hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \int_0^{\frac \pi 2} \cos^2 x \rd x       | r = \frac \pi 4       | c = Definite Integral from 0 to Half Pi of Square of Cosine x }} {{eqn | r = \dfrac 1 2 \times \dfrac \pi 2       | c =  }} {{eqn | r = \dfrac 2 {\paren {2 \times 1}^2} \dfrac \pi 2       | c =  }} {{eqn | r = \dfrac {\paren {2 \times 1}!} {\paren {2^1 \times 1!}^2} \dfrac \pi 2       | c =  }} {{end-eqn}} Thus $\map P 1$ 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: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 k} x \rd x = \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2$ from which it is to be shown that: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} } x \rd x = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2$ === Induction Step === This is the induction step: Let $I_k = \displaystyle \int_0^{\frac \pi 2} \cos^{2 k} x \rd x$. {{begin-eqn}} {{eqn | l = I_{k + 1}       | r = \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } I_k       | c = Reduction Formula for Definite Integral of Power of Cosine }} {{eqn | r = \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2       | c = Induction Hypothesis }} {{eqn | r = \frac {2 \paren {k + 1} \paren {2 \paren {k + 1} - 1} } {2^2 \paren {k + 1}^2} \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2       | c =  }} {{eqn | r = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{> 0}: \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \int_0^{\frac \pi 2} \cos x \rd x       | r = \left[{\sin x}\right]_0^{\frac \pi 2}       | c = Primitive of $\cos x$ }} {{eqn | r = \sin \frac \pi 2 - \sin 0       | c =  }} {{eqn | r = 1       | c = Sine of Right Angle, Sine of Zero is Zero }} {{eqn | r = \frac {\left({2^0 \times 1!}\right)^2} {\left({2 \times 0 + 1}\right)!}       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\left({2^k k!}\right)^2} {\left({2 k + 1}\right)!}$ from which it is to be shown that: :$\displaystyle \int_0^{\frac \pi 2} \cos^{2 \left({k + 1}\right) + 1} x \rd x = \dfrac {\left({2^{k + 1} \left({k + 1}\right)!}\right)^2} {\left({2 \left({k + 1}\right) + 1}\right)!}$ === Induction Step === This is the induction step: Let $I_k = \displaystyle \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x$. {{begin-eqn}} {{eqn | l = I_{k + 1}       | r = \frac {2 \left({k + 1}\right)} {2 \left({k + 1}\right) + 1} I_k       | c = Reduction Formula for Definite Integral of Power of Cosine }} {{eqn | r = \frac {2 \left({k + 1}\right)} {2 \left({k + 1}\right) + 1} \dfrac {\left({2^k k!}\right)^2} {\left({2 k + 1}\right)!}       | c = Induction Hypothesis }} {{eqn | r = \frac {\left({2 \left({k + 1}\right)}\right)^2} {\left({2 \left({k + 1}\right) + 1}\right) \left({2 \left({k + 1}\right)}\right)} \dfrac {\left({2^k k!}\right)^2} {\left({2 k + 1}\right)!}       | c = multiplying top and bottom by $2 \left({k + 1}\right)$ }} {{eqn | r = \dfrac {\left({2^{k + 1} \left({k + 1}\right)!}\right)^2} {\left({2 \left({k + 1}\right) + 1}\right)!}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$ {{qed}}
By definition, $\arctan$ is the inverse of the tangent function's restriction to $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.  By Tangent of Zero: :$\tan 0 = 0$ As $0 \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$, we have $\arctan 0 = 0$ by the definition of an inverse function. {{qed}} Category:Arctangent Function saoxb2rrn1mo9nd574xdklbg0ug78cl
{{AimForCont}}: :$\order G = k p^n: p \nmid k$ where $\order G$ denotes the order of $G$. By Divisors of Power of Prime: :$k \nmid p^n$ From the First Sylow Theorem: :$\exists H \le G: \order H = k$ where $H \le G$ denotes that $H$ is a subgroup of $G$. Thus: :$\exists h \in H: \order h \divides k \implies \order h \nmid p$ where $\divides$ denotes divisibility. Thus: :$\exists h \in G: \order h \ne p^n: n \in \Z$ Thus by Proof by Contradiction, $\order G$ must be a power of $p$. {{Qed}}
Let every element of $G$ be a $p$-element. Let $q$ be a prime number which is a divisor of the order $\order G$ of $G$. By Cauchy's Lemma (Group Theory), there exists an element of $G$ whose order is a divisor of $q$. But as the order of all elements of $G$ divide $p^n$ it follows that $q = p$. Thus $G$ is a group whose order is $p^n$ for some $n \in \Z_{>0}$. {{qed}}
Let $N$ be a positive integer.  Then:  {{begin-eqn}} {{eqn	| l = \paren {N + 1}^{k + 1} - N^{k + 1} 	| r = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i N^i - N^{k + 1} 	| c = Binomial Theorem: Integral Index }} {{eqn	| r = \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i + \binom {k + 1} {k + 1} N^{k + 1} - N^{k + 1} }} {{eqn	| r = 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i 	| c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} Summing from $N = 1$ to $N = n$, we have on the {{LHS}}: {{begin-eqn}} {{eqn	| l = \sum_{N \mathop = 1}^n \paren {\paren {N + 1}^{k + 1} - N^{k + 1} }  	| r = \paren {n + 1}^{k + 1} - 1^{k + 1} 	| c = Telescoping Series: Example 1 }} {{eqn	| r = \paren {n + 1}^{k + 1} - 1 }} {{end-eqn}} So, we have:  {{begin-eqn}} {{eqn	| l = \paren {n + 1}^{k + 1} - 1 	| r = \sum_{N \mathop = 1}^n \paren {1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i} }} {{eqn	| r = \sum_{N \mathop = 1}^n 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i \sum_{N \mathop = 1}^n N^i }} {{eqn	| r = n + \sum_{i \mathop = 1}^k \binom {k + 1} i S_i }} {{end-eqn}} giving:  :$\displaystyle \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 1} 	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n} \rd x 	| c = Primitive of Power }} {{eqn	| r = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^3}^n} \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \int_0^1 \frac 1 {1 + x^3} \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \intlimits {\frac 1 6 \ln \size {\frac {\paren {x + 1}^2} {x^2 - x + 1} } + \frac 1 {\sqrt 3} \map \arctan {\frac {2 x - 1} {\sqrt 3} } } 0 1 	| c = Primitive of $\dfrac 1 {x^3 + a^3}$ }} {{eqn	| r = \frac 1 6 \ln 4 + \frac 1 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} } - \frac 1 6 \ln 1 - \frac 1 {\sqrt 3} \map \arctan {-\frac 1 {\sqrt 3} } }} {{eqn	| r = \frac 1 3 \ln 2 + \frac 2 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} }  	| c = Logarithm of Power, Arctangent Function is Odd, Natural Logarithm of 1 is 0 }} {{eqn	| r = \frac 1 3 \ln 2 + \frac 2 {\sqrt 3} \times \frac \pi 6 	| c = Arctangent of $\dfrac {\sqrt 3} 3$ }} {{eqn	| r = \frac {\pi \sqrt 3} 9 + \frac {\ln 2} 3 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} 	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{4 n} \rd x 	| c = Primitive of Power }} {{eqn	| r = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^4}^n} \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \int_0^1 \frac 1 {1 + x^4} \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \intlimits {\frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } } 0 1 	| c = Primitive of $\dfrac 1 {1 + x^4}$ }} {{eqn	| r = \frac 1 {2 \sqrt 2} \intlimits {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1 + \frac 1 {4 \sqrt 2} \intlimits {\ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } 0 1 }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \intlimits {\frac 1 {2 \sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1 	| r = \frac 1 {2 \sqrt 2} \arctan 0 - \frac 1 {2 \sqrt 2} \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {x^2 - 1} x} } }} {{eqn	| r = -\frac 1 {2 \sqrt 2} \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {x^2 - 1} x} } 	| c = Arctangent of Zero is Zero }} {{eqn	| r = -\frac 1 {2 \sqrt 2} \lim_{u \mathop \to -\infty} \map \arctan {\frac 1 {\sqrt 2} u} 	| c = setting $u = x - \dfrac 1 x$ }} {{eqn	| r = \frac \pi {4 \sqrt 2} 	| c = Limit to Infinity of Arctangent Function }} {{eqn	| r = \frac {\pi \sqrt 2} 8 }} {{end-eqn}} and:  {{begin-eqn}} {{eqn	| l = \frac 1 {4 \sqrt 2} \intlimits {\ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } 0 1 	| r = \frac 1 {4 \sqrt 2} \paren {\ln \size {\frac {2 + \sqrt 2} {2 - \sqrt 2} } - \ln 1} }} {{eqn	| r = \frac 1 {4 \sqrt 2} \ln \size {\frac {2 + \sqrt 2} {2 - \sqrt 2} } 	| c = Natural Logarithm of 1 is 0 }} {{eqn	| r = \frac 1 {4 \sqrt 2} \ln \size {\frac {\paren {2 + \sqrt 2}^2} {\paren {2 - \sqrt 2} \paren {2 + \sqrt 2} } } }} {{eqn	| r = \frac {\sqrt 2} 8 \ln \size {\paren {\frac {2 + \sqrt 2} {\sqrt 2} }^2} 	| c = Difference of Two Squares }} {{eqn	| r = \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4 	| c = Logarithm of Power }} {{end-eqn}} Putting these together gives:  :$\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \int_0^1 \frac 1 {1 + x^4} \rd x = \frac {\pi \sqrt 2} 8 + \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4$   {{qed}}
{{begin-eqn}} {{eqn	| l = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 2} 	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n + 1} \rd x 	| c = Primitive of Power }} {{eqn	| r = \int_0^1 x \paren {\sum_{n \mathop = 0}^\infty \paren {-x^3}^n} \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \int_0^1 \frac x {x^3 + 1} \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \intlimits {\frac 1 6 \ln \size {\frac {x^2 - x + 1} {\paren {x + 1}^2} } + \frac 1 {\sqrt 3} \map \arctan {\frac {2 x - 1} {\sqrt 3} } } 0 1 	| c = Primitive of $\dfrac x {x^3 + a^3}$ }} {{eqn	| r = \frac 1 6 \map \ln {\frac 1 4} + \frac 1 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} } - \frac 1 6 \ln 1 - \frac 1 {\sqrt 3} \map \arctan {-\frac 1 {\sqrt 3} } }} {{eqn	| r = \frac 2 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} } -\frac 1 3 \ln 2 	| c = Logarithm of Power, Natural Logarithm of 1 is 0, Arctangent Function is Odd }} {{eqn	| r = \frac 2 {\sqrt 3} \times \frac \pi 6 - \frac 1 3 \ln 3 	| c = Arctangent of $\dfrac {\sqrt 3} 3$ }} {{eqn	| r = \frac {\pi \sqrt 3} 9 - \frac {\ln 2} 3 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^4}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4} - \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^4}       | c = separating odd positive terms from even negative terms }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^4} - \frac 1 {16} \sum_{n \mathop = 1}^\infty \frac 1 {n^4} }} {{eqn | r = \frac {\pi^4} {96} - \frac 1 {16} \sum_{n \mathop = 1}^\infty \frac 1 {n^4}       | c = Sum of Reciprocals of Fourth Powers of Odd Integers }} {{eqn | r = \frac {\pi^4} {96} - \frac 1 {16} \times \frac {\pi^4} {90}       | c = Riemann Zeta Function of 4 }} {{eqn | r = \frac {\pi^4 \paren {15 - 1} } {1440} }} {{eqn | r = \frac {7 \pi^4} {720} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^6}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6} - \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^6}       | c = separating odd positive terms from even negative terms }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^6} - \frac 1 {64} \sum_{n \mathop = 1}^\infty \frac 1 {n^6} }} {{eqn | r = \frac {\pi^6} {960} - \frac 1 {64} \sum_{n \mathop = 1}^\infty \frac 1 {n^6}       | c = Sum of Reciprocals of Sixth Powers of Odd Integers }} {{eqn | r = \frac {\pi^6} {960} - \frac 1 {64} \times \frac {\pi^6} {945}       | c = Riemann Zeta Function of 6 }} {{eqn | r = \frac {\pi^6 \paren {63 - 1} } {60 \, 480} }} {{eqn | r = \frac {31 \pi^6} {30 \, 240} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^4}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n}\right)^4} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^4}       | c =  }} {{eqn | r = \frac 1 {16} \sum_{n \mathop = 1}^\infty \frac 1 {n^4} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^4}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\pi^4} {90}       | r = \frac 1 {16} \times \dfrac {\pi^4} {90} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^4}       | c = Riemann Zeta Function of 4 }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^4}       | r = \dfrac {\pi^4} {90} - \frac 1 {16} \times \dfrac {\pi^4} {90} }}  {{eqn | r = \dfrac {\left({16 - 1}\right) \pi^4} {90 \times 16} }} {{eqn | r = \dfrac {\pi^4} {96} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^6}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n}\right)^6} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^6}       | c =  }} {{eqn | r = \frac 1 {64} \sum_{n \mathop = 1}^\infty \frac 1 {n^6} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^6}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\pi^6} {945}       | r = \frac 1 {64} \times \dfrac {\pi^6} {945} + \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^6}       | c = Riemann Zeta Function of 6 }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^6}       | r = \dfrac {\pi^6} {945} - \frac 1 {64} \times \dfrac {\pi^6} {945} }}  {{eqn | r = \dfrac {\left({64 - 1}\right) \pi^6} {945 \times 64} }} {{eqn | r = \dfrac {\pi^6} {960} }} {{end-eqn}} {{qed}}
By definition of half-range Fourier sine series: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$ where for all $n \in \Z_{> 0}$: :$b_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$ Thus by definition of $f$: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x       | c =  }} {{eqn | r = \frac 2 \pi \int_0^\pi x \paren {\pi - x} \sin n x \rd x       | c = Definition of $f$ }} {{eqn | r = 2 \int_0^\pi x \sin n x \rd x - \frac 2 \pi \int_0^\pi x^2 \sin n x \rd x       | c = Linear Combination of Definite Integrals and simplifying }} {{end-eqn}} Splitting it up into two: {{begin-eqn}} {{eqn | l = 2 \int_0^\pi x \sin n x \rd x       | r = 2 \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi       | c = Primitive of $x \sin n x$ }} {{eqn | r = 2 \paren {\paren {\frac {\sin n \pi} {n^2} - \frac {\pi \cos n \pi} n} - \paren {\frac {\sin 0} {n^2} - \frac {0 \cos 0} n} }       | c =  }} {{eqn | r = -2 \frac {\pi \cos n \pi} n       | c = Sine of Multiple of Pi and removing vanishing terms }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \frac 2 \pi \int_0^\pi x^2 \sin n x \rd x       | r = 2 \intlimits {\frac {2 x \sin n x} {n^2} + \paren {\frac 2 {n^3} - \frac {x^2} n} \cos n x} 0 \pi       | c = Primitive of $x^2 \sin n x$ }} {{eqn | r = \frac 2 \pi \paren {\paren {\frac {2 \pi \sin n \pi} {n^2} + \paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi} - \paren {\frac {0 \sin 0} {n^2} + \paren {\frac 2 {n^3} - \frac {0^2} n} \cos 0} }       | c =  }} {{eqn | r = \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} \cos 0}       | c = Sine of Multiple of Pi and removing vanishing terms }} {{eqn | r = \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} }       | c = Cosine of Zero is One }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = b_n       | r = -2 \frac {\pi \cos n \pi} n - \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} }       | c =  }} {{eqn | r = \frac {4 \paren {1 - \cos n \pi} } {\pi n^3} - 2 \frac {\pi \cos n \pi} n + \frac {2 \pi \cos n \pi} n       | c = simplifying }} {{eqn | r = \frac {4 \paren {1 - \paren {-1}^n} } {\pi n^3}       | c = Cosine of Multiple of Pi and simplifying }} {{end-eqn}} When $n$ is even, $\paren {-1}^n = 1$ and so $\dfrac {4 \paren {1 - \paren {-1}^n} } {\pi n^3} = 0$. When $n$ is odd, $n$ can be expressed as $n = 2 r + 1$ for $r \ge 0$. Thus: {{begin-eqn}} {{eqn | l = b_n       | r = \frac {4 \paren {1 - \paren {-1}^{2 r + 1} } } {\pi \paren {2 r + 1}^3}       | c =  }} {{eqn | r = \frac {4 \paren {1 + 1} } {\pi \paren {2 r + 1}^3}       | c = as $\paren {-1}^{2 r + 1} = -1$ }} {{eqn | r = \frac 8 {\pi \paren {2 r + 1}^3}       | c =  }} {{end-eqn}} Hence: :$\displaystyle \map f x \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 r + 1} x} {\paren {2 r + 1}^3}$ {{qed}}
By Half-Range Fourier Sine Series for $x \paren {\pi - x}$ over $\openint 0 \pi$: :$\displaystyle x \paren {\pi - x} = \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} x} {\paren {2 n + 1}^3}$ for $x \in \openint 0 \pi$. Setting $x = \dfrac \pi 2$: {{begin-eqn}} {{eqn | l = \frac \pi 2 \paren {\pi - \frac \pi 2}       | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} \frac \pi 2} {\paren {2 n + 1}^3} }} {{eqn | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {n + \frac 1 2} \pi} {\paren {2 n + 1}^3}       | c =  }} {{eqn | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^3}       | c = Sine of Half-Integer Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 4 \times \frac \pi 8        | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^3}       | c =  }} {{end-eqn}} whence the result. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty x^{s - 1} \map \sech x \rd x 	| r = 2 \int_0^\infty \frac { x^{s - 1} } {e^x + e^{-x} } \rd x 	| c = {{Defof|Hyperbolic Secant}}  }} {{eqn	| r = 2 \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {- e^{-2 x} } } \rd x }} {{eqn	| r = 2 \int_0^\infty x^{s - 1} e^{-x} \sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-2 n x} \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x 	| c = Fubini's Theorem }} {{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty \paren {\frac t {2 n + 1} }^{s - 1} e^{-t} \frac {\rd t} {2 n + 1} 	| c = substituting $t = \paren {2 n + 1} x$ }} {{eqn	| r = 2 \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} \int_0^\infty t^{s - 1} e^{-t} \rd t }} {{eqn	| r = 2 \map \Gamma s \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} 	| c = {{Defof|Gamma Function}} }}  {{end-eqn}} So:  :$\displaystyle \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}$ {{qed}} Category:Hyperbolic Secant Function Category:Definite Integrals Category:Sums of Sequences jwfpp7u2qgwm1uxknt52a8db1tiey6s
By Half-Range Fourier Sine Series for $x \paren {\pi - x}$ over $\openint 0 \pi$: :$\displaystyle x \paren {\pi - x} = \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} x} {\paren {2 n + 1}^3}$ for $x \in \openint 0 \pi$. Setting $x = \dfrac \pi 4$: {{begin-eqn}} {{eqn | l = \frac \pi 4 \paren {\pi - \frac \pi 4}       | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} \frac \pi 4} {\paren {2 n + 1}^3} }} {{eqn | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\map \sin {\frac {n \pi} 2 + \frac \pi 4} } {\paren {2 n + 1}^3} }} {{eqn | r = \frac 8 \pi \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 \cos \frac \pi 4 + \cos \frac {n \pi} 2 \sin \frac \pi 4} {\paren {2 n + 1}^3} }} {{eqn | r = \frac 8 \pi \times \frac 1 {\sqrt 2} \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}       | c = Cosine of $45^\circ$, Sine of $45^\circ$ }} {{eqn | ll= \leadsto       | l = \frac {3 \pi^2} {16} \times \frac \pi 8 \times \sqrt 2       | r = \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}       | c =  }} {{eqn | r = \frac {3 \pi^3 \sqrt 2} {128} }} {{end-eqn}} Now we need to show that: :$\displaystyle \frac 1 {1^3} + \frac 1 {3^3} - \frac 1 {5^3} - \frac 1 {7^3} + \cdots = \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}$ For $n \equiv 0 \pmod 4$: :$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \sin 2 k \pi + \cos 2 k \pi = 1$ For $n \equiv 1 \pmod 4$: :$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac \pi 2} + \map \cos {2 k \pi + \dfrac \pi 2} = 1$ For $n \equiv 2 \pmod 4$: :$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \pi} + \map \cos {2 k \pi + \pi} = -1$ For $n \equiv 3 \pmod 4$: :$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac {3 \pi} 2} + \map \cos {2 k \pi + \dfrac {3 \pi} 2} = -1$ This shows that the {{RHS}} sum is alternating in sign in pairs. Hence the result. {{qed}}
We observe that: {{begin-eqn}} {{eqn | l = \frac 1 {2 j + 1} - \frac 1 {2 j + 3}       | r = \frac {\paren {2 j + 3} - \paren{2 j + 1} } {\paren {2 j + 1} \paren {2 j + 3} }       | c =  }} {{eqn | r = \frac 2 {\paren {2 j + 1} \paren {2 j + 3} }       | c =  }} {{end-eqn}} and that $\displaystyle \sum_{j \mathop = 0}^n \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3}}$ is a telescoping series.  Therefore: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^n \frac 1 {\paren {2 j + 1} \paren {2 j + 3} }       | r = \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }       | c = Telescoping Series: Example 1 }} {{eqn | r = \frac 1 2 \paren {\frac 1 {2 \times 0 + 1} - \frac 1 {2 n + 3} }       | c =  }} {{eqn | r = \frac 1 2 - \frac 1 {4 n + 6}       | c =  }} {{eqn | r = \frac {\paren {2 n + 3} - 1} {4 n + 6}       | c =  }} {{eqn | r = \frac {2 n + 2} {4 n + 6}       | c =  }} {{eqn | r = \frac {n + 1} {2 n + 3}       | c =  }} {{end-eqn}} {{qed}}
=== 1 iff 2 === Follows from Image is Subset iff Subset of Preimage. {{qed|lemma}} === 2 implies 3 === Let $f^{-1}(\mathfrak n) \supseteq \mathfrak m$. We have to show that $f^{-1}(\mathfrak n) \subseteq \mathfrak m$. By Preimage of Proper Ideal of Ring is Proper Ideal, $f^{-1}(\mathfrak n)$ is a proper ideal. By Proper Ideal of Ring is Contained in Maximal Ideal, $f^{-1}(\mathfrak n)$ is contained in some maximal ideal of $A$. Because $A$ is a commutative local ring, $\mathfrak m$ is its only maximal ideal. {{qed|lemma}} === 3 implies 2 === Follows by definition of set equality. {{qed}} Category:Local Rings m8hf14x23hdt1mm4hpgfbmqy4u9umpm