hcju/proofwikiqa huggingface dataset:
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as: :$\map f x = \cos \lambda x$ From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as: :$\displaystyle \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$ Because of the nature of this expansion, we have that: :$\map f {0^+} = \map f {0^-}$ and so the expansion holds for $x = 0$. So, setting $x = 0$: {{begin-eqn}} {{eqn | l = \cos 0       | r = \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\lambda^2 - n^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac \pi {\sin \lambda \pi}       | r = 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac 1 {\lambda^2 - n^2} }       | c = Cosine of Zero is One and rearrangement }} {{eqn | ll= \leadsto       | l = \pi \csc \lambda \pi       | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda} {\lambda^2 - n^2}       | c = {{Defof|Real Cosecant Function}} }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\lambda + n + \lambda - n} {\paren {\lambda + n} \paren {\lambda - n} }       | c = Difference of Two Squares }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\lambda - n} {\paren {\lambda + n} \paren {\lambda - n} } + \frac {\lambda + n} {\paren {\lambda + n} \paren {\lambda - n} } }       | c =  }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }       | c =  }} {{end-eqn}} This expression is fine as is, but to obtain the form we set out to prove, we observe: {{begin-eqn}} {{eqn | o =       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} } }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \paren {\frac 1 {n - 1 - \lambda} }       | c = Both converge by Power Series Expansion for Logarithm of 1 + x }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} }       | c = Translation of Index Variable of Summation }} {{eqn | r = - \paren {-1}^0 \paren {\frac 1 {0 - \lambda} } }} {{eqn | r = \frac 1 \lambda }} {{end-eqn}} and thus: {{begin-eqn}} {{eqn | o =       | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} } }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} } + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} } }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} } }} {{end-eqn}} The result follows. {{qed}}
Let $p = \deg f$ and $q = \deg g$ be their degrees. By Degree of Product of Polynomials over Integral Domain, $\map \deg {f g} = p + q$. For a natural number $n \ge 0$, let: :$a_n$ be the coefficient of the monomial $x^n$ in $f$. :$b_n$ be the coefficient of the monomial $x^n$ in $g$. By Coefficients of Product of Two Polynomials, the coefficient of $x^{p + q}$ in $f g$ is the sum: :$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k}$ By Indexed Summation over Interval of Length One: :$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k} = a_p b_q = c d$ {{qed}} Category:Polynomial Theory i4jft09o5r7jccya4040z7whofeekm4
As $D$ is a principal ideal domain, every ideal of $D$ is a principal ideal $\ideal r$ generated by some $r \in D$. So, let $\ideal p$ be an arbitrary prime ideal of $D$ generated by $p$, where $p \ne 0_R$. As $\ideal p$ is prime, $p$ is irreducible. {{explain|Prove the above}} The result follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal. {{qed}}
By Intersection of Ideals of Ring contains Product, $\mathfrak a \mathfrak b \subseteq \mathfrak a \cap \mathfrak b$. It remains to show that $\mathfrak a \mathfrak b \supseteq \mathfrak a \cap \mathfrak b$. Let $c \in \mathfrak a \cap \mathfrak b$. Because $\mathfrak a$ and $\mathfrak b$ are coprime, there exist $x \in \mathfrak a$, $y \in \mathfrak b$ with $x+y = 1$. Then $c = cx + cy$. Because: :$c \in \mathfrak b$ and $x \in \mathfrak a$ :$c \in \mathfrak a$ and $y \in \mathfrak b$ we have $c \in \mathfrak a \mathfrak b$. {{qed}} Category:Coprime Ideals 7h9lrmtswmws7oipm2o41dmvsxhw377
The proof goes by induction on $n$. For $n = 1$, the statement is trivial. Let $n > 1$. Suppose that the statement holds for $n - 1$ prime ideals. It is to be shown that the statement holds for $n$ prime ideals. {{AimForCont}} suppose that $\mathfrak a \nsubseteq \mathfrak p_i$ for all $i \in \set {1, \ldots, n}$. By the induction hypothesis, for all $i$ we have: :$\mathfrak a \nsubseteq \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$ For all $i$, let $x_i \in \mathfrak a \setminus \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$. Suppose there exists $i$ such that $x_i \notin \mathfrak p_i$. Then: :$x_i \in \mathfrak a \setminus \ds \bigcup_j \mathfrak p_j$ and we conclude that: :$\mathfrak a \nsubseteq \ds \bigcup_j \mathfrak p_j$ which is a contradiction. {{qed|lemma}} The other possibility is that for all $i$: :$x_i \in \mathfrak p_i$ Let: :$y = \ds \sum_{i \mathop = 1}^n \prod_{\substack{j \mathop = 1 \\ j \neq i}}^n x_j$ be defined as a summation of products. All terms except the $i$th are in $\mathfrak p_i$. Hence for all $i$: :$y \notin \mathfrak p_i$ {{explain|this needs to cite some results about summations}} Thus: :$y \in \mathfrak a \setminus \bigcup_j \mathfrak p_j$ This also leads to a contradiction. {{qed}}
By Euler's Reflection Formula: :$\map \Gamma {i t} \, \map \Gamma {1 - i t} = \pi \, \map \csc {\pi i t}$ From Gamma Difference Equation:  :$-i t \, \map \Gamma {i t} \, \map \Gamma {-i t} = \pi \, \map \csc {\pi i t}$ Then: {{begin-eqn}} {{eqn | l = \cmod {-i t} \cmod {\map \Gamma {i t} } \cmod {\map \Gamma {-i t} }       | r = \cmod t \cmod {\map \Gamma {i t} } \cmod {\overline {\map \Gamma {i t} } }       | c = Complex Conjugate of Gamma Function }} {{eqn | r = \cmod t \cmod {\map \Gamma {i t} }^2       | c = $\cmod z = \cmod {\overline z}$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \cmod {\pi \, \map \csc {\pi i t} }       | r = \cmod {-i \pi \, \map \csch {\pi t} }       | c = Hyperbolic Sine in terms of Sine }} {{eqn | r = \pi \, \map \csch {\pi \cmod t}       | c = Hyperbolic Sine Function is Odd }} {{end-eqn}} So: :$\cmod {\map \Gamma {i t} }^2 = \dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t}$ As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root and write: :$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t} }$ However, by Hyperbolic Sine Function is Odd:  :$\dfrac {\pi \, \map \csch {-\pi t} } {-t} = \dfrac {-\pi \, \map \csch {\pi t} } {-t} = \dfrac {\pi \, \map \csch {\pi t} } t$ Hence we can remove the modulus and simply write:  :$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \csch \pi t} t}$ {{qed}}
Let $\cdot : A \times B \to B$ the ring action of $B$. === 1 implies 2 === Let $C$ equal the underlying ring of $B$ and $f : A \to C$ equal the canonical mapping to the unital algebra $B$. We show that $B$ is the algebra defined by $f$. ==== Addition ==== By definition of the underlying ring of $B$, the addition of $C$ is the addition of $B$, say $+$. By definition of the module defined by $f$, the addition of the algebra defined by $f$ is also $+$. ==== Multiplication ==== By definition of the underlying ring of $B$, the multiplication of $C$ is the ring product of $B$, say $\times$. By definition of the algebra defined by $f$, its multiplication is also $\times$. ==== Ring action ==== It remains to show that the ring action $\cdot$ of $B$ is the ring action $*$ of the module defined by $f$. We have, for $a \in A$ and $b \in B$: {{begin-eqn}} {{eqn | l = a * b       | r = f(a) \times b       | c = {{defof|Module Defined by Ring Homomorphism}} }} {{eqn | r = (a \cdot 1_B) \times b       | c = {{defof|Canonical Homomorphism from Ring to Unital Algebra}} }} {{eqn | r = a \cdot (1_B \times b)       | c = {{defof|Bilinear Mapping}} }} {{eqn | r = a \cdot b       | c = {{defof|Unit of Algebra}} }} {{end-eqn}} {{qed|lemma}} === 2 implies 1 === Let $B$ equal the algebra defined by $f$. ==== Addition ==== By definition of the module defined by $f$, the addition of $B$ is the addition of $C$, say $+$. By definition of the underlying ring of $B$, its addition is also $+$. ==== Multiplication ==== By definition of the algebra defined by $f$, the multiplication of $B$ is the ring product of $C$, say $\times$. By definition of the underlying ring of $B$, its multiplication is also $\times$. Thus $C$ is the underlying ring of $B$. ==== Homomorphism ==== By Identity is Unique, the unit $1_B$ of $B$ equals the unity $1_C$ of $C$. Let $g : A \to B$ be the canonical mapping. We show that $g=f$. We have, for $a \in A$: {{begin-eqn}} {{eqn | l = g(a)       | r = a \cdot 1_B       | c = {{defof|Canonical Homomorphism from Ring to Unital Algebra}} }} {{eqn | r = a \cdot 1_C       | c = $1_B = 1_C$ }} {{eqn | r = f(a) \times 1_C       | c = {{defof|Module Defined by Ring Homomorphism}} }} {{eqn | r = f(a)       | c = {{defof|Unity of Ring}} }} {{end-eqn}} {{qed}} Unital Associative Commutative Algebra 67ke3vq45ececccbro88t31mv49533m
We have that: :$\pi^2 - \paren {-x}^2 = \pi^2 - x^2$ and so $\map f x$ is even on $\openint {-\pi} \pi$. It follows from Fourier Series for Even Function over Symmetric Range: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x$ Thus by definition of $f$: {{begin-eqn}} {{eqn | l = a_0       | r = \frac 2 \pi \int_0^\pi \map f x \rd x       | c = Cosine of Zero is One }} {{eqn | r = \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \pi \intlimits {\pi^2 x - \frac {x^3} 3} 0 \pi       | c = Primitive of Power }} {{eqn | r = \frac 2 \pi \paren {\paren {\pi^3 - \frac {\pi^3} 3} - \paren {0 - \frac 0 3} }       | c =  }} {{eqn | r = \frac 2 \pi \paren {\frac {2 \pi^3} 3}       | c =  }} {{eqn | r = \frac {4 \pi^2} 3       | c =  }} {{end-eqn}} {{qed|lemma}} Then for $n > 0$: {{begin-eqn}} {{eqn | l = a_n       | r = \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x       | c =  }} {{eqn | r = \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \cos n x \rd x       | c = Definition of $f$ }} {{eqn | r = 2 \pi \int_0^\pi \cos n x \rd x - \frac 2 \pi \int_0^\pi x^2 \cos n x \rd x       | c = Linear Combination of Definite Integrals }} {{end-eqn}} Splitting this up into two: {{begin-eqn}} {{eqn | o =        | r = 2 \pi \int_0^\pi \cos n x \rd x       | c =  }} {{eqn | r = 2 \pi \intlimits {\frac {\sin n x} n} 0 \pi       | c = Primitive of $\cos a x$ }} {{eqn | r = 2 \pi \paren {\frac {\sin n \pi} n - \frac {\sin 0} n}       | c =  }} {{eqn | r = 2 \pi \paren {0 - 0}       | c = Sine of Multiple of Pi }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = -\frac 2 \pi \int_0^\pi x^2 \cos n x \rd x       | c =  }} {{eqn | r = -\frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi       | c = Primitive of $x^2 \cos n x$ }} {{eqn | r = -\frac 2 \pi \paren {\paren {\frac {2 \pi \cos n \pi} {n^2} + \paren {\frac {\pi^2} n - \frac 2 {n^3} } \sin n \pi} - \paren {\frac {0 \cos 0} {n^2} + \paren {\frac {0^2} n - \frac 2 {n^3} } \sin 0} }       | c =  }} {{eqn | r = -\frac 2 \pi \paren {\frac {2 \pi \cos n \pi} {n^2} }       | c = Sine of Multiple of Pi and removing vanishing terms }} {{eqn | r = -\frac {4 \cos n \pi} {n^2}       | c =  }} {{eqn | r = -\frac {4 \paren {-1}^n} {n^2}       | c = Cosine of Multiple of Pi }} {{eqn | r = \paren {-1}^{n - 1} \frac 4 {n^2}       | c = simplification }} {{end-eqn}} {{qed|lemma}} Finally: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x       | c =  }} {{eqn | r = \frac 1 2 \frac {4 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 4 {n^2} \cos n x       | c = substituting for $a_0$ and $a_n$ from above }} {{eqn | r = \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}       | c = simplifying }} {{end-eqn}} {{qed}}
=== 1 implies 2 === Let there exist an $A$-algebra homomorphism $A_f \to A_g$. By Existence of Homomorphism between Localizations of Ring, $f$ is contained in the saturation of the set of powers $\set {g^n : n \in \N}$. That is, $f$ divides some power of $g$. {{qed|lemma}} === 2 implies 1 === Let $f$ divide some power of $g$. Say $f d = g^n$. Then for all $m \in \N$, $f^m d^m = g^{nm}$. Thus $f^m$ divides some power of $g$. Thus the set of powers $\set {f^m : m \in \N}$ is contained in the saturation of $\set {g^n : n \in \N}$. By Existence of Homomorphism between Localizations of Ring, there exists an $A$-algebra homomorphism $A_f \to A_g$. {{qed|lemma}} === 3 iff 4 === By definition, a principal open subset $\map D f$ is the complement in the spectrum $\Spec A$ of the vanishing set $\map V f$. By Relative Complement inverts Subsets, $\map V f \subseteq \map V g$ {{iff}} $\map D f \supseteq \map D g$. {{qed|lemma}} === 2 implies 3 === Let $I$ be a prime ideal containing $f$. Let $f$ divide some power of $g$. Say $f d = g^n$. By definition of ideal, $g^n = f d \in I$. By definition of prime ideal, $g \in I$. So every prime ideal containing $f$ also contains $g$. {{qed|lemma}} === 3 implies 2 === {{proof wanted}} Category:Localization of Rings ij451qc4p6k2jezeki27pu6q472klvj
{{missingLinks}} By Noether Normalization Lemma, we find a finite and injective morphism: :$\alpha: k \left[{x_1, \ldots, x_n}\right] \to L$ If we can prove that $n = 0$, the proof is complete. Let $n > 0$. Then: :$x_1 \in k \left[{x_1, \dotsc, x_n}\right]$ and: :$\alpha \left({x_1}\right) \ne 0$ We have that $\alpha \left({x_1}\right)^{-1}$ is integral over $k \left[{x_1, \dotsc, x_n}\right]$. Thus there exists a $m \in \N$ and $a_0, \dotsc, a_{m-1} \in k \left[{x_1, \dotsc, x_n}\right]$ such that: :$\displaystyle \alpha \left({x_1}\right)^{-m} + \sum_{i \mathop = 0}^{m-1} \alpha \left({a_i}\right) \alpha \left({x_1}\right)^{-i} = 0$ If we multiply this by $\alpha \left({x_1}\right)^m$, we find that: {{begin-eqn}} {{eqn | l = 0       | r = 1 + \sum_{i \mathop = 0}^{m - 1} \alpha \left({a_i}\right) \alpha \left({x_1}\right)^{m - i} }} {{eqn | r = \alpha\left({1 +  x_1 \left({\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }\right)}\right) }} {{end-eqn}} and thus, since $\alpha$ is injective, we find that: :$\displaystyle 1 = x_1 \left({- \sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }\right)$ which means that $x_1$ is invertible.  This contradiction shows that $n = 0$. {{qed}} {{Namedfor|Oscar Zariski}} Category:Commutative Algebra Category:Field Extensions 6yb9f0ademarqp30nbp22rgs0xsbytb
Taking $\phi = N$ in the proof of Cayley-Hamilton Theorem for Finitely Generated Modules we see that $N$ satisfies: :$p_N \left({x}\right) = \det \left({x \cdot I_n - N}\right) = 0$ Take $\mathfrak a$ to be the ideal generated by the entries of $N$. {{qed}} {{explain}} Category:Linear Algebra 12a2gfrxh7ezx78uzydbd8w2iqbcxjm
Let $m_1, \ldots, m_n$ be a generating set for $M$. Then for each $i$, $\phi \left({m_i}\right) \in \mathfrak a M$, say: :$\displaystyle \phi \left({m_i}\right) = \sum_{j \mathop = 1}^n a_j m_j$ for $i = 1, \ldots, n$. {{explain|Is it significant that $\phi \left({m_i}\right)$ shares the same symbol as $\phi$, or would it be clearer to use a separate symbol?}} Thus for each $i$: :$(1): \quad \displaystyle \sum_{j \mathop = 1}^n \left[{\delta_{ij} \phi - a_{ij} }\right] m_i = 0$ where $\delta_{ij}$ is the Kronecker delta. Now let $\Delta$ be the matrix defined as: : $\Delta := \left({\phi \delta_{ij} - a_{ij} }\right)$ {{explain|$\phi$ and $\delta_{ij}$ have changed places -- is this significant? Otherwise consistency in presentation is to be aimed for.}} Let $\operatorname{adj} \left({\Delta}\right)$ be the adjugate matrix of $\Delta$. Recall Cramer's Rule: {{begin-eqn}} {{eqn | l = \operatorname{adj} \left({\Delta}\right) \cdot \Delta       | r = \Delta \cdot \operatorname{adj} \left({\Delta}\right)       | c =  }} {{eqn | r = \det \left({\Delta}\right) \cdot \mathbf I_n       | c =  }} {{end-eqn}} {{Explain|Explanation needed as to what $\cdot$ means in this context, why it is significant to commute the factors of the expression, and how the final expression arises (probably explained once Cramer's rule is posted up.}} Multiplying through by $\operatorname{adj} \left({\Delta}\right)$ in $(1)$ and applying Cramer's Rule: :$\displaystyle \sum_{j \mathop = 1}^n \det \left({\Delta}\right) m_i = 0$ Therefore $\det \left({\Delta}\right)$ annihilates each $m_i$ and is the zero endomorphism of $M$. But $\det \left({\phi \delta_{ij} - a_{ij}}\right)$ is a monic polynomial in $\phi$ with coefficients in $\mathfrak a$. Thus we have an equation of the required form. {{qed}} {{Namedfor|Arthur Cayley|cat = Cayley|name2 = William Rowan Hamilton|cat2 = Hamilton}} Category:Commutative Algebra 29rl82viqzs2luy06wxty7atbibni97
{{missingLinks}} We have an injective morphism: :$\dfrac A {f^{-1} \left({\mathfrak m}\right)} \to \dfrac B {\mathfrak m}$ We have that $\dfrac B {\mathfrak m}$ is a field extension of $k$ which is finitely generated Thus, by Zariski's Lemma, $\dfrac B {\mathfrak m}$ is a finite field extension.  By Subalgebra of Finite Field Extension is Field, $\dfrac A {f^{-1} \left({\mathfrak m}\right)}$ is a field. Thus $f^{-1} \left({\mathfrak m}\right)$ is a maximal ideal. {{qed}}
By Integral Ring Extension is Integral over Intermediate Ring, $E$ is integral over $A$. Let $a \in A$ be nonzero. Because $E$ is a field, $a$ is a unit of $E$. By Ring Element is Unit iff Unit in Integral Extension, $a$ is a unit of $A$. Thus $A$ is a field. {{qed}}
Consider the function: :$\map f x = \begin{cases} 0 & : x = a \\ \dfrac 1 {\sqrt{x - a} } & : x \in \hointl a b \end{cases}$ Since $\dfrac 1 {\sqrt{x - a} }$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$. Therefore, $f$ satisfies $(1)$ in the requirements of a piecewise continuous function with improper integrals for the subdivision $\set {a, b}$ of $\closedint a b$. Also:  {{begin-eqn}} {{eqn | l = \int_{a+}^{b-} \map f x \rd x       | r = \int_{a+}^{b-} \dfrac 1 {\sqrt{x - a} } \rd x }} {{eqn | r = \bigintlimits {2 \sqrt {x - a} } {a+} {b-} }} {{eqn | r = \lim_{x \mathop \to b-} 2 \sqrt{x - a} - \lim_{x \mathop \to a+} 2 \sqrt{x - a} }} {{eqn | r = 2 \sqrt {b - a} - 2 \sqrt {a - a}       | c = as $2 \sqrt{x - a}$ is left-continuous at $b$ and right-continuous at $a$ }} {{eqn | r = 2 \sqrt{b - a} }} {{end-eqn}} Hence $\displaystyle \int_{a+}^{b-} \map f x \rd x$ exists. Thus $f$ is a piecewise continuous function with improper integrals. However, we have that $\map f x$ approaches $\infty$ as $x$ approaches $a$ from above. Thus $f$ is not bounded. Therefore $f$ is not piecewise continuous and bounded. Hence the result. {{qed}}
{{begin-eqn}} {{eqn	| l = \cmod {\map \Gamma {\frac 1 2 + i t} }^2 	| r = \map \Gamma {\frac 1 2 + i t} \overline {\map \Gamma {\frac 1 2 + i t} } 	| c = Modulus in Terms of Conjugate }} {{eqn	| r = \map \Gamma {\frac 1 2 + i t} \map \Gamma {\frac 1 2 - i t} 	| c = Complex Conjugate of Gamma Function }} {{eqn	| r = \map \Gamma {\frac 1 2 + i t} \map \Gamma {1 - \paren {\frac 1 2 + i t} } 	| c = applying some algebra }} {{eqn	| r = \pi \map \csc {\pi \paren {\frac 1 2 + i t} } 	| c = Euler's Reflection Formula }} {{eqn	| r = \pi \map \sec {\pi i t} 	| c = Sine of Complement equals Cosine }} {{eqn	| r = \pi \map \sech {\pi t} 	| c = Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root of both sides and write: :$\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$ {{qed}} Category:Gamma Function Category:Hyperbolic Secant Function 2livusfxejek9s219i2q3olje9w7s1s
Let $p > 0$. We have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin p x} x \rd x 	| r = \frac 1 p \int_0^\infty \frac {\sin t} { \frac 1 p t} \rd t 	| c = substituting $t = p x$ }} {{eqn	| r = \int_0^\infty \frac {\sin t} t \rd t }} {{eqn	| r = \frac \pi 2 	| c = Dirichlet Integral }} {{end-eqn}} Then: {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin \left({- p x}\right)} x \rd x 	| r = -\int_0^\infty \frac {\sin p x} x \rd x 	| c = Sine Function is Odd }} {{eqn	| r = -\frac \pi 2 	| c = per above computation }} {{end-eqn}} For $p = 0$, we have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin 0 x} x \rd x  	| r = \int_0^\infty \frac 0 x \rd x 	| c = Sine of Zero is Zero }} {{eqn	| r = 0 }} {{end-eqn}} Hence the result.  {{qed}}
The inclusion $\operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right) \subseteq \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$ follows from: * Ideal of Ring is Contained in Radical: : $\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$ * Sum of Larger Ideals is Larger: : $\mathfrak a + \mathfrak b \subseteq \operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)$ * Radical of Ideal Preserves Inclusion: : $\operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right) \subseteq \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$ For the other inclusion, let $x \in \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$. Then there exists $n \in \N$ such that the power $x^n \in \operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)$. By definition of sum of ideals, there exist $a \in \operatorname{Rad} \left({\mathfrak a}\right)$ and $b \in \operatorname{Rad} \left({\mathfrak b}\right)$ with $x^n = a + b$. Let $p, q \in \N$ with $a^p \in \mathfrak a$ and $b^q \in \mathfrak b$. By the Binomial Theorem: : $\left({a + b}\right)^{p + q - 1} \in \mathfrak a + \mathfrak b$ {{explain|a link needed here}} Thus $x^{n \cdot \left({p + q - 1}\right)} \in \mathfrak a + \mathfrak b$. That is: :$x \in \operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right)$. {{qed}} Category:Radical of Ideals cemw7ua76cs2kx2t62r8avtbr022rwi
Let $a \in \mathfrak a$. By definition of power: : $a^1 = a$ Thus: : $a \in \operatorname{Rad} \left({\mathfrak a}\right)$ {{qed}} Category:Radical of Ideals 893sgr06yi4pmfepysm4qb70v66010g
Let $x \in \operatorname{Rad} \left({\mathfrak a}\right)$. We show that $x \in \operatorname{Rad} \left({\mathfrak b}\right)$. By definition of radical, there exists $n \in \N$ such that the power $x^n \in \mathfrak a$. Because $\mathfrak a \subseteq \mathfrak b$, also $x^n \in \mathfrak b$. Thus $x \in \operatorname{Rad} \left({\mathfrak b}\right)$. {{qed}} Category:Radical of Ideals k3nevcqozzdlobgpyk4rf16r31o9smw
We have: {{begin-eqn}} {{eqn | l = \operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right)       | r = \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)       | c = Radical of Sum of Ideals }} {{eqn | r = \operatorname{Rad} \left({\left({1}\right)}\right)       | c = $\operatorname{Rad} \left({\mathfrak a}\right)$ and $\operatorname{Rad} \left({\mathfrak b}\right)$ are coprime }} {{eqn | r = \left({1}\right)       | c = Radical of Unit Ideal }} {{end-eqn}} By Unit Ideal iff Radical is Unit Ideal: : $\mathfrak a + \mathfrak b = \left({1}\right)$ That is, $\mathfrak a$ and $\mathfrak b$ are coprime. {{qed}} Category:Radical of Ideals aoq7mk0hcvnnhw4vsczqcag1xuo85ha