hcju/proofwikiqa huggingface dataset:
From Absolute Value Function is Even Function, $\map f x$ is an even function. By Fourier Series for Even Function over Symmetric Range, we have: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$ where: {{begin-eqn}} {{eqn | l = a_n       | r = \frac 2 \lambda \int_0^\pi \map f x \cos \frac {n \pi x} \lambda \rd x       | c =  }} {{eqn | r = \frac 2 \lambda \int_0^\pi x \cos \frac {n \pi x} \lambda \rd x       | c = Definition of $\map f x$ }} {{end-eqn}} Hence from Half-Range Fourier Cosine Series for Identity Function: :$\map f x \sim \dfrac \lambda 2 - \displaystyle \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$ Hence the result. {{qed}} Category:Absolute Value Function Category:Fourier Series for Absolute Value Function phtrxwceggby7p0s2kbw9bfd6cvvtbs
By definition, the absolute value function is an even function: :$\size {-x} = x = \size x$ Thus by Fourier Series for Even Function over Symmetric Range, $\size x$ can be expressed as: :$\displaystyle \size x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$ where for all $n \in \Z_{\ge 0}$: :$a_n = \displaystyle \frac 2 \pi \int_0^\pi \size x \cos n x \rd x$ On the real interval $\openint 0 \pi$: :$\size x = x$ and so for all $n \in \Z_{\ge 0}$: :$a_n = \displaystyle \frac 2 \pi \int_0^\pi x \cos n x \rd x$ Thus Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$ can be applied directly. So for $x \in \openint {-\pi} \pi$: :$\displaystyle \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1}  x} {\paren {2 n - 1}^2}$ {{qed}}
By Fourier Series for Absolute Value Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as: :$\forall x \in \openint {-\lambda} \lambda: \map f x = \size x$ has a Fourier series: :$\map f x \sim \dfrac \lambda 2 - \displaystyle \dfrac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$ Substituting for $\lambda = \pi$ gives: :$\size x = \dfrac \pi 2 - \displaystyle \dfrac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1}  x} {\paren {2 n - 1}^2}$ as required.  {{qed}}
{{begin-eqn}} {{eqn | l = \map \vers {-\theta}       | r = 1 - \map \cos {-\theta}       | c = {{Defof|Versed Sine}} }} {{eqn | r = 1 - \cos \theta       | c = Cosine Function is Even }} {{eqn | r = \vers \theta       | c = {{Defof|Versed Sine}} }} {{end-eqn}} {{qed}} Category:Versed Sines Category:Examples of Even Functions g73g61r5fhzw1rsyohcg8mt45yih5e5
Let $\alpha \in I$. Let $x, y \in S_\alpha$. By the Axiom of Choice, there exists $z \in S$. Define $x', y' \in S$ by: :$x'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ x & \beta = \alpha \end{cases}$ and :$y'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ y & \beta = \alpha \end{cases}$ By definition of a $T_2$ (Hausdorff) space: :$\exists U, V \in \tau : x' \in U, y' \in V$ and $U \cap V = \O$ Recall that the topology on the product space $T$ is the Tychonoff topology. By definition of the Tychonoff topology the natural basis is a basis for the Tychonoff topology. Then: :$\exists U', V' \in \BB : x' \in U' \subseteq U, y' \in V' \subseteq V$ where $\BB$ is the natural basis for $\tau$. From Subsets of Disjoint Sets are Disjoint: :$U' \cap V' = \O$ From Natural Basis of Tychonoff Topology: :$\displaystyle U' = \prod_{\beta \mathop \in I} U_\beta$ where: ::for all $\beta \in I : U_\beta \in \tau_\beta$ ::for all but finitely many indices $\beta : U_\beta = X_\beta$ Similarly: :$\displaystyle V' = \prod_{\beta \mathop \in I} V_\beta$ where: ::for all $\beta \in I : V_\beta \in \tau_\beta$ ::for all but finitely many indices $\beta : V_\beta = X_\beta$ From General Case of Cartesian Product of Intersections: :$\displaystyle \paren{\prod_{\beta \mathop \in I} U_\beta} \bigcap \paren{\prod_{\beta \mathop \in I} V_\beta} = \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta}$ Thus: :$\displaystyle \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta} = \O$ Now for all $\beta \in I$ such that $\beta \ne \alpha$: :$z_\beta = x'_\beta \in U_\beta$ and  :$z_\beta = y'_\beta \in V_\beta$ So: :$\forall \beta \in I, \beta \ne \alpha : U_\beta \cap V_\beta \ne \O$ From Cartesian Product of Family is Empty iff Factor is Empty it follows that: :$U_\alpha \cap V_\alpha = \O$ Also we have: :$x = x'_\alpha \in U_\alpha$ and :$y = y'_\alpha \in V_\alpha$ Since $U_\alpha$ and $V_\alpha$ are open in $\struct {S_\alpha, \tau_\alpha}$ then $x$ and $y$ are separated by open sets. Since $x$ and $y$ were arbitrary, then $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space by definition. Since $\alpha \in I$ was arbitrary, then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space. {{qed}} Category:Hausdorff Spaces Category:Product Spaces io4k9m95vn3ut5ocdmh2u52yo5v4aql
=== Necessary Condition === By the axiom of choice, the contrapositive statement holds: :if $S_i \ne \O$ for all $i \in I$ then $S \ne \O$ By the Rule of Transposition, the converse holds: :if $S = \O$  then $S_i = \O$ for some $i \in I$ {{qed|lemma}} === Sufficient Condition === Let $S_j = \O$ for some $j \in I$. By the definition of the Cartesian product:definition 2: :$\displaystyle \prod_{i \mathop \in I} S_i := \set {f: \paren {f: I \to \bigcup_{i \mathop \in I} S_i} \land \paren {\forall i \in I: \paren {\map f i \in S_i} } }$ For any $f: I \to \bigcup_{i \mathop \in I} S_i$ we have: :$\map f j \notin S_j$ Thus: :$f \notin \displaystyle \prod_{i \mathop \in I} S_i$ It follows that: :$\displaystyle \prod_{i \mathop \in I} S_i = \O$ {{qed}} Category:Cartesian Product Category:Empty Set lhqesm1rrvtrhy2d4nxacllts8cis8s
We begin with the natural logarithm of $\norm {\mathbf x}_p$: {{begin-eqn}} {{eqn | l = \dfrac \d {\d p} \map \ln {\norm {\bf x}_p}       | r = \frac {\dfrac \d {\d p} \norm {\bf x}_p} {\norm {\bf x}_p} }} {{eqn | r = \map {\dfrac \d {\d p} } {\frac 1 p \map \ln {\sum_{n \mathop = 0}^\infty \size {x_n}^p} } }} {{eqn | r = -\frac 1 {p^2} \map \ln {\sum_{n \mathop = 0}^\infty \size {x_n}^p} + \frac 1 p \frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map \ln {\size {x_n} } } {\sum_{n \mathop = 0}^\infty \size {x_n}^p}       | c = Derivative of Power of Constant }} {{eqn | r = -\frac 1 {p^2} \map \ln {\norm {\bf x}_p^p} + \frac 1 p \frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map \ln {\size {x_n} } } {\norm {\bf x}_p^p} }} {{eqn | r = \frac 1 p \paren {\frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map {\ln} {\size {x_n} } } {\norm {\bf x}_p^p} - \map \ln {\norm {\bf x}_p} } }} {{end-eqn}} Multiplication by $\norm {\mathbf x}_p$ completes the proof. {{qed}} {{NoSources}} Category:P-Norms 7awcc9mqw5r3j7przrsv2wt2atlinh7
{{begin-eqn}} {{eqn | l = \paren {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p }^{\frac 1 p}       | o = \ge        | r = \paren { {\size {x_i}^p }_{i \mathop \in \N_0} }^{\frac 1 p} }} {{eqn | r = {\size {x_i} }_{i \mathop \in \N_0} }} {{eqn | ll= \leadsto       | l = \frac 1 p \map \ln {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p}       | o = \ge       | r = \map \ln {\size {x_i}_{i \mathop \in \N_0} } }} {{eqn | ll= \leadsto       | l = \frac 1 p \map \ln {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p} \sum_{i \mathop = 0}^\infty {\size {x_i} }^p       | o = \ge       | r = \sum_{i \mathop = 0}^\infty {\size {x_i} }^p \map \ln {\size {x_i} }       | c = Multiply both sides by $\size {x_i}$ and sum over $i \in \N_0$ }} {{end-eqn}} By derivative of p-norm {{WRT}} $p$: {{begin-eqn}} {{eqn | l = \dfrac \d {\d p} \norm {\mathbf x}_p       | r = \frac {\norm {\mathbf x}_p} p \paren {\frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map \ln {\size {x_n} } } {\norm {\bf x}_p^p} - \map \ln {\norm {\bf x}_p} } }} {{eqn | r = \frac 1 {p \norm {\mathbf x}_p^{p \mathop - 1} } \paren {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p \map \ln {\size {x_n} } - \sum_{n \mathop = 0}^{\infty} {\size {x_n} }^p \frac 1 p \sum_{i \mathop = 0}^\infty \map \ln {\size {x_i} } } }} {{end-eqn}} By P-Norm is Norm, $\norm {\mathbf x}_p > 0$ for $\mathbf x \ne \sequence 0$. By previously derived inequality, the term in parenthesis is negative. Hence: :$\forall p \ge 1: \dfrac \d {\d p} \norm {\mathbf x}_p < 0$ {{qed}} Category:P-Norms tinom8aldd9i9ojcszzph3derdxyc3t
It is apparent by inspection that: :$(1): \quad g$ is an extension of $f$ :$(2): \quad g$ is an odd function. Let $\map T x$ be the Fourier series representing $g$: :$\map g x \sim \map T x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \cos \frac {n \pi x} \lambda \rd x$ :$b_n = \displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \sin \frac {n \pi x} \lambda \rd x$ From Fourier Cosine Coefficients for Odd Function over Symmetric Range: :$\forall n \in \Z_{\ge 0}: a_n = 0$ and from Fourier Sine Coefficients for Odd Function over Symmetric Range: :$\forall n \in \Z_{>0}: b_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map g x \sin \frac {n \pi x} \lambda \rd x$ But on $\openint 0 \lambda$, by definition: :$\forall x \in \openint 0 \lambda: \map g x = \map f x$ Hence: :$\map T x = \displaystyle \sum_{n \mathop = 1}^\infty \paren b_n \sin \frac {n \pi x} \lambda$ where: :$b_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$ That is, $\map T x$ is the same as $\map S x$. Hence the result. {{qed}} Category:Half-Range Fourier Series f8yrsjeo4qkda53fgo1p4mkfp8kpmj5
It is apparent by inspection that: :$(1): \quad g$ is an extension of $f$ :$(2): \quad g$ is an even function. Let $\map T x$ be the Fourier series representing $g$: :$\map g x \sim \map T x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \cos \frac {n \pi x} \lambda \rd x$ :$b_n = \displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \sin \frac {n \pi x} \lambda \rd x$ From Fourier Sine Coefficients for Even Function over Symmetric Range: :$\forall n \in \Z_{> 0}: b_n = 0$ and from Fourier Cosine Coefficients for Even Function over Symmetric Range: :$\forall n \in \Z_{>0}: a_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map g x \cos \frac {n \pi x} \lambda \rd x$ But on $\openint 0 \lambda$, by definition: :$\forall x \in \openint 0 \lambda: \map g x = \map f x$ Hence: :$\map T x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren a_n \cos \frac {n \pi x} \lambda$ where: :$a_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$ That is, $\map T x$ is the same as $\map S x$. Hence the result. {{qed}} Category:Half-Range Fourier Series 4jkesnhoecrlce0p4zm4l919iuqymah
=== Definition 1 implies Definition 2 === Let $A, B \subseteq S$ such that: :$\exists N_A, N_B \subseteq S: \exists U, V \in \tau: A \subseteq U \subseteq N_A, B \subseteq V \subseteq N_B: N_A \cap N_B = \O$ From Subsets of Disjoint Sets are Disjoint then: :$U \cap V = \O$ Thus: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V: U \cap V = \O$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $A, B \subseteq S$ such that: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V: U \cap V = \O$ Let $N_A = U$ and $N_B = V$. From Set is Subset of Itself then: :$\exists N_A, N_B \subseteq S: \exists U, V \in \tau: A \subseteq U \subseteq N_A, B \subseteq V \subseteq N_B: N_A \cap N_B = \O$ {{qed}} Category:Separation Axioms to0dy94kbu5sreh806mo3ke4x4nc3md
Let $x, y \in S$. From Singleton of Element is Subset: :$x$ and $y$ are separated as points by neighborhoods {{iff}} the singletons $\set x$ and $\set y$ are separated as sets by neighborhoods. From Equivalence of Definitions of Sets Separated by Neighborhoods: :the singletons $\set x$ and $\set y$ are separated as sets by neighborhoods {{iff}} the singletons $\set x$ and $\set y$ are separated as sets by open sets. From Singleton of Element is Subset: :the singletons $\set x$ and $\set y$ are separated as sets by open sets {{iff}} $x$ and $y$ are separated as points by open sets. {{qed}} Category:Separation Axioms 2cvik2cs1kk1t7a8q7a5rp4yotk6nln
Suppose $T$ is a $T_3$ space. As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice. Let $\alpha \in I$. From Subspace of Product Space Homeomorphic to Factor Space, $\struct{S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$. From $T_3$ property is hereditary, $T_\alpha$ is $T_3$. From T3 Space is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is $T_3$. As $\alpha \in I$ was arbitrary, the result follows.
Let $U$ be open in $T$ and $x \in U$. From Natural Basis of Tychonoff Topology, there exists $\displaystyle U' = \prod_{\alpha \mathop \in I} U'_\alpha$ such that: :for all $\alpha \in I : U_\alpha \in \tau_\alpha$ :$J = \set {\alpha \in I: U_\alpha \ne X_\alpha}$ is finite :$x \in U'$ :$U' \subseteq U$ By definition of a $T_3$ space, for all $\alpha \in J$: :$\exists V_\alpha \in \tau_\alpha : x_\alpha \in V_\alpha$ and $V^-_\alpha \subseteq U'_\alpha$ For $\alpha \notin J$ let: :$V_\alpha = X_\alpha$ and let: :$\displaystyle V = \prod_{\alpha \mathop \in I} V_\alpha$ By definition of the Cartesian product V: :$x \in V$ From Natural Basis of Tychonoff Topology: :$V$ open in $T$. Now consider the Cartesian product of closed sets: :$\displaystyle \prod_{\alpha \mathop \in I} V^-_\alpha$ From Product of Closed Sets is Closed, $\displaystyle \prod_{\alpha \mathop \in I} V^-_\alpha$ is closed. From Underlying Set of Topological Space is Closed, for $\alpha \notin J$: :$V^-_\alpha = X^-_\alpha = X_\alpha$ From Cartesian Product of Family of Subsets, it follows that: :$\displaystyle V = \prod_{\alpha \mathop \in I} V_\alpha \subseteq \prod_{\alpha \mathop \in I} V^-_\alpha \subseteq \prod_{\alpha \mathop \in I} U'_\alpha \subseteq U$ It follows that $T$ is a $T_3$ space by definition.
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$: :$\forall x \in \openint 0 \lambda: \map f x = x$ From Half-Range Fourier Cosine Series for Identity Function, the half-range Fourier cosine series for $\map f x$ can be expressed as: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda       | c =  }} {{eqn | r = \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \paren {\cos \dfrac {\pi x} \lambda + \frac 1 {3^2} \cos \dfrac {3 \pi x} \lambda + \frac 1 {5^2} \cos \dfrac {5 \pi x} \lambda + \dotsb}       | c =  }} {{end-eqn}} The result follows by setting $\lambda = \pi$. {{qed}}
We have: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}       | r = \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}       | c =  }} {{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}       | c =  }} {{eqn | n = 1       | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }       | c = Sum of Geometric Sequence }} {{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }       | c = extracting factors }} {{eqn | r = e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }       | c = Exponential of Sum and some algebra }} {{eqn | r = e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }       | c = Sine Exponential Formulation }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}       | r = \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }       | c = Euler's Formula and simplifying }} {{eqn | n = 2       | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \cos \paren {n + \alpha} \theta       | r = \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }       | c = equating real parts }} {{end-eqn}} Note that the {{RHS}} at $(1)$ is not defined when $e^{i \theta} = 1$. This happens when $\theta = 2 k \pi$ for $k \in \Z$. For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold. Then: {{begin-eqn}} {{eqn | l = \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta       | r = \intlimits {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } {\theta \mathop = 0} {\theta \mathop = \theta}       | c = Primitive of $\cos a x$ }} {{eqn | r = \paren {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {\sin \paren {n + \alpha} 0} {n + \alpha} }       | c =  }} {{eqn | r = \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}       | c = Sine of Zero is Zero }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}       | r = \sum_{n \mathop = -m}^m \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta       | c =  }} {{eqn | r = \int_0^\theta \sum_{n \mathop = -m}^m \cos \paren {\alpha + n} \theta \rd \theta       | c = Linear Combination of Definite Integrals }} {{eqn | r = \int_0^\theta \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} } \rd \theta       | c = from $(2)$ }} {{end-eqn}} Hence the result. {{qed}}
From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$: :$\displaystyle \cos \alpha x \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\alpha^2 - n^2} }$ Setting $x = 0$: {{begin-eqn}} {{eqn | l = \cos 0       | r = \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\alpha^2 - n^2} }       | c =  }} {{eqn | ll= \leadsto       | l = 1       | r = \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {\alpha^2 - n^2} }       | c = Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = \frac \pi {2 \alpha \sin \alpha \pi}       | r = \frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {\alpha^2 - n^2}       | c =  }} {{eqn | ll= \leadsto       | l = \pi \cosec \pi \alpha       | r = \dfrac 1 \alpha + \displaystyle 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}       | c = {{Defof|Cosecant|subdef = Real Function}} and rearranging }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}       | r = \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}       | c =  }} {{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}       | c =  }} {{eqn | n = 1       | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }       | c = Sum of Geometric Sequence }} {{eqn | r = e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }       | c = extracting factors }} {{eqn | r = e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }       | c = Exponential of Sum and some algebra }} {{eqn | r = e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }       | c = Sine Exponential Formulation }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}       | r = \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }       | c = Euler's Formula and simplifying }} {{eqn | n = 2       | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \sin \paren {n + \alpha} \theta       | r = \map \sin {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta} } {\map \sin {\theta / 2} }       | c = equating imaginary parts }} {{end-eqn}} Note that the {{RHS}} at $(1)$ is not defined when $e^{i u} = 1$. This happens when $u = 2 k \pi$ for $k \in \Z$. For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold. Then: {{begin-eqn}} {{eqn | l = \int_0^\theta \sin \paren {\alpha + n} u \rd u       | r = \intlimits {\dfrac {-\cos \paren {n + \alpha} u} {n + \alpha} } {u \mathop = 0} {u \mathop = \theta}       | c = Primitive of $\sin a x$ }} {{eqn | r = \paren {\dfrac {-\cos \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {-\cos \paren {n + \alpha} 0} {n + \alpha} }       | c =  }} {{eqn | r = \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha}       | c = Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = -m}^m \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha}       | r = \sum_{n \mathop = -m}^m \int_0^\theta \sin \paren {\alpha + n} u \rd u       | c =  }} {{eqn | r = \int_0^\theta \sum_{n \mathop = -m}^m \sin \paren {\alpha + n} u \rd u       | c = Linear Combination of Definite Integrals }} {{eqn | r = \int_0^\theta \map \sin {\alpha u} \frac {\map \sin {\paren {m + \frac 1 2} u} } {\map \sin {u / 2} } \rd u       | c = from $(2)$, changing the variable name }} {{end-eqn}} Hence the result. {{qed}}