Diferencia entre revisiones de «Lagrange identity»
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== Lagrange's identity obtained from product identity == | |||
Lagrange's identity obtained from product identity | |||
We present an identity of products that reduces to Lagrange's identity | We present an identity of products that reduces to Lagrange's identity | ||
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derivation has induced us to present it for complex numbers. | derivation has induced us to present it for complex numbers. | ||
== Lagrange's identity for complex numbers == | |||
Lagrange's identity for complex numbers | |||
Let <math>a_{i},b_{i}\in\mathbb{C}</math> be complex numbers and the overbar | Let <math>a_{i},b_{i}\in\mathbb{C}</math> be complex numbers and the overbar | ||
Línea 43: | Línea 41: | ||
Expand the product on the LHS of the product identity in terms of | Expand the product on the LHS of the product identity in terms of | ||
series | series up to fourth order | ||
Recall that products of the form <math>\left(1+x_{i}\right)</math> can be expanded | |||
in terms of sums as | in terms of sums as | ||
<math> | <math> | ||
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where <math>\mathcal{O}^{3+}(x)</math> means terms with order three or higher | where <math>\mathcal{O}^{3+}(x)</math> means terms with order three or higher | ||
in <math>x</math>. | in <math>x</math>. | ||
<math> | <math> | ||
\prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}-b_{i}\bar{b}_{i}+a_{i}\bar{a}_{i}b_{i}\bar{b}_{i}\right)=1-\sum_{i=1}^{n}\left(a_{i}\bar{a}_{i}+b_{i}\bar{b}_{i}\right)+\sum_{i=1}^{n}a_{i}\bar{a}_{i}b_{i}\bar{b}_{i} | \prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}-b_{i}\bar{b}_{i}+a_{i}\bar{a}_{i}b_{i}\bar{b}_{i}\right)=1-\sum_{i=1}^{n}\left(a_{i}\bar{a}_{i}+b_{i}\bar{b}_{i}\right)+\sum_{i=1}^{n}a_{i}\bar{a}_{i}b_{i}\bar{b}_{i} | ||
+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)+\mathcal{O}^{5+}. | +\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)+\mathcal{O}^{5+}. | ||
</math> | </math> | ||
The two factors on the RHS are also written in terms of series | The two factors on the RHS are also written in terms of series | ||
<math> | <math> | ||
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<math> | <math> | ||
\prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}\right)\prod_{i=1}^{n}\left(1-b_{i}\bar{b}_{i}\right)=1-\sum_{i=1}^{n}\left(a_{i}\bar{a}_{i}+b_{i}\bar{b}_{i}\right) | \prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}\right)\prod_{i=1}^{n}\left(1-b_{i}\bar{b}_{i}\right)=1-\sum_{i=1}^{n}\left(a_{i}\bar{a}_{i}+b_{i}\bar{b}_{i}\right) | ||
+\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}\right)+\mathcal{O}^{5+}. | +\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}\right)+\mathcal{O}^{5+}. | ||
</math> | </math> | ||
Substitution of | Substitution of these two results in the product identity give | ||
in the product identity give | |||
<math> | <math> | ||
\sum_{i=1}^{n}a_{i}\bar{a}_{i}b_{i}\bar{b}_{i}+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | \sum_{i=1}^{n}a_{i}\bar{a}_{i}b_{i}\bar{b}_{i}+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | ||
</math> | </math> | ||
The product of two conjugates series can be expressed as series involving | The product of two conjugates series can be expressed as series involving | ||
the product of conjugate terms | the product of conjugate terms | ||
The conjugate series product is <math>\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\bar{x}_{i}\right)=\sum_{i=1}^{n}x_{i}\bar{x}_{i}+\sum_{i<j}^{n}\left(x_{i}\bar{x}_{j}+\bar{x}_{i}x_{j}\right)</math>.% | |||
}, thus | }, thus | ||
<math> | <math> | ||
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=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | =\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | ||
</math> | </math> | ||
The terms of the last two series on the LHS are grouped as | The terms of the last two series on the LHS are grouped as | ||
<math> | <math> | ||
a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}-a_{i}b_{i}\bar{a}_{j}\bar{b}_{j}-\bar{a}_{i}\bar{b}_{i}a_{j}b_{j}=\left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\bar{a}_{i}b_{j}-\bar{a}_{j}b_{i}\right), | a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}-a_{i}b_{i}\bar{a}_{j}\bar{b}_{j}-\bar{a}_{i}\bar{b}_{i}a_{j}b_{j}=\left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\bar{a}_{i}b_{j}-\bar{a}_{j}b_{i}\right), | ||
</math> | </math> | ||
in order to obtain the complex Lagrange's identity | in order to obtain the complex Lagrange's identity | ||
<math> | <math> | ||
\left(\sum_{i=1}^{n}a_{i}b_{i}\right)\left(\sum_{i=1}^{n}\overline{a_{i}b_{i}}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\overline{a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | \left(\sum_{i=1}^{n}a_{i}b_{i}\right)\left(\sum_{i=1}^{n}\overline{a_{i}b_{i}}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\overline{a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right). | ||
</math> | </math> | ||
In terms of the modulii, | In terms of the modulii, | ||
<math> | <math> | ||
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</math> | </math> | ||
== other identities == | |||
other identities | |||
The non trivial identities for real numbers obtained to sixth order | The non trivial identities for real numbers obtained to sixth order | ||
series expansion of the product identity <math>\prod_{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=\prod_{i=1}^{n}\left(1-a_{i}^{2}\right)\prod_{i=1}^{n}\left(1-b_{i}^{2}\right)</math> | series expansion of the product identity <math>\prod_{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=\prod_{i=1}^{n}\left(1-a_{i}^{2}\right)\prod_{i=1}^{n}\left(1-b_{i}^{2}\right)</math> | ||
are | are | ||
<math> | <math> | ||
\sum_{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)\right]+\sum_{i<j<k}^{n}\left[a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right]=\left(\sum_{i=1}^{n}b_{i}^{2}\right)\left(\sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\right) | \sum_{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)\right]+\sum_{i<j<k}^{n}\left[a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right]=\left(\sum_{i=1}^{n}b_{i}^{2}\right)\left(\sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\right) | ||
</math> | </math> | ||
and its counterpart, obtained by interchanging the | |||
and its counterpart, obtained by interchanging the variables <math>a</math> and <math>b</math>. | |||
Expand the product identity in series up to sixth order. The LHS is | Expand the product identity in series up to sixth order. The LHS is | ||
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involving <math>a^{4}b^{2}</math> and <math>a^{2}b^{4}</math> give the non trivial sixth | involving <math>a^{4}b^{2}</math> and <math>a^{2}b^{4}</math> give the non trivial sixth | ||
order identities | order identities | ||
<math> | <math> | ||
\sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+ | \sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+ \sum_{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)= & \left(\sum_{i=1}^{n}b_{i}^{2}\right)\left(\sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\right) | ||
\sum_{i<j}^{n}b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)+ | </math> | ||
<math> | |||
\sum_{i<j}^{n}b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)+ \sum_{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)= & \left(\sum_{i=1}^{n}a_{i}^{2}\right)\left(\sum_{i<j}^{n}b_{i}^{2}b_{j}^{2}\right). | |||
</math> | </math> | ||
=== conclusions === | |||
Lagrange's identity for complex numbers has been obtained from a straight-forward | Lagrange's identity for complex numbers has been obtained from a straight-forward | ||
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proof is yet another way to obtain the CS inequality. | proof is yet another way to obtain the CS inequality. | ||
<references/> | |||
<ref> P.~Fjelstad and S.~G. Gal., ''n-dimensional hyperbolic complex numbers'', Adv. Appl. Clifford Alg., 1998 [8(1), pag.47–68] </ref> | |||
P.~Fjelstad and S.~G. Gal. | P.~Fjelstad and S.~G. Gal. | ||
\newblock {n-dimensional hyperbolic complex numbers}. | \newblock {n-dimensional hyperbolic complex numbers}. |
Revisión del 12:16 19 ago 2012
Lagrange's identity obtained from product identity
We present an identity of products that reduces to Lagrange's identity when a series expansion to fourth order terms are considered. Sixth and higher order terms produce other series identities.
Normed division algebras require that the norm of the product is equal to the product of the norms. Lagrange's identity exhibits this equality. Due to Hurwitz theorem, it admits this interpretation only for algebras isomorphic to the real numbers, complex numbers, quaternions and octonions . If divisors of zero are allowed, many other algebraic structures in are possible. Two such possibilities for hyperbolic numbers has been introduced by Fjelstad and Gal \cite{fjelstad+gal1998} and more recently by Catoni et al. \cite{catoni2005}. Another approach has been presented in the context of a deformed Lorentz metric. This latter proposal is based on a transformation stemming from the product operation and magnitude definition in hyperbolic scator algebra \cite{fernandez-guasti2011_2}. The product identity used as a starting point here, is a consequence of the equality for scator algebras.
Lagrange's identity can be proved in a variety of ways \cite{Steele2004}. Most derivations use the identity as a starting point and prove in one way or another that the equality is true. In the present approach, Lagrange's identity is actually derived without assuming it \emph{a priori}. The pseudo-norm of the product identity used in the derivation has the strength to imply an infinite number of identities. An example when sixth order terms are retained is shown here. The ease of the derivation has induced us to present it for complex numbers.
Lagrange's identity for complex numbers
Let be complex numbers and the overbar represents complex conjugate.
The product identity reduces to the complex Lagrange's identity when fourth order terms, in a series expansion, are considered.
Expand the product on the LHS of the product identity in terms of series up to fourth order
Recall that products of the form can be expanded in terms of sums as where means terms with order three or higher in .
The two factors on the RHS are also written in terms of series The product of this expression up to fourth order is Substitution of these two results in the product identity give The product of two conjugates series can be expressed as series involving the product of conjugate terms
The conjugate series product is .% }, thus
The terms of the last two series on the LHS are grouped as
in order to obtain the complex Lagrange's identity
In terms of the modulii,
other identities
The non trivial identities for real numbers obtained to sixth order series expansion of the product identity are
and its counterpart, obtained by interchanging the variables and .
Expand the product identity in series up to sixth order. The LHS is Consider only the sixth order terms The RHS of the product identity is similarly expanded in series up to sixth order
and only sixth order terms retained These two results are equated for equal powers of . The terms and give trivial identities whereas the terms involving and give the non trivial sixth order identities
Error al representar (error de sintaxis): {\displaystyle \sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+ \sum_{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)= & \left(\sum_{i=1}^{n}b_{i}^{2}\right)\left(\sum_{i<j}^{n}a_{i}^{2}a_{j}^{2}\right) }
Error al representar (error de sintaxis): {\displaystyle \sum_{i<j}^{n}b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)+ \sum_{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)= & \left(\sum_{i=1}^{n}a_{i}^{2}\right)\left(\sum_{i<j}^{n}b_{i}^{2}b_{j}^{2}\right). }
conclusions
Lagrange's identity for complex numbers has been obtained from a straight-forward product identity. The procedure is elementary and very economical. A derivation for the reals is obviously even more succinct. In a wider context, this product identity can be seen as a consequence of the relationship for scator algebras. Since the Cauchy–Schwarz inequality is a particular case of Lagrange's identity \cite{Steele2004}, this proof is yet another way to obtain the CS inequality.
[1] P.~Fjelstad and S.~G. Gal. \newblock {n-dimensional hyperbolic complex numbers}. \newblock {\em Adv. Appl. Clifford Alg.}, 8(1):47–68, 1998.
F.~Catoni, R.~Cannata, E.~Nichelatti, and P.~Zampetti.
\newblock {Commmutative hypercomplex numbers and functions of hypercomplex
variable: a matrix study}.
\newblock {\em Advances in Applied Clifford Algebras}, 15(2):183–212, 2005.
M.~Fernández-Guasti.
\newblock {Alternative realization for the composition of relativistic
velocities}.
\newblock In {\em {Optics and Photonics 2011}}, volume 8121 of {\em {The nature
of light: What are photons? IV}}, page 812108–1–11. SPIE, 2011.
J.~Michael Steele.
\newblock {\em {Cauchy-Schwarz master class: an introduction to the art of
mathematical inequalities}}.
\newblock CUP, 2004.
- ↑ P.~Fjelstad and S.~G. Gal., n-dimensional hyperbolic complex numbers, Adv. Appl. Clifford Alg., 1998 [8(1), pag.47–68]