hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\map {\omega_f} x = 0$. Let $\epsilon > 0$. Suppose that $\forall I: I \in N_x: \omega_f \left({I} \right) \ge \epsilon$. Then by definition, $\map {\omega_f} x \ge \epsilon$.  This contradicts $\map {\omega_f} x = 0$. From this contradiction we deduce that: :$\exists I: I \in N_x: \omega_f \left({I}\right) < \epsilon$ For this particular $I$, there is an open set $O \subset I$ by the definition of open subset neighborhood. Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a subset of $O$. So for our specific $x$, if $y$ satisfies: :$\size {x - y} < \delta$ Then: :$y \in I$ and, if $y \in D$: :$\size {\map f x - \map f y} \le \map {\omega_f} I$ Since $\map {\omega_f} I < \epsilon$ it follows by the definition of continuity that $f$ is continuous at $x$. {{qed|lemma}} === Sufficient Condition === Let $f$ be continuous at $x$. Then $\forall \epsilon > 0: \exists \delta \in \R_{>0}$ such that, if $y \in D$: :$\size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$ Let the interval $I_\delta$ be defined as: :$I_\delta := \openint {x - \delta} {x + \delta}$ $I_\delta$ is an element of $N_x$ as $I_\delta$ is a neighborhood of $x$. We have: {{begin-eqn}} {{eqn | l = \map {\omega_f} {I_\delta}       | r = \sup \set {\size {\map f y - \map f z}: y, z \in I_\delta \cap D}       | c = Definition of $\map {\omega_f} {I_\delta}$ }} {{eqn | r = \sup \set {\size {\map f y - \map f x + \map f x - \map f z}: y, z \in I_\delta \cap D}       | c =  }} {{eqn | o = \le       | r = \sup \set {\size {\map f y - \map f x} + \size {\map f x - \map f z}: y, z \in I_\delta \cap D}       | c = Triangle Inequality for Vectors in Euclidean Space and Supremum of Function is less than Supremum of Greater Function }} {{eqn | r = \sup \set {\size {\map f y - \map f x}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f z}: z \in I_\delta \cap D}       | c = Supremum of Sum equals Sum of Suprema }} {{eqn | r = \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D} }} {{eqn | r = 2 \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D}       | c =  }} {{eqn | o = <       | r = 2 \epsilon       | c = as $\size {\map f x - \map f y} < \epsilon$ whenever $y \in I_\delta \cap D$ }} {{end-eqn}} This gives: {{begin-eqn}} {{eqn | l = \map {\omega_f} x       | r = \inf \set {\map {\omega_f} I: I \in N_x}       | c = Definition of $\map {\omega_f} x$ }} {{eqn | o = \le       | r = \map {\omega_f} {I_\delta}       | c = {{Defof|Infimum of Subset of Real Numbers}} and that $I_\delta \in N_x$ }} {{eqn | o = <       | r = 2 \epsilon       | c =  }} {{end-eqn}} This holds true for any value of $\epsilon$. Thus $\map {\omega_f} x$ must be $0$. Hence the result. {{qed}} Category:Real Analysis 5g3boxs69sq27ql4ga0qr7ixu7lklh9
We start from the result Numbers of Type Integer a plus b root 2 form Subdomain of Reals. We have from that result that $1 + 0 \sqrt 2$ is the unity. Now consider a representative element $a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}$. From Difference of Two Squares we have: :$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$ which leads to: :$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1$ so demonstrating that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$. But for any two given integers $a$ and $b$, the values $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$ are not in general integers. For instance, when $a = 3$ and $b = 4$ we get: :$\dfrac a {a^2 - 2 b^2} = \dfrac 3 {23}$ and $\dfrac b {a^2 - 2 b^2} = \dfrac 4 {23}$ (unless I've got my sums wrong). So not every element of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ has a product inverse, demonstrating that $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ is not a field. {{qed}}
From Ring of Integers Modulo Prime is Field we have that $\struct {\Z_p, +, \times}$ is a field. So Characteristic of Finite Ring with No Zero Divisors applies, and so the characteristic of $\struct {\Z_p, +, \times}$ is prime. The result follows. {{qed}}
For $n = 0$: {{begin-eqn}} {{eqn | l = \map \Gamma 1       | r = \int_0^\infty e^{-t} \rd t       | c = {{Defof|Integral Form of Gamma Function|Gamma Function}} }} {{eqn | r = \bigintlimits {-e^{-t} } 0 \infty       | c = }} {{eqn | r = 0 - \paren {-1} }} {{eqn | r = 1 }} {{end-eqn}} Then by Gamma Difference Equation: :$\forall z \in \Z_{> 0}: \map \Gamma {z + 1} = z \, \map \Gamma z$ Hence the result. {{qed}}
By Equivalence of Local Uniform Convergence and Compact Convergence, $f_n$ converges to $f$ locally uniformly on $U$. Then for any $z \in U$, there is an $\epsilon > 0$ so that: :$B_\epsilon \left({z}\right) \subset U$ and $f_n$ converges uniformly on $B_\epsilon \left({z}\right)$. Let $\gamma$ be any simple closed curve in $B_\epsilon \left({z}\right)$. Since $f_n \to f$ uniformly on $\gamma$ (because $\gamma \subset B_\epsilon \left({z}\right)$), we have: :$\displaystyle \lim_{n \to \infty} \int_\gamma f_n \left({z}\right) \, \mathrm d z = \int_\gamma f \left({z}\right) \, \mathrm d z$ Since each $f_n$ is analytic, we have that: :$\displaystyle \forall n \in \N:  \int_\gamma f_n \left({z}\right) \, \mathrm d z = 0$ So we conclude also that :$\displaystyle \int_\gamma f \left({z}\right) \, \mathrm d z = 0$ Since $\gamma$ was arbitrary, we have by Morera's Theorem that $f$ is analytic in $B_\epsilon \left({z}\right)$. Since $z$ was arbitrary, $f$ is analytic on all of $U$. {{qed}}
Define $z \in \C$ as the value of the complex Riemann integral: :$z = \displaystyle \int_a^b f \left({t}\right) \rd t$ Define $r \in \left[{0 \,.\,.\, \infty }\right)$ as the modulus of $z$, and $\theta \in \left[{0 \,.\,.\, 2 \pi}\right)$ as the argument of $z$. From Modulus and Argument of Complex Exponential, we have $z = re^{i \theta}$. Then: {{begin-eqn}} {{eqn | l = r       | r = ze^{-i \theta}       | c = Reciprocal of Complex Exponential }} {{eqn | r = \int_a^b e^{-i \theta} f \left({t}\right) \rd t       | c = Linear Combination of Complex Integrals }} {{eqn | r = \int_a^b \operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t + i \int_a^b \operatorname{Im} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t       | c = {{Defof|Complex Riemann Integral}} }} {{end-eqn}} As $r$ is wholly real, we have: : $\displaystyle 0 = \operatorname{Im}\left({r}\right) = \int_a^b \operatorname{Im} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t$ Then: {{begin-eqn}} {{eqn | l = r       | r = \int_a^b \operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t }} {{eqn | o = \le       | r = \int_a^b \left\vert{\operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) }\right\vert \rd t       | c = Absolute Value of Definite Integral }} {{eqn | o = \le       | r = \int_a^b \left\vert{e^{-i \theta} f \left({t}\right) }\right\vert \rd t       | c = Modulus Larger than Real Part }} {{eqn | r =  \int_a^b \left\vert{e^{-i \theta} }\right\vert \left\vert{f \left({t}\right) }\right\vert \rd t }} {{eqn | r = \int_a^b \left\vert{f \left({t}\right) }\right\vert \rd t       | c = Modulus of Exponential of Imaginary Number }} {{end-eqn}} As $\displaystyle r = \left|{\int_a^b f \left({t}\right) \rd t}\right|$ by its definition, the result follows. {{qed}}
By definition of contour, $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves. Let $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$. Then: {{begin-eqn}} {{eqn | l = \size {\int_C \map f z \rd z}       | r = \size {\sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}       | c = {{Defof|Complex Contour Integral}} }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \size {\int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}       | c = Triangle Inequality for Complex Numbers }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map f {\map {\gamma_i} t} } \size {\map {\gamma_i'} t} \rd t       | c = Modulus of Complex Integral }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \max_{t \mathop \in \closedint {a_i} {b_i} } \size {\map f {\map {\gamma_i} t} } \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t       | c = Linear Combination of Integrals }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \max_{z \mathop \in \Img C} \size {\map f z} \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t       | c = as $\map {\gamma_i} t \in \Img C$ }} {{eqn | r = \max_{z \mathop \in \Img C} \size {\map f z} \map L C       | c = {{Defof|Length of Contour (Complex Plane)}} }} {{end-eqn}} {{qed}}
Write $f \left({z}\right) = \left({z - \rho_1}\right) \cdots \left({z - \rho_n}\right) g \left({z}\right)$, so $g \left({z}\right) \ne 0$ for $z \in D_r$. It is sufficient to check the equality for each factor of $f$ in this expansion. First let $h \left({z}\right) = z - \rho_k$ for some $k \in \left\{{1, \ldots, n}\right\}$. Making use of the substitution $u = r e^{i \theta} - \rho_k$ we find that: :$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{h \left({r e^{i \theta} }\right)}\right\vert \ \mathrm d \theta = \frac 1 {2 \pi i} \int_\gamma \frac {\ln \left\vert{u}\right\vert} {u + \rho_k} \, \mathrm d u$ where $\gamma$ is a circle of radius $r$ centred at $-\rho_k$, traversed anticlockwise. {{mistake|I think this last statement is False! The circle is not centered at 0 so $\left\vert{u}\right\vert$ is not constant.}} On this circle, $\ln \left\vert{u}\right\vert = \ln r$ is constant, and we have that: :$\displaystyle \int_\gamma \frac 1 {u + \rho_k} \ \mathrm d u = \int_{\left\vert{z}\right\vert \mathop = r} \frac {\mathrm d u} u = 2 \pi i$ Therefore the left hand side of $(1)$ is $\ln r$ as required. To show equality for $g \left({z}\right)$, first observe that by the Residue Theorem: :$\displaystyle \int_{\left\vert{z}\right\vert = r} \frac {\ln g \left({z}\right)} z \, \mathrm d z = 2 \pi i \ln g \left({0}\right)$ Therefore substituting $z = r e^{i \theta}$ we have :$\displaystyle 2 \pi i \ln g \left({0}\right) = i \int_0^{2 \pi} \ln g \left({r e^{i \theta} }\right) \, \mathrm d \theta$ Comparing the imaginary parts of this equality we see that: :$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta = \ln \left\vert{g \left({0}\right)}\right\vert$ as required. {{qed}}
Write :$\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$ so $\map g z \ne 0$ for $z \in D_r$. It is sufficient to check equality for each factor of $f$ in this expansion. When $\cmod z = r$, we have: :$\dfrac 1 z = \dfrac {\overline z} {r^2}$ and: :$\cmod {\dfrac z r} = 1$ where $\overline z$ denotes the complex conjugate of $z$. So: {{begin-eqn}} {{eqn | l = \cmod {\frac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} } }       | r = \frac {\cmod {r^2 - \overline z \rho_k} } {\cmod {r \paren {z - \rho_k} } }       | c =  }} {{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {z / r - \rho_k / r} }       | c =  }} {{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \rho_k / z} }       | c =  }} {{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \overline z \rho_k / r^2} }       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} so the {{LHS}} is $0$. Moreover: :$\\ln \cmod {\dfrac {r^2 - 0 \overline {\rho_k} } {r \paren {0 - \rho_k} } } = \ln \cmod {\rho_k} - \ln r$ so the {{RHS}} is $0$. Therefore, the formula holds for $\dfrac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} }$. Since: :$r^2 - z \overline {\rho_i} = 0 \implies \cmod z = r^2 / \cmod {\overline {\rho_i} } > r$ it follows that $\map g z$ is holomorphic without zeroes on $D_r$. So the {{RHS}} is $\ln \size {\map g 0}$. On the other hand, by the mean value property: :$\displaystyle \dfrac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \map g 0$ as required. {{qed}}
Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$. Let $U_i \cap U_j = \varnothing$ for all $i \ne j$. By Existence of Laurent Series, around each $a_k$ there is an expansion: :$\displaystyle \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$  convergent in $U_k$. Write:  :$\displaystyle X = \bigcup_{i \mathop = 1}^n U_i$ Then, by Contour Integral of Concatenation of Contours:  :$\displaystyle \oint_L \map f z \rd z = \oint_{\partial \paren {U \setminus X} } \map f z \rd z + \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$  As all poles of $f$ in $U$ are contained in $X$, $f$ is holomorphic on $U \setminus X$, and so by the Cauchy-Goursat Theorem:  :$\displaystyle \oint_{\partial \paren {U \setminus X} } \map f z \rd z = 0$ Giving:  :$\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$ We have:  {{begin-eqn}} {{eqn	| l = \oint_{\partial U_k} \map f z \rd z 	| r = \oint_{\partial U_k} \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j \rd z }} {{eqn	| r = \oint_{\partial U_k} \paren {\sum_{j \mathop = -\infty}^{-2} c_j \paren {z - a_k}^j + \frac {c_{-1} } {z - a_k} + \sum_{j \mathop = 0}^\infty c_j \paren {z - a_k}^j} \rd z }} {{eqn	| r = \sum_{j \mathop = -\infty}^{-2} \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z + \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z + \sum_{j \mathop = 0}^\infty \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z }} {{eqn	| r = \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z  	| c = by Contour Integral of Power, terms where $j \ge 0$ and $j < -1$ vanish }} {{eqn	| r = 2 \pi i c_{-1} 	| c = Contour Integral of Power  }} {{eqn	| r = 2 \pi i \Res f {a_k} 	| c = {{Defof|Residue}} }} {{end-eqn}} So: :$\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$ {{qed}} Category:Complex Analysis Category:Named Theorems moyv3ydyrhwo2lmkoqzgr51623zznif
{{Proofread}} Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is analytic. Let $z_0$ be any point in the region $R$ such that: :$\dfrac {\map f z} {z - z_0}$ is analytic everywhere except at $z_0$. We draw a circle $C_1$ with center at $z_0$ and radius $r$ such that $r \to 0$. This makes $C$ and $C_1$ a multiply connected region. {{explain|A diagram at this point would be useful.}} According to Cauchy's Integral Theorem for a multiply connected region: {{begin-eqn}} {{eqn | l = I       | o = :=       | r = \oint_C \frac {\map f z} {z - z_0} \rd z       | c =  }} {{eqn | r = \oint_{C_1} \frac {\map f z} {z - z_0} \rd z       | c =  }} {{eqn | r = \oint_{C_1} \frac {\map f {z_0} + \paren {\map f z - \map f {z_0} } } {z - z_0} \rd z       | c =  }} {{eqn | r = \map f {z_0} \oint_{C_1} \frac {\rd z} {z - z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z       | c =  }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z - z_0       | r = r e^{i \theta}       | c =  }} {{eqn | ll= \leadsto       | l = \d z       | r = i r e^{i \theta} \rd \theta       | c =  }} {{eqn | ll= \leadsto       | l = \oint_{C_1} \frac {\rd z} {z - z_0}       | r = \int_0^{2 \pi} \frac {i r e^{i \theta} } {r e^{i \theta} } \rd \theta       | c =  }} {{eqn | r = i \int_0^{2 \pi} \rd \theta       | c =  }} {{eqn | r = 2 \pi i       | c =  }} {{end-eqn}} Now: :$\displaystyle I = 2 \pi i \map f {z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$ According to Epsilon-Delta definition of limit, for every $\left|{z - z_0}\right| < \delta$ there exists a $\epsilon \in \R_{>0}$ such that: :$\cmod {\map f z - \map f {z_0} } < \epsilon$ Hence: {{begin-eqn}} {{eqn | l = \cmod {\oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}       | o = \le       | r = \oint_{C_1} \frac {\cmod {\map f z - \map f {z_0} } } {z - z_0} \cmod {\d z}       | c =  }} {{eqn | o = \le       | r = \frac {\epsilon} {\delta} \oint_{C_1} \cmod {\d z}       | c =  }} {{eqn | r = 2 \pi \epsilon       | c =  }} {{end-eqn}} {{explain|The precise meaning of $\cmod {\d z}$}} As $\epsilon \to 0$: :$\displaystyle \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$ So: {{begin-eqn}} {{eqn | l = I       | r = \map f {z_0} \oint_{C_1} \frac {\rd z} {z - z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z       | c =  }} {{eqn | r = 2 \pi i \map f {z_0} + 0       | c =  }} {{eqn | ll= \leadsto       | l = \oint_C \frac {\map f z} {z - z_0} \rd z       | r = 2 \pi i \, \map f {z_0}       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Augustin Louis Cauchy|cat = Cauchy}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$ === Basis for the Induction === $\map P 0$ holds, as this is: :$\displaystyle \map f a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}} \rd z$ which is Cauchy's Integral Formula. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map {f^{\paren k} } a = \frac {k!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z$ Then we need to show: :$\displaystyle \map {f^{\paren {k + 1} } } a = \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \frac {\rd} {\rd a} \map {f^{\paren k} } a       | r = \frac {k!} {2 \pi i} \int_{\partial D} \frac {\rd} {\rd a} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z       | c = }} {{eqn | r = \frac {k!} {2 \pi i}\int_{\partial D} \frac {\paren {k + 1} \map f z} {\paren {z - a}^{k + 2} } \rd z       | c = }} {{eqn | r = \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z       | c = }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$ {{qed}}
Suppose to the contrary. Let $K$ be a subfield of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$. Then: :$\exists a \in K: n \cdot a = 0$ But as $K$ is a subfield of $\C$ it follows that $K \subseteq \C$ which means: :$\exists a \in \C: n \cdot a = 0$ Thus, by definition of characteristic: :$0 < \Char \C \le n$ But $\C$ is infinite and so $\Char \C = 0$. From that contradiction follows the result. {{qed}}
If $a = 0_R$ then $n \cdot a = 0_R$ is immediate. So let $a \in R: a \ne 0_R$. Then: {{begin-eqn}} {{eqn | l = n \cdot a       | r = n \cdot \paren {1_R \circ a}       | c = {{Defof|Unity of Ring}} }} {{eqn | r = \paren {n \cdot 1_R} \circ a       | c = Powers of Ring Elements }} {{eqn | r = 0_R \circ a       | c = {{Defof|Characteristic of Ring}} }} {{eqn | r = 0_R       | c = {{Defof|Ring Zero}} }} {{end-eqn}} {{qed}}
We have that $\struct {R, +, \circ}$ is finite, so its additive group $\struct {R, +}$ is likewise finite. The result follows by Element of Finite Group is of Finite Order and the definition of characteristic. {{qed}}
By Convergence Exponent is Less Than Order, $f$ has finite exponent of convergence $\tau \leq \omega$. By Relation Between Rank and Exponent of Convergence, $f$ has finite rank $p\leq\omega$. {{ProofWanted}}
From Gamma Function is Unique Extension of Factorial: {{begin-eqn}} {{eqn | n = 1       | l = \left({y + n}\right)^y n!       | r = \left({y + n}\right)^y \Gamma \left({n + 1}\right)       | c = Gamma Function Extends Factorial }} {{eqn | o = \le       | r = \Gamma \left({y + n + 1}\right)       | c =  }} {{eqn | o = \le       | r = \left({n + 1}\right)^y \Gamma \left({n + 1}\right)       | c =  }} {{eqn | r = \left({n + 1}\right)^y n!       | c =  }} {{end-eqn}} for $0 < y \le 1$ and $n \in \N$. Let $x$ be given. Let $n + 1$ be the largest natural number such that $n + 1 \le x$. Let $x = y + n + 1$, and thus $0 < y \le 1$. Then: {{begin-eqn}} {{eqn | l = \frac {\Gamma \left({x}\right)}{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }       | o = \le       | r = \frac {\left({n + 1}\right)^y n!}{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }       | c = from $(1)$ }} {{eqn | o = \sim       | r = \frac {\left({n + 1}\right)^y n! \sqrt{\left({2 \pi}\right)} n^n n^{1/2} e^{-n} }{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }       | c = Stirling's Formula }} {{eqn | r = \frac {\left({n + 1}\right)^y n! n^n n^{1/2} e^{-n} }{\left({y + n + 1}\right)^{y + n} \left({y + n + 1}\right)^{1/2} e^{-y-n-1} }       | c =  }} {{eqn | r = \left({\frac {n + 1} {y + n + 1} }\right)^y \left({\frac n {y + n + 1} }\right)^{1/2} \left({1 + \frac {y + 1} n }\right)^{-n} e^{y+1}       | c =  }} {{eqn | o = \to       | r = 1 \cdot 1 \cdot \frac 1 {e^{y + 1} } e^{y+1} \text { as } n \to \infty       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Similarly for the {{RHS}}. The result follows from Gamma Function Extends Factorial. {{qed}}
Let: :$\displaystyle \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$ Then $\map F x$ is $1$-periodic (because of absolute convergence), and has Fourier coefficients: {{begin-eqn}} {{eqn | l = \hat F_k       | r = \int_0^1 \sum_{n \mathop \in \Z} \map f {x + n} e^{-2\pi i k x} \rd x       | c =  }} {{eqn | r = \sum_{n \mathop \in \Z} \int_0^1 \map f {x + n} e^{-2\pi i k x} \rd x       | c = because $f$ is Schwarz, so convergence is uniform }} {{eqn | r = \sum_{n \mathop \in \Z} \int_n^{n + 1} \map f x e^{-2\pi i k x} \rd x       | c =  }} {{eqn | r = \int_\R \map f x e^{-2\pi i k x} \rd x       | c =  }} {{eqn | r = \map {\hat f} k       | c = where $\hat f$ is the Fourier transform of $f$ }} {{end-eqn}} Therefore by the definition of the Fourier series of $f$: :$\displaystyle \map F x = \sum_{k \mathop \in \Z} \map {\hat f} k e^{i k x}$ Choosing $x = 0$ in this formula: :$\displaystyle \sum_{n \mathop \in \Z} \map f n = \sum_{k \mathop \in \Z} \map {\hat f} k$ as required. {{qed}} {{Namedfor|Siméon-Denis Poisson|cat = Poisson}}
We have the Weierstrass products: :$\displaystyle \map \sin {\pi z} = \pi z \prod_{n \mathop \ne 0} \paren {1 - \frac z n} \map \exp {\frac z n}$ From the Weierstrass form of the Gamma function: :$\displaystyle \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \map \exp {-\frac z n}$ from which: {{begin-eqn}} {{eqn   | l = \dfrac 1 {-z \, \map \Gamma z \, \map \Gamma {-z} } 	| r = \frac {-z^2 \, \map \exp {\gamma z} \, \map \exp {-\gamma z} } {-z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \paren {1 - \frac z n} \, \map \exp {\frac z n} \, \map \exp {-\frac z n} }} {{eqn   | r = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2} } }} {{eqn   | r = \dfrac {\map \sin {\pi z} } \pi  	| c = Euler Formula for Sine Function }} {{end-eqn}} whence: {{begin-eqn}} {{eqn | l = \map \Gamma z \, \map \Gamma {1 - z}       | r = -z \, \map \Gamma z \, \map \Gamma {-z}       | c = Gamma Difference Equation }} {{eqn | r = \frac \pi {\map \sin {\pi z} } }} {{end-eqn}} {{qed}} {{Namedfor|Leonhard Paul Euler|cat = Euler}}