Rate of Convergence for g(x) Limit at x=0

[ductiletoaste]

Ok i have a question I am have been stuck on for a while.
g(x) = f(x)^(1/3) and f(x)^(1/3) = (e^x-cos(x)-x-x^2)^(1/3)

What is the rate of convergence for lim x->0 of g(x)?

Now to make it easier i took the taylor poly of the crazy function to degree 3. Which is (x^3)/6

The part I am confused about is what our prof told us in class...
lim h->0 of G(h) = 0, and lim h->0 of F(h) = L
We say that F(h) converges to L with a Rate of Convergence O(G(h)).

So what is the rate of convergence? and what is o(G(h))?
Thank you for any direction you can provide!

 

[ductiletoaste]

Can anyone give me any advice?

 

[mathman]

The description you gave of the professor's explanation is very confusing.

 

[ductiletoaste]

ur telling me. I have been lost and that's a not a description its a direct quote from one of his lectures! But for the sake of helping me figure this out let's forget the whole confusing explantion. how would u normally find the Rate of convergence of g(x) = f(x)^(1/3) and f(x)^(1/3) = (e^x-cos(x)-x-x^2)^(1/3)

What is the rate of convergence for lim x->0 of g(x)?

 

[mathman]

Could you define "rate of convergence"?

 

[ductiletoaste]

no offense but if u don't know what rate of convergence is then u can't hope to answer this question.

 

[mathman]

I've been a mathematician all my life. I never heard that particular term used in an absolute sense, only relative to something else, like x³ -> 0 faster than x² when x -> 0.

 

[ductiletoaste]

Well this was an example problem given to me by a Phd professor. So i can assure you there is nothing wrong with the question.

 

[mathman]

We say that F(h) converges to L with a Rate of Convergence O(G(h)).

So what is the rate of convergence? and what is o(G(h))?

The first statement says that |F(h)-L| -> 0 at the same rate as G(h)->0.
The second statement means that |F(h)-L| -> 0 faster than G(h) -> 0.

Did you edit the statement after first posting?