avewright/proofwiki-math huggingface dataset:
Theorem: Power Series Expansion for Logarithm of x/Formulation 1

{{begin-eqn}}
{{eqn | l = \ln x 
      | r = 2 \paren {\sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \paren {\frac {x - 1} {x + 1} }^{2 n + 1} }
}}
{{eqn | r = 2 \paren {\frac {x - 1} {x + 1} + \frac 1 3 \paren {\frac {x - 1} {x + 1} }^3 + \frac 1 5 \paren {\frac {x - 1} {x + 1} }^5 + \cdots}
}}
{{end-eqn}}
valid for all $x \in \R_{>0}$.

Proof:
From Power Series Expansion for $\dfrac 1 2 \map \ln {\dfrac {1 + x} {1 - x} }$:
:$(1): \quad \ds \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}$
for $-1 < x < 1$.
Let $z = \dfrac {1 + x} {1 - x}$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \dfrac {1 + x} {1 - x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z \paren {1 - x}
      | r = 1 + x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z - z x
      | r = 1 + x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z - 1 
      | r = z x + x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \paren {z + 1}
      | r = z - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {z - 1} {z + 1}
      | c = 
}}
{{end-eqn}}
Then we have:
:$\ds \lim_{x \mathop \to 1^-} \dfrac {1 + x} {1 - x} \to +\infty$
and:
:$\ds \lim_{x \mathop \to -1^+} \dfrac {1 + x} {1 - x} \to 0$
Thus when $x \in \openint {-1} 1$ we have that $z \in \openint 0 \to$.
Thus, substituting $z$ for $\dfrac {1 + x} {1 - x}$ in $(1)$ gives the result.
{{qed}}
Theorem: Power Series Expansion for Logarithm of x/Formulation 2

{{begin-eqn}}
{{eqn | l = \ln x 
      | r = \sum_{n \mathop = 1}^\infty \dfrac 1 n \paren {\frac {x - 1} x}^n
}}
{{eqn | r = \frac {x - 1} x + \frac 1 2 \paren {\frac {x - 1} x}^2 + \frac 1 3 \paren {\frac {x - 1} x}^3 + \cdots
}}
{{end-eqn}}
valid for all $x \in \R$ such that $x \ge \dfrac 1 2$.

Proof:
From Power Series Expansion for $\map \ln {1 - x}$:
{{begin-eqn}}
{{eqn | l = \map \ln {1 - x}
      | r = -\sum_{n \mathop = 1}^\infty \frac {x^n} n
      | c = for $-1 \le x < 1$.
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \map \ln {\frac 1 {1 - x} }
      | r = \sum_{n \mathop = 1}^\infty \frac {x^n} n
      | c = Logarithm of Reciprocal
}}
{{end-eqn}}
Let $z = \dfrac 1 {1 - x}$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \dfrac 1 {1 - x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z \paren {1 - x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z - z x
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z x
      | r = z - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {z - 1} z
      | c = 
}}
{{end-eqn}}
Then we have:
:$\ds \lim_{x \mathop \to 1^-} \dfrac 1 {1 - x} \to +\infty$
and:
:$\ds \lim_{x \mathop \to -1^+} \dfrac 1 {1 - x} \to \frac 1 2$
Thus when $x \in \hointr {-1} 1$ we have that $z \in \hointr {\dfrac 1 2} \to$.
Thus, substituting $z$ for $\dfrac 1 {1 - x}$ in $(1)$ gives the result.
{{qed}}
Theorem: Power Series Expansion for Secant Function

The (real) secant function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \sec x
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {E_{2 n} x^{2 n} } {\paren {2 n}!}
}}
{{eqn | r = 1 + \frac {x^2} 2 + \frac {5 x^4} {24} + \frac {61 x^6} {720} + \dfrac {1385 x^8} {40320} + \cdots
}}
{{end-eqn}}
where $E_{2 n}$ denotes the Euler numbers.
This converges for $\size x < \dfrac \pi 2$.

Proof:
{{begin-eqn}}
{{eqn | l = \sec x
      | r = \map \sech {i x}
      | c = Secant in terms of Hyperbolic Secant
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {E_n \paren {i x}^n} {n!}
      | c = {{Defof|Euler Numbers}}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {E_{2 n} \paren {i x}^{2 n} } {\paren {2 n}!}
      | c = Odd terms vanish
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} x^{2 n} } {\paren {2 n}!}
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Real Arccotangent Function

The arccotangent function has a Taylor series expansion:
:<nowiki>$\arccot x = \begin {cases} \ds \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1 \\ \\
\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \ge 1 \\ \\
\ds \pi + \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \le -1
\end {cases}$</nowiki>
That is:
:<nowiki>$\arccot x = \begin {cases} \dfrac \pi 2 - \paren {x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \cdots} & : -1 \le x \le 1 \\ \\
\dfrac 1 x - \dfrac 1 {3 x^3} + \dfrac 1 {5 x^5} - \cdots & : x \ge 1 \\ \\
\pi + \dfrac 1 x - \dfrac 1 {3 x^3} + \dfrac 1 {5 x^5} - \cdots & : x \le -1
\end {cases}$</nowiki>

Proof:
From Sum of Arctangent and Arccotangent:
:$\arccot x = \dfrac \pi 2 - \arctan x$
The result follows from Power Series Expansion for Real Arctangent Function.
{{qed}}
Theorem: Power Series Expansion for Real Arcsecant Function

The arcsecant function has a Taylor Series expansion:
{{begin-eqn}}
{{eqn | l = \arcsec x
      | r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} x^{2 n + 1} }
      | c = 
}}
{{eqn | r = \frac \pi 2 - \paren {\frac 1 x + \frac 1 {2 \times 3 x^3} + \frac {1 \times 3} {2 \times 4 \times 5 x^5} + \frac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 7 x^7} + \cdots}
      | c = 
}}
{{end-eqn}}
which converges for $\size x \ge 1$.

Proof:
From Arccosine of Reciprocal equals Arcsecant:
:$\arcsec x = \arccos \dfrac 1 x$
From Power Series Expansion for Real Arccosine Function:
:$\ds \arccos x = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}$
which is converges for $\size x \le 1$.
The result follows by subtituting $\dfrac 1 x$ for $x$.
This converges for $\size {\dfrac 1 x} \le 1$, that is, for $size x \ge 1$
{{qed}}
Theorem: Power Series Expansion for Real Arccosecant Function

The arccosecant function has a Taylor Series expansion:
{{begin-eqn}}
{{eqn | l = \arccsc x
      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} x^{2 n + 1} }
      | c = 
}}
{{eqn | r = \frac 1 x + \frac 1 {2 \times 3 x^3} + \frac {1 \times 3} {2 \times 4 \times 5 x^5} + \frac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 7 x^7} + \cdots
      | c = 
}}
{{end-eqn}}
which converges for $\size x \ge 1$.

Proof:
From Arcsine of Reciprocal equals Arccosecant:
:$\arccsc x = \arcsin \dfrac 1 x$
From Power Series Expansion for Real Arcsine Function:
:$\ds \arcsin x = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}$
which is converges for $\size x \le 1$.
The result follows by subtituting $\dfrac 1 x$ for $x$.
This converges for $\size {\dfrac 1 x} \le 1$, that is, for $\size x \ge 1$.
{{qed}}
Theorem: Power Series Expansion for Hyperbolic Sine Function

The hyperbolic sine function has the power series expansion:
{{begin-eqn}}
{{eqn | l = \sinh x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
      | c = 
}}
{{eqn | r = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} + \cdots
      | c = 
}}
{{end-eqn}}
valid for all $x \in \R$.

Proof:
From Derivative of Hyperbolic Sine:
:$\dfrac \d {\d x} \sinh x = \cosh x$
From Derivative of Hyperbolic Cosine:
:$\dfrac \d {\d x} \cosh x = \sinh x$
Hence:
{{begin-eqn}}
{{eqn | l = \dfrac {\d^2} {\d x^2} \sinh x
      | r = \sinh x
      | c = 
}}
{{end-eqn}}
and so for all $m \in \N$:
{{begin-eqn}}
{{eqn | ll= m = 2 k:
      | l = \dfrac {\d^m} {\d x^m} \sinh x
      | r = \sinh x
      | c = 
}}
{{eqn | ll= m = 2 k + 1:
      | l = \dfrac {\d^m} {\d x^m} \sinh x
      | r = \cosh x
      | c = 
}}
{{end-eqn}}
where $k \in \Z$.
This leads to the Maclaurin series expansion:
{{begin-eqn}}
{{eqn | l = \sinh x
      | r = \sum_{r \mathop = 0}^\infty \paren {\frac {x^{2 k} } {\paren {2 k}!} \map \sinh 0 + \frac {x^{2 k + 1} } {\paren {2 k + 1}!} \map \cosh 0}
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 0}^\infty \frac {x^{2 k + 1} } {\paren {2 k + 1}!}
      | c = $\map \sinh 0 = 0$, $\map \cosh 0 = 1$
}}
{{end-eqn}}
From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$.
{{qed}}
Theorem: Power Series Expansion for Hyperbolic Cosine Function

The hyperbolic cosine function has the power series expansion:
{{begin-eqn}}
{{eqn | l = \cosh x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \frac {x^6} {6!} + \cdots
      | c = 
}}
{{end-eqn}}
valid for all $x \in \R$.

Proof:
From Derivative of Hyperbolic Cosine:
:$\dfrac \d {\d x} \cosh x = \sinh x$
From Derivative of Hyperbolic Sine:
:$\dfrac \d {\d x} \sinh x = \cosh x$
Hence:
:$\dfrac {\d^2} {\d x^2} \cosh x = \cosh x$
and so for all $m \in \N$:
{{begin-eqn}}
{{eqn | ll= m = 2 k:
      | l = \dfrac {\d^m} {\d x^m} \cosh x
      | r = \cosh x
      | c = 
}}
{{eqn | ll= m = 2 k + 1:
      | l = \dfrac {\d^m} {\d x^m} \cosh x
      | r = \sinh x
      | c = 
}}
{{end-eqn}}
where $k \in \Z$.
This leads to the Maclaurin series expansion:
{{begin-eqn}}
{{eqn | l = \cosh x
      | r = \sum_{r \mathop = 0}^\infty \paren {\frac {x^{2 k} } {\paren {2 k}!} \map \cosh 0 + \frac {x^{2 k + 1} } {\paren {2 k + 1}!} \map \sinh 0}
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 0}^\infty \frac {x^{2 k} } {\paren {2 k}!}
      | c = $\map \sinh 0 = 0$, $\map \cosh 0 = 1$
}}
{{end-eqn}}
From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$.
{{qed}}
Theorem: Power Series Expansion for Hyperbolic Cotangent Function

The hyperbolic cotangent function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \coth x
      | r = \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = \frac 1 x + \frac x 3 - \frac {x^3} {45} + \frac {2 x^5} {945} - \frac {x^7} {4725} + \cdots
      | c = 
}}
{{end-eqn}}
where $B_{2 n}$ denotes the Bernoulli numbers.
This converges for $0 < \size x < \pi$.

Proof:
{{begin-eqn}}
{{eqn | l = \coth x
      | r = \frac {e^x + e^{-x} } {e^x - e^{-x} }
      | c = {{Defof|Hyperbolic Cotangent|index = 1}}
}}
{{eqn | r = \frac {e^{2 x} + 1} {e^{2 x} - 1}
      | c =
}}
{{eqn | r = 1 + \frac 2 {e^{2 x} - 1}
      | c =
}}
{{eqn | r = 1 + \frac 1 x \frac {2 x} {e^{2 x} - 1}
      | c =
}}
{{eqn | r = 1 + \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_n \paren {2 x}^n} {n!}
      | c = {{Defof|Bernoulli Numbers}}
}}
{{eqn | r = \frac 1 x + \frac 1 x \sum_{n \mathop = 2}^\infty \frac {B_n \paren {2 x}^n} {n!}
      | c = as $B_0 = 1$ and $B_1 = -\dfrac 1 2$
}}
{{eqn | r = \frac 1 x + \frac 1 x \sum_{n \mathop = 1}^\infty \frac {B_{2 n} \paren {2 x}^{2 n} } {\paren {2 n}!}
      | c = Odd Bernoulli Numbers Vanish
}}
{{eqn | r = \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_{2 n} \paren {2 x}^{2 n} } {\paren {2 n}!}
      | c =
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c =
}}
{{end-eqn}}
By Combination Theorem for Limits of Real Functions we can deduce the following:
{{begin-eqn}}
{{eqn | o = 
      | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {2^{2 n + 2} B_{2 n + 2} \, x^{2 n + 1} } {\paren {2 n + 2}!} } {\frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} } }
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi (n + 1)} \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 2 x^2
      | c = Asymptotic Formula for Bernoulli Numbers
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 2 {\pi^2 e^2} x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac {x^2} {\pi^2 e^2}
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac {x^2} {\pi^2 e^2}
      | c =
}}
{{eqn | r = \frac {e^2 x^2} {\pi^2 e^2}
      | c = {{Defof|Euler's Number as Limit of Sequence}}
}}
{{eqn | r = \frac {x^2} {\pi^2} 
      | c =
}}
{{end-eqn}}
This is less than $1$ {{iff}}:
:$\size x < \pi$
Hence by the Ratio Test, the series converges for $\size x < \pi$.
{{qed}}
Theorem: Power Series Expansion for Hyperbolic Secant Function

The hyperbolic secant function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \sech x
      | r = \sum_{n \mathop = 0}^\infty \frac {E_{2 n} x^{2 n} } {\paren {2 n}!} 
      | c = 
}}
{{eqn | r = 1 - \frac {x^2} 2 + \frac {5 x^4} {24} - \frac {61 x^6} {720} + \cdots
      | c = 
}}
{{end-eqn}}
where $E_{2 n}$ denotes the Euler numbers.
This converges for $\size x < \dfrac \pi 2$.

Proof:
By definition of the Euler numbers:
:$\ds \sech x = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}$
From Odd Euler Numbers Vanish:
:$E_{2 k + 1} = 0$
for $k \in \Z$.
Hence the result.
{{qed}}
Theorem: Power Series Expansion for Hyperbolic Cosecant Function

The hyperbolic cosecant function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \csch x
      | r = \sum_{n \mathop = 0}^\infty \dfrac {2 \paren {1 - 2^{2 n - 1} } B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = \frac 1 x - \frac x 6 + \frac {7 x^3} {360} - \frac {31 x^5} {15 \, 120} + \cdots
      | c = 
}}
{{end-eqn}}
where $B_n$ denotes the Bernoulli numbers.
This converges for $0 < \size x < \pi$.

Proof:
{{begin-eqn}}
{{eqn | l = \sinh x
      | r = 2 \sinh \dfrac x 2 \cosh \dfrac x 2
      | c = Double Angle Formula for Hyperbolic Sine
}}
{{eqn | ll= \leadstoandfrom
      | l = \dfrac 1 {\sinh x}
      | r = \dfrac 1 {2 \sinh \dfrac x 2 \cosh \dfrac x 2}
      | c = taking the reciprocal of both sides
}}
{{eqn | ll= \leadstoandfrom
      | l = \csch x
      | r = \dfrac 1 2 \csch \dfrac x 2 \sech \dfrac x 2
      | c = Cosecant is Reciprocal of Sine, Secant is Reciprocal of Cosine
}}
{{eqn | r = \dfrac 1 2 \csch \dfrac x 2 \dfrac 2 {e^{\frac x 2} + e^{\frac x 2} }
      | c = {{Defof|Hyperbolic Secant|index = 1}}
}}
{{eqn | r = \csch \dfrac x 2 \dfrac {e^{-\frac x 2} } {1 + e^{-x} }
}}
{{eqn | ll= \leadstoandfrom
      | l = \csch x \paren {1 + e^{-x} }
      | r = \csch \dfrac x 2 e^{-\frac x 2}
      | c = multiplying both sides by $1 + e^{-x}$
}}
{{eqn | ll= \leadstoandfrom
      | l = \csch x
      | r = \csch \dfrac x 2 e^{-\frac x 2} - \csch x e^{-x}
      | c = subtracting $\csch x e^{-x}$ from both sides
}}
{{eqn | r = \dfrac 2 {e^{\frac x 2} - e^{-\frac x 2} } e^{-\frac x 2} - \dfrac 2 {e^x - e^{-x} } e^{-x}
      | c = {{Defof|Hyperbolic Cosecant|index = 1}}
}}
{{eqn | r = 2 \dfrac 1 {e^x - 1} - \dfrac 2 {e^{2 x} - 1}
}}
{{eqn | r = \dfrac 1 x \paren {2 \dfrac x {e^x - 1} - \dfrac {2 x} {e^{2 x} - 1} }
}}
{{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \paren {2 \dfrac {x^n B_n} {n!} - \dfrac {\paren {2 x}^n B_n} {n!} }
      | c = {{Defof|Bernoulli Numbers}}
}}
{{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \dfrac {B_n \paren {2 x^n - \paren {2 x}^n} } {n!}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {B_n x^{n - 1} 2 \paren {1 - 2^{n - 1} } } {n!}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {2 \paren {1 - 2^{2 n - 1} } B_{2 n} x^{2 n - 1} } {\paren {2 n}!}
      | c = Odd Bernoulli Numbers Vanish and the term $n = 1$ vanishes
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Hyperbolic Tangent and Cotangent

:$\tanh x + \coth x = 2 \coth 2 x$

Proof:
{{begin-eqn}}
{{eqn | l = \tanh x + \coth x
      | r = \frac {\sinh x} {\cosh x} + \frac {\cosh x} {\sinh x}
      | c = {{Defof|Hyperbolic Tangent}} and Hyperbolic Cotangent
}}
{{eqn | r = \frac {\cosh^2 x + \sinh^2 x} {\sinh x \cosh x}
      | c = 
}}
{{eqn | r = 2 \frac {\cosh^2 x + \sinh^2 x} {2 \sinh x \cosh x}
      | c = 
}}
{{eqn | r = 2 \frac {\cosh 2 x} {\sinh 2 x}
      | c = Double Angle Formula for Hyperbolic Sine and Double Angle Formula for Hyperbolic Cosine
}}
{{eqn | r = 2 \coth 2 x
      | c = {{Defof|Hyperbolic Cotangent}}
}}
{{end-eqn}}
{{qed}}
Category:Hyperbolic Tangent Function
Category:Hyperbolic Cotangent Function
ftf1rt5bh2yse6tendewnhhm3d88cb1
Theorem: Power Series Expansion for Real Area Hyperbolic Cosine

The (real) area hyperbolic cosine function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \arcosh x
      | r = \map \ln {2 x} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } }
      | c = 
}}
{{eqn | r = \map \ln {2 x} - \paren {\dfrac 1 {2 \times 2 x^2} + \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} + \cdots}
      | c = 
}}
{{end-eqn}}
for $x \ge 1$.

Proof:
=== Lemma 1 ===
{{:Power Series Expansion for Real Area Hyperbolic Cosine/Lemma 1}}{{qed|lemma}}
We have that the (real) area hyperbolic cosine is defined for $x \ge 1$.
Let $z = \dfrac 1 x$.
Then we have:
:$0 < \dfrac 1 z \le 1$
Now we consider:
{{begin-eqn}}
{{eqn | l = \map \arcosh {\dfrac 1 z} + \map \ln {2 z}
      | r = \map \ln {2 z} + \map \ln {\dfrac 1 z + \sqrt {\dfrac 1 {z^2} - 1} }
      | c = {{Defof|Real Area Hyperbolic Cosine}}
}}
{{eqn | r = \map \ln {2 z \paren {\dfrac 1 z + \dfrac {\sqrt {1 - z^2} } z} }
      | c = Sum of Logarithms and simplification
}}
{{eqn | r = \map \ln {2 \paren {1 + \sqrt {1 - z^2} } }
      | c = simplifying
}}
{{eqn | r = \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }
      | c = Sum of Logarithms
}}
{{eqn | ll= \leadsto
      | l = \map \arcosh {\dfrac 1 z}
      | r = -\map \ln {2 z} + \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }
      | c = rearranging
}}
{{eqn | r = \map \ln {\dfrac 1 {2 z} } + \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \map \ln {\dfrac 1 z} + \map \ln {1 + \sqrt {1 - z^2} }
      | c = Sum of Logarithms
}}
{{eqn | r = \map \ln {\dfrac 1 z} + \ln 2 - \paren {\dfrac 1 2 \cdot \dfrac {z^2} 2 + \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {z^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {z^6} 6 + \cdots}
      | c = Lemma $2$
}}
{{eqn | r = \map \ln {\dfrac 2 z} - \paren {\dfrac 1 2 \cdot \dfrac {z^2} 2 + \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {z^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {z^6} 6 + \cdots}
      | c = Sum of Logarithms
}}
{{eqn | ll= \leadsto
      | l = \arcosh x
      | r = \map \ln {2 x} - \paren {\dfrac 1 {2 \times 2 x^2} + \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} + \cdots}
      | c = substituting $x \gets \dfrac 1 z$
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Real Area Hyperbolic Tangent

The (real) area hyperbolic tangent function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \artanh x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}
      | c = 
}}
{{eqn | r = x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots
      | c = 
}}
{{end-eqn}}
for $\size x < 1$.

Proof:
From Sum of Infinite Geometric Sequence:
:$(1): \quad \ds \frac 1 {1 - x^2} = \sum_{n \mathop = 0}^\infty \paren {x^2}^n$
for $-1 < x < 1$.
From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:
{{begin-eqn}}
{{eqn | l = \int_0^x \frac 1 {1 - t^2} \rd t
      | r = \sum_{n \mathop = 0}^\infty \int_0^x \paren {t^2}^n \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \artanh x 
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}
      | c = Primitive of Reciprocal of $\dfrac 1 {1 - t^2}$, Integral of Power
}}
{{end-eqn}}
For $-1 < x < 1$, the sequence $\sequence {\dfrac {x^{2 n + 1} } {2 n + 1} }$ is decreasing and converges to zero.
Hence the result.
{{qed}}
Theorem: Power Series Expansion for Real Area Hyperbolic Cotangent

The (real) area hyperbolic cotangent function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \arcoth x
      | r = \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1} x^{2 n + 1} }
      | c = 
}}
{{eqn | r = \frac 1 x + \frac 1 {3 x^3} + \frac 1 {5 x^5} + \frac 1 {7 x^7} + \cdots
      | c = 
}}
{{end-eqn}}
for $\size x > 1$.

Proof:
From Power Series Expansion for Real Area Hyperbolic Tangent:
{{begin-eqn}}
{{eqn | l = \artanh x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}
      | c = 
}}
{{eqn | r = x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots
      | c = 
}}
{{end-eqn}}
for $\size x < 1$.
From Real Area Hyperbolic Tangent of Reciprocal equals Real Area Hyperbolic Cotangent:
:$\map \artanh {\dfrac 1 x} = \arcoth x$
So:
{{begin-eqn}}
{{eqn | l = \arcoth x
      | r = \sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \times \frac 1 {x^{2 n + 1} }
      | c = 
}}
{{eqn | r = \frac 1 x + \frac 1 3 \times \frac 1 {x^3} + \frac 1 5 \times \frac 1 {x^5} + \frac 1 7 \times \frac 1 {x^7} + \cdots
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Power Series Expansion for Exponential of x by Sine of x

{{begin-eqn}}
{{eqn | l = e^x \sin x
      | r = \sum_{n \mathop = 1}^\infty \frac {2^{n / 2} \, \map \sin {n \pi / 4} x^n} {n!}
      | c = 
}}
{{eqn | r = x + x^2 + \frac {x^3} 3 - \frac {x^5} {30} - \frac {x^6} {90} + \cdots
      | c = 
}}
{{end-eqn}}
for all $x \in \R$.

Proof:
Let $\map f x = e^x \sin x$.
By definition of Maclaurin series:
:$(1): \quad \map f x \sim \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0$
where $\map {f^{\paren n} } 0$ denotes the $n$th derivative of $f$ {{WRT|Differentiation}} $x$ evaluated at $x = 0$.
It remains to be shown that:
:$\map {f^{\paren n} } 0 = 2^{n / 2} \map \sin {\dfrac {n \pi} 4}$
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {f^{\paren n} } x = 2^{n / 2} e^x \, \map \sin {x + \dfrac {n \pi} 4}$
Theorem: Sine of x plus Cosine of x/Sine Form

:$\sin x + \cos x = \sqrt 2 \sin \left({x + \dfrac \pi 4}\right)$

Proof:
{{begin-eqn}}
{{eqn | l = \sin x + \cos x
      | r = \sqrt 2 \cos \left({x - \dfrac \pi 4}\right)
      | c = Sine of x plus Cosine of x: Cosine Form
}}
{{eqn | r = \sqrt 2 \sin \left({x - \frac \pi 4 + \frac \pi 4}\right)
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \sqrt 2 \sin \left({x + \frac \pi 4}\right)
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Cosine Function
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Theorem: Cosine of x minus Sine of x/Sine Form

:$\cos x - \sin x = \sqrt 2 \, \map \sin {x + \dfrac {3 \pi} 4}$

Proof:
{{begin-eqn}}
{{eqn | l = \cos x - \sin x
      | r = \cos x - \map \cos {\frac \pi 2 - x}
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = -2 \, \map \sin {\frac {x + \paren {\frac \pi 2 - x} } 2} \map \sin {\frac {x - \paren {\frac \pi 2 - x} } 2}
      | c = Cosine minus Cosine
}}
{{eqn | r = -2 \sin \frac \pi 4 \, \map \sin {x - \frac \pi 4}
      | c = simplifying
}}
{{eqn | r = -\sqrt 2 \, \map \sin {x - \frac \pi 4}
      | c = Sine of $\dfrac \pi 4$
}}
{{eqn | r = \sqrt 2 \, \map \sin {x - \frac \pi 4 + \pi}
      | c = Sine of Angle plus Straight Angle
}}
{{eqn | r = \sqrt 2 \, \map \sin {x + \dfrac {3 \pi} 4}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Cosine of x minus Sine of x
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