(New page: <math>g(x+y) = \frac{g(x)+g(y)}{1-g(x)g(y)}</math> <math>\lim_{h \to 0} g(h) = 0</math> <math>\lim_{h \to 0} \frac{g(h)}{h}= 1</math> a. Show that <math> g(0) = 0</math>. b. Show that ...) |
|||
| Line 16: | Line 16: | ||
--[[User:Jmason|Jmason]] 15:28, 5 October 2008 (UTC) | --[[User:Jmason|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. | ||
Revision as of 17:43, 5 October 2008
$ 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.
