Understanding Function Expansion: Mathematically Explained by Non-Mathematicians

[Niles]

Homework Statement

Hi

I have sometimes seen a function f being written as
[tex]
f \approx f_0 + \varepsilon f_1(x)+ \varepsilon^2 f_2(x) + \ldots
[/tex]
where [itex]f_0[/itex] is an equlibrium value and all higher-order terms are non-equilibrium values (not derivates!). The assumption has always been that [itex]\varepsilon \ll 1[/itex].

My question is: Mathematically, I guess we are expanding the function f around its equlibrium value. But when I look at the expression for a Taylor expansion, I can't make this fit with anything.

Are we non-mathematicians even allowed to write the function like this?

 

[Mandelbroth]

Homework Statement

Hi

I have sometimes seen a function f being written as
[tex]
f \approx f_0 + \varepsilon f_1(x)+ \varepsilon^2 f_2(x) + \ldots
[/tex]
where [itex]f_0[/itex] is an equlibrium value and all higher-order terms are non-equilibrium values (not derivates!). The assumption has always been that [itex]\varepsilon \ll 1[/itex].

My question is: Mathematically, I guess we are expanding the function f around its equlibrium value. But when I look at the expression for a Taylor expansion, I can't make this fit with anything.

Are we non-mathematicians even allowed to write the function like this?

Thank you for using the (much prettier looking) \varepsilon for your epsilons.

Well, us math-folk sometimes have a dislike for approximations and what you have written doesn't specify a domain or codomain (so we wouldn't technically call it a function by strict definition), but I see no other reason why you couldn't do that.

If your ##\{f_i\}## follow some sort of pattern, we can probably solve the limit $$f(x)=\lim_{n\rightarrow+\infty}\sum_{0\leq i\leq n}\varepsilon^if_i(x).$$

Can you give us more information? I'm interested in understanding what you're doing here.

 

[pasmith]

Homework Statement

Hi

I have sometimes seen a function f being written as
[tex]
f \approx f_0 + \varepsilon f_1(x)+ \varepsilon^2 f_2(x) + \ldots
[/tex]
where [itex]f_0[/itex] is an equlibrium value and all higher-order terms are non-equilibrium values (not derivates!). The assumption has always been that [itex]\varepsilon \ll 1[/itex].

My question is: Mathematically, I guess we are expanding the function f around its equlibrium value. But when I look at the expression for a Taylor expansion, I can't make this fit with anything.

That looks like an asymptotic expansion, not a Taylor series. In such expansions we usually don't care whether
[tex]
\lim_{N \to \infty} \sum_{n=0}^N \epsilon^n f_n(x)
[/tex]
even exists; what we're interested in is whether [itex]\sum_{n=0}^N \epsilon^n f_n(x)[/itex] for some finite [itex]N[/itex] is a sufficiently good approximation to some [itex]F(x,\epsilon)[/itex] when [itex]|\epsilon| \ll 1[/itex].

Generally the idea is to exploit a small parameter to turn a problem we can't solve analytically for [itex]F(x,\epsilon)[/itex] into a series of problems we can solve for the [itex]f_n[/itex].

 

[Niles]

Thanks for the help so far, both of you.

Generally the idea is to exploit a small parameter to turn a problem we can't solve analytically for [itex]F(x,\epsilon)[/itex] into a series of problems we can solve for the [itex]f_n[/itex].

That is exactly how it is used in my case (a Chapman-Enskog expansion). But doesn't this require that the various terms [itex]f_n[/itex] are somewhat independent, so we can solve for each order independently?

 

[Chestermiller]

Suppose you have a function of the two variables ε and x, f(ε,x) and you expand in a Taylor series in ε about ε=0. Then you get:

[tex]f(ε,x)=f(0,x)+ε\left(\frac{\partial f(ε,x)}{\partial ε}\right)_{ε=0}+\frac{ε^2}{2}\left(\frac{\partial^2 f(ε,x)}{\partial ε^2}\right)_{ε=0}+ ...[/tex]
Then you can identify the functions in your summations with the partial derivative terms in this series.

 

[pasmith]

Thanks for the help so far, both of you.



That is exactly how it is used in my case (a Chapman-Enskog expansion). But doesn't this require that the various terms [itex]f_n[/itex] are somewhat independent, so we can solve for each order independently?

The problem for [itex]f_n[/itex] should depend only on [itex]f_0, \dots, f_{n-1}[/itex], so that one can work forward from [itex]f_0[/itex].