Lagrange identity

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. An extended version of these results are available in an open source journal .

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 $\mathbb {R} ^{n}$ are possible , . One 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 . The product identity used as a starting point here, is a consequence of the product of the norms equality for scator algebras.

Lagrange's identity can be proved in a variety of ways . 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 a priori.

Lagrange's identity for complex numbers

Let $a_{i},b_{i}\in \mathbb {C}$ be complex numbers and the overbar represents complex conjugate.

The product identity $\prod _{i=1}^{n}\left(1-a_{i}{\bar {a}}_{i}-b_{i}{\bar {b}}_{i}+a_{i}{\bar {a}}_{i}b_{i}{\bar {b}}_{i}\right)=\prod _{i=1}^{n}\left(1-a_{i}{\bar {a}}_{i}\right)\prod _{i=1}^{n}\left(1-b_{i}{\bar {b}}_{i}\right)$ 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. To this end, recall that products of the form $\left(1+x_{i}\right)$ can be expanded in terms of sums as $\prod _{i=1}^{n}\left(1+x_{i}\right)=1+\sum _{i=1}^{n}x_{i}+\sum _{i<j}^{n}x_{i}x_{j}+{\mathcal {O}}^{3+}(x),$ where ${\mathcal {O}}^{3+}(x)$ means terms with order three or higher in $x$.

$\prod _{i=1}^{n}\left(1-a_{i}{\bar {a}}_{i}-b_{i}{\bar {b}}_{i}+a_{i}{\bar {a}}_{i}b_{i}{\bar {b}}_{i}\right)=1-\sum _{i=1}^{n}\left(a_{i}{\bar {a}}_{i}+b_{i}{\bar {b}}_{i}\right)+\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}b_{i}{\bar {b}}_{i}+\sum _{i<j}^{n}\left(a_{i}{\bar {a}}_{i}a_{j}{\bar {a}}_{j}+b_{i}{\bar {b}}_{i}b_{j}{\bar {b}}_{j}\right)+\sum _{i<j}^{n}\left(a_{i}{\bar {a}}_{i}b_{j}{\bar {b}}_{j}+a_{j}{\bar {a}}_{j}b_{i}{\bar {b}}_{i}\right)+{\mathcal {O}}^{5+}.$

The two factors on the RHS are also written in terms of series $\prod _{i=1}^{n}\left(1-a_{i}{\bar {a}}_{i}\right)\prod _{i=1}^{n}\left(1-b_{i}{\bar {b}}_{i}\right)=\left(1-\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}+\sum _{i<j}^{n}a_{i}{\bar {a}}_{i}a_{j}{\bar {a}}_{j}+{\mathcal {O}}^{5+}\right)\left(1-\sum _{i=1}^{n}b_{i}{\bar {b}}_{i}+\sum _{i<j}^{n}b_{i}{\bar {b}}_{i}b_{j}{\bar {b}}_{j}+{\mathcal {O}}^{5+}\right).$

The product of this expression up to fourth order is

$\prod _{i=1}^{n}\left(1-a_{i}{\bar {a}}_{i}\right)\prod _{i=1}^{n}\left(1-b_{i}{\bar {b}}_{i}\right)=1-\sum _{i=1}^{n}\left(a_{i}{\bar {a}}_{i}+b_{i}{\bar {b}}_{i}\right)+\left(\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}\right)\left(\sum _{i=1}^{n}b_{i}{\bar {b}}_{i}\right)+\sum _{i<j}^{n}\left(a_{i}{\bar {a}}_{i}a_{j}{\bar {a}}_{j}+b_{i}{\bar {b}}_{i}b_{j}{\bar {b}}_{j}\right)+{\mathcal {O}}^{5+}.$

Substitution of these two results in the product identity give $\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}b_{i}{\bar {b}}_{i}+\sum _{i<j}^{n}\left(a_{i}{\bar {a}}_{i}b_{j}{\bar {b}}_{j}+a_{j}{\bar {a}}_{j}b_{i}{\bar {b}}_{i}\right)=\left(\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}\right)\left(\sum _{i=1}^{n}b_{i}{\bar {b}}_{i}\right).$

The product of two conjugates series can be expressed as series involving the product of conjugate terms

The conjugate series product is $\left(\sum _{i=1}^{n}x_{i}\right)\left(\sum _{i=1}^{n}{\bar {x}}_{i}\right)=\sum _{i=1}^{n}x_{i}{\bar {x}}_{i}+\sum _{i<j}^{n}\left(x_{i}{\bar {x}}_{j}+{\bar {x}}_{i}x_{j}\right)$.% }, thus

$\left(\sum _{i=1}^{n}a_{i}b_{i}\right)\left(\sum _{i=1}^{n}{\overline {a_{i}b_{i}}}\right)-\sum _{i<j}^{n}\left(a_{i}b_{i}{\bar {a}}_{j}{\bar {b}}_{j}+{\bar {a}}_{i}{\bar {b}}_{i}a_{j}b_{j}\right)+\sum _{i<j}^{n}\left(a_{i}{\bar {a}}_{i}b_{j}{\bar {b}}_{j}+a_{j}{\bar {a}}_{j}b_{i}{\bar {b}}_{i}\right)=\left(\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}\right)\left(\sum _{i=1}^{n}b_{i}{\bar {b}}_{i}\right).$

The terms of the last two series on the LHS are grouped as $a_{i}{\bar {a}}_{i}b_{j}{\bar {b}}_{j}+a_{j}{\bar {a}}_{j}b_{i}{\bar {b}}_{i}-a_{i}b_{i}{\bar {a}}_{j}{\bar {b}}_{j}-{\bar {a}}_{i}{\bar {b}}_{i}a_{j}b_{j}=\left(a_{i}{\bar {b}}_{j}-a_{j}{\bar {b}}_{i}\right)\left({\bar {a}}_{i}b_{j}-{\bar {a}}_{j}b_{i}\right),$

in order to obtain the complex Lagrange's identity

$\left(\sum _{i=1}^{n}a_{i}b_{i}\right)\left(\sum _{i=1}^{n}{\overline {a_{i}b_{i}}}\right)+\sum _{i<j}^{n}\left(a_{i}{\bar {b}}_{j}-a_{j}{\bar {b}}_{i}\right)\left({\overline {a_{i}{\bar {b}}_{j}-a_{j}{\bar {b}}_{i}}}\right)=\left(\sum _{i=1}^{n}a_{i}{\bar {a}}_{i}\right)\left(\sum _{i=1}^{n}b_{i}{\bar {b}}_{i}\right).$

In terms of the modulii,

$\left|\sum _{i=1}^{n}a_{i}b_{i}\right|^{2}+\sum _{i<j}^{n}\left|a_{i}{\bar {b}}_{j}-a_{j}{\bar {b}}_{i}\right|^{2}=\left(\sum _{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum _{i=1}^{n}\left|b_{i}\right|^{2}\right).$

other identities

The non trivial identities for real numbers obtained to sixth order series expansion of the product identity $\prod _{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=\prod _{i=1}^{n}\left(1-a_{i}^{2}\right)\prod _{i=1}^{n}\left(1-b_{i}^{2}\right)$ are

$\sum _{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)\right]+\sum _{i<j<k}^{n}\left[a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right]=\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right)$

and its counterpart, obtained by interchanging the variables $a$ and $b$.

Expand the product identity in series up to sixth order. The LHS is $\prod _{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=1+\sum _{i=1}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)+\sum _{i<j}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)\left(-a_{j}^{2}-b_{j}^{2}+a_{j}^{2}b_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)\left(-a_{j}^{2}-b_{j}^{2}+a_{j}^{2}b_{j}^{2}\right)\left(-a_{k}^{2}-b_{k}^{2}+a_{k}^{2}b_{k}^{2}\right)+{\mathcal {O}}^{7+}.$ Consider only the sixth order terms ${\mathcal {O}}^{6}\left({\textrm {LHS}}\right)=-\sum _{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)\right]-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}+b_{i}^{2}b_{j}^{2}b_{k}^{2}-\sum _{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)-\sum _{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)$ The RHS of the product identity is similarly expanded in series up to sixth order

$\prod _{i=1}^{n}\left(1-a_{i}^{2}\right)\prod _{i=1}^{n}\left(1-b_{i}^{2}\right)=\left(1-\sum _{i=1}^{n}a_{i}^{2}+\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}+{\mathcal {O}}^{7+}\right)\left(1-\sum _{i=1}^{n}b_{i}^{2}+\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}-\sum _{i<j<k}^{n}b_{i}^{2}b_{j}^{2}b_{k}^{2}+{\mathcal {O}}^{7+}\right),$ and only sixth order terms retained ${\mathcal {O}}^{6}\left({\textrm {RHS}}\right)=-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}-\sum _{i<j<k}^{n}b_{i}^{2}b_{j}^{2}b_{k}^{2}-\left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\right)-\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right).$ These two results are equated for equal powers of $a^{n}b^{m}$. The terms $a^{6}$ and $b^{6}$ give trivial identities whereas the terms involving $a^{4}b^{2}$ and $a^{2}b^{4}$ give the non trivial sixth order identities

$\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)=\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right)$

$\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)=\left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\right).$

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 $\left\Vert \mathbf {ab} \right\Vert =\left\Vert \mathbf {a} \right\Vert \left\Vert \mathbf {b} \right\Vert$ relationship for scator algebras. Since the Cauchy–Schwarz inequality is a particular case of Lagrange's identity , this proof is yet another way to obtain the CS inequality.

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