Primitive of Reciprocal of x cubed by a squared minus x squared squared/Partial Fraction Expansion 0 34068 394763 188738 2019-03-05T17:00:36Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 \paren {a^2 - x^2}^2$ == :$\dfrac 1 {x^3 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} + \dfrac {2 x} {a^6 \paren {a^2 - x^2} } + \dfrac x {a^4 \paren {a^2 - x^2}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^3 \paren {a^2 - x^2}^2}       | r = \frac 1 {x^3 \paren {a + x}^2 \paren {a - x}^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {a + x} + \frac B {\paren {a + x}^2} + \frac C {a - x} + \frac D {\paren {a - x}^2} + \frac E x + \frac F {x^2} + \frac G {x^3}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x^3 \paren {a^2 - x^2} \paren {a - x} + B x^3 \paren {a - x}^2       | c = multiplying through by $x^3 \paren {a^2 - x^2}^2$ }} {{eqn | o =        | ro= +       | r = C x^3 \paren {a^2 - x^2} \paren {a + x} + D x^3 \paren {a + x}^2       | c =  }} {{eqn | o =        | ro= +       | r = E x^2 \paren {a^2 - x^2}^2 + F x \paren {a^2 - x^2}^2 + G \paren {a^2 - x^2}^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^6 - A a x^5 - A a^2 x^4 + A a^3 x^3 + B x^5 - 2 B a x^4 + B a^2 x^3       | c = multiplying out }} {{eqn | o =        | ro= -       | r = C x^6 - C a x^5 + C a^2 x^4 + C a^3 x^3 + D x^5 + 2 D a x^4 + D a^2 x^3       | c =  }} {{eqn | o =        | ro= +       | r = E x^6 - 2 E a^2 x^4 + E a^4 x^2 + F x^5 - 2 F a^2 x^3 + F x a^4       | c =  }} {{eqn | o =        | ro= +       | r = G x^4 - 2 G a^2 x^2 + G a^4       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = G a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = G       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a^3 \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {4 a^5}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \paren {-a}^3 \paren {2 a}^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac {-1} {4 a^5}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = F       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = E a^4 - 2 G a^2        | c =  }} {{eqn | ll= \leadsto       | l = E a^4       | r = 2 \frac 1 {a^4} a^2        | c =  }} {{eqn | ll= \leadsto       | l = E       | r = \frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^6$ in $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = 0       | r = A - C + E       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = A - C + \frac 2 {a^6}       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = C - \frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^3 + B a^2 + C a^3 + D a^2 - 2 F a^2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A a^3 + \frac {-1} {4 a^5} a^2 + C a^3 + \frac 1 {4 a^5} a^2       | r = 0       | c = substituting for $B$, $D$ and $F$ }} {{eqn | ll= \leadsto       | l = A + C       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = -C       | c =  }} {{eqn | r = \frac {-1} {a^6}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^6} }} {{eqn | l = B       | r = \frac {-1} {4 a^5} }} {{eqn | l = C       | r = \frac 1 {a^6} }} {{eqn | l = D       | r = \frac 1 {4 a^5} }} {{eqn | l = E       | r = \frac 2 {a^6} }} {{eqn | l = F       | r = 0 }} {{eqn | l = G       | r = \frac 1 {a^4} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x^3 \paren {a^2 - x^2}^2}       | o = \equiv       | r = \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac 1 {a^6 \paren {a + x} } + \dfrac 1 {a^6 \paren {a - x} } - \dfrac 1 {4 a^5 \paren {a + x}^2} + \dfrac 1 {4 a^5 \paren {a - x}^2}       | c =  }} {{eqn | o = \equiv       | r = \frac {\paren {a + x} - \paren {a - x} } {a^6 \paren {a + x} \paren {a - x} } + \frac {\paren {a + x}^2 - \paren {a - x}^2} {4 a^5 \paren {a + x}^2 \paren {a - x}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}       | c = common denominators }} {{eqn | o = \equiv       | r = \frac {2 x} {a^6 \paren {a^2 - x^2} } + \frac {\paren {a^2 + 2 a x + x^2} - \paren {a^2 - 2 a x + x^2} } {4 a^5 \paren {a^2 - x^2}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac {2 x} {a^6 \paren {a^2 - x^2} } + \frac {4 a x} {4 a^5 \paren {a^2 - x^2}^2} + \frac 1 {a^4 x^3} + \frac 2 {a^6 x}       | c = simplifying }} {{eqn | o = \equiv       | r = \frac 1 {a^4 x^3} + \frac 2 {a^6 x} + \frac {2 x} {a^6 \paren {a^2 - x^2} } + \frac x {a^4 \paren {a^2 - x^2}^2}       | c = simplifying and rearranging }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving a squared minus x squared/Lemmata mg0zd2ych3vpf10mlszcsga1v0g3v4h


Primitive of x squared over Root of a squared minus x squared 0 34139 188942 2014-06-12T20:57:02Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$ </only..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$ </onlyinclude> == Proof == With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u}{\mathrm d x}       | r = 1       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v}{\mathrm d x}       | r = \frac x {\sqrt {a^2 - x^2} }       | c =  }} {{eqn | ll= \implies       | l = v       | r = -\sqrt {a^2 - x^2}       | c = Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} }       | r = \int x \frac {x \ \mathrm d x} {\sqrt {a^2 - x^2} }       | c =  }} {{eqn | r = -x \sqrt {a^2 - x^2} - \int \left({-\sqrt {a^2 - x^2} }\right) \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = -x \sqrt {a^2 - x^2} + \int \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d x       | c = simplifying }} {{eqn | r = -x \sqrt {a^2 - x^2} + \left({\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a}\right) + C       | c = Primitive of $\sqrt {a^2 - x^2}$ }} {{eqn | r = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C       | c =  }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\dfrac {x^2} {\sqrt {x^2 + a^2} }$ * Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of x over Root of a squared minus x squared|next=Primitive of x cubed over Root of a squared minus x squared}}: $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.239$ Category:Primitives involving Root of a squared minus x squared jqy62n9bfz203t5e1r181cpnio7poo3


Sum of Infinite Geometric Sequence/Corollary 2 0 34268 456673 456640 2020-03-20T13:09:22Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Infinite Geometric Sequence == Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$. Let $z \in S$. Let $\size z < 1$, where $\size z$ denotes: :the absolute value of $z$, for real and rational $z$ :the complex modulus of $z$ for complex $z$. Then: <onlyinclude> :$\displaystyle \sum_{n \mathop = 0}^\infty a z^n = \frac a {1 - z}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty a z^n       | r = a \sum_{n \mathop = 0}^\infty z^n       | c = }} {{eqn | r = a \frac 1 {1 - z}       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \frac a {1 - z}       | c = }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Handbook of Mathematical Functions|1964|Milton Abramowitz|author2 = Irene A. Stegun|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Arithmetic Mean}}: $3.1.10$: Sum of Geometric Progression to $n$ Terms * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Arithmetic-Geometric Series}}: $\S 19$: Geometric Series: $19.5$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Sum of Geometric Sequence/Corollary 1|next = Sum of Infinite Geometric Sequence/Corollary 2/Mistake|entry = geometric series}} * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Geometric Series|next = Definition:Geometry|entry = geometric series}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Geometric Series|next = Definition:Geometry|entry = geometric series}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Geometry|entry = geometric series}} Category:Sum of Infinite Geometric Sequence ptukb8u6dds04fc9iy9h57wenboim4t


Perpendicular in Right-Angled Triangle makes two Similar Triangles/Porism 0 34269 286669 196141 2017-03-02T06:58:12Z Prime.mover 59 wikitext text/x-wiki == Porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles == <onlyinclude>{{:Euclid:Proposition/VI/8/Porism}}</onlyinclude> == Proof == Follows directly from Perpendicular in Right-Angled Triangle makes two Similar Triangles. {{qed}} {{Euclid Note|8|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Perpendicular in Right-Angled Triangle makes two Similar Triangles|next = Construction of Part of Line}}: Book $\text{VI}$. Propositions Category:Right Triangles r54mxsgaiufhrpyzuf5f7djwbbpd39l


Primitive of x over Root of a x squared plus b x plus c 0 34340 189802 2014-07-06T05:51:32Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $a \in \R_{\ne 0}$.  Then: :$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $a \in \R_{\ne 0}$. Then: :$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ </onlyinclude> == Proof == First: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | r = \frac 1 {2 a} \int \frac {2 a x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} } - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }        | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2 a} \int \frac {\mathrm d z} {\sqrt z} - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }        | c = Integration by Substitution }} {{eqn | r = \frac 1 {2 a} 2 \sqrt z - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = Primitive of Power }} {{eqn | r = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = substituting for $z$ }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Reciprocal of Root of a x squared plus b x plus c|next=Primitive of x squared over Root of a x squared plus b x plus c}}: $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.281$ Category:Primitives involving Root of a x squared plus b x plus c tld44ihdhusqw66zxp0tyyqvk373rwz


Primitive of Sine Function/Corollary

Primitive of Sine Function/Corollary 0 34448 485084 484694 2020-09-03T20:32:38Z Prime.mover 59 wikitext text/x-wiki == Corollary to Primitive of Sine Function == <onlyinclude> :$\displaystyle \int \sin a x \rd x = - \frac {\cos a x} a + C$ </onlyinclude> where $a$ is a non-zero constant. == Proof == {{begin-eqn}} {{eqn | l = \int \sin x \rd x       | r = -\cos x + C       | c = Primitive of $\sin x$ }} {{eqn | ll= \leadsto       | l = \int \sin a x \rd x       | r = \frac 1 a \paren {-\cos a x} + C       | c = Primitive of Function of Constant Multiple }} {{eqn | r = -\frac {\cos a x} a + C       | c = simplifying }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\cos a x$ * Primitive of $\tan a x$ * Primitive of $\cot a x$ * Primitive of $\sec a x$ * Primitive of $\csc a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of Power of x over Odd Power of x minus Odd Power of a|next = Primitive of x by Sine of a x}}: $\S 14$: Integrals involving $\sin a x$: $14.339$ * {{BookReference|Calculus|1983|K.G. Binmore|prev = Primitive of Cosine of a x|next = Primitive of Reciprocal of x squared plus a squared/Arctangent Form}}: $9$ Sums and Integrals: $9.8$ Standard Integrals Category:Primitives involving Sine Function t41fjeldr3kefxenh7brlgw05q2l91m


Primitive of Cosine Function/Corollary

Primitive of Cosine Function/Corollary 0 34467 484693 484690 2020-09-02T06:31:44Z Prime.mover 59 wikitext text/x-wiki == Corollary to Primitive of Cosine Function == <onlyinclude> :$\displaystyle \int \cos a x \rd x = \frac {\sin a x} a + C$ </onlyinclude> where $a$ is a non-zero constant. == Proof == {{begin-eqn}} {{eqn | l = \int \cos x \rd x       | r = \sin x + C       | c = Primitive of $\cos x$ }} {{eqn | ll= \leadsto       | l = \int \cos a x \rd x       | r = \frac 1 a \paren {\sin a x} + C       | c = Primitive of Function of Constant Multiple }} {{eqn | r = \frac {\sin a x} a + C       | c = simplifying }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\sin a x$ * Primitive of $\tan a x$ * Primitive of $\cot a x$ * Primitive of $\sec a x$ * Primitive of $\csc a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of x over Power of Sine of a x|next = Primitive of x by Cosine of a x}}: $\S 14$: Integrals involving $\cos a x$: $14.369$ * {{BookReference|Calculus|1983|K.G. Binmore|prev = Primitive of Exponential of a x|next = Primitive of Sine of a x}}: $9$ Sums and Integrals: $9.8$ Standard Integrals Category:Primitives involving Cosine Function huni9yvmkvxlhknwmr1qvrlgdn37axi


Weierstrass Substitution

Weierstrass Substitution 0 34494 484485 419749 2020-09-01T06:33:47Z Prime.mover 59 wikitext text/x-wiki {{POTW Candidate}} == Proof Technique == The '''Weierstrass Substitution''' is an application of Integration by Substitution. The substitution is: :$u \leftrightarrow \tan \dfrac \theta 2$ for $-\pi < \theta < \pi$, $u \in \R$. It yields: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \frac {2 u} {1 + u^2} }} {{eqn | l = \cos \theta       | r = \frac {1 - u^2} {1 + u^2} }} {{eqn | l = \frac {\d \theta} {\d u}       | r = \frac 2 {1 + u^2} }} {{end-eqn}} This can be stated: :$\displaystyle \int \map F {\sin \theta, \cos \theta} \rd \theta = 2 \int \map F {\frac {2 u} {1 + u^2}, \frac {1 - u^2} {1 + u^2} } \frac {d u} {1 + u^2}$ where $u = \tan \dfrac \theta 2$. == Proof == Let $u = \tan \dfrac \theta 2$ for $-\pi < \theta < \pi$. From Shape of Tangent Function, this substitution is valid for all real $u$. Then: {{begin-eqn}} {{eqn | l = u       | r = \tan \dfrac \theta 2       | c =  }} {{eqn | ll= \leadsto       | l = \theta       | r = 2 \tan^{-1} u       | c = Definition of Inverse Tangent }} {{eqn | ll= \leadsto       | l = \dfrac {\d \theta} {\d u}       | r = \dfrac 2 {1 + u^2}       | c = Derivative of Arctangent Function and Derivative of Constant Multiple }} {{eqn | l = \sin \theta       | r = \dfrac {2 u} {1 + u^2}       | c = Tangent Half-Angle Substitution for Sine }} {{eqn | l = \cos \theta       | r = \dfrac {1 - u^2} {1 + u^2}       | c = Tangent Half-Angle Substitution for Cosine }} {{end-eqn}} The result follows from Integration by Substitution. {{qed}} == Also known as == This technique is also known as '''tangent half-angle subsitution'''. == Also see == * Hyperbolic Tangent Half-Angle Substitution {{Namedfor|Karl Theodor Wilhelm Weierstrass|cat = Weierstrass}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of Function of Arcsine|next = Primitive of Reciprocal of a x + b}}: $\S 14$: Important Transformations: $14.58$ * {{Planetmath|url = weierstrasssubstitutionformulas|title = Weierstrass substitution formulas}} * {{MathWorld|Weierstrass Substitution|WeierstrassSubstitution}} Category:Proof Techniques Category:Integral Substitutions Category:Primitives involving Sine Function Category:Primitives involving Cosine Function h31fkbztzhk6dycqjlbbqutnziypfx6


Primitive of x by Root of a x squared plus b x plus c/Lemma 0 34503 190359 2014-07-19T10:53:22Z Prime.mover 59 Created page with "== Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ == <onlyinclude> Let $a \in \R_{\ne 0}$.  Then: :$\displaystyle..." wikitext text/x-wiki == Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ == <onlyinclude> Let $a \in \R_{\ne 0}$. Then: :$\displaystyle \int x \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | o =        | r = \int x \sqrt {a x^2 + b x + c} \ \mathrm d x }} {{eqn | r = \int \frac {2 a x \sqrt {a x^2 + b x + c} \ \mathrm d x} {2 a}       | c = multiplying top and bottom by $2 a$ }} {{eqn | r = \int \frac {\left({2 a x + b - b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x} {2 a} }} {{eqn | n = 1       | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = Linear Combination of Integrals }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x       | r = \int \sqrt z \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {2 \left({\sqrt z}\right)^3} 3       | c = Primitive of Power }} {{eqn | n = 2       | r = \frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3       | c = substituting for $z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | o =        | r = \int x \sqrt {a x^2 + b x + c} \ \mathrm d x }} {{eqn | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = from $(1)$ }} {{eqn | r = \frac 1 {2 a} \left({\frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3}\right) - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = from $(2)$ }} {{eqn | r = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving Root of a x squared plus b x plus c/Lemmata 4u1cbx5qhf0q9wmlflbfd0jq9ms9cch


Primitive of Reciprocal of x cubed plus a cubed/Lemma 0 34512 481683 295964 2020-08-14T22:14:42Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 + a^3$ == <onlyinclude> :$\displaystyle \int \frac {\d x} {x^2 - a x + a^2} = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }$ </onlyinclude> == Proof == The discriminant of $x^2 - a x + a^2$ is: {{begin-eqn}} {{eqn | l = \map {\mathrm {Disc} } {x^2 - a x + a^2}       | r = \paren {-a}^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = a^2 - 4 a^2       | c =  }} {{eqn | r = - 3 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a x + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \paren {-a}^2} } \map \arctan {\frac {2 x - a} {\sqrt {4 a^2 - \paren {-a}^2} } }       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving x cubed plus a cubed/Lemmata 8jvvghwfpa4szi3v1kui5n4qqhao8o7


Primitive of Reciprocal of x cubed plus a cubed/Partial Fraction Expansion 0 34514 481681 190410 2020-08-14T22:05:46Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^3 + a^3$ == :$\dfrac 1 {x^3 + a^3} = \dfrac 1 {3 a^2 \paren {x + a} } - \dfrac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^3 + a^3}       | o = \equiv       | r = \frac 1 {\paren {x + a} \paren {x^2 - a x + a^2} }       | c = Sum of Two Cubes }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac {B x + C} {x^2 - a x + a^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {x^2 - a x + a^2} + \paren {B x + C} \paren {x + a}       | c = multiplying through by $x^3 + a^3$ }} {{eqn | n = 1       | o = \equiv       | r = A x^2 - A a x + A a^2 + B x^2 + B a x + C x + C a       | c = multiplying out }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = A + B       | r = 0       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = -A       | r = B       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a + B a + C       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -A a + -A a + C       | r = 0       | c = from $(2)$ }} {{eqn | n = 3       | ll= \leadsto       | l = 2 A a       | r = C }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2 + C a       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A a^2 + \paren {2 A a} a       | r = 1       | c = from $(3)$ }} {{eqn | ll= \leadsto       | l = A a^2 + \paren {2 A a} a       | r = \frac 1 {3 a^2}       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = 1 - \frac a {3 a^2}       | c = from $(3)$ }} {{eqn | r = \frac {2 a} {3 a^2}       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac {-1} {3 a^2}       | c = from $(2)$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {3 a^2} }} {{eqn | l = B       | r = -\frac 1 {3 a^2} }} {{eqn | l = C       | r = \frac {2 a} {3 a^2} }} {{end-eqn}} Hence the result. {{qed}} Category:Primitives involving x cubed plus a cubed/Lemmata 9lbkep0ytd8y4bd8lal1r9n6lm2o0ie


Primitive of Reciprocal of x fourth plus a fourth/Partial Fraction Expansion 0 34521 407615 220600 2019-06-12T20:44:32Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\dfrac 1 {x^4 + a^4} = \dfrac {x + a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)} - \dfrac {x - a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^4 + a^4}       | o = \equiv       | r = \frac 1 {\left({x^2 + a x \sqrt 2 + a^2}\right) \left({x^2 - a x \sqrt 2 + a^2}\right)}       | c = Sum of Two Fourth Powers }} {{eqn | o = \equiv       | r = \frac {A x + B} {x^2 + a x \sqrt 2 + a^2} + \frac {C x + D} {x^2 - a x \sqrt 2 + a^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = \left({A x + B}\right) \left({x^2 - a x \sqrt 2 + a^2}\right) + \left({C x + D}\right) \left({x^2 + a x \sqrt 2 + a^2}\right)       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^3 + B x^2 - A a x^2 \sqrt 2 - B a x \sqrt 2 + A x a^2 + B a^2       | c =  }} {{eqn | o =        | ro= +       | r = C x^3 + D x^2 + C a x^2 \sqrt 2 + D a x \sqrt 2 + C x a^2 + D a^2       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = A + C       | r = 0       | c =  }} {{eqn | n = 2       | ll= \implies       | l = -A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a \sqrt 2 + B + C a \sqrt 2 + D       | r = 0       | c =  }} {{eqn | n = 3       | ll= \implies       | l = 2 C a \sqrt 2 + B + D       | r = 0       | c = substituting for $A$ from $(2)$ }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = -B a \sqrt 2 + A a^2 + D a \sqrt 2 + C a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -B a \sqrt 2 - C a^2 + D a \sqrt 2 + C a^2       | r = 0       | c = substituting for $A$ from $(2)$ }} {{eqn | ll= \implies       | l = -B + D       | r = 0 }} {{eqn | n = 4       | ll= \implies       | l = B       | r = D }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2 + D a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = 2 D a^2       | r = 1       | c = substituting for $B$ from $(4)$ }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {2 a^2}       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {2 a^2}       | c = from $(4)$ }} {{eqn | ll= \implies       | l = 2 C a \sqrt 2 + \frac 1 {a^2}       | r = 0       | c = substituting for $B$ and $D$ in $(3)$ }} {{eqn | ll= \implies       | l = C       | r = \frac {-1} {2 a^3 \sqrt 2}       | c = substituting for $B$ and $D$ in $(3)$ }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {2 a^3 \sqrt 2}       | c = from $(2)$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a^3 \sqrt 2} }} {{eqn | l = B       | r = \frac 1 {2 a^2} }} {{eqn | l = C       | r = \frac {-1} {2 a^3 \sqrt 2} }} {{eqn | l = D       | r = \frac 1 {2 a^2} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x^4 + a^4}       | r = \frac {\frac 1 {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 + a x \sqrt 2 + a^2} + \frac {\frac {-1} {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 - a x \sqrt 2 + a^2} }} {{eqn | r = \frac {x + a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)} - \frac {x - a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)} }} {{end-eqn}} {{qed}} Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata j8cyzt8lqhoaai6e99d6zbcf1e2lln2


Primitive of Reciprocal of x fourth plus a fourth/Lemma 1 0 34524 436976 436975 2019-11-27T16:02:34Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\displaystyle \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2} = \frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a}$ == Proof == The discriminant of $x^2 + a x \sqrt 2 + a^2$ is: {{begin-eqn}} {{eqn | l = \map {\operatorname {Disc} } {x^2 + a x \sqrt 2 + a^2}       | r = \paren {a \sqrt 2}^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = 2 a^2 - 4 a^2       | c =  }} {{eqn | r = - 2 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \paren {a \sqrt 2}^2} } \map \arctan {\frac {2 x + a \sqrt 2} {\sqrt {4 a^2 - \paren {a \sqrt 2}^2} } }       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 2} \, \map \arctan {\frac {2 x + a \sqrt 2} {a \sqrt 2} }       | c = simplifying }} {{eqn | r = \frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata me6as1uajdq23e9etlwa77ztuuqvnce


Primitive of Reciprocal of x fourth plus a fourth/Lemma 2 0 34525 295966 190446 2017-05-02T21:35:43Z Barto 3079 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\displaystyle \int \frac {\mathrm d x} {x^2 - a x \sqrt 2 + a^2} = \frac {-\sqrt 2} a \arctan \left({1 - \frac {x \sqrt 2} a}\right)$ == Proof == The discriminant of $x^2 - a x \sqrt 2 + a^2$ is: {{begin-eqn}} {{eqn | l = \operatorname{Disc} \left({x^2 - a x \sqrt 2 + a^2}\right)       | r = \left({-a \sqrt 2}\right)^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = 2 a^2 - 4 a^2       | c =  }} {{eqn | r = - 2 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 - a x \sqrt 2 + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \left({-a \sqrt 2}\right)^2} } \arctan \left({\frac {2 x - a \sqrt 2} {\sqrt {4 a^2 - \left({-a \sqrt 2}\right)^2} } }\right)       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 2} \arctan \left({\frac {2 x - a \sqrt 2} {a \sqrt 2} }\right)       | c = simplifying }} {{eqn | r = \frac {\sqrt 2} a \arctan \left({\frac {x \sqrt 2} a - 1}\right)       | c = simplifying }} {{eqn | r = \frac {-\sqrt 2} a \arctan \left({1 - \frac {x \sqrt 2} a}\right)       | c = Arctangent is Odd Function }} {{end-eqn}} {{qed}} Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata 81h5r7qmelpci09qmsm9pr12wm50mp2


Primitive of x squared over x fourth plus a fourth/Partial Fraction Expansion 0 34526 190454 2014-07-22T20:43:09Z Prime.mover 59 Created page with "== Lemma for Primitive of $\dfrac {x^2} {x^4 + a^4}$ ==  :$\dfrac {x^2} {x^4 + a^4} = \dfrac x {2 a \sqrt 2 \left({x^2 -..." wikitext text/x-wiki == Lemma for Primitive of $\dfrac {x^2} {x^4 + a^4}$ == :$\dfrac {x^2} {x^4 + a^4} = \dfrac x {2 a \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)} - \dfrac x {2 a \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac {x^2} {x^4 + a^4}       | o = \equiv       | r = \frac {x^2} {\left({x^2 + a x \sqrt 2 + a^2}\right) \left({x^2 - a x \sqrt 2 + a^2}\right)}       | c = Sum of Two Fourth Powers }} {{eqn | o = \equiv       | r = \frac {A x + B} {x^2 + a x \sqrt 2 + a^2} + \frac {C x + D} {x^2 - a x \sqrt 2 + a^2}       | c =  }} {{eqn | ll= \implies       | l = x^2       | o = \equiv       | r = \left({A x + B}\right) \left({x^2 - a x \sqrt 2 + a^2}\right) + \left({C x + D}\right) \left({x^2 + a x \sqrt 2 + a^2}\right)       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^3 + B x^2 - A a x^2 \sqrt 2 - B a x \sqrt 2 + A x a^2 + B a^2       | c =  }} {{eqn | o =        | ro= +       | r = C x^3 + D x^2 + C a x^2 \sqrt 2 + D a x \sqrt 2 + C x a^2 + D a^2       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = A + C       | r = 0       | c =  }} {{eqn | n = 2       | ll= \implies       | l = -A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a \sqrt 2 + B + C a \sqrt 2 + D       | r = 1       | c =  }} {{eqn | n = 3       | ll= \implies       | l = 2 C a \sqrt 2 + B + D       | r = 1       | c = substituting for $A$ from $(2)$ }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = -B a \sqrt 2 + A a^2 + D a \sqrt 2 + C a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -B a \sqrt 2 - C a^2 + D a \sqrt 2 + C a^2       | r = 0       | c = substituting for $A$ from $(2)$ }} {{eqn | ll= \implies       | l = -B + D       | r = 0 }} {{eqn | n = 4       | ll= \implies       | l = B       | r = D }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2 + D a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = 2 D a^2       | r = 0       | c = substituting for $B$ from $(4)$ }} {{eqn | ll= \implies       | l = D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = B       | r = 0       | c = from $(4)$ }} {{eqn | ll= \implies       | l = 2 C a \sqrt 2       | r = 1       | c = substituting for $B$ and $D$ in $(3)$ }} {{eqn | ll= \implies       | l = C       | r = \frac 1 {2 a \sqrt 2}       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-1} {2 a \sqrt 2}       | c = from $(2)$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {2 a \sqrt 2} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac 1 {2 a \sqrt 2} }} {{eqn | l = D       | r = 0 }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x^4 + a^4}       | r = \frac {\frac {-1} {2 a \sqrt 2} x} {x^2 + a x \sqrt 2 + a^2} + \frac {\frac 1 {2 a \sqrt 2} x} {x^2 - a x \sqrt 2 + a^2} }} {{eqn | r = \frac x {2 a \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)} - \frac x {2 a \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)} }} {{end-eqn}} {{qed}} Category:Primitives involving x to the fourth plus or minus a to the fourth/Lemmata r1y6egprou63bvbk2vptndw85mpst2f


Primitive of Reciprocal of p plus q by Sine of a x/Weierstrass Substitution 0 34543 191926 191924 2014-08-22T21:43:43Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $p + q \sin a x$ == The Weierstrass Substitution of $\displaystyle \int \frac {\mathrm d x} {p + q \sin a x}$ is: :$\displaystyle \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u + p}$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {p + q \sin a x}       | r = \frac 1 a \int \frac {\mathrm d z} {p + q \sin z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {p + q \frac {2 u} {u^2 + 1} } \frac {2 \ \mathrm d u} {u^2 + 1}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \ \mathrm d u} {\left({u^2 + 1}\right) \frac {p \left({u^2 + 1}\right) + 2 q u} {u^2 + 1} }       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u + p}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function Category:Weierstrass Substitutions 217lo91fbifomc7bs74shv7co8s8a71


Primitive of Reciprocal of square of p plus q by Sine of a x/Weierstrass Substitution 0 34545 454008 192065 2020-03-12T08:10:10Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $\paren {p + q \sin a x}^2$ == The Weierstrass Substitution of $\displaystyle \int \frac {\d x} {\paren {p + q \sin a x}^2}$ is: :$\displaystyle \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p + q \sin a x}^2}       | r = \frac 1 a \int \frac {\d z} {\paren {p + q \sin z}^2}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {\paren {p + q \frac {2 u} {u^2 + 1} }^2} \frac {2 \rd u} {u^2 + 1}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \rd u} {\paren {u^2 + 1} \paren {\frac {p \paren {u^2 + 1} + 2 q u} {u^2 + 1} }^2}       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function Category:Weierstrass Substitutions 25g9i93frf69kzc9w7ufgd6qeccuw85


Primitive of Exponential of a x by Sine of b x/Lemma 0 34772 405304 191960 2019-05-19T08:26:14Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \sin b x$ == <onlyinclude> :$\displaystyle \int e^{a x} \sin b x \rd x = \frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x$ </onlyinclude> == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \sin b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = b \cos b x       | c = Derivative of $\sin a x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\ d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of $e^{a x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int e^{a x} \sin b x \rd x       | r = \sin b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \paren {b \cos b x} \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}} Category:Primitive of Exponential of a x by Sine of b x 6osx4jjp5kxtrrj01u72mbjop8nfqsi


Primitive of Exponential of a x by Cosine of b x/Lemma 0 34773 405305 191965 2019-05-19T08:27:37Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \cos b x$ == <onlyinclude> :$\displaystyle \int e^{a x} \cos b x \rd x = \frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x$ </onlyinclude> == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v}{\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \cos b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = -b \sin b x       | c = Derivative of $\cos a x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of $e^{a x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int e^{a x} \cos b x \rd x       | r = \cos b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \paren {-b \sin b x} \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}} Category:Primitive of Exponential of a x by Cosine of b x j0ddqk8949wn098gpmtnh5cdyjfilus


Primitive of Exponential of a x by Power of Sine of b x/Lemma 1 0 34784 447531 289822 2020-02-08T19:50:21Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \sin^n b x \cos b x$ == :$\displaystyle \int e^{a x} \sin^{n - 1} b x \cos b x \rd x = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + n b^2} + \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \sin^n b x \rd x - \int e^{a x} \sin^{n - 2} b x \rd x} + C$ == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \sin^{n - 1} b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \paren {n - 1} b \sin^{n - 2} b x \cos b x       | c = Derivative of $\sin a x$, Derivative of Power, Chain Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x} \cos b x       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2}       | c = Primitive of $e^{a x} \cos b x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | r = \int e^{a x} \sin^{n - 1} b x \cos b x \rd x       | o =        | c =  }} {{eqn | r = \int \paren {\sin^{n - 1} b x} \paren {e^{a x} \cos b x} \rd x       | c =  }} {{eqn | r = \paren {\sin^{n - 1} b x} \paren {\frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} }       | c = Integration by Parts }} {{eqn | o =        | ro= -       | r = \int \paren {\frac {e^{a x} \paren {a \cos b x + b \sin b x} } {a^2 + b^2} } \paren {\paren {n - 1} b \sin^{n - 2} b x \cos b x} \rd x + C       | c =  }} {{eqn | r = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} a b} {a^2 + b^2} \int e^{a x} \sin^{n - 2} b x \cos^2 b x \rd x + C       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} b^2} {a^2 + b^2} \int e^{a x} \sin^{n - 1} b x \cos b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \paren {1 + \frac {\paren {n - 1} b^2} {a^2 + b^2} } \int e^{a x} \sin^{n - 1} b x \cos b x \rd x       | o =        | c = gathering terms }} {{eqn | r = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} a b} {a^2 + b^2} \int e^{a x} \sin^{n - 2} b x \cos^2 b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \paren {a^2 + n b^2} \int e^{a x} \sin^{n - 1} b x \cos b x \rd x       | o =        | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}       | c = simplifying }} {{eqn | o =        | ro= -       | r = \paren {n - 1} a b \int e^{a x} \sin^{n - 2} b x \cos^2 b x \rd x + C       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}       | c =  }} {{eqn | o =        | ro= -       | r = \paren {n - 1} a b \int e^{a x} \sin^{n - 2} b x \paren {1 - \sin^2 b x} \rd x + C       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x}       | c = simplifying }} {{eqn | o =        | ro= -       | r = \paren {n - 1} a b \int e^{a x} \sin^{n - 2} b x \rd x       | c =  }} {{eqn | o =        | ro= +       | r = \paren {n - 1} a b \int e^{a x} \sin^n b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \int e^{a x} \sin^{n - 1} b x \cos b x \rd x       | o =        | c =  }} {{eqn | r = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + n b^2}       | c = simplifying }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} a b} {a^2 + n b^2} \int e^{a x} \sin^{n - 2} b x \rd x       | c =  }} {{eqn | o =        | ro= +       | r = \frac {\paren {n - 1} a b} {a^2 + n b^2} \int e^{a x} \sin^n b x \rd x + C       | c =  }} {{end-eqn}} and so rearranging: :$\displaystyle \int e^{a x} \sin^{n - 1} b x \cos b x \rd x = \frac {e^{a x} \sin^{n - 1} b x \paren {a \cos b x + b \sin b x} } {a^2 + n b^2} + \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \sin^n b x \rd x - \int e^{a x} \sin^{n - 2} b x \rd x} + C$ {{qed}} Category:Primitive of Exponential of a x by Power of Sine of b x t1hfy192myllxo19rvpgybts5swykcq


Primitive of Exponential of a x by Power of Sine of b x/Lemma 2 0 34788 191953 191154 2014-08-23T12:28:51Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \sin^n b x$ == :$\dfrac {a^2 + n b^2} a e^{a x} \sin^n b x - \dfrac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right) = e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)$ == Proof == {{begin-eqn}} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)       | c =  }} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \sin^{n - 1} b x \sin b x - \frac {n b} a e^{a x} \sin^{n - 1} b x a \cos b x - \frac {n b} a e^{a x} \sin^{n - 1} b x b \sin b x       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({\frac {a^2 + n b^2} a \sin b x - \frac {n b} a a \cos b x - \frac {n b} a b \sin b x}\right)       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({a \sin b x + \frac {n b^2} a \sin b x - n b \cos b x - \frac {n b^2} a \sin b x}\right)       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)       | c =  }} {{end-eqn}} {{qed}} Category:Primitive of Exponential of a x by Power of Sine of b x j3w4z35t9a8n9n75vht7p4s0dj5af6j


Primitive of Exponential of a x by Power of Cosine of b x/Lemma 1 0 34789 447530 289821 2020-02-08T19:43:36Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \cos b x$ == :$\displaystyle \int e^{a x} \cos^{n - 1} b x \sin b x \rd x = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + n b^2} - \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \cos^n b x \rd x - \int e^{a x} \cos^{n - 2} b x \rd x} + C$ == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \cos^{n - 1} b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = -\paren {n - 1} b \cos^{n - 2} b x \sin b x       | c = Derivative of $\cos a x$, Derivative of Power, Chain Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x} \sin b x       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2}       | c = Primitive of $e^{a x} \sin b x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | r = \int e^{a x} \cos^{n - 1} b x \sin b x \rd x       | o =        | c =  }} {{eqn | r = \int \paren {\cos^{n - 1} b x} \paren {e^{a x} \sin b x} \rd x       | c =  }} {{eqn | r = \paren {\cos^{n - 1} b x} \paren {\frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2} }       | c = Integration by Parts }} {{eqn | o =        | ro= -       | r = \int \paren {\frac {e^{a x} \paren {a \sin b x - b \cos b x} } {a^2 + b^2} } \paren {-\paren {n - 1} b \cos^{n - 2} b x \sin b x} \rd x + C       | c =  }} {{eqn | r = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {\paren {n - 1} a b} {a^2 + b^2} \int e^{a x} \cos^{n - 2} b x \sin^2 b x \rd x + C       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} b^2} {a^2 + b^2} \int e^{a x} \cos^{n - 1} b x \sin b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \paren {1 + \frac {\paren {n - 1} b^2} {a^2 + b^2} } \int e^{a x} \cos^{n - 1} b x \sin b x \rd x       | o =        | c = gathering terms }} {{eqn | r = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {\paren {n - 1} a b} {a^2 + b^2} \int e^{a x} \cos^{n - 2} b x \sin^2 b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \paren {a^2 + n b^2} \int e^{a x} \cos^{n - 1} b x \sin b x \rd x       | o =        | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x}       | c = simplifying }} {{eqn | o =        | ro= +       | r = \paren {n - 1} a b \int e^{a x} \cos^{n - 2} b x \sin^2 b x \rd x + C       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x}       | c =  }} {{eqn | o =        | ro= +       | r = \paren {n - 1} a b \int e^{a x} \cos^{n - 2} b x \paren {1 - \cos^2 b x} \rd x + C       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x}       | c = simplifying }} {{eqn | o =        | ro= +       | r = \paren {n - 1} a b \int e^{a x} \cos^{n - 2} b x \rd x       | c =  }} {{eqn | o =        | ro= -       | r = \paren {n - 1} a b \int e^{a x} \cos^n b x \rd x + C       | c =  }} {{eqn | ll= \leadsto       | r = \int e^{a x} \cos^{n - 1} b x \sin b x \rd x       | o =        | c =  }} {{eqn | r = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + n b^2}       | c = simplifying }} {{eqn | o =        | ro= +       | r = \frac {\paren {n - 1} a b} {a^2 + n b^2} \int e^{a x} \cos^{n - 2} b x \rd x       | c =  }} {{eqn | o =        | ro= -       | r = \frac {\paren {n - 1} a b} {a^2 + n b^2} \int e^{a x} \cos^n b x \rd x + C       | c =  }} {{end-eqn}} and so rearranging: :$\displaystyle \int e^{a x} \cos^{n - 1} b x \sin b x \rd x = \frac {e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} } {a^2 + n b^2} - \frac {\paren {n - 1} a b} {a^2 + n b^2} \paren {\int e^{a x} \cos^n b x \rd x - \int e^{a x} \cos^{n - 2} b x \rd x} + C$ {{qed}} Category:Primitive of Exponential of a x by Power of Cosine of b x a68b4cw9jmb8vmt4983mb5qbxkyqqnh


Primitive of Exponential of a x by Power of Cosine of b x/Lemma 2 0 34790 438886 191959 2019-12-10T17:12:58Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of $e^{a x} \cos^n b x$ == :$\dfrac {a^2 + n b^2} a e^{a x} \cos^n b x + \dfrac {n b} a e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + n b \sin b x}$ == Proof == {{begin-eqn}} {{eqn | r = \dfrac {a^2 + n b^2} a e^{a x} \cos^n b x + \dfrac {n b} a e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x}       | c =  }} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \cos^{n - 1} b x \cos b x + \frac {n b} a e^{a x} \cos^{n - 1} b x a \sin b x - \frac {n b} a e^{a x} \cos^{n - 1} b x b \cos b x       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {\frac {a^2 + n b^2} a \cos b x + \frac {n b} a a \sin b x - \frac {n b} a b \cos b x}       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + \frac {n b^2} a \cos b x + n b \sin b x - \frac {n b^2} a \cos b x}       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + n b \sin b x}       | c =  }} {{end-eqn}} {{qed}} Category:Primitive of Exponential of a x by Power of Cosine of b x dn1an85z8kzr7d5l6xopz30ave4yqhd


Primitive of Power of Hyperbolic Tangent of a x 0 34887 191348 2014-08-12T07:05:34Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \int \tanh^n a x \ \mathrm d x = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C$ </..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \int \tanh^n a x \ \mathrm d x = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \int \tanh^n a x \ \mathrm d x       | r = \int \tanh^{n - 2} a x \tanh^2 a x \ \mathrm d x       | c =  }} {{eqn | r = \int \tanh^{n - 2} a x \left({1 - \operatorname{sech}^2 a x}\right) \ \mathrm d x       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | r = -\int \tanh^{n - 2} a x \operatorname{sech}^2 a x \ \mathrm d x + \int \tanh^{n - 2} \ \mathrm d x       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C       | c = Primitive of $\tanh^n a x \operatorname{sech}^2 a x$ }} {{end-eqn}} {{qed}} == Also see == * Primitive of $\sinh^n a x$ * Primitive of $\cosh^n a x$ * Primitive of $\coth^n a x$ * Primitive of $\operatorname{sech}^n a x$ * Primitive of $\operatorname{csch}^n a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Reciprocal of p plus q by Hyperbolic Tangent of a x|next=Primitive of Hyperbolic Cotangent of a x}}: $\S 14$: Integrals involving $\tanh a x$: $14.614$ Category:Primitives involving Hyperbolic Tangent Function cv5oudgmta7w63zco5lo5gycx68zy61


Derivative of Tangent Function/Corollary 0 34889 492829 485862 2020-10-06T06:37:48Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Tangent Function == <onlyinclude> :$\map {\dfrac \d {\d x} } {\tan a x} = a \sec^2 a x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\tan x}       | r = \sec^2 x       | c = Derivative of $\tan x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\tan a x}       | r = a \sec^2 a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}} {{qed}} == Also see == * Derivative of $\sin a x$ * Derivative of $\cos a x$ * Derivative of $\cot a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$ Category:Derivatives of Trigonometric Functions Category:Tangent Function qknl3w7pk5kotj74ifd5vefncx0gpv4


Derivative of Cotangent Function/Corollary 0 34891 485863 417400 2020-09-06T19:35:35Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Cotangent Function == <onlyinclude> :$\map {\dfrac \d {\d x} } {\cot a x} = -a \csc^2 a x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\cot x}       | r = -\csc^2 x       | c = Derivative of $\cot x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\cot a x}       | r = -a \csc^2 a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}} {{qed}} == Also see == * Derivative of $\sin a x$ * Derivative of $\cos a x$ * Derivative of $\tan a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$ Category:Derivatives of Trigonometric Functions Category:Cotangent Function j58vp1ogd2cewdu3mxdgtpjdh36ayc7


Hyperbolic Tangent Half-Angle Substitution for Cosine 0 35045 494645 494644 2020-10-14T06:21:02Z Prime.mover 59 wikitext text/x-wiki == Corollary to Double Angle Formula for Hyperbolic Cosine == <onlyinclude> :$\cosh 2 x = \dfrac {1 + \tanh^2 x}{1 - \tanh^2 x}$ </onlyinclude> where $\cosh$ and $\tanh$ denote hyperbolic cosine and hyperbolic tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \cosh 2 x       | r = \cosh^2 x + \sinh^2 x       | c = Double Angle Formula for Hyperbolic Cosine }} {{eqn | r = \paren {\cosh^2 x + \sinh^2 x} \frac {\cosh^2 x} {\cosh^2 x}       | c =  }} {{eqn | r = \paren {1 + \tanh^2 x} \cosh^2 x       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {1 + \tanh^2 x} {\sech^2 x}       | c = {{Defof|Hyperbolic Secant|index = 2}} }} {{eqn | r = \frac {1 + \tanh^2 x} {1 - \tanh^2 x}       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{end-eqn}} {{qed}} Category:Hyperbolic Cosine Function Category:Hyperbolic Tangent Half-Angle Substitutions cypg11rly023i8qefiqt1phsmly7j2a


Hyperbolic Tangent Half-Angle Substitution for Sine 0 35046 494642 494641 2020-10-14T06:18:15Z Prime.mover 59 wikitext text/x-wiki == Corollary to Double Angle Formula for Hyperbolic Sine == <onlyinclude> :$\sinh 2 x = \dfrac {2 \tanh x} {1 - \tanh^2 x}$ </onlyinclude> where $\sin$ and $\tan$ denote hyperbolic sine and hyperbolic tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \sinh 2 x       | r = 2 \sinh x \cosh x       | c = Double Angle Formula for Hyperbolic Sine }} {{eqn | r = 2 \sinh x \cosh x \frac {\cosh x} {\cosh x}       | c =  }} {{eqn | r = 2 \tanh x \cosh^2 x       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {2 \tanh x} {\operatorname{sech}^2 x}       | c = {{Defof|Hyperbolic Secant|index = 2}} }} {{eqn | r = \frac {2 \tanh x} {1 - \tanh^2 x}       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{end-eqn}} {{qed}} Category:Hyperbolic Sine Function Category:Hyperbolic Tangent Half-Angle Substitutions o62ilw2z556idlwxcm8zfgjl0zsc3qb


Square Modulo 3/Corollary 1

Square Modulo 3/Corollary 1 0 35054 439743 367192 2019-12-16T16:15:20Z Prime.mover 59 wikitext text/x-wiki == Corollary to Square Modulo 3 == <onlyinclude> Let $x, y \in \Z$ be integers. Then: :$3 \divides \paren {x^2 + y^2} \iff 3 \divides x \land 3 \divides y$ where $3 \divides x$ denotes that $3$ divides $x$. </onlyinclude> == Proof == === Sufficient Condition === Let $3 \divides x \land 3 \divides y$. Then by definition of divisibility: :$x \equiv 0 \pmod 3$ and :$y \equiv 0 \pmod 3$ From Square Modulo 3: :$x^2 \equiv 0 \pmod 3$ and :$y^2 \equiv 0 \pmod 3$ Then from Modulo Addition is Well-Defined: :$\paren {x^2 + y^2} \equiv 0 \pmod 3$ Thus: :$3 \divides \paren {x^2 + y^2}$ {{qed|lemma}} === Necessary Condition === Now suppose $3 \divides \paren {x^2 + y^2}$. Then by definition of divisibility: :$\paren {x^2 + y^2} \pmod 3$ From Square Modulo 3: : $x^2 \equiv 0 \pmod 3$ or $x^2 \equiv 1 \pmod 3$ and : $y^2 \equiv 0 \pmod 3$ or $y^2 \equiv 1 \pmod 3$ Thus the only way $\paren {x^2 + y^2} \equiv 0 \pmod 3$ is for: : $x^2 \equiv 0 \pmod 3$ and: : $y^2 \equiv 0 \pmod 3$ {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Square Modulo 3|next = Square Modulo 3/Corollary 2}}: Chapter $2$: Some Properties of $\Z$: Exercise $2.9$ Category:Square Modulo 3 1ypv1hvnzyj2k4taolsech0kh83dxlj


Primitive of Reciprocal of p plus q by Cosine of a x/Weierstrass Substitution 0 35065 435007 222169 2019-11-12T16:54:27Z Prime.mover 59 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $p + q \cos a x$ == The Weierstrass Substitution of $\displaystyle \int \frac {\d x} {p + q \cos a x}$ is: :$\displaystyle \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \cos a x}       | r = \frac 1 a \int \frac {\d z} {p + q \cos z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {p + q \paren {\frac {1 - u^2} {1 + u^2} } } \frac {2 \rd u} {1 + u^2}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \rd u} {\paren {1 + u^2} \paren {\frac {p \paren {1 + u^2} + q \paren {1 - u^2} } {1 + u^2} } }       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\d u} {\paren {p - q} u^2 + \paren {p + q} }       | c = simplifying }} {{eqn | r = \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }       | c = simplifying further }} {{end-eqn}} {{qed}} Category:Primitives involving Cosine Function Category:Weierstrass Substitutions a8btjt59amzygnx0tcakkfjvju57f9u


Primitive of Reciprocal of p plus q by Hyperbolic Sine of a x/Hyperbolic Tangent Half-Angle Substitution 0 35075 191968 2014-08-23T15:33:05Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $p + q \sinh a x$ ==  The Hyperbolic Tangent Half-Angle Substitutio..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $p + q \sinh a x$ == The Hyperbolic Tangent Half-Angle Substitution of $\displaystyle \int \frac {\mathrm d x} {p + q \sinh a x}$ is: :$\displaystyle \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u - p}$ where $u = \tanh \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {p + q \sinh a x}       | r = \frac 1 a \int \frac {\mathrm d z} {p + q \sinh z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac {\dfrac {2 \ \mathrm d u} {u^2 - 1} } {p + q \dfrac {2 u} {u^2 - 1} }       | c = Hyperbolic Tangent Half-Angle Substitution: $u = \tanh \dfrac z 2 = \tanh \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \ \mathrm d u} {p \left({u^2 - 1}\right) + 2 q u}       | c = multiplying top and bottom by $u^2 - 1$ }} {{eqn | r = \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u - p}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving Hyperbolic Sine Function Category:Hyperbolic Tangent Half-Angle Substitutions 2jhgnn99npdwlb6twjpyt584qgwpeso


Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x/Lemma 0 35117 257910 192217 2016-06-05T13:48:49Z Z423x5c6 2799 wikitext text/x-wiki == Lemma for Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x == :$\displaystyle \frac 1 2 \left({\arctan \dfrac {-p} q}\right) + \frac \pi 4 = \frac {\arctan \dfrac q p} 2$ == Proof == {{begin-eqn}} {{eqn | l = y       | r = \frac 1 2 \left({\arctan \dfrac {-p} q}\right) + \frac \pi 4       | c = making a definition }} {{eqn | ll= \implies       | l = 2 y - \frac \pi 2       | r = \arctan \dfrac {-p} q       | c = rearranging }} {{eqn | ll= \implies       | l = \tan \left({2 y - \frac \pi 2}\right)       | r = \dfrac {-p} q       | c = Definition of Arctangent }} {{eqn | ll= \implies       | l = \tan \left({2 y + \frac {3 \pi} 2}\right)       | r = \dfrac {-p} q       | c = Tangent Function is Periodic on Reals }} {{eqn | ll= \implies       | l = -\cot \left({2 y}\right)       | r = \dfrac {-p} q       | c = Tangent of Angle plus Three Right Angles }} {{eqn | ll= \implies       | l = \tan \left({2 y}\right)       | r = \dfrac q p       | c = Cotangent is Reciprocal of Tangent }} {{eqn | ll= \implies       | l = y       | r = \frac {\arctan \dfrac q p} 2       | c = Definition of Arctangent }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function and Cosine Function iax7us9hl47oqc56p5un76hncihps5n


Cartesian Product of Countable Sets is Countable/Corollary/Proof 2 0 35214 340928 331093 2018-01-31T22:28:40Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cartesian Product of Countable Sets is Countable == {{:Cartesian Product of Countable Sets is Countable/Corollary}} == Proof == <onlyinclude> Proof by induction: === Basis for the Induction === When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable. So shown for basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$ where $f_k$ is an injection. Now we need to show that for $n = k + 1$: :$\exists f_{k+1}: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$ where $f_{k+1}$ is an injection. === Induction Step === This is our induction step: By the induction hypothesis: :$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$ where $f_k$ is an injection. Thus by definition, $S_1 \times S_2 \times \cdots \times S_k$ is countable. By hypothesis $S_{k + 1}$ is countable. So by the basis for the induction: :$\exists g: \left({S_1 \times S_2 \times \cdots \times S_k}\right) \times S_{k+1} \to \N \times \N$ where $g$ is an injection. By Cartesian Product of Countable Sets is Countable, :$\exists r: \N \times \N \to \N$ where $r$ is an injection. Therefore, by Composite of Injections is Injection: :$f_{k+1} = r \circ g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$ is an injection. The result follows by induction. {{qed}} </onlyinclude> == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Cartesian Product of Countable Sets is Countable|next = }}: $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Theorem $7.6$ Category:Cartesian Product of Countable Sets is Countable fcwzd4m1hzib8tl1g1is5ax6aiuy7cb


Finding Center of Circle/Porism

Finding Center of Circle/Porism 0 35372 376817 196554 2018-11-15T20:19:14Z Caliburn 3218 wikitext text/x-wiki == Porism to Finding Center of Circle == <onlyinclude> {{:Euclid:Proposition/III/1/Porism}} </onlyinclude> == Proof == See Perpendicular Bisector of Chord Passes Through Center, where the proof has been extracted from the proof given by Euclid. == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Finding Center of Circle/Proof 1|next=Perpendicular Bisector of Chord Passes Through Center}}: Book $\text{III}$. Propositions Category:Finding Center of Circle 1j7b6jitih9ygbfsdl0qz4cynob7yev


Ratio of Areas of Similar Triangles/Porism 0 35434 475770 197161 2020-06-24T06:09:11Z Prime.mover 59 wikitext text/x-wiki == Porism to Ratio of Areas of Similar Triangles == <onlyinclude>{{:Euclid:Proposition/VI/19/Porism}}</onlyinclude> == Proof == Follows immediately from Ratio of Areas of Similar Triangles. {{qed}} {{Euclid Note|19|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Ratio of Areas of Similar Triangles|next=Similar Polygons are composed of Similar Triangles}}: Book $\text{VI}$. Propositions Category:Ratio of Areas of Similar Triangles f018i7q1se5o6mrq7n38ui9pdmd5epx


Euclidean Algorithm/Euclid's Proof/Porism 0 35502 267938 197317 2016-08-28T09:04:24Z Prime.mover 59 wikitext text/x-wiki == Porism to Euclidean Algorithm == <onlyinclude>{{:Euclid:Proposition/VII/2/Porism}}</onlyinclude> == Proof == An algebraic proof of this is given in Common Divisor Divides GCD. {{qed}} {{Euclid Note|2|VII}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Euclidean Algorithm/Euclid's Proof|next = Greatest Common Divisor of Three Numbers}}: Book $\text{VII}$. Propositions Category:Euclidean Algorithm pvrckxq7tdbvicpi0q1bef5u23jmjho


Axiom:Euclid's Common Notion 1

Axiom:Euclid's Common Notion 1 100 35775 430024 227425 2019-10-08T12:22:09Z Prime.mover 59 wikitext text/x-wiki == Common Notion == <onlyinclude> {{EuclidCommonNotionStatement|1|Things which are equal to the same thing are also equal to each other.}} </onlyinclude> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Fifth Postulate|next = Axiom:Euclid's Common Notion 2}}: Book $\text{I}$. Common Notions * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Axiom:Euclid's Postulates|next = Axiom:Euclid's Fourth Postulate}}: Chapter $2$: The Logic of Shape: Euclid Category:Euclid's Common Notions h0nhjl1y9uz9fo7c996us571zkruoef


Axiom:Euclid's Common Notion 4

Axiom:Euclid's Common Notion 4 100 35778 196479 196470 2014-10-12T11:02:04Z Prime.mover 59 wikitext text/x-wiki == Common Notion == <onlyinclude> {{EuclidCommonNotionStatement|4|Things which coincide with one another are equal to one another.}} </onlyinclude> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Axiom:Euclid's Common Notion 3|next=Axiom:Euclid's Common Notion 5}}: Book $\text{I}$. Common Notions Category:Euclid's Common Notions 66tjpc0xsmt6pe35ctcrjtz4brpd1y6


Similar Polygons are composed of Similar Triangles/Porism 0 35954 197167 2014-10-17T22:56:20Z Prime.mover 59 Created page with "== Porism to Similar Polygons are composed of Similar Triangles ==  <onlyinclude>{{:Euclid:Proposition/VI/20/Porism}}</onlyinclude>  == Proof ==  Follows immediately from..." wikitext text/x-wiki == Porism to Similar Polygons are composed of Similar Triangles == <onlyinclude>{{:Euclid:Proposition/VI/20/Porism}}</onlyinclude> == Proof == Follows immediately from Similar Polygons are composed of Similar Triangles. {{qed}} {{Euclid Note|20|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Similar Polygons are composed of Similar Triangles|next=Similarity of Polygons is Equivalence Relation}}: Book $\text{VI}$. Propositions Category:Triangles jnr9ony766y8a4etcrnn63xcl4dp3q5


Commensurability of Squares on Proportional Straight Lines/Lemma 0 36119 207209 207023 2015-02-01T13:01:40Z Prime.mover 59 wikitext text/x-wiki == Lemma to Commensurability of Squares on Proportional Straight Lines == <onlyinclude>{{:Euclid:Proposition/X/14/Lemma}}</onlyinclude> == Proof == :400px Let $AB$ and $C$ be two unequal straight lines. Let $AB > C$. Let the semicircle $ADB$ be described with $AB$ as the diameter. Using Fitting Chord Into Circle, let $AD$ be fitted into $ADB$ equal to $C$. Let $DB$ be joined. From Relative Sizes of Angles in Segments, $\angle ADB$ is a right angle. From Pythagoras's Theorem: :$AB^2 = AD^2 + DB^2$ and so $AB^2$ is greater than $AD^2$, that is, $C^2$, by $DB^2$. Conversely, given two straight lines $A$ and $B$ the same technique can be used to find the straight line the square of whose length equals the sum of the squares on $A$ and $B$. {{qed}} {{Euclid Note|14|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Commensurable Magnitudes are Incommensurable with Same Magnitude|next=Commensurability of Squares on Proportional Straight Lines}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory fqppdzd3vhirugs7oq6dpubculyyg5o


Straight Line Commensurable with Medial Straight Line is Medial/Porism 0 36145 479396 207087 2020-07-22T15:23:21Z Prime.mover 59 wikitext text/x-wiki == Porism to Straight Line Commensurable with Medial Straight Line is Medial == <onlyinclude>{{:Euclid:Proposition/X/23/Porism}}</onlyinclude> == Proof == {{EuclidSaid}} :''And in the same way as was explained in the case of the rationals it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called '''''medial''''' and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.'' :''But, if any straight line be commensurable in square with a medial straight line , then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.'' {{Euclid Note|23|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Straight Line Commensurable with Medial Straight Line is Medial|next = Rectangle Contained by Medial Straight Lines Commensurable in Length is Medial}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory tghou6sxgavec5f2u5avpf2nrx2iu9q


Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 1 0 36151 207098 207097 2015-01-28T07:00:36Z Prime.mover 59 wikitext text/x-wiki == Lemma to Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater == <onlyinclude>{{:Euclid:Proposition/X/29/Lemma 1}}</onlyinclude> == Proof == Let $a$ and $b$ be two natural numbers such that $a > b$. Let $a$ and $b$ be either both even or both odd. From: :Even Number minus Even Number is Even and: :Odd Number minus Odd Number is Even their difference $c = a - b$ is even. Let $d = \dfrac c 2$. Let $a$ and $b$ be similar plane numbers. By definition of similar plane numbers it is noted that $a$ and $b$ may both be square. Then by Square of Sum less Square: :$a b + d^2 = \left({a - d}\right)^2$ From Product of Similar Plane Numbers is Square, $a b$ is a square number. So two square numbers $a b$ and $d^2$ have been found whose sum is square. Similarly: :$a b = \left({a - d}\right)^2 - d^2$ from which it is noted that two square numbers $\left({a - d}\right)^2$ and $d^2$ have been found whose difference is square. {{qed}} {{Euclid Note|29|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Construction of Components of Second Bimedial|next=Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 2}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory 3mec0xytbdrz345iqxzu3w1ctfr23pm


Axiom:Peano's Axioms/Formulation 1

Axiom:Peano's Axioms/Formulation 1 100 36585 488563 475174 2020-09-17T07:35:37Z Prime.mover 59 wikitext text/x-wiki == Axioms == '''Peano's Axioms''' are intended to reflect the intuition behind $\N$, the mapping $s: \N \to \N: \map s n = n + 1$ and $0$ as an element of $\N$. Let there be given a set $P$, a mapping $s: P \to P$, and a distinguished element $0$. Historically, the existence of $s$ and the existence of $0$ were considered the first two of '''Peano's Axioms''': {{begin-axiom}} {{axiom | n = \text P 1         | q =          | m = 0 \in P         | t = $0$ is an element of $P$ }} {{axiom | n = \text P 2         | q = \forall n \in P         | m = \map s n \in P         | t = For all $n \in P$, its successor $\map s n$ is also in $P$ }} {{end-axiom}} The other three are as follows: <onlyinclude> {{begin-axiom}} {{axiom | n = \text P 3         | q = \forall m, n \in P         | m = \map s m = \map s n \implies m = n         | t = $s$ is injective }} {{axiom | n = \text P 4         | q = \forall n \in P         | m = \map s n \ne 0         | t = $0$ is not in the image of $s$ }} {{axiom | n = \text P 5         | q = \forall A \subseteq P         | m = \paren {0 \in A \land \paren {\forall z \in A: \map s z \in A} } \implies A = P         | t = Principle of Mathematical Induction: }} {{axiom | t = Any subset $A$ of $P$, containing $0$ and }} {{axiom | t = closed under $s$, is equal to $P$ }} {{end-axiom}} </onlyinclude> == Also defined as == {{:Axiom:Peano's Axioms/Also defined as}} == Also known as == {{:Axiom:Peano's Axioms/Also known as}} == Also see == * Equivalence of Formulations of Peano's Axioms {{LinkToCategory|Peano's Axioms|Peano's axioms}} {{NamedforAxiom|Giuseppe Peano}} == Historical Note == {{:Axiom:Peano's Axioms/Historical Note}} == Sources == * {{BookReference|Naive Set Theory|1960|Paul R. Halmos|prev = Definition:Transitive Set|next = Axiom:Peano's Axioms/Historical Note}}: $\S 12$: The Peano Axioms * {{BookReference|Number Theory|1964|J. Hunter|prev = Definition:Successor Mapping|next = Axiom:Peano's Axioms/Also defined as}}: Chapter $\text {I}$: Number Systems and Algebraic Structures: $2$. The positive integers * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Mathematician:Giuseppe Peano|next = Definition:Successor Mapping}}: $\S 16$ * {{BookReference|A Handbook of Terms used in Algebra and Analysis|1972|A.G. Howson|prev = Definition:Natural Numbers|next = Principle of Finite Induction}}: $\S 4$: Number systems $\text{I}$: Peano's Axioms * {{BookReference|Introduction to Topology|1975|Bert Mendelson|ed = 3rd|edpage = Third Edition|prev = Definition:Natural Numbers|next = Principle of Mathematical Induction for Peano Structure}}: Chapter $1$: Theory of Sets: $\S 1$: Introduction * {{BookReference|Numbers, Sets and Axioms|1982|Alan G. Hamilton|prev = Natural Number Multiplication is Cancellable|next = Axiom:Peano's Axioms/Historical Note}}: $\S 1$: Numbers: $1.1$ Natural Numbers and Integers * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Peano Curve|next = Mathematician:Karl Pearson|entry = Peano's postulates}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Peano Curve|next = Mathematician:Karl Pearson|entry = Peano's postulates}} * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Basic Universe|next = Axiom:Peano's Axioms/Historical Note}}: Chapter $3$: The Natural Numbers: $\S 1$ Preliminaries * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Axiom:Peano's Axioms/Historical Note|next = Definition:Peano Curve|entry = Peano's axioms}} Category:Axioms/Peano's Axioms 3vdnds24cohvzd8e49ex12rsgr7lw5e


Axiom:Peano's Axioms/Formulation 2

Axiom:Peano's Axioms/Formulation 2 100 36586 493102 475173 2020-10-07T21:53:59Z Prime.mover 59 wikitext text/x-wiki == Axioms == '''Peano's Axioms''' are intended to reflect the intuition behind $\N$, the mapping $s: \N \to \N: \map s n = n + 1$ and $0$ as an element of $\N$. Let there be given a set $P$, a mapping $s: P \to P$, and a distinguished element $0$. Historically, the existence of $s$ and the existence of $0$ were considered the first two of '''Peano's Axioms''': {{begin-axiom}} {{axiom | n = \text P 1         | q =          | m = 0 \in P         | t = $0$ is an element of $P$ }} {{axiom | n = \text P 2         | q = \forall n \in P         | m = \map s n \in P         | t = For all $n \in P$, its successor $\map s n$ is also in $P$ }} {{end-axiom}} The other three are as follows: <onlyinclude> {{begin-axiom}} {{axiom | n = \text P 3         | q = \forall m, n \in P         | m = \map s m = \map s n \implies m = n         | t = $s$ is injective }} {{axiom | n = \text P 4         | m = \Img s \ne P         | t = $s$ is not surjective }} {{axiom | n = \text P 5         | q = \forall A \subseteq P         | m = \paren {\paren {\exists x \in A: \neg \exists y \in P: x = \map s y} \land \paren {\forall z \in A: \map s z \in A} }         | t = Principle of Mathematical Induction: }} {{axiom | m = \implies A = P         | t = Any subset $A$ of $P$, containing an element not }} {{axiom | t = in the image of $s$ and closed under $s$, }} {{axiom | t = is equal to $P$ }} {{end-axiom}} </onlyinclude> == Also defined as == {{:Axiom:Peano's Axioms/Also defined as}} == Also see == * Equivalence of Formulations of Peano's Axioms {{LinkToCategory|Peano's Axioms|Peano's axioms}} {{NamedforAxiom|Giuseppe Peano|name2 = Richard Dedekind}} == Historical Note == {{:Axiom:Peano's Axioms/Historical Note}} == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Definition:Natural Numbers|next = Axiom:Peano's Axioms/Also defined as}}: Introduction $\S 4$: The natural numbers Category:Axioms/Peano's Axioms adsnjp6vlat6hxl3jpkxiaeu2moj83e


Greatest Power of Two not Divisor

Greatest Power of Two not Divisor 0 37121 401274 255679 2019-04-18T22:22:41Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $S = \set {1, 2, 3, 4, \ldots, n}$ be a set of integers from $1$ through $n$. Let $2^t$ be the greatest power of $2$ in $S$. Then $2^t$ does not divide any other integer in $S$. That is, no other member of $S$ is a multiple of $2^t$. == Proof == Let $k$ be a multiple of $2^t$ in $S$. Then $k = 2^t \times \ell$ for some $\ell \in \Z, \ell \ge 1$. If $\ell = 2$, then $k = 2 \times 2^t = 2^{t + 1}$, which would contradict $2^t$ being the highest power of $2$ in $S$. Otherwise, if $\ell > 2$, then we would have $k = \ell \times 2^t > 2\times 2^t = 2^{t+1}$. As $S$ contains all integers up to $k$, we would have $2^{t+1} \in S$, contradicting $2^t$ being the highest power of $2$ in $S$. So $\ell = 1$, that is, $2^t$ is the only multiple of $2^t$ in $S$. {{qed}}  == Also see == * Harmonic Number is not Integer Category:Number Theory 56o43b5ryyhluqucd74ea6gduw1e9ii


Bertrand-Chebyshev Theorem/Lemma 1

Bertrand-Chebyshev Theorem/Lemma 1 0 37634 310909 204675 2017-08-15T20:51:08Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> For all $n \in \N$: :$\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n + 1}$ where $\dbinom {2 n} n$ denotes a binomial coefficient. </onlyinclude> == Proof == From Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index, $\dbinom n k$ increases for $k < \dfrac n 2$. From Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index, $\dbinom n k$ decreases for $k > \dfrac n 2$. Therefore, $\dbinom {2 n} n$ is the largest term in the sequence $\dbinom {2 n} 0, \dbinom {2 n} 1, \ldots, \dbinom {2 n} {2 n}$. Finally, observe that the mean of the terms in the sequence is $\dfrac {2^{2 n}} {2 n + 1}$, by Sum of Binomial Coefficients over Lower Index. {{qed|lemma}} Category:Bertrand-Chebyshev Theorem cg5y0b16819x7n1i2i5r70t3ldykyww


Bertrand-Chebyshev Theorem/Lemma 2

Bertrand-Chebyshev Theorem/Lemma 2 0 37635 310912 204653 2017-08-15T20:52:31Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> For all $m \in \N$: :$\displaystyle \prod_{p \mathop \le m} p \le 2^{2 m}$ where the product is taken over all prime numbers $p \le m$. </onlyinclude> == Proof == It is plainly true for $m \le 2$; we will proceed by strong induction. If $m > 2$ is even then: {{begin-eqn}} {{eqn | l = \prod_{p \mathop \le m} p       | r = \prod_{p \mathop \le m - 1} p       | c =  }} {{eqn | r = 2^{2 \left({m - 1}\right)}       | o = \le        | c = by the induction hypothesis }} {{eqn | r = 2^{2 m}       | o = <       | c =  }} {{end-eqn}} By Sum of Binomial Coefficients over Lower Index: :$\displaystyle \sum_{r \mathop = 0}^{2 k + 1} \binom {2 k + 1} r = 2^{2 k + 1}$ Therefore: {{begin-eqn}} {{eqn | l = \binom {2 k + 1} k + \binom {2 k + 1} {k + 1}        | r = 2^{2 k + 1}       | o = \le       | c =  }} {{eqn | ll = \implies       | n = 1       | l = \binom {2 k + 1} k        | r = 2^{2 k}       | o = \le       | c = since $\displaystyle \binom {2 k + 1} k = \binom {2 k + 1} {k + 1}$ }} {{end-eqn}} If $m = 2 k + 1$ is odd, then all the prime numbers $k + 2 \le p \le 2 k + 1$ divide: :$\dbinom {2 k + 1} k = \dfrac {\left({2 k + 1}\right)!} {k! \left({k + 1}\right)!}$ Thus: {{begin-eqn}} {{eqn | l = \prod_{p \mathop \le m} p       | r = \prod_{p \mathop \le k + 1} p \prod_{k + 2 \mathop \le p \mathop \le 2 k + 1} p       | c =  }} {{eqn | r = \binom {2 k + 1} k \prod_{p \mathop \le k + 1} p        | o = \le       | c =  }} {{eqn | r = 2^{2 k} \prod_{p \mathop \le k + 1} p       | o = \le       | c = by $(1)$ }} {{eqn | r = 2^{2 k} 2^{2 \left({k + 1}\right)}       | o = \le       | c = by the induction hypothesis }} {{eqn | r = 2^{2 m}       | c =  }} {{end-eqn}} {{qed|lemma}} Category:Bertrand-Chebyshev Theorem gloahkyqwuyc7t0w9lrh07b5nugudqi


Bertrand-Chebyshev Theorem/Lemma 3

Bertrand-Chebyshev Theorem/Lemma 3 0 37636 310913 298898 2017-08-15T20:52:55Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> If $p$ is a prime number and $p^k \mathrel \backslash \dbinom {2 n} n$, then $p^k \le 2 n$. </onlyinclude> == Proof  == Let $l$ be the largest power of $p$ with $p^l \le 2 n$. By De Polignac's Formula, the largest power of $p$ dividing $n!$ is $\displaystyle \sum_{i \mathop \ge 1} \left \lfloor{\frac n {p^i}}\right \rfloor$. So: {{begin-eqn}} {{eqn | l = k       | r = \sum_{i \mathop \ge 1} \left \lfloor{\frac {2 n} {p^i} }\right \rfloor - 2 \sum_{i \mathop \ge 1} \left \lfloor{\frac n {p^i} }\right \rfloor       | o = \le       | c = Largest power of $p$ dividing $\dbinom {2 n} n$ }} {{eqn | r = \sum_{i \mathop = 1}^l \left({\left \lfloor{\frac {2 n} {p^i} }\right \rfloor - 2 \left \lfloor{\frac n {p^i} }\right \rfloor}\right)       | c = as terms with $i > l$ are zero }} {{eqn | r = \sum_{i \mathop = 1}^l 1       | o = \le       | c = since $\left \lfloor{2 x}\right \rfloor - 2 \left \lfloor{x}\right \rfloor \le 1$ }} {{eqn | r = l       | c =  }} {{end-eqn}} {{qed|lemma}} Category:Bertrand-Chebyshev Theorem fyirzym68wzdvkpndgukykiboi5d1w7


Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index 0 37642 204674 2014-12-20T15:04:02Z Prime.mover 59 Created page with "== Corollary to Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index == <onlyinclude> Let $n \in \Z_{>0}$ be a Definition:Strictly Positive Integ..." wikitext text/x-wiki == Corollary to Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index == <onlyinclude> Let $n \in \Z_{>0}$ be a strictly positive integer. Let $\dbinom n k$ be the binomial coefficient of $n$ over $k$ for a positive integer $k \in \Z_{\ge 0}$. Let $S_n = \left\langle{x_k}\right\rangle$ be the sequence defined as: :$x_k = \dbinom n k$ Then $S_n$ is strictly decreasing exactly where $\dfrac n 2 < k \le n$. </onlyinclude> == Proof == If $k > \dfrac n 2$ then it follows that $n - k < \dfrac n 2$. Thus: {{begin-eqn}} {{eqn | l = \binom n {k + 1}       | r = \binom n {n - \left({k + 1}\right)}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \binom n {n - k - 1}       | c =  }} {{eqn | o = <       | r = \binom n {n - k}       | c = Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index }} {{eqn | r = \binom n k       | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} Hence the result. {{Qed}} Category:Binomial Coefficients oiqxsersvz5a2zobgc675r70q44aeop


Isometry between Metric Spaces is Continuous/Corollary 0 37658 440595 204984 2019-12-23T11:19:29Z Prime.mover 59 wikitext text/x-wiki == Corollary to Isometry between Metric Spaces is Continuous == <onlyinclude> Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces. Let $\phi: M_1 \to M_2$ be an isometry. Then its inverse $\phi^{-1}: M_2 \to M_1$ is a continuous mapping. </onlyinclude> == Proof == From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry. The result follows from Isometry between Metric Spaces is Continuous. {{qed}} == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev = Isometry between Metric Spaces is Continuous|next = Definition:Homeomorphism/Metric Spaces/Definition 1}}: $\S 2.7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.5$ Category:Isometries fdtfx2mvzpjttuy7bpupoee5lcwjuj6


Axiom:Neighborhood Space Axioms

Axiom:Neighborhood Space Axioms 100 38011 206189 206186 2015-01-08T06:15:51Z Prime.mover 59 wikitext text/x-wiki == Axioms == A neighborhood space is a set $S$ such that, for each $x \in S$, there exists a set of subsets $\mathcal N_x$ of $S$ satisfying the following conditions: <onlyinclude> {{begin-axiom}} {{axiom|n = N1        |lc= There exists at least one element in $\mathcal N_x$        |q = \forall x \in S        |m = \mathcal N_x \ne \varnothing }} {{axiom|n = N2        |lc= Each element $x$ is in its own $\mathcal N_x$        |q = \forall x \in S        |m = \forall N \in \mathcal N_x: x \in N }} {{axiom|n = N3        |lc= Each superset of $N \in \mathcal N_x$ is also in $\mathcal N_x$        |q = \forall x \in S: \forall N \in \mathcal N_x        |m = N' \supseteq N \implies N' \in \mathcal N_x }} {{axiom|n = N4        |lc= Intersection of $2$ elements of $\mathcal N_x$ is also in $\mathcal N_x$        |q = \forall x \in S: \forall M, N \in \mathcal N_x        |m = M \cap N \in N_x }} {{axiom|n = N5        |lc= Exists $N' \subseteq N \in \mathcal N_x$ which is $\mathcal N_y$ of each $y \in N'$        |q = \forall x \in S: \forall N \in \mathcal N_x        |m = \exists N' \in \mathcal N_x, N' \subseteq N: \forall y \in N': N' \in \mathcal N_y }} {{end-axiom}} </onlyinclude> These stipulations are called the '''neighborhood space Axioms'''. Each element of $\mathcal N_x$ is called a '''neighborhood''' of $x$. == Also see == * Basic Properties of Neighborhood in Topological Space == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev=Definition:Neighborhood Space|next=Definition:Neighborhood (Neighborhood Space)}}: $\S 3.3$: Neighborhoods and Neighborhood Spaces: Definition $3.4$ Category:Axioms hfetij0zdmhwht9wcsvkgiyegltpx9t


Finite Intersection of Open Sets of Neighborhood Space is Open 0 38022 206235 206234 2015-01-09T18:40:52Z Prime.mover 59 wikitext text/x-wiki == Corollary to Intersection of two Open Sets of Neighborhood Space is Open == <onlyinclude> Let $\left({S, \mathcal N}\right)$ be a neighborhood space. Let $n \in \N_{>0}$ be a natural number. Let $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\left({S, \mathcal N}\right)$. Then $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$. </onlyinclude> == Proof == Proof by induction: Let $U_1, U_2, \ldots$ be open sets of $\left({S, \mathcal N}\right)$. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$. $P \left({1}\right)$ is true, as this just says: :$U_1$ is an open set of $\left({S, \mathcal N}\right)$. === Basis for the Induction === $P \left({2}\right)$ is the case: :$U_1 \cap U_2$ is an open set of $\left({S, \mathcal N}\right)$ which has been proved above. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$. Then we need to show: :$\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i$ is an open set of $\left({S, \mathcal N}\right)$. === Induction Step === This is our induction step: We have that: :$\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i = \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ From the induction hypothesis: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$. From the basis for the induction: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ is an open set of $\left({S, \mathcal N}\right)$. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$ \displaystyle\forall n \in \N_{>0}: \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$. {{qed}} == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev=Intersection of two Open Sets of Neighborhood Space is Open|next=Union of Open Sets of Neighborhood Space is Open}}: $\S 3.3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$ Category:Neighborhood Spaces f9v794ni1r6prl1imt2ae75s9q9wv77


Form of Geometric Sequence of Integers/Corollary 0 38081 456119 456010 2020-03-19T08:16:00Z Prime.mover 59 wikitext text/x-wiki == Corollary to Form of Geometric Sequence of Integers == <onlyinclude> Let $p$ and $q$ be integers. Then the finite sequence $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ defined as: :$a_j = p^j q^{n - j}$ is a geometric sequence whose common ratio is $\dfrac p q$. </onlyinclude> == Proof == Let the greatest common divisor of $p$ and $q$ be $d$. Then by Integers Divided by GCD are Coprime: :$p = d r$ :$q = d s$ where $r$ and $s$ are coprime integers. Thus: :$a_j = p^j q^{n - j}$ {{begin-eqn}} {{eqn | l = a_j       | r = p^j q^{n - j}       | c =  }} {{eqn | r = \paren {d r}^j \paren {d s}^{n - j}       | c =  }} {{eqn | r = d^n r^j s^{n - j}       | c =  }} {{end-eqn}} and so by Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence whose common ratio is $\dfrac r s$. Then: {{begin-eqn}} {{eqn | l = \dfrac r s       | r = \paren {\dfrac p d} / \paren {\dfrac q d}       | c =  }} {{eqn | r = \dfrac p d \dfrac d q       | c =  }} {{eqn | r = \dfrac p q       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Geometric Sequences of Integers 71grpt00w5ulx7b185qefx22vp26lp5


Greatest Common Measure of Commensurable Magnitudes/Porism 0 38120 206969 206915 2015-01-22T22:57:40Z Prime.mover 59 wikitext text/x-wiki == Porism to Greatest Common Measure of Commensurable Magnitudes == <onlyinclude>{{:Euclid:Proposition/X/3/Porism}}</onlyinclude> == Proof == Apparent from the construction. {{qed}} {{Euclid Note|3|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Greatest Common Measure of Commensurable Magnitudes|next=Greatest Common Measure of Three Commensurable Magnitudes}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory kexhfffj0zodfj5xyoh3174xcbplrsm


Magnitudes with Rational Ratio are Commensurable/Porism 0 38127 367972 206965 2018-09-29T08:04:39Z Prime.mover 59 wikitext text/x-wiki == Porism to Magnitudes with Rational Ratio are Commensurable == <onlyinclude>{{:Euclid:Proposition/X/6/Porism}}</onlyinclude> == Proof == Apparent from the construction. {{qed}} {{Euclid Note|6|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Magnitudes with Rational Ratio are Commensurable|next = Incommensurable Magnitudes have Irrational Ratio}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory ee42pvp66zptdd39xh88xd5afjam95d


Ratios of Equal Magnitudes/Porism

Ratios of Equal Magnitudes/Porism 0 38129 206964 206949 2015-01-22T22:53:35Z Prime.mover 59 wikitext text/x-wiki == Porism to Ratios of Equal Magnitudes == <onlyinclude>{{:Euclid:Proposition/V/7/Porism}}</onlyinclude> == Proof == Apparent from the construction. {{qed}} {{Euclid Note|7|V}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Ratios of Equal Magnitudes|next=Relative Sizes of Ratios on Unequal Magnitudes}}: Book $\text{V}$. Propositions Category:Ratios 2q9s5dp5m68bw4wissdq5bt1y0kl4bl


Construction of Incommensurable Lines/Lemma 0 38133 242124 206984 2015-12-30T12:36:11Z Prime.mover 59 wikitext text/x-wiki == Lemma to Construction of Incommensurable Lines == <onlyinclude>{{:Euclid:Proposition/X/10/Lemma}}</onlyinclude> == Proof == The propositions referred to are: :{{EuclidPropLink|book = VIII|prop = 26|title = Similar Plane Numbers have Same Ratio as between Two Squares}} and its converse, which {{AuthorRef|Euclid}} does not explicitly prove. {{EuclidSaid}} :''For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.'' :''Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.'' {{qed}} {{Euclid Note|10|X|It was suggested by {{AuthorRef|Johan Ludvig Heiberg|Heiberg}} that this lemma is a later interpolation, and not part of the original work by {{AuthorRef|Euclid}}. The theorem Construction of Incommensurable Lines that it supports is itself also questionable.

Moreover, there is no reason why numbers which are not similar plane numbers should be introduced here.}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage =Second Edition|Sir Thomas L. Heath|prev = Commensurability of Squares/Porism|next = Construction of Incommensurable Lines}}: Book $\text{X}$. Propositions Category:Euclidean Number Theory k1cfmilxg0xpqwkrpdlgs4cttxdw5gs


Sum of Möbius Function over Divisors/Lemma 0 38589 457776 209811 2020-03-27T07:31:28Z Prime.mover 59 wikitext text/x-wiki == Lemma to Sum of Möbius Function over Divisors == Let $n \in \Z_{>0}$, i.e. let $n$ be a strictly positive integer. Let $\displaystyle \sum_{d \mathop \divides n}$ denote the sum over all of the divisors of $n$. Let $\map \mu d$ be the Möbius function. Then: <onlyinclude> :$\displaystyle \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$ That is: :$\mu * u = \iota$ where $u$ and $\iota$ are the unit arithmetic function and identity arithmetic function respectively. </onlyinclude> == Proof == The lemma is clearly true if $n=1$. Assume, then, that $n > 1$ and write, by the Fundamental Theorem of Arithmetic: :$n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ In the sum $\displaystyle \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the divisors of $n$ which are products of distinct primes. Thus: {{begin-eqn}} {{eqn | l = \sum_{d \mathop \divides n} \map \mu d       | r = \map \mu 1 + \map \mu {p_1} + \dotsb + \map \mu {p_k} + \map \mu {p_1 p_2} + \dotsb + \map \mu {p_{k - 1} p_k} + \dotsb + \map \mu {p_1 p_2 \dotsm p_k}       | c =  }} {{eqn | r = \dbinom k 0 + \dbinom k 1 \paren {-1} + \dbinom k 2 \paren {-1}^2 + \cdots + \dbinom k k \paren {-1}^k       | c =  }} {{eqn | r = 0       | c = Alternating Sum and Difference of Binomial Coefficients for Given n }} {{end-eqn}} {{Handwaving|The jump from the expression in $\mu$ to the one in binomial coefficients is far too big. Intermediate steps need to be included.}} Hence, the sum is $1$ for $n = 1$, and $0$ for $n > 1$, which are precisely the values of $\floor {\dfrac 1 n}$. {{qed}} Category:Euler Phi Function Category:Möbius Function cx9ok1t5lpk4hncdanse65yjsjy18ma


Symmetry Group of Square/Cayley Table

Symmetry Group of Square/Cayley Table 0 38643 385558 372598 2018-12-31T12:06:43Z Prime.mover 59 wikitext text/x-wiki == Cayley Table of Symmetry Group of Square == <onlyinclude> The Cayley table of the symmetry group of the square can be written: :$\begin{array}{c|cccccc}        & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ \hline e      & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ r      & r & r^2 & r^3 & e & t_{AC} & t_{BD} & t_y & t_x \\ r^2    & r^2 & r^3 & e & r & t_y & t_x & t_{BD} & t_{AC} \\ r^3    & r^3 & e & r & r^2 & t_{BD} & t_{AC} & t_x & t_y \\ t_x    & t_x & t_{BD} & t_y & t_{AC} & e & r^2 & r^3 & r \\ t_y    & t_y & t_{AC} & t_x & t_{BD} & r^2 & e & r & r^3 \\ t_{AC} & t_{AC} & t_x & t_{BD} & t_y & r & r^3 & e & r^2 \\ t_{BD} & t_{BD} & t_y & t_{AC} & t_x & r^3 & r & r^2 & e\\ \end{array}$ </onlyinclude> where the various symmetry mappings of the square $\mathcal S = ABCD$ are: :the identity mapping $e$ :the rotations $r, r^2, r^3$ of $90^\circ, 180^\circ, 270^\circ$ counterclockwise respectively :the reflections $t_x$ and $t_y$ are reflections about the $x$ and $y$ axis respectively :the reflection $t_{AC}$ is a reflection about the diagonal through vertices $A$ and $C$ :the reflection $t_{BD}$ is a reflection about the diagonal through vertices $B$ and $D$. == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Composition of Symmetries is Associative|next = Count of Binary Operations on Set}}: $\S 2$: Example $2.5$ * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Right Operation is Associative|next = Symmetry Group of Equilateral Triangle/Cayley Table}}: Exercise $2.10$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Symmetry Group of Equilateral Triangle/Cayley Table|next = All Elements Self-Inverse then Abelian}}: $\S 1.4$: Exercise $2.1$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Dihedral Group D4/Matrix Representation/Formulation 2/Cayley Table|next = Definition:Symmetry Group of Rectangle}}: Chapter $1$: Definitions and Examples: Exercise $3$ Category:Examples of Cayley Tables qcis72b8caopz6ltd4xlp826o8ww80r