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

[ArcanaNoir]

Homework Statement

I'm trying to show that f(x,y)=x\sqrt{1+y^2}+y\sqrt{1+x^2} is associative, or provide a counter example. (All variables are real numbers.)

Homework Equations

The Attempt at a Solution

I have it down to needing to show that x\sqrt{(1+y^2)(1+z^2)}+z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} is equal to z\sqrt{(1+x^2)(1+y^2)}+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Playing with actual numbers I have always been able to see that they are indeed quite equal, yet algebraically I cannot prove it with arbitrary variables. I have been working on it for several days, and have tried expanding, completing the square, and everything else I can think of, but no luck

 

[SammyS]

Homework Statement

I'm trying to show that f(x,y)=x\sqrt{1+y^2}+y\sqrt{1+x^2} is associative, or provide a counter example. (All variables are real numbers.)

Homework Equations

The Attempt at a Solution

I have it down to needing to show that x\sqrt{(1+y^2)(1+z^2)}+z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} is equal to z\sqrt{(1+x^2)(1+y^2)}+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Playing with actual numbers I have always been able to see that they are indeed quite equal, yet algebraically I cannot prove it with arbitrary variables. I have been working on it for several days, and have tried expanding, completing the square, and everything else I can think of, but no luck

It is clearly commutative. Can that fact help you?

 

[ArcanaNoir]

It is clearly commutative. Can that fact help you?

Not that I can see, did you have something specific in mind? Commutativity doesn't imply associativity by itself. Based on your suggestion I tried changing some things like (y*x)*z and z*(y*x) and messing with order in that way but was unable to get anywhere with it.

 

[micromass]

Homework Statement

I'm trying to show that f(x,y)=x\sqrt{1+y^2}+y\sqrt{1+x^2} is associative, or provide a counter example. (All variables are real numbers.)

Homework Equations

The Attempt at a Solution

I have it down to needing to show that x\sqrt{(1+y^2)(1+z^2)}+z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} is equal to z\sqrt{(1+x^2)(1+y^2)}+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Playing with actual numbers I have always been able to see that they are indeed quite equal, yet algebraically I cannot prove it with arbitrary variables. I have been working on it for several days, and have tried expanding, completing the square, and everything else I can think of, but no luck

We need to prove that

x\sqrt{(1+y^2)(1+z^2)}+z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} =z\sqrt{(1+x^2)(1+y^2)}+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Assume that $x,y,z\geq 0$ first.

There are two terms with square roots inside square roots. Let's focus on those first. Isolate one of the terms:


z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} =(z\sqrt{(1+x^2)(1+y^2)}-x\sqrt{(1+y^2)(1+z^2)})+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Square both sides of the equations. You will end up with one term having a square root inside a square root. Isolate that term. Square both sides of the equation. You will end up with only terms which have square roots. Now you might be able to see more clearly why equality holds. If not, we need to keep on isolating terms and squaring them.

This is tedious, but it works.

 

[ArcanaNoir]

We need to prove that

x\sqrt{(1+y^2)(1+z^2)}+z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} =z\sqrt{(1+x^2)(1+y^2)}+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Assume that $x,y,z\geq 0$ first.

There are two terms with square roots inside square roots. Let's focus on those first. Isolate one of the terms:


z\sqrt{1+(x\sqrt{1+y^2}+y\sqrt{1+x^2})^2} =(z\sqrt{(1+x^2)(1+y^2)}-x\sqrt{(1+y^2)(1+z^2)})+x\sqrt{1+(z\sqrt{1+y^2}+y\sqrt{1+z^2})^2}

Square both sides of the equations. You will end up with one term having a square root inside a square root. Isolate that term. Square both sides of the equation. You will end up with only terms which have square roots. Now you might be able to see more clearly why equality holds. If not, we need to keep on isolating terms and squaring them.

This is tedious, but it works.

Okay, I will try this. Why do we have to assume they are greater than 0 though? I have to show it holds for the whole real line.

 

[micromass]

Because $x^2 = y^2$ and $x=y$ are not equivalent. They are only equivalent for positive numbers. So you should show it for positive numbers first, and then find an argument why it also holds for negative numbers.

 

[ArcanaNoir]

Because $x^2 = y^2$ and $x=y$ are not equivalent. They are only equivalent for positive numbers. So you should show it for positive numbers first, and then find an argument why it also holds for negative numbers.

Okay, I'll get working on it straight away. Thanks for the tips!

 

[ArcanaNoir]

Square both sides of the equations. You will end up with one term having a square root inside a square root. Isolate that term. Square both sides of the equation. You will end up with only terms which have square roots. Now you might be able to see more clearly why equality holds. If not, we need to keep on isolating terms and squaring them.

This is tedious, but it works.

Are you sure this is possible to do by hand? Like in 20 pages or less...

 

[micromass]

Are you sure this is possible to do by hand? Like in 20 pages or less...

Maybe you want to use a computer algebra system...

 

[ArcanaNoir]

Maybe you want to use a computer algebra system...

I was trying to use maple, but it was doing a horrible job. It won't expand things when I tell it to.

 

[ArcanaNoir]

holy crap. Just figured out how to expand it completely, it's equally as useless. It's ten lines in maple for the right hand side.

 

[LCKurtz]

So say you have rhs := right hand side. Calculate lhs and let Maple calculate lhs-rhs and see if it gives 0.

 

[Ray Vickson]

So say you have rhs := right hand side. Calculate lhs and let Maple calculate lhs-rhs and see if it gives 0.

I have tried this in Maple, and here is what happens: if we let
F(x,y,z) = f(x,f(y,z))^2-f(f(x,y),z)^2
and use 3d plotting to graph F(x0,y,z) for various fixed values of x0 and reasonable ranges of y and z, we always get a surface plot of random roundoff errors, with values of order 10^-(13) or less. So, numerically we do have confirmation.

We can also use Maple to get a multivariate Taylor expansion of F(x,y,z) at (0,0,0) up to various orders. For orders <= 30 the expansion is zero. Going much beyond order 30 takes patience, as the computation time increases rapidly: using Maple 11 on an HP Probook 4310s, the cpu times are 0.266 sec for order 20, 1.966 sec for order 25 and 9.313 sec for order 30. Anyway, we have some type of semi-confirmation that F = 0, since its multivariate derivatives of order <= 30 all vanish at (0,0,0).

What does not appear easy is to show F = 0 symbolically. I wonder if this is one of those cases where proving that A = B (for given symbolic A and B) is essentially un-doable; I read somewhere that there are some general results in symbolic computation that deal with such problems and that there are, indeed, such examples.

 

[ArcanaNoir]

I have tried this in Maple, and here is what happens: if we let
F(x,y,z) = f(x,f(y,z))^2-f(f(x,y),z)^2
and use 3d plotting to graph F(x0,y,z) for various fixed values of x0 and reasonable ranges of y and z, we always get a surface plot of random roundoff errors, with values of order 10^-(13) or less. So, numerically we do have confirmation.

We can also use Maple to get a multivariate Taylor expansion of F(x,y,z) at (0,0,0) up to various orders. For orders <= 30 the expansion is zero. Going much beyond order 30 takes patience, as the computation time increases rapidly: using Maple 11 on an HP Probook 4310s, the cpu times are 0.266 sec for order 20, 1.966 sec for order 25 and 9.313 sec for order 30. Anyway, we have some type of semi-confirmation that F = 0, since its multivariate derivatives of order <= 30 all vanish at (0,0,0).

What does not appear easy is to show F = 0 symbolically. I wonder if this is one of those cases where proving that A = B (for given symbolic A and B) is essentially un-doable; I read somewhere that there are some general results in symbolic computation that deal with such problems and that there are, indeed, such examples.

Thank you for looking at the plot, I had wondered about that but wasn't sure how to do it. Maybe I need to just admit to my professor that I'm stumped on this one. I suppose if I get all the rest of the problems it won't hurt my pride too bad. I wonder why he even wrote such an exercise!

 

[Ray Vickson]

Thank you for looking at the plot, I had wondered about that but wasn't sure how to do it. Maybe I need to just admit to my professor that I'm stumped on this one. I suppose if I get all the rest of the problems it won't hurt my pride too bad. I wonder why he even wrote such an exercise!

There must be some clever, alternative way to do it. What is the problem context (i.e., surrounding course content)? Maybe f(x,y) has some type of geometric interpretation that makes things clearer, or maybe such functions appear naturally in some types of applications. Do you have any suggestions?

 

[gerben]

Realize that f(x, y) = g(x, y) + g(y, x), then write both f(f(x, y), z) and f(z, f(x, y)) in terms of g and see that they are the equal.

 

[Dick]

Realize that f(x, y) = g(x, y) + g(y, x), then write both f(f(x, y), z) and f(z, f(x, y)) in terms of g and see that they are the equal.

Oh, that doesn't work. Did you try it? Put g(x,y)=x+y. That doesn't even work, even though f(x,y)=x+y does. I'd be really interested to know if there is an easy proof.

 

[Dick]

You can make it a little easier by noticing f(f(kx,ky),kz))-f(kx,f(ky,kz))=0 iff f(f(x,y),z)=f(x,f(y,z)) for k!=0. So you can choose say, k=1/z and now you only have to prove f(f(x,y),1)=f(x,f(y,1)). That's a pretty modest simplification, I'll admit. I also found a paper that claims to show that the only rational functions f(x,y) that are associative are f(x,y)=x,y,x+y,xy, up to mobius transformations of f. But that doesn't help here since your f(x,y) isn't rational. But still, shows associative functions f are pretty special.

 

[Ray Vickson]

Thank you for looking at the plot, I had wondered about that but wasn't sure how to do it. Maybe I need to just admit to my professor that I'm stumped on this one. I suppose if I get all the rest of the problems it won't hurt my pride too bad. I wonder why he even wrote such an exercise!

Success at last! Letting F1 = f(x,f(y,z)) and F2 = f(f(x,y),z), F1 becomes the sum of two terms with square-roots inside square roots. In Maple, we can extract the first one as L1 = op(1,F1). Since we are trying to prove F1 = F2, let's follow the suggestion by 'micromass' and compare L1 with R1 = F2 - op(2,F1); these ought to be equal, because in Maple we have F1 = op(1,F1)+op(2,F1). Leaving off the question of signs, square and expand both L1 and R1. We find that RR1 = R1^2 consists of 11 separate terms, of which two involve the same square-roots-within square-roots, but with different outside factors---both negative. The exact way to extract these depends on where they occur in the expression tree, but one can list all 11 terms through S = seq(op(i,RR1),i=1..11); in one session the two terms of interest were S[5] and S[11]. Writing RR1 = RR1a -S[5]-S[11], we want to prove that S[5]+S[11] = RR1a - L1^2. Now the right-hand-side no longer has square-roots inside square-roots. When we again square and compare (S[5]+S[11])^2 with (RR1a-L1^2)^2, we find the difference = 0!

Note that the exact way of extracting the wanted terms may be session-dependent and/or machine-dependent, etc., so success depends on a mix of human and machine processing, with the human determining what terms to group together. Getting a machine to do this unaided might be almost beyond the capabilities of current software.

In principle, all the work above could be done manually, if you have a couple of free days to spare dong algebra.

Of course, since we compared squares with squares, there are still sign considerations to worry about, but those seem comparatively simple to deal with.

 

[Dick]

Success at last! Letting F1 = f(x,f(y,z)) and F2 = f(f(x,y),z), F1 becomes the sum of two terms with square-roots inside square roots. In Maple, we can extract the first one as L1 = op(1,F1). Since we are trying to prove F1 = F2, let's follow the suggestion by 'micromass' and compare L1 with R1 = F2 - op(2,F1); these ought to be equal, because in Maple we have F1 = op(1,F1)+op(2,F1). Leaving off the question of signs, square and expand both L1 and R1. We find that RR1 = R1^2 consists of 11 separate terms, of which two involve the same square-roots-within square-roots, but with different outside factors---both negative. The exact way to extract these depends on where they occur in the expression tree, but one can list all 11 terms through S = seq(op(i,RR1),i=1..11); in one session the two terms of interest were S[5] and S[11]. Writing RR1 = RR1a -S[5]-S[11], we want to prove that S[5]+S[11] = RR1a - L1^2. Now the right-hand-side no longer has square-roots inside square-roots. When we again square and compare (S[5]+S[11])^2 with (RR1a-L1^2)^2, we find the difference = 0!

Note that the exact way of extracting the wanted terms may be session-dependent and/or machine-dependent, etc., so success depends on a mix of human and machine processing, with the human determining what terms to group together. Getting a machine to do this unaided might be almost beyond the capabilities of current software.

In principle, all the work above could be done manually, if you have a couple of free days to spare dong algebra.

Of course, since we compared squares with squares, there are still sign considerations to worry about, but those seem comparatively simple to deal with.

That's nice to know. Still be nice to know what the instructor had in mind and what makes that f(x,y) so special. Grrr.

 

[SammyS]

Not that I can see, did you have something specific in mind? Commutativity doesn't imply associativity by itself. Based on your suggestion I tried changing some things like (y*x)*z and z*(y*x) and messing with order in that way but was unable to get anywhere with it.

Nothing specific.

I saw that your thread had been up for over 6 hours without a reply, so I thought I'd suggest something.

I see it turns out to be a very difficult problem.

 

[micromass]

If you want an elegant solution, try this:

sinh(a+b) = cosh(a)sinh(b) + sinh(a)cosh(b)

 

[Dick]

If you want an elegant solution, try this:

sinh(a+b) = cosh(a)sinh(b) + sinh(a)cosh(b)

I'm sure that's elegant in some way, but f(a,b)=sinh(a+b) isn't associative, is it?

 

[micromass]

I'm sure that's elegant in some way, but f(a,b)=sinh(a+b) isn't associative, is it?

Sure it's not, but that's not what I was talking about.

 

[Dick]

Sure it's not, but that's not what I was talking about.

Ok, I'll keep guessing. It's suggestive.

 

[micromass]

This will probably give it away, but you can find an isomorphism between the structure in the OP and $(\mathbb{R},+)$.

 

[Dick]

This will probably give it away, but you can find an isomorphism between the structure in the OP and $(\mathbb{R},+)$.

Yes, that gave it away. And I was almost there. Thanks. Well, maybe I'm overexaggerating. It seemed clear for a minute. Now I'm thrashing around again. I'll give it more thought tomorrow. I see you've reparametrized the same problem, but I'm a little blind right now how it helps. Obviously, I don't need hints, I'm not the OP.

 

[ArcanaNoir]

This will probably give it away, but you can find an isomorphism between the structure in the OP and $(\mathbb{R},+)$.

Not sure what you mean. Feel free to be more explicit, this isn't coursework, it's for some solutions I'm writing for my professor's book. This problem is from an appendix on group theory.

 

[ArcanaNoir]

If you want an elegant solution, try this:

sinh(a+b) = cosh(a)sinh(b) + sinh(a)cosh(b)

Are you suggesting I change coordinate systems or something like that?

 

[Dick]

Are you suggesting I change coordinate systems or something like that?

Something like that. Notice if you put x=sinh(a) and y=sinh(b) then sinh(a+b) is equal to your f(x,y). So f(x,y)=sinh(asinh(x)+asinh(y)).

 

[ArcanaNoir]

Success at last! Letting F1 = f(x,f(y,z)) and F2 = f(f(x,y),z), F1 becomes the sum of two terms with square-roots inside square roots. In Maple, we can extract the first one as L1 = op(1,F1). Since we are trying to prove F1 = F2, let's follow the suggestion by 'micromass' and compare L1 with R1 = F2 - op(2,F1); these ought to be equal, because in Maple we have F1 = op(1,F1)+op(2,F1). Leaving off the question of signs, square and expand both L1 and R1. We find that RR1 = R1^2 consists of 11 separate terms, of which two involve the same square-roots-within square-roots, but with different outside factors---both negative. The exact way to extract these depends on where they occur in the expression tree, but one can list all 11 terms through S = seq(op(i,RR1),i=1..11); in one session the two terms of interest were S[5] and S[11]. Writing RR1 = RR1a -S[5]-S[11], we want to prove that S[5]+S[11] = RR1a - L1^2. Now the right-hand-side no longer has square-roots inside square-roots. When we again square and compare (S[5]+S[11])^2 with (RR1a-L1^2)^2, we find the difference = 0!

Note that the exact way of extracting the wanted terms may be session-dependent and/or machine-dependent, etc., so success depends on a mix of human and machine processing, with the human determining what terms to group together. Getting a machine to do this unaided might be almost beyond the capabilities of current software.

In principle, all the work above could be done manually, if you have a couple of free days to spare dong algebra.

Of course, since we compared squares with squares, there are still sign considerations to worry about, but those seem comparatively simple to deal with.

Wow that's epic, thank you for taking such a deep look at it.

 

[Dick]

Ah, I get it, finally! All of the squares and square roots were really just hiding the fact that your $f(x,y)$ can be written as $g(g^{-1}(x)+g^{-1}(y))$ for some other function $g(x)$. This confers the magic 'associative' property.

 

[Dick]

Hmm. Has everyone stopped following this thread? There's a really easy way and there's really hard way. The constrast beween the two is pretty severe.

 

[LCKurtz]

I haven't had the time or inclination to work very much on this, but I admit I am curious.

 

[Dick]

I haven't had the time or inclination to work very much on this, but I admit I am curious.

Just writing it as f(x,y)=sinh(arcsinh(x)+arcsinh(y)) (following micromass's hint) makes it immediately fall apart. The only part that's even a little hard is keeping track of the parentheses.

 

[Dick]

I guess what I'm really after here is asking whether anybody else realizes how fantastically brilliant micromass's suggestion is, especially following Ray Vickson's heroically hard but straightforward solution. It also let's you construct lots of other associative functions that would be equally hard to prove the direct way. Using micromass's suggestion, the first time I worked it out the proof fell out so quickly I didn't even see how it happened, it was that quick. I had to replay it a few times before I really understood it. No real 'calculation' was even needed, much less an intensive computer algebra project. It's that good.

 

[LCKurtz]

Just got back with a bit of time to look at it again. I fully agree with Dick at the brilliance of Micromass's observation. And good work to you Dick to observe observation about the general $g(g^{-1}(x)+g^{-1}(y))$ form.

 

[ArcanaNoir]

Something like that. Notice if you put x=sinh(a) and y=sinh(b) then sinh(a+b) is equal to your f(x,y). So f(x,y)=sinh(asinh(x)+asinh(y)).

Hmm. Has everyone stopped following this thread? There's a really easy way and there's really hard way. The constrast beween the two is pretty severe.

Ah, I get it, finally! All of the squares and square roots were really just hiding the fact that your $f(x,y)$ can be written as $g(g^{-1}(x)+g^{-1}(y))$ for some other function $g(x)$. This confers the magic 'associative' property.

Ah thanks for the tips! I tend not to have internet access over the weekends as I'm at work from Friday afternoon through Sunday night, and I don't have internet there. Also, I've been afraid to put more time on this problem as I have 5 other problems I have not yet solved and I was afraid I'd have nothing to show my professor! I will definitely get back to these hints as soon as I complete the other problems!

 

[Dick]

Ah thanks for the tips! I tend not to have internet access over the weekends as I'm at work from Friday afternoon through Sunday night, and I don't have internet there. Also, I've been afraid to put more time on this problem as I have 5 other problems I have not yet solved and I was afraid I'd have nothing to show my professor! I will definitely get back to these hints as soon as I complete the other problems!

Definitely come back to this. Once you've wrapped your head around the hints, you might find it's the easiest of the problems. Wouldn't that be a nice surprise?

 

[ArcanaNoir]

Just writing it as f(x,y)=sinh(arcsinh(x)+arcsinh(y)) (following micromass's hint) makes it immediately fall apart. The only part that's even a little hard is keeping track of the parentheses.

Okay maybe I'm dense, but I don't see how f(x,y)=sinh(arcsinh(x)+arcsinh(y)). Here is what I did to try to get it to come out:

\sinh (\arcsinh (x) + \arcsinh (y) )\\<br /> = \sinh (\ln (x+\sqrt{1+x^2}) + \ln (y+\sqrt{1+y^2} ) ) \\<br /> = \sinh (\ln [(x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )]) \\<br /> =\frac{e^{\ln \text{stuff}} - e^{-\ln \text{stuff}}}{2} \\<br /> =\frac{1}{2} ((x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )-\frac{1}{(x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )})

And I have tried some manipulations after that but never get it to simplify to be f(x,y).
Am I using the wrong identity?

 

[LCKurtz]

Definitely come back to this. Once you've wrapped your head around the hints, you might find it's the easiest of the problems. Wouldn't that be a nice surprise?

Dick, you might be interested in this:

http://www.researchgate.net/publication/2121594_On_the_Possible_Monoid_Structures_of_the_Natural_Numbers_N_or_Finding_All_Associative_Binary_Operations_on_N

Look at equation 3.2 and proposition 3.1. As is usual, someone has thought about this stuff before. It looks to me like if you use R instead of N, let $f(x,y) = x+y$ and $\omega=\sinh$, the result follows from the associativity of addition. And, as you have observed, there are other bijections $\omega$.

 

[Dick]

Okay maybe I'm dense, but I don't see how f(x,y)=sinh(arcsinh(x)+arcsinh(y)). Here is what I did to try to get it to come out:

\sinh (\arcsinh (x) + \arcsinh (y) )\\<br /> = \sinh (\ln (x+\sqrt{1+x^2}) + \ln (y+\sqrt{1+y^2} ) ) \\<br /> = \sinh (\ln [(x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )]) \\<br /> =\frac{e^{\ln \text{stuff}} - e^{-\ln \text{stuff}}}{2} \\<br /> =\frac{1}{2} ((x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )-\frac{1}{(x+\sqrt{1+x^2})(y+\sqrt{1+y^2} )})

And I have tried some manipulations after that but never get it to simplify to be f(x,y).
Am I using the wrong identity?

Use micromass's hint. And that sinh(arcsinh(x))=x and cosh(arcsinh(x))=sqrt(1+x^2).

 

[Dick]

Dick, you might be interested in this:

http://www.researchgate.net/publication/2121594_On_the_Possible_Monoid_Structures_of_the_Natural_Numbers_N_or_Finding_All_Associative_Binary_Operations_on_N

Look at equation 3.2 and proposition 3.1. As is usual, someone has thought about this stuff before. It looks to me like if you use R instead of N, let $f(x,y) = x+y$ and $\omega=\sinh$, the result follows from the associativity of addition. And, as you have observed, there are other bijections $\omega$.

Thanks for the reference! I'll try and look at it once I get past ResearchGate's signup procedure...

 

[ArcanaNoir]

Thanks for the reference! I'll try and look at it once I get past ResearchGate's signup procedure...

Close the pop up and you should see the PDF link on the right hand side of the page.

 

[Dick]

Close the pop up and you should see the PDF link on the right hand side of the page.

Thanks, but I don't see it. I think I may have to have an account and be logged in. But that's ok, I found the same paper on arXiv.org.

 

[SammyS]

Okay maybe I'm dense, but I don't see how f(x,y)=sinh(arcsinh(x)+arcsinh(y)).
...

And I have tried some manipulations after that but never get it to simplify to be f(x,y).
Am I using the wrong identity?

Try looking at it this way:

Suppose that given x and y, there exist real numbers, a and b, such that

x = sinh(a)

and

y = sinh(b) .​

Then f(x,y) is becomes the following:

$\displaystyle f(x,\,y)=f(\sinh(a),\,\sinh(b))$

$\displaystyle = \sinh(a)\sqrt{1+\sinh^2(b)}+\sinh(b)\sqrt{1+\sinh^2(a)}$​

Then use one of the Pythagorean identities to get rid of the radical.

After that use an argument addition identity (analogous to angle addition for trig functions).

 

[Dick]

Try looking at it this way:

Suppose that given x and y, there exist real numbers, a and b, such that

x = sinh(a)

and

y = sinh(b) .​

Then f(x,y) is becomes the following:

$\displaystyle f(x,\,y)=f(\sinh(a),\,\sinh(b))$

$\displaystyle = \sinh(a)\sqrt{1+\sinh^2(b)}+\sinh(b)\sqrt{1+\sinh^2(a)}$​

Then use then one of the Pythagorean identities to get rid of the radical.

After that use an argument addition identity (analogous to angle addition for trig functions).

I'm not sure that's terribly illuminating, I'd just say following micromass's hint, $\sinh(arcsinh(x)+arcsinh(y))=\sinh(arcsinh(x)) \cosh(arcsinh(y))+\cosh(arcsinh(x)) \sinh(arcsinh(y))$. Using $\cosh(arcsinh(x))=\sqrt{1+x^2}$ this seems pretty straightforward. Sorry I can't find a texy way to do the inverse hyperbolics.

 

[haruspex]

http://www.researchgate.net/publication/2121594_On_the_Possible_Monoid_Structures_of_the_Natural_Numbers_N_or_Finding_All_Associative_Binary_Operations_on_N

Look at equation 3.2 and proposition 3.1. As is usual, someone has thought about this stuff before. It looks to me like if you use R instead of N, let $f(x,y) = x+y$ and $\omega=\sinh$, the result follows from the associativity of addition. And, as you have observed, there are other bijections $\omega$.

I mused as to whether there's an entire field (functions on operators) that could open up here (or maybe already has). As has been noted in this thread, if f is invertible S→S and ° is an operator S×S→S then f induces an operator °_f by x °_f y = f((f^-1(x)°f^-1(y)). If ° is associative, so is °_f. Likewise commutativity. More surprisingly perhaps, if ° distributes across another operator °' then °_f distributes across °'_f.
If f is linear then it's rather trivial: °_f = °. Some other examples:
f(x) = e^x:

+_f = × (or, by abuse of notation, e⁺ = ×). Using the same shorthand:
x e^× y = x^ln(y) = y^ln(x)​

f(x) = √x:

x √+ y = √(x²+y²)​

f(x) = tan(x):

x tan(+) y = (x+y)/(1-xy)​

It's suggestive, but needs a deeper result to make it interesting.

 

[Dick]

I mused as to whether there's an entire field (functions on operators) that could open up here (or maybe already has). As has been noted in this thread, if f is invertible S→S and ° is an operator S×S→S then f induces an operator °_f by x °_f y = f((f^-1(x)°f^-1(y)). If ° is associative, so is °_f. Likewise commutativity. More surprisingly perhaps, if ° distributes across another operator °' then °_f distributes across °'_f.
If f is linear then it's rather trivial: °_f = °. Some other examples:
f(x) = e^x:

+_f = × (or, by abuse of notation, e⁺ = ×). Using the same shorthand:
x e^× y = x^ln(y) = y^ln(x)​

f(x) = √x:

x √+ y = √(x²+y²)​

f(x) = tan(x):

x tan(+) y = (x+y)/(1-xy)​

It's suggestive, but needs a deeper result to make it interesting.

Probably already has been thought about, that's what the monoids paper was about. I'm more interested in whether ArcanaNoir can figure out that this is a pretty easy problem with the right clues.

 

[SammyS]

I'm not sure that's terribly illuminating, I'd just say following micromass's hint, $\sinh(arcsinh(x)+arcsinh(y))=\sinh(arcsinh(x)) \cosh(arcsinh(y))+\cosh(arcsinh(x)) \sinh(arcsinh(y))$. Using $\cosh(arcsinh(x))=\sqrt{1+x^2}$ this seems pretty straightforward. Sorry I can't find a texy way to do the inverse hyperbolics.

Well, it does give $\displaystyle \ f(x,y)=\sinh(\text{arcsinh}(x)+\text{arcsinh}(y)) \$ rather easily.

To get texy inverse hyperbolics I use

\text{arcsinh} .