Slater 37 - 편집 역사

2020년 12월 28일 (월) 12:15에 Pythagoras0님의 편집

2020-12-28T12:15:24Z

Pythagoras0: /* articles */

2020-11-16T10:28:04Z

articles

2020년 11월 16일 (월) 10:27에 Pythagoras0님의 편집

2020-11-16T10:27:37Z

2020년 11월 14일 (토) 06:41에 Pythagoras0님의 편집

2020-11-14T06:41:23Z

2020년 11월 13일 (금) 14:03에 imported>Pythagoras0님의 편집

2020-11-13T14:03:24Z

imported>Pythagoras0: /* articles */

2020-11-13T14:03:16Z

articles

imported>Pythagoras0: /* books */

2020-11-13T14:03:02Z

books

imported>Pythagoras0: 찾아 바꾸기 – “4909919” 문자열을 “” 문자열로

2012-11-02T12:40:47Z

찾아 바꾸기 – “4909919” 문자열을 “” 문자열로

2012년 10월 29일 (월) 17:55에 imported>Pythagoras0님의 편집

2012-10-29T17:55:55Z

imported>Pythagoras0: 찾아 바꾸기 – “\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}” 문자열을 “\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}

2012-10-29T14:43:54Z

찾아 바꾸기 – “\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}” 문자열을 “\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}

@@ 3번째 줄: / 3번째 줄: @@
 *  not checked
-−
-−
 ==type of identity==
@@ 12번째 줄: / 12번째 줄: @@
 *  I(17)
-−
-−
 ==Bailey pair 1==
-*  Use the folloing<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
+*  Use the folloing<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
 *  Specialize<math>x=q^{3}, y=-q, z\to\infty</math>.
 *  Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>
-−
-−
 ==Bailey pair 2==
-*  Use the following '''[Slater52-1] '''(4.2)
+*  Use the following '''[Slater52-1] '''(4.2)
 *  Specialize<math>a=q^{2},d=q^2,e=q</math>
-*  Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
+*  Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
-−
-−
-==Bailey pair ==
+==Bailey pair ==
-*  Bailey pairs  <math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>  <math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
+*  Bailey pairs  <math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>  <math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
-−
-−
 ==q-series identity==
@@ 48번째 줄: / 48번째 줄: @@
 <math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math>
-* [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math>
+* [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math>
 * [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 ** http://www.research.att.com/~njas/sequences/?q=
-−
-−
 ==Bethe type equation (cyclotomic equation)==
-Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
+Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
   \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>.
-Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math>  has a unique root <math>0<\mu<1</math>. We get
+Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math>  has a unique root <math>0<\mu<1</math>. We get
 <math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math>
-−
 a=1,d=1,e=1
-The equation  becomes <math>1-x=x</math>.
+The equation  becomes <math>1-x=x</math>.
 <math>4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2</math>
-−
-−
 ==dilogarithm identity==
@@ 82번째 줄: / 82번째 줄: @@
 <math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math>
-−
-−
 ==related items==
@@ 90번째 줄: / 90번째 줄: @@
 * [[asymptotic analysis of basic hypergeometric series]]
-−
-−
 ==articles==
 * [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]
 **  McLaughlin, 2008
 * [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]
-**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167
+**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167
 * [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]
 **  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475

@@ 94번째 줄: / 94번째 줄: @@
 ==articles==
 * [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]
 **  McLaughlin, 2008
@@ 103번째 줄: / 101번째 줄: @@
 * [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]
 **  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475
 [[분류:개인노트]]
 [[분류:math and physics]]
 [[분류:migrate]]

← 이전 판		2020년 11월 16일 (월) 10:27 판
89번째 줄:		89번째 줄:

	* [[asymptotic analysis of basic hypergeometric series]]		* [[asymptotic analysis of basic hypergeometric series]]
−
−
−
−
−
−	~~==books==~~
−
−
−
−	* [[2010년 books and articles]]
−	* http://gigapedia.info/1/
−	* http://gigapedia.info/1/
−	* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
−
−

@@ 1번째 줄: / 1번째 줄: @@
 ==Note==
-*  not checked<br>
+*  not checked
@@ 10번째 줄: / 10번째 줄: @@
 * [[Slater list|Slater's list]]
-*  I(17)<br>
+*  I(17)
@@ 18번째 줄: / 18번째 줄: @@
 ==Bailey pair 1==
-*  Use the folloing<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math><br>
+*  Use the folloing<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
-*  Specialize<br><math>x=q^{3}, y=-q, z\to\infty</math>.<br>
+*  Specialize<math>x=q^{3}, y=-q, z\to\infty</math>.
-*  Bailey pair<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><br><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math><br>
+*  Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>
@@ 28번째 줄: / 28번째 줄: @@
 ==Bailey pair 2==
-*  Use the following '''[Slater52-1] '''(4.2)<br>  <br>
+*  Use the following '''[Slater52-1] '''(4.2)
-*  Specialize<br><math>a=q^{2},d=q^2,e=q</math><br>
+*  Specialize<math>a=q^{2},d=q^2,e=q</math>
-*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><br><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math><br>
+*  Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
@@ 38번째 줄: / 38번째 줄: @@
 ==Bailey pair ==
-*  Bailey pairs<br>  <br><math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><br><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math><br>  <br><math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><br><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math><br>
+*  Bailey pairs  <math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>  <math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
@@ 48번째 줄: / 48번째 줄: @@
 <math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math>
-* [[Bailey pair and lemma|Bailey's lemma]]<br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<br><math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math><br>
+* [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math>
-* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
+* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 ** http://www.research.att.com/~njas/sequences/?q=
@@ 88번째 줄: / 88번째 줄: @@
 ==related items==
-* [[asymptotic analysis of basic hypergeometric series]]<br>
+* [[asymptotic analysis of basic hypergeometric series]]
@@ 98번째 줄: / 98번째 줄: @@
-* [[2010년 books and articles]]<br>
+* [[2010년 books and articles]]
 * http://gigapedia.info/1/
 * http://gigapedia.info/1/
@@ 111번째 줄: / 111번째 줄: @@
 ==articles==
-*   <br>
-* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]<br>
+* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]
-**  McLaughlin, 2008<br>
+**  McLaughlin, 2008
-* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>
+* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]
-**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167<br>
+**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167
-* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]<br>
+* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]
-**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475<br>
+**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475
-*   <br>
 * http://www.ams.org/mathscinet
 * [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
 * [http://arxiv.org/ ]http://arxiv.org/
 * http://pythagoras0.springnote.com/
-* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
 * http://dx.doi.org/
 [[분류:개인노트]]
 [[분류:math and physics]]
 [[분류:migrate]]

@@ 89번째 줄: / 89번째 줄: @@
 * [[asymptotic analysis of basic hypergeometric series]]<br>
@@ 103번째 줄: / 118번째 줄: @@
 * [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]<br>
 **  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475<br>
 [[분류:개인노트]]
 [[분류:math and physics]]

← 이전 판		2020년 11월 13일 (금) 14:03 판
89번째 줄:		89번째 줄:

	* [[asymptotic analysis of basic hypergeometric series]]<br>		* [[asymptotic analysis of basic hypergeometric series]]<br>
−
−
−
−
−
−	~~==books==~~
−
−
−
−	* [[2010년 books and articles]]<br>
−	* http://gigapedia.info/1/
−	* http://gigapedia.info/1/
−	* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
−
−

← 이전 판		2012년 11월 2일 (금) 12:40 판
103번째 줄:		103번째 줄:
	* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=		* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=

−	~~[[4909919\|4909919]]~~	+

← 이전 판		2012년 10월 29일 (월) 17:55 판
126번째 줄:		126번째 줄:
	* http://dx.doi.org/		* http://dx.doi.org/
	[[분류:개인노트]]		[[분류:개인노트]]
		+	[[분류:math and physics]]