hcju/proofwikiqa huggingface dataset:
From Power Series Expansion for $\map \ln {1 + x}$: :$(1): \quad \displaystyle \map \ln {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$ for $-1 < x \le 1$. From Power Series Expansion for $\map \ln {1 + x}$: Corollary: :$(2): \quad \displaystyle \map \ln {1 - x} = - \sum_{n \mathop = 1}^\infty \frac {x^n} n$ for $-1 < x < 1$. Then we have: {{begin-eqn}} {{eqn | l = \map \ln {1 + x} - \map \ln {1 - x}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n + \sum_{n \mathop = 1}^\infty \frac {x^n} n       | c = subtracting $(2)$ from $(1)$ }} {{eqn | r = \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r - 1} \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}       | c = separating out into odd and even indices }} {{eqn | r = \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} - \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r} } {2 r}       | c = $\paren {-1}^{2 r} = 1$, $\paren {-1}^{2 r - 1} = -1$ }} {{eqn | r = \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1} + \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}       | c = even indices cancel }} {{eqn | ll= \leadsto       | l = \map \ln {\frac {1 + x} {1 - x} }       | r = 2 \sum_{r \mathop = 1}^\infty \frac {x^{2 r + 1} } {2 r + 1}       | c = Difference of Logarithms }} {{end-eqn}} Hence the result. {{qed}}
From Power Series Expansion for $\dfrac 1 2 \map \ln {\dfrac {1 + x} {1 - x} }$: :$(1): \quad \displaystyle \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}$ 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: :$\displaystyle \lim_{x \mathop \to 1^-} \dfrac {1 + x} {1 - x} \to +\infty$ and: :$\displaystyle \lim_{x \mathop \to -1^+} \dfrac {1 + x} {1 - x} \to 0$ 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}}
From the corollary to 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 $-1 < x \ge 1$ }} {{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: :$\displaystyle \lim_{x \mathop \to 1^-} \dfrac 1 {1 - x} \to +\infty$ and: :$\displaystyle \lim_{x \mathop \to -1^+} \dfrac 1 {1 - x} \to \frac 1 2$ Thus when $x \in \openint {-1} 1$ we have that $z \in \hointr {\dfrac 1 2} \to$. Thus, substituting $z$ for $\dfrac 1 {1 - x}$ in $(1)$ gives the result. {{qed}}
{{begin-eqn}} {{eqn | l = \cot x       | r = i \frac {e^{i x} + e^{- i x} } {e^{i x} - e^{- i x} }       | c = Cotangent Exponential Formulation }} {{eqn | r = i \frac {e^{2 i x} + 1 } {e^{2 i x} - 1 }       | c = }} {{eqn | r = i \left({1 + \frac 2 {e^{2 i x} - 1} }\right)       | c = }} {{eqn | r = i + \frac {2 i} {e^{2 i x} - 1}       | c = }} {{eqn | r = i + \frac 1 x \frac {2 i x} {e^{2 i x} - 1}       | c = }} {{eqn | r = i + \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_n \left({2 i x}\right)^n} {n!}       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \frac 1 x + \frac 1 x \sum_{n \mathop = 2}^\infty \frac {B_n \left({2 i x}\right)^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} \left({2 i x}\right)^{2 n} } {\left({2 n}\right)!}       | c = Odd Bernoulli Numbers Vanish }} {{eqn | r = \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_{2 n} \left({2 i x}\right)^{2 n} } {\left({2 n}\right)!}       | c = }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = }} {{end-eqn}} By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | o =        | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\frac {\left({-1}\right)^n 2^{2 n + 2} B_{2 n + 2} \, x^{2 n + 1} } {\left({2 n + 2}\right)!} } {\frac {\left({-1}\right)^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} } } \right \rvert       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({2 n + 2}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 4 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 2 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {\left({- 1}\right)^{n + 2} 4 \sqrt {\pi (n + 1)} \left({\frac {n + 1} {\pi e} }\right)^{2 n + 2} } {\left({- 1}\right)^{n + 1} 4 \sqrt {\pi n} \left({\frac n {\pi e} }\right)^{2 n} } } \right \rvert 2 x^2       | c = Asymptotic Formula for Bernoulli Numbers }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\left({n + 1}\right)^2} {\left({2 n + 1}\right) \left({n + 1}\right)} \sqrt {\frac {n + 1} n } \left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac 2 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac {x^2} {\pi^2 e^2}       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\left({1 + \frac 1 n}\right)^n}\right)^2} \right \rvert \frac {x^2} {\pi^2 e^2}       | c = }} {{eqn | r = \frac {e^2 x^2} {\pi^2 e^2}       | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}} }} {{eqn | r = \frac {x^2} {\pi^2}        | c = }} {{end-eqn}} This is less than $1$ {{iff}}: : $\left|{x}\right| < \pi$ Hence by the Ratio Test, the series converges for $\left|{x}\right| < \pi$. {{qed}}
{{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}}
From Sum of Arctangent and Arccotangent: :$\operatorname {arccot} x = \dfrac \pi 2 - \arctan x$ The result follows from Power Series Expansion for Real Arctangent Function. {{qed}}
From Arccosine of Reciprocal equals Arcsecant: :$\operatorname {arcsec} x = \arccos \dfrac 1 x$ From Power Series Expansion for Real Arccosine Function: :$\displaystyle \arccos x = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \frac {\left({2 n}\right)!} {2^{2 n} \left({n!}\right)^2} \frac {x^{2 n + 1} } {2 n + 1}$ which is converges for $\left\lvert{x}\right\rvert \le 1$. The result follows by subtituting $\dfrac 1 x$ for $x$. This converges for $\left\lvert{\dfrac 1 x}\right\rvert \le 1$, that is, for $\left\lvert{x}\right\rvert \ge 1$ {{qed}}
From Arcsine of Reciprocal equals Arccosecant: :$\operatorname {arccsc} x = \arcsin \dfrac 1 x$ From Power Series Expansion for Real Arcsine Function: :$\displaystyle \arcsin x = \sum_{n \mathop = 0}^\infty \frac {\left({2 n}\right)!} {2^{2 n} \left({n!}\right)^2} \frac {x^{2 n + 1} } {2 n + 1}$ which is converges for $\left\lvert{x}\right\rvert \le 1$. The result follows by subtituting $\dfrac 1 x$ for $x$. This converges for $\left\lvert{\dfrac 1 x}\right\rvert \le 1$, that is, for $\left\lvert{x}\right\rvert \ge 1$ {{qed}}
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}}
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}}
{{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 \left({2 x}\right)^n} {n!}       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \frac 1 x + \frac 1 x \sum_{n \mathop = 2}^\infty \frac {B_n \left({2 x}\right)^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} \left({2 x}\right)^{2 n} } {\left({2 n}\right)!}       | c = Odd Bernoulli Numbers Vanish }} {{eqn | r = \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_{2 n} \left({2 x}\right)^{2 n} } {\left({2 n}\right)!}       | c = }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = }} {{end-eqn}} By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | o =        | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\frac {2^{2 n + 2} B_{2 n + 2} \, x^{2 n + 1} } {\left({2 n + 2}\right)!} } {\frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} } } \right \rvert       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({2 n + 2}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 4 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 2 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {\left({- 1}\right)^{n + 2} 4 \sqrt {\pi (n + 1)} \left({\frac {n + 1} {\pi e} }\right)^{2 n + 2} } {\left({- 1}\right)^{n + 1} 4 \sqrt {\pi n} \left({\frac n {\pi e} }\right)^{2 n} } } \right \rvert 2 x^2       | c = Asymptotic Formula for Bernoulli Numbers }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\left({n + 1}\right)^2} {\left({2 n + 1}\right) \left({n + 1}\right)} \sqrt {\frac {n + 1} n } \left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac 2 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac {x^2} {\pi^2 e^2}       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\left({1 + \frac 1 n}\right)^n}\right)^2} \right \rvert \frac {x^2} {\pi^2 e^2}       | c = }} {{eqn | r = \frac {e^2 x^2} {\pi^2 e^2}       | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}} }} {{eqn | r = \frac {x^2} {\pi^2}        | c = }} {{end-eqn}} This is less than $1$ {{iff}}: : $\left|{x}\right| < \pi$ Hence by the Ratio Test, the series converges for $\left|{x}\right| < \pi$. {{qed}}
By definition of the Euler numbers: :$\displaystyle \sech x = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}$ From Odd Euler Numbers Vanish: :$E_{2 k + 1} = 0$ for $k \in \Z$. Hence the result. {{qed}}
{{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
From Sum of Infinite Geometric Sequence: :$(1): \quad \displaystyle \frac 1 {1 - x^2} = \sum_{n \mathop = 0}^\infty \paren {x^2}^n$ 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 = \tanh^{-1} 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}}
From Power Series Expansion for Real Inverse Hyperbolic Tangent: {{begin-eqn}} {{eqn | l = \tanh^{-1} 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 Inverse Hyperbolic Tangent of Reciprocal equals Inverse Hyperbolic Cotangent :$\map {\tanh^{-1} } {\dfrac 1 x} = \coth^{-1} x$ So: {{begin-eqn}} {{eqn | l = \coth^{-1} 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}}
A result in combinatorics known as the Exponential formula states that if  :$f(x) = \sum_{n=1}^{\infty} \frac{c_n}{n!} x^n $ then  :$e^{f(x)} = \sum_{k=0}^{\infty} \frac{B_k(c_1,c_2,...c_k)}{k!}x^k $ Where $B_k(c_1,c_2,...c_k)$ is the k-th complete Bell Polynomial. From the Power Series Expansion for Cosine Function, we get :$ \cos(x)-1 = \sum_{p=1}^{\infty} \frac{(-1)^p}{2p!} x^{2p} $ Using this we may plug the sequence 0,-1,0,1,0,-1 into the first 6 complete Bell Polynomials an arbitrary even Complete Bell Polynomial, will take the from :$ B_n(x_1,x_2,x_3,....,x_n)=\sum \frac{n!}{k_1!k_2!...k_n!}(\frac{x_1}{1!})^{c_1}(\frac{x_2}{2!})^{c_2}....(\frac{x_n}{n!})^{c_n}$ where the sum is taken over all n-tuples $(c_1,....,c_n)$ such that $c_1+2c_2....+nc_n=n$ in other words, it is taken over every integer partion of n. Here we have it that :$x_{4n+0}=1$ :$x_{4n+1}=0$ :$x_{4n+2}=-1$ :$x_{4n+3}=0$ Because there is no way to form a partition of an odd number without using an odd number, all the summands in the odd Complete Bell polynomials contain a 0, thus they equal 0. Explicitly calculating the first 6 even complete Bell Polynomials we get :$ B_0(0)=1$ :$ B_2(0,-1)=0^2-1=-1$ :$ B_4(0,-1,0,1)=0^4+6(0)^2(-1)+4(0)(0)+3(-1)^2 + 1=4$ :$ B_6(0,-1,0,1,0,-1)=0^6+6(-1)(0)+15(-1)(1)+10(0)^2+15(0)^2(1)+15(-1)^3+60(0)(-1)(0)+20(0)^3(0)+45(0)^2(-1)^2+15(0)^4(-1)+(-1)=-31$ Note that in all the summands for the above polynomials, they are either 0 or of the same sign. This is provable to always be the case, as if $n\equiv 2 (mod 4)$, then an odd number of integers congruent to 2 mod 4 must be chosen for every partition, thus all summands will be a negative. A similar argument holds for the case $n\equiv 0 (mod 4)$ Thus for our given values of x_k, we are only summing over the partitions of 2m such that all summands in the partition are even, and because all the summands in the polynomial are the same sign we have $B_2m(0,-1,0....,\pm 1)=(-1)^m B_2m(0,1,0,1,....,1).$ Thus from the definition of Bell Polynomials as a sum over all incomplete Bell polynomials we have :$B_2m(0,-1,0....,\pm x_2m)=(-1)^m P_{2}(2m)$,  where $P_{2}(2m)$ is all partitions of a set of size 2m into even blocks. Thus we have :$ e^{\cos(x)}=e(e^{cos(x)-1})= e(\sum_{n=0}^{\infty} \frac{(-1)^m P_2(2m)}{2m!}x^{2m})$ where $P_2(2m)$ is the partition of the set of size 2m into even blocks. {{qed}}
Let $\map f x = e^x \sin x$. By definition of Maclaurin series: :$(1): \quad \map f x \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0$ 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: :$\map {f^{\paren n} } x = 2^{n / 2} e^x \, \map \sin {x + \dfrac {n \pi} 4}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \map f x       | r = e^x \sin x       | c =  }} {{eqn | r = 2^{0 / 2} e^x \, \map \sin {x + \dfrac {0 \pi} 4}       | 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 $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\map {f^{\paren k} } x = 2^{k / 2} e^x \, \map \sin {x + \dfrac {k \pi} 4}$ from which it is to be shown that: :$\map {f^{\paren {k + 1} } } x = 2^{\paren {k + 1} / 2} e^x \, \map \sin {x + \dfrac {\paren {k + 1} \pi} 4}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \map {f^{\paren {k + 1} } } x       | r = \frac \d {\d x} \map {f^{\paren k} } x       | c =  }} {{eqn | r = \frac \d {\d x} 2^{k / 2} e^x \, \map \sin {x + \dfrac {k \pi} 4}       | c = Induction Hypothesis }} {{eqn | r = 2^{k / 2} \paren {e^x \frac \d {\d x} \map \sin {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}       | c = Product Rule }} {{eqn | r = 2^{k / 2} \paren {e^x \, \map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} \frac \d {\d x} e^x}       | c = Derivative of Sine Function, Derivatives of Function of $a x + b$ }} {{eqn | r = 2^{k / 2} e^x \paren {\map \cos {x + \dfrac {k \pi} 4} + \map \sin {x + \dfrac {k \pi} 4} }       | c = Derivative of Exponential Function and simplification }} {{eqn | r = 2^{k / 2} e^x \sqrt 2 \, \map \sin {x + \dfrac {k \pi} 4 + \dfrac \pi 4}       | c = Sine of x plus Cosine of x: Sine Form }} {{eqn | r = 2^{\paren {k + 1} / 2} e^x \, \map \sin {x + \dfrac {\paren {k + 1} \pi} 4}       | c = simplification }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and by the Principle of Mathematical Induction: :$\forall n \in \Z_{\ge 0}: \map {f^{\paren n} } x = 2^{n / 2} e^x \, \map \sin {x + \dfrac {n \pi} 4}$ The result follows by setting $x = 0$ and substituting for $\map {f^{\paren n} } 0$ in $(1)$. {{qed}}
{{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 50kr9m6reez8222mlptqy1iuc99l55t
{{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 = Prosthaphaeresis Formula for 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:Sine Function Category:Cosine Function 8dj4j8ytof7v7jtv48ura2l69qgog6q
{{begin-eqn}} {{eqn | l = \cos x - \sin x       | r = \sqrt 2 \, \map \sin {x + \dfrac {3 \pi} 4}       | c = Cosine of x minus Sine of x: Sine Form }} {{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - \paren {x + \dfrac {3 \pi} 4} }       | c = Cosine of Complement equals Sine }} {{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - x - \dfrac {3 \pi} 4}       | c = simplifying }} {{eqn | r = \sqrt 2 \, \map \cos {-\frac \pi 4 - x}       | c = simplifying }} {{eqn | r = \sqrt 2 \, \map \cos {x + \frac \pi 4}       | c = Cosine Function is Even }} {{end-eqn}} {{qed}} Category:Sine Function Category:Cosine Function 5j3k07mkmk3fwukvbwtabh1elybtg0l