ArcanaNoir said:What identity are you using?
Dick said:The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.
ArcanaNoir said:but is it true that sinh(a+b)= sinh(a)+sinh(b)?
Dick said:The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.
Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.Dick said:arcsinh(sinh( arcsinh(x)+arcsinh(y) ))=arcsinh(x)+arcsinh(y) is what I and ILS mean.
ArcanaNoir said:Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.
Dick said: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 said: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 \\ <br /> = \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.
Ray Vickson said: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) .