Condition that forces taylor series to converge in C
in [Math]
|
Prev: THE PRIMORDIAL PROPORTION OF A CIRCLE PROPORTION,AND SUBSQUENT MATHEMATICAL ACTIVATION OF THE CIRCLE PROPORTION BY -1
Next: The hypersphere-universe brane
From: John on 18 Jun 2010 12:32 Let f be an analytic function in D (D is domain). Suppose sum{i=0,...,oo} f^(n) (a) converges (a in D). I need to prove that there exists an analytic extension of f to C. I think the way is to prove that the taylor series of f at the point a is converges in C and hence it's the desired analytic extension. I don't know how the fact sum{i=0,...,oo} f^(n) converges should help... Any hints ? Thanks.
From: TCL on 18 Jun 2010 14:48 On Jun 18, 12:32 pm, John <to1m...(a)yahoo.com> wrote: > Let f be an analytic function in D (D is domain). Suppose > sum{i=0,...,oo} f^(n) (a) converges (a in D). > I need to prove that there exists an analytic extension of f to C. > > I think the way is to prove that the taylor series of f at the point a > is converges in C and hence it's the desired analytic extension. I > don't know how the fact sum{i=0,...,oo} f^(n) converges should help... > > Any hints ? > > Thanks. c_n = f^(n)(a) is bounded.
From: Chip Eastham on 18 Jun 2010 14:56 On Jun 18, 12:32 pm, John <to1m...(a)yahoo.com> wrote: > Let f be an analytic function in D (D is domain). Suppose > sum{i=0,...,oo} f^(n) (a) converges (a in D). > I need to prove that there exists an analytic extension of f to C. > > I think the way is to prove that the taylor series of f at the point a > is converges in C and hence it's the desired analytic extension. I > don't know how the fact sum{i=0,...,oo} f^(n) converges should help... > > Any hints ? > > Thanks. Radius of convergence --c
From: Rob Johnson on 18 Jun 2010 15:08 In article <0309fb75-fadd-4f53-9657-4d952dfdb03d(a)x21g2000yqa.googlegroups.com>, John <to1mmy2(a)yahoo.com> wrote: >Let f be an analytic function in D (D is domain). Suppose >sum{i=0,...,oo} f^(n) (a) converges (a in D). >I need to prove that there exists an analytic extension of f to C. > >I think the way is to prove that the taylor series of f at the point a >is converges in C and hence it's the desired analytic extension. I >don't know how the fact sum{i=0,...,oo} f^(n) converges should help... > >Any hints ? Write f(x) in terms of f^(n)(a) using a Taylor series. Using the summation condition you've cited, can you determine the radius of convergence of the series you've just written? Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: John on 18 Jun 2010 17:10
On Jun 18, 9:48 pm, TCL <tl...(a)cox.net> wrote: > On Jun 18, 12:32 pm, John <to1m...(a)yahoo.com> wrote: > > > Let f be an analytic function in D (D is domain). Suppose > > sum{i=0,...,oo} f^(n) (a) converges (a in D). > > I need to prove that there exists an analytic extension of f to C. > > > I think the way is to prove that the taylor series of f at the point a > > is converges in C and hence it's the desired analytic extension. I > > don't know how the fact sum{i=0,...,oo} f^(n) converges should help.... > > > Any hints ? > > > Thanks. > > c_n = f^(n)(a) is bounded. Thank you. Now the calculation of the radius of convergence becomes easy. |