: \map f z = \dfrac {z^2 + 1} z$
Then $f$ has:
:a simple pole at $z = 0$.
== Sources ==
* 
Category:Poles (Complex Analysis)
Pole at Infinity implies Polynomial Function

== Theorem == 
 
Let $f : \C \to \C$ be an entire function.
Let $f$ have a pole of order $N$ at $\infty$. 
Then $f$ is a polynomial of degree $N$. 
 
== Proof == 
By Complex Function is Entire iff it has Everywhere Convergent Power Series, there exists a power series: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$ 
convergent for all $z \in \C$, where $\sequence {a_n}$ is a sequence of complex coefficients. 
This gives: 
:$\ds \map f {\frac 1 z} = \sum_{n \mathop = 0}^\infty \frac {a_n} {z^n}$ 
It is given that $\map f z$ has a pole of order $N$ at $\infty$, so $\map f {\dfrac 1 z}$ has a pole of order $N$ at $0$. 
So $N$ is the least positive integer such that:
:$\ds z^N \map f {\frac 1 z} = \sum_{n \mathop = 0}^\infty a_n z^{N - n}$ 
is holomorphic at $0$, with $a_N \ne 0$. 
Therefore, all exponents of $z$, with non-zero coefficients, in this series must be non-negative. 
So $a_n = 0$ for $n > N$. 
Hence: 
:$\ds \map f z = \sum_{n \mathop = 0}^N a_n z^n$
with $a_N \ne 0$. 
That is, $f$ is a polynomial of degree $N$. 
Category:Complex Analysis
Poles and Zeroes of Meromorphic Function are of Finite Order

== Theorem ==
Let $f: D \to \C$ be a meromorphic function on the open subset $D$ of $\C$.
=== Zeroes of Meromorphic Function are of Finite Order ===
=== Poles of Meromorphic Function are of Finite Order ===
== Sources ==
* 
Category:Meromorphic Functions
Poles of Gamma Function

== Theorem ==
The gamma function $\Gamma: \C \to \C$ is analytic throughout the complex plane except at $\set {0, -1, -2, -3, \dotsc}$ where it has simple poles.
== Proof ==
First we examine the location of the poles.
We examine the Weierstrass form of the Gamma function:
:$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \ds \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$
The factors of the continued product clearly do not tend to zero.
So the continued product is zero only if one of the factors is zero.
This occurs when $1 + \dfrac z n = 0$.
This occurs at $z \in \set {-1, -2, \dotsc}$.
We also have the expression outside the continued product to consider.
From Exponential Tends to Zero and Infinity, the exponential function is never $0$.
So this expression is zero whenever $z = 0$.
Furthermore, if some factor in the continued product is zero, this excludes the possibility that any other factor is zero.
So the zeros are simple.
Hence:
:$\dfrac 1 {\map \Gamma z} = 0 \iff z \in \set {0, -1, -2, -3, \ldots}$
Therefore $\Gamma$ has simple poles at these points.
Next we show that $\Gamma$ is analytic on $\map \Re z > 0$. 
By Gamma Difference Equation this proves that $\Gamma$ is analytic on $\C \setminus \set {0, -1, -2, -3, \dotsc}$.
Let the functions $\Gamma_n$ be defined by
:$\ds \map {\Gamma_n} z = \int_0^n t^{z - 1} e^{-t} \rd t$
Clearly each $\Gamma_n$ is analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show that $\Gamma_n \to \Gamma$ locally uniformly.
By Modulus of Complex Integral:
:$\ds \cmod {\map \Gamma z - \map {\Gamma_n} Z } \le \int_n^\infty t^{x - 1} e^{-t} \rd t$
where $x = \map \Re z$. 
Let $a \in \C$ with $\map \Re a > 0$.
Let $D$ be an open disk of radius $r$ about $a$.
By definition of exponential function, we have the expansion:
:$e^z = 1 + z + \dfrac {z^2} 2 + \dotsb$
from which we see that:
:$\forall \alpha \in \R_{>0}: z^\alpha = \map \oo {e^z}$
where $\oo$ denotes little-$\oo$ notation.
In particular:
:$\exists c \in \R_{>0}: \forall t \ge 1: e^{-t} \le c t^{-\map \Re a - r - 1}$
Then:
So $\Gamma_n \to \Gamma$ uniformly in $D$, and the proof is complete.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33$
Category:Gamma Function
Poles of Rational Function are Zeroes of Denominator

== Theorem ==
Let $\map f x = \dfrac {\map P x} {\map Q x}$ be a rational function.
Let $\map f x$ be such that $\map P x$ and $\map Q x$ have no common divisors.
Then the zeroes of $\map Q x$ are the poles of $f$.
== Proof ==
== Sources ==
* 
* 
Category:Rational Functions
Poles of Riemann Zeta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Then $\zeta$ has a simple pole at $s = 1$ with residue $1$, and no other poles.
== Proof ==
By Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part:
:$\ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
is meromorphic for $\map \Re s > 0$, and the integral converges to a finite value for fixed $s$ in this region.
Therefore in this region the only pole of $\zeta$ is at $s = 1$, with residue:
By Unsymmetric Functional Equation for Riemann Zeta Function:
:$\map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\dfrac 1 2 s \pi} \map \Gamma s \map \zeta s$
Therefore, for $\map \Re s \le 0$:
By Complex Exponential Function is Entire, the factor $\ds 2^{1 - t} \pi^{-t}$ has no poles when $\map \Re t \ge 1$.
By Poles of Gamma Function, $\map \Gamma t$ has no poles when $\map \Re t \ge 1$.
By Complex Cosine Function is Entire, $\map \cos {\dfrac 1 2 t \pi}$ also has no poles in this region.
Therefore, the only possible pole is a simple pole at $t = 1$ from the factor $\map \zeta t$.
But at this point:
:$\map \cos {\dfrac 1 2 t \pi} = \map \cos {\dfrac \pi 2} = 0$
which cancels the pole of $\zeta$.
Category:Riemann Zeta Function
Polydivisible Number/Examples/3,608,528,850,368,400,786,036,725

== Theorem ==
The largest polydivisible number has $25$ digits:
:$3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036 \, 725$
== Proof ==
From No Polydivisible Number with 26 Digits Exists, there are no polydivisible numbers with more digits.
== Also see ==
* Pandigital Polydivisible Number: $381 \, 654 \, 729$
== Sources ==
* 
* : $3,608,528,850,368,400,786,036,725$
Category:Polydivisible Numbers
Category:3,608,528,850,368,400,786,036,725
Polydivisible Number/Examples/381,654,729

== Theorem ==
The integer $381 \, 654 \, 729$ is the only polydivisible number which is penholodigital.
== Proof ==
First it is demonstrated that indeed $381 \, 654 \, 729$ has this property:
It remains to be demonstrated that it is the only such number.
Let $\sqbrk {abcdefghi}$ be a polydivisible number which is penholodigital.
By Divisibility by 5:
:$e = 5$
By Divisibility by 2:
:$b, d, f, h$ are even.
Hence:
:$a, c, g, i$ are odd.
By Divisibility by 3:
:$a + b + c$, $d + 5 + f$, $g + h + i$ are divisible by $3$.
By Divisibility by 8:
:$\sqbrk {fgh}$ is divisible by $8$.
Because $f$ is even:
:$\sqbrk {f00}$ is divisible by $8$.
We must therefore have:
:$\sqbrk {gh}$ divisible by $8$.
Thus:
:$\sqbrk {gh}$ must be one of $16, 32, 72, 96$.
Because $i$ is odd and $g + h + i$ are divisible by $3$:
:$\sqbrk {ghi}$ must be one of $321, 327, 723, 729, 963$.
By Divisibility by 4:
:$\sqbrk {cd}$ is divisible by $4$.
Because $c$ is odd:
:$d$ must be $2$ or $6$.
Suppose $h = 6$.
Then $\sqbrk {ghi} = 963$ and $d = 2$.
Since $d + 5 + f$ is divisible by $3$ and $f$ is even:
:$f = 8$
The remaining even number $b$ must be $4$.
So:
:$\sqbrk {abcdefghi} = \sqbrk {a4c258963}$
Thus the only remaining options for $a$ and $c$ are $1$ and $7$.
However, we find that neither $1472589$ nor $7412689$ is divisible by $7$.
Therefore we must have:
:$h = 2$
It follow therefore that:
:$d = 6$
Because $d + 5 + f$ is divisible by $3$, and $f$ is even:
:$f = 4$
The remaining even number $b$ must be $8$.
So:
:$\sqbrk {abcdefghi} = \sqbrk {a8c654g2i}$
Since $a + 8 + c$ is divisible by $3$:
:$\sqbrk {a8c}$ must be one of $183, 189, 381, 387, 783, 789, 981, 987$.
Suppose $g = 3$.
Then:
:$\sqbrk {a8c}$ must be one of $189, 789, 981, 987$.
Inspection of the possible numbers reveals that none of $1896543$, $7896543$, $9816543$, $9876543$ is divisible by $7$.
Suppose $g = 7$.
Then:
:$\sqbrk {a8c}$ must be one of $183, 189, 381, 981$.
Among $1836547$, $1896547$, $3816547$, $9816547$, only $3816547$ is divisible by $7$.
The results follow.
== Historical Note ==
== Sources ==
* : $381,654,729$
Category:Polydivisible Numbers
Category:Penholodigital Integers
Category:381,654,729
Polydivisible Number/Examples/381,654,729/Historical Note

== Historical Note on Pandigital Polydivisible Number $381 \, 654 \, 729$ ==
 reports in his $1997$ work that this factoid was posted in the The Sunday Times as their brainteaser no. $1040$, by a contributor going by the name Upton, but further details have not been uncovered.
== Sources ==
* : $381,654,729$
Category:Historical Notes
Polydivisible Number/Examples/381,654,729/Mistake

== Source Work ==
:
:The Dictionary
::$381,654,729$
== Mistake ==
:The unique integer such that the number formed by the first $n$ digits is divisible by the digit $n$.
That should read:
:The unique pandigital integer (in the sense that zero is excluded) such that the number formed by the first $n$ digits is divisible by the digit $n$.
It is a trivial task to create a $9$-digit integer with this property that is not pandigital. Straight off the top of my head:
:$222 \, 456 \, 564$
It is also worth pointing out that $3 \, 816 \, 547 \, 290$ is also pandigital (but in the sense that zero is included) and also has the same property -- although $10$ is not a digit as such.
Numbers with this property are referred to as polydivisible.
== Sources ==
* : $381,654,729$
Category:Mistakes/Curious and Interesting Numbers
Polygamma Function in terms of Hurwitz Zeta Function

== Theorem ==
:$\map {\psi_n} z = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}$
where:
:$\psi_n$ is the polygamma function
:$\Gamma$ is the gamma function
:$\zeta$ is the Hurwitz zeta function
:$z \in \C_{>0}$
:$n \in \Z_{\ge 1}$.
== Proof ==
== Sources == 
* 
Category:Hurwitz Zeta Function
Category:Polygamma Function
Polygamma Reflection Formula

== Theorem ==
Let $z \in \C \setminus \Z$.
Let $\psi_n$ denote the $n$th polygamma function.
Then:
:$\map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z} = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z$
== Lemma ==
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Euler's Reflection Formula
* Digamma Reflection Formula
== Sources == 
* 
Category:Polygamma Reflection Formula
Category:Polygamma Function
Category:Reflection Formulas
Polygamma Reflection Formula/Lemma

== Polygamma Reflection Formula: Lemma ==
Let $\psi$ denote the digamma function.
The expression:
:$\map \psi z - \map \psi {1 - z}$
is defined on the domain $\C \setminus \Z$.
That is, on the set of complex numbers but specifically excluding the integers.
== Proof ==
From the definition of the digamma function:
:$\map \psi z$ is defined for $z \in \C \setminus \Z_{\le 0}$
and:
:$\map \psi {1 - z}$ is defined for $\paren {1 - z} \in \C \setminus \Z_{\le 0}$.
Therefore, $\map \psi z - \map \psi {1 - z}$ is defined for $z \in \C \setminus \Z$.
Category:Polygamma Reflection Formula
Polygamma Reflection Formula/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
By definition:
:$\map {\psi_n} z = \dfrac {\d^n} {\d z^n} \map \psi z$
where:
:$\psi$ denotes the digamma function
:$z \in \C \setminus \Z_{\le 0}$.
Then:
Finally, from the , we note that:
:$\map \psi z - \map \psi {1 - z}$
is defined on the domain $\C \setminus \Z$.
Category:Polygamma Reflection Formula
Polygon Triangulation Theorem

== Theorem ==
Let $P$ be a polygon with $n$ sides, where $n \in \N_{ \ge 3 }$.
Then there exists a triangulation of $P$ that fulfills this condition:
* If $AB$ is a side of a triangle in the triangulation of $P$, then $AB$ is either a side of $P$, or a chord of $P$ that lies completely in the interior of $P$.
All triangulations of $P$ that fulfill this condition consist of exactly $n-2$ triangles.
== Proof ==
We show existence by using strong induction over $n$, the number of sides of $P$.
=== Basis for the Induction ===
Let $n = 3$.
Then $P$ is a triangle, and $\family P$ is a triangulation of $P$ that consists of $1 = n-2$ triangles.
All sides of $P$ are equal to sides of $P$.
This is our basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
For each $i$ such that $3 \le i \le n$, a polygon $P$ with $i$ sides has a triangulation that fulfills the above condition, and consists of $i-2$ triangles.
Then we need to show: 
A polygon $P$ with $n+1$ sides has a triangulation that fulfills the above condition, and consists of $n-1$ triangles.
=== Induction Step ===
This is our induction step:
As $P$ has $n+1$ sides with $n \in \N_{ \ge 3}$, it follows that $P$ is not a triangle.
Polygon not Equal to Triangle has Chord shows that there exists a chord of $P$ that lies completely in the interior of $P$.
Let $AB$ be such an arbitrary chord of $P$.
The chord $AB$ dissects $P$ into two polygons $P_1$ and $P_2$.
Let $P_1$ have $n_1$ sides, and let $P_2$ have $n_2$ sides.
All sides of $P$ are sides of either $P_1$ or $P_2$, and $AB$ is a side of both $P_1$ and $P_2$.
It follows that the combined number of sides of $P_1$ and $P_2$ are equal to:
:$n_1 + n_2 = \paren { n+1 } + 2 = n+3$
As $n_1 , n_2 \in \N_{ \ge 3}$, we have:
:$n_1 \le n + 3 - 3 = n$
and symmetrically, $n_2 \le n$.
By principle of strong induction, there exists a triangulation of $P_1$ that consists of $n_1 - 2$ triangles, and a triangulation of $P_2$ that consists of $n_2 - 2$ triangles.
The triangles of these two triangulation combine to form a triangulation of $P$.
These triangles have sides that are sides and chords of $P_1$ and $P_2$, which means they are sides and chords of $P$.
The total number of triangles in this triangulation of $P$ is:
:$\paren { n_1 - 2 } + \paren { n_2 -2 } = n - 1$
which proves the induction hypothesis.
In the induction step, the chord $AB$ could be chosen arbitrarily.
It follows that for any triangulation of $P$ that fulfills the above condition, its triangles can be constructed by this inductive process.
These triangles cannot be dissected without creating a side that is not a chord of $P$.
It follows that the triangulation consists of exactly $n-2$ triangles.
== Sources ==
* : $\S 1.3.1$
Category:Triangles
Polygon has Salient Angle

== Theorem ==
Let $P$ be a polygon.
Then $P$ has at least one salient angle.
== Proof ==
Recall the definition of salient angle:
:A salient angle is an internal angle which is less than $180 \degrees$.
An internal angle which is not a salient angle is a re-entrant angle.
Let $C$ be a circle such that all vertices of $P$ lie within the interior of $C$.
Let $A$ be the vertex of $P$ that has the minimal distance to the circumference of $C$.
If more than one vertex of $P$ has the same minimal distance to the circumference, choose $A$ among them arbitrarily.
Let $\angle A$ denote the internal angle at $A$.
 $\angle A$ is a re-entrant angle.
Let $H$ be the point on the circumference of $C$ that lies closest to $A$.
Let $r$ be the line segment with endpoints $A$ and $H$.
Let $t$ be tangent to $C$ at $H$, which can be constructed by Construction of Tangent from Point to Circle.
Let $l$ be the straight line parallel with $t$ through $A$, which can be constructed by Construction of Parallel Line.
Suppose one of the two adjacent sides of $A$ forms an acute or right angle with $r$.
Then that side would either intersect $C$, or have an endpoint that lies closer to $C$ than $A$.
But :
:$P$ lies completely inside $C$
and
:$A$ is the closest vertex of $P$ to $C$.
Hence neither of the two adjacent sides of $A$ forms an acute or right angle with $r$.
It follows that the adjacent sides of $A$ lie on the same side of $l$.
Then $r$ is an intersecting line of the re-entrant angle formed by the two sides.
As $r$ ends in $H$ in the exterior of $P$, it follows that this re-entrant angle is not the internal angle of the vertex $A$.
This contradicts our definition of $A$.
Hence it follows that the internal angle of the vertex $A$ is salient angle.
Category:Polygons
Category:Internal Angles