let -1,1) \rightarrow R" alt="f-1,1) \rightarrow R" /> be . Prove that is real analytic in some neighborhood of 0 if and only if there is a nonempty interval and a constant such that for all and all
please help
A function is "real analytic" in a region if and only if, at each point in that region, its Taylor series exists and converges to the value of the function in some neighborhood of the point.
Look at the error formula for Taylor polynomials. If you can show that the error goes to 0 as n goes to infinity, you have it.
HallsofIvy's hint gives the "if" part. Let's consider the "only if", i.e. show that a real analytic function statisfies the given bound.
The classical proof uses Cauchy's inequality for complex analytic functions; it's rather simple but you definitely can't think about it if you've not studied complex analysis. Let me sketch a proof that "stays in the real line".
I write the proof for . It is simply adaptable to other points. We have for and . Choose . We have, for all , (the power series converges absolutely inside the interval of convergence). Thus, for , where .
I had an answer ready for that one, but it turns out it doesn't work... Since I can't fix it, there remains the "Cauchy inequality" method. Do you know analytic continuation?
If so, you can make sense of
,
for all such that is analytic in (hence on in after continuation).
The proof is just term-by-term integration of the series .
Then the inequality is straightforward for all , letting for some such that is analytic on (choosing small enough).
There may be a way to fix the previous method, but it's a bit late here...