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[SOLVED] Associativity puzzle

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Klaas van Aarsen

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Mar 5, 2012
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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?
 

Amer

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Mar 1, 2012
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what do you mean by associative function ? I know the associative property
 

Bacterius

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what do you mean by associative function ? I know the associative property
I think in this case the associative property is $f(f(a, b), c) = f(a, f(b, c))$? My best guess :confused:
 
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Klaas van Aarsen

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Yep. That's the one.

Does $f(x,f(y,z)) = f(f(x,y),z)$ hold for all $x, y, z \in \mathbb R$?
 

Evgeny.Makarov

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Jan 30, 2012
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As the famous math joke goes, "Yes, it IS obvious!" (Notices of the AMS, 52:1, PDF).

Use hyperbolic functions.
 
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Klaas van Aarsen

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That's right!
But how? :rolleyes:
 

Amer

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Mar 1, 2012
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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?
[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]

[tex] f(f(u,v),w) = \left( \sinh u \mid \cosh v \mid + \sinh v \mid \cosh u \mid\right) \mid \cosh w \mid + \sinh w \sqrt{1 + \left( \sinh u \mid \cosh v \mid + \sinh v \mid \cosh u \mid \right)^2 } [/tex]

the other one
[tex] f(x, f(y,z)) = x \sqrt{ 1 + f^2} + f \sqrt{1 + x^2 } [/tex]
as the previous
[tex]f(y,z) = y \sqrt{ 1 + z^2} + z\sqrt{1+ y^2} [/tex]
[tex] f(v,w) = \sinh u \mid \cosh w \mid + \sinh w \mid \cosh v \mid [/tex]

[tex] f(u,f(v,w)) = \sinh u \sqrt{ 1 + \left( \sinh v \mid \cosh w \mid + \sinh w \mid \cosh v\mid \right)^2 } + \left(\sinh u \mid \cosh w \mid + \sinh w \mid \cosh v \mid \right) \mid \cosh u \mid [/tex]

i cant see how they are the same, if i did it right at the first place :D
 

Opalg

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Feb 7, 2012
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[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]
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 $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.
 

Evgeny.Makarov

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Jan 30, 2012
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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 $\langle\mathbb{R},f,0,{-}\rangle$.
 

Klaas van Aarsen

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Mar 5, 2012
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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 $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.
Yep!
That takes care of it.


Follow-up extended solution:
$$f(x,y) = \sinh(\sinh^{-1}x+\sinh^{-1}y)$$
So:
\begin{aligned}
f\big(x,f(y,z)\big) &= \sinh \Big(\sinh^{-1}x+\sinh^{-1}\big(\sinh(\sinh^{-1}y+\sinh^{-1}z)\big)\Big) \\
&= \sinh\big(\sinh^{-1}x+(\sinh^{-1}y+\sinh^{-1}z)\big) \\
&= \sinh\big((\sinh^{-1}x+\sinh^{-1}y)+\sinh^{-1}z)\big) \\
&= f\big(f(x,y),z\big) & \blacksquare
\end{aligned}


More generally, any $f(x,y)$ of the form
$$f(x,y) = g\big(g^{-1}(x) + g^{-1}(y)\big)$$
is associative.

- - - Updated - - -

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 $\langle\mathbb{R},f,0,{-}\rangle$.
Nice insight!
 
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