Question

(iii) Find the most general function that, for all x and y, satisfies the identity

t(x) + t(y) = t(z)

where z = (x+y)/(1-xy)

I can't think of how to do this. I can see some relation to the t=tan((1/2)A) substitution where sin(A) = (2t)/(1-t^2) but don't know how to use it in this context.

Answer 1

tan(a+b) = ( tan a + tan b )/(1 - tan a.tan b)

Let t = arctan, x = tan a and y = tan b
and you now have an example of this function t.

Now, the find the most general form, you're probably going to have to have a look at 2-dimensional Taylor series.

Answer 2

t(x) = arctan(x) ? Now I'm confused because I can't see how this satisfies t(x) + t(y) = t(z) still? :S I think it's the manipulation of these general functions which I am messing up on, it's something I'm unfamiliar with..
Thanks so much for your help James.

Answer 3

Nevermind I understand it now, you have no idea how much this is going to help me James haha.

Answer 4

2-dimensional Taylor series? Is there any other way? I've only looked at single dimensional taylor series

Answer 5

(as far as I know)

Answer 6

It's subtle. I would dread doing the calculations to find the most general function. I think this class of function is good enough.

Answer 7

would you say t(x) = c . arctan(x) would be the general solution? since the constants just cancel out in the equation

Answer 8

Yes.

Answer 9

Thanks

Answer 10

Or at least that's one family of solutions. Technically, we should prove there are no other such functions. But I hypothesize that this family is sufficient.

Answer 11

Hmm, how would we do that..

Answer 12

Multi-dimensional Taylor series. I would ask your lecturer/tutor what they want you to do here.

Answer 13

Ok I will, thanks again.

Answer 14

The trivial solution will work. Let t(s)=0 for all s in R.

Answer 15

Just what I was about to add. z is undefined if xy=-1