hcju/proofwikiqa huggingface dataset:
Let $x, y, z \in S$ be arbitrary elements of $S$ such that $x \prec y \prec z$. Let $x, z \in \displaystyle \bigcup \CC$. First let $x, z \in C$ where $C \in \CC$. Then as $C$ is convex, $y \in C$. Hence, by definition of union, $y \in \displaystyle \bigcup \CC$. Now let $x \in C_1, z \in C_2$ where $C_1, C_2 \in \CC$. We have that $\displaystyle \bigcap \CC \ne \O$. Thus $C_1 \cap C_2 \ne \O$. Then $\exists a \in C_1 \cap C_2: x < a < z$. Hence one of the following cases holds: :$(1): \quad x < y < a < z$, whence $y \in C_1$, by convexity of $C_1$ :$(2): \quad x < a < y < z$, whence $y \in C_2$, by convexity of $C_2$ :$(3): \quad y = a$, whence $y \in C_1$ and $y \in C_2$, by definition of $a$. Thus in all cases $y \in \displaystyle \bigcup \CC$. Thus $\displaystyle \bigcup \CC$ is convex by definition. {{qed}}
From the lemma: :$A \subseteq A^*$ :$B \subseteq B^*$ :$A^* \cap B^* = \O$ Let $p \notin A^* \cup A^-$. Thus $p \notin A^*$ and $p \notin A^-$. Then there exists an open interval $\openint s t$ which is disjoint from $A$ such that $p \in \openint s t$. Now $\openint s t$ can only intersect $A^*$ only if it intersects some $\closedint a b \subseteq A^*$ where $a, b \in A$. But we have: :$\openint s t \cap A = \O$ and as $a, b \in A$ it follows that: :$\openint s t \subseteq \openint a b$ That means $p \in A^*$. But we have $p \notin A^*$. Therefore: :$\openint s t \cap A^* = \O$ Thus: :$p \notin \paren {A^*}^-$ Hence: {{begin-eqn}} {{eqn | l = \paren {A^*}^-       | o = \subseteq       | r = \paren {A^* \cup A^-} \cap B^*       | c =  }} {{eqn | r = \paren {A^* \cap B^*} \cup \paren {A^- \cap B^*}       | c =  }} {{eqn | r = \O       | c =  }} {{end-eqn}} Hence the result. {{explain|Fill in the justification for the above chain of reasoning}} {{qed}}
Let $A_\alpha \cap {S_\alpha}^- \ne \O$. Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$. This belongs to the complement in $S$ of the closed set $\paren {B^*}^-$. Hence there exists a neighborhood $\openint x y$ of $p$ which is disjoint from $\paren {B^*}^-$. Then: :$\openint x y \cap S_\alpha \ne \O$ and so: :$\openint p y \ne \O$ But $\openint p y$ is disjoint from both $A^*$ and $B^*$. Thus there must exist some $C_\gamma$ which contains $\openint p y$. {{qed}}
Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \displaystyle \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \displaystyle \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$: :$\displaystyle A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$ where $\relcomp S X$ denotes the complement of $X$ with respect to $S$. Let $M$ be the linearly ordered set: :$M = \set {A_\alpha, B_\beta, C_\gamma}$ as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set. Let: :$S_\alpha$ be the set of strict upper bounds for $A_\alpha$. :$S_\beta$ be the set of strict upper bounds for $B_\beta$. From Successor Sets of Linearly Ordered Set Induced by Convex Component Partition: :each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$ :each of the sets $B_\beta \in M$ has an immediate successor in $M$ if $B_\beta$ intersects the closure of $S_\beta$ :that immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$. For each $\gamma$, let $k_\gamma \in C_\gamma$ be selected and fixed. {{explain|We probably need AoC for this.}} Let $A_\alpha \cap {S_\alpha}^- \ne \O$. Then there exists a unique ${k_\alpha}^+ \in {C_\alpha}^+$. In such cases, let $I_\alpha = \hointr p { {k_\alpha}^+}$ where $p \in A \cap {S_\alpha}^-$. The other case is when $A_\alpha \cap {S_\alpha}^- = \O$. In this case, let $I_\alpha = \O$. Let $J_\alpha$ be defined similarly for the strict lower bounds of $A_\alpha$, using the same set of points $k_\gamma \in C_\gamma$. Then: :For each $\alpha$, let $U_\alpha = J_\alpha \cup A_\alpha \cup I_\alpha$. :For each $\beta$, let $V_\beta = J_\beta \cup A_\beta \cup I_\beta$. Each $U_\alpha$ and $V_\beta$ is a convex open set in $S$ containing $A_\alpha$ and $B_\beta$ respectively. Thus $\displaystyle U = \bigcup U_\alpha$ and $\displaystyle V = \bigcup V_\beta$ are open sets in $T$ containing $A^*$ and $B^*$ respectively. We have that: :no $A_\alpha$ intersects no $B_\beta$ :the same $K_\gamma$ being used throughout implies that no $J_\beta$ or $I_\beta$ can intersect any $J_\alpha$ or $I_\alpha$. Thus no $U_\alpha$ can intersect any $V_\beta$. Thus $U \cap V = \O$. So we have two disjoint open sets $U, V$ in $S$ containing the two separated sets $A$ and $B$ respectively. As $A$ and $B$ are arbitrary, it follows that such $U$ and $V$ can be found whatever $A$ and $B$ are, provided they are separated. Hence the result by definition of $T_5$ space. {{qed}}
Let $p \in S$. By definition of linearly ordered space, the rays: :$R_1 := \set {x \in S: x \prec p}$ :$R_2 := \set {x \in S: p \prec x}$ are open in $T$. Thus their union: :$R_1 \cup R_2 = \set {x \in S: x \prec p \lor p \prec x}$ is also open in $T$. Consider the complement in $S$ of $R_1 \cup R_2$: :$\relcomp S {R_1 \cup R_2} = \set {y \in S: y \notin R_1 \cup R_2}$ By definition of closed set, $\relcomp S {R_1 \cup R_2}$ is closed in $T$ But $\relcomp S {R_1 \cup R_2} = \set p$. Thus $p$ is a closed point of $T$. Hence the result, by definition of $T_1$ (Fréchet) space. {{qed}}
By Linearly Ordered Space is $T_1$, $T$ is a $T_1$ (Fréchet) space. By Linearly Ordered Space is $T_5$, $T$ is a $T_5$ space. Hence the result, by definition of completely normal space. {{qed}}
=== Necessary Condition === Let $T$ be a compact space. Let $A \subseteq S$. {{AimForCont}} $A$ has no supremum. Consider the sets: :$P_\alpha = \set {x \in S: x \prec \alpha}$ for $\alpha \in A$ :$B_\beta = \set {x \in S: \beta \prec x}$ for $\beta$ an upper bound of $A$. We have that $P_\alpha$ and $B_\beta$ cover $S$ but contain no finite subcover. Thus $T$ is not a compact space. Similarly if $A$ has no infimum. It follows that $A$ has both a supremum and infimum. As $A$ is arbitrary, it follows that $S$ is a complete ordered set. {{qed}} === Sufficient Condition === Let $T$ be a complete ordered set. Let $\UU$ be an open cover of $S$. Let $a$ be the infimum of $S$. Let $B$ be the set of the elements $y \in S$ for which $\hointr a y$ have a finite cover from elements of $\UU$. Let $\alpha$ be the supremum of $B$. If $\alpha \in U \in \UU$, then $U \subseteq B$. So unless $\alpha$ is the supremum of $S$ itself, there exists an open interval $\openint x y \subseteq U$ such that $\alpha \in \openint x y$. But we have that $\alpha$ be the supremum of $B$. Hence $\openint x y = \O$. But that means $y \in B$ which is impossible. Thus $B = S$. That is, the set of elements of $S$ which have a finite subcover of $\UU$ is $S$ itself. Hence $T$ is a compact space by definition. {{qed}}
We have that $A \setminus \set p$ is separated by $\set {x \in A: x \prec p}$ and $\set {x \in A: p \prec x}$. If $p$ is an endpoint of $A$, then either: :$\set {x \in A: x \prec p} = \O$ or: :$\set {x \in A: p \prec x} = \O$ {{qed}}
The complement relative to $\closedint 0 \Omega$ of $\set \Omega$ is $\hointr 0 \Omega$, which is open in $\closedint 0 \Omega$. Hence, by definition, $\set \Omega$ is a closed set of $\closedint 0 \Omega$. Let $G_i$ be a countable set of open sets of $\closedint 0 \Omega$ which contain $\Omega$. Then we can find a set of basis elements of $\closedint 0 \Omega$ of the form $\hointl {\alpha_i} \Omega \subseteq G_i$ for each $i$. Each $\hointr 0 {\alpha_i}$ is countable. We also have Countable Union of Countable Sets is Countable. Thus the infimum of the $\alpha_i$ is an ordinal $\gamma$ which is less than $\Omega$. Therefore: :$\displaystyle \bigcap G_i \supseteq \hointl \gamma \Omega \ne \set \Omega$ That is, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$. {{qed}}
From Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$. Thus, by definition, $\Omega$ does not have a countable local basis. Hence the result by definition of first-countable space. {{qed}}
Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable. Let $H \subseteq \hointr 0 \Omega$ be a countable subset of $\hointr 0 \Omega$. Let $\sigma$ be the supremum of $H$. As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable. Thus $\sigma$ strictly precedes $\Omega$. The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$. By definition of the closure $H^-$ of $H$ as the smallest closed set of $\hointr 0 \Omega$ containing $H$, it follows that $H^- \subseteq \closedint 0 \sigma$. Therefore, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$. Thus the closure of $H$ does not equal $\hointr 0 \Omega$. Thus $H$ is not everywhere dense in $\hointr 0 \Omega$. Hence, by definition, $\hointr 0 \Omega$ is not separable. {{qed}}
Let $H \subseteq \closedint 0 \Omega$ be a countable subset of $\closedint 0 \Omega$. From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$. Thus the closure of $H$ in $\closedint 0 \Omega$ does not equal $\closedint 0 \Omega$. Thus $H$ is not everywhere dense in $\closedint 0 \Omega$. Hence, by definition, $\closedint 0 \Omega$ is not separable. {{qed}}
By definition, $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both linearly ordered spaces. The result follows from Linearly Ordered Space is Completely Normal. {{qed}}
From Omega is Closed in Uncountable Closed Ordinal Space but not $G_\delta$ Set, $\set \Omega$ is not a $G_\delta$ set. From Ordinal Space is Completely Normal, $\closedint 0 \Omega$ is a $T_1$ (Fréchet) space. Thus by definition $\set \Omega$ is closed in $\closedint 0 \Omega$. Thus we have that $\set \Omega$ is a closed set of $\closedint 0 \Omega$ which is not a $G_\delta$ set. The result follows by definition of perfectly normal space. {{qed}}
=== 1 implies 2 === Follows from Ring Homomorphism Preserves Units. {{qed|lemma}} === 2 implies 1 === Let $a$ be a unit of $B$. Let $P \in A \sqbrk x$ be a monic polynomial with $\map P {1 / a} = 0$. Let $n$ be its degree and $\map P x = x^n + \map Q x$. Then $1 + a^n \map Q {1 / a} = 0$. Note that $a^{n - 1} \map Q {1 / a} \in A$. Thus $a$ is a unit of $A$, with inverse $-a^{n - 1} \map Q {1 / a}$. {{qed}} Category:Integral Ring Extensions bb33y4znlaafmq9bnk4dlyrjd8b2njg
=== Upper extension === We prove that $E/K$ is separable. Let $\alpha \in E$. Let $f$ be its minimal polynomial over $F$. Let $g$ be its minimal polynomial over $K$. Then $f$ is separable. Then $f \in K[x]$ and $f(\alpha) = 0$, hence by definition $g$ divides $f$. By Divisor of Separable Polynomial is Separable, $g$ is separable. {{qed|lemma}} === Lower extension === It follows immediately by definition of a separable extension that $K/F$ is. {{qed}}
Let $\Lambda$ be the orbit of $\alpha$ under $G$. By: :Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit the product: :$\map p x = \displaystyle \prod_{\beta \in \Lambda} \paren {x - \beta}$ is the minimal polynomial of $\alpha$ over $F$. By Product of Distinct Monic Linear Polynomials is Separable, $p$ is separable. Thus $\alpha$ is separable over $F$. {{qed}} Category:Field Extensions bxlw60sqmofc7ph69vzt15v415lzzn5
From Basis for Open Ordinal Topology, the set $\mathcal B$ of subsets of $\closedint 0 \Gamma$ of the form: :$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$ for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\closedint 0 \Gamma$. As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\closedint 0 \Gamma$. Each point in $\closedint 0 \Gamma$ has a countable basis. From Countable Union of Countable Sets is Countable, it follows that $\closedint 0 \Gamma$ has a countable basis. {{qed}}
From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form: :$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$ for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\hointr 0 \Gamma$. As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\hointr 0 \Gamma$. Each point in $\hointr 0 \Gamma$ has a countable basis. From Countable Union of Countable Sets is Countable, it follows that $\hointr 0 \Gamma$ has a countable basis. {{qed}}
By Product over Finite Set with Zero Factor, we have $\map p \alpha = 0$. By definition, $p \in E \sqbrk x$. === $p$ has coefficients in $F$ === We show that $p \in F \sqbrk x$. Let $\sigma \in G$, and denote the induced automorphism of $E[x]$ still by $\sigma$. We show that $\map \sigma p = p$. We have: {{begin-eqn}} {{eqn | l = \map \sigma p       | r = \prod_{\beta \mathop \in \Lambda} \map \sigma {x - \beta}       | c = Ring Homomorphism Commutes with Product over Finite Set }} {{eqn | r = \prod_{\beta \mathop \in \Lambda} \paren {x - \map \sigma \beta}       | c = {{Defof|Induced Automorphism of Polynomial Ring}} }} {{eqn | r = \prod_{\beta \mathop \in \Lambda} \paren {x - \beta}       | c = Group Element Permutes Orbit under Group of Permutations, Change of Variables in Product over Finite Set }} {{eqn | r = p }} {{end-eqn}} Because this is true for all $\sigma \in G$, indeed the coefficients of $p$ are in $\Fix G = F$. Thus $p \in F \sqbrk x$. === $p$ is the minimal polynomial === By Product of Monic Polynomials is Monic, $p$ is monic. In particular, $p$ is nonzero. Thus $\alpha$ is algebraic over $F$. Let $f$ be its minimal polynomial over $F$. We show that $f = p$. By definition and because $\map p \alpha = 0$, $f$ divides $p$. By Group of Automorphisms is Contained in Automorphism Group over Fixed Field, $G \le \Aut {E/F}$. By Automorphism Group of Field Extension Permutes Roots of Minimal Polynomial, each $\beta \in \Lambda$ is a root of $f$. By Polynomial Factor Theorem, each $x - \beta$ divides $f$. By Product of Pairwise Coprime Divisors of Polynomial over Field is Divisor, $p$ divides $f$. By Monic Polynomials are Equal iff Divide Each Other, $p = f$. {{qed}} Category:Field Extensions ipul9olktcc8b2ho7gt3tdf5j6dk9nt
From Countable Closed Ordinal Space is Second-Countable, $\closedint 0 \Gamma$ has a basis which is $\sigma$-locally finite. From Ordinal Space is Completely Normal, $\closedint 0 \Gamma$ is a completely normal space. From Sequence of Implications of Separation Axioms it follows that $\closedint 0 \Gamma$ is a regular space. The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis. {{qed}}
From Countable Open Ordinal Space is Second-Countable, $\hointr 0 \Gamma$ has a basis which is $\sigma$-locally finite. From Ordinal Space is Completely Normal, $\hointr 0 \Gamma$ is a completely normal space. From Sequence of Implications of Separation Axioms it follows that $\hointr 0 \Gamma$ is a regular space. The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis. {{qed}}
Let $H$ be a subset of an ordinal space. Then $H$ has an infimum: its first element. Let $H$ be a subset of $\closedint 0 \Gamma$. Then $H$ has a supremum. Therefore $\closedint 0 \Gamma$ is a complete order space. {{qed}}
By definition, $\closedint 0 \Gamma$ is a linearly ordered space. The result follows from Linearly Ordered Space is Compact iff Complete. {{qed}}
Consider the set: :$\set {\hointr 0 \Gamma: \alpha < \Gamma}$ This is an open cover of $\hointr 0 \Gamma$. But because $\Gamma$ is a limit ordinal, it has no finite subcover. Hence the result by definition of compact. {{qed}}
Let $p$ be fixed. In this context, $\map f x \map K {p, x}$ is the pointwise product of the functions $\map f x$ and $\map K {p, x}$. From Pointwise Operation is Well-Defined, it follows that $\map f x \map K {p, x}$ is a real function on $x$. We have that both $\map f x$ and $\map K {p, x}$ are integrable. It follows from Pointwise Product of Integrable Real Functions is Integrable that $\map f x \map K {p, x}$ is an integrable function. From Definite Integral is Unique, for a given $p$ there is exactly one real number $\map F p$ for which: :$\map F p = \displaystyle \int_a^b \map f x \map K {p, x} \rd x$ {{qed}} {{explain|A lot of stuff in analysis and calculus has been glossed over, as can be seen from the redlinks above. This will need to be tightened up.}}
Let $T$ be expressed in its full form as an integral fransform: :$\map T f := \displaystyle \int_a^b \map f x \map K {p, x} \rd x$ for some integrable function $\map K {p, x}$. Then: {{begin-eqn}} {{eqn | l = \map T {\alpha f + \beta g}       | r = \int_a^b \paren {\alpha \map f x + \beta \map g x} \map K {p, x} \rd x       | c =  }} {{eqn | r = \int_a^b \paren {\alpha \map f x \map K {p, x} + \beta \map g x \map K {p, x} } \rd x       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \alpha \int_a^b \map f x \map K {p, x} \rd x + \beta \int_a^b \map g x \map K {p, x} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \alpha \map T f + \beta \, \map T g       | c = {{Defof|Integral Operator}} }} {{end-eqn}} {{Qed}} Category:Integral Transforms Category:Integral Operator is Linear cbv0r9frhaeyot7kcm79cyuy75oqfw0
From Integral Operator is Linear: :$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$ The result follows by setting $\alpha = \beta = 1$. {{Qed}}
From Integral Operator is Linear: :$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$ The result follows by setting $\beta = 0$. {{Qed}}
Let: :$x_1 = \map {T^{-1} } F$ :$x_2 = \map {T^{-1} } G$ Thus: :$F = \map T {x_1}$ :$G = \map T {x_2}$ Then for all $p, q \in \R$: {{begin-eqn}} {{eqn | l = \map T {p x_1 + q x_2}       | r = p \map T {x_1} + q \map T {x_2}       | c = Integral Operator is Linear }} {{eqn | r = p F + q G       | c =  }} {{end-eqn}} and so $x = p F + q G$ is a solution to the equation: :$\map T x = p F + q G$ But this equation has only one solution: :$x = \map {T^{-1} } {p F + q G}$ Thus $p F + q G$ must coincide with the above: :$p \map {T^{-1} } F + q \map {T^{-1} } G = \map {T^{-1} } {p F + q G}$ which proves that $T^{-1}$ is a linear operator. {{qed}}
Let $\map S x$ converge to some $L \in \R$. Let $r \in \Z$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map S {x + 2 r \pi}       | r = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n \paren {x + 2 r \pi} + b_n \sin n \paren {x + 2 r \pi} }       | c = Definition of $\map S {x + 2 r \pi}$ }} {{eqn | r = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \map \cos {n x + 2 r n \pi} + b_n \map \sin {n x + 2 r n \pi} }       | c =  }} {{eqn | r = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = \map S x       | c = Definition of $\map S x$ }} {{end-eqn}} {{qed}}
Let $\map S x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$. Set: {{begin-eqn}} {{eqn | l = \cos n x       | r = \frac {e^{i n x} + e^{- i n x} } 2       | c = Cosine Exponential Formulation }} {{eqn | l = \sin n x       | r = \frac {e^{i n x} - e^{- i n x} } {2 i}       | c = Sine Exponential Formulation }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map S x       | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \frac {e^{i n x} + e^{- i n x} } 2 + b_n \frac {e^{i n x} - e^{- i n x} } {2 i} }       | c =  }} {{eqn | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\paren {\frac {a_n} 2 + \frac {b_n} {2 i} } e^{i n x} + \paren {\frac {a_n} 2 - \frac {b_n} {2 i} } e^{- i n x} }       | c =  }} {{eqn | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\paren {\frac {a_n - i b_n} 2} e^{i n x} + \paren {\frac {a_n + i b_n} 2} e^{- i n x} }       | c =  }} {{eqn | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}       | c =  }} {{end-eqn}} Let $b_0 := 0$. Then let: :$\forall n \in \N: c_n = \dfrac {a_n - i b_n} 2$ Thus it follows that: :$c_0 = \dfrac {a_0 - 0 \times i} 2 = \dfrac {a_0} 2$ Then let: :$c_{-n} = \dfrac {a_n + i b_n} 2$ from which it follows that: :$c_{-n} = \overline {\paren {\dfrac {a_n - i b_n} 2} }$ and again: :$c_0 = \dfrac {a_0 + 0 \times i} 2 = \dfrac {a_0} 2$ Then it is noted that: :$e^{i \times 0 \times x} = e^0 = 1$ and it follows that: {{begin-eqn}} {{eqn | l = \map S x       | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}       | c = from above }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\frac {a_n - i b_n} 2} e^{i n x} + \sum_{n \mathop = 1}^\infty \paren {\frac {a_n + i b_n} 2} e^{- i n x}       | c = subsuming $a_0$ into first summation }} {{eqn | r = \sum_{n \mathop = 0}^\infty c_n e^{i n x} + \sum_{n \mathop = 1}^\infty c_{-n} e^{- i n x}       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty c_n e^{i n x} + \sum_{n \mathop = -\infty}^{-1} c_n e^{i n x}       | c =  }} {{eqn | r = \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}       | c =  }} {{end-eqn}} where it is noted that $c_{-n} = \overline {c_n}$ as required. {{qed}}
From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^\infty r^k \sin k x       | r = \map \Im {\sum_{k \mathop = 1}^\infty r^k e^{i k x} }       | c =  }} {{eqn | r = \map \Im {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}       | c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$ }} {{eqn | r = \map \Im {\frac 1 {1 - r e^{i x} } }       | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$ }} {{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }       | c =  }} {{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }       | c =  }} {{eqn | r = \map \Im {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + r^2} }       | c = Corollary to Euler's Formula }} {{eqn | r = \dfrac {r \sin x} {1 - 2 r \cos x + r^2}       | c = after simplification }} {{end-eqn}} It is noted that when $x$ is a multiple of $2 \pi$ then: :$1 - 2 r \cos x + r^2 = 1 - 2 + 1 = 0$ leaving the {{RHS}} undefined. {{qed}}