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en>Eyesnore at 16:42, 12 October 2014

2014-10-12T16:42:18Z

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	~~In [[mathematics]], the '''formal derivative''' is an operation on elements~~ of ~~a [[polynomial ring]] or a ring of [[formal power series]] that mimics the form of the derivative from [[derivative\|calculus]]. Though they appear similar,~~ the ~~algebraic advantage of a formal derivative~~ is ~~that~~ it does not rely on the notion of a [[limit (mathematics)\|limit]], which is in general impossible to define for a [[ring (mathematics)\|ring]]. Many of the properties of the derivative are true of the formal derivative, but some, especially those that make numerical statements, are not. The primary use of formal differentiation in algebra is to test for [[multiple roots of a polynomial]].		The title of the writer is Figures but it's not the most masucline name out there. One of the issues he loves most is ice skating but he is struggling to discover time for it. For years he's been working as a receptionist. Puerto Rico is where he's been residing for many years and he will by no means move.<br><br>Feel free to visit my page :: [http://www.lankaclipstv.com/blog/165665 at home std testing]

	~~==Definition==~~
	~~The definition of a formal derivative is as follows: fix a ring~~ '~~'R'' (~~not ~~necessarily commutative) and let ''A'' = ''R''[''x''] be~~ the ~~ring of polynomials over ''R''~~. ~~Then the formal derivative is an operation on elements~~ of ~~''A'', where if~~

	~~:<math>f(x)\,=\,a_n x^n + \cdots + a_1 x + a_0</math>~~

	~~then its formal derivative is~~

	~~:<math>f'(x)\,=\,Df(x) = n a_n x^{n - 1} + \cdots + 2 a_2 x + a_1</math>~~

	~~just as for polynomials over~~ the ~~[[real numbers\|real]] or [[complex numbers\|complex]] numbers.~~

	~~Note that <math>n a_n</math> does not mean multiplication in the ring, but rather <math>\sum_{k=1}^n a_n,</math> where <math>k</math>~~ is ~~never used inside the sum.~~

	~~One should mention that there is a problem with this definition for noncommutative rings. The formula itself is correct,~~ but ~~there~~ is ~~no standard form of a polynomial. Therefore using this definition it's difficult~~ to ~~prove <math>(f(x)\cdot b)'=f'(x)\cdot b</math>.~~

	~~==Alternative definition well suited for noncommutative rings==~~
	~~Let~~ for ~~<math>r\in R</math> holds <math>r'=0,</math> let <math>x'=1.</math>~~
	~~Let us define derivative for expressions, such that <math>(a+b)'=a'+b',</math> and <math>(a\cdot b)'=a'\cdot b+a\cdot b'.</math>~~

	~~We must prove that this definition gives the same result for an expression independent on the method the expression was evaluated, therefore that~~ it ~~is compatible with the axioms of equality~~.
	*<math>(a+b)'=a'+b'=b'+a'=(b+a)',</math>
	*<math>((a+b)+c)'=(a+b)'+c'=(a'+b')+c'=a'+(b'+c')=a'+(b+c)'=(a+(b+c))',</math>
	*<math>(a(bc))'=a'(bc)+a(bc)'=a'bc+a(b'c+bc')=a'bc+ab'c+abc'=</math>
	~~:<math>=(a'b+ab')c+(ab)c'=(ab)'c+(ab)c'=((ab)c)',</math>~~
	*<math>((a+b)c)'=(a+b)'c+(a+b)c'=(a'+b')c+(a+b)c'=(a'c+b'c)+(ac'+bc')=</math>
	~~:<math>=\cdots=(a'c+(b'c+ac'))+bc'=(a'c+(ac'+b'c))+bc'=\cdots=(a'c+ac')+(b'c+bc')=</math>~~
	~~:<math>=(ac)'+(bc)'=(ac+bc)',</math>~~
	~~and the distributivity from the other side from symmetry.~~

	~~Linearity naturally follows from the definition.~~

	~~Formula for derivative of a polynomial (in standard shape for commutative rings) is direct consequence of the definition:~~
	~~<math>(\sum_i a_ix^i)'=\sum_i (a_ix^i)'=\sum_i ((a_i)'x^i+a_i(x^i)')=\sum_i(0x^i+a_i(\sum_{j=1}^ix^{j-1}(x')x^{i-j}))=\sum_i\sum_{j=1}^i a_ix^{i-1}.</math>~~

	~~==Properties==~~
	~~It can be verified that:~~

	* Formal differentiation is linear: for any two polynomials ''f''(''x''), ''g''(''x'') and elements ''r'', ''s~~'' of ''R'', we have~~

	~~::<math>(r \cdot f + s \cdot g)'(x) = r \cdot f'(x) + s \cdot g'(x).</math>~~

	:When ''R'' is not commutative there is another, different linearity property in which ''r'' and ''s'' appear on the right rather than on the left. When ''R'' does not contain an identity element then neither of these reduces to the case of simply a sum of polynomials or the sum of a polynomial with a multiple of another polynomial, which must also be included as a ~~"linearity" property.~~

	* The formal derivative satisfies the [[Product rule\|Leibniz rule]], or product rule:

	~~::<math>(f \cdot g)'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x).</math>~~

	~~:Note the order of the factors; when ''R'' is not commutative this is important.~~

	~~These two properties make ''D'' a [[derivation (abstract algebra)\|derivation]] on ''A'' (see also [[module of relative differential forms]] for a discussion of a generalization)~~.

	~~==Application to finding repeated factors==~~
	~~As in calculus, the derivative detects multiple roots: if ''R'' is a field then ''R''[''x'']~~ is a [[Euclidean domain]], and in this situation we can define multiplicity of roots; namely, for every polynomial ''f''(''x'') and every element ''r'' of ''R'', there exists a nonnegative integer ''m<sub>r</sub>'' and a polynomial ''g''(''x'') such that

	~~:<math>f(x) = (x - r)^{m_r} g(x)</math>~~

	where ''g''(''r'') is not equal to ''0''. ''m<sub>r</sub>'' is the multiplicity of ''r'' as a root of ''f''. It follows from the Leibniz rule that in this situation, ''m<sub>r</sub>'' is also the number of differentiations that must be performed on ''f''(''x'') before ''r'' is not a root of the resulting polynomial. The utility of this observation is that although in general not every polynomial of degree ''n'' in ''R''[''x''] has ''n'' roots counting multiplicity (this is the maximum, by the above theorem), we may pass to [[field extension]]s in which this is true (namely, [[algebraic closure]]s). Once we do, we may uncover a multiple root that was not a root at all simply over ''R''. For example, if ''R'' is the field with three elements, the polynomial

	~~:<math>f(x)\,=\,x^6 + 1</math>~~

	~~has no roots in ''R''; however, its formal derivative is zero since 3 = 0 in ''R''~~ and in any extension of ''R'', so when we pass to the algebraic closure it has a multiple root that could not have been detected by factorization in ''R'' itself. Thus, formal differentiation allows an [[computability theory (computer science)\|effective]] notion of multiplicity. This is important in [[Galois theory]], where the distinction is made between [[separable field extension]]s (defined by ~~polynomials with~~ no ~~multiple roots) and inseparable ones.~~

	~~==Correspondence to analytic derivative==~~
	~~When the ring ''R'' of scalars is commutative, there is an alternative and equivalent definition of the formal derivative, which resembles the one seen in differential calculus~~. ~~The element Y-X of the ring ''R''[X,Y] divides Y~~<~~sup~~>n<~~/sup~~> ~~- X<sup>n</sup> for any nonnegative integer ''n'', and therefore divides ''f''(Y) - ''f''(X) for any polynomial ''f'' in one indeterminate. If we denote the quotient (in ''R''[X,Y]) by ''g''~~:

	:~~<math>g(X,Y) = \frac{f(Y) - f(X)}{Y - X}</math>~~

	~~then it is not hard to verify that ''g''(X,X) (in ''R''[X]) coincides with the formal derivative of ''f'' as it was defined above.~~

	~~This formulation of the derivative works equally well for a formal power series, assuming only that the ring of scalars is commutative.~~

	Actually, if the division in this definition is carried out in the class of functions of <math>Y</math> continuous at <math>X</math>, it will recapture the classical definition of the derivative. If it is carried out in the class of functions continuous in both <math>X</math> and <math>Y</math>, we get uniform differentiability, and our function <math>f</math> will be continuously differentiable. Likewise, by choosing different classes of functions (say, the Lipschitz class), we get different flavors of differentiability. This way differentiation becomes a part of algebra of functions.

	~~==See also==~~

	* [[Derivative]]
	* [[Euclidean domain]]
	* [[Module of relative differential forms]]
	* [[Galois theory]]
	* [[Formal power series]]
	* [[Pincherle derivative]]

	~~==References==~~
	* {{Lang Algebra\|edition=3r}}
	* [http://~~arxiv~~.~~org~~/~~abs~~/~~0905.3611 Michael Livshits, You could simplify calculus, arXiv:0905.3611v1 ]~~

	~~[[Category:Abstract algebra]~~]

en>Addbot: Bot: Migrating 2 interwiki links, now provided by Wikidata on d:q4516348

2013-03-15T04:04:19Z

Bot: Migrating 2 interwiki links, now provided by Wikidata on d:q4516348

← Older revision		Revision as of 06:04, 15 March 2013
Line 1:		Line 1:
	~~They contact me Emilia~~. ~~Body building~~ is ~~one~~ of the ~~things I love most~~. ~~Bookkeeping~~ is ~~what I~~ do. ~~California~~ is ~~our birth place~~.<br><br>~~Look at my web site~~ - [~~http~~://~~xrambo~~.~~com~~/~~blog~~/~~192034~~ http://~~xrambo~~.~~com~~/~~blog~~/~~192034~~]		In [[mathematics]], the '''formal derivative''' is an operation on elements of a [[polynomial ring]] or a ring of [[formal power series]] that mimics the form of the derivative from [[derivative\|calculus]]. Though they appear similar, the algebraic advantage of a formal derivative is that it does not rely on the notion of a [[limit (mathematics)\|limit]], which is in general impossible to define for a [[ring (mathematics)\|ring]]. Many of the properties of the derivative are true of the formal derivative, but some, especially those that make numerical statements, are not. The primary use of formal differentiation in algebra is to test for [[multiple roots of a polynomial]].

			==Definition==
			The definition of a formal derivative is as follows: fix a ring ''R'' (not necessarily commutative) and let ''A'' = ''R''[''x''] be the ring of polynomials over ''R''. Then the formal derivative is an operation on elements of ''A'', where if

			:<math>f(x)\,=\,a_n x^n + \cdots + a_1 x + a_0</math>

			then its formal derivative is

			:<math>f'(x)\,=\,Df(x) = n a_n x^{n - 1} + \cdots + 2 a_2 x + a_1</math>

			just as for polynomials over the [[real numbers\|real]] or [[complex numbers\|complex]] numbers.

			Note that <math>n a_n</math> does not mean multiplication in the ring, but rather <math>\sum_{k=1}^n a_n,</math> where <math>k</math> is never used inside the sum.

			One should mention that there is a problem with this definition for noncommutative rings. The formula itself is correct, but there is no standard form of a polynomial. Therefore using this definition it's difficult to prove <math>(f(x)\cdot b)'=f'(x)\cdot b</math>.

			==Alternative definition well suited for noncommutative rings==
			Let for <math>r\in R</math> holds <math>r'=0,</math> let <math>x'=1.</math>
			Let us define derivative for expressions, such that <math>(a+b)'=a'+b',</math> and <math>(a\cdot b)'=a'\cdot b+a\cdot b'.</math>

			We must prove that this definition gives the same result for an expression independent on the method the expression was evaluated, therefore that it is compatible with the axioms of equality.
			*<math>(a+b)'=a'+b'=b'+a'=(b+a)',</math>
			*<math>((a+b)+c)'=(a+b)'+c'=(a'+b')+c'=a'+(b'+c')=a'+(b+c)'=(a+(b+c))',</math>
			*<math>(a(bc))'=a'(bc)+a(bc)'=a'bc+a(b'c+bc')=a'bc+ab'c+abc'=</math>
			:<math>=(a'b+ab')c+(ab)c'=(ab)'c+(ab)c'=((ab)c)',</math>
			*<math>((a+b)c)'=(a+b)'c+(a+b)c'=(a'+b')c+(a+b)c'=(a'c+b'c)+(ac'+bc')=</math>
			:<math>=\cdots=(a'c+(b'c+ac'))+bc'=(a'c+(ac'+b'c))+bc'=\cdots=(a'c+ac')+(b'c+bc')=</math>
			:<math>=(ac)'+(bc)'=(ac+bc)',</math>
			and the distributivity from the other side from symmetry.

			Linearity naturally follows from the definition.

			Formula for derivative of a polynomial (in standard shape for commutative rings) is direct consequence of the definition:
			<math>(\sum_i a_ix^i)'=\sum_i (a_ix^i)'=\sum_i ((a_i)'x^i+a_i(x^i)')=\sum_i(0x^i+a_i(\sum_{j=1}^ix^{j-1}(x')x^{i-j}))=\sum_i\sum_{j=1}^i a_ix^{i-1}.</math>

			==Properties==
			It can be verified that:

			* Formal differentiation is linear: for any two polynomials ''f''(''x''), ''g''(''x'') and elements ''r'', ''s'' of ''R'', we have

			::<math>(r \cdot f + s \cdot g)'(x) = r \cdot f'(x) + s \cdot g'(x).</math>

			:When ''R'' is not commutative there is another, different linearity property in which ''r'' and ''s'' appear on the right rather than on the left. When ''R'' does not contain an identity element then neither of these reduces to the case of simply a sum of polynomials or the sum of a polynomial with a multiple of another polynomial, which must also be included as a "linearity" property.

			* The formal derivative satisfies the [[Product rule\|Leibniz rule]], or product rule:

			::<math>(f \cdot g)'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x).</math>

			:Note the order of the factors; when ''R'' is not commutative this is important.

			These two properties make ''D'' a [[derivation (abstract algebra)\|derivation]] on ''A'' (see also [[module of relative differential forms]] for a discussion of a generalization).

			==Application to finding repeated factors==
			As in calculus, the derivative detects multiple roots: if ''R'' is a field then ''R''[''x''] is a [[Euclidean domain]], and in this situation we can define multiplicity of roots; namely, for every polynomial ''f''(''x'') and every element ''r'' of ''R'', there exists a nonnegative integer ''m<sub>r</sub>'' and a polynomial ''g''(''x'') such that

			:<math>f(x) = (x - r)^{m_r} g(x)</math>

			where ''g''(''r'') is not equal to ''0''. ''m<sub>r</sub>'' is the multiplicity of ''r'' as a root of ''f''. It follows from the Leibniz rule that in this situation, ''m<sub>r</sub>'' is also the number of differentiations that must be performed on ''f''(''x'') before ''r'' is not a root of the resulting polynomial. The utility of this observation is that although in general not every polynomial of degree ''n'' in ''R''[''x''] has ''n'' roots counting multiplicity (this is the maximum, by the above theorem), we may pass to [[field extension]]s in which this is true (namely, [[algebraic closure]]s). Once we do, we may uncover a multiple root that was not a root at all simply over ''R''. For example, if ''R'' is the field with three elements, the polynomial

			:<math>f(x)\,=\,x^6 + 1</math>

			has no roots in ''R''; however, its formal derivative is zero since 3 = 0 in ''R'' and in any extension of ''R'', so when we pass to the algebraic closure it has a multiple root that could not have been detected by factorization in ''R'' itself. Thus, formal differentiation allows an [[computability theory (computer science)\|effective]] notion of multiplicity. This is important in [[Galois theory]], where the distinction is made between [[separable field extension]]s (defined by polynomials with no multiple roots) and inseparable ones.

			==Correspondence to analytic derivative==
			When the ring ''R'' of scalars is commutative, there is an alternative and equivalent definition of the formal derivative, which resembles the one seen in differential calculus. The element Y-X of the ring ''R''[X,Y] divides Y<sup>n</sup> - X<sup>n</sup> for any nonnegative integer ''n'', and therefore divides ''f''(Y) - ''f''(X) for any polynomial ''f'' in one indeterminate. If we denote the quotient (in ''R''[X,Y]) by ''g'':

			:<math>g(X,Y) = \frac{f(Y) - f(X)}{Y - X}</math>

			then it is not hard to verify that ''g''(X,X) (in ''R''[X]) coincides with the formal derivative of ''f'' as it was defined above.

			This formulation of the derivative works equally well for a formal power series, assuming only that the ring of scalars is commutative.

			Actually, if the division in this definition is carried out in the class of functions of <math>Y</math> continuous at <math>X</math>, it will recapture the classical definition of the derivative. If it is carried out in the class of functions continuous in both <math>X</math> and <math>Y</math>, we get uniform differentiability, and our function <math>f</math> will be continuously differentiable. Likewise, by choosing different classes of functions (say, the Lipschitz class), we get different flavors of differentiability. This way differentiation becomes a part of algebra of functions.

			==See also==

			* [[Derivative]]
			* [[Euclidean domain]]
			* [[Module of relative differential forms]]
			* [[Galois theory]]
			* [[Formal power series]]
			* [[Pincherle derivative]]

			==References==
			* {{Lang Algebra\|edition=3r}}
			* [http://arxiv.org/abs/0905.3611 Michael Livshits, You could simplify calculus, arXiv:0905.3611v1 ]

			[[Category:Abstract algebra]]

en>Qetuth: more specific stub type + linked maths

2011-12-18T11:43:44Z

more specific stub type + linked maths

New page

They contact me Emilia. Body building is one of the things I love most. Bookkeeping is what I do. California is our birth place.<br><br>Look at my web site - [http://xrambo.com/blog/192034 http://xrambo.com/blog/192034]