$ g(x+y) = \frac{g(x)+g(y)}{1-g(x)g(y)} $
$ \lim_{h \to 0} g(h) = 0 $
$ \lim_{h \to 0} \frac{g(h)}{h}= 1 $
a. Show that $ g(0) = 0 $.
b. Show that $ g'(x) = 1 + [g(x)]^2 $
c. Find $ g(x) $ by solving the differential equation in part (b).
Anyone know where to start? I'm defeated at every turn; I can't break the function into even/odd portion that have any use and none of the laws of exponentials/logarithms seem to be very useful. The only fact I can pull out is that $ g'(0)=1 $ which can be determined through L'Hopitals.
--Jmason 15:28, 5 October 2008 (UTC)
- You can show g(0) = 0 by solving for g(x) (Yes, you can do it. No, it's not that hard), and then plugging 0 in for x. As for the other parts, I haven't got that far yet. I'll see what I get. And wow, I've been working on this problem a half hour already, I think.Jhunsber
- Part b is a lot trickier. Remember that $ g'(x)=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} $ If you solved for g(x) for part a, you can plug that in substituting h for y since if h=y, then g(x+y) will = g(x+h). From there, just simplify and reverse distribute until you get the answer you need. Jhunsber
- Okay, part c really tripped me up. It took me an hour of finding consistently that g(x) = g(x) (I guess that's a good thing, right?) before I realized that I should look at g'(x) as dy/dx and g(x) as y. Then it looked very easy. in fact, perhaps too easy. I'm going to go check to see if this works or not. I really hope it does. Okay, checked it and it is true for all x. So yes, it really is easy if you think about it in the right way... Jhunsber 23:58, 5 October 2008 (UTC)
