hcju/proofwikiqa huggingface dataset:
Define $B = \left\{ {\complement_S\left({x^\succeq}\right): x \in S}\right\}$ By definition of lower topology: :$B$ is sub-basis of $T$. By definitions of sub-basis and basis: :$\displaystyle \mathcal B = \left\{ {\bigcap \mathcal F: \mathcal F \subseteq B, \, \mathcal F \text{ is finite}}\right\}$ is a basis. By definition of basis: :$\displaystyle \tau \subseteq \left\{ {\bigcup X: X \subseteq \mathcal B}\right\}$ Let $x \in A$, $y \in S$ such that :$y \preceq x$ By definition of open set: :$A \in \tau$ By definition of subset: :$\exists Y \subseteq \mathcal B: A = \bigcup Y$ By definition of union: :$\exists Z \in Y: x \in Z$ By definition of subset: :$\exists X \subseteq B: Z = \bigcap X \land X$ is finite. we will prove that :$\forall Q \in X: y \in Q$ Let $Q \in X$. By definition of subset: :$Q \in B$ By definition of $B$: :$\exists z \in S: Q = \complement_S\left({z^\succeq}\right)$ By definition of intersection: :$x \in Q$ By Upper Closure is Upper Set: :$z^\succeq$ is an upper set. By Complement of Upper Set is Lower Set: :$Q$ is a lower set. Thus by definition of lower set: :$y \in Q$ {{qed|lemma}} By definition of intersection: :$y \in Z$ Thus by definition of union: :$y \in A$ {{qed}}
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$. Then there exists $N \in \N$ such that: :$\size {a_m - a_n} < 1$ for all $m, n \ge N$. In particular, by the Triangle Inequality, for all $m \ge N$: {{begin-eqn}} {{eqn | l = \size {a_m}       | r = \size {a_N + a_m - a_N} }} {{eqn | o = \le        | r = \size {a_N} + \size {a_m - a_N} }} {{eqn | o = \le        | r = \size {a_N} + 1 }} {{end-eqn}} So $\sequence {a_n}$ is bounded, as claimed. {{qed}}
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$. Then there exists $N \in \N$ such that: :$\size {a_m - a_n} < 1$ for all $m, n \ge N$. Note that for $m \le N$: {{begin-eqn}} {{eqn | l = \size {a_m}       | o = \le       | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } }} {{eqn |o = <        | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1 }} {{eqn-intertext|and for $m > N$:}} {{eqn | l = \size {a_m}        | r = \size {a_N + a_m - a_N}  }} {{eqn | o = \le       | r = \size {a_N} + \size {a_m - a_N} }} {{eqn | o = <        | r = \size {a_N} + 1 }} {{eqn | o = \le        | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1 }} {{end-eqn}} Hence for all $m \in \N$: :$\size {a_m} < \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1$ Therefore $\sequence {a_n}$ is bounded, as claimed. {{qed}}
{{begin-eqn}} {{eqn	| l = \int \map {\tan^n} x \rd x 	| r = \int \map {\tan^{n - 2} } x \, \map {\tan^2} x \rd x }} {{eqn	| r = \int \map {\tan^{n - 2} } x \paren {\map {\sec^2} x - 1} \rd x 	| c = Sum of Squares of Sine and Cosine }} {{eqn	| r = \int \map {\tan^{n - 2} } x \, \map {\sec^2} x \rd x - \int \map {\tan^{n - 2} } x \rd x 	| c = Linear Combination of Integrals }} {{end-eqn}} Let: {{begin-eqn}} {{eqn 	| l = t 	| r = \map \tan x }} {{eqn	| ll= \leadsto 	| l = \frac {\d t} {\d x} 	| r = \map {\sec^2} x 	| c = Derivative of Tangent Function }} {{end-eqn}} Then:   {{begin-eqn}} {{eqn	| l = \int \map {\tan^{n - 2} } x \, \map {\sec^2} x \rd x - \int \map {\tan^{n - 2} } x \rd x 	| r = \int t^{n - 2} \rd t - \int \map {\tan^{n - 2} } x \rd x 	| c = Integration by Substitution }} {{eqn	| r = \frac {t^{n - 1} } {n - 1} - \int \map {\tan^{n - 2} } x \rd x 	| c = Primitive of Power }} {{eqn	| r = \frac {\map {\tan^{n - 1} } x} {n - 1} - \int \map {\tan^{n - 2} } x \rd x  	| c = substituting back $t \to \map \tan x$ }} {{end-eqn}} {{qed}}
By definition of closed set: :$S \setminus A$ is open. By Open Subset is Lower in Lower Topology: :$S \setminus A$ is lower. Thus by Complement of Lower Set is Upper Set and Relative Complement of Relative Complement: :$A$ is upper. {{qed}}
{{begin-eqn}} {{eqn	| l = \sinh 2 x 	| r = \frac 1 2 \paren {e^{2 x} - e^{-2 x} } 	| c = {{Defof|Hyperbolic Sine}} }} {{eqn	| r = \frac 1 2 \paren {e^x + e^{-x} } \paren {e^x - e^{-x} } 	| c = Difference of Two Squares }} {{eqn	| r = 2 \paren {\frac{e^x + e^{-x} } 2 \cdot \frac {e^x - e^{-x} } 2} }} {{eqn	| r = 2 \sinh x \cosh x 	| c = {{Defof|Hyperbolic Sine}}, {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \sinh 2 x 	| r = -i \sin 2 i x 	| c = Hyperbolic Sine in terms of Sine }} {{eqn	| r = -2 i \sin i x \cos i x 	| c = Double Angle Formula for Sine }} {{eqn	| r = 2 \sinh x \cosh x 	| c = Hyperbolic Sine in terms of Sine, Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sinh 2 x       | r = \map \sinh {x + x} }} {{eqn | r = \sinh x \cosh x + \cosh x \sinh x       | c = Hyperbolic Sine of Sum }} {{eqn | r = 2 \sinh x \cosh x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \sin 2\theta 	| r = \frac{1}{2i} \left(e^{2i\theta}-e^{-2i\theta}\right) 	| c = Sine Exponential Formulation }} {{eqn	| r = \frac{1}{2i} \left(e^{i\theta}+e^{-i\theta}\right) \left(e^{i\theta}-e^{-i\theta}\right) 	| c = Difference of Two Squares }} {{eqn	| r = 2 \left(\frac{e^{i\theta}-e^{-i\theta} }{2i} \cdot \frac{e^{i\theta} + e^{-i\theta} }{2}\right) }} {{eqn	| r = 2 \sin \theta \cos \theta 	| c = Sine Exponential Formulation, Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 2 \cosh^2 x - 1       | r = \cosh 2 x       | c = Double Angle Formula for Hyperbolic Cosine:Corollary 1 }} {{eqn | l = \cosh^2 x       | r = \frac {\cosh 2 x + 1} 2       | c = solving for $\cosh^2 x$ }} {{end-eqn}} {{qed}}
Follows from Factorial as Product of Two Factorials: :$10! = 6! \times 7!$ and so: :$\dfrac {10!} {6!} = 10 \times 9 \times 8 \times 7 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ Hence the result. {{qed}}
From 720 is Product of Consecutive Numbers in Two Ways: :$720 = 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8$ From 5040 is Product of Consecutive Numbers in Two Ways: :$5040 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8 \times 7$ Then: :$-720 = \left({-6}\right) \left({-5}\right) \left({-4}\right) \left({-3}\right) \left({-2}\right) = \left({-10}\right) \left({-9}\right) \left({-8}\right)$ The same trick cannot be used for $-5040$ because there are four divisors in $10 \times 9 \times 8 \times 7$, and negating them makes the product positive. Hence the result. {{qed}} Category:Factorials htuoyjqsfoqjoxcauiwslkjcxp6eitx
{{begin-eqn}} {{eqn | l = \cosh 3 x       | r = 4 \cosh^3 x - 3 \cosh x       | c = Triple Angle Formula for Hyperbolic Cosine }} {{eqn | ll= \leadsto       | l = 4 \cosh^3 x       | r = \cosh 3 x + 3 \cosh x       | c = rearranging }} {{eqn | ll= \leadsto       | l = \cosh^3 x       | r = \dfrac {\cosh 3 x + 3 \cosh x} 4       | c = dividing both sides by $4$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 2 \sinh^2 x + 1       | r = \cosh 2 x       | c = Double Angle Formula for Hyperbolic Cosine: Corollary 2 }} {{eqn | ll= \leadsto       | l = \sinh^2 x       | r = \frac {\cosh 2 x - 1} 2       | c = solving for $\sinh^2 x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \cosh^4 x 	| r = \frac 1 {2^4}\left(e^{x} + e^{-x}\right)^4 	| c = {{Defof|Hyperbolic Cosine}} }} {{eqn	| r = \frac 1 {16} \left({e^{4 x} + 4 e^{2 x} + 6 e^{0 x} + 4 e^{-2 x} + e^{-4 x} }\right) 	| c = Binomial Theorem }} {{eqn	| r = \frac 1 8 \left({\frac{e^{4 x} + e^{-4 x} } 2}\right) + \frac 4 8 \left({\frac{e^{2 x} + e^{-2 x} } 2 }\right) + \frac 6 {16} }} {{eqn	| r = \frac {3 + 4 \cosh 2 x + \cosh 4 x} 8 	| c = {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh 4 x       | r = \left({\cosh^2 x}\right)^2 }} {{eqn | r = \left({\frac {\cosh 2 x + 1} 2}\right)^2       | c = Square of Hyperbolic Cosine }} {{eqn | r = \frac {\cosh^2 2 x + 2 \cosh 2 x + 1} 4       | c = multiplying out }} {{eqn | r = \frac {\frac {\cosh 4 x + 1} 2 + 2 \cosh 2 x + 1} 4       | c = Square of Hyperbolic Cosine }} {{eqn | r = \frac {\cosh 4 x + 1 + 4 \cosh 2 x + 2} 8       | c = multiplying top and bottom by $2$ }} {{eqn | r = \frac {3 + 4 \cosh 2 x + \cosh 4 x} 8       | c = rearrangement }} {{end-eqn}} {{qed}}
:$5906 = \paren {\dfrac {149} {17} }^4 + \paren {\dfrac {25} {17} }^4$ Suppose $5906$ is a sum of two integer $4$th powers. We have: :$9^4 = 6561 > 5906$ which shows that no $4$th power greater than $8^4$ is in the sum. :$7^4 + 7^4 = 4802 < 5906$ which shows that some $4$th power greater than $7^4$ is in the sum. So the sum must contain $8^4$. We have: :$5906 - 8^4 = 1810$ but $1810$ is not an integer $4$th power. Therefore $5906$ is not a sum of two integer $4$th powers. {{finish|It remains to show that it is the smallest such. It is, acccording to A NEW CHARACTERIZATION OF THE INTEGER 5906 by A. Bremner and P. Morton}}
Let $\tau_K$ be the subspace topology on $K$ induced by $\tau$. Let $\tau’_K$ be the subspace topology on $K$ induced by $\tau_H$. Then {{begin-eqn}} {{eqn | l = V \in \tau’_K       | o = \leadstoandfrom       | r = \exists U’ \in \tau_H : V = U’ \cap K        | c = {{Defof|Subspace Topology}} $\tau’_K$ }} {{eqn | o = \leadstoandfrom       | r = \exists U \in \tau : V = \paren {U \cap H} \cap K        | c = {{Defof|Subspace Topology}} $\tau_H$ }} {{eqn | o = \leadstoandfrom       | r = \exists U \in \tau : V = U \cap \paren {H \cap K}        | c = Intersection is Associative }} {{eqn | o = \leadstoandfrom       | r = \exists U \in \tau : V = U \cap K        | c = Intersection with Subset is Subset }} {{eqn | o = \leadstoandfrom       | r = V \in \tau_K        | c = {{Defof|Subspace Topology}} $\tau_K$ }} {{end-eqn}} Category:Topological Subspaces mrfeh7eskyccopbrjwm7cr98zmshv6h
By assumption: :for all non-empty subsets $Y$ of $S$: $f$ preserves the infimum of $Y$. Thus by Mapping Preserves Non-Empty Infima implies Mapping is Continuous in Lower Topological Lattice: :$f$ is continuous mapping. {{qed}}
Suppose that $H$ is compact in $T'$. Let $\set {W_i}_{i \mathop \in J}$ be an open cover of $H$ in $T$. Then $\displaystyle H \subseteq \bigcup_{i \mathop \in J} W_i$. Then:  {{begin-eqn}} {{eqn | l = H       | r = H \cap K       | c = Intersection with Subset is Subset: from $H \subseteq K$ }} {{eqn | o = \subseteq       | r = \paren {\bigcup_{i \mathop \in J} W_i} \cap K       | c = Set Intersection Preserves Subsets: Corollary: from $\displaystyle H \subseteq \bigcup_{i \in J} W_i$ }} {{eqn | r = \bigcup_{i \mathop \in J} \paren {W_i \cap K}       | c = Intersection Distributes over Union: Family of Sets }} {{end-eqn}} Since $W_i \cap K \in \tau_K$, $\set {W_i \cap K}_{i \mathop \in J}$ is an open cover of $H$ in $T'$. Since $H$ is compact in $T'$, $\set {W_i \cap K}_{i \mathop \in J}$ has some finite subcover $\set {W_i \cap K}_{i \mathop = 1}^r$. Therefore: {{begin-eqn}} {{eqn | l = H       | o = \subseteq       | r = \set {W_i \cap K}_{i \mathop = 1}^r       | c = {{Defof|Finite Subcover}} }} {{eqn | o = \subseteq       | r = \paren {\bigcup_{i \mathop = 1}^r W_i} \cap K       | c = Intersection Distributes over Union: Family of Sets }} {{eqn | r = \bigcup_{i \mathop = 1}^r W_i       | c = Intersection is Subset }} {{end-eqn}} So $\set {W_i}_{i \mathop = 1}^r$ is an open cover of $H$ in $T$, which is a finite subcover. As $\set {W_i}_{i \mathop \in J}$ is arbitrary: Any open cover of $H$ has a finite subcover in $T$. So $H$ is compact in $T$. {{qed}} Category:Compact Spaces Category:Topological Subspaces n15rln3t7hfy09w6nb2fy174c5pehnf
We will prove that :$\forall A \in B, x \in F \cap A, y \in F: y \preceq x \implies y \in A$ Let $A \in B$, $x \in F \cap A$, $y \in F$. By definition of sub-basis: :$A$ is open. By Open Subset is Lower in Lower Topology: :$A$ is lower. By definition of intersection: :$x \in A$. Thus by definition of lower set: :$y \preceq x \implies y \in A$. {{qed|lemma}} Define $H := \left\{ {\bigcap G: G \subseteq B, G \text{ is finite} }\right\}$ By definitions of sub-basis and basis: :$H$ is basis of $T$. We will prove that :$\forall A \in H: \inf F \in A \implies F \cap A \ne \varnothing$ Let $A \in H$ such that :$\inf F \in A$ By definition of $H$: :$\exists G \subseteq B: A = \bigcap G \land G$ is finite. By definition of intersection: :$\forall C \in G: \inf F \in C \in B$ By assumption: :$\forall C \in G: F \cap C \ne \varnothing$ By definition of non-empty set: :$\forall C \in G: \exists x: x \in F \cap C$ By Axiom of Choice: :$\exists f:G \to F: \forall C \in G: f\left({C}\right) \in F \cap C$ By Image of Mapping from Finite Set is Finite: :$f\left[{G}\right]$ is finite. By Filtered iff Finite Subsets have Lower Bounds: :$\exists h \in F: \forall x \in f\left[{G}\right]: h \preceq x$ Then :$\forall C \in G: h \in C$ By definition of intersection: :$h \in \bigcap G$ Thus by definitions of intersection and non-empty set: :$F \cap A \ne \varnothing$ {{qed|lemma}} Thus by Characterization of Closure by Basis: :$\inf F \in F^-$ {{qed}}
Let $n$ be a tri-automorphic number with $d$ digits. Let $n = 10 a + b$. Then: :$3 n^2 = 300a^2 + 60 a b + 3 b^2$ As $n$ is tri-automorphic, we have: :$(1): \quad 300 a^2 + 60 a b + 3 b^2 = 1000 z + 100 y + 10 a + b$ and: :$(2): \quad 3 b^2 - b = 10 x$ where $x$ is an integer. This condition is only satisfied by $b = 2$, $b = 5$, or $b = 7$ {{ProofWanted|Guess: Try proving for $n {{=}} 10 a + b$ and then by induction.}} Substituting $b = 2$ in equation $(1)$: :$a = 9$ Substituting $b = 5$ in equation $(1)$: :$a = 7$ Substituting $b = 7$ in equation $(1)$: :$a = 6$ {{qed}} <!-- Corollary:  For di-automorphic numbers, equation (5) becomes 2b^2-b=10x and b=8  For tetra-automorphic numbers, equation (5) becomes 4b^2-b=10x and b=4  For penta-automorphic numbers, equation (5) becomes 5b^2-b=10x and b=5 -->
From Even Power is Even Function, $x^2$ is an even function. By Fourier Series for Even Function over Symmetric Range, we have: :$\displaystyle x^2 \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$ where: {{begin-eqn}} {{eqn | l = a_n       | r = \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x }} {{eqn | ll= \leadsto       | l = \frac 1 2 a_0       | r = \frac 2 {2 \pi} \int_0^\pi x^2 \rd x       | c = Cosine of Zero is One }} {{eqn | r = \frac 1 \pi \cdot \frac {\pi^3} 3       | c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn | r = \frac {\pi^2} 3 }} {{end-eqn}} Then:  {{begin-eqn}} {{eqn | l = \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x       | r = \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi       | c = Primitive of $x^2 \cos a x$, Fundamental Theorem of Calculus }} {{eqn | r = \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} } 0 \pi       | c = Sine of Multiple of Pi }} {{eqn | r = \frac 2 \pi \cdot \frac {2 \pi \cos n \pi} {n^2} }} {{eqn | r = \frac {4 \cos n \pi} {n^2} }} {{eqn | r = \paren {-1}^n \frac 4 {n^2}       | c = Cosine of Multiple of Pi }} {{end-eqn}} Substituting for $a_n$ in $(1)$:  :$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$ as required. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | o =        | m = 6999 + 9996       | mo= =       | r = 16995 }} {{eqn | n = 2       | o =        | m = 16995 + 59961       | mo= =       | r = 76956 }} {{eqn | n = 3       | o =        | m = 76956 + 65967       | mo= =       | r = 142923 }} {{eqn | n = 4       | o =        | m = 142923 + 329241       | mo= =       | r = 472164 }} {{eqn | n = 5       | o =        | m = 472164 + 461274       | mo= =       | r = 933438 }} {{eqn | n = 6       | o =        | m = 933438 + 834339       | mo= =       | r = 1767777 }} {{eqn | n = 7       | o =        | m = 1767777 + 7777671       | mo= =       | r = 9545448 }} {{eqn | n = 8       | o =        | m = 9545448 + 8445459       | mo= =       | r = 17990907 }} {{eqn | n = 9       | o =        | m = 17990907 + 70909971       | mo= =       | r = 88900878 }} {{eqn | n = 10       | o =        | m = 88900878 + 87800988       | mo= =       | r = 176701866 }} {{eqn | n = 11       | o =        | m = 176701866 + 668107671       | mo= =       | r = 844809537 }} {{eqn | n = 12       | o =        | m = 844809537 + 735908448       | mo= =       | r = 1580717985 }} {{eqn | n = 13       | o =        | m = 1580717985 + 5897170851       | mo= =       | r = 7477888836 }} {{eqn | n = 14       | o =        | m = 7477888836 + 6388887747       | mo= =       | r = 13866776583 }} {{eqn | n = 15       | o =        | m = 13866776583 + 38567766831       | mo= =       | r = 52434543414 }} {{eqn | n = 16       | o =        | m = 52434543414 + 41434543425       | mo= =       | r = 93869086839 }} {{eqn | n = 17       | o =        | m = 93869086839 + 93868096839       | mo= =       | r = 187737183678 }} {{eqn | n = 18       | o =        | m = 187737183678 + 876381737781       | mo= =       | r = 1064118921459 }} {{eqn | n = 19       | o =        | m = 1064118921459 + 9541298114601       | mo= =       | r = 10605417036060 }} {{eqn | n = 20       | o =        | m = 10605417036060 + 06063071450601       | mo= =       | r = 16668488486661 }} {{end-eqn}} which is palindromic. $7998$ and its reversal converge on the same sequence immediately: {{begin-eqn}} {{eqn | l = 7998 + 8997       | r = 16995 }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the longest.}}
Since $x^4 = \paren {-x}^4$, $x^4$ is an even function. By Fourier Series for Even Function over Symmetric Range, the Fourier series of $\map f x$ can be expressed as: :$x^4 \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty a_n \cos n x$ where for all $n \in \Z_{> 0}$: {{begin-eqn}} {{eqn | l = a_n        | r = \dfrac 2 \pi \displaystyle \int_0^\pi x^4 \cos n x \  \d x }} {{eqn | r = \dfrac 2 \pi \paren {\intlimits {\frac {\sin n x} n x^4 + \frac {4 \cos n x} {n^2} x^3 - \frac {12 \sin n x} {n^3} x^2 - \frac {24 \cos n x} {n^4} x + \frac {24 \sin n x} {n^5} } {x \mathop = 0} {x \mathop = \pi} }       | c = Primitive of $x^4 \cos a x$ }} {{eqn | r = \dfrac 2 \pi \paren {\frac {4 \pi^3 \cos n \pi} {n^2} - \frac {24 \pi \cos n \pi} {n^4}  }       | c =  }} {{eqn | r = \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_0       | r = \dfrac 2 \pi \int_0^\pi x^4 \  \d x }} {{eqn | r = \dfrac 2 \pi \intlimits {\frac {x^5} 5} {x \mathop = 0} {x \mathop = \pi}       | c = Primitive of Power }} {{eqn | r = \frac {2 \pi^4} 5 }} {{end-eqn}} This gives: :$x^4 \sim \dfrac {\pi^4} 5 + \displaystyle \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi \cos n x$ {{qed}} Category:Examples of Fourier Series 4corhmbfraqiwoqpxj7c5d4d2uajbe6
By definition of Fourier series: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ where: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac 1 \pi \int_0^{2 \pi} \map f x \cos n x \rd x }} {{eqn | l = b_n       | r = \dfrac 1 \pi \int_0^{2 \pi} \map f x \sin n x \rd x }} {{end-eqn}} for all $n \in \Z_{>0}$. Thus: {{begin-eqn}} {{eqn | l = a_0       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \rd x       | c = Cosine of Zero is One }} {{eqn | r = \frac 1 \pi \int_0^{2 \pi} \pi - x \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 1 \pi \intlimits {\pi x - \frac 1 2 x^2} 0 {2 \pi}       | c = Primitive of Power }} {{eqn | r = 2 \pi^2 - \frac 4 2 \pi^2       | c =  }} {{eqn | r = 0 }} {{end-eqn}} {{qed|lemma}} For $n > 0$: {{begin-eqn}} {{eqn | l = a_n       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \cos n x \rd x       | c =  }} {{eqn | r = \frac 1 \pi \int_0^{2 \pi} \paren {\pi - x} \cos n x \rd x       | c = Definition of $f$ }} {{eqn | r = \pi \int_0^{2 \pi} \cos n x \rd x - \frac 1 \pi \int_0^{2 \pi} x \cos n x \rd x       | c =  }} {{eqn | r = - \frac 1 \pi \int_0^{2 \pi} x \cos n x \rd x       | c = Integral over $2 \pi$ of $\cos n x$ }} {{eqn | r = - \frac 1 \pi \intlimits {\frac {\cos n x} {n^2} + \frac {x \sin n x} n} 0 {2 \pi}       | c = Primitive of $x \cos n x$ }} {{eqn | r = - \frac 1 \pi \paren {\cos 0 - \cos 2 \pi}       | c = Sine of Integer Multiple of Pi }} {{eqn | r = 0       | c = Sine and Cosine are Periodic on Reals }} {{end-eqn}} {{qed|lemma}} Now for the $\sin n x$ terms: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin n x \rd x       | c =  }} {{eqn | r = \frac 1 \pi \int_0^{2 \pi} \paren {\pi - x} \sin n x \rd x       | c = Definition of $f$ }} {{eqn | r = \int_0^{2 \pi} \sin n x \rd x - \frac 1 \pi \int_0^{2 \pi} x \sin n x \rd x       | c =  }} {{eqn | r = - \frac 1 \pi \int_0^{2 \pi} x \sin n x \rd x       | c = Integral over $2 \pi$ of $\sin n x$ }} {{eqn | r = - \frac 1 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 {2 \pi}       | c = Primitive of $x \sin n x$ }} {{eqn | r = \frac 1 \pi \frac {2 \pi \cos 2 n \pi} n       | c = Sine of Multiple of Pi }} {{eqn | r = \frac 2 n       | c = Cosine of Multiple of Pi }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 2 n \sin n x       | c = substituting for $a_0$, $a_n$ and $b_n$ from above }} {{eqn | r = 2 \sum_{n \mathop = 1}^\infty \frac {\sin n x} n        | c = rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^{14} k^3       | r = \paren {\dfrac {14 \paren {14 + 1} } 2}^2       | c = Sum of Sequence of Cubes }} {{eqn | r = 11 \, 025       | c =  }} {{eqn | l = \sum_{k \mathop = 1}^{10} k^3       | r = \paren {\dfrac {10 \paren {10 + 1} } 2}^2       | c = Sum of Sequence of Cubes }} {{eqn | r = 3025       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 11}^{14} k^3       | r = \sum_{k \mathop = 1}^{14} k^3 - \sum_{k \mathop = 1}^{10} k^3       | c =  }} {{eqn | r = 11 \, 025 - 3025       | c =  }} {{eqn | r = 8000       | c =  }} {{eqn | r = 20^3       | c =  }} {{end-eqn}} {{qed}}