hcju/proofwikiqa huggingface dataset:
From Derivative of Arcsecant Function: :$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {\left|{x}\right| \sqrt {x^2 - 1} }$ Since for all $x \in \R$, we have $\left|{x}\right| = \sqrt{x^2}$, we can write: :$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {\sqrt {x^2} \sqrt {x^2 - 1}}$ Multiplying the denominator by $1 = \dfrac {\sqrt{x^2}} {\sqrt{x^2}}$ yields: :$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2}}}$ {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac {B_{2 n} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} }        | r = \lim_{n \mathop \to \infty} \frac {\paren {-1}^{n + 1} \paren {2 n}! \map \zeta {2 n} } {2^{2 n - 1} \pi^{2 n} \paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} }       | c = Riemann Zeta Function at Even Integers }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {\paren {2 n}! \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} } }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {2 \sqrt {\pi n} \paren {\frac {2 n} e}^{2 n} \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} }       | c = Stirling's Formula }} {{eqn | r = \lim_{n \mathop \to \infty} \map \zeta {2 n}       | c =  }} {{eqn | r = 1       | c = }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot x - \tan x       | r = \frac {\cos x} {\sin x} - \frac {\sin x} {\cos x}       | c = {{Defof|Tangent Function|Tangent}} and Cotangent }} {{eqn | r = \frac {\cos^2 x - \sin^2 x} {\sin x \cos x}       | c =  }} {{eqn | r = 2 \frac {\cos^2 x - \sin^2 x} {2 \sin x \cos x}       | c =  }} {{eqn | r = 2 \frac {\cos 2 x} {\sin 2 x}       | c = Double Angle Formula for Sine and Double Angle Formula for Cosine }} {{eqn | r = 2 \cot 2 x       | c = {{Defof|Cotangent}} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities ahdgo2n5ikx91lw9qfq4541fnvulqgr
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\coth u}       | r = \map {\frac \d {\d u} } {\coth u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\csch^2 u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Cotangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\csch u}       | r = \map {\frac \d {\d u} } {\csch u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\csch u \coth u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Cotangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \sinh^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \sinh y       | c = {{Defof|Real Inverse Hyperbolic Sine}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \cosh y       | c = Derivative of Hyperbolic Sine }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {\cosh y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \pm \frac 1 {\sqrt {\sinh^2 y + 1} }       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{end-eqn}} For all $x \in \R$ we have that $\cosh y \ge 1$. Thus it follows that it is necessary to take the positive root of $\sqrt {\sinh^2 y + 1}$. So: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = \frac 1 {\sqrt {\sinh^2 y + 1} }       | c = }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {\sinh^{-1} x}       | r = \frac 1 {\sqrt {x^2 + 1} }       | c = Definition of $x$ and $y$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \cosh^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \cosh y       | c = {{Defof|Real Inverse Hyperbolic Cosine}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \sinh y       | c = Derivative of Hyperbolic Cosine }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {\sinh y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \pm \frac 1 {\sqrt {\cosh^2 y - 1} }       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{end-eqn}} Note that when $y = 0$, $\cosh y$ is defined and equals $1$. But from Derivative of Hyperbolic Cosine: :$\valueat {\dfrac \d {\d y} \cosh y} {y \mathop = 0} = \sinh 0 = 0$ Thus $\dfrac {\d y} {\d x} = \dfrac 1 {\sinh y}$ is not defined at $y = 0$. Hence the limitation of the domain of $\map {\dfrac \d {\d x} } {\cosh^{-1} x}$ to exclude $x = 1$. Now it is necessary to determine the sign of $\dfrac {\d y} {\d x}$. From: :Real Inverse Hyperbolic Cosine is Strictly Increasing :Derivative of Strictly Increasing Real Function is Strictly Positive it follows that $\map {\dfrac \d {\d x} } {\cosh^{-1} x} > 0$ on $\R_{>1}$. Thus: :$\dfrac {\d y} {\d x} = \dfrac 1 {\sqrt {\cosh^2 y - 1} }$ where $\sqrt {\cosh^2 y - 1}$ denotes the positive square root of $\cosh^2 y - 1$. Hence by definition of $x$ and $y$ above: :$\map {\dfrac \d {\d x} } {\cosh^{-1} x} = \dfrac 1 {\sqrt {x^2 - 1} }$ {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \tanh^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \tanh y       | c = {{Defof|Real Inverse Hyperbolic Tangent}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \sech^2 y       | c = Derivative of Hyperbolic Tangent }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {\sech^2 y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {1 - \tanh^2 y}       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {\tanh^{-1} x}       | r = \frac 1 {1 - x^2}       | c = Definition of $x$ and $y$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \coth^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \coth y       | c = {{Defof|Real Inverse Hyperbolic Cotangent}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = -\csch^2 y       | c = Derivative of Hyperbolic Cotangent }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {-1} {\csch^2 y}       | c = Derivative of Inverse Function }} {{eqn | r = \frac {-1} {\coth^2 y - 1}       | c = Difference of Squares of Hyperbolic Cotangent and Cosecant }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {\coth^{-1} x}       | r = \frac {-1} {x^2 - 1}       | c = Definition of $x$ and $y$ }} {{end-eqn}} {{qed}}
300pxright $\sech^{-1} x$ is defined only on the half-open real interval $\hointl 0 1$. Thus on $\hointl 0 1$: {{begin-eqn}} {{eqn | l = y       | r = \sech^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \sech y       | c = where $y \in \R_{>0}$ }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = -\sech y \tanh y       | c = Derivative of Hyperbolic Secant }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \dfrac {-1} {\sech y \ \tanh y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {-1} {\sech y \sqrt {1 - \sech^2 y} }       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {\sech^{-1} x}       | r = \frac {-1} {x \sqrt {1 - x^2} }       | c = Definition of $x$ and $y$ }} {{end-eqn}} When $x = 1$, however, $\sqrt{1 - x^2} = 0$ and so $\dfrac {-1} {x \sqrt {1 - x^2} }$ is undefined. Hence $\sech^{-1} x$ can be defined only on $\openint 0 1$. {{explain|Explain why positive square root is taken at Sum of Squares of Hyperbolic Secant and Tangent}} {{qed}} Category:Derivatives of Inverse Hyperbolic Functions Category:Inverse Hyperbolic Secant 0i0nkjd488kcr56ejx2a18l91hr94ww
600pxright Let $x > 1$. Then we have: {{begin-eqn}} {{eqn | l = y       | r = \csch^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \csch y       | c = where $y \ne 0$ }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = -\csch y \coth y       | c = Derivative of Hyperbolic Cosecant }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \dfrac {-1} {\csch y \coth y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {-1} {\csch y \sqrt {1 + \csch^2 y} }       | c = Difference of Squares of Hyperbolic Cotangent and Cosecant }} {{eqn | ll= \leadsto       | l = \map {\frac \d {\d x} } {\csch^{-1} x}       | r = \frac {-1} {x \sqrt {1 + x^2} }       | c = Definition of $x$ and $y$ }} {{eqn | r = \frac {-1} {\size x \sqrt {1 + x^2} }       | c = as $x > 0$ }} {{end-eqn}} We have that Inverse Hyperbolic Cosecant is Odd Function. Hence from Derivative of Odd Function is Even, $\map {\dfrac \d {\d x} } {\csch^{-1} x}$ is even. Hence for $x < -1$ we have that: :$\map {\dfrac \d {\d x} } {\csch^{-1} x} = \dfrac {-1} {\paren {-x} \sqrt {1 + x^2} }$ and so for $x < -1$: :$\map {\dfrac \d {\d x} } {\csch^{-1} x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$ and the result follows. {{qed}} Category:Derivatives of Inverse Hyperbolic Functions Category:Inverse Hyperbolic Cosecant q0u3omwylgs3hn10jhsv7daovgjgqwd
{{begin-eqn}} {{eqn |l = \csc x + \cot x       |r = \frac 1 {\sin x} + \frac {\cos x} {\sin x}       |c = by definition of Cosecant and Cotangent }} {{eqn |r = \frac {1 + \cos x} {\sin x}       |c = }} {{eqn |r = \frac {2 \cos^2 {\frac x 2} } {2 \sin {\frac x 2} \cos {\frac x 2} }       |c = Double Angle Formula for Sine and Double Angle Formula for Cosine }} {{eqn |r = \frac {\cos {\frac x 2} } {\sin {\frac x 2} }       |c = }} {{eqn |r = \cot {\frac x 2}       |c = by definition of Cotangent }} {{end-eqn}} {{qed}} Category:Trigonometric Identities cxkxef0xrf4mz2sgewslcm4tji8i49q
From Leibniz's Rule: :$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$ where $\paren n$ denotes the order of the derivative. Setting $n = 2$: {{begin-eqn}} {{eqn | l = \paren {\map f x \, \map g x}''       | r = \paren {\map f x \, \map g x}^{\paren 2}       | c = {{Defof|Nth Derivative}} }} {{eqn | r = \sum_{k \mathop = 0}^2 \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x       | c = Leibniz's Rule }} {{eqn | r = \binom 2 0 \map {f^{\paren 0} } x \, \map {g^{\paren 2} } x + \binom 2 1 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \binom 2 2 {f^{\paren 2} } x \, \map {g^{\paren 0} } x       | c =  }} {{eqn | r = \map {f^{\paren 0} } x \, \map {g^{\paren 2 } } x + 2 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \map {f^{\paren 2} } x \, \map {g^{\paren 0} } x       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \map f x \, \map {g''} x + 2 \map {f'} x \, \map {g'} x + \map {f''} x \, \map g x       | c = {{Defof|Nth Derivative}} }} {{end-eqn}} {{qed}} {{Namedfor|Gottfried Wilhelm von Leibniz}}
From Leibniz's Rule: :$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$ where $\paren n$ denotes the order of the derivative. Setting $n = 3$: {{begin-eqn}} {{eqn | l = \paren {\map f x \, \map g x}'''       | r = \paren {\map f x \, \map g x}^{\paren 3}       | c = {{Defof|Nth Derivative}} }} {{eqn | r = \sum_{k \mathop = 0}^3 \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x       | c = Leibniz's Rule }} {{eqn | r = \binom 3 0 \map {f^{\paren 0} } x \, \map {g^{\paren 3} } x + \binom 3 1 \map {f^{\paren 1} } x \, \map {g^{\paren 2} } x + \binom 3 2 \map {f^{\paren 2} } x \map {g^{\paren 1} } x + \binom 3 3 \map {f^{\paren 3} } x \, \map {g^{\paren 0} } x       | c =  }} {{eqn | r = \map {f^{\paren 0} } x \, \map {g^{\paren 3} } x + 3 \map {f^{\paren 1} } x \, \map {g^{\paren 2} } x + 3 \map {f^{\paren 2} } x \, \map {g^{\paren 1} } x + \map {f^{\paren 3} } x \, \map {g^{\paren 0} } x       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \map f x \, \map {g'''} x + 3 \map {f'} x \, \map {g''} x + 3 \map {f''} x \, \map {g'} x + \map {f'''} x \, \map g x       | c = {{Defof|Nth Derivative}} }} {{end-eqn}} {{qed}} {{Namedfor|Gottfried Wilhelm von Leibniz}}
By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | o =        | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {\paren {-1}^n 2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} } {\paren {2 n + 2}!} x^{2 n + 1} } {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1} } }       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \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 {2^{2 n + 2} - 1} {2^{2 n} - 1} } \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 {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \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 {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \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 {B_{2 n + 2} } { B_{2 n} } } 8 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 \paren {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} } } 8 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 8 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 4 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \frac {4 e^2} {\pi^2 e^2} x^2       | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}} }} {{eqn | r = \frac 4 {\pi^2} x^2       | c = }} {{end-eqn}} This is less than $1$ {{iff}}: :$\size x < \dfrac \pi 2$ Hence by the Ratio Test, the series converges for $\size x < \dfrac \pi 2$.
From Power Series Expansion for Cotangent Function: :$(1): \quad \cot x = \displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$ Then: {{begin-eqn}} {{eqn | l = \tan x       | r = \cot x - 2 \cot 2 x       | c = Cotangent Minus Tangent }} {{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)!} - 2 \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n 2^{2 n} B_{2 n} \, \left({2 x}\right)^{2 n - 1} } {\left({2 n}\right)!}        | c = by $(1)$ }} {{eqn | r =  \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} \left({1 - 2^{2 n} }\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = }} {{eqn | r =  \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^{n - 1} 2^{2 n} \left({2^{2 n} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = }} {{end-eqn}} {{qed|lemma}}
We have: {{begin-eqn}} {{eqn | l = \frac x {e^x - 1}       | r = \frac x 2 \left({\frac 2 {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {e^x - e^x + 2} {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x -  1} }\right)       | c =  }} {{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)       | c =  }} {{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)       | c =  }} {{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)       | c = multiplying top and bottom by $e^{-x/2}$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | n = 1       | l = \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)       | r = \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} x^{2 n}       | c = {{Defof|Bernoulli Numbers/Generating Function|Bernoulli numbers}} }} {{end-eqn}} Replacing $x$ with $2 i x$ in the {{LHS}} $(1)$: {{begin-eqn}} {{eqn | o =        | r = \frac {2 i x} 2 \left({\frac {e^{2 i x / 2} + e^{-2 i x / 2} } {e^{2 i x / 2} - e^{-2 i x / 2} } }\right)       | c =  }} {{eqn | r = i x \left({\frac {e^{i x} + e^{-i x} } {e^{i x} - e^{-i x} } }\right)       | c =  }} {{eqn | r = x \cot x       | c = Cotangent Exponential Formulation }} {{end-eqn}} Replacing $x$ with $2 i x$ in the {{RHS}} $(1)$: {{begin-eqn}} {{eqn | o =        | r = \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} \left({2 i x}\right)^{2 n}       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}       | c =  }} {{eqn | ll= \leadsto       | l = x \cot x       | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}       | c =  }} {{eqn | ll= \leadsto       | l = \cot x       | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n - 1}       | c =  }} {{end-eqn}} Then from Cotangent Minus Tangent: :$\tan x = \cot x - 2 \cot 2 x$ from which: {{begin-eqn}} {{eqn | l = \tan x       | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1} - 2 \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} \left({2 x}\right)^{2 n - 1}       | c = by $(1)$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} (1 - 2^{2 n}) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}       | c = }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^{n - 1} 2^{2 n} (2^{2 n} - 1) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}       | c = }} {{end-eqn}} {{qed|lemma}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$ $\map P 1$ is true, as this just says: :$\ds \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$ === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \int \map {\paren {f_1 \pm f_2} } x \rd x       | r = \int \paren {\map {f_1} x \pm \map {f_2} x} \rd x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \int \map {f_1} x \rd x \pm \map {f_2} x \rd x       | c = Linear Combination of Primitives }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$ Then we need to show: :$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | o =        | r = \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x       | c =  }} {{eqn | r = \int \paren {\map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x + \map {f_{k + 1} } x} \rd x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x + \int \map {f_{k + 1} } x \rd x       | c = Basis for the Induction }} {{eqn | r = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x + \int \map {f_{k + 1} } x \rd x       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$ {{qed}}
=== Lemma 1 === {{:Hurwitz's Theorem (Number Theory)/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Hurwitz's Theorem (Number Theory)/Lemma 2}}{{qed|lemma}} There are an infinite number of convergents to $\xi$. Taking these in sets of $3$ at a time, it can be seen from Lemma 2 that there are an infinite number of approximations that satisfy the given inequality. From Lemma 1 it is seen that this inequality is the best possible. {{qed}} {{Namedfor|Adolf Hurwitz|cat = Hurwitz}}
Let $u = c x$. By Derivative of Identity Function: Corollary: :$\dfrac {\d u} {\d x} = c$ Thus: {{begin-eqn}} {{eqn | l = \int \map f {c x} \rd x       | r = \int \frac {\map f u} c \rd u       | c = Primitive of Composite Function }} {{eqn | r = \frac 1 c \int \map f u \rd u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map F x       | r = \int \map {\paren {g \circ f} } x \rd x }} {{eqn | r = \int \map g {\map f x} \rd x       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \int \map g u \rd x       | c = where $u = \map f x$ }} {{eqn | ll= \leadsto       | l = \frac {\d F} {\d x}       | r = \map g u       | c = {{Defof|Primitive (Calculus)}} }} {{eqn | ll= \leadsto       | l = \frac {\d F} {\d u} \frac {\d u} {\d x}       | r = \map g u       | c = Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d F} {\d u} \frac \d {\d x} \map f x       | r = \map g u       | c = Definition of $u$ }} {{eqn | ll= \leadsto       | l = \frac {\d F} {\d u} \map {f'} x       | r = \map g u       | c = {{Defof|Derivative}} }} {{eqn | ll= \leadsto       | l = \frac {\d F} {\d u}       | r = \frac {\map g u} {\map {f'} x}       | c =  }} {{eqn | ll= \leadsto       | l = F       | r = \int \frac {\map g u} {\map {f'} x} \rd u       | c = {{Defof|Primitive (Calculus)}} }} {{eqn | r = \int \map g u \frac {\d x} {\d u} \rd u       | c = Derivative of Inverse Function: $\dfrac {\d x} {\d u} = \dfrac 1 {\d u / \d x}$ }} {{end-eqn}} {{qed}}
For $n = 1$ we have: :$1 - a_1 \ge 1 - a_1$ which is clearly true.  Suppose the proposition is true for $n = k$, that is: :$\displaystyle \prod_{i \mathop = 1}^k \left({1 - a_i}\right) \ge 1 - \sum_{i \mathop = 1}^k a_i$ Then: {{begin-eqn}} {{eqn | l = \prod_{i \mathop = 1}^{k + 1} \left({1 - a_i}\right)        | r = \left({1 - a_{k + 1} }\right) \prod_{i \mathop = 1}^k \left({1 - a_i}\right)       | c =  }} {{eqn | o = \ge       | r = \left({1 - a_{k + 1} }\right) \left({1 - \sum_{i \mathop = 1}^k a_i}\right)       | c = as $1 - a_{k + 1} \ge 0$ }} {{eqn | r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i + a_{k + 1} \sum_{i \mathop = 1}^k a_i       | c =  }} {{eqn | o = \ge       | r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i       | c = as $a_i \ge 0$ }} {{eqn | r = 1 - \sum_{i \mathop = 1}^{k + 1} a_i       | c =  }} {{end-eqn}} Thus, by Principle of Mathematical Induction, the proof is complete.  {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\dfrac {x^{n + 1} } {n + 1} }       | r = \paren {n + 1} \paren {\dfrac {x^{\paren {n + 1} - 1} } {n + 1} }       | c = Power Rule for Derivatives }} {{eqn | r = x^n       | c =  }} {{eqn | ll= \leadsto       | l = \int x^n \rd x       | r = \frac {x^{n + 1} } {n + 1} + C       | c = {{Defof|Primitive (Calculus)}} }} {{end-eqn}} When $n = -1$ we have $n + 1 = 0$, and $\dfrac {x^{n + 1} } {n + 1} = \dfrac {x^0} 0$ is undefined. {{qed}}