i need to prove that if f,g are analytic in (-a,a) and g(x) isnt zero for all x in (-a,a) then f/g is also analytic in some interval (-b,b). my defintiion for analytic function is that, f is analytic if there exists r>0 such that it equlas its power series in (-r,r). now in order to prove this i thought to set the next things: first, to prove that 1/g is analytic in (-1/a,1/a) and then bacuase either a>=1/a or vice versa we have by another statement that if f,g are analytic in some interval then f*g is analytic in the same interval. my problem is, if this is correct then i need a formula for n-th derivative of 1/g, in order to show that the limit of the lagrange's residue of the taylor series approaches 0 in this interval? my question what is this formula? thanks in advance.
Why did you choose 1/a? 1/g(1/a) does not equal g(a), if you were hoping to pull something slick like that. In fact, if 1/a > a, you're screwed, because not even g is analytic in (-1/a, 1/a), so you have no reason to expect 1/g to be analytic in that region
so what should i do here? i just meant that if 1/g is analytic in (-1/a,1/a) and 1/a>a then obviously 1/g is analytic in (-a,a) and so is f so i can pull it off, i didnt think that 1/g(1/a)=g(a) i don't see really how you can do it unless we are given that g is isomorphism and that g(1)=1, which we aren't given.
But 1/g might not be analytic in (-1/a, 1/a), for example if g is log(1+x), and a=1/2. can you prove 1/g is analytic in (-a,a) without actually going about deriving a formula for each term of the power series?
Can't you use the fact that f(z) is analytic iff it satisfies the Cauchy Reimann differential equations? I think this would avoid an argument based around from limits.
i havent yet learned cauchy riemann DE. you can say that this definition came from calculus 2 class (which means basically, power series, series, etc). and mathwonk, i think iv'e already proven that prodcuts of two analaytic functions is analytic, it basically uses cauchy's product. and i said already that in order to prove this i need to use this theorem. but obviously the domain of 1/g isn't (-a,a), so how to solve it?
well usually one attacks f/g by attacking products and reciprocals separately. and what do you mean "obviously"? what is your definition of domain?
the definition of a function f analytic in a domain is such that there exists r>0 such that in (-r,r) the function equals its power series. well i don't really know, what is the domain of 1/g and if it's analytic, but then this is what i need to prove, that's analytic in some interval, and to use it to prove that f/g analytic in some open interval. i already prove that products of two analytic function is analytic in the same interval, now this...
i be;lieve the definition of analytic is that it equals its powere series locally in the region. i.e. for every point of the region there is such an r. but the domain of analyticity of the reciprocal of an analytic function g is at least as large as the domain of g where g is non zero. it can also include places where g was infinite, i.e. had a pole. e.g. 1/z is analytic except at the origin, but z is also analytic there.
Well... actually, I feel a bit ignorant, but I have tried to prove that, unsuccesfuly. I have tried to use Cauchy's multiplication of series: "The product of two absolutely convergent series absolutely converges as well." - but that only proves that if f,g are analytic in (-a,a), than (fg) is analytic at least in (-a,a) - but why not in a larger interval? Same thing for (f+g): Easy to prove that (f+g) converges in (-a,a) - but how can I prove that it DOES NOT converge in x=b>a ? Any ideas? Thanks...
I have no practical experience with complex analysis yet, but I seem to remember that mathwonk's definition of analyticity being correct(from Penrose's road to reality) That said, you said you were able to prove that if f,g are analytic in (-a,a) then (fg) would be as well, but you were asking why not for some larger interval? Why would you need a larger interval, if this is true for a general (-a,a) wouldn't this cover every neighborhood near an a? No clue on how to prove it does not converge in b > a, but good luck, I find this very alluring.
actually... Actually... actually it is just not true; I cannot think now of an example, but it might be that f,g are analytic ONLY in (-a,a), but f+g or fg or both are analytic in a larger interval... Oh, here I have an example: take f(x) = x/(1+x), g(x) = 1/(1+x). Their sum is h(x)==1, which is analytic all over R... Peleg.