hcju/proofwikiqa huggingface dataset:
By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by:  :$\displaystyle \map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$ for $t < \beta$.  From Variance as Expectation of Square minus Square of Expectation:  :$\displaystyle \var X = \expect {X^2} - \paren {\expect X}^2$ From Moment in terms of Moment Generating Function, we also have:  :$\expect {X^2} = \map {M''_X} 0$ We have: {{begin-eqn}} {{eqn	| l = \map {M''_X} t 	| r = \frac {\d^2} {\d t^2} \paren {1 - \frac t \beta}^{-\alpha} }} {{eqn	| r = \beta^\alpha \frac {\d^2} {\d t^2} \paren {\frac 1 {\paren {\beta - t}^\alpha} } }} {{eqn	| r = \beta^\alpha \map {\frac \d {\d t} } {\map {\frac \d {\map \d {\beta - t} } } {\frac 1 {\paren {\beta - t}^\alpha} } \cdot \map {\frac \d {\d t} } {\beta - t} } 	| c = Chain Rule for Derivatives }} {{eqn	| r = \paren {-1}^2 \beta^\alpha \alpha \map {\frac \d {\d t} } {\frac 1 {\paren {\beta - t}^{\alpha + 1} } } 	| c = Derivative of Power }} {{eqn	| r = \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} } 	| c = Chain Rule for Derivatives, Derivative of Power }} {{end-eqn}} Setting $t = 0$, we obtain the second moment:  {{begin-eqn}} {{eqn 	| l = \map {M''_X} 0  	| r = \expect {X^2} }} {{eqn	| r = \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - 0}^{\alpha + 2} } }} {{eqn	| r = \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\beta^{\alpha + 2} } }} {{eqn	| r = \frac {\alpha \paren {\alpha + 1} } {\beta^2} }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \var X 	| r = \frac {\alpha \paren {\alpha + 1} } {\beta^2} - \frac {\alpha^2} {\beta^2} }} {{eqn	| r = \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2} }} {{eqn	| r = \frac \alpha {\beta^2} }} {{end-eqn}} {{qed}}
Let $B_k$ denote the kth Bernoulli number Let $B_k \left(x\right)$ denote the kth Bernoulli polynomial By Value of Odd Bernoulli Polynomial at One Half: {{begin-eqn}} {{eqn | l = B_{2 n + 1} \left({\frac 1 2}\right)       | r = 0 }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}       | r = 0 }} {{eqn | ll= \leadsto       | l = 2^{2 n + 1} \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}       | r = 0 }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{-k}       | r = 0 }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k       | r = 0 }} {{end-eqn}} By Odd Bernoulli Numbers Vanish: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k       | r = \left({2 n + 1}\right) 2 B_1 + \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k} }} {{eqn | r = -\left({2 n + 1}\right) + \sum_{k \mathop = 0}^n \binom {2 n + 1 } {2 k} B_{2 k} 2^{2 k}       | c = $B_1 = -\frac 1 2$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}       | r = 2 n + 1 }} {{eqn | ll = \leadsto       | l = \sum_{k \mathop = 1}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}       | r = 2 n + 1 - B_0 }} {{eqn | r = 2 n       | c = $B_0 = 1$ }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \binom {2 \times 1 + 1} 2 2^2 B_2       | r = \frac {2 \times 3} 2 \times 2^2 \times \frac 1 6       | c = Binomial Coefficient with Two, {{Defof|Bernoulli Numbers}} }} {{eqn | r = 2       | c = after simplification }} {{eqn | r = 2 \times 1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} = 2 \left({r + 1}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k}       | r = \sum_{k \mathop = 1}^r \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} + \dbinom {2 \left({r + 1}\right) + 1} {2 \left({r + 1}\right)} 2^{2 \left({r + 1}\right)} B_{2 \left({r + 1}\right)}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^r \dbinom {2 r + 3} {2 k} 2^{2 k} B_{2 k} + \left({2 r + 3}\right) 2^{2 r + 2} B_{2 r + 2}       | c = Binomial Coefficient with Self minus One and simplification }} {{end-eqn}} {{finish|This approach may not be workable. May be easier / better to start with Definition:Bernoulli Numbers/Recurrence Relation and take it from there.}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
{{begin-eqn}} {{eqn	| l = \map {M^{\paren n}_X} t 	| r = \frac {\d^n} {\d t^n} \expect {e^{t X} } 	| c = {{Defof|Moment Generating Function}} }} {{eqn	| r = \frac {\d^n} {\d t^n} \expect {\sum_{m \mathop = 0}^\infty \frac {t^m X^m} {m!} } 	| c = Power Series Expansion for Exponential Function }} {{eqn	| r = \frac {\d^n} {\d t^n} \sum_{m \mathop = 0}^\infty \expect {\frac {t^m X^m} {m!} } 	| c = Linearity of Expectation Function }} {{eqn	| r = \sum_{m \mathop = 0}^\infty \frac {\d^n} {\d t^n} \paren {\frac {t^m} {m!} } \expect {X^m} 	| c = Linearity of Expectation Function, Power Series is Termwise Differentiable within Radius of Convergence }} {{eqn	| r = \sum_{m \mathop = n}^\infty \frac {m^{\underline n} t^{m - n} } {m!} \expect {X^m} 	| c = Nth Derivative of Mth Power }} {{eqn	| r = \sum_{m \mathop = n}^\infty \frac { m! t^{m - n} } {m! \paren {m - n}!} \expect {X^m} 	| c = Falling Factorial as Quotient of Factorials }} {{eqn	| r = \frac {t^{n - n} } {\paren {n - n}!} \expect {X^n} + \sum_{m \mathop = n + 1}^\infty \frac {t^{m - n} } {\paren {m - n}!} \expect {X^m} }} {{eqn	| r = \expect {X^n} + \sum_{m \mathop = n + 1}^\infty \frac {t^{m - n} } {\paren {m - n}!} \expect {X^m} }} {{end-eqn}} Setting $t = 0$ yields the result.  {{qed}} Category:Moment Generating Functions k4iyncnkkrzhkym2p9x38ngku446zqh
From the definition of the continuous uniform distribution, $X$ has probability density function:  :$\displaystyle f_X \left({x}\right) = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$ From the definition of a moment generating function: :$\displaystyle M_X \left({t}\right) = \mathbb E \left[{ e^{t X} }\right] = \int_{-\infty}^\infty e^{tx} f_X \left({x}\right) \rd x$ First, consider the case $t \ne 0$.  Then:  {{begin-eqn}} {{eqn	| l = M_X \left({t}\right) 	| r = \int_{-\infty}^a 0 e^{t x} \rd x + \int_a^b \frac{ e^{t x} } {b - a} \rd x + \int_b^\infty 0 e^{t x} \rd x }} {{eqn	| r = \left[{ \frac { e^{t x} } {t \left({b - a}\right)} }\right]_a^b 	| c = Primitive of $e^{a x}$, Fundamental Theorem of Calculus }} {{eqn	| r = \frac{ e^{t b} - e^{t a} } {t \left({b - a}\right)} }} {{end-eqn}} In the case $t = 0$, we have $\mathbb E \left[{X^0}\right] = \mathbb E \left[{1}\right] = 1$.  {{qed}} Category:Moment Generating Functions Category:Continuous Uniform Distribution nervbj8ujvtk5o86llw1mblr0sq999h
From the definition of the Gaussian distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect { e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$ So:  {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x }} {{eqn	| r = \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {\paren {\sqrt 2 \sigma u + \mu} t - u^2} \rd u 	| c = substituting $u = \dfrac {x - \mu} {\sqrt 2 \sigma}$ }} {{eqn	| r = \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u^2 - \sqrt 2 \sigma u t} } \rd u }} {{eqn	| r = \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u - \frac {\sqrt 2} 2 \sigma t}^2 + \frac 1 2 \sigma^2 t^2} \rd u }} {{eqn	| r = \frac {\map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-v^2} \rd v 	| c = substituting $v = u - \dfrac {\sqrt 2} 2 \sigma t$ }} {{eqn	| r = \frac {\sqrt \pi \map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi} 	| c = Gaussian Integral }} {{eqn	| r = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Gaussian Distribution kitsn4swc477kgusg7n8yl4qv0ucdu8
{{begin-eqn}} {{eqn | l = \paren {-1}^{n + 1} \dfrac {\pi^{2 n + 1} E_{2 n} } {2^{2 n + 2} \paren {2 n}!}       | r = \sum_{j \mathop = 0}^\infty \dfrac {\paren {-1}^j} {\paren {2 j + 1}^{2 n + 1} }       | c = Sum of Reciprocals of Odd Powers of Odd Integers Alternating in Sign }} {{eqn | ll= \leadsto       | l = \paren {-1}^{n + 1} E_{2 n}       | r = \dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} } \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {\paren {2 j + 1}^{2 n + 1} }       | c = multiplying both sides by $\dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} }$ }} {{eqn | ll= \leadsto       | l = \paren {-1}^{2 n + 2} E_{2 n}       | r = \paren {-1}^{n + 1} \dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} } \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {\paren {2 j + 1}^{2 n + 1} }       | c = multiplying both sides by $\paren {-1}^{n + 1}$ }} {{eqn | ll= \leadsto       | l = E_{2 n}       | r = \paren {-1}^{n + 1} \dfrac {2^{2 n + 2} \paren {2 n}!} {\pi^{2 n + 1} } \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {\paren {2 j + 1}^{2 n + 1} }       | c = $\paren {-1}^{2 n + 2} = 1$ as $2 n + 2$ is even }} {{end-eqn}} {{qed}}
Let $\mathbf a$ and $\mathbf b$ be expressed in component form: {{begin-eqn}} {{eqn | l = \mathbf a       | r = a_1 \mathbf e_1 + a_2 \mathbf e_2 + \cdots + a_n \mathbf e_n       | c =  }} {{eqn | l = \mathbf b       | r = b_1 \mathbf e_1 + b_2 \mathbf e_2 + \cdots + b_n \mathbf e_n       | c =  }} {{end-eqn}} where $\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n$ denote the unit vectors in the positive directions of the coordinate axes of the Cartesian coordinate space into which $\mathbf a$ has been embedded. Thus $\mathbf a$ and $\mathbf b$ can be expressed as: {{begin-eqn}} {{eqn | l = \mathbf a       | r = \tuple {a_1, a_2, \ldots, a_n}       | c =  }} {{eqn | l = \mathbf b       | r = \tuple {b_1, b_2, \ldots, b_n}       | c =  }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = \size {\mathbf a}       | r = \size {\tuple {a_1, a_2, \ldots, a_n} }       | c =  }} {{eqn | r = \sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} }       | c =  }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = \size {\mathbf b}       | r = \size {\tuple {b_1, b_2, \ldots, b_n} }       | c =  }} {{eqn | r = \sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} }       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \hat {\mathbf a}       | r = \widehat {\tuple {a_1, a_2, \ldots, a_n} }       | c =  }} {{eqn | r = \dfrac 1 {\sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} } } \mathbf a       | c =  }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = \hat {\mathbf b}       | r = \widehat {\tuple {b_1, b_2, \ldots, b_n} }       | c =  }} {{eqn | r = \dfrac 1 {\sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} } }       | c =  }} {{end-eqn}} Let $\mathbf a = \mathbf b$. Then by Equality of Ordered Tuples: :$(1): \quad a_1 = b_1, a_2 = b_2, \ldots a_n = b_n$ Then: {{begin-eqn}} {{eqn | l = \size {\mathbf a}       | r = \sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} }       | c =  }} {{eqn | r = \sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} }       | c = from $(1)$ }} {{eqn | r = \size {\mathbf b}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \hat {\mathbf a}       | r = \dfrac 1 {\sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} } } \mathbf a       | c =  }} {{eqn | r = \dfrac 1 {\sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} } } \mathbf b       | c = from $(1)$ }} {{eqn | r = \hat {\mathbf b}       | c =  }} {{end-eqn}} {{Finish|The other direction now needs to be attended to.}}
:400px Let $\mathbf a = \vec {OP}$ and $\mathbf b = \vec {PQ}$. Then: :$\vec {OQ} = \mathbf a + \mathbf b$ Let $P'$ and $Q'$ be points on $OP$ and $OQ$ respectively so that: :$OP' : OP = OQ' : OQ = m$ Then $P'Q'$ is parallel to $PQ$ and $m$ times it in length. Thus: :$\vec {P'Q'} = m \mathbf b$ which shows that: {{begin-eqn}} {{eqn | l = m \paren {\mathbf a + \mathbf b}       | r = \vec {OQ'}       | c =  }} {{eqn | r = \vec {OP} + \vec {OP'}       | c =  }} {{eqn | r = m \mathbf a + m \mathbf b       | c =  }} {{end-eqn}} {{qed}}
By definition of vector cross product: :$\mathbf a \times \mathbf b = \left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta \hat {\mathbf n}$ where: :$\left\vert{a}\right\vert$ denotes the length of $\mathbf a$ :$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction :$\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right hand rule. As $\hat {\mathbf n}$ is the unit vector: :$\left\vert{\left({\left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta \hat {\mathbf n} }\right)}\right\vert = \left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta$ By Area of Parallelogram, the area of the parallelogram equals the product of one of its bases and the associated altitude. Let $\mathbf a$ denote the base of the parallelogram. Then its altitude is $\left\vert{\mathbf b}\right\vert \sin \theta$. The result follows. {{qed}}
From the definition of the Gaussian distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$ From the definition of differential entropy:  :$\displaystyle \map h X = -\int_{-\infty}^\infty \map {f_X} x \ln \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \map h X 	| r = - \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \map \ln {\frac 1 {\sigma \sqrt {2 \pi} } \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } } \rd x }} {{eqn	| r = \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \map \ln {\sigma \sqrt {2 \pi} \, \map \exp {\frac {\paren {x - \mu}^2} {2 \sigma^2} } } \rd x 	| c = Logarithm of Reciprocal }} {{eqn	| r = \frac {\sqrt 2 \sigma} { \sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {-t^2} \map \ln {\sigma \sqrt {2 \pi} \, \map \exp {t^2} } \rd t 	| c = substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$ }} {{eqn	| r = \frac 1 {\sqrt \pi} \int_{-\infty}^\infty \paren {\map \ln {\sigma \sqrt {2 \pi} } + \map \ln {\map \exp {t^2} } } \, \map \exp {-t^2} \rd t 	| c = Sum of Logarithms }} {{eqn	| r = \frac {\map \ln {\sigma \sqrt {2 \pi} } } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t + \frac 1 {\sqrt \pi} \int_{-\infty}^\infty t^2 \map \exp {-t^2} \rd t }} {{eqn	| r = \frac {\sqrt \pi \, \map \ln {\sigma \sqrt {2 \pi} } } {\sqrt \pi} + \frac 1 {\sqrt \pi} \paren {\sqbrk {-\frac t 2 \, \map \exp {-t^2} }_{-\infty}^\infty + \frac 1 2 \int_{-\infty}^\infty \map \exp {-t^2} \rd t} 	| c = Gaussian Integral, Integration by Parts, Fundamental Theorem of Calculus }} {{eqn	| r = \map \ln {\sigma \sqrt {2 \pi} } + \frac 1 {2 \sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t 	| c = Exponential Tends to Zero and Infinity }} {{eqn	| r = \map \ln {\sigma \sqrt {2 \pi} } + \frac {\sqrt \pi} {2 \sqrt \pi} 	| c = Gaussian Integral }} {{eqn	| r = \map \ln {\sigma \sqrt {2 \pi} } + \frac 1 2 }} {{end-eqn}} {{qed}} Category:Gaussian Distribution Category:Differential Entropy rpz9fs51gc011cywgniful9jyzj31c5
{{ProofWanted|from Scalar Triple Product equals Determinant, and a result about $3 \times 3$ determinants.}}
{{begin-eqn}} {{eqn | l = \mathbf c \times \left({\mathbf a \times \mathbf b}\right)       | r = \left({\mathbf{c \cdot b} }\right) \mathbf a - \left({\mathbf{c \cdot a} }\right) \mathbf b       | c = Lagrange's Formula }} {{eqn | ll= \leadsto       | l = \left({\mathbf a \times \mathbf b}\right) \times \mathbf c       | r = -\left({\left({\mathbf{c \cdot b} }\right) \mathbf a - \left({\mathbf{c \cdot a} }\right) \mathbf b}\right)       | c = Vector Cross Product is Anticommutative }} {{eqn | r = -\left({\mathbf{b \cdot c} }\right) \mathbf a + \left({\mathbf{a \cdot c} }\right) \mathbf b       | c = Dot Product Operator is Commutative }} {{eqn | r = \left({\mathbf{a \cdot c} }\right) \mathbf b - \left({\mathbf{b \cdot c} }\right) \mathbf a       | c = Real Addition is Commutative }} {{end-eqn}} {{qed}} {{Namedfor|Joseph Louis Lagrange}}
From the definition of the continuous uniform distribution, $X$ has probability density function:  :$\displaystyle \map {f_X} x = \begin{cases} \frac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$ From the definition of differential entropy:  :$\displaystyle \map h X = - \int_{-\infty}^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$ So:  {{begin-eqn}} {{eqn	| l = \map h X 	| r = -\int_a^b \frac 1 {b - a} \map \ln {\frac 1 {b - a} } \rd x }} {{eqn	| r = \frac {\map \ln {b - a} } {b - a} \int_a^b \rd x 	| c = Logarithm of Reciprocal }} {{eqn	| r = \frac {\map \ln {b - a} } {b - a} \sqbrk x_a^b  	| c = Primitive of Constant, Fundamental Theorem of Calculus }} {{eqn	| r = \frac {\paren {b - a} \map \ln {b - a} } {b - a} }} {{eqn	| r = \map \ln {b - a} }} {{end-eqn}} {{qed}} Category:Continuous Uniform Distribution Category:Differential Entropy ivkumneo9vu1k0mykfx78n8qf6zda3k
{{begin-eqn}} {{eqn | l = \left({\mathbf a \cdot \mathbf c}\right) \left({\mathbf b \cdot \mathbf d}\right)       | r = \left({\sum_{i \mathop = 1}^3 a_i c_i}\right) \left({\sum_{j \mathop = 1}^3 b_j d_j}\right)       | c = {{Defof|Dot Product}} }} {{eqn | r = \left({\sum_{i \mathop = 1}^3 a_i d_i}\right) \left({\sum_{j \mathop = 1}^3 b_j c_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le 3} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)       | c = Binet-Cauchy Identity }} {{eqn | r = \left({\mathbf a \cdot \mathbf d}\right) \left({\mathbf b \cdot \mathbf c}\right) + \left({a_1 b_2 - a_2 b_1}\right) \left({c_1 d_2 - c_2 d_1}\right)       | c = {{Defof|Dot Product}} and expanding {{RHS}} }} {{eqn | o =        | ro= +       | r = \left({a_1 b_3 - a_3 b_1}\right) \left({c_1 d_3 - c_3 d_1}\right)       | c =  }} {{eqn | o =        | ro= +       | r = \left({a_2 b_3 - a_3 b_2}\right) \left({c_2 d_3 - c_3 d_2}\right)       | c =  }} {{eqn | r = \left({\mathbf a \cdot \mathbf d}\right) \left({\mathbf b \cdot \mathbf c}\right) + \left({a_1 b_2 - a_2 b_1}\right) \left({c_1 d_2 - c_2 d_1}\right)       | c =  }} {{eqn | o =        | ro= +       | r = \left({-\left({a_1 b_3 - a_3 b_1}\right)}\right) \left({-\left({c_1 d_3 - c_3 d_1}\right)}\right)       | c = two sign changes which cancel each other out }} {{eqn | o =        | ro= +       | r = \left({a_2 b_3 - a_3 b_2}\right) \left({c_2 d_3 - c_3 d_2}\right)       | c =  }} {{eqn | r = \left({\mathbf a \cdot \mathbf d}\right) \left({\mathbf b \cdot \mathbf c}\right) + \left({a_1 b_2 - a_2 b_1}\right) \left({c_1 d_2 - c_2 d_1}\right)       | c =  }} {{eqn | o =        | ro= +       | r = \left({a_3 b_1 - a_1 b_3}\right) \left({c_3 d_1 - c_1 d_3}\right)       | c = rearranging }} {{eqn | o =        | ro= +       | r = \left({a_2 b_3 - a_3 b_2}\right) \left({c_2 d_3 - c_3 d_2}\right)       | c =  }} {{eqn | r = \left({\mathbf a \cdot \mathbf d}\right) \left({\mathbf b \cdot \mathbf c}\right) + \left({\mathbf a \times \mathbf b}\right) \cdot \left({\mathbf c \times \mathbf d}\right)       | c = {{Defof|Vector Cross Product}} }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \left({\mathbf f \cdot \left({\mathbf g \times \mathbf h}\right)}\right)       | r = \dfrac {\d \mathbf f} {\d x} \cdot \left({\mathbf g \times \mathbf h}\right) + \mathbf f \cdot \dfrac \d {\d x} \left({\mathbf g \times \mathbf h}\right)       | c = Derivative of Dot Product of Vector-Valued Functions }} {{eqn | r = \dfrac {\d \mathbf f} {\d x} \cdot \left({\mathbf g \times \mathbf h}\right) + \mathbf f \cdot \left({\dfrac {\d \mathbf g} {\d x} \times \mathbf h + \mathbf g \times \dfrac {\d \mathbf h} {\d x} }\right)       | c = Derivative of Vector Cross Product of Vector-Valued Functions }} {{eqn | r = \dfrac {\d \mathbf f} {\d x} \cdot \left({\mathbf g \times \mathbf h}\right) + \mathbf f \cdot \left({\dfrac {\d \mathbf g} {\d x} \times \mathbf h}\right) + \mathbf f \cdot \left({\mathbf g \times \dfrac {\d \mathbf h} {\d x} }\right)       | c = Dot Product Distributes over Addition }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf f \left({x}\right) \cdot \dfrac {\d \mathbf f \left({x}\right)} {\d x}       | r = \mathbf f \left({x}\right) \cdot \sum_{k \mathop = 0}^n \dfrac {\d f_k \left({x}\right)} {\d x} \mathbf e_k       | c = {{Defof|Derivative of Vector-Valued Function}} }} {{eqn | r = \sum_{k \mathop = 0}^n f_k \left({x}\right) \dfrac {\d f_k \left({x}\right)} {\d x}       | c = {{Defof|Dot Product}} }} {{eqn | n = 1       | r = \sum_{k \mathop = 0}^n f_k \left({x}\right) \dfrac {\d f_k \left({x}\right)} {\d x}       | c = {{Defof|Dot Product}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \dfrac {\d \left\lvert{\mathbf f \left({x}\right)}\right\rvert} {\d x}       | r = \dfrac \d {\d x} \sqrt {\sum_{k \mathop = 0}^n \left({f_k \left({x}\right)}\right)^2}       | c = {{Defof|Vector Length}} }} {{eqn | r = \frac 1 {2 \sqrt {\displaystyle \sum_{k \mathop = 0}^n \left({f_k \left({x}\right)}\right)^2} } \dfrac \d {\d x} \sum_{k \mathop = 0}^n \left({f_k \left({x}\right)}\right)^2       | c = Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = \frac 1 {\left\lvert{\mathbf f \left({x}\right)}\right\rvert} \frac 1 2 \sum_{k \mathop = 0}^n \dfrac \d {\d x} \left({\left({f_k \left({x}\right)}\right)^2}\right)       | c = {{Defof|Vector Length}}, Sum Rule for Derivatives: General Result }} {{eqn | r = \frac 1 {\left\lvert{\mathbf f \left({x}\right)}\right\rvert} \frac 1 2 \sum_{k \mathop = 0}^n 2 f_k \left({x}\right) \dfrac {\d f_k \left({x}\right)} {\d x}       | c = Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = \frac 1 {\left\lvert{\mathbf f \left({x}\right)}\right\rvert} \mathbf f \left({x}\right) \cdot \dfrac {\d \mathbf f \left({x}\right)} {\d x}       | c = simplification, and from $(1)$ }} {{end-eqn}} Hence the result. {{qed}}
From the definition of the beta distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \expect X = \int_0^1 x \, \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^\alpha \paren {1 - x}^{\beta - 1} \rd x }} {{eqn	| r = \frac {\map \Beta {\alpha + 1, \beta} } {\map \Beta {\alpha, \beta} }  	| c = {{Defof|Beta Function|index = 1}} }} {{eqn	| r = \frac {\map \Gamma {\alpha + 1} \, \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 1} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta} 	| c = {{Defof|Beta Function|index = 3}} }} {{eqn	| r = \frac \alpha {\alpha + \beta} \cdot \frac {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } 	| c = Gamma Difference Equation }} {{eqn	| r = \frac \alpha {\alpha + \beta} }} {{end-eqn}} {{qed}}