dpopchev
- 25
- 0
Homework Statement
If I have the following expansion
[tex] f(r,t) \approx g(r) + \varepsilon \delta g(r,t) + O(\varepsilon^2)[/tex]
This means for other function U(f(r,t))
[tex] U(f(r,t)) = U( g(r) + \varepsilon \delta g(r,t)) \approx U(g) + \varepsilon \delta g \dfrac{dU}{dg} + O(\varepsilon^2)[/tex]
Then up to linear order in ε how to calculate
[tex] \dfrac{dU}{df} = \ldots O(\varepsilon^2)?[/tex]
The attempt at a solution
No idea how to approach this:
[tex] \dfrac{dU}{df} = \dfrac {d U}{d g}\dfrac{dg}{df} = \dfrac{dU/dg}{df/dg} = ?[/tex]
But then again not sure how to calculate this, if I try
[tex] \dfrac {d U}{d g} = (U(f(r,t)) - U(g))\dfrac{1}{\varepsilon \delta g}[/tex]
This will lead me again to dU/dg
Additionally how to calculate dg/df term
EDIT: added details to expression
If I have the following expansion
[tex] f(r,t) \approx g(r) + \varepsilon \delta g(r,t) + O(\varepsilon^2)[/tex]
This means for other function U(f(r,t))
[tex] U(f(r,t)) = U( g(r) + \varepsilon \delta g(r,t)) \approx U(g) + \varepsilon \delta g \dfrac{dU}{dg} + O(\varepsilon^2)[/tex]
Then up to linear order in ε how to calculate
[tex] \dfrac{dU}{df} = \ldots O(\varepsilon^2)?[/tex]
The attempt at a solution
No idea how to approach this:
[tex] \dfrac{dU}{df} = \dfrac {d U}{d g}\dfrac{dg}{df} = \dfrac{dU/dg}{df/dg} = ?[/tex]
But then again not sure how to calculate this, if I try
[tex] \dfrac {d U}{d g} = (U(f(r,t)) - U(g))\dfrac{1}{\varepsilon \delta g}[/tex]
This will lead me again to dU/dg
Additionally how to calculate dg/df term
EDIT: added details to expression
Last edited: