Show this function is associative (or provide a counter example)

[Dick]

What identity are you using?

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

 

[ArcanaNoir]

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

but is it true that sinh(a+b)= sinh(a)+sinh(b)?

 

[Dick]

but is it true that sinh(a+b)= sinh(a)+sinh(b)?

Definitely not! That's not what I'm doing. Look at it again.

 

[ArcanaNoir]

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

arcsinh(sinh( arcsinh(x)+arcsinh(y) ))=arcsinh(x)+arcsinh(y) is what I and ILS mean.

Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.

 

[Dick]

Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.

A=arcsinh(x)+arcsinh(y). This might be a little hard to see because it's really simple and you are expecting something complicated.

 

[ArcanaNoir]

A=arcsinh(x)+arcsinh(y). This might be a little hard to see because it's really simple and you are expecting something complicated.

ohhhh. thank you :) I was trying to work from the inside out.

 

[ArcanaNoir]

Holy crap, here it goes! Maybe I can be done with this problem now!

A:
(x*y)*z = sinh[arcsinh(x)+arcsinh(y)]*z
=sinh[arcsinh(sinh[arcsinh(x)+arcsinh(y)])+arcsinh(z)]
=sinh[arcsinh(x)+arcsinh(y)+arcsinh(z)]

B:
x*(y*z) = sinh(arcsinh(x)+arcsinh[y*z])
=sinh(arcsinh(x)+arcsinh[sinh(arcsinh(y)+arcsinh(z))])
=sinh(arcsinh(x)+arcsinh(y)+arcsinh(z))

Yeah?

 

[Ray Vickson]

Okay, you guys have been wonderful, especially Dick and Micro. Although I'm beginning to suspect Micro is a genius, in which case he sneezes and answers pop out of him, so it's not like he has to try really hard. :P j/k I appreciate your tireless efforts Micro! (not kidding about thinking you're a genius...)
Anyway I'm trying to use Micro's hint but I feel like I'm slipping past the part where I'm supposed to rearrange the expressions to show they are equivalent. I will show my simplification for the expressions, maybe someone can point out where I was supposed to do something trig-y.

I have verified that $\cosh (\sinh ^{-1}(x))=\sqrt{1+x^2}$ and that $f(x,y)=\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))$

A:
$$(x\ast y)\ast z = \sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))\ast z \\$$ $$= \sinh (\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]+\sinh ^{-1} (z)) \\<br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\cosh (\sinh ^{-1} (z))+\\<br /> \cosh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\sinh (\sinh ^{-1} (z)) \\ <br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} (y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot \sqrt{1+z^2} +\\<br /> \cosh [\sinh ^{-1} [ \sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1}(y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot z \\<br /> \sinh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]\sqrt{1+z^2}+\cosh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]z$$

B:
$$x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh [\sinh ^{-1} (x) + \sinh ^{-1} [\sinh (\sinh ^{-1}(y))\cosh (\sinh ^{-1} (z))+\\<br /> \cosh (\sinh ^{-1} (y))\sinh (\sinh ^{-1} (z))]] \\<br /> =\sinh [ \sinh ^{-1}(x)+\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}]] \\<br /> =\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}])+\\<br /> \cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1}[y\sqrt{1+z^2}+z\sqrt{1+y^2}]) \\<br /> =x\sqrt{1+(y\sqrt{1+z^2}+z\sqrt{1+y^2})^2}+[y\sqrt{1+z^2}+z\sqrt{1+y^2}]\sqrt{1+x^2}$$

So it seems to me that this isn't going to end any better than when I didn't use hyp. trig, Hence why I think I'm missing the critical point.

No! From $f(u,v) = g\left( g^{-1}(u) + g^{-1}(v) \right)$, where $g(.) = \sinh(.)$,
it follows that
$$f(f(x,y),z) = g\left( g^{-1}(f(x,y)) + g^{-1}(z)\right) = g \left( g^{-1}(x) + g^{-1}(y) + g^{-1}(z) \right) .$$

 

[ArcanaNoir]

No! From $f(u,v) = g\left( g^{-1}(u) + g^{-1}(v) \right)$, where $g(.) = \sinh(.)$,
it follows that
$$f(f(x,y),z) = g\left( g^{-1}(f(x,y)) + g^{-1}(z)\right) = g \left( g^{-1}(x) + g^{-1}(y) + g^{-1}(z) \right) .$$

Yeah... I got that finally.

 

[ArcanaNoir]

Another huge thank you to everyone who helped with this problem, especially Ray, Dick, and Micro, who invested quite some time in helping me. I really really appreciate it, I couldn't have done it without all the outstanding help!