The discussion revolves around the function f(x,y)=x√(1+y²)+y√(1+x²) and the task of demonstrating its associativity or providing a counterexample. Participants are exploring the properties of this function within the context of real numbers.
The discussion is ongoing, with participants sharing various strategies and insights. Some have attempted numerical verification using software like Maple, leading to observations about roundoff errors and Taylor expansions. There is a recognition of the complexity involved in proving the associativity symbolically.
Participants have noted the need to consider cases where x, y, and z are non-negative, while also expressing the desire to extend the proof to all real numbers. There is an acknowledgment of the potential challenges in symbolic computation related to this problem.
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) .