hcju/proofwikiqa huggingface dataset:
Let $\left({X, \tau}\right)$ be a topological space. Let $U_1, U_2 \in \tau$ such that: : $U_1 \cup U_2 = X$ : $U_1 \cap U_2 \ne \varnothing$ is connected Let $\ast \in U_1 \cap U_2$. Let: : $i_k : U_1 \cap U_2 \hookrightarrow U_k$ : $j_k : U_k \hookrightarrow U_1 \cup U_2$ be inclusions. For the sake of simplicity let: :$\pi_1 \left({X}\right) = \pi_1 \left({X, \ast}\right)$ It is to be shown that $\pi_1 \left(X\right)$ is the amalgamated free product: :$\pi_1 \left({U_1}\right) *_{\pi_1 \left({U_1 \cap U_2}\right)} \pi_1 \left({U_2}\right)$ {{ProofWanted}} {{Namedfor|Karl Johannes Herbert Seifert|name2 = Egbert Rudolf van Kampen|cat = Seifert|cat2 = van Kampen}} Category:Category Theory jwn0lbhl81z1yclqkyw9y203rdwherz
Let $A$ and $B$ be subsets of the real numbers $\R$. Let $g^{-1} \sqbrk A$ and $h^{-1} \sqbrk B$ denote the preimages of $A$ and $B$ under $g$ and $h$ respectively.  Applying the definition of independent random variables: {{begin-eqn}} {{eqn | l = \map \Pr {\map g X \in A, \map h Y \in B}       | r = \map \Pr {X \in g^{-1} \sqbrk A, Y \in h^{-1} \sqbrk B}       | c = {{Defof|Preimage of Subset under Mapping}} }} {{eqn | r = \map \Pr {X \in g^{-1} \sqbrk A} \map \Pr {Y \in h^{-1} \sqbrk B}       | c = {{Defof|Independent Random Variables}} }} {{eqn | r = \map \Pr {\map g X \in A} \map \Pr {\map h Y \in B}       | c = {{Defof|Preimage of Subset under Mapping}} }} {{end-eqn}} Hence $\map g X$ and $\map h Y$ are independent random variables. {{qed}}
Let $i = \sqrt{-1}$. Let $\expect X$ denote the expectation of $X$. {{begin-eqn}} {{eqn | l = \map {\phi_{X + Y} } t       | r = \expect {e^{i t \paren {X + Y} } }       | c = {{Defof|Characteristic Function of Random Variable}} }} {{eqn | r = \expect {e^{i t X} e^{i t Y} }       | c =  }} {{eqn | r = \expect {e^{i t X} } \expect {e^{ i t Y} }       | c = Functions of Independent Random Variables are Independent, Expected Value of Product is Product of Expected Value }} {{eqn | r = \map {\phi_X} t \map {\phi_Y} t       | c =  }} {{end-eqn}} Hence:  :$\phi_{X + Y} = \phi_X \phi_Y$ {{qed}}
Let $\NN_{\bar s}$ denote the neighborhood filter of $\bar s$. By definition of the lower limit, there exists a sequence of open neighborhoods $\sequence {V_k}_{k \mathop \in \N} \in \NN_{\bar s}$ such that: :$\displaystyle \lim_{k \mathop \to \infty} \set {\inf_{s \mathop \in V_k} \map f s} = \liminf_{s \mathop \to \bar s} \map f s$ This implies that $\forall \epsilon > 0 \exists k_\epsilon \in \N$ such that: :$\displaystyle \inf_{s \mathop \in V_{k_\epsilon} } \map f s \ge \liminf_{s \mathop \to \bar s} \map f s - \epsilon$ By our hypothesis $s_n \to \bar s$ and because $V_{k_\epsilon} \in N_{\bar s}$ there exists $\map N {k_\epsilon} \in \N$ such that: :$\displaystyle \forall n \ge \map N {k_\epsilon}: s_n \in V_{k_\epsilon}$ Consequently: {{begin-eqn}} {{eqn | o = \le       | l = \inf_{s \mathop \in V_{k_\epsilon} } \map f s       | r = \inf_{n \mathop \ge \map N {k_\epsilon} } \map f {s_n}       | c = because $\set {s_n : n \ge \map N {k_\epsilon} } \subseteq V_{k_\epsilon}$ }} {{eqn | o = \le       | r = \sup_{N \mathop \in \N} \set {\inf_{n \mathop \ge N} \map f {s_n} }       | c = {{Defof|Supremum}} }} {{eqn | r = \liminf_{n \mathop \to \infty} \map f {s_n}       | c = {{Defof|Limit Inferior}} }} {{end-eqn}} Combining these estimates, it follows that for all $\epsilon > 0$: :$\displaystyle \liminf_{s \mathop \to \bar s} \map f s \le \liminf_{n \mathop \to \infty} \map f {s_n} + \epsilon$ Hence the result. {{qed}} Category:Topology gegmqs4w4gguk4x28m54zv8634zj0k9
By definition of opposite group: :$(1): \quad \forall a, b \in G : a \circ b = b * a$ === Necessary Condition === Let $\left({G, \circ}\right)$ be Abelian.  Then: {{begin-eqn}} {{eqn | lo= \forall a, b \in G:       | l = a \circ b       | r = b \circ a }} {{eqn | ll= \implies       | l = a \circ b       | r = a * b       | c = Definition of Opposite Group }} {{eqn | ll= \implies       | l = \left({G, \circ}\right)       | r = \left({G, *}\right)       | c = Equality of Algebraic Structures }} {{end-eqn}} {{qed}} === Sufficient Condition === Let $\left({G, \circ}\right) = \left({G, *}\right)$. Then: {{begin-eqn}} {{eqn | lo= \forall a, b \in G:       | l = a \circ b       | r = a * b       | c = Equality of Algebraic Structures }} {{eqn | ll= \implies       | l = a \circ b       | r = b \circ a       | c = Definition of Opposite Group }} {{end-eqn}} Thus by definition $\left({G, \circ}\right)$ is an Abelian group. {{qed}} Category:Abelian Groups Category:Opposite Groups oryghvkg723af266g8uf3p8bh2rn1a8
=== Necessary Condition === Let $x_n \to x$. Let $k \in \set {1, \ldots, N}$. From Projection from Product Topology is Continuous it follows that $\pr_k$ is continuous. By Continuous Mapping is Sequentially Continuous, $\pr_k$ is also sequentially continuous. Hence $\map {\pr_k} {x_n} \to \map {\pr_k} x$. {{qed|lemma}} === Sufficient Condition === Let $\map {\pr_k} {x_n} \to \map {\pr_k} x$ for all $k \in \set {1, \ldots, N}$. Let $U \in \tau$ be an open neighborhood of $x$. By definition of the product topology and Synthetic Basis and Analytic Basis are Compatible it follows that: :$\BB := \set {U_1 \times U_2 \times \cdots \times U_N : \forall k \in \set {1, \ldots, N} : U_k \in \tau_k}$ is an analytic basis for $\tau$. Hence there exists an index set $I$ such that: :$\displaystyle U = \bigcup_{i \mathop \in I} \paren {U_{1, i} \times \cdots \times U_{N, i} }$ where $U_{k, i} \in \tau_k$ for all $i \in I, k \in \set {1, \ldots, N}$. As $x \in U$ it follows that there exists $i_0 \in I$ such that: :$\displaystyle x \in U_{1, i_0} \times \cdots \times U_{N, i_0}$  By our hypothesis $\map {\pr_k} {x_n} \to \map {\pr_k} x$ it follows that: :$\forall k \in \set{1, \ldots, N}: \exists M_k \in \N : \forall n \ge M_k: \map {\pr_k} {x_n} \in U_{k, i_0}$ Thus for all $n \ge M := \max \set {M_1, \dotsc, M_N}$ it holds that: :$ x_n = \tuple {\map {\pr_1} {x_n}, \dotsc, \map {\pr_N} {x_n} } \in U_{1,i_0} \times \cdots \times U_{N, i_0} \subset U$ Hence the result. {{qed}}
=== Necessary Condition === This is demonstrated in Cauchy Sequence Converges on Real Number Line. {{qed|lemma}} === Sufficient Condition === This is demonstrated in Real Convergent Sequence is Cauchy Sequence. {{qed}}
=== Lemma === {{:Complex Sequence is Cauchy iff Convergent/Lemma 1}} Let $\sequence {x_n}$ be a real sequence where: :$x_n = \Re \paren {z_n}$ for every $n$ :$\Re \paren {z_n}$ is the real part of $z_n$ Let $\sequence {y_n}$ be a real sequence where :$y_n = \Im \paren {z_n}$ for every $n$ :$\Im \paren {z_n}$ is the imaginary part of $z_n$ We find: :$\sequence {z_n}$ is a Cauchy sequence :$\iff \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma :$\iff \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Real Sequence is Cauchy iff Convergent :$\iff \sequence {z_n}$ is convergent by definition of convergent complex sequence {{qed}}
=== Lemma === {{:Complex Sequence is Cauchy iff Convergent/Lemma 1}} Let $\sequence {x_n}$ be a real sequence where: :$x_n = \Re \paren {z_n}$ for every $n$ :$\Re \paren {z_n}$ is the real part of $z_n$ Let $\sequence {y_n}$ be a real sequence where: :$y_n = \Im \paren {z_n}$ for every $n$ :$\Im \paren {z_n}$ is the imaginary part of $z_n$ === Necessary Condition === Let $\sequence {z_n}$ be a Cauchy sequence. We aim to prove that $\sequence {z_n}$ is convergent. We find: :$\sequence {z_n}$ is a Cauchy sequence :$\implies \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma :$\implies \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Real Sequence is Cauchy iff Convergent :$\implies \sequence {z_n}$ is convergent by definition of convergent complex sequence. {{qed|lemma}} === Sufficient Condition === Let $\sequence {z_n}$ be convergent. We aim to prove that $\sequence {z_n}$ is a Cauchy sequence. We find: :$\sequence {z_n}$ is convergent :$\implies \sequence {x_n}$ and $\sequence {y_n} $ are convergent by definition of convergent complex sequence :$\implies \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Real Sequence is Cauchy iff Convergent :$\implies \sequence {z_n}$ is a Cauchy sequence by Lemma {{qed}}
Let $x = \arctan a$ and $y = \arctan b$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \tan x       | r = a       | c =  }} {{eqn | n = 2       | l = \tan y       | r = b       | c =  }} {{eqn | l = \tan \left({\arctan a + \arctan b}\right)       | r = \tan \left({x + y}\right)       | c = }} {{eqn | r = \frac {\tan x + \tan y} {1 - \tan x \tan y}       | c = Tangent of Sum }} {{eqn | r = \frac {a + b} {1 - a b}       | c = by $(1)$ and $(2)$ }} {{eqn | ll = \implies       | l = \arctan a + \arctan b       | r = \arctan \frac {a + b} {1 - a b}       | c =  }} {{end-eqn}} {{qed}}
Let $x = \arctan a$ and $y = \arctan b$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \tan x       | r = a       | c =  }} {{eqn | n = 2       | l = \tan y       | r = b       | c =  }} {{eqn | l = \map \tan {\arctan a - \arctan b}       | r = \map \tan {x - y}       | c = }} {{eqn | r = \frac {\tan x - \tan y} {1 + \tan x \tan y}       | c = Tangent of Difference }} {{eqn | r = \frac {a - b} {1 + a b}       | c = by $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = \arctan a - \arctan b       | r = \arctan \frac {a - b} {1 + a b}       | c =  }} {{end-eqn}} {{qed}}
Let $x = \arccot a$ and $y = \arccot b$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \cot x       | r = a       | c =  }} {{eqn | n = 2       | l = \cot y       | r = b       | c =  }} {{eqn | l = \map \cot {\arccot a + \arccot b}       | r = \map \cot {x + y}       | c = }} {{eqn | r = \frac {\cot x \cot y - 1} {\cot x + \cot y}       | c = Cotangent of Sum }} {{eqn | r = \frac {a + b} {1 - a b}       | c = by $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = \arccot a + \arccot b       | r = \arccot \frac {a b - 1} {a + b}       | c =  }} {{end-eqn}} {{qed}}
Let $x = \arccot a$ and $y = \arccot b$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \cot x       | r = a       | c =  }} {{eqn | n = 2       | l = \cot y       | r = b       | c =  }} {{eqn | l = \map \cot {\arccot a - \arccot b}       | r = \map \cot {x - y}       | c = }} {{eqn | r = \frac {\cot x \cot y + 1} {\cot x - \cot y}       | c = Cotangent of Difference }} {{eqn | r = \frac {a - b} {1 + a b}       | c = by $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = \arccot a - \arccot b       | r = \arccot \frac {a b + 1} {a - b}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\laptrans {e^{a t} } } s       | r = \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^{\to +\infty} e^{\paren {a - s} t} \rd t       | c = Exponential of Sum }} {{eqn | r = \lim_{L \mathop \to \infty} \int_0^L e^{\paren {a - s} t} \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{L \mathop \to \infty} \intlimits {\frac 1 {a - s} e^{\paren {a - s} t} } 0 L       | c = Primitive of Exponential Function, Integration by Substitution }} {{eqn | r = \lim_{L \mathop \to \infty} \frac 1 {a - s} \paren {e^{\paren {a - s} L} - e^{\paren {a - s} 0} }       | c =  }} {{eqn | r = \lim_{L \mathop \to \infty} \frac 1 {s - a} \paren {1 - e^{\paren {a - s} L} }       | c = Exponential of Zero and rearranging }} {{end-eqn}} Because $s > a$, we have that $a - s < 0$. Hence: :$\displaystyle \lim_{L \mathop \to \infty} \paren {a - s} L \to -\infty$ So: {{begin-eqn}} {{eqn | l = \lim_{L \mathop \to \infty} \frac 1 {s - a} \paren {1 - e^{\paren {a - s} L} }       | r = \frac 1 {s - a} \paren {1 - 0}       | c = Exponential Tends to Zero }} {{eqn | r = \frac 1 {s - a} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\laptrans {e^{a t} } } s       | r = \map {\laptrans {1 \times e^{a t} } } s       | c =  }} {{eqn | r = \map {\laptrans 1} {s - a}       | c = Laplace Transform of Exponential times Function }} {{eqn | r = \frac 1 {s - a}       | c = Laplace Transform of Constant Mapping }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map \delta {t - c} \, \map f t}       | r = \int^{\to+\infty}_0 e^{-s t} \map \delta {t - c} \, \map f t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int^{c^+}_{c^-} e^{-s t} \map \delta {t - c} \, \map f t \rd t       | c = Integrand elsewhere zero by {{Defof|Dirac Delta Function}} }} {{eqn | r = \int^{c^+}_{c^-} e^{-s c} \map \delta {t - c} \, \map f c \rd t       | c = $e^{-s t}$ and $\map f t$ are constant in interval $\closedint {c^-} {c^+}$ }} {{eqn | r = e^{-s c} \, \map f c \int^{c^+}_{c^-} \map \delta {t - c} \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = e^{-s c} \, \map f c \int^{0^+}_{0^-} \map \delta {t - c} \rd \paren {t - c}       | c = Integration by Substitution }} {{eqn | r = e^{-s c} \, \map f c       | c = {{Defof|Dirac Delta Function}} }} {{end-eqn}} {{qed}} Category:Laplace Transforms Category:Dirac Delta Function g5ejsz48akqz5vky3llm1lvp3fvsg3r
Consider the differential equation: :$(1): \quad D^2_x \map f x = -\map f x$ subject to the initial conditions: :$(2): \quad \map f 0 = 1$ :$(3): \quad D_x \map f 0 = 0$ === Step 1 === We will prove that $y = \cos x$ is a particular solution of $(1)$. {{begin-eqn}} {{eqn | l = y       | r = \cos x       | c =  }} {{eqn | l = D^2_x y       | r = D^2_x \cos x       | c = taking second derivative of both sides }} {{eqn | r = \map {D_x} {-\sin x}       | c = Derivative of Cosine Function }} {{eqn | r = -\map {D_x} {\sin x}       | c = Derivative of Constant Multiple }} {{eqn | r = -\cos x       | c = Derivative of Sine Function }} {{eqn | r = -y       | c =  }} {{end-eqn}} Thus $y = \cos x$ fulfils $(1)$. Then from Cosine of Zero is One: :$\cos 0 = 1$ Thus $y = \cos x$ fulfils $(2)$. Then: {{begin-eqn}} {{eqn | l = D_x \cos 0       | r = -\sin 0       | c = Derivative of Cosine Function }} {{eqn | r = 0       | c = Sine of Zero is Zero }} {{end-eqn}} Thus $y = \cos x$ fulfils $(3)$. So $y = \cos x$ is a particular solution of $(1)$. {{qed|lemma}} === Step 2 === We will prove that $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$. {{begin-eqn}} {{eqn | l = z       | r = \frac {e^{i x} + e^{-i x} } 2       | c =  }} {{eqn | l = D^2_x z       | r = \map {D^2_x} {\frac {e^{i x} + e^{-i x} } 2}       | c = taking second derivative of both sides }} {{eqn | r = \frac 1 2 \paren {D^2_x e^{i x} + D^2_x e^{-i x} }       | c = Linear Combination of Derivatives }} {{eqn | r = \frac 1 2 \paren {i D_x e^{i x} - i D_x e^{-i x} }       | c = Derivative of Exponential Function }} {{eqn | r = \frac 1 2 \paren {i^2 e^{i x} - i \paren {-i} e^{-i x} }       | c = Derivative of Exponential Function }} {{eqn | r = \frac 1 2 \paren {- e^{i x} - e^{-i x} }       | c = $i^2 = -1$ }} {{eqn | r = -\frac {e^{i x} + e^{-i x} }2       | c =  }} {{eqn | r = -z       | c =  }} {{end-eqn}} Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(1)$. Then: {{begin-eqn}} {{eqn | l = \frac {e^{i \times 0} + e^{-i \times 0} } 2       | r = \frac {1 + 1} 2       | c = Exponential of Zero }} {{eqn | r = 1       | c =  }} {{end-eqn}} Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(2)$. Then: {{begin-eqn}} {{eqn | l = \intlimits {D_x \frac {e^{i x} + e^{-i x} } 2} {x \mathop = 0} {}       | r = \intlimits {\frac {i e^{i x} - i e^{-i x} } 2} {x \mathop = 0} {}       | c = Derivative of Exponential Function }} {{eqn | r = \frac {i - i} 2       | c = Exponential of Zero }} {{eqn | r = 0       | c =  }} {{end-eqn}} Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(3)$. So $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$. {{qed|lemma}} We have shown that $y$ and $z$ are both particular solutions of $(1)$. But a particular solution to a differential equation is unique. {{MissingLinks|This result definitely needs backing up with an appropriate link}} Therefore $y = z$. That is: :$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$ {{qed}}
Let $z = x + y i$ be a complex number and $O = 0 + 0 i$ be the origin on the complex plane. We have its modulus: {{begin-eqn}} {{eqn | l = \left\vert{z}\right\vert       | r = \left\vert{x + y i}\right\vert       | c =  }} {{eqn | r = \sqrt {x^2 + y^2}       | c = Definition of Modulus }} {{end-eqn}} and its distance from the origin on the complex plane: {{begin-eqn}} {{eqn | l = d \left({z, O}\right)       | r = d \left({\left({x, y}\right), \left({0, 0}\right)}\right)       | c =  }} {{eqn | r = \sqrt{\left({x - 0}\right)^2 + \left({y - 0}\right)^2}       | c = Distance Formula }} {{eqn | r = \sqrt {x^2 + y^2}       | c =  }} {{end-eqn}} The two are seen to be equal. {{qed}} {{link wanted|There may be a more directly relevant link to the distance formula based on the fact (which we've proved somewhere) that the complex plane is a metric space, and that the distance formula is that metric.}} Category:Complex Analysis qonf9e72mw078j2ed6wty07774jop2t
By Peano's Axioms, $\N_0 \sim \O$, where $\N_n$ denotes the initial segment of natural number $n$. By definition, $\O$ is finite. By definition, $\O$ is a countable set. {{qed}} Category:Set Theory Category:Empty Set Category:Countable Sets ly0tg8512aadn5cgyxnqkjdd9niomfe
=== Lemma === {{:Lindelöf's Lemma/Lemma|Lemma}} {{qed|lemma}} We have that $S$ is covered by $C$. This means that $S$ is a subset of $\displaystyle \bigcup_{O \mathop \in C} O$. From the lemma: :$\displaystyle \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$ where $D$ is a countable subset of $C$. Hence $S$ is also a subset of $\displaystyle \bigcup_{O \mathop \in D} O$. In other words, $S$ is covered by $D$. That is, $S$ is covered by a countable subset of $C$. This finishes the proof of the theorem. {{qed}} {{Namedfor|Ernst Leonard Lindelöf|cat = Lindelöf}} Category:Real Analysis ikufg8dfndbd59rgeksn2inb9wt94g5
Starting from the right, we have: {{begin-eqn}} {{eqn | l = \cos^2 \theta - \sin^2\theta       | r = \left({\frac 1 2 \left({e^{i \theta} + e^{-i \theta} }\right)}\right)^2 - \left({\frac 1 {2 i} \left({e^{i \theta} - e^{-i \theta} }\right)}\right)^2       | c = Cosine Exponential Formulation, Sine Exponential Formulation }} {{eqn | r = \frac 1 4 \left({e^{i \theta} + e^{-i \theta} }\right)^2 + \frac 1 4 \left({e^{i \theta} - e^{-i \theta} }\right)^2       | c = $i$ is the imaginary unit }} {{eqn | r = \frac 1 4 \left({e^{2 i \theta} + 2 + e^{-2 i \theta} + e^{2 i \theta} - 2 + e^{-2 i \theta} }\right)       | c = Square of Sum, Square of Difference }} {{eqn | r = \frac 1 2 \left({e^{2 i \theta} + e^{-2 i \theta} }\right)       | c = Simplifying }} {{eqn | r = \cos 2 \theta       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
Starting from the right, we have {{begin-eqn}} {{eqn | l = \dfrac {2 \tanh x} {1 + \tanh^2 x}       | r = \dfrac {2 \left({\dfrac {e^x - e^{-x} } {e^x + e^{-x} } }\right)} {1 + \left({\dfrac{e^x - e^{-x} } {e^x + e^{-x} } }\right)^2}       | c = {{Defof|Hyperbolic Tangent|index = 1}} }} {{eqn | r = \dfrac {2 \left({e^x + e^{-x} }\right) \left({e^x - e^{-x} }\right)} {\left({e^x + e^{-x} }\right)^2 + \left({e^x - e^{-x} }\right)^2}       | c = multiplying both numerator and denominator by $\left({e^x + e^{-x} }\right)^2$ }} {{eqn | r = \dfrac {2 \left({e^{2 x} - e^{-2 x} }\right)} {e^{2 x} + 2 + e^{-2 x} + e^{2x} - 2 + e^{-2x} }       | c = Difference of Two Squares, Square of Sum, Square of Difference }} {{eqn | r = \dfrac {2 \left({e^{2 x} - e^{-2 x} }\right)} {2 e^{2 x} + 2 e^{-2 x} }       | c = simplifying }} {{eqn | r = \dfrac {e^{2 x} - e^{-2 x} } {e^{2 x} + e^{-2 x} }       | c = simplifying }} {{eqn | r = \tanh {2 x}       | c = {{Defof|Hyperbolic Tangent|index = 1}} }} {{end-eqn}} {{qed}}
:300px Let $h$ be the height of $\triangle ABC$. {{begin-eqn}} {{eqn | l = \mathcal A       | r = \frac 1 2 b h       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac b 2 \sqrt {r^2 - \left({\frac b 2}\right)^2}       | c = Pythagoras's Theorem }} {{eqn | r = \frac b 2 \sqrt {\frac {4 r^2 - b^2} 4}       | c = simplification }} {{eqn | r = \frac b 4 \sqrt {4 r^2 - b^2}       | c = simplification }} {{end-eqn}} {{qed}} Category:Areas of Triangles Category:Isosceles Triangles reh3cdnhcur5jglo3ano21wpmtummtc
{{begin-eqn}} {{eqn | l = \frac {2 \tan \theta} {1 - \tan^2 \theta}       | r = \displaystyle \dfrac {2 i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } } {1 - \paren {i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } }^2}       | c = Tangent Exponential Formulation }} {{eqn | r = \dfrac {2 i \paren {1 - e^{2 i \theta} } \paren {1 + e^{2 i \theta} } } {\paren {1 + e^{2 i \theta} }^2 + \paren {1 - e^{2 i \theta} }^2}       | c = multiplying both numerator and denominator by $\left({1 + e^{2 i \theta} }\right)^2$; $i$ is the imaginary unit }} {{eqn | r = \dfrac {2 i \paren {1 - e^{4 i \theta} } } {1 + 2 e^{2 i \theta} + e^{4 i \theta} + 1 - 2 e^{2 i \theta} + e^{4 i \theta} }       | c = Difference of Two Squares, Square of Sum, Square of Difference }} {{eqn | r = \dfrac {2 i \paren {1 - e^{4 i \theta} } } {2 + 2 e^{4 i \theta} }       | c = simplifying }} {{eqn | r = i \dfrac {1 - e^{4 i \theta} } {1 + e^{4 i \theta} }       | c = simplifying }} {{eqn | r = \map \tan {2 \theta}       | c = Tangent Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\paren {f \cdot g} } {-x}       | r = \map f {-x} \cdot \map g {-x}       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{eqn | r = \map f x \cdot \map g x       | c = {{Defof|Even Function}} }} {{eqn | r = \map {\paren {f \cdot g} } x       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{end-eqn}} Thus, by definition, $\paren {f \cdot g}$ is an even function. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\paren {f \cdot g} } {-x}       | r = \map f {-x} \cdot \map g {-x}       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{eqn | r = \paren {-\map f x} \cdot \map g x       | c = {{Defof|Odd Function}} and {{Defof|Even Function}} }} {{eqn | r = -\map f x \cdot \map g x       | c =  }} {{eqn | r = -\map {\paren {f \cdot g} } x       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{end-eqn}} Thus, by definition, $\paren {f \cdot g}$ is an odd function. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\paren {f \cdot g} } {-x}       | r = \map f {-x} \cdot \map g {-x}       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{eqn | r = \paren {-\map f x} \cdot \paren {-\map g x}       | c = {{Defof|Odd Function}} }} {{eqn | r = \map f x \cdot \map g x       | c =  }} {{eqn | r = \map {\paren {f \cdot g} } x       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{end-eqn}} Thus, by definition, $\paren {f \cdot g}$ is an even function. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\MM \set {e^{-a t} } } s       | r = \int_0^{\to +\infty} t^{s - 1} e^{-a t} \rd t       | c = {{Defof|Mellin Transform}} }} {{eqn | r = \int_0^{\to +\infty} \paren {\dfrac t a}^{s - 1} e^{-a \paren {\frac t a} } \frac {\d t} a       | c = Integration by Substitution, $t \mapsto \dfrac t a$, $\d t \mapsto \dfrac {\d t} a$ }} {{eqn | r = a^{-s} \int_0^{\to +\infty} t^{s - 1} e^{-t} \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = a^{-s} \, \map \Gamma s       | c = {{Defof|Integral Form of Gamma Function|Gamma Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathcal M \left\{ { \delta_c \left({t}\right)} \right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s-1} \delta_c \left({t}\right) \rd t       | c = Definition of Mellin Transform }} {{eqn | r = \int_{c^-}^{c^+} t^{s-1} \delta_c \left({t}\right) \rd t       | c = Definition of Dirac delta function: integrand is elsewhere zero }} {{eqn | r = \int_{c^-}^{c^+} c^{s-1} \delta_c \left({t}\right) \rd t       | c = $t$ is constant in interval $\left[{c^- \,.\,.\, c^+}\right]$ }} {{eqn | r = c^{s-1} \int_{c^-}^{c^+} \delta_c \left({t}\right) \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = c^{s-1}       | c = Definition of Dirac Delta Function }} {{end-eqn}} {{qed}}