C.E Messages 100 Reaction score 0 May 19, 2009 #31 I got if from when you said "[itex](v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma[/itex]" in post 8.
Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 May 19, 2009 #32 Ah, that was my own typo. Sorry about that. So you should now have [tex]F=-kx=-\frac{2\gamma p A^2}{V}x[/tex] which doesn't quite match the given answer, but I suspect we're right and the given answer wrong. Please update if you find another solution.
Ah, that was my own typo. Sorry about that. So you should now have [tex]F=-kx=-\frac{2\gamma p A^2}{V}x[/tex] which doesn't quite match the given answer, but I suspect we're right and the given answer wrong. Please update if you find another solution.
C.E Messages 100 Reaction score 0 May 19, 2009 #33 I have read about Taylor expansions on wikipedia and am still a little confused. What are the Taylor expansions I should be using for (v2-Ax)^gamma and (v1+Ax)^gamma? How did you find them?
I have read about Taylor expansions on wikipedia and am still a little confused. What are the Taylor expansions I should be using for (v2-Ax)^gamma and (v1+Ax)^gamma? How did you find them?
Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 May 19, 2009 #34 The idea here is that [tex]f(a+b)\approx f(a)+b\,f^\prime(a)[/tex] for small b. In this case [tex]f(a)=a^\gamma[/tex] So [tex]f(a+b)\approx a^\gamma+\gamma b a^{\gamma-1}=a^\gamma\left(1+\frac{\gamma b}{a}\right)[/tex] as you figured out from your knowledge that [itex](1+b)^\gamma\approx 1+\gamma b[/tex] (and I messed up).[/itex]
The idea here is that [tex]f(a+b)\approx f(a)+b\,f^\prime(a)[/tex] for small b. In this case [tex]f(a)=a^\gamma[/tex] So [tex]f(a+b)\approx a^\gamma+\gamma b a^{\gamma-1}=a^\gamma\left(1+\frac{\gamma b}{a}\right)[/tex] as you figured out from your knowledge that [itex](1+b)^\gamma\approx 1+\gamma b[/tex] (and I messed up).[/itex]