hcju/proofwikiqa huggingface dataset:
Let $U \in \upsilon_i$. Let $x \in \map {p_i^\to} U$. Then by definition of the direct image mapping: :$\exists y \in U : x = \map {p_i} y$ By the definition of the subspace topology: :$\exists U' \in \tau: U = U' \cap Y_i$ For all $k \in I$ let $\pr_k$ denote the projection from $X$ to $X_k$. By definition of the natural basis of the Tychonoff topology $\tau$: : there exists a finite subset $J$ of $I$ and: : for each $k \in J$, there exists a $V_k \in \tau_k$ such that: : $\displaystyle y \in \bigcap_{k \in J} \map{\pr_k^\gets} {V_k} \subseteq U'$ Then: : $\displaystyle y \in \paren{\bigcap_{k \in J} \map{\pr_k^\gets} {V_k}} \cap Y_i \subseteq U' \cap Y_i = U$ By definition of direct image mapping: :$\displaystyle x = \map {p_i} y \in \map {p_i^\to} {\paren{\bigcap_{k \in J} \map{\pr_k^\gets} {V_k}} \cap Y_i} \subseteq \map {p_i^\to} U$ Recall that $p_i$ is an injection. Then: {{begin-eqn}} {{eqn | l = \map {p_i^\to} {\paren{\bigcap_{k \in J} \map{\pr_k^\gets } {V_k} } \cap Y_i}       | r = \map {p_i^\to} {\bigcap_{k \in J} \paren{\map{\pr_k^\gets } {V_k} \cap Y_i} }       | c = Intersection Distributes over Intersection }} {{eqn | r = \bigcap_{k \in J} \map {p_i^\to} {\map{\pr_k^\gets } {V_k} \cap Y_i}       | c = Image of Intersection under Injection }} {{end-eqn}} Let $k \in J$. ==== Lemma ==== {{:Subspace of Product Space Homeomorphic to Factor Space/Proof 2/Lemma 1}}{{qed|lemma}} By open set axiom $\text O 2$, then: :$\bigcap_{k \in J} \map {p_i^\to} {\map{\pr_k^\gets } {V_k} \cap Y_i}$ is open in $\struct{X_i, \tau_i}$ Since $x \in \map {p_i^\to} U$ was arbitrary then it has been shown that: :$\forall x \in \map {p_i^\to} U : \exists V \in \tau_i : x \in V \subseteq \map {p_i^\to} U$ Then by definition: $\forall x \in \map {p_i^\to} U: \map {p_i^\to} U$ is a neighborhood of $x$. From Set is Open iff Neighborhood of all its Points then $\map {p_i^\to} U$ is open in $\tau_i$. Since $U \in \upsilon_i$ was arbitrary, then $p_i$ is an open mapping.
Let $Y \sim \Exponential \lambda$. We aim to show that: :$\map \Pr {Y < -\beta \ln x} = \map \Pr {X > x}$ for all $x \in \hointl 0 1$. We have: {{begin-eqn}} {{eqn	| l = \map \Pr {Y < -\beta \ln x} 	| r = \frac 1 \beta \int_0^{-\beta \ln x} \map \exp {-\frac u \beta} \rd u 	| c = {{Defof|Exponential Distribution}} }} {{eqn	| r = \intlimits {-\map \exp {-\frac u \beta} } 0 {-\beta \ln x} 	| c = Primitive of $\exp a x$ }} {{eqn	| r = -\map \exp {-\frac {-\beta \ln x} {\beta} } + \map \exp 0 }} {{eqn	| r = 1 - x 	| c = {{Defof|Natural Logarithm}}, Exponential of Zero }} {{eqn	| r = \frac 1 {1 - 0} \int_x^1 \rd u 	| c = Primitive of Constant }} {{eqn	| r = \map \Pr {X > x} 	| c = {{Defof|Continuous Uniform Distribution}} }} {{end-eqn}} {{qed}} Category:Exponential Distribution Category:Continuous Uniform Distribution nx8t2kla4otxkne6p4xe84b3qk9vniv
Let: :$Y \sim \BetaDist {\dfrac 1 n} 1$ We aim to show that:  :$\map \Pr {Y < x^n} = \map \Pr {X < x}$ for all $x \in \closedint 0 1$.  We have: {{begin-eqn}} {{eqn	| l = \map \Pr {Y < x^n} 	| r = \int_0^{x^n} \frac 1 {\map \Beta {\frac 1 n, 1} } u^{\frac 1 n - 1} \paren {1 - u}^{1 - 1} \rd u 	| c = {{Defof|Beta Distribution}} }} {{eqn	| r = \frac {\map \Gamma {\frac 1 n + 1} } {\map \Gamma {\frac 1 n} \map \Gamma 1} \intlimits {n u^{\frac 1 n} } 0 {x^n} 	| c = {{Defof|Gamma Function}}, Primitive of Power }} {{eqn	| r = \frac n n x 	| c = Gamma Difference Equation }} {{eqn	| r = x }} {{eqn	| r = \frac 1 {1 - 0} \int_0^x \rd u 	| c = Primitive of Constant }} {{eqn	| r = \map \Pr {X < x} 	| c = {{Defof|Continuous Uniform Distribution}} }} {{end-eqn}} {{qed}} Category:Continuous Uniform Distribution Category:Beta Distribution qohnmnficfpp575bvekyr81helo7pqy
=== Discrete Random Variable === {{:Expectation of Non-Negative Random Variable is Non-Negative/Discrete}} === Continuous Random Variable === {{:Expectation of Non-Negative Random Variable is Non-Negative/Continuous}} Category:Expectation Category:Expectation of Non-Negative Random Variable is Non-Negative bfd9eculp7we3axjo3wfok1n8nc0w8c
Let $\map {\operatorname {supp} } X$ be the support of $X$.  Note that since $X$ is discrete, its sample space and hence support is countable.  Therefore, there exists some sequence $\sequence {x_i}_{i \in I}$ such that:  :$\map {\operatorname {supp} } X = \set {x_i \mid i \in I}$ for some $I \subseteq \N$.  By the definition of a sample space, we have:  :$\map \Pr {X = x_i} \ge 0$ for all $i \in I$.  Note that since $\map \Pr {X \ge 0} = 1$, we have $\map \Pr {X < 0} = 0$.  So, for any $x < 0$ we necessarily have $\map \Pr {X = x} = 0$, meaning that $x \not\in \map {\operatorname {supp} } X$.  We therefore have that any elements of $\map {\operatorname {supp} } X$ are non-negative. That is:  :$x_i \ge 0$ for all $i \in I$. Therefore:  :$x_i \map \Pr {X = x_i} \ge 0$ for all $i \in I$.  Summing over all $i \in I$, we have:  :$\displaystyle \sum_{i \in I} x_i \map \Pr {X = x_i} \ge 0$ Hence, by the definition of expectation: :$\expect X \ge 0$ {{qed}}
Note that since $\map \Pr {X = c} = 1$, we have $\map \Pr {X \ne c} = 0$ from Probability of Event not Occurring. Therefore:  :$\map {\mathrm {supp} } X = \set c$ {{MissingLinks|supp}} We therefore have: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \sum_{x \mathop \in \map {\mathrm {supp} } X} x \map \Pr {X = x} 	| c = {{Defof|Expectation|subdef = Discrete}} }} {{eqn	| r = c \map \Pr {X = c} }} {{eqn	| r = c }} {{end-eqn}} {{qed}} Category:Expectation jjxmtlo4bax6yhrwixi5r35sm5q59c0
Note that we have:  :$\map \Pr {X - Y \ge 0} = 1$ From Expectation of Non-Negative Random Variable is Non-Negative, we then have:  :$\expect {X - Y} \ge 0$ From Sum of Expectations of Independent Trials, we have:  :$\expect X + \expect {-Y} \ge 0$ From Expectation of Linear Transformation of Random Variable, we have: :$\expect X - \expect Y \ge 0$ That is:  :$\expect X \ge \expect Y$ {{qed}} Category:Expectation 1lj57fi74c5hffhuwnorm00zc1dydog
{{begin-eqn}} {{eqn	| l = \cov {X, Y} 	| r = \expect {\paren {X - \expect X} \paren {Y - \expect Y} } 	| c = {{Defof|Covariance}} }} {{eqn	| r = \expect {\paren {Y - \expect Y} \paren {X - \expect X} } 	| c = Real Multiplication is Commutative }} {{eqn	| r = \cov {Y, X} 	| c = {{Defof|Covariance}} }} {{end-eqn}} {{qed}} Category:Covariance g10n0xxgx2xzb7jpqdkdcfthwb0zd4s
{{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \cov {X_i, Y_j} 	| r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n \cov {Y_j, X_i} } 	| c = Covariance is Symmetric }} {{eqn	| r = \sum_{i \mathop = 1}^n \cov {\sum_{j \mathop = 1}^n Y_j, X_i} 	| c = Covariance of Sums of Random Variables: Lemma }} {{eqn	| r = \sum_{i \mathop = 1}^n \cov {X_i, \sum_{j \mathop = 1}^n Y_j} 	| c = Covariance is Symmetric }} {{eqn	| r = \cov {\sum_{i \mathop = 1}^n X_i, \sum_{j \mathop = 1}^n Y_j} 	| c = Covariance of Sums of Random Variables: Lemma }} {{end-eqn}} {{qed}} Category:Covariance Category:Covariance of Sums of Random Variables 3sewc8mvmg5k18dbkwk83xyhdmviws6
Let $\map {\operatorname {supp} } X$ be the support of $X$.  Then: {{begin-eqn}} {{eqn	| l = \expect {a X + b} 	| r = \sum_{x \mathop \in \map {\operatorname {supp} } X} \paren {a x + b} \map \Pr {X = x} 	| c = Expectation of Function of Discrete Random Variable }} {{eqn	| r = a \sum_{x \mathop \in \map {\operatorname {supp} } X} x \map \Pr {X = x} + b \sum_{x \mathop \in \map {\operatorname {supp} } X} \map \Pr {X = x} }} {{eqn	| r = a \expect X + b \times 1 	| c = {{Defof|Expectation/Discrete|Expectation: Discrete Random Variable}}, {{Defof|Probability Mass Function}} }} {{eqn	| r = a \expect X + b }} {{end-eqn}} {{qed}}
Let $\map {\operatorname {supp} } X$ be the support of $X$.  Let $f_X : \map {\operatorname {supp} } X \to \R$ be the probability density function of $X$. Then: {{begin-eqn}} {{eqn	| l = \expect {a X + b} 	| r = \int_{x \in \map {\operatorname {supp} } X} \paren {a x + b} \map {f_X} x \rd x 	| c = Expectation of Function of Continuous Random Variable }} {{eqn	| r = a \int_{x \in \map {\operatorname {supp} } X} x \map {f_X} x \rd x + b \int_{x \in \map {\operatorname {supp} } X} \map {f_X} x \rd x 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = a \expect X + b \times 1 	| c = {{Defof|Expectation of Continuous Random Variable}} }} {{eqn	| r = a \expect X + b }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn	| l = \var {a X + b} 	| r = \expect {\paren {a X + b - \expect {a X + b} }^2} 	| c = {{Defof|Variance}} }} {{eqn	| r = \expect {\paren {a X + b - a \expect X - b}^2} 	| c = Expectation of Linear Transformation of Random Variable }} {{eqn	| r = \expect {a^2 \paren {X - \expect X}^2} }} {{eqn	| r = a^2 \expect {\paren {X - \expect X}^2} 	| c = Expectation of Linear Transformation of Random Variable }} {{eqn	| r = a^2 \var X 	| c = {{Defof|Variance}} }} {{end-eqn}} {{qed}} Category:Variance 2wnjot8um78hb1la52xi5uluf0gf3mw
Let $\map f x$ be the function defined as: :$\forall x \in \openint {-l} l: \begin{cases} -1 & : -l < x < 0 \\ 1 & : 0 < x < l \end {cases}$ By inspection we see that $\map f x$ is an odd function. Hence from Fourier Series for Odd Function over Symmetric Range we can express $f$ by a half-range Fourier sine series: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n \pi x} l$ where for all $n \in \Z_{> 0}$: :$b_n = \displaystyle \frac 2 l \int_0^l \map f x \sin \dfrac {n \pi x} l \rd x$ over the interval $\openint 0 l$. In that interval $\openint 0 l$, we have: :$\map S x = 1$ Hence: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 l \int_0^l \sin \dfrac {n \pi x} l \rd x       | c =  }} {{eqn | r = \frac 2 l \intlimits {\frac l {n \pi} \paren {-\cos \dfrac {n \pi x} l} } 0 c       | c = Primitive of $\sin a x$ }} {{eqn | r = \frac 2 {n \pi} \paren {-\cos n \pi - \paren {-\cos 0} }       | c =  }} {{eqn | r = \frac 2 {n \pi} \paren {1 - \cos n \pi}       | c = Cosine of Zero is One and simplifying }} {{eqn | r = \frac 2 {n \pi} \paren {1 - \paren {-1}^n}       | c = Cosine of Multiple of Pi }} {{eqn | r = \begin {cases} 0 & : \text {$n$ even} \\ \dfrac 4 {n \pi}  & : \text {$n$ odd} \end {cases}       | c =  }} {{eqn | ll= \leadsto       | l = b_{2 r + 1}       | r = \dfrac 4 {\paren {2 r + 1} \pi}       | c = for $r \in \Z_{\ge 0}$: substituting $n = 2 r + 1$ }} {{end-eqn}} The result follows. {{qed}}
Let $\map f x: \openint {-l} l \to \R$ denote the absolute value function on the open interval $\openint {-l} l$: :$\map f x = \size x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$ From Fourier Series for Absolute Value Function over Symmetric Range, $\map f x$ can immediately be expressed as: :$\displaystyle \map f x \sim \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$ {{finish|We need a result that the Fourier series over an interval is the restriction of the resulting periodic function.}}
Let $\map f x: \openint {-l} l \to \R$ denote the identity function on the open interval $\openint {-l} l$: :$\map f x = x$ From Fourier Series for Identity Function over Symmetric Range, $\map f x$ can immediately be expressed as: :$\displaystyle \map f x \sim \frac {2 l} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \dfrac {n \pi x} l$ {{finish|We need a result that the Fourier series over an interval is the restriction of the resulting periodic function.}}
From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty. The conclusions follow immediately from Subspace of Product Space Homeomorphic to Factor Space. {{qed}}
The conclusions are symmetrical. {{WLOG}}, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$. Let $f: T_1 \to T_1 \times \set b$ be defined as: :$\map f x = \tuple {x, b}$ === Lemma === {{:Subspace of Product Space Homeomorphic to Factor Space/Product with Singleton/Lemma}}{{qed|lemma}} $f^{-1}$ is a restriction to the subspace $T_1 \times \set b$ of the projection $\pr_1$ of $T_1 \times T_2$ onto $T_1$. From Projection from Product Topology is Continuous , $\pr_1$ is a continuous. From Restriction of Continuous Mapping is Continuous, $f^{-1}$ is a continuous. It is to be shown that $f$ is continuous. Let $x \in T_1$. Let $U$ be an open set in $T_1 \times \set b$ such that: :$\map f x \in U$ By definition of the subspace topology then for some $U'$ open in $T_1 \times T_2$: :$U' \cap \paren {T_1 \times \set b} = U$ By the definition of the Natural Basis of Tychonoff Topology, there exist open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that: :$\map f x = \tuple {x, b} \in V_1 \times V_2 \subseteq U'$ Then for any $y \in V_1$: :$\map f y = \tuple {y, b} \in V_1 \times V_2 \subseteq U'$ But: :$\tuple {y, b} \in T_1 \times \set b$ so: :$\tuple {y, b} \in U$ Thus if $y \in V_1$, it follows that $\map f y \in U$. Then $f$ is continuous by definition. {{qed}}
=== $f$ is an Injection === Let $x, y \in T_1$. {{begin-eqn}} {{eqn | o =       | r = \map f x = \map f y }} {{eqn | ll = \leadstoandfrom       | o =        | r = \tuple {x, b} = \tuple {y, b}       | c = Definition of $f$ }} {{eqn | ll = \leadstoandfrom       | o =        | r = x = y       | c = Equality of ordered pairs }} {{end-eqn}} Then $f$ is an injection by definition. {{qed|lemma}} === $f$ is a Surjection === Let $t \in T_1 \times \set b$ Then: :$\exists x \in T_1: t = \tuple {x, b}$ So: :$\map f x = \tuple {x, b} = t$ Then $f$ is a surjection by definition. {{qed|lemma}} It follows that $f$ is a bijection by definition. {{qed}} Category:Subspace of Product Space Homeomorphic to Factor Space 27x54hb7usfyinkkzgmz4wxk5mkuj4v
Note that the support of $\operatorname U \openint 0 1$ is $\openint 0 1$. It is therefore sufficient to show that for $0 < z < 1$:  :$\map \Pr {\dfrac X {X + Y} \le z} = z$ Note that if $x, y > 0$ then:  :$0 < \dfrac x {x + y} < 1$ Note also that:  :$\dfrac x {x + y} \le z$  with $0 < z < 1$ is equivalent to:  :$x \le z x + z y$ which is in turn equivalent to:  :$\paren {1 - z} x \le z y$ That is:  :$x \le \dfrac z {1 - z} y$ We therefore have:  :$\map \Pr {\dfrac X {X + Y} \le z} = \map \Pr {X \le \dfrac z {1 - z} Y}$ Let $f_{X, Y}$ be the joint probability density function of $X$ and $Y$.  Let $f_X$ and $f_Y$ be the probability density function of $X$ and $Y$ respectively.  From Condition for Independence from Joint Probability Density Function, we have for each $x, y \in \R_{> 0}$: :$\map {f_{X, Y} } {x, y} = \map {f_X} x \map {f_Y} y$ We therefore have:  {{begin-eqn}} {{eqn	| l = \map \Pr {\dfrac X {X + Y} \le z} 	| r = \map \Pr {X \le \frac z {1 - z} Y} }} {{eqn	| r = \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_{X, Y} } {x, y} \rd x \rd y }} {{eqn	| r = \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_X} x \map {f_Y} y \rd x \rd y }} {{eqn	| r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {\int_0^{\frac z {1 - z} y} \frac 1 \beta \map \exp {-\frac x \beta} \rd x} \rd y 	| c = {{Defof|Exponential Distribution}} }} {{eqn	| r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \intlimits {-\map \exp {-\frac x \beta} } 0 {\frac z {1 - z} y} \rd y 	| c = Primitive of Exponential of $a x$ }} {{eqn	| r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {1 - \map \exp {-\frac z {\beta \paren {1 - z} } y} } \rd y 	| c = Exponential of Zero }} {{eqn	| r = 1 - \frac 1 \beta \int_0^\infty \map \exp {-y \paren {\frac z {\beta \paren {1 - z} } + \frac 1 \beta} } \rd y }} {{eqn	| r = 1 - \frac 1 \beta \int_0^\infty \map \exp {-\frac y {\beta \paren {1 - z} } } \rd y }} {{eqn	| r = 1 + \frac 1 \beta \intlimits {\beta \paren {1 - z} \map \exp {-\frac y {\beta \paren {1 - z} } } } 0 \infty 	| c = Primitive of Exponential of $a x$ }} {{eqn	| r = 1 + \frac 1 \beta \paren {0 - \beta \paren {1 - z} } 	| c = Exponential Tends to Zero and Infinity, Exponential of Zero }} {{eqn	| r = 1 - \paren {1 - z} }} {{eqn	| r = z }} {{end-eqn}} {{qed}} Category:Continuous Uniform Distribution Category:Exponential Distribution qmtahlp9cgrdrdpa5pd4st5ndbi06gi
Let $j \in I$. Let $T \subseteq S_j$. Let $\family {T_i}_{i \mathop \in I}$ be the family of sets defined by: :$T_i = \begin {cases} T & : i = j \\ S_i & : i \ne j \end {cases}$ Then: {{begin-eqn}} {{eqn | l = \map {\pr_j^\gets} T       | r = \set{x \in S: \map {\pr_j} x \in T}       | c = {{Defof|Inverse Image Mapping of Mapping}} }} {{eqn | r = \set{x \in S: x_j \in T}       | c = {{Defof|Projection on Family of Sets}} }} {{eqn | r = \set{x \in S: x_j \in T, \forall i \ne j: x_i \in S_i}       | c = as $x \in S \implies \paren {\forall i \in I: x_i \in S_i}$ }} {{eqn | r = \set{x \in S: \forall i \in I: x_i \in T_i}       | c = Definition of $T_i$ for all $i \in I$ }} {{eqn | r = \prod_{i \in I} T_i       | c = {{Defof|Cartesian Product of Family}} }} {{end-eqn}} {{qed}} Category:Projections Category:Inverse Image Mappings el7zzfkser4b4rt4npuyj4kxhjn95pl
{{begin-eqn}} {{eqn | l = \map {I_\R} {-x}       | r = -x       | c = {{Defof|Identity Function}} }} {{eqn | r = -\map {I_\R} x       | c =  }} {{end-eqn}} Hence the result by definition of odd function. {{qed}} Category:Identity Mappings Category:Examples of Odd Functions oxam7nnhno8vncvpprywkit3jqbco7m
From Odd Power is Odd Function, $x$ is a Odd Function. By Fourier Series for Odd Function over Symmetric Range, we have: :$\displaystyle x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$ where: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 \pi \int_0^\pi x \sin n x \rd x       | c =  }} {{eqn | r = \frac 2 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi       | c = Primitive of $x \sin n x$, Fundamental Theorem of Calculus }} {{eqn | r = -\frac 2 \pi \intlimits {\frac{x \cos n x} n} 0 \pi       | c = Sine of Multiple of Pi }} {{eqn | r = -\frac 2 \pi \paren {\frac{\pi \cos n \pi} n - \frac {0 \cos 0} n}       | c =  }} {{eqn | r = -\frac {2 \pi \cos n \pi} {\pi n} }} {{eqn | r = -\frac {2 \paren {-1}^n} n       | c = Cosine of Multiple of Pi }} {{eqn | r = \frac {2 \paren {-1}^{n + 1} } n }} {{end-eqn}} Substituting for $b_n$ in $(1)$:  :$\displaystyle x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$ as required.  {{qed}}
By definition of half-range Fourier cosine series: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \dfrac {n \pi x} \lambda$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map f x \cos \dfrac {n \pi x} \lambda \rd x$ Thus by definition of $f$: {{begin-eqn}} {{eqn | l = a_0       | r = \frac 2 \lambda \int_0^\lambda \map f x \rd x       | c = Cosine of Zero is One }} {{eqn | r = \frac 2 \lambda \int_0^\lambda x \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \lambda \intlimits {\frac {x^2} 2} 0 \lambda       | c = Primitive of Power }} {{eqn | r = \frac 2 \lambda \paren {\frac {l^2} 2 - \frac {0^2} 2}       | c =  }} {{eqn | r = \lambda       | c = simplification }} {{end-eqn}} {{qed|lemma}} For $n > 0$: {{begin-eqn}} {{eqn | l = a_n       | r = \frac 2 \lambda \int_0^l \map f x \cos \dfrac {n \pi x} \lambda \rd x       | c =  }} {{eqn | r = \frac 2 \lambda \int_0^l x \cos \dfrac {n \pi x} \lambda \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \lambda \intlimits {\frac {l^2} {\pi^2 n^2} \cos \dfrac {n \pi x} \lambda + \frac \lambda {\pi n} x \sin \dfrac {n \pi x} \lambda} 0 \lambda       | c = Primitive of $x \cos n x$ }} {{eqn | r = \frac 2 \lambda \paren {\paren {\frac {\lambda^2} {\pi^2 n^2} \cos n \pi + \frac {\lambda^2} {\pi n} \sin n \pi} - \paren {\frac {\lambda^2} {\pi^2 n^2} \cos 0 + \frac \lambda {\pi n} 0 \sin 0} }       | c =  }} {{eqn | r = \frac {2 \lambda} {\pi^2 n^2} \paren {\cos n \pi - \cos 0}       | c = Sine of Multiple of Pi and simplification }} {{eqn | r = \frac {2 \lambda} {\pi^2 n^2} \paren {\paren {-1}^n - 1}       | c = Cosine of Multiple of Pi }} {{end-eqn}} When $n$ is even, $\paren {-1}^n = 1$. We can express $n = 2 r$ for $r \ge 1$. Hence in that case: {{begin-eqn}} {{eqn | l = a_{2 r}       | r = \frac {2 \lambda} {\pi^2 n^2} \paren {\paren {-1}^n - 1}       | c =  }} {{eqn | r = \frac {2 \lambda} {\pi^2 n^2} \paren {1 - 1}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} When $n$ is odd, $\paren {-1}^n = -1$. We can express $n = 2 r + 1$ for $r \ge 0$. Hence in that case: {{begin-eqn}} {{eqn | l = a_{2 r - 1}       | r = \frac {2 \lambda} {\pi^2 \paren {2 r + 1}^2} \paren {-1 - 1}       | c =  }} {{eqn | r = -\frac {4 \lambda} {\pi^2} \paren {\frac 1 {\paren {2 r + 1}^2} }       | c = rearranging }} {{end-eqn}} 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 \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{r \mathop = 0}^\infty \frac 1 {\paren {2 r + 1}^2} \cos \dfrac {\paren {2 r + 1} \pi x} \lambda       | c = substituting for $a_0$ and $a_n$ from above }} {{eqn | r = \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda       | c = changing the name of the variable }} {{end-eqn}} {{qed}} Category:Half-Range Fourier Series for Identity Function 6ban72q3brqfa31nsk4okvynnqfo5s8
By definition of half-range Fourier sine series: :$(1): \quad \map f x \sim \displaystyle \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n \pi x} \lambda$ where for all $n \in \Z_{> 0}$: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 \lambda \int_0^\lambda \map f x \sin \dfrac {n \pi x} \lambda \rd x       | c =  }} {{eqn | r = \frac 2 \lambda \int_0^\lambda x \sin \frac {n \pi x} \lambda \rd x       | c = Definition of $\map f x$ }} {{eqn | r = \frac 2 \lambda \intlimits {\frac {\lambda^2} {n^2 \pi^2} \sin \dfrac {n \pi x} \lambda - \frac \lambda {n \pi} x \cos \dfrac {n \pi x} \lambda} 0 \lambda       | c = Primitive of $x \sin a x$, Fundamental Theorem of Calculus }} {{eqn | r = \frac 2 \lambda \paren {\paren {\frac {\lambda^2} {n^2 \pi^2} \sin n \pi - \frac {\lambda^2} {n \pi} \cos n \pi} - \paren {\frac {\lambda^2} {n^2 \pi^2} \sin 0 - \frac \lambda {n \pi} 0 \cos 0} }       | c =  }} {{eqn | r = -\frac {2 \lambda} {n \pi} \cos n \pi       | c = Sine of Multiple of Pi and simplification }} {{eqn | r = -\frac {2 \lambda} {n \pi} \paren {-1}^n       | c = Cosine of Multiple of Pi }} {{eqn | r = \frac {2 \lambda} {n \pi} \paren {-1}^{n + 1} }} {{end-eqn}} Substituting for $b_n$ in $(1)$:  :$\displaystyle \map f x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$ as required. {{Qed}} Category:Half-Range Fourier Series for Identity Function o6g77z023fc25hdl5u9cbgvh8h4djq8
Recall the definition of the absolute value function: :$\size x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$ Testing the $3$ cases in turn: {{begin-eqn}} {{eqn | l = x       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = -x       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = \size {-x}       | r = x       | c = {{Defof|Absolute Value}} }} {{eqn | r = \size x       | c = {{Defof|Absolute Value}} }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | r = 0 }} {{eqn | ll= \leadsto       | l = -x       | r = 0 }} {{eqn | ll= \leadsto       | l = \size {-x}       | r = 0       | c = {{Defof|Absolute Value}} }} {{eqn | r = \size x       | c = {{Defof|Absolute Value}} }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = -x       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \size {-x}       | r = -x       | c = {{Defof|Absolute Value}}: as $-x > 0$ }} {{eqn | r = \size x       | c = {{Defof|Absolute Value}}: as $x < 0$ }} {{end-eqn}} {{qed|lemma}} Hence the result by definition of even function. {{qed}} Category:Absolute Value Function Category:Examples of Even Functions a3h0lzjlyt6aah5vi3qsfm5hn1qw61a