hcju/proofwikiqa huggingface dataset:
Let $f \left({x}\right) = e^x \cos x$. By definition of Maclaurin series: :$(1): \quad f \left({x}\right) \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} f^{\left({n}\right)} \left({0}\right)$ where $f^{\left({n}\right)} \left({0}\right)$ denotes the $n$th derivative of $f$ {{WRT|Differentiation}} $x$ evaluated at $x = 0$. It remains to be shown that: :$f^{\left({n}\right)} \left({0}\right) = 2^{n / 2} \cos \left({\dfrac {n \pi} 4}\right)$ The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$f^{\left({n}\right)} \left({x}\right) = 2^{n / 2} e^x \cos \left({x + \dfrac {n \pi} 4}\right)$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = e^x \cos x       | c =  }} {{eqn | r = 2^{0 / 2} e^x \cos \left({x + \dfrac {0 \pi} 4}\right)       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$f^{\left({k}\right)} \left({x}\right) = 2^{k / 2} e^x \cos \left({x + \dfrac {k \pi} 4}\right)$ from which it is to be shown that: :$f^{\left({k + 1}\right)} \left({x}\right) = 2^{\left({k + 1}\right) / 2} e^x \cos \left({x + \dfrac {\left({k + 1}\right) \pi} 4}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = f^{\left({k + 1}\right)} \left({x}\right)       | r = \frac \d {\d x} f^{\left({k}\right)} \left({x}\right)       | c =  }} {{eqn | r = \frac \d {\d x} 2^{k / 2} e^x \cos \left({x + \dfrac {k \pi} 4}\right)       | c = Induction Hypothesis }} {{eqn | r = 2^{k / 2} \left({\cos \left({x + \dfrac {k \pi} 4}\right) \frac \d {\d x} e^x + e^x \frac \d {\d x} \cos \left({x + \dfrac {k \pi} 4}\right)}\right)       | c = Product Rule }} {{eqn | r = 2^{k / 2} \left({\cos \left({x + \dfrac {k \pi} 4}\right) \frac \d {\d x} e^x - e^x \sin \left({x + \dfrac {k \pi} 4}\right)}\right)       | c = Derivative of Cosine Function, Derivatives of Function of $a x + b$ }} {{eqn | r = 2^{k / 2} e^x \left({\cos \left({x + \dfrac {k \pi} 4}\right) - \sin \left({x + \dfrac {k \pi} 4}\right)}\right)       | c = Derivative of Exponential Function and simplification }} {{eqn | r = 2^{k / 2} e^x \sqrt 2 \cos \left({x + \dfrac {k \pi} 4 + \dfrac \pi 4}\right)       | c = Cosine of x minus Sine of x: Cosine Form }} {{eqn | r = 2^{\left({k + 1}\right) / 2} e^x \cos \left({x + \dfrac {\left({k + 1}\right) \pi} 4}\right)       | c = simplification }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and by the Principle of Mathematical Induction: :$\forall n \in \Z_{\ge 0}: f^{\left({n}\right)} \left({x}\right) = 2^{n / 2} e^x \cos \left({x + \dfrac {n \pi} 4}\right)$ The result follows by setting $x = 0$ and substituting for $f^{\left({n}\right)} \left({0}\right)$ in $(1)$. {{qed}}
From Power Series Expansion for Cotangent Function: {{begin-eqn}} {{eqn | n = 1       | l = \cot x       | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}       | c =  }} {{eqn | r = \frac 1 x - \frac x 3 - \frac {x^3} {45} - \frac {2 x^5} {945} + \cdots       | c =  }} {{end-eqn}} for $0 < \size x < \pi$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int_0^x \cot x \rd x       | r = \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} \rd x       | c =  }} {{eqn | r = \int_0^x \frac 1 x \rd x + \sum_{n \mathop = 1}^\infty \int_0^x \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} \rd x       | c = extracting the zeroth term }} {{eqn | ll= \leadsto       | l = \ln \size {\sin x}       | r = \ln \size x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n} } {\paren {2 n} \paren {2 n}!}       | c = Primitive of $\cot x$, Integral of Power, Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = \ln \size {\sin x}       | r = \ln \size x - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n 2^{2 n - 1} B_{2 n} \, x^{2 n} } {n \paren {2 n}!}       | c = simplification }} {{end-eqn}} {{qed}}
From Power Series Expansion for Tangent Function: {{begin-eqn}} {{eqn | n = 1       | l = \tan x       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}       | c =  }} {{eqn | r = x + \frac {x^3} 3 + \frac {2 x^5} {15} + \frac {17 x^7} {315} + \cdots       | c =  }} {{end-eqn}} for $\size x < \dfrac \pi 2$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int_0^x \tan x \rd x       | r = \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = -\ln \size {\cos x}       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n} } {\paren {2 n} \paren {2 n}!}       | c = Primitive of $\tan x$: Cosine Form, Integral of Power }} {{eqn | ll= \leadsto       | l = \ln \size {\cos x}       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n 2^{2 n - 1} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n} } {n \paren {2 n}!}       | c = simplification }} {{end-eqn}} {{qed}}
By Moment Generating Function of Gaussian Distribution, the moment generating function of $X$ is given by:  :$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$ From Moment in terms of Moment Generating Function:  :$\expect X = \map {M'_X} 0$ We have:  {{begin-eqn}} {{eqn	| l = \map {M_X'} t 	| r = \frac \d {\d t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} }} {{eqn	| r = \map {\frac \d {\d t} } {\mu t + \frac 1 2 \sigma^2 t^2} \frac \d {\map \d {\mu t + \dfrac 1 2 \sigma^2 t^2} } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} 	| c = Chain Rule for Derivatives }} {{eqn	| r = \paren {\mu + \sigma^2 t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} 	| c = Derivative of Power, Derivative of Exponential Function }} {{end-eqn}} Setting $t = 0$:  {{begin-eqn}} {{eqn	| l = \map {M_X'} 0 	| r = \paren {\mu + 0\sigma^2} \map \exp {0\mu + 0 \sigma^2} }} {{eqn	| r = \mu \exp 0  }} {{eqn	| r = \mu 	| c = Exponential of Zero }} {{end-eqn}} {{qed}}
From the definition of the Gaussian distribution, $X$ has probability density function: :$f_X \left({x}\right) = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \exp \left({-\dfrac { \left({x - \mu}\right)^2} {2 \sigma^2} }\right)$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \mathbb E \left[{X}\right] = \int_{-\infty}^\infty x f_X \left({x}\right) \rd x$ So: {{begin-eqn}} {{eqn	| l = \mathbb E \left[{X}\right]  	| r = \frac 1 { \sigma \sqrt{2 \pi} } \int_{-\infty}^\infty x \exp \left({- \frac {\left({x - \mu}\right)^2} {2 \sigma^2} }\right) \rd x }} {{eqn	| r = \frac {\sqrt 2 \sigma} { \sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \left({\sqrt 2 \sigma t + \mu}\right) \exp \left({-t^2}\right) \rd t 	| c = substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$ }} {{eqn	| r = \frac 1 {\sqrt \pi} \left({\sqrt 2 \sigma \int_{-\infty}^\infty t \exp \left({-t^2}\right) \rd t + \mu \int_{-\infty}^\infty \exp \left({-t^2}\right) \rd t}\right) }} {{eqn	| r = \frac 1 {\sqrt \pi} \left({\sqrt 2 \sigma \left[{-\frac 1 2 \exp \left({-t^2}\right)}\right]_{-\infty}^\infty + \mu \sqrt \pi}\right) 	| c = Fundamental Theorem of Calculus, Gaussian Integral }} {{eqn	| r = \frac {\mu \sqrt \pi} {\sqrt \pi} 	| c = Exponential Tends to Zero and Infinity }} {{eqn	| r = \mu }} {{end-eqn}} {{qed}}
From Power Series Expansion for Cosecant Function: {{begin-eqn}} {{eqn | l = 2 \csc 2 x       | r = 2 \sum_{n \mathop = 0}^\infty \dfrac {\left({-1}\right)^{n - 1} 2 \left({2^{2 n - 1} - 1}\right) B_{2 n} \,  \left({2 x}\right)^{2 n - 1} } {\left({2 n}\right)!}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 2 \csc 2 x       | r = \sum_{n \mathop = 0}^\infty \dfrac {\left({-1}\right)^{n - 1} 2^{2 n + 1} \left({2^{2 n - 1} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c =  }} {{end-eqn}} for $0 < \left\lvert{2 x}\right\rvert < \pi$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int 2 \csc 2 x \rd x       | r = \sum_{n \mathop = 0}^\infty \int \dfrac {\left({-1}\right)^{n - 1} 2^{2 n + 1} \left({2^{2 n - 1} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} \rd x       | c =  }} {{eqn | l = \int 2 \csc 2 x \rd x       | r = \int \frac 1 x \rd x + \sum_{n \mathop = 1}^\infty \int \dfrac {\left({-1}\right)^{n - 1} 2^{2 n + 1} \left({2^{2 n - 1} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} \rd x       | c = extracting the zeroth term }} {{eqn | ll= \implies       | l = \ln \left \vert {\tan x} \right \vert       | r = \ln \left\lvert{x}\right\rvert + \sum_{n \mathop = 1}^\infty \dfrac {\left({-1}\right)^{n - 1} 2^{2 n + 1} \left({2^{2 n - 1} - 1}\right) B_{2 n} \, x^{2 n} } {\left({2 n}\right) \left({2 n}\right)!}       | c = Primitive of $\csc a x$, Integral of Power, Primitive of Reciprocal }} {{eqn | r = \ln \left\lvert{x}\right\rvert + \sum_{n \mathop = 1}^\infty \dfrac {\left({-1}\right)^{n - 1} 2^{2 n} \left({2^{2 n - 1} - 1}\right) B_{2 n} \, x^{2 n} } {n \left({2 n}\right)!}       | c = simplification }} {{end-eqn}} for $0 < \left\lvert{x}\right\rvert < \dfrac \pi 2$. {{qed}}
Let $\map f x = \dfrac {\map \ln {1 + x} } {1 + x}$. By definition of Maclaurin series: :$(1): \quad \map f x \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0$ where $\map {f^{\paren n} } 0$ denotes the $n$th derivative of $f$ {{WRT|Differentiation}} $x$ evaluated at $x = 0$. From Nth Derivative of Natural Logarithm by Reciprocal: :$\dfrac {\d^n} {\d x^n} \dfrac {\map \ln {1 + x} } {1 + x} = \paren {-1}^{n + 1} n! \dfrac {H_n - \map \ln {1 + x} } {\paren {1 + x}^{n + 1} }$ The result follows by setting $x = 0$ and substituting for $\map {f^{\paren n} } 0$ in $(1)$. {{finish|Prove the range of convergence}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \dfrac {\ln x} {x^n}       | r = \dfrac \d {\d x} x^{-n} \ln x       | c =  }} {{eqn | r = \ln x \dfrac \d {\d x} x^{-n} + x^{-n} \dfrac \d {\d x} \ln x       | c = Product Rule }} {{eqn | r = - n x^{-n - 1} \ln x + x^{-n} \dfrac 1 x       | c = Primitive of Reciprocal, Primitive of Power }} {{eqn | r = -n x^{-n - 1} \ln x + x^{-n - 1}       | c =  }} {{eqn | r = \frac {1 - n \ln x} {x^{n + 1} }       | c =  }} {{end-eqn}} {{qed}} Category:Derivatives Category:Logarithms f3acwgqgevuihr93igx14zxwrvjif43
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$ === Basis for the Induction === $\map P 1$ is the case: :$\dfrac \d {\d x} \dfrac {\ln x} x = \paren {-1}^n n! \dfrac {H_n - \ln x} {x^{n + 1} }$ {{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \dfrac {\ln x} x       | r = \dfrac {1 - \ln x} {x^2}       | c = Derivative of Logarithm over Power }} {{eqn | r = \paren {-1}^2 \times 1! \times \dfrac {H_1 - \ln x} {x^{1 + 1} }       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\dfrac {\d^k} {\d x^k} \dfrac {\ln x} x = \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }$ from which it is to be shown that: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x = \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x       | r = \dfrac \d {\d x} \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \paren {-1}^{k + 1} k! \map {\dfrac \d {\d x} } {\dfrac {H_k} {x^{k + 1} } - \dfrac {\ln x} {x^{k + 1} } }       | c = Linear Combination of Derivatives }} {{eqn | r = \paren {-1}^{k + 1} k! \paren {\dfrac {-\paren {k + 1} H_k} {x^{k + 2} } - \dfrac {1 - \paren {k + 1} \ln x} {x^{k + 2} } }       | c = Derivative of Power, Derivative of Logarithm over Power }} {{eqn | r = \paren {-1}^{k + 2} \paren {k + 1}! \paren {\dfrac {H_k} {x^{k + 2} } + \dfrac 1 {\paren {k + 1} x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }       | c = extracting $-\paren {k + 1}$ as a factor }} {{eqn | r = \paren {-1}^{k + 2} \paren {k + 1}! \paren {\paren {H_k + \dfrac 1 {\paren {k + 1} } } \dfrac 1 {x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }       | c =  }} {{eqn | r = \paren {-1}^{k + 2} \paren {k + 1}! \paren {\dfrac {H_{k + 1} } {x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }       | c =  }} {{eqn | r = \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: \dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$ {{qed}} Category:Derivatives Category:Natural Logarithms cm4r7tcoruidcpnlxpy8ugbr0dga9x6
By Moment Generating Function of Gaussian Distribution, the moment generating function of $X$ is given by:  :$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$ From Variance as Expectation of Square minus Square of Expectation:  :$\var X = \expect {X^2} - \paren {\expect X}^2$ 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} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} }} {{eqn	| r = \map {\frac \d {\d t} } {\map {\frac \d {\d t} } {\mu t + \frac 1 2 \sigma^2 t^2} \frac \d {\map \d {\mu t + \dfrac 1 2 \sigma^2 t^2} } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} } 	| c = Chain Rule for Derivatives }} {{eqn	| r = \map {\frac \d {\d t} } {\paren {\mu + \sigma^2 t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} } 	| c = Derivative of Power, Derivative of Exponential Function }} {{eqn	| r = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} \map {\frac \d {\d t} } {\mu + \sigma^2 t} + \paren {\mu + \sigma^2 t} \frac \d {\d t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} 	| c = Product Rule }} {{eqn	| r = \sigma^2 \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} + \paren {\mu + \sigma^2 t}^2 \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2} }} {{end-eqn}} Setting $t = 0$, we obtain the second moment:  {{begin-eqn}} {{eqn 	| l = \map {M''_X} 0  	| r = \expect {X^2} }} {{eqn	| r = \sigma^2 \map \exp {0 \mu + 0 \sigma^2} + \paren {\mu + 0 \sigma^2}^2 \map \exp {0 \mu + 0 \sigma^2} }} {{eqn	| r = \sigma^2 \exp 0 + \mu^2 \exp 0  }} {{eqn	| r = \sigma^2 + \mu^2 	| c = Exponential of Zero }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \var X 	| r = \sigma^2 + \mu^2 - \paren {\expect X}^2 }} {{eqn	| r = \sigma^2 + \mu^2 - \mu^2 	| c = Expectation of Gaussian Distribution }} {{eqn	| r = \sigma^2 }} {{end-eqn}} {{qed}}
Let $\mu = \expect X$.  Let $X$ have probability density function $f_X$.  As $f_X$ is a probability density function:  :$\displaystyle \int_{-\infty}^\infty f_X \paren x \rd x = \Pr \paren {-\infty < X < \infty} = 1$ Then: {{begin-eqn}} {{eqn	| l = \var X 	| r = \expect {\paren {X - \mu}^2} 	| c = {{Defof|Variance of Continuous Random Variable}} }} {{eqn	| r = \int_{-\infty}^\infty \paren {X - \mu}^2 f_X \paren x \rd x 	| c = {{Defof|Expectation of Continuous Random Variable}} }} {{eqn	| r = \int_{-\infty}^\infty \paren {x^2 - 2 \mu x + \mu^2} f_X \paren x \rd x }} {{eqn	| r = \int_{-\infty}^\infty x^2 f_X \paren x \rd x - 2 \mu \int_{-\infty}^\infty x f_X \paren x \rd x + \mu^2 \int_{-\infty}^\infty f_X \paren x \rd x }} {{eqn	| r = \expect {X^2} - 2 \mu^2 + \mu^2 	| c = {{Defof|Expectation of Continuous Random Variable}} }} {{eqn	| r = \expect {X^2} - \mu^2 }} {{eqn	| r = \expect {X^2} - \paren {\expect X}^2 }} {{end-eqn}} {{qed}}
We let $\mu = \expect X$, and take the expression for variance: :$\var X := \displaystyle \sum_{x \mathop \in \Img X} \paren {x - \mu}^2 \Pr \paren {X = x}$ Then: {{begin-eqn}} {{eqn | l = \var X       | r = \sum_x \paren {x^2 - 2 \mu x + \mu^2} \Pr \paren {X = x}       | c =  }} {{eqn | r = \sum_x x^2 \Pr \paren {X = x} - 2 \mu \sum_x x \Pr \paren {X = x} + \mu^2 \sum_x \Pr \left({X = x}\right)       | c =  }} {{eqn | r = \sum_x x^2 \Pr \paren {X = x} - 2 \mu \sum_x x \Pr \left({X = x}\right) + \mu^2       | c = {{Defof|Probability Mass Function}}: $\displaystyle \sum_x \Pr \paren {X = x} = 1$ }} {{eqn | r = \sum_x x^2 \Pr \paren {X = x} - 2 \mu^2  + \mu^2       | c = {{Defof|Expectation}}: $\displaystyle \sum_x x \Pr \paren {X = x} = \mu$ }} {{eqn | r = \sum_x x^2 \Pr \paren {X = x} - \mu^2       | c =  }} {{end-eqn}} Hence the result, from $\mu = \expect X$. {{qed}}
From the definition of the continuous uniform distribution, $X$ has probability density function:  :$\map {f_X} x = \begin {cases} \dfrac 1 {b - a} & : a \le x \le b \\ 0 & : \text {otherwise} \end {cases}$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \int_{-\infty}^a 0 x \rd x + \int_a^b \frac x {b - a} \rd x + \int_b^\infty 0 x \rd x }} {{eqn	| r = \intlimits {\frac {x^2} {2 \paren {b - a} } } a b 	| c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn	| r = \frac {b^2 - a^2} {2 \paren {b - a} } }} {{eqn	| r = \frac {\paren {b - a} \paren {b + a} } {2 \paren {b - a} } 	| c = Difference of Two Squares }} {{eqn	| r = \frac {a + b} 2 }} {{end-eqn}} {{qed}}
From the definition of the continuous uniform distribution, $X$ has probability density function:  :$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$ From Variance as Expectation of Square minus Square of Expectation:  :$\displaystyle \var X = \int_{-\infty}^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$ So: {{begin-eqn}} {{eqn	| l = \var X 	| r = \int_{-\infty}^a 0 x^2 \rd x + \int_a^b \frac {x^2} {b - a} \rd x + \int_b^\infty 0 x^2 \rd x - \frac {\paren {a + b}^2} 4 	| c = Expectation of Continuous Uniform Distribution }} {{eqn	| r = \intlimits {\frac {x^3} {3 \paren {b - a} } } a b - \frac {\paren {a + b}^2} 4  	| c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn	| r = \frac {b^3 - a^3} {3 \paren {b - a} } - \frac {\paren {a + b}^2} 4  }} {{eqn	| r = \frac {4 \paren {b - a} \paren {a^2 + a b + b^2} } {12 \paren {b - a} } - \frac {3 \paren {a + b}^2} {12} 	| c = Difference of Two Cubes }} {{eqn	| r = \frac {4 a^2 + 4 a b + 4 b^2 - 3 a^2 - 6 a b - 3 b^2} {12} 	| c = Square of Sum }} {{eqn	| r = \frac {b^2 - 2 a b + a^2} {12} }} {{eqn	| r = \frac {\paren {b - a}^2} {12} 	| c = Square of Difference }} {{end-eqn}} {{qed}}
From the definition of the Gaussian distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$ Note that $f_X$ is non-zero, sufficient to ensure a unique median. By the definition of a median, to prove that $\mu$ is the median of $X$ we must verify:  :$\displaystyle \map \Pr {X < \mu} = \int_{-\infty}^\mu \map {f_X} x \rd x = \frac 1 2$ We have: {{begin-eqn}} {{eqn	| l = \int_{-\infty}^\mu \map {f_X} x \rd x 	| r = \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\mu \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x }} {{eqn	| r = \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^{\frac {\mu - \mu} {\sqrt 2 \sigma} } \map \exp {-t^2} \rd t 	| c = substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$ }} {{eqn	| r = \frac 1 {\sqrt \pi} \int_{-\infty}^0 \map \exp {-t^2} \rd t }} {{eqn	| r = \frac 1 {2 \sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t 	| c = Definite Integral of Even Function }} {{eqn	| r = \frac {\sqrt \pi} {2 \sqrt \pi} 	| c = Gaussian Integral }} {{eqn	| r = \frac 1 2 }} {{end-eqn}} {{qed}} Category:Gaussian Distribution 0259aldxcj8yy3k46arwrat8zv8r931
From the definition of the Gamma distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \expect X = \int_0^\infty x \, \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^\alpha e^{-\beta x} \rd x }} {{eqn	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \left({\frac t \beta}\right)^\alpha e^{-t} \frac {\rd t} \beta 	| c = substituting $t = \beta x$ }} {{eqn	| r = \frac {\beta^\alpha} {\beta^{\alpha + 1} \map \Gamma \alpha} \int_0^\infty t^\alpha e^{-t} \rd t }} {{eqn	| r = \frac {\map \Gamma {\alpha + 1} } {\beta \, \map \Gamma \alpha}  	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \frac {\alpha \, \map \Gamma \alpha} {\beta \, \map \Gamma \alpha}  	| c = Gamma Difference Equation }} {{eqn	| r = \frac \alpha \beta }} {{end-eqn}} {{qed}}
By Moment Generating Function of Gaussian Distribution, the moment generating function of $X$ is given by:  :$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$ for $t < \beta$.  From Moment in terms of Moment Generating Function:  :$\expect X = \map {M'_X} 0$ We have:  {{begin-eqn}} {{eqn	| l = \map {M'_X} t 	| r = \map {\frac \d {\d t} } {1 - \frac t \beta}^{-\alpha} }} {{eqn	| r = \beta^\alpha \map {\frac \d {\d t} } {\frac 1 {\paren {\beta - t}^\alpha} } }} {{eqn	| r = \beta^\alpha \, \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 \frac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }  	| c = Derivative of Power }} {{eqn	| r = \frac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} } }} {{end-eqn}} Setting $t = 0$: {{begin-eqn}} {{eqn	| l = \map {M'_X} 0 	| r = \frac {\beta^\alpha \alpha} {\paren {\beta - 0}^{\alpha + 1} } }} {{eqn	| r = \frac {\beta^\alpha \alpha} {\beta^{\alpha + 1} } }} {{eqn	| r = \frac \alpha \beta }} {{end-eqn}} {{qed}}
From the definition of the Gamma distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$ From Variance as Expectation of Square minus Square of Expectation:  :$\displaystyle \var X = \int_0^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$ So: {{begin-eqn}} {{eqn	| l = \var X 	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha + 1} e^{-\beta x} \rd x - \paren {\frac \alpha \beta}^2 	| c = Expectation of Gamma Distribution }} {{eqn	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac t \beta}^{\alpha + 1} e^{-t} \frac {\d t} \beta - \frac {\alpha^2} {\beta^2} 	| c = substituting $t = \beta x$ }} {{eqn	| r = \frac {\beta^\alpha} {\beta^{\alpha + 2} \, \map \Gamma \alpha} \int_0^\infty t^{\alpha + 1} e^{-t} \rd t - \frac {\alpha^2} {\beta^2} }} {{eqn	| r = \frac {\map \Gamma {\alpha + 2} } {\beta^2 \, \map \Gamma \alpha} - \frac {\alpha^2} {\beta^2} 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \frac {\map \Gamma {\alpha + 2} - \alpha^2 \, \map \Gamma \alpha} {\beta^2 \, \map \Gamma \alpha} }} {{eqn	| r = \frac {\alpha \paren {\alpha + 1} \map \Gamma \alpha - \alpha^2 \, \map \Gamma \alpha} {\beta^2 \, \map \Gamma \alpha} 	| c = Gamma Difference Equation }} {{eqn	| r = \frac {\alpha \, \map \Gamma \alpha \paren {\alpha + 1 - \alpha} } {\beta^2 \, \map \Gamma \alpha} }} {{eqn	| r = \frac \alpha {\beta^2} }} {{end-eqn}} {{qed}}