hcju/proofwikiqa huggingface dataset:
Let $\epsilon \in \R_{>0}$. === Lemma === {{:Closed Balls Centered on P-adic Number is Countable/Lemma}} From Closed Ball contains Smaller Closed Ball: :$\map {B^-_{p^{-n}}} a \subseteq \map {B^-_\epsilon} a$ From Open Ball contains Strictly Smaller Closed Ball: :$\map {B^-_\epsilon} a \subseteq \map {B_{p^{-\paren {n - 1} } } } a$ From Open Ball in P-adic Numbers is Closed Ball :$\map {B_{p^{-\paren{n-1}}}} a = \map {B^-_{p^{-n} } } a$ Hence: :$\map {B^-_\epsilon} a \subseteq \map {B^-_{p^{-n} } } a$ By definition of set equality: :$\map {B^-_\epsilon} a = \map {B^{\, -}_{p^{-n} } } a$ Since $\epsilon \in \R_{>0}$ was arbitrary then: :$\forall \epsilon \in \R_{>0} : \exists n \in \Z : \map {B^-_\epsilon} a = \map {B^-_{p^{-n} } } a$ Hence the set of all closed balls centered on $a$ is: :$\BB^- = \set {\map {B^-_{p^{-n} } } a : n \in \Z}$ From Surjection from Countably Infinite Set iff Countable, it follows that $\BB^-$ is a countable set. {{qed}} Category:Topology of P-adic Numbers Category:Closed Balls Centered on P-adic Number is Countable qxwgv5j5qqatseakf75a81t2fm2673p
Let $\epsilon \in \R_{\ge 0}$. === Lemma === {{:Closed Balls Centered on P-adic Number is Countable/Open Balls/Lemma}} From Open Ball contains Smaller Open Ball: :$\map {B_\epsilon} a \subseteq \map {B_{p^{-n} } } a$ From Open Ball contains Strictly Smaller Closed Ball: :$\map {B^-_{p^{-\paren {n + 1} } } } a \subseteq \map {B_\epsilon} a$ From Open Ball in P-adic Numbers is Closed Ball :$\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$ Hence: :$\map {B_{p^{-n} } } a \subseteq \map {B_\epsilon} a$ By definition of set equality: :$\map {B_\epsilon} a = \map {B_{p^{-n} } } a$ Since $\epsilon \in \R_{>0}$ was arbitrary then: :$\forall \epsilon \in \R_{>0} : \exists n \in \Z : \map {B_\epsilon} a = \map {B_{p^{-n} } } a$ Hence the set of all open balls centered on $a$ is: :$\BB = \set {\map {B_{p^{-n} } } a : n \in \Z}$ From Surjection from Countably Infinite Set iff Countable, it follows that $\BB$ is a countable set. {{qed}} Category:Topology of P-adic Numbers Category:Closed Balls Centered on P-adic Number is Countable 83loljw38qt8v4l4yzf3c0pmabfggps
From Sequence of Powers of Reciprocals is Null Sequence: :$\exists n_1 \in \N: \forall k \ge n_1 : p^{-k} < \epsilon$ Similarly: :$\exists n_2 \in \N: \forall k \ge n_2 : p^{-k} < \dfrac 1 \epsilon$ Hence:  :$p^{-n_1} < \epsilon$ and $ p^{-n_2} < \dfrac 1 \epsilon$ That is: :$p^{-n_1} < \epsilon < p^{n_2}$ From Power Function on Integer between Zero and One is Strictly Decreasing: :$-n_2 < n_1$. Let: :$n = \min \set {k : -n_2 \mathop \le k \mathop \le n_1 \text{ and } p^{-k} \le \epsilon}$ By choice of $n$: :$p^{-n} \le \epsilon < p^{n_2}$ From Power Function on Integer between Zero and One is Strictly Decreasing: :$n > -n_2$ Also by the choice of $n$: :$n \le n_1$ So: :$n_1 > n - 1 \ge -n_2$ {{AimForCont}}: :$p^{-\paren {n - 1} } \le \epsilon$ Then: :$n - 1 \in \set {k: -n_2 \le k \le n_1 \text{ and } p^{-k} \le \epsilon}$   So: :$n = \min \set {k: -n_2 \le k \le n_1 \text { and } p^{-k} \le \epsilon} \le n - 1$ which is a contradiction. Hence: :$\epsilon < p^{-\paren {n - 1} }$ The result follows. {{qed}} Category:Closed Balls Centered on P-adic Number is Countable fvc5hobq613kwnxyja592u2yufpzmhy
From Lemma for Closed Balls: :$\exists m \in \Z : p^{-m} \le \epsilon < p^{-\paren {m - 1} }$ Suppose $\epsilon \ne p^{-m}$. Then: :$p^{-m} < \epsilon < p^{-\paren {m - 1} }$ and the theorem is proved with $n = m - 1$. Now suppose $\epsilon = p^{-m}$.   From Power Function on Integer between Zero and One is Strictly Decreasing: :$p^{-\paren{m + 1}} < p^{-m}$ So: :$p^{-\paren{m + 1}} < \epsilon \le p^{-m}$ and the theorem is proved with $n = m$. {{qed}} Category:Closed Balls Centered on P-adic Number is Countable 0h67ftslpo0m83dphe84sbcbyw5vlza
{{begin-eqn}} {{eqn | l = x \in \map {B_\epsilon} a        | o = \leadstoandfrom       | r = \map d {x, a} < \epsilon       | c = {{defof|Open Ball|open ball}} }} {{eqn | o = \leadsto       | r = \map d {x, a} < \delta       | c = As $\epsilon \le \delta$ }} {{eqn | o = \leadstoandfrom       | r = x \in \map {B_\delta} a        | c = {{defof|Open Ball|open ball}} }} {{end-eqn}} By definition of subset: :$\map {B_\epsilon} a \subseteq \map {B_\delta} a$ {{qed}} Category:Open Balls 7w688nz78wvhfuywyxx20ked0x27cnw
{{begin-eqn}} {{eqn | l = x \in \map {B^-_\epsilon} a        | o = \leadsto       | r = \map d {x, a} \le \epsilon       | c = {{Defof|Closed Ball}} }} {{eqn | o = \leadsto       | r = \map d {x, a} \le \delta       | c = As $\epsilon \le \delta$ }} {{eqn | o = \leadsto       | r = x \in \map {B^-_\delta} a        | c = {{Defof|Closed Ball}} }} {{end-eqn}} By definition of subset: :$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$ {{qed}} Category:Metric Spaces i3895kegu0xmjo9zyhdjqdtsqt5p3y1
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {B^-_\epsilon} a        | c = {{Defof|Closed Ball}} }} {{eqn | ll= \leadsto       | l = \map d {x, a}       | o = \le       | r = \epsilon       | c = {{Defof|Closed Ball}} }} {{eqn | ll= \leadsto       | l = \map d {x, a}       | o = <       | r = \delta       | c = As $\epsilon < \delta$ }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \map {B_\delta} a       | c = {{Defof|Open Ball}} }} {{end-eqn}} By definition of subset: :$\map {B^-_\epsilon} a \subseteq \map {B_\delta} a$ {{qed}} Category:Open Balls Category:Closed Balls 3s9zktusrpgqeqjklteinaazlyp8r8b
{{begin-eqn}} {{eqn | l = x \in \map {B_\epsilon} a        | o = \leadsto       | r = \map d {x, a} < \epsilon       | c = {{Defof|Open Ball}} }} {{eqn | o = \leadsto       | r = \map d {x, a} < \delta       | c = As $\epsilon \le \delta$ }} {{eqn | o = \leadsto       | r = \map d {x, a} \le \delta       | c = {{Defof|Total Ordering}} }} {{eqn | o = \leadsto       | r = x \in \map {B^-_\delta} a        | c = {{Defof|Closed Ball}} }} {{end-eqn}} By definition of subset: :$\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$ {{qed}} Category:Metric Spaces nfzy95wgtx4yur4j4l4l8hiemv6gr3g
Let's apply the definition of linear independence. Assume a linear combination of the functions $f_1,\ldots,f_n$ is the zero function: {{begin-eqn}} {{eqn | n = 1       | l = \displaystyle \sum_{i \mathop = 1}^n c_i \, \map {f_i} x        | r = 0       | c = for all $x$ }} {{end-eqn}} Let $\vec c$ have components $c_1, \ldots, c_n$. For $i = 1, \ldots, n$ replace $x = x_i$ in (1). There are $n$ linear homogeneous algebraic equations, written as: :$S \vec c = \vec 0$ Because $S$ is invertible, then $\vec c = \vec 0$. The functions are linearly independent. {{qed}}
By definition the $p$-adic integers $\Z_p$ is the closed ball $\map {B^-_1} 0$. From Open and Closed Balls in P-adic Numbers are Compact Subspaces, $\map {B^-_1} 0$ is compact. {{qed}}
Let $U$ be an open set of $M$ which has $x$ as an element. Then by definition of an open set: :$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$ From Open Ball of Metric Space is Open Set, $\BB_x$ is a set of open set which have $x$ as an element. By definition of a local basis, $\BB_x$ is a local basis of $x$. {{qed}} Category:Open Balls n2gebf0i8dwl8gaa0qxbfvgnozetg1q
From Local Basis of P-adic Number the set $\set { \map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of clopen sets. From Open Balls of P-adic Number: :$\set {\map {B_{p^{-n} } } a : n \in \Z} = \set {a + p^{n + 1} \Z_p : n \in \Z} = \set {a + p^n \Z_p : n \in \Z}$ The result follows. {{qed}}
From Local Basis of P-adic Number the set $\set {\map {B_{p^{-n } } } a: n \in \Z}$ is a local basis of clopen sets. From Open Ball in P-adic Numbers is Closed Ball: :$\set {\map {B_{p^{-n} } } a: n \in \Z} = \set {\map {B^-_{p^{-\paren{n + 1} } } } a: n \in \Z} = \set {\map {B^-_{p^{-n} } } a : n \in \Z}$ The result follows. {{qed}} Category:Topology of P-adic Numbers sa9vv0afr1gs87t6jfdivj12cgp2ufn
By definition of a coherent sequence: :$\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$ From Zero Padded Basis Representation, for all $n \in \N$ there exists a sequence $\sequence{b_{j,n}}_{0 \le j \le n} :$ :$(1) \quad \displaystyle \alpha_n = \sum_{j \mathop = 0}^{n} b_{j,n} p^j$ :$(2) \quad \forall j \in \closedint 0 n : 0 \le b_{j,n} < p$ === Lemma === {{:Coherent Sequence is Partial Sum of P-adic Expansion/Lemma}}{{qed|lemma}} For all $n \in \N$, let: :$d_n = b_{n, n}$ Then: :$\forall n \in \N : 0 \le d_n < p$ By definition: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$  is a $p$-adic expansion. From the lemma: :$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$ {{qed}}
By definition of a coherent sequence: :$\forall n \in \N_{>0}: \alpha_n \equiv \alpha_{n - 1} \pmod {p^n}$ That is: :$\forall n \in \N_{>0}: \exists c_n \in \Z : \alpha_n - \alpha_{n - 1} = c_n p^n$ So it remains to show that: :$\forall n \in \N_{>0} : 0 \le c_n < p$ {{AimForCont}} for some $N \in \N_{>0}$: :$c_N \ge p$ Then: :$c_N p^N \ge p^{N + 1}$ By definition of a coherent sequence: :$\alpha_{N - 1} \ge 0$ Then: :$\alpha_N = c_N p^N + \alpha_{N - 1} \ge p^{N + 1}$ This contradicts the coherent sequence condition: :$\alpha_N \le p^{N + 1} - 1$ So: :$\forall n \in \N_{>0} : c_n < p$ {{AimForCont}} for some $M \in \N_{>0}$: :$c_M < 0$ Then: :$-c_M \ge 1$ Now: :$c_M p^M + \alpha_{M - 1} = \alpha_M \ge 0$ So: :$\alpha_{M - 1} \ge -c_M p^M \ge p^M$ This contradicts the coherent sequence condition: :$\alpha_{M - 1} < p^M$ So: :$\forall n \in \N_{>0} : c_n \ge 0$ {{qed}} Category:P-adic Number Theory npzgu5n3mwvv2o5977ffbrj5a3dez3t
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero. For $1\le i \le n$ express polynomial $p_i$ as: :$\displaystyle \map {p_i} {x} = \sum_{k \mathop = 1}^n a_{ik} x^{k-1}$ Then: {{begin-eqn}} {{eqn   | l = \left( \map {p_i} {x_j} \right)    | r = \left( a_{ij} \right)V_x   | c =  Definition:Matrix Multiplication }} {{eqn   | l = P   | r = \left( a_{ij} \right) V_x   | c = Because $\map {p_i} {x_j} = 0$ for $i \ne j$. }} {{eqn   | l = \left( a_{ij} \right)    | r = P V_x^{-1}   | c = Solve for matrix $\left( a_{ij} \right)$ }} {{eqn   | l = \left( \map {p_i} {y_j} \right)   | r = \left( a_{ij} \right)V_y   | c = Definition:Matrix Multiplication }} {{eqn   | l = \left( p_i\left( y_j \right) \right)   | r = P V_x^{-1} V_y   | c = Substitute $\left( a_{ij} \right) = P V_x^{-1}$. }} {{end-eqn}} Use second equation $\map {p_i} {y_j} = \dfrac{ \map {p} {y_j} }{y_j - x_i}$: {{begin-eqn}} {{eqn   | l = \left( \map {p_i} {y_j} \right)   | r = -CQ   | c = Definition:Matrix Multiplication }} {{eqn   | l = -CQ   | r = P V_x^{-1} V_y   | c = Equate competing equations for $\left( \map {p_i} {y_j} \right)$. }} {{eqn   | l = C   | r = -P V_x^{-1} V_y Q^{-1}   | c = Solve for $C$. }} {{end-eqn}} {{qed}}
Let $\pi \in S_n$. From Existence and Uniqueness of Cycle Decomposition, $\pi$ can be uniquely expressed as a cycle decomposition, up to the order of factors. From K-Cycle can be Factored into Transpositions, each one of the cyclic permutations that compose this cycle decomposition can be expressed as a product of transpositions. The result follows. {{qed}}
Apply Vandermonde Matrix Identity for Cauchy Matrix and Hilbert Matrix is Cauchy Matrix. Matrices $V_x$ and $V_y$ are invertible by Inverse of Vandermonde Matrix. Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero. {{qed}}
Let $\N_k$ denote the set $\set {1, 2, \ldots, k}$. Let $\pi$ be a $k$-cycle in $S_k$. By definition, $\pi$ moves all $k$ elements of $\set {1, 2, \ldots, k}$. Let $j \in \N_k$ be an arbitrary element of $\N_k$. By definition of $k$-cycle, $k$ is the smallest integer such that $\map {\pi^k} j = j$. Thus for each $r \in \set {1, 2, \ldots, k}$, $\map {\pi^r} j$ is a different element of $\N_k$. That is: :$\set {\map {\pi^r} j: r \in \N_k} = \N_k$ Let $\pi$ and $\rho$ be $k$-cycles in $S_k$ such that $\pi \ne \rho$. From the above, the distinguishing feature of $\rho$ and $\pi$ is the order of elements of $\N_k \setminus \set j$ that $\map {\pi^r} j$ and $\map {\rho^r} j$ move $j$ to for $1 \le r < k$. Each different ordered $\paren {k - 1}$-tuple of $\N_k \setminus \set j$ corresponds to a different $k$-cycle of $S_k$ Thus there are as many $k$-cycles of $S_k$ as there are different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$. From Number of Permutations, there are $\paren {k - 1}!$ different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$. Hence the result. {{qed}} Category:Cyclic Permutations rfb5xxt0n8pdnr09d64h4ybcvmr8j4l
If ones appear in a row of $\mathbf A$, then replace $\mathbf A$ by $\mathbf A^T$ and $\mathbf B$ by $\mathbf B^T$. Assume $\mathbf A$ has a column of ones. Apply Sum of Elements of Invertible Matrix to the inverse $\mathbf B = \mathbf A^{-1}$: :$\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$ where $\mathbf J_n$ denotes the square ones matrix of order $n$. If $\mathbf A = \mathbf B^{-1}$ has a column of ones, then $\mathbf B^{-1} - \mathbf J_n$ has a column of zeros, implying determinant zero. Substitute $\map \det {\mathbf B^{-1} - \mathbf J_n} = 0$ in Sum of Elements of Invertible Matrix: :$\displaystyle \sum_{i \mathop  = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - 0$ which implies the statement. {{qed}}
The theorem is proved by induction: === Basis for the Induction === $n = 0$  By definition: $\alpha_0 = b_{0, 0}$ This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\alpha_k = \displaystyle \sum_{i \mathop = 0}^k b_{i, i} p^i$ Now we need to show true for $n = k + 1$: :$\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$ === Induction Step === This is our induction step. From Difference of Consecutive terms of Coherent Sequence there exists $c_{k + 1} \in \Z$: :$(a) \quad 0 \le c_{k + 1} < p$ :$(b) \quad \alpha_{k + 1} = c_{k + 1} p^{k + 1} + \alpha_k$ From the induction hypothesis: :$\alpha_{k + 1} = c_{k + 1} p^{k + 1} + \sum_{j \mathop = 0}^k b_{j, j} p^j$ From the Zero Padded Basis Representation it follows that the two sequences: :$\sequence {b_{j, k + 1} }_{0 \le j \le k + 1}$ and :$\sequence {b_{0, 0}, b_{1, 1}, \dots, b_{k - 1, k - 1}, b_{k, k}, c_{k + 1}}$ are equal. Then: :$\forall j \in \closedint 0 {k + 1} : b_{j, k + 1} = b_{j, j}$ It follows that: :$\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$ By induction: :$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n b_{i, i} p^i$ {{qed}}
We see the cases $0$ and $1$ above hold. Suppose $N > 1$ is equal to the sum of the squares of their digits when expressed in base $10$. Since $N^2 > N$, $N$ cannot be a $1$-digit integer. Suppose $N$ is a $2$-digit integer. Write $N = \sqbrk {a b} = 10 a + b$. Then we have $a^2 + b^2 = 10 a + b$. This can be reduced to $b \paren {b - 1} = a \paren {10 - a}$. Since $b \paren {b - 1}$ is even, $a$ must be even as well. The list of possible values of $a \paren {10 - a}$ are: :$2 \paren {10 - 2} = 8 \paren {10 - 8} = 16$ :$4 \paren {10 - 4} = 6 \paren {10 - 6} = 24$ The list of possible values of $b \paren {b - 1}$ are: :$b \paren {b - 1} \le 4 \paren {4 - 1} = 12$ for $b \le 4$ :$5 \paren {5 - 1} = 20$ :$b \paren {b - 1} \ge 6 \paren {6 - 1} = 30$ for $b \ge 6$ We see that they do not coincide. Thus $N$ cannot be a $2$-digit integer. Suppose $100 \le N \le 199$. Write $N = \sqbrk {1 a b} = 100 + 10 a + b$. Then we have $1^2 + a^2 + b^2 = 100 + 10 a + b$. This can be reduced to $b \paren {b - 1} = a \paren {10 - a} + 99$. But we have $b \paren {b - 1} < 9 \times 8 = 72 < 99 \le a \paren {10 - a} + 99$. So $N$ cannot be in this range. Suppose $200 \le N \le 299$. Then the sum of the squares of their digits cannot exceed $2^2 + 9^2 + 9^2 = 186 < 200$. So $N$ cannot be in this range. Suppose $300 \le N \le 999$. Then the sum of the squares of their digits cannot exceed $9^2 + 9^2 + 9^2 = 243 < 300$. So $N$ cannot be in this range. Suppose $N$ is a $k$-digit integer with $k \ge 4$. Then the sum of the squares of their digits cannot exceed $9^2 \times k$. We have: {{begin-eqn}} {{eqn | l = N       | o = \ge       | r = 10^{k - 1}       | c = The smallest $k$-digit number }} {{eqn | o = \ge       | r = 100 \paren {1 + 9 \paren {k - 3} }       | c = Bernoulli's Inequality }} {{eqn | o = >       | r = 81 \paren {k + 8 k - 26} }} {{eqn | o = >       | r = 9^2 \times k       | c = From $k \ge 4$ }} {{end-eqn}} So $N$ cannot be greater than $1$. {{qed}}
From P-adic Metric on P-adic Numbers is Non-Archimedean Metric the $p$-adic metric is a metric on $\Q_p$. By definition, the sequence $\sequence {x_n}$ converges in $\Q_p$ {{iff}}: :$\sequence {x_n}$ converges in the $p$-adic metric.  The result then follows from Convergent Sequence in Metric Space has Unique Limit. {{qed}} Category:Definitions/Sequences Category:Definitions/Convergence Category:Definitions/P-adic Number Theory 3slwirzf4i3l5txg7zadqvy22emxzlf
We find the probability that the two random numbers not less than $n$ is coprime. We pick randomly $x, y \in \set {1, \dots, n}$. There are $n^2$ ways to do this. There are three cases: If $x > y$, for each $x$, there are $\map \phi x$ values of $y$ so that $x \perp y$. Since $1 \le x \le n$, there are a total of $\map \phi 1 + \dots + \map \phi n = \map \Phi n$ valid values of $y$. If $x < y$, the result is similar: there are $\map \Phi n$ valid values of $x$. If $x = y$, $x \perp y$ only if $x = y = 1$. Therefore the probability that the two random numbers not less than $n$ is coprime is: :$\dfrac {\map \Phi n + \map \Phi n + 1} {n^2}$ Taking limit to infinity and using Probability of Two Random Integers having no Common Divisor: {{begin-eqn}} {{eqn | l = \frac 6 {\pi^2}       | r = \lim_{n \mathop \to \infty} \frac {\map \Phi n + \map \Phi n + 1} {n^2} }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {2 \map \Phi n} {n^2} + \lim_{n \mathop \to \infty} \frac 1 {n^2} }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {2 \map \Phi n} {n^2}       | c = Basic Null Sequences }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \frac {\map \Phi n} {n^2}       | r = \frac 3 {\pi^2} }} {{end-eqn}} {{qed}} {{expand|Add an analytic number theoretic proof using Möbius Inversion Formula, or otherwise}}
By Complex Sine Function is Entire, we have that $\sin$ is a entire function.  {{AimForCont}} that $\sin$ was a bounded function. Then, by Liouville's Theorem, we would have that $\sin$ is a constant function. However we have, for instance, by Sine of Zero is Zero:  :$\sin 0 = 0$ and by Sine of 90 Degrees:  :$\sin \dfrac \pi 2 = 1$ Therefore, $\sin$ is clearly not a constant function, a contradiction. We hence conclude, by Proof by Contradiction, that $\sin$ is unbounded. {{qed}} Category:Sine Function seuz2itjkzi3bv5b5rvhne7kz8jmf3d
{{begin-eqn}} {{eqn	| l = \arccsc x 	| r = \map \arcsin {\frac 1 x} 	| c = Arccosecant of Reciprocal equals Arcsine }} {{eqn	| r = -i \map \Ln {\sqrt {1 - \paren {\frac 1 x}^2} + i \times \frac 1 x} 	| c = Arcsine Logarithmic Formulation }} {{eqn	| r = -i \map \Ln {\sqrt {1 - \frac 1 {x^2} } + \frac i x} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \arcsec x  	| r = \map \arccos {\frac 1 x} 	| c = Arcsecant of Reciprocal equals Arccosine }} {{eqn	| r = -i \map \Ln {i \sqrt {1 - \paren {\frac 1 x}^2} + \frac 1 x} 	| c = Arccosine Logarithmic Formulation }} {{eqn	| r = -i \map \Ln {i \sqrt {1 - \frac 1 {x^2} } + \frac 1 x} }} {{end-eqn}} {{qed}}
The Abi-Khuzam Inequality states that :$\sin A \cdot \sin B \cdot \sin C \le  \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 A \cdot B \cdot C$ The maximum value of $A B C - 8 \omega^3$ occurs when two of the angles are equal. So taking $A = B$, and using $A + B + C = \pi$, the maximum occurs at the maximum of: :$\map f A = A^2 \paren {\pi - 2 A} - 8 \paren {\map \arccot {2 \cot A - \cot 2 A} }^3$ 	 which occurs when: :$2 A \paren {\pi - 3 A} - \dfrac {48 \paren {\map \arccot {\frac 1 2 \paren {3 \cot A + \tan A} } }^2 \paren {1 + 2 \cos 2 A} } {5 + 4 \cos 2 A} = 0$ {{finish|Needs expanding and completing}}
{{begin-eqn}} {{eqn | l = \lim_{x \mathop \to \infty} \paren {x - \map \Gamma {\frac 1 x} }       | r = \lim_{x \mathop \to 0} \paren {\frac 1 x - \map \Gamma x} }} {{eqn | r = \lim_{x \mathop \to 0} \paren {\frac 1 x - \frac {\map \Gamma {x + 1} } x}       | c = Gamma Difference Equation }} {{eqn | r = \lim_{x \mathop \to 0} \paren {\frac {\map \Gamma 1 - \map \Gamma {1 + x} } x}       | c = $\map \Gamma 1 = 1$ as Gamma Function Extends Factorial }} {{eqn | r = -\map {\Gamma'} 1       | c = Definition of Derivative }} {{eqn | r = \gamma       | c = Derivative of Gamma Function at 1 }} {{end-eqn}} {{qed}}
This is an instance of Jung's Theorem, setting $n = 2$. {{qed}} {{Namedfor|Heinrich Wilhelm Ewald Jung}}
{{begin-eqn}} {{eqn	| l = {}_2 \map {F_1} {1, p; p; x} 	| r = \sum_{n \mathop = 0}^\infty \frac {1^{\bar n} p^{\bar n} } {p^{\bar n} } \frac {x^n} {n!} 	| c = {{Defof|Gaussian Hypergeometric Function}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty n! \frac {x^n} {n!} 	| c = One to Integer Rising is Integer Factorial }} {{eqn	| r = \sum_{n \mathop = 0}^\infty x^n }} {{eqn	| r = \frac 1 {1 - x} 	| c = Sum of Infinite Geometric Sequence, justified as $\size x < 1$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = x \, {}_2 \map {F_1} {\frac 1 2, 1; \frac 3 2; -x^2} 	| r = x \sum_{n \mathop = 0}^\infty \frac {\paren {\frac 1 2}^{\bar n} 1^{\bar n} } {\paren {\frac 3 2}^{\bar n} } \frac {\paren {-x^2}^n} {n!} 	| c = {{Defof|Gaussian Hypergeometric Function}} }} {{eqn	| r = x \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {\frac 1 2 + n} } {\map \Gamma {\frac 1 2} } \times \frac {\map \Gamma {\frac 3 2} } {\map \Gamma {\frac 3 2 + n} } n! \frac {\paren {-x^2}^n} {n!} 	| c = Rising Factorial as Quotient of Factorials, One to Integer Rising is Integer Factorial }} {{eqn	| r = x \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {\frac 1 2 + n} } {\map \Gamma {\frac 1 2 + n} } \times \frac {\map \Gamma {\frac 1 2} } {\map \Gamma {\frac 1 2} } \times \frac 1 {2 \paren {n + \frac 1 2} } \paren {-1}^n x^{2 n} 	| c = Gamma Difference Equation }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1}  }} {{eqn	| r = \arctan x 	| c = Power Series Expansion for Real Arctangent Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = x \, {}_2 \map {F_1} {\frac 1 2, \frac 1 2; \frac 3 2; x^2} 	| r = x \sum_{n \mathop = 0}^\infty \frac {\paren {\paren {\frac 1 2}^{\bar n} }^2} {\paren {\frac 3 2}^{\bar n} } \frac {x^{2 n} } {n!} 	| c = {{Defof|Gaussian Hypergeometric Function}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {\map \Gamma {\frac 1 2 + n} }^2 \map \Gamma {\frac 3 2} } {\paren {\map \Gamma {\frac 1 2} }^2 \map \Gamma {\frac 3 2 + n} } \frac {x^{2 n + 1} } {n!} 	| c = Rising Factorial as Quotient of Factorials }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {\map \Gamma {\frac 1 2 + n} }^2 \map \Gamma {\frac 1 2} } {2 \paren {\map \Gamma {\frac 1 2} }^2 \paren {\frac 1 2 + n} \map \Gamma {\frac 1 2 + n} } \frac {x^{2 n + 1} } {n!} 	| c = Gamma Difference Equation }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {\frac 1 2 + n} } {\paren {2 n + 1} \sqrt \pi} \frac {x^{2 n + 1} } {n!} 	| c = Gamma Function of One Half }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {\paren {\frac {\paren {2 n}!} {2^{2 n} n!} \sqrt \pi} \times \frac 1 {\paren {2 n + 1} \sqrt \pi} \times \frac 1 {n!} x^{2 n + 1} } 	| c = Gamma Function of Positive Half-Integer }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1} }} {{eqn	| r = \arcsin x 	| c = Power Series Expansion for Real Arcsine Function }} {{end-eqn}} {{qed}}