avewright/proofwiki-math huggingface dataset:
Theorem: Fourier Transform of Exponential Function

Let $\map f x$ be defined as the real exponential function where the absolute value of the input is used in the exponent and the exponent is scaled by a factor of $-2 \pi a$:
:$\map f x = e^{-2 \pi a \size x}:    \R \to \R$
Then:
:$\map {\hat f} s = \dfrac 1 \pi \dfrac a {a^2 + s^2}$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.

Proof:
By the definition of a Fourier transform:
{{begin-eqn}}
{{eqn | l = \map {\hat f} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-2 \pi a \size x} \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^0 e^{-2 \pi i x s} e^{-2 \pi a \size x} \rd x + \int_0^\infty e^{-2 \pi i x s} e^{-2 \pi a \size x} \rd x
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_{-\infty}^0 e^{-2 \pi i x s} e^{-2 \pi a \paren {-x} } \rd x + \int_0^\infty e^{-2 \pi i x s} e^{-2 \pi a \paren x} \rd x
      | c = {{Defof|Absolute Value|index = 1}}
}}
{{eqn | r = \int_{-\infty}^0 e^{-x \paren {2 \pi i s - 2 \pi a} } \rd x + \int_0^\infty e^{-x \paren {2 \pi i s + 2 \pi a} } \rd x
      | c = Product of Powers
}}
{{eqn | r = \paren {\lim_{\gamma \mathop \to -\infty} \intlimits {-\frac 1 {2 \pi \paren{i s - a} } e^{-x \paren{2 \pi i s - 2 \pi a } } } \gamma 0} + \paren {\lim_{\gamma \mathop \to +\infty} \intlimits {-\frac 1 {2 \pi \paren {a + i s} } e^{-x \paren {2 \pi i s + 2 \pi a} } } 0 \gamma}
      | c = Primitive of Exponential of a x
}}
{{eqn | r = \dfrac 1 {2 \pi} \paren {\dfrac 1 {a - i s} + \dfrac 1 {a + i s} }
      | c = 
}}
{{eqn | r = \dfrac 1 {2 \pi} \paren {\dfrac {a + i s} {a^2 + s^2} + \dfrac {a - i s} {a^2 + s^2} }
      | c = {{Defof|Rational Addition}}
}}
{{eqn | r = \dfrac 1 \pi \paren {\dfrac a {a^2 + s^2} }
      | c = 
}}
{{end-eqn}}
Therefore:
:$\map {\hat f} s = \dfrac 1 \pi \dfrac a {a^2 + s^2 }$
{{qed}}
Theorem: Bisectors of Angles between Two Straight Lines/General Form

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in general form as:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
The angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
:$\dfrac {l_1 x + m_1 y + n_1} {\sqrt { {l_1}^2 + {m_1}^2} } = \pm \dfrac {l_2 x + m_2 y + n_2} {\sqrt { {l_2}^2 + {m_2}^2} }$

Proof:
First we convert $\LL_1$ and $\LL_2$ into normal form:
{{begin-eqn}}
{{eqn | l = \dfrac {l_1 x + m_1 y + n_1} {\sqrt { {l_1}^2 + {m_1}^2} }
      | r = 0
}}
{{eqn | l = \dfrac {l_2 x + m_2 y + n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = x \cos \alpha + y \sin \alpha
      | r = -\dfrac {n_1} {\sqrt { {l_1}^2 + {m_1}^2} }
      | c = where $\cos \alpha = \dfrac {l_1} {\sqrt { {l_1}^2 + {m_1}^2} }$ and $\sin \alpha = \dfrac {m_1} {\sqrt { {l_1}^2 + {m_1}^2} }$
}}
{{eqn | l = x \cos \beta + y \sin \beta
      | r = -\dfrac {n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
      | c = where $\cos \beta = \dfrac {l_2} {\sqrt { {l_2}^2 + {m_2}^2} }$ and $\sin \beta = \dfrac {m_2} {\sqrt { {l_2}^2 + {m_2}^2} }$
}}
{{end-eqn}}
Then from Bisectors of Angles between Two Straight Lines: Normal Form, the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
{{begin-eqn}}
{{eqn | l = x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta}
      | r = -\dfrac {n_1} {\sqrt { {l_1}^2 + {m_1}^2} } + \dfrac {n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
}}
{{eqn | l = x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta}
      | r = -\dfrac {n_1} {\sqrt { {l_1}^2 + {m_1}^2} } - \dfrac {n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
}}
{{eqn | ll= \leadsto
      | l = \paren {x \cos \alpha + y \sin \alpha + \dfrac {n_1} {\sqrt { {l_1}^2 + {m_1}^2} } } - \paren {x \cos \beta + y \sin \beta + \dfrac {n_2} {\sqrt { {l_2}^2 + {m_2}^2} } }
      | r = 0
}}
{{eqn | l = \paren {x \cos \alpha + y \sin \alpha + \dfrac {n_1} {\sqrt { {l_1}^2 + {m_1}^2} } } + \paren {x \cos \beta + y \sin \beta + \dfrac {n_2} {\sqrt { {l_2}^2 + {m_2}^2} } }
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \dfrac {l_1 x + m_1 y + n_1} {\sqrt { {l_1}^2 + {m_1}^2} }
      | r = \dfrac {l_2 x + m_2 y + n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
}}
{{eqn | l = \dfrac {l_1 x + m_1 y + n_1} {\sqrt { {l_1}^2 + {m_1}^2} }
      | r = -\dfrac {l_2 x + m_2 y + n_2} {\sqrt { {l_2}^2 + {m_2}^2} }
      | c = substituting back for $\cos \alpha$ and $\sin \alpha$
}}
{{end-eqn}}
{{qed}}
Theorem: Quadratic Representation of Pair of Straight Lines

Consider the general quadratic equation in $2$ variables:
:$(1): \quad a x^2 + b x y + c y^2 + d x + e y + f = 0$
Then $(1)$ is the locus of $2$ straight lines in the Cartesian plane {{iff}} it can be expressed in the form:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
where $l_1$, $m_1$, $n_1$, $l_2$, $m_2$ and $n_2$ are real numbers.

Proof:
=== Sufficient Condition ===
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in general form as:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
Let $\tuple {x, y}$ be a point on either $\LL_1$ or $\LL_2$.
Then because either $l_1 x + m_1 y + n_1 = 0$ or $l_2 x + m_2 y + n_2 = 0$, it is certainly the case that:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
If $\tuple {x, y}$ is not on either $\LL_1$ or $\LL_2$, then neither $l_1 x + m_1 y + n_1 = 0$ nor $l_2 x + m_2 y + n_2 = 0$ hold.
Hence $l_1 x + m_1 y + n_1 \ne 0$ and $l_2 x + m_2 y + n_2 \ne 0$, and so:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} \ne 0$
Hence two straight lines embedded in $\CC$ can be expressed by an equation in the form:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
as required.
{{qed|lemma}}
Theorem: Pointwise Limit of Increasing Functions is Increasing

Let $X \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $X \to \R$ converging pointwise to a function $f : X \to \R$. 
Let $f_n$ be a increasing function for each $n$.
Then $f$ is increasing.

Proof:
Suppose that $f$ is not increasing.
That is:
:there exists $x, y \in X$ such that $x < y$ and $\map f x > \map f y$. 
Let: 
:$r = \map f x - \map f y > 0$
Since $\sequence {f_n}$ converges to $f$ converging pointwise, the sequence $\sequence {\map {f_n} x}$ converges to $\map f x$. 
That is, there exists $N_1 \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac r 3$
for $n > N_1$.
Similarly, $\sequence {\map {f_n} y}$ converges to $\map f y$.
That is, there exists $N_2 \in \N$ such that: 
:$\size {\map {f_n} y - \map f y} < \dfrac r 3$
for $n > N_2$.
Take $n > \max \set {N_1, N_2}$.
Then, we have: 
:$-\dfrac r 3 < \map {f_n} x - \map f x < \dfrac r 3$
and:
:$-\dfrac r 3 < \map {f_n} y - \map f y < \dfrac r 3$
so that: 
:$-\dfrac {2 r} 3 < \paren {\map {f_n} x - \map f x} - \paren {\map {f_n} y - \map f y} < \dfrac {2 r} 3$
Note that: 
{{begin-eqn}}
{{eqn	| l = \paren {\map {f_n} x - \map f x} - \paren {\map {f_n} y - \map f y}
	| r = \paren {\map {f_n} x - \map {f_n} y} - \paren {\map f x - \map f y}
}}
{{eqn	| r = \paren {\map {f_n} x - \map {f_n} y} - r
}}
{{end-eqn}}
Then, we have: 
:$\map {f_n} x - \map {f_n} y > \dfrac r 3 > 0$
But $x < y$, so this contradicts that $f_n$ is increasing.
So $f$ is increasing.
{{qed}}
Category:Real Analysis
jjc0xl0dnfp2r1h3vmpedlf8caaxvpp
Theorem: Entire Function Bounded by Polynomial is Polynomial

Let $f : \C \to \C$ be an entire function such that: 
:$\cmod {\map f z} \le M {\cmod z}^k$
for all $z \in \C$, for some $k \in \N$ and real $M > 0$. 
Then $f$ is a polynomial of degree at most $k$.

Proof:
Let $r > 0$ be a real number.
Let $D = \map {B_r} 0$ be the open ball with centre $0$ of radius $r$.
By Holomorphic Function is Analytic, we have that: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$
for all $z \in \C$, where: 
:$\ds a_n = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f t} {t^{n + 1} } \rd t$
Note that for $t \in \partial D$, we have: 
:$\cmod {\map f t} \le M {\cmod t}^k = M r^k$
so for all $t \in \partial D$:
:$\cmod {\dfrac {\map f t} {t^{n + 1} } } \le \dfrac {M r^k} {r^{n + 1} }$
From the Estimation Lemma for Contour Integrals::
{{begin-eqn}}
{{eqn | l = \cmod {a_n}
      | r = \cmod {\frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f t} {t^{n + 1} } \rd t}
}}
{{eqn | o = \le
      | r = \frac 1 {2 \pi} \times 2 \pi r \times \frac {M r^k} {r^{n + 1} }
      | c = Estimation Lemma for Contour Integrals
}}
{{eqn | r = \frac M {r^{n - k} }
}}
{{end-eqn}}
Then for $n > k$, we have: 
:$\dfrac M {r^{n - k} } \to 0$
as $r \to \infty$.
Since $r$ was arbitrary, we have: 
:$a_n = 0$
for all $n > k$. 
That is: 
:$\ds \map f z = \sum_{n \mathop = 0}^k a_n z^n$
for all $z \in \C$.
That is, $f$ is a polynomial of degree at most $k$.
{{qed}}
Category:Complex Analysis
pwg9jue6hh5tjvaa0w11jq6vze3z2yu
Theorem: Homogeneous Quadratic Equation represents Two Straight Lines through Origin

Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2 = 0$
Then $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$

Proof:
From Characteristic of Quadratic Equation that Represents Two Straight Lines, $E$ represents $2$ straight lines in the Cartesian plane {{iff}}
:$a b c + 2 f g h - a f^2 - b g^2 - c h^2 = 0$
where in this case $c = f = g = 0$, giving:
:$a b \times 0 + 2 \times 0 \times 0 \times h - a \times 0^2 - b \times 0^2 - 0 \times h^2 = 0$
The result follows from using the Quadratic Formula on $E$.
Setting $x = 0$ gives $y = 0$, confirming that $\tuple {0, 0}$ is a point on both straight lines.
{{qed}}
Theorem: Homogeneous Quadratic Equation representing Imaginary Straight Lines

Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Let $h^2 - a b < 0$.
Then 
Then $E$ represents a conjugate pair of imaginary straight lines in the Cartesian plane:

Proof:
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$
But when $h^2 - a b$, $\sqrt {h^2 - a b}$ is not defined as a real number.
Instead we have:
:$y = \dfrac {h \pm i \sqrt {h^2 - a b} } b x$
{{qed}}
Theorem: Homogeneous Quadratic Equation representing Coincident Straight Lines

Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Let $h^2 - a b = 0$.
Then 
Then $E$ represents $2$ straight lines in the Cartesian plane which completely coincide:

Proof:
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$
But when $h^2 - a b = 0$, we get:
:$y = \dfrac h b x$
{{qed}}
Theorem: Angle Between Two Straight Lines described by Homogeneous Quadratic Equation

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a homogeneous quadratic equation $E$ in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Then the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$

Proof:
Let us rewrite $E$ as follows:
:$b y^2 + 2 h x y + a x^2 = b \paren {y - \mu_1 x} \paren {y - \mu_2 x}$
Thus from Homogeneous Quadratic Equation represents Two Straight Lines through Origin:
:$\LL_1$ and $\LL_2$ are represented by the equations $y = \mu_1 x$ and $y = \mu_2 x$ respectively.
From Sum of Roots of Quadratic Equation:
:$\mu_1 + \mu_2 = -\dfrac {2 h} b$
From Product of Roots of Quadratic Equation:
:$\mu_1 \mu_2 = \dfrac a b$
From Angle between Straight Lines in Plane:
:$\tan \psi = \dfrac {\mu_1 - \mu_2} {1 + \mu_1 \mu_2}$
We have that:
{{begin-eqn}}
{{eqn | l = \paren {\mu_1 - \mu_2}^2
      | r = \paren {\mu_1 + \mu_2}^2 - 4 \mu_1 \mu_2
      | c = 
}}
{{eqn | r = \paren {-\dfrac {2 h} b}^2 - 4 \dfrac a b
      | c = 
}}
{{eqn | r = \dfrac {4 \paren {h^2 - a b} } {b^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tan \psi
      | r = \dfrac {\sqrt {4 \paren {h^2 - a b} / b^2} } {1 + a / b}
      | c = 
}}
{{eqn | r = \dfrac {2 \sqrt {\paren {h^2 - a b} } } {a + b}
      | c = multiplying top and bottom by $b$ and simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Homogeneous Quadratic Equation to describe Perpendicular Straight Lines

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a homogeneous quadratic equation $E$ in two variables.
Let $\LL_1$ and $\LL_2$ be perpendicular.
Then $E$ is of the form:
:$a x^2 + 2 h x y - a y^2$

Proof:
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ is of the form:
:$a x^2 + 2 h x y + b y^2$
From Angle Between Two Straight Lines described by Homogeneous Quadratic Equation, the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
When $\psi = \dfrac \pi 2$, $\tan \psi$ is undefined.
Hence:
:$a + b = 0$
and so $b = -a$.
{{qed}}
Theorem: Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Then $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$

Proof:
From Characteristic of Quadratic Equation that Represents Two Straight Lines we have the conditions in which $E$ does indeed represent $2$ straight lines.
Let $E$ be written as:
{{begin-eqn}}
{{eqn | l = b \paren {y - \mu_1 x - b_1} \paren {y - \mu_2 x - b_2}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b y^2 - b \paren {\mu_1 + \mu_2} x y - b \paren {b_1 + b_2} y + b \mu_1 \mu_2 x^2 + b \paren {b_1 \mu_2 + b_2 \mu_2} + b b_1 b_2
      | r = 0
      | c = multiplying out
}}
{{end-eqn}}
Comparing coefficients of equivalent terms:
{{begin-eqn}}
{{eqn | l = b \mu_1 \mu_2
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mu_1 \mu_2
      | r = \dfrac a b
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = -b \paren {\mu_1 + \mu_2}
      | r = 2 h
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mu_1 + \mu_2
      | r = \dfrac {-2 h} b
      | c = 
}}
{{end-eqn}}
Hence:
:$a x^2 + 2 h x y + b y^2 = b \paren {y - \mu_1 x} \paren {y - \mu_2 x}$
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, it follows that:
:$y = \mu_1 x$
:$y = \mu_2 x$
are two straight lines through the origin represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
By definition of slope of a straight line, these have the same slope as the straight lines $\LL_1$ and $\LL_2$:
:$y = \mu_1 x + b_1$
:$y = \mu_2 x + b_2$
which are described by $E$.
{{qed}}
Theorem: Angle Between Two Straight Lines described by Quadratic Equation

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Then the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$

Proof:
From Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
From Angle Between Two Straight Lines described by Homogeneous Quadratic Equation, the angle $\psi$ between $\LL'_1$ and $\LL'_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
As:
:$\LL'_1$ is parallel to $\LL_1$
:$\LL'_2$ is parallel to $\LL_2$
it follows that the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
{{qed}}
Theorem: Condition for Quadratic Equation to describe Perpendicular Straight Lines

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Let $\LL_1$ and $\LL_2$ be perpendicular.
Then:
:$a + b = 0$
That is, $E$ is of the form:
:$a \paren {x^2 - y^2} + 2 h x y + 2 g x + 2 f y + c = 0$

Proof:
From Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
As $\LL_1$ and $\LL_2$ are parallel respectively to $\LL'_1$ and $\LL'_2$, it follows that $\LL'_1$ and $\LL'_2$ are themselves perpendicular.
Hence from Condition for Homogeneous Quadratic Equation to describe Perpendicular Straight Lines, $\LL'_1$ and $\LL'_2$ are represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y - a y^2$
That is:
:$a + b = 0$
The result follows.
{{qed}}
Theorem: Quadratic Equation for Parallel Straight Lines

Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Let $\LL_1$ and $\LL_2$ be parallel.
Then:
:$h^2 - a b = 0$

Proof:
From Homogeneous Quadratic Equation representing Coincident Straight Lines, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ coincident straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
where:
:$h^2 - a b = 0$
Hence $\LL_1$ and $\LL_2$ are represented by the homogeneous quadratic equation:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
where:
:$h^2 - a b = 0$
{{qed}}
Theorem: Harmonic Conjugacy is Symmetric

Let $AB$ and $PQ$ be line segments on a straight line.
Let $P$ and $Q$ be '''harmonic conjugates''' with respect to $A$ and $B$.
Then $A$ and $B$ are '''harmonic conjugates''' with respect to $P$ and $Q$ .

Proof:
By definition of harmonic conjugates, $\tuple {AB, PQ}$ is a harmonic range.
:400px
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {AP} {PB}
      | r = -\dfrac {AQ} {QB}
      | c = {{Defof|Harmonic Range}}
}}
{{eqn | ll= \leadsto
      | l = -\dfrac {PA} {PB}
      | r = -\paren {-\dfrac {AQ} {BQ} }
      | c = reversing the direction of $AP$ and $QB$
}}
{{eqn | r = \dfrac {AQ} {BQ}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -PA
      | r = \dfrac {AQ \times PB} {BQ}
      | c = multiplying by $PB$
}}
{{eqn | ll= \leadsto
      | l = -\dfrac {PA} {AQ}
      | r = \dfrac {PB} {BQ}
      | c = dividing by $BQ$
}}
{{end-eqn}}
Hence, by definition, $\tuple {PQ, AB}$ is a harmonic range.
Hence the result, by definition of harmonic conjugates.
{{qed}}
Theorem: Condition for Pairs of Lines through Origin to be Harmonic Conjugates

Consider $4$ lines $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$ through the origin $O$ whose equations embedded in the Cartesian plane are as follows:
{{begin-eqn}}
{{eqn | ll= (\LL_1):
      | l = y
      | r = \lambda x
}}
{{eqn | ll= (\LL_2):
      | l = y
      | r = \mu x
}}
{{eqn | ll= (\LL_3):
      | l = y
      | r = \lambda' x
}}
{{eqn | ll= (\LL_4):
      | l = y
      | r = \mu' x
}}
{{end-eqn}}
Then the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates {{iff}}:
:$\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$

Proof:
=== Sufficient Condition ===
Let $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ be harmonic conjugates as asserted.
A straight line in the plane which does not pass through $O$ will either:
:intersect all four of $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$
or:
:be parallel to one such straight line and intersect the other three.
Hence, let a straight line be drawn parallel to $\LL_1$ so as to intersect each of $\LL_2$, $\LL_3$, $\LL_4$ at $L'$, $M$ and $M'$.
It would cut $\LL_1$ at $L$, but $L$ is the point at infinity of $\LL_1$.
500px
From Straight Line which cuts Harmonic Pencil forms Harmonic Range, the points of intersection define a harmonic range.
We have that $L'M'$ is parallel to $\LL_1$.
From Harmonic Range with Unity Ratio, the harmonic range defined by $L'M'$  is such that $M$ is the midpoint of $L'M'$.
$L'M'$ can be expressed in slope-intercept form as $y = \lambda x + c$ for some $c \in \R_{\ne 0}$.
Hence the abscissae of the points $L'$, $m$ and $M'$ are respectively:
:$\dfrac c {\lambda' - \lambda}$, $\dfrac c {\mu - \lambda}$, $\dfrac c {\mu' - \lambda}$
Hence the condition that the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates is:
:$\dfrac 2 {\mu - \lambda} = \dfrac 1 {\lambda' - \lambda} + \dfrac 1 {\mu' - \lambda}$
which is more usually expressed as:
:$\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$
{{qed|lemma}}
Theorem: Nth Derivative of Exponential of Minus One over x

Let $n \in \N$ be a natural number.
Let $\map {P_n} x$ be a real polynomial of degree $n$.
Then:
:$\ds \dfrac {\d^n} {\d x^n} \map \exp {- \frac 1 x} = \frac {\map {P_n} x}{x^{2n}} \map \exp {- \frac 1 x}$
where $\map {P_n} x$ satisfies the following recurrence relation:
:$\map {P_{n + 1}} x = x^2 \dfrac \d {\d x} \map {P_n} x - \paren {2 n x - 1} \map {P_n} x$
and $\map {P_0} x = 1$.

Proof:
Proof by induction:
Theorem: Condition for Pairs of Lines through Origin to be Harmonic Conjugates/Homogeneous Quadratic Equation Form

Consider the two homogeneous quadratic equations:
{{begin-eqn}}
{{eqn | n = E1
      | l = a_1 x^2 + 2 h_1 x y + b_1 y^2
      | r = 0
}}
{{eqn | n = E2
      | l = a_2 x^2 + 2 h_2 x y + b_2 y^2
      | r = 0
}}
{{end-eqn}}
each representing two straight lines through the origin.
Then the two straight lines represented by $(\text E1)$ are harmonic conjugates of the two straight lines represented by $(\text E2)$ {{iff}}:
:$a_1 b_2 + a_2 b_1 - 2 h_1 h_2 = 0$

Proof:
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $(\text E1)$ and $(\text E2)$ represent straight lines through the origin {{iff}}:
{{begin-eqn}}
{{eqn | l = h_1^2 - a_1 b_1
      | o = >
      | r = 0
}}
{{eqn | l = h_2^2 - a_2 b_2
      | o = >
      | r = 0
}}
{{end-eqn}}
Let the two straight lines represented by $(\text E1)$ be defined by the equations:
{{begin-eqn}}
{{eqn | l = y
      | r = \lambda_1 x
}}
{{eqn | l = y
      | r = \mu_1 x
}}
{{end-eqn}}
Let the two straight lines represented by $(\text E2)$ be defined by the equations:
{{begin-eqn}}
{{eqn | l = y
      | r = \lambda_2 x
}}
{{eqn | l = y
      | r = \mu_2 x
}}
{{end-eqn}}
Then we can write the Condition for Pairs of Lines through Origin to be Harmonic Conjugates as:
:$(1): \quad 2 \paren {\lambda_1 \mu_1 + \lambda_2 \mu_2} = \paren {\lambda_1 + \mu_1} \paren {\lambda_2 + \mu_2}$
We can express $a_1 x^2 + 2 h_1 x y + b_1 y^2 = 0$ as:
:$b_1 \paren {y - \lambda_1 x} \paren {y - \mu_1 x} = 0$
and $a_2 x^2 + 2 h_2 x y + b_2 y^2 = 0$ as:
:$b_2 \paren {y - \lambda_2 x} \paren {y - \mu_2 x} = 0$
from which we obtain via Sum of Roots of Quadratic Equation and Product of Roots of Quadratic Equation:
{{begin-eqn}}
{{eqn | l = \lambda_1 + \mu_1
      | r = -\dfrac {2 h_1} {b_1}
}}
{{eqn | l = \lambda_2 + \mu_2
      | r = -\dfrac {2 h_2} {b_2}
}}
{{eqn | l = \lambda_1 \mu_1
      | r = \dfrac {a_1} {b_1}
}}
{{eqn | l = \lambda_2 \mu_2
      | r = \dfrac {a_2} {b_2}
}}
{{end-eqn}}
Therefore $(1)$ reduces to:
{{begin-eqn}}
{{eqn | l = 2 \paren {\dfrac {a_1} {b_1} + \dfrac {a_2} {b_2} }
      | r = \dfrac {4 h_1 h_2} {b_1 b_2}
}}
{{eqn | ll= \leadsto
      | l = a_1 b_2 + a_2 b_1 - 2 h_1 h_2
      | r = 0
}}
{{end-eqn}}
{{qed}}