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* 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.[[User:Jhunsber|Jhunsber]] | * 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.[[User:Jhunsber|Jhunsber]] | ||
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| + | * Part b is a lot trickier. Remember that <math>g'(x)=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}</math> 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. [[User:Jhunsber|Jhunsber]] | ||
Revision as of 18:16, 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.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
