avewright/proofwiki-math huggingface dataset:
Theorem: Particular Values of Chebyshev Polynomials of the First Kind/0/Even Order

Let $\map {T_n} x$ denote the Chebyshev polynomial of the first kind of order $n$.
Let $n = 2 m$ for some $m \in \N$.
Then:
:$\map {T_n} 0 = \paren {-1}^m$

Proof:
Let $n = 2 m$ for some $m \in \N$.
The proof proceeds by induction.
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\map {T_{2 m} } 0 = \paren {-1}^m$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \map {T_0} 0
      | r = \paren {-1}
      | c = Chebyshev Polynomial of the First Kind: $T_0$
}}
{{eqn | r = \paren {-1}^0
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\map {T_{2 k} } 0 = \paren {-1}^k$
from which it is to be shown that:
:$\map {T_{2 \paren {k + 1} } } 0 = \paren {-1}^{k + 1}$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \map {T_{2 \paren {k + 1} } } 0
      | r = 2 \times 0 \times \map {T_{2 k + 1} } 0 - \map {T_{2 k} } 0
      | c = Recurrence Formula for Chebyshev Polynomials of the First Kind, setting $n = 2 \paren {k + 1} = 2 k + 2$
}}
{{eqn | r = \paren {-1} \times \map {T_{2 k} } 0
      | c = simplifying
}}
{{eqn | r = \paren {-1} \times \paren {-1}^k
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {-1}^{k + 1}
      | c = simplifyng
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m \in \Z_{\ge 0}: \map {T_{2 m} } 0 = \paren {-1}^m$
{{qed}}
Theorem: Particular Values of Chebyshev Polynomials of the First Kind/0/Odd Order

Let $\map {T_n} x$ denote the Chebyshev polynomial of the first kind of order $n$.
Let $n = 2 k + 1$ for some $k \in \N$.
Then:
:$\map {T_n} 0 = 0$

Proof:
Let $n = 2 k + 1$ for some $k \in \N$.
{{begin-eqn}}
{{eqn | q = \forall x \in \Dom {T_n}
      | l = \map {T_n} {-x}
      | r = \paren {-1}^n \map {T_n} x
      | c = Particular Values of Chebyshev Polynomials of the First Kind: $-x$
}}
{{eqn | ll= \leadsto
      | l = \map {T_n} 0
      | r = \paren {-1}^n \map {T_n} 0
      | c = 
}}
{{eqn | r = -\map {T_n} 0
      | c = as $n$ is odd: $\paren {-1}^n = -1$
}}
{{end-eqn}}
The only number $x$ such that $x = -x$ is $x = 0$.
Hence the result.
{{qed}}
Theorem: Recurrence Formula for Chebyshev Polynomials of the Second Kind

Let $\map {U_n} x$ denote the Chebyshev polynomials of the second kind of order $n$.
Then:
:$\map {U_n} x = \begin {cases} 1 & : n = 0 \\ 2 x & : n = 1 \\ 2 x \map {U_{n - 1} } x - \map {U_{n - 2} } x & : n > 1 \end {cases}$

Proof:
From the {{Defof|Chebyshev Polynomial of the Second Kind}}, we have:
:$\map {U_n} x = \dfrac {\map \sin {\paren {n + 1} \arccos x} } {\map \sin {\arccos x} }$
For $n = 0$, we have:
{{begin-eqn}}
{{eqn | l = \map {U_0} x
      | r = \dfrac {\map \sin {\paren {0 + 1} \arccos x} } {\map \sin {\arccos x} }
      | c = {{Defof|Chebyshev Polynomial of the Second Kind}} at $n = 0$
}}
{{eqn | ll = \leadsto
      | l = \map {U_0} x
      | r = 1
      | c = 
}}
{{end-eqn}}
For $n = 1$, we have:
{{begin-eqn}}
{{eqn | l = \map {U_1} x
      | r = \dfrac {\map \sin {\paren {1 + 1} \arccos x} } {\map \sin {\arccos x} }
      | c = {{Defof|Chebyshev Polynomial of the Second Kind}} at $n = 1$
}}
{{eqn | ll = \leadsto
      | l = \map {U_1} x
      | r = \dfrac {\map \sin {2 \arccos x} } {\map \sin {\arccos x} }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map {U_1} x
      | r = \dfrac {2 \map \sin {\arccos x} \map \cos {\arccos x} } {\map \sin {\arccos x} }
      | c =  Double Angle Formula for Sine
}}
{{eqn | ll = \leadsto
      | l = \map {U_1} x
      | r = 2 x
      | c = 
}}
{{end-eqn}}
For $n > 1$, let $x = \map \cos \theta$.
Then:
{{begin-eqn}}
{{eqn | l = \map {U_n} x
      | r = \dfrac {\map \sin {\paren {n + 1} \arccos x} } {\map \sin {\arccos x} }
      | c = {{Defof|Chebyshev Polynomial of the Second Kind}}
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} {\map \cos \theta}
      | r = \dfrac {\map \sin {\paren {n + 1} \arccos {\map \cos \theta} } } {\map \sin {\arccos {\map \cos \theta} } }
      | c = $x \to \map \cos \theta$
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} {\map \cos \theta}
      | r = \dfrac {\map \sin {\paren {n + 1} \theta} } {\map \sin \theta}
      | c = 
}}
{{end-eqn}}
From {{Formulation|Sine of Integer Multiple of Argument|4}}, we have:
:$\ds \map \sin {n \theta} = \paren {2 \cos \theta} \map \sin {\paren {n - 1} \theta}  - \map \sin {\paren {n - 2} \theta}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {U_n} {\map \cos \theta}
      | r = \dfrac {\map \sin {\paren {n + 1} \theta} } {\map \sin \theta}
      | c = 
}}
{{eqn | r = \frac {\paren {2 \cos \theta} \map \sin {\paren n \theta}  - \map \sin {\paren {n - 1} \theta} } {\map \sin \theta}
      | c = {{Formulation|Sine of Integer Multiple of Argument|4}}
}}
{{eqn | r = \paren {2 \cos \theta} \map {U_{n - 1} } {\map \cos \theta}  - \map {U_{n - 2} } {\map \cos \theta} 
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} x
      | r = 2 x \map {U_{n - 1} } x - \map {U_{n - 2} } x
      | c = $\map \cos \theta \to x$
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Chebyshev Polynomial of the Second Kind

The $n$th '''Chebyshev polynomial of the second kind''' can be expressed as a power series expansion in the form:
{{begin-eqn}}
{{eqn | l = \map {U_n} x
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k
}}
{{eqn | r = \dbinom {n + 1} 1 x^n - \dbinom {n + 1} 3 x^{n - 2} \paren {1 - x^2} + \dbinom {n + 1} 5 x^{n - 4} \paren {1 - x^2}^2 - \dbinom {n + 1} 7 x^{n - 6} \paren {1 - x^2}^3 + \cdots
}}
{{end-eqn}}
where $n \in \N$.

Proof:
From the Definition of Chebyshev Polynomial of the Second Kind, we have:
:$\map {U_n} {\cos \theta} \sin \theta = \map \sin {\paren {n + 1} \theta}$
From De Moivre's Formula, we have:
:$\map \cos {\paren {n + 1} \theta} + i \map \sin {\paren {n + 1} \theta} = \paren {\cos \theta + i \sin \theta}^{n + 1}$
As $n \in \Z_{>0}$, we use the Binomial Theorem on the {{RHS}}, resulting in:
:$\ds \map \cos {\paren {n + 1} \theta} + i \map \sin {\paren {n + 1} \theta} = \sum_{k \mathop \ge 0} \binom {n + 1} k \paren {\cos^{n + 1 - k} \theta} \paren {i \sin \theta}^k$
When $k$ is odd, the expression being summed is imaginary.
Equating the imaginary parts of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ odd, gives:
{{begin-eqn}}
{{eqn | l = \map \sin {\paren {n + 1} \theta}
      | r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n + 1 - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}
      | c = 
}}
{{eqn | r = \sin \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n - \paren {2 k} } \theta} \paren {\sin^2 \theta}^k
      | c = 
}}
{{eqn | r = \sin \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n - \paren {2 k} } \theta} \paren {1 - \cos^2 \theta}^k
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sin \theta \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k
      | c = $\cos \theta \to x$
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} x
      | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Particular Values of Chebyshev Polynomials of the Second Kind/-x

:$\map {U_n} {-x} = \paren {-1}^n \map {U_n} x$

Proof:
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {U_n} {-x} = \paren {-1}^n \map {U_n} x$
First we note that:
{{begin-eqn}}
{{eqn | l = \map {U_n} {-x}
      | r = \paren {-1}^n \map {U_n} x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \dfrac 1 {\paren {-1}^n} \map {U_n} {-x}
      | r = \map {U_n} x
      | c = dividing both sides by $\paren {-1}^n$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {-1}^n \map {U_n} {-x}
      | r = \map {U_n} x
      | c = as $\paren {-1}^n = \dfrac 1 {\paren {-1}^n}$
}}
{{end-eqn}}
That is:
:$(1): \quad \map {U_n} {-x} = \paren {-1}^n \map {U_n} x \iff \map {U_n} x = \paren {-1}^n \map {U_n} {-x}$
Theorem: Particular Values of Chebyshev Polynomials of the Second Kind/1

:$\map {U_n} 1 = n + 1$

Proof:
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {U_n} 1 = n + 1$
Theorem: Particular Values of Chebyshev Polynomials of the Second Kind/-1

:$\map {U_n} {-1} = \paren {-1}^n \paren {n + 1}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {U_n} {-x}
      | r = \paren {-1}^n \map {U_n} x
      | c = Particular Values of Chebyshev Polynomials of the Second Kind: $-x$
}}
{{eqn | ll= \leadsto
      | l = \map {U_n} {-1}
      | r = \paren {-1}^n \map {U_n} 1
      | c = setting $x = 1$
}}
{{eqn | r = \paren {-1}^n \paren {n + 1}
      | c = Particular Values of Chebyshev Polynomials of the Second Kind: $1$
}}
{{end-eqn}}
{{qed}}
Theorem: Particular Values of Chebyshev Polynomials of the Second Kind/0

:$\map {U_n} 0 = \begin {cases} \paren {-1}^{n / 2} & : \text {$n$ even} \\ 0 & : \text {$n$ odd} \end {cases}$

Proof:
=== Even Order ===
{{:Particular Values of Chebyshev Polynomials of the Second Kind/0/Even Order}}
Theorem: Particular Values of Chebyshev Polynomials of the Second Kind/0/Even Order

Let $\map {U_n} x$ denote the Chebyshev polynomial of the second kind of order $n$.
Let $n = 2 m$ for some $m \in \N$.
Then:
:$\map {U_n} 0 = \paren {-1}^m$

Proof:
Let $n = 2 m$ for some $m \in \N$.
The proof proceeds by induction.
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\map {U_{2 m} } 0 = \paren {-1}^m$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \map {U_0} 0
      | r = \paren {-1}
      | c = Chebyshev Polynomial of the Second Kind: $U_0$
}}
{{eqn | r = \paren {-1}^0
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\map {U_{2 k} } 0 = \paren {-1}^k$
from which it is to be shown that:
:$\map {U_{2 \paren {k + 1} } } 0 = \paren {-1}^{k + 1}$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \map {U_{2 \paren {k + 1} } } 0
      | r = 2 \times 0 \times \map {U_{2 k + 1} } 0 - \map {U_{2 k} } 0
      | c = Recurrence Formula for Chebyshev Polynomials of the First Kind, setting $n = 2 \paren {k + 1} = 2 k + 2$
}}
{{eqn | r = \paren {-1} \times \map {U_{2 k} } 0
      | c = simplifying
}}
{{eqn | r = \paren {-1} \times \paren {-1}^k
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {-1}^{k + 1}
      | c = simplifyng
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m \in \Z_{\ge 0}: \map {U_{2 m} } 0 = \paren {-1}^m$
{{qed}}
Theorem: General Solution to Chebyshev's Differential Equation

Consider '''Chebyshev's differential equation''':
:$(1): \quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} + n^2 y = 0$
where $n \in \N$.
The general solution to $(1)$ is given by:
:$y = \begin {cases} A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} & : n = 1, 2, 3, \ldots \\ \\ A \arccos x + B \arcsin x + C & : n = 0 \end {cases}$
where:
:$\map {T_n} x$ denotes the Chebyshev polynomial of the first kind of order $n$
:$\map {U_n} x$ denotes the Chebyshev polynomial of the second kind of order $n$
:$\size x < 1$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = \cos \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d x
      | r = -\sin \theta \rd \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d \theta} {\d x}
      | r = -\frac 1 {\sin \theta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2 \theta} {\d x \d \theta}
      | r = \frac {\cos \theta} {\sin^2 \theta}
      | c = Derivative of Cosecant Function
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \paren {1 - \cos^2 \theta} \frac {\d } {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = substituting $x = \cos \theta$ and Chain Rule for Derivatives into Chebyshev's differential equation
}}
{{eqn | r = \paren {1 - \cos^2 \theta} \frac {\d \theta} {\d x} \map {\frac \d {\d \theta} } {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \paren {1 - \cos^2 \theta} \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x \d \theta} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = Product Rule for Derivatives
}}
{{eqn | r = \paren {\sin^2 \theta} \paren {-\frac 1 {\sin \theta} } \paren {\frac {\d^2 y} {\d \theta^2} \paren {-\frac 1 {\sin \theta} } + \frac {\d y} {\d \theta} \frac {\cos \theta} {\sin^2 \theta} } - \cos \theta \frac {\d y} {\d \theta} \paren {-\frac 1 {\sin \theta} } + n^2 y
      | c = Sum of Squares of Sine and Cosine and substituting $\ds \frac {\d \theta} {\d x} = -\frac 1 {\sin \theta}$ and $\ds \frac {\d^2 \theta} {\d x \d \theta} = \frac {\cos \theta} {\sin^2 \theta}$
}}
{{eqn | r = \frac {\d^2 y} {\d \theta^2} - \frac {\d y} {\d \theta} \frac {\cos \theta } {\sin \theta} + \frac {\d y} {\d \theta} \frac {\cos \theta } {\sin \theta} + n^2 y
      | c = simplification: cancelling $\sin^2 \theta$ from top and bottom
}}
{{eqn | ll = \leadsto
      | l = 0
      | r = \frac {\d^2 y} {\d \theta^2} + n^2 y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: General Solution to Chebyshev's Differential Equation

Consider '''Chebyshev's differential equation''':
:$(1): \quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} + n^2 y = 0$
where $n \in \N$.
The general solution to $(1)$ is given by:
:$y = \begin {cases} A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} & : n = 1, 2, 3, \ldots \\ \\ A \arccos x + B \arcsin x + C & : n = 0 \end {cases}$
where:
:$\map {T_n} x$ denotes the Chebyshev polynomial of the first kind of order $n$
:$\map {U_n} x$ denotes the Chebyshev polynomial of the second kind of order $n$
:$\size x < 1$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = \sin \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d x
      | r = \cos \theta \rd \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d \theta} {\d x}
      | r = \frac 1 {\cos \theta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2 \theta} {\d x \d \theta}
      | r = \frac {\sin \theta} {\cos^2 \theta}
      | c = Derivative of Secant Function
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \paren {1 - \sin^2 \theta} \frac {\d } {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = substituting $x = \sin \theta$ and Chain Rule for Derivatives into Chebyshev's differential equation
}}
{{eqn | r = \paren {1 - \sin^2 \theta} \frac {\d \theta} {\d x} \map {\frac \d {\d \theta} } {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \paren {1 - \sin^2 \theta} \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x \d \theta} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y
      | c = Product Rule for Derivatives
}}
{{eqn | r = \paren {\cos^2 \theta} \paren {\frac 1 {\cos \theta} } \paren {\frac {\d^2 y} {\d \theta^2} \paren {\frac 1 {\cos \theta} } + \frac {\d y} {\d \theta} \frac {\sin \theta} {\cos^2 \theta} } - \sin \theta \frac {\d y} {\d \theta} \paren {\frac 1 {\cos \theta} } + n^2 y
      | c = Sum of Squares of Sine and Cosine and substituting $\ds \frac {\d \theta} {\d x} = \frac 1 {\cos \theta}$ and $\ds \frac {\d^2 \theta} {\d x \d \theta} = \frac {\sin \theta} {\cos^2 \theta}$
}}
{{eqn | r = \frac {\d^2 y} {\d \theta^2} - \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta} + \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta} + n^2 y
      | c = simplification: cancelling $\cos^2 \theta$ from top and bottom
}}
{{eqn | ll = \leadsto
      | l = 0
      | r = \frac {\d^2 y} {\d \theta^2} + n^2 y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Relationship between Chebyshev Polynomial of the First and Second Kind/Formulation 2

:$\paren {1 - x^2} \map {U_{n - 1} } x = x \map {T_n} x - \map {T_{n + 1} } x$

Proof:
Let $x = \cos \theta$.
Then:
{{begin-eqn}}
{{eqn | l = \map {T_n} x
      | r = \map \cos {n \arccos x}
      | c = {{Defof|Chebyshev Polynomial of the First Kind}}
}}
{{eqn | ll = \leadsto
      | l = \map {T_n} {\cos \theta}
      | r = \map \cos {n \arccos \cos \theta}
      | c = $x \to \cos \theta$
}}
{{eqn | ll = \leadsto
      | l = \map {T_n} {\cos \theta}
      | r = \map \cos {n \theta}
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {U_n} x
      | r = \dfrac {\map \sin {\paren {n + 1} \arccos x} } {\map \sin {\arccos x} }
      | c = {{Defof|Chebyshev Polynomial of the Second Kind}}
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} {\map \cos \theta}
      | r = \dfrac {\map \sin {\paren {n + 1} \arccos {\map \cos \theta} } } {\map \sin {\arccos {\map \cos \theta} } }
      | c = $x \to \map \cos \theta$
}}
{{eqn | ll = \leadsto
      | l = \map {U_n} {\map \cos \theta}
      | r = \dfrac {\map \sin {\paren {n + 1} \theta} } {\map \sin \theta}
      | c = 
}}
{{end-eqn}}
From Cosine of Sum, we have:
:$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta + \theta}
      | r = \cos n \theta \cos \theta - \sin n \theta \sin \theta
      | c = Cosine of Sum
}}
{{eqn | ll = \leadsto
      | l = \map \cos {\paren {n + 1} \theta }
      | r = \cos n \theta \cos \theta - \sin n \theta \sin \theta
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map {T_{n + 1} } {\cos \theta}
      | r = \map {T_n} {\cos \theta} \cos \theta - \map {U_{n - 1} } {\map \cos \theta} \sin^2 \theta
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map {T_{n + 1} } {\cos \theta}
      | r = \map {T_n} {\cos \theta} \cos \theta - \map {U_{n - 1} } {\map \cos \theta} \paren {1 - \cos^2 \theta }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll = \leadsto
      | l = \map {T_{n + 1} } x
      | r = x \map {T_n} x - \paren {1 - x^2} \map {U_{n - 1} } x
      | c = $\cos \theta \to x$
}}
{{eqn | ll = \leadsto
      | l = \paren {1 - x^2} \map {U_{n - 1} } x
      | r = x \map {T_n} x - \map {T_{n + 1} } x
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Legendre Polynomial in terms of Gaussian Hypergeometric Function

:$\map {P_n} x = \map F {n + 1, -n; 1; \dfrac {1 - x} 2}$

Proof:
From Solution to Hypergeometric Differential Equation, we have:
{{:Solution to Hypergeometric Differential Equation}}
Inputting $\map F {n + 1, -n; 1; \dfrac {1 - x} 2}$ into the hypergeometric differential equation, we obtain:
{{begin-eqn}}
{{eqn | l = 0
      | r = x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {c - \paren {a + b + 1} x} \dfrac {\d y} {\d x} - a b y
      | c = Solution to Hypergeometric Differential Equation
}}
{{eqn | r = x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {1 - \paren {\paren {n + 1} + \paren {-n} + 1} x} \dfrac {\d y} {\d x} - \paren {n + 1} \paren {-n} y
      | c = Setting $a := n + 1$, $b := -n$ and $c := 1$
}}
{{eqn | r = x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {1 - 2 x} \dfrac {\d y} {\d x} + n \paren {n + 1} y
      | c = 
}}
{{eqn | r = \frac {\paren {1 - z} } 2 \paren {1 - \frac {\paren {1 - z} } 2 } \dfrac {\d^2 y} {\paren {-\dfrac {\d z} 2 }^2} + \paren {1 - \dfrac {2 \paren {1 - z} } 2 } \dfrac {\d y} {\paren {-\dfrac {\d z} 2 } } + n \paren {n + 1} y
      | c = Setting $x := \dfrac {\paren {1 - z} } 2$ and $\d x = -\dfrac {\d z} 2$
}}
{{eqn | r = \frac {\paren {1 - z} } 2 \frac {\paren {1 + z} } 2 \dfrac {\d^2 y} {\d z^2} \paren 4 + z  \dfrac {\d y} {\d z} \paren {-2} + n \paren {n + 1} y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = \paren {1 - z^2} \dfrac {\d^2 y} {\d z^2} - 2 z \dfrac {\d y} {\d z} + n \paren {n + 1} y
      | c = Legendre's differential equation
}}
{{end-eqn}}
We now see that $\map F {n + 1, -n; 1; \dfrac {1 - x} 2}$ simultaneously solves both the hypergeometric differential equation '''AND''' Legendre's differential equation!
We also have that the general solution to Legendre's differential equation is given by:
show general solution here
{{improve|need a page for general solution of Legendre differential equation}}
Therefore,  we obtain:
:$\map {P_n} x = \map F {n + 1, -n; 1; \dfrac {1 - x} 2}$
{{qed}}
Theorem: Laplace Transform of Exponential times Hyperbolic Sine

:$\map {\laptrans {e^{b t} \sinh a t} } s = \dfrac a {\paren {s - b}^2 - a^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {e^{b t} \sinh a t} } s
      | r = \map {\laptrans {\sinh a t} } {s - b}
      | c = First Translation Property of Laplace Transforms
}}
{{eqn | r = \frac a {\paren {s - b}^2 - a^2}
      | c = Laplace Transform of Hyperbolic Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Exponential times Hyperbolic Cosine

:$\map {\laptrans {e^{b t} \cosh a t} } s = \dfrac {s - b} {\paren {s - b}^2 - a^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {e^{b t} \cosh a t} } s
      | r = \map {\laptrans {\cosh a t} } {s - b}
      | c = First Translation Property of Laplace Transforms
}}
{{eqn | r = \frac {s - b} {\paren {s - b}^2 - a^2}
      | c = Laplace Transform of Hyperbolic Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Difference between Exponentials

:$\map {\laptrans {\dfrac {e^{b t} - e^{a t} } {b - a} } } s = \dfrac 1 {\paren {s - a} \paren {s - b} }$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {\dfrac {e^{b t} - e^{a t} } {b - a} } } s
      | r = \dfrac 1 {b - a} \paren {\map {\laptrans {e^{b t} } } s - \map {\laptrans {e^{a t} } } s}
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac 1 {s - b} - \dfrac 1 {s - a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac {\paren {s - a} - \paren {s - b} } {\paren {s - a} \paren {s - b} } }
      | c = common denominator
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac {-a + b} {\paren {s - a} \paren {s - b} } }
      | c = simplification
}}
{{eqn | r = \dfrac 1 {\paren {s - a} \paren {s - b} }
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of b e^bt - a e^at over b - a

:$\map {\laptrans {\dfrac {b e^{b t} - a e^{a t} } {b - a} } } s = \dfrac s {\paren {s - a} \paren {s - b} }$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {\dfrac {b e^{b t} - a e^{a t} } {b - a} } } s
      | r = \dfrac 1 {b - a} \paren {b \map {\laptrans {e^{b t} } } s - a \map {\laptrans {e^{a t} } } s}
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac b {s - b} - \dfrac a {s - a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac {b \paren {s - a} - a \paren {s - b} } {\paren {s - a} \paren {s - b} } }
      | c = common denominator
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac {b s - b a - a s + a b} {\paren {s - a} \paren {s - b} } }
      | c = multiplying out
}}
{{eqn | r = \dfrac 1 {b - a} \paren {\dfrac {\paren {b - a} s} {\paren {s - a} \paren {s - b} } }
      | c = simplifying
}}
{{eqn | r = \dfrac s {\paren {s - a} \paren {s - b} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}