hcju/proofwikiqa huggingface dataset:
First we establish the following, as they will be needed later. {{begin-eqn}} {{eqn | o =        | r = \cos \paren {\alpha + n} \theta + \cos \paren {\alpha - n} \theta       | c =  }} {{eqn | r = 2 \map \cos {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \cos {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}       | c = Prosthaphaeresis Formula for Cosine plus Cosine }} {{eqn | n = 1       | r = 2 \cos \alpha \theta \cos n \theta       | c = simplification }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = \cos \paren {\alpha + n} \theta - \cos \paren {\alpha - n} \theta       | c =  }} {{eqn | r = -2 \map \sin {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \sin {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}       | c = Prosthaphaeresis Formula for Cosine minus Cosine }} {{eqn | n = 2       | r = -2 \sin \alpha \theta \sin n \theta       | c = simplification }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | o =        | r = \sum_{n \mathop = -m}^m \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}       | c =  }} {{eqn | r = \dfrac {\cos \paren {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -m}^{-1} \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\cos \paren {-n + \alpha} \theta} {-n + \alpha}       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\cos \paren {\alpha + n} \theta} {\alpha + n} + \dfrac {\cos \paren {\alpha - n} \theta} {\alpha - n} }       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\paren {\alpha - n} \paren {\cos \paren {\alpha + n} \theta} + \paren {\alpha + n} \paren {\cos \paren {\alpha - n} \theta} }{\alpha^2 - n^2} }       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \cos \paren {\alpha + n} \theta - n \cos \paren {\alpha + n} \theta + \alpha \cos \paren {\alpha - n} \theta + n \cos \paren {\alpha - n} \theta} {\alpha^2 - n^2} }       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \paren {\cos \paren {\alpha + n} \theta + \cos \paren {\alpha - n} \theta} - n \paren {\cos \paren {\alpha + n} \theta - \cos \paren {\alpha - n} \theta} } {\alpha^2 - n^2} }       | c =  }} {{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {2 \alpha \cos \alpha \theta \cos n \theta + 2 n \sin \alpha \theta \sin n \theta} {\alpha^2 - n^2} }       | c = from $(1)$ and $(2)$ }} {{end-eqn}} {{finish}}
From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$: :$\displaystyle \cos \alpha x \sim \frac {2 \alpha \sin \pi \alpha} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\alpha^2 - n^2} }$ Setting $x = \pi$: {{begin-eqn}} {{eqn | l = \cos \pi \alpha       | r = \frac {2 \alpha \sin \pi \alpha} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \pi} {\alpha^2 - n^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \cos \pi \alpha       | r = \frac {2 \alpha \sin \pi \alpha} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {-1}^n} {\alpha^2 - n^2} }       | c = Cosine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi \cos \pi \alpha} {\sin \pi \alpha}       | r = \frac 1 \alpha + \sum_{n \mathop = 1}^\infty \frac {2 \alpha} {\alpha^2 - n^2}       | c =  }} {{eqn | ll= \leadsto       | l = \pi \cot \pi \alpha       | r = \frac 1 \alpha + \sum_{n \mathop = 1}^\infty \frac {2 \alpha} {\alpha^2 - n^2}       | c = {{Defof|Cotangent|subdef = Real Function}} and rearranging }} {{end-eqn}} {{qed}}
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$. {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} }       | r = e^{i \theta} \sum_{k \mathop = 0}^n e^{i k \alpha}       | c = factorising $e^{i \theta}$ }} {{eqn | r = e^{i \theta} \paren {\frac {e^{i \paren {n + 1} \alpha} - 1} {e^{i \alpha} - 1} }       | c = Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$ }} {{eqn | r = \frac {e^{i \theta} e^{i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }       | c = extracting factors }} {{eqn | r = e^{i \paren {\theta + n \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }       | c = Exponential of Sum and some algebra }} {{eqn | r = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }       | c = Euler's Formula }} {{eqn | r = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}} Category:Exponential Function 7iscusmh9wldjzguoor0u6h1z6rccxl
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$. {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} }       | r = e^{i \theta} e^{i \alpha} \sum_{k \mathop = 0}^{n - 1} e^{i k \alpha}       | c = factorising $e^{i \theta} e^{i \alpha}$ }} {{eqn | r = e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha} - 1} {e^{i \alpha} - 1} }       | c = Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$ }} {{eqn | r = e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha / 2} \paren {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } } {e^{i \alpha / 2} \paren {e^{i \alpha / 2} - e^{-i \alpha / 2} } } }       | c = extracting factors }} {{eqn | r = e^{i \paren {\theta + \paren {n + 1} \alpha / 2} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }       | c = Exponential of Sum and some algebra }} {{eqn | r = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }       | c = Euler's Formula }} {{eqn | r = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1: :$\displaystyle \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } =  \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$ It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined. From Euler's Formula, this can be expressed as: :$\displaystyle \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$ Equating imaginary parts: :$\displaystyle \sum_{k \mathop = 0}^n \map \sin {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \sin {\theta + \frac {n \alpha} 2}$ {{qed}}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2: :$\displaystyle \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined. From Euler's Formula, this can be expressed as: :$\displaystyle \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ Equating imaginary parts: :$\displaystyle \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha} = \map \sin {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ {{qed}}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1: :$\displaystyle \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } =  \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$ It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined. From Euler's Formula, this can be expressed as: :$\displaystyle \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$ Equating real parts: :$\displaystyle \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}$ {{qed}}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2: :$\displaystyle \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined. From Euler's Formula, this can be expressed as: :$\displaystyle \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ Equating real parts: :$\displaystyle \sum_{k \mathop = 1}^n \map \cos {\theta + k \alpha} = \map \cos {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$ {{qed}}
From the binomial theorem: {{begin-eqn}} {{eqn | l = \paren {x + y}^n       | r = \sum_{k \mathop = 0}^n \binom n k y^k x^{n - k} }} {{eqn | ll = \leadsto       | l = 1       | r = \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}       | c = $y = t, ~x = 1 - t$ }} {{eqn | l = 0       | r = \sum_{k \mathop = 0}^n \binom n k \paren {k t^{k - 1} \paren {1 - t}^{n - k} - t^k \paren{n - k} \paren {1 - t}^{n - k - 1} }       | c = Derivative {{WRT}} $t$ }} {{eqn | r = \sum_{k \mathop = 0}^n k p_{n,k} \paren {\frac 1 t + \frac 1 {1 - t} } - \frac n {1 - t} }} {{eqn | ll = \leadsto       | l = n t       | r = \sum_{k \mathop = 0}^n k p_{n,k} }} {{end-eqn}} {{qed}}
From the binomial theorem: :$\displaystyle 1 = \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}$ From Lemma 1: {{begin-eqn}} {{eqn | l = n t       | r = \sum_{k \mathop = 0}^n \binom n k k t^k \paren {1 - t}^{n - k} }} {{eqn | ll = \leadsto       | l = n       | r = \sum_{k \mathop = 0}^n \binom n k k \paren {k t^{k - 1} \paren {1 - t}^{n - k} - t^k \paren {n - k} \paren {1 - t}^{n - k - 1} }       | c = Derivative {{WRT|Differentiation}} $t$ }} {{eqn | r = \frac 1 t \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t - \frac n {1 - t} \sum_{k \mathop = 0}^n k \map {p_{n, k} } t + \frac 1 {1 - t} \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t  }} {{eqn | r = \frac 1 {t \paren {1 - t} } \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t - \frac n {1 - t} n t }} {{eqn | ll = \leadsto        | l = \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t       | r = n t \paren {1 - t} + n^2 t^2 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n \paren {k - nt}^2 \map {p_{n,k} } t       | r = \sum_{k \mathop = 0}^n \paren {k^2 - 2 n t k + n^2 t^2} \map {p_{n,k} } t }} {{eqn | r = n t \paren {1 - t} + n^2 t^2 - 2 n t n t + n^2 t^2 }} {{eqn | r = n t \paren {1 - t} }} {{end-eqn}} {{qed}}
{{AimForCont}} $\displaystyle \lim_{n \to \infty} x_n = L_1$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = L_2$, where $L_1 \ne L_2$. Let $\displaystyle \epsilon = \frac {\norm {L_1 - L_2} } 3$. From properties of norm it follows that $\epsilon > 0$. By definition: :$\exists N_1 \in \N : \forall n > N_1 : \norm {x_n - L_1} < \epsilon$ :$\exists N_2 \in \N : \forall n > N_2 : \norm {x_n - L_2} < \epsilon$ Choose $n > N_1 + N_2$. Then: {{begin-eqn}} {{eqn | l = \norm {L_1 - L_2}       | r = \norm {L_1 - x_n + x_n - L_2} }} {{eqn | o = \le       | r = \norm {L_1 - x_n} + \norm {x_n - L_2}       | c = Triangle Inequality }} {{eqn | o = <        | r = \epsilon + \epsilon }} {{eqn | r = \frac 2 3 \norm {L_1 - L_2} }} {{end-eqn}} which implies that: :$\displaystyle 1 < \frac 2 3$ This is a contradiction. Hence $L_1 = L_2$. {{qed}}
Suppose $T$ is a $T_{3 \frac 1 2}$ space. Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice. Let $\alpha \in I$ be arbitrary. From Subspace of Product Space Homeomorphic to Factor Space: :$\struct{S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$. From $T_{3 \frac 1 2}$ Property is Hereditary: :$T_\alpha$ is $T_{3 \frac 1 2}$. From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism: :$\struct {S_\alpha, \tau_\alpha}$ is $T_{3 \frac 1 2}$. Because $\alpha \in I$ was arbitrary then the result follows.
Let $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space for each $\alpha \in I$.  Let $x \in S$. Let $F$ be a closed subset of $S$ such that $x \notin F$. By definition of a closed subset: :$S \setminus F \in \tau$ By definition of the Tychonoff topology there exists an open set $B$ of the natural basis containing $x$ which is disjoint from $F$. By definition of the natural basis, $B$ is of the form: :$\map {\pr_{\alpha_1}^\gets} {U_1} \cap \dotsb \cap \map {\pr_{\alpha_n}^\gets} {U_n}$ where: :$\pr_{\alpha_k}$ is the $\alpha_k$-th projection from $S$  :$U_k$ is open in $S_{\alpha_k}$ for all $1 \le k \le n$ By definition of a $T_{3 \frac 1 2}$ space, for each $1 \le k \le n$ there exists a continuous mapping: :$f_k: S_{\alpha_k} \to \closedint 0 1$ such that: :$\map {f_k} {x_{\alpha_k} } = 1$ and: :$\map {f_k} {S_{\alpha_k} \setminus U_k} = 0$ Let $g_k = f_k \circ \pr_{\alpha_k}$ be the composite mapping of $f_k$ with $\pr_{\alpha_k}$ for each $1 \le k \le n$. From Composite of Continuous Mappings is Continuous each $g_k: S \to \closedint 0 1$ is a continuous mapping. We define $g: S \to \closedint 0 1$ by setting: :$\map g y = \min \set {\map {g_k} {y_{\alpha_k} }: k = 1, \dotsc, n}$ From Minimum Rule for Continuous Functions: :$g$ is continuous.  Now: {{begin-eqn}} {{eqn | l = \map g x       | r = \min \set{\map {g_k} x : k = 1, \dotsc, n}       | c = Definition of $g$ }} {{eqn | r = \min \set {\map {f_k} {\map {\pr_{\alpha_k} } x} : k = 1, \dotsc, n}       | c = {{Defof|Composite Mapping}} }} {{eqn | r = \min \set {\map {f_k} {x_{\alpha_k} } : k = 1, \dotsc, n}       | c = {{Defof|Projection}} }} {{eqn | r = \min \set {1 : k = 1, \dotsc, n}       | c = Definition of $f_k$ }} {{eqn | r = 1       | c = {{Defof|Minimum}} }} {{end-eqn}} Let $y \in F$. By definition of disjoint sets: :$\exists j \in \set {1, \dotsc, n} : y \notin \map {\pr_{\alpha_j}^\gets} {U_j}$ By definition of the inverse image mapping of $\pr_{\alpha_j}$: :$y_{\alpha_j} \notin U_j$ Thus: :$\map {f_j} {y_{\alpha_j} } = 0$ So: {{begin-eqn}} {{eqn | l = \map g y       | r = \min \set {\map {g_k} y : k = 1, \dotsc, n}       | c = Definition of $g$ }} {{eqn | r = \min \set {\map {f_k} {\map {\pr_{\alpha_k} } y} : k = 1, \dotsc, n}       | c = {{Defof|Composite Mapping}} }} {{eqn | r = \min \set {\map {f_k} {y_{\alpha_k} } : k = 1, \dotsc, n}       | c = {{Defof|Projection}} }} {{eqn | r = 0       | c = {{Defof|Minimum}} and $\map {f_k} {y_{\alpha_k} } = 0$ }} {{end-eqn}} Therefore $T$ is a $T_{3 \frac 1 2}$ space.
By definition of half-range Fourier sine series: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \sin n x$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$ Because $\lambda \notin \Z$ we have that $\lambda \ne n$ for all $n$. Thus for $n > 0$: {{begin-eqn}} {{eqn | l = a_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 \sin \lambda x \sin n x \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \pi \intlimits {\frac {\map \sin {\lambda - n} x} {2 \paren {\lambda - n} } - \frac {\map \sin {\lambda + n} x} {2 \paren {\lambda + n} } } 0 \pi       | c = Primitive of $\sin \lambda x \sin n x$ }} {{eqn | r = \frac 2 \pi \paren {\paren {\frac {\sin \paren {\lambda - n} \pi} {2 \paren {\lambda - n} } - \frac {\sin \paren {\lambda + n} \pi} {2 \paren {\lambda + n} } } - \paren {\frac {\sin 0} {2 \paren {\lambda - n} } - \frac {\sin 0} {2 \paren {\lambda + n} } } }       | c =  }} {{eqn | r = \frac 1 \pi \paren {\frac {\sin \paren {\lambda - n} \pi} {\lambda - n} - \frac {\sin \paren {\lambda + n} \pi} {\lambda + n} }       | c = Sine of Multiple of Pi and simplification }} {{eqn | r = \frac 1 \pi \paren {\frac {\sin \lambda \pi \cos n \pi - \cos \lambda \pi \sin n \pi} {\lambda - n} - \frac {\sin \lambda \pi \cos n \pi + \cos \lambda \pi \sin n \pi} {\lambda + n} }       | c = Sine of Sum }} {{eqn | r = \frac {\sin \lambda \pi \cos n \pi} \pi \paren {\frac 1 {\lambda - n} - \frac 1 {\lambda + n} }       | c = Sine of Multiple of Pi and simplification }} {{eqn | r = \frac {\paren {-1}^n \sin \lambda \pi} \pi \frac {\lambda + n - \lambda + n} {\paren {\lambda - n} \paren {\lambda + n} }       | c = Cosine of Multiple of Pi and manipulation }} {{eqn | r = \paren {-1}^n \frac {2 \sin \lambda \pi} \pi \frac n {\lambda^2 - n^2}       | c = Difference of Two Squares }} {{end-eqn}} {{qed|lemma}} Finally: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \sum_{n \mathop = 1}^\infty b_n \sin n x       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \sin \pi} \pi \frac n {\lambda^2 - n^2} \sin n x       | c = substituting for $b_n$ from above }} {{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac n {\lambda^2 - n^2} \sin n x}       | c = simplifying }} {{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n x} {\lambda^2 - n^2} }       | c = further manipulation }} {{end-eqn}} {{qed}}
From the definition of min: :$\map \min {a, b} = \begin{cases} a: & a \le b \\ b: & b \le a \end{cases}$ Let $a < b$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 2 \paren {a + b - \size {a - b} }       | r = \dfrac 1 2 \paren {a + b - \paren {b - a} }       | c = {{Defof|Absolute Value}} }} {{eqn | r = \dfrac 1 2 \paren {a + b - b + a} }} {{eqn | r = \dfrac 1 2 \paren {2 a} }} {{eqn | r = a }} {{eqn | r = \min \set {a, b}       | c = {{Defof|Min Operation}} }} {{end-eqn}} {{qed|lemma}} Let $a \ge b$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 2 \paren {a + b - \size {a - b} }       | r = \dfrac 1 2 \paren {a + b - \paren {a - b} }       | c = {{Defof|Absolute Value}} }} {{eqn | r = \dfrac 1 2 \paren {a + b - a + b} }} {{eqn | r = \dfrac 1 2 \paren {2 b} }} {{eqn | r = b }} {{eqn | r = \min \set {a, b}       | c = {{Defof|Min Operation}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === {{:Continuity Test for Real-Valued Functions/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Continuity Test for Real-Valued Functions/Sufficient Condition}}{{qed}} Category:Continuous Mappings Category:Continuity Test for Real-Valued Functions 1o8c32wdhvtd4kvxbdmbfr4awob4krp
By definition, $f$ is everywhere continuous {{iff}} $f$ is continuous at every point $x \in S$. From Continuity Test for Real-Valued Functions, $f$ is continuous at every point $x \in S$ {{iff}}: :$\forall x \in S : \forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$ {{qed}} Category:Continuity Test for Real-Valued Functions bb6rmo9cdjehak1ju8nm3if157p2jru
From Definite Integral on Zero Interval: {{begin-eqn}} {{eqn | l = \map F {-\pi}       | r = \int_{-\pi}^{-\pi} \map f t \rd t - \dfrac {a_0} 2 \paren {-\pi}       | c =  }} {{eqn | r = 0 - \dfrac {a_0} 2 \paren {-\pi}       | c =  }} {{eqn | r = \dfrac {a_0 \pi} 2       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map F \pi       | r = \int_{-\pi}^\pi \map f t \rd t - \dfrac {a_0} 2 \pi       | c =  }} {{eqn | r = \int_{-\pi}^\pi \paren {\dfrac {\map {f^+} t + \map {f^-} t} 2} \rd t - \dfrac {a_0} 2 \pi       | c = Fourier's Theorem }} {{end-eqn}} {{finish}}
To prove $G$ is isomorphic to $\struct {\R, +}$, we need to find a bijective homorphism $\phi: \openint {-1} 1 \to \R$: :$\forall x, y \in G: \map \phi {x \circ y} = \map \phi x + \map \phi y$ From Group Examples: $\dfrac {x + y} {1 + x y}$: :the identity element of $G$ is $0$ :the inverse of $x$ in $G$ is $-x$. This also holds for $\struct {\R, +}$. This hints at the structure of a possible isomorphism, namely: :$(1): \quad$ that it is an odd function :$(2): \quad$ that it passes through $0$ :$(3): \quad$ that it is defined on the open interval $\openint {-1} 1$. One such function is the inverse hyperbolic arctangent function $\tanh^{-1}$, which is indeed defined on $\openint {-1} 1$ and has the above properties. Hence: {{begin-eqn}} {{eqn | l = \map {\tanh^{-1} } x       | r = \frac 1 2 \map \ln {\frac {x + 1} {x - 1} }       | c = {{Defof|Inverse Hyperbolic Tangent|subdef = Real|index = 2}} }} {{eqn | ll= \leadsto       | l = \map {\tanh^{-1} } {x \circ y}       | r = \frac 1 2 \map \ln {\frac {\frac {x + y} {1 + x y} + 1} {\frac {x + y} {1 + x y} - 1} }       | c = Definition of $\struct {G, \circ}$ }} {{eqn | r = \frac 1 2 \map \ln {\frac {x + y + 1 + x y} {x + y - 1 - x y} }       | c =  }} {{eqn | r = \frac 1 2 \map \ln {\frac {\paren {x + 1} \paren {y + 1} } {\paren {x - 1} \paren {y - 1} } }       | c =  }} {{eqn | r = \frac 1 2 \map \ln {\frac {x + 1} {x - 1} } + \frac 1 2 \map \ln {\frac {y + 1} {y - 1} }       | c =  }} {{eqn | r = \tanh^{-1} x + \tanh^{-1} y       | c =  }} {{end-eqn}} This demonstrates the homorphism between $\struct {G, \circ}$ and $\struct {\R,  +}$. We have that Real Inverse Hyperbolic Tangent Function is Strictly Increasing over $\openint {-1} 1$. Hence from Strictly Monotone Real Function is Bijective, $\tanh^{-1}: \openint {-1} 1 \to \R$ is a bijection. Hence the result that $\struct {G, \circ} \cong \struct {\R,  +}$. {{Qed}}