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- #1

- Mar 5, 2012

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Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?

- Thread starter Klaas van Aarsen
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- Thread starter
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- #1

- Mar 5, 2012

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Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?

- Jan 26, 2012

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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

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- #4

- Mar 5, 2012

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Yep. That's the one.

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

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

- Jan 30, 2012

- 2,492

Use hyperbolic functions.

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- #6

- Mar 5, 2012

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That's right!

But how?

But how?

[tex] f(f(x,y),z) = f \sqrt{ 1 + z^2} + z \sqrt{1 + f^2 } [/tex]

Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?

[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

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- #8

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

- Jan 30, 2012

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- #10

- Mar 5, 2012

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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 $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.

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.

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Nice insight!

Last edited: