What is an example of a real-valued function f whose Taylor series has radius of convergence 0?

Question

Several references assure me that there are functions f which are real-valued, defined on an open interval (a, b), and infinitely differentiable at some c in (a, b); and which have a Taylor series at c with radius of convergence 0.

What is an example of such a function? (If the function is hard to construct or if it's hard to prove it has radius of convergence 0, then citing a reference where I can read about it would satisfy me.)

(Here, I mean infinitely differentiable in the real sense only. I already know R = 0 is not possible for a function with a complex derivative.)

Answer 1

Dr. D gives a pretty good answer. To make his solution take real values, all it is needed is to replace the complex Fourier series with its real part (say, instead of summing over e^(i λ_k x), sum over cos(λ_k x) ). The series Dr. D gives converges because λ_k > k^(2k) and c_k = λ_k^(1-k) < 1/(k^(2k^2 - 2k)) < C 1/k^2 is absolutely summable.

It is also simple to check that the derivatives when evaluated at 0

(d/dx)^m f(0) = Σ c_k λ_k^m

grows sufficiently fast.

Remember.Kelly's solution doesn't work because the sequence of functions do not have uniform convergence: the pointwise limit of the sequence of the functions is the function that takes 0 everywhere except at the origin, where it takes the value 1: the limit is not even continuous, nevermind smooth.

Dr. D's example, and its analogues, are as simple an answer you are going to get, for several reasons: i) since you are trying to show, explicitly, that the Taylor series for the function cannot converge, you cannot define the function using a power series expansion. ii) it *has* to be a infinite series sum, since *most* everyday functions we come across that has a simple description all turns out to be real analytic, so a finite linear combination must also be real analytic. iii) In view of the second point and the Stone-Weierstrass type theorems, basically we need to find another descriptor of the function using another normal family of functions that can approximate, at least, the continuous functions. iv) In view of the third point and the fact that the non-convergence of the Taylor series requires really fast growth on the size of the derivatives, a natural candidate to consider is the Fourier series, where, due to the duality between derivation in physical space and multiplication (by the frequency) in frequency space, construction of functions with larger and larger derivatives is not too difficult. Lastly, v) since you asked for a function to have radius of convergence 0 of Taylor series, it cannot be too smooth: it must necessarily be highly oscillatory (but at a very small amplitude) everywhere, this also makes a Fourier series the ideal descriptor of the function.

Answer 2

If you want a reference, try Real and Complex Analysis by Walter Rudin. This is reference 5 and 6 on the wikipedia article on taylor series. They briefly mention what you're talking about. I'm supposing that the reference goes into more detail.

http://en.wikipedia.org/wiki/Taylor_seri...

You can download that ebook (illegally) at 4shared.com. It's the section on holomorphic fourier transforms.

This is way out of my league, but the reference gives the following suggestion on how to construct f(x).

Write f(x) = ∑{k=1,∞} c_k * exp(i*λ_k*x)

Then choose c_k and λ_k like this:

c_k = λ_k^(1-k)

then choose λ_k such that

λ_k > 2∑{j=1,k-1} c_j*λ_j ^k

and λ_k > k^(2k)

This does look unlikely to give you a real valued function, but see if you can make anything of this.

Answer 3

To ksoileau:

The function exp(-1/x^2) doesn't have a Taylor series at x=0.

How about this function (if something like that is allowed):

f=1/(1-x/ε)  =1+x/ε+(x/ε)^2+......   provided |x|<ε

The radius of convergence is R=ε, which approaches zero as ε→0.

Answer 4

Try exp(-1/x^2)