The function f(x,y) = x√(1+y²) + y√(1+x²) is under scrutiny for its associativity. The discussion reveals that proving its associativity algebraically is complex, despite numerical evidence suggesting equality for specific values. Participants suggest using algebraic manipulation techniques such as isolating terms and squaring both sides to simplify the proof. Additionally, numerical methods using Maple 11 confirm that the function behaves consistently under certain conditions, although a symbolic proof remains elusive.
PREREQUISITESMathematicians, students in advanced algebra or calculus courses, and anyone interested in exploring the properties of complex functions and their proofs.
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) .