Schottky defect - Revision history

en>Logical1004: /* External links */

2014-11-12T11:27:06Z

External links

← Older revision		Revision as of 13:27, 12 November 2014
Line 1:		Line 1:
	~~{{Distinguish2\|the Tsallis [[Tsallis statistics#q-exponential\|q-exponential]]}}~~		Alyson Meagher is the name her parents gave her but she doesn't like when individuals use her full name. Credit authorising is exactly where my primary income comes from. For many years she's been living in Kentucky but her husband desires them to transfer. What me and my family adore is to climb but I'm thinking on starting some thing new.<br><br>my weblog online psychics, [http://Kjhkkb.net/xe/notice/374835 http://Kjhkkb.net],

	~~{{Lowercase}}In [[combinatorics\|combinatorial]] [[mathematics]],~~ the '~~''q-exponential'''~~ is ~~a [[q-analog]] of the [[exponential function]],~~
	~~namely the eigenfunction of the [[q-derivative]]~~

	~~==Definition==~~
	~~The q-exponential <math>e_q(z)</math> is defined as~~
	~~:<math>e_q(z)=~~
	~~\sum_{n=0}^\infty \frac{z^n}{[n]_q!} =~~
	~~\sum_{n=0}^\infty \frac{z^n (1-q)^n}{(q;q)_n} =~~
	~~\sum_{n=0}^\infty z^n\frac{(1-q)^n}{(1-q^n)(1-q^{n-1}) \cdots (1-q)}</math>~~

	where ~~<math>[n]_q!</math> is the [[q-factorial]] and~~
	~~:<math>(q;q)_n=(1-q^n)(1-q^{n-1})\cdots (1-q)</math>~~

	~~is the [[q-Pochhammer symbol]]. That this is the q-analog of the exponential follows~~ from ~~the property~~

	~~:<math>\left(\frac{d}{dz}\right)_q e_q(z) = e_q(z)</math>~~

	~~where the derivative on the left is the [[q-derivative]]~~. ~~The above is easily verified by considering the q-derivative of the [[monomial]]~~

	~~:<math>\left(\frac{d}{dz}\right)_q z^n = z^{n-1} \frac{1-q^n}{1-q}~~
	~~=[n]_q z^{n-1}~~.~~</math>~~

	~~Here, <math>[n]_q</math>~~ is ~~the [[q-bracket]]~~.

	~~==Properties==~~
	~~For real~~ <~~math>q~~>1<~~/math~~>, ~~the function <math>e_q(z)</math> is an~~ [~~[entire function]] of ''z''. For <math>q<1<~~/~~math>, <math>e_q(z)<~~/~~math> is regular in the disk <math>\|z\|<1/(1-q)</math>~~.

	~~Note the inverse, <math>~e_q(z) ~ e_{1~~/~~q} (-z) =1<~~/~~math>.~~

	~~==Relations==~~
	~~For <math>q<1<~~/~~math>, a function that is closely related is~~

	:~~<math>e_q(z) = E_q(z(1-q)).<~~/~~math>~~

	~~Here, <math>E_q(t)<~~/~~math> is a special case of the [[basic hypergeometric series]~~]:

	~~:<math>E_q(z) = \;_{1}\phi_0 (0;q,z) = \prod_{n=0}^\infty~~
	~~\frac {1}{1-q^n z} ~. </math>~~

	~~==References==~~
	* F. H. Jackson (1908), "On q-functions and a certain difference operator", ''Trans. Roy. Soc. Edin.'', '''46''' 253-281.

	* Exton, H. (1983), ''q-Hypergeometric Functions and Applications'', New York: Halstead Press, Chichester: Ellis Horwood, 1983, ISBN 0853124914, ISBN 0470274530, ISBN 978-0470274538

	* Gasper G., and Rahman, M. (2004), ''Basic Hypergeometric Series'', ~~Cambridge University Press, 2004, ISBN 0521833574~~

	~~{{DEFAULTSORT:Q-Exponential}}~~
	~~[[Category:Q-analogs]]~~
	~~[[Category:Exponentials]]~~

76.120.175.135: fix link

2014-02-04T02:34:56Z

fix link

← Older revision		Revision as of 04:34, 4 February 2014
Line 1:		Line 1:
	~~The author~~'~~s title~~ is ~~Christy Brookins. Distributing manufacturing~~ is where ~~my primary income arrives from~~ and ~~it's something I truly appreciate~~. ~~Ohio~~ is ~~exactly~~ where ~~my house~~ is ~~but my husband wants us to move~~. ~~To climb~~ is ~~some thing I really enjoy performing~~.<br><br>~~Stop by my blog post~~ [~~http:~~//~~www~~.~~weddingwall~~.~~com~~.au/~~groups~~/~~easy~~-~~advice~~-~~for~~-~~successful~~-~~personal~~-~~development~~-~~today/ best psychic~~]		{{Distinguish2\|the Tsallis [[Tsallis statistics#q-exponential\|q-exponential]]}}

			{{Lowercase}}In [[combinatorics\|combinatorial]] [[mathematics]], the '''q-exponential''' is a [[q-analog]] of the [[exponential function]],
			namely the eigenfunction of the [[q-derivative]]

			==Definition==
			The q-exponential <math>e_q(z)</math> is defined as
			:<math>e_q(z)=
			\sum_{n=0}^\infty \frac{z^n}{[n]_q!} =
			\sum_{n=0}^\infty \frac{z^n (1-q)^n}{(q;q)_n} =
			\sum_{n=0}^\infty z^n\frac{(1-q)^n}{(1-q^n)(1-q^{n-1}) \cdots (1-q)}</math>

			where <math>[n]_q!</math> is the [[q-factorial]] and
			:<math>(q;q)_n=(1-q^n)(1-q^{n-1})\cdots (1-q)</math>

			is the [[q-Pochhammer symbol]]. That this is the q-analog of the exponential follows from the property

			:<math>\left(\frac{d}{dz}\right)_q e_q(z) = e_q(z)</math>

			where the derivative on the left is the [[q-derivative]]. The above is easily verified by considering the q-derivative of the [[monomial]]

			:<math>\left(\frac{d}{dz}\right)_q z^n = z^{n-1} \frac{1-q^n}{1-q}
			=[n]_q z^{n-1}.</math>

			Here, <math>[n]_q</math> is the [[q-bracket]].

			==Properties==
			For real <math>q>1</math>, the function <math>e_q(z)</math> is an [[entire function]] of ''z''. For <math>q<1</math>, <math>e_q(z)</math> is regular in the disk <math>\|z\|<1/(1-q)</math>.

			Note the inverse, <math>~e_q(z) ~ e_{1/q} (-z) =1</math>.

			==Relations==
			For <math>q<1</math>, a function that is closely related is

			:<math>e_q(z) = E_q(z(1-q)).</math>

			Here, <math>E_q(t)</math> is a special case of the [[basic hypergeometric series]]:

			:<math>E_q(z) = \;_{1}\phi_0 (0;q,z) = \prod_{n=0}^\infty
			\frac {1}{1-q^n z} ~. </math>

			==References==
			* F. H. Jackson (1908), "On q-functions and a certain difference operator", ''Trans. Roy. Soc. Edin.'', '''46''' 253-281.

			* Exton, H. (1983), ''q-Hypergeometric Functions and Applications'', New York: Halstead Press, Chichester: Ellis Horwood, 1983, ISBN 0853124914, ISBN 0470274530, ISBN 978-0470274538

			* Gasper G., and Rahman, M. (2004), ''Basic Hypergeometric Series'', Cambridge University Press, 2004, ISBN 0521833574

			{{DEFAULTSORT:Q-Exponential}}
			[[Category:Q-analogs]]
			[[Category:Exponentials]]

en>JYBot: r2.7.1) (Robot: Adding pl:Defekt Schottky'ego

2012-07-20T13:21:29Z

r2.7.1) (Robot: Adding pl:Defekt Schottky'ego

New page

The author's title is Christy Brookins. Distributing manufacturing is where my primary income arrives from and it's something I truly appreciate. Ohio is exactly where my house is but my husband wants us to move. To climb is some thing I really enjoy performing.<br><br>Stop by my blog post [http://www.weddingwall.com.au/groups/easy-advice-for-successful-personal-development-today/ best psychic]