Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am wondering if there are any functions $f(x)$ such that it cannot be expressed as a power series of $x$? This might turn out to be a silly question, but I can't think of one at the moment.

Thanks!

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

Yes, there are many. One dramatic example is $$f(x) = \begin{cases} 0 & x\in\mathbb Q \\ 1 & x\in\mathbb R\setminus\mathbb Q\end{cases}$$

but there are also more subtle ones, such as $$g(x) = \begin{cases} 0 & x=0 \\ e^{-1/x^2} & x\ne 0 \end{cases}$$ which is even differentiable arbitrarily many times everywhere but cannot be expressed as a power series in any interval that includes $0$.

You can get a $\mathcal C^\infty(\mathbb R)$ function that has a power series nowhere by letting $(q_n)_{n\in\mathbb N}$ enumerate the rational numbers and forming the infinite sum $$ h(x) = \sum_{n=1}^{\infty} \frac{g(x-q_n)}{2^n} $$

share|improve this answer
    
And while you're at it, one might add that in a precise sense almost all functions, and even almost all $\mathcal C^\infty(\mathbb R)$ functions, are nowhere given by a power series. –  Marc van Leeuwen Oct 5 '12 at 7:21
add comment

For example, $f(x)=e^{-\frac{1}{x^{2}}}$ for $x\neq0$ and $f(0)=0$, you can see that $f(x)$ is continuous and differentiable of infinite types, but $f(x)$ cannot be expressed as a power series of $x$ at $0$.

share|improve this answer
add comment

If a function $f(x)$ can be expressed as a power series of $x$,it is called analytic function.You can read more about them by searching.

There are many examples of non-analytic functions,one you may consider is given by http://en.wikipedia.org/wiki/Non-analytic_smooth_function.

Edit:Sorry,I did not see the answers posted during the time I was typing..

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.