- Thread starter
- Admin
- #1
- Mar 5, 2012
- 9,006
There was a thread on another forum that I'd like to share.
Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?
Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?
I think in this case the associative property is $f(f(a, b), c) = f(a, f(b, c))$? My best guesswhat do you mean by associative function ? I know the associative property
[tex] f(f(x,y),z) = f \sqrt{ 1 + z^2} + z \sqrt{1 + f^2 } [/tex]There was a thread on another forum that I'd like to share.
Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?
Also, $\cosh a$ is always positive, so we can write $\sqrt{\sinh^2 a + 1} = \cosh a$. Thus $$f(x,y) = \sinh u\cosh v + \cosh u\sinh v = \sinh(u+v).$$ It follows that $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.[tex] f(f(x,y),z) = f \sqrt{ 1 + z^2} + z \sqrt{1 + f^2 } [/tex]
[tex] f(f(x,y),z) = \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right) \sqrt{ 1 + z^2} + z \sqrt{1 + \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right)^2 } [/tex]
Let
[tex] x = \sinh u , y = \sinh v , z = \sinh w [/tex]
and note
[tex] \sinh^2 a + 1 = \cosh^2 a [/tex]
Yep!Also, $\cosh a$ is always positive, so we can write $\sqrt{\sinh^2 a + 1} = \cosh a$. Thus $$f(x,y) = \sinh u\cosh v + \cosh u\sinh v = \sinh(u+v).$$ It follows that $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.
Nice insight!Yes, in other words, $\sinh$ is a surjective homomorphism (in fact, an isomorphism) from $\mathbb{R}$ with $+$ to $\mathbb{R}$ with $f$. For this problem, it is only important that $\sinh$ is surjective and respects the operations, but it can be viewed as an isomorphism of abelian groups $\langle\mathbb{R},{+},0,{-}\rangle$ and $\langle\mathbb{R},f,0,{-}\rangle$.