# A positive smooth function with all derivatives vanishing at zero

Let’s consider the set $$\mathcal C^\infty(\mathbb R)$$ of real smooth functions, i.e. functions that have derivatives of all orders on $$\mathbb R$$.

Does a positive function $$f \in \mathcal C^\infty(\mathbb R)$$ with all derivatives vanishing at zero exists?

Such a map $$f$$ cannot be expandable in power series around zero, as it would vanish in a neighborhood of zero. However, the answer to our question is positive and we’ll prove that $f(x) = \left\{\begin{array}{lll} e^{-\frac{1}{x^2}} &\text{if} &x \neq 0\\ 0 &\text{if} &x = 0 \end{array}\right.$ provides an example.

$$f$$ is well defined and positive for $$x \neq 0$$. As $$\lim\limits_{x \to 0} -\frac{1}{x^2} = -\infty$$, we get $$\lim\limits_{x \to 0} f(x) = 0$$ proving that $$f$$ is continuous on $$\mathbb R$$. Let’s prove by induction that for $$x \neq 0$$ and $$n \in \mathbb N$$, $$f^{(n)}(x)$$ can be written as $f^{(n)}(x) = \frac{P_n(x)}{x^{3n}}e^{-\frac{1}{x^2}}$ where $$P_n$$ is a polynomial function. The statement is satisfied for $$n = 1$$ as $$f^\prime(x) = \frac{2}{x^3}e^{-\frac{1}{x^2}}$$. Suppose that the statement is true for $$n$$ then $f^{(n+1)}(x)=\left[\frac{P_n^\prime(x)}{x^{3n}} – \frac{3n P_n(x)}{x^{3n+1}}+\frac{2 P_n(x)}{x^{3n+3}}\right] e^{-\frac{1}{x^2}}$ hence the statement is also true for $$n+1$$ by taking $$P_{n+1}(x)= x^3 P_n^\prime(x) – 3n x^2 P_n(x) + 2 P_n(x)$$. Which concludes our induction proof.

Finally, we have to prove that for all $$n \in \mathbb N$$, $$\lim\limits_{x \to 0} f^{(n)}(x) = 0$$. For that, we use the power expansion of the exponential map $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$. For $$x \neq 0$$, we have $\left\vert x \right\vert^{3n} e^{\frac{1}{x^2}} \ge \frac{\vert x \vert^{3n}}{(2n)! \vert x \vert ^{4n}} = \frac{1}{(2n)! \vert x \vert^n}$ Therefore $$\lim\limits_{x \to 0} \left\vert x \right\vert^{3n} e^{\frac{1}{x^2}} = \infty$$ and $$\lim\limits_{x \to 0} f^{(n)}(x) = 0$$ as $$f^{(n)}(x) = \frac{P_n(x)}{x^{3n} e^{\frac{1}{x^2}}}$$ with $$P_n$$ a polynomial function.

# A power series converging everywhere on its circle of convergence defining a non-continuous function

Consider a complex power series $$\displaystyle \sum_{k=0}^\infty a_k z^k$$ with radius of convergence $$0 \lt R \lt \infty$$ and suppose that for every $$w$$ with $$\vert w \vert = R$$, $$\displaystyle \sum_{k=0}^\infty a_k w^k$$ converges.

We provide an example where the power expansion at the origin $\displaystyle f(z) = \sum_{k=0}^\infty a_k z^k$ is discontinuous on the closed disk $$\vert z \vert \le R$$.

The function $$f$$ is constructed as an infinite sum $\displaystyle f(z) = \sum_{n=1}^\infty f_n(z)$ with $$f_n(z) = \frac{\delta_n}{a_n-z}$$ where $$(\delta_n)_{n \in \mathbb N}$$ is a sequence of positive real numbers and $$(a_n)$$ a sequence of complex numbers of modulus larger than one and converging to one. Let $$f_n^{(r)}(z)$$ denote the sum of the first $$r$$ terms in the power series expansion of $$f_n(z)$$ and $$\displaystyle f^{(r)}(z) \equiv \sum_{n=1}^\infty f_n^{(r)}(z)$$.

We’ll prove that:

1. If $$\sum_n \delta_n \lt \infty$$ then $$\sum_{n=1}^\infty f_n^{(r)}(z)$$ converges and $$f(z) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(z)$$ for $$\vert z \vert \le 1$$ and $$z \neq 1$$.
2. If $$a_n=1+i \epsilon_n$$ and $$\sum_n \delta_n/\epsilon_n < \infty$$ then $$\sum_{n=1}^\infty f_n^{(r)}(1)$$ converges and $$f(1) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(1)$$
3. If $$\delta_n/\epsilon_n^2 \to \infty$$ then $$f(z)$$ is unbounded on the disk $$\vert z \vert \le 1$$.

First, let’s recall this corollary of Lebesgue’s dominated convergence theorem:

Let $$(u_{n,i})_{(n,i) \in \mathbb N \times \mathbb N}$$ be a double sequence of complex numbers. Suppose that $$u_{n,i} \to v_i$$ for all $$i$$ as $$n \to \infty$$, and that $$\vert u_{n,i} \vert \le w_i$$ for all $$n$$ with $$\sum_i w_i < \infty$$. Then for all $$n$$ the series $$\sum_i u_{n,i}$$ is absolutely convergent and $$\lim_n \sum_i u_{n,i} = \sum_i v_i$$.
Continue reading A power series converging everywhere on its circle of convergence defining a non-continuous function

# Root test

The root test is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number. The root test was developed first by Augustin-Louis Cauchy.

We denote $l = \limsup\limits_{n \to \infty} \sqrt[n]{\vert a_n \vert}.$ $$l$$ is a non-negative real number or is possibly equal to $$\infty$$. The root test states that:

• if $$l < 1$$ then the series converges absolutely;
• if $$l > 1$$ then the series diverges.

The root test is inconclusive when $$l = 1$$.

### A case where $$l=1$$ and the series diverges

The harmonic series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$$ is divergent. However $\sqrt[n]{\frac{1}{n}} = \frac{1}{n^{\frac{1}{n}}}=e^{- \frac{1}{n} \ln n}$ and $$\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n}} = 1$$ as $$\lim\limits_{n \to \infty} \frac{\ln n}{n} = 0$$.

### A case where $$l=1$$ and the series converges

Consider the series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$. We have $\sqrt[n]{\frac{1}{n^2}} = \frac{1}{n^{\frac{2}{n}}}=e^{- \frac{2}{n} \ln n}$ Therefore $$\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n^2}} = 1$$, while the series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$ is convergent as we have seen in the ratio test article. Continue reading Root test

# Ratio test

The ratio test is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number and is nonzero when $$n$$ is large. The test is sometimes known as d’Alembert’s ratio test.

Suppose that $\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = l$ The ratio test states that:

• if $$l < 1$$ then the series converges absolutely;
• if $$l > 1$$ then the series diverges.

What if $$l = 1$$? One cannot conclude in that case.

### Cases where $$l=1$$ and the series diverges

Consider the harmonic series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$$. We have $$\lim\limits_{n \to \infty} \frac{n+1}{n} = 1$$. It is well know that the harmonic series diverges. Recall that one proof uses the Cauchy’s convergence test based for $$k \ge 1$$ on the inequalities: $\sum_{n=2^k+1}^{2^{k+1}} \frac{1}{n} \ge \sum_{n=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = \frac{2^{k+1}-2^k}{2^{k+1}} \ge \frac{1}{2}$

An even simpler case is the series $$\displaystyle \sum_{n=1}^\infty 1$$.

### Cases where $$l=1$$ and the series converges

We also have $$\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = 1$$ for the infinite series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$. The series is however convergent as for $$n \ge 1$$ we have:$0 \le \frac{1}{(n+1)^2} \le \frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}$ and the series $$\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n} – \frac{1}{n+1} \right)$$ obviously converges.

Another example is the alternating series $$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}$$.

# Radius of convergence of power series

We look here at the radius of convergence of the sum and product of power series.

Let’s recall that for a power series $$\displaystyle \sum_{n=0}^\infty a_n x^n$$ where $$0$$ is not the only convergence point, the radius of convergence is the unique real $$0 < R \le \infty$$ such that the series converges whenever $$\vert x \vert < R$$ and diverges whenever $$\vert x \vert > R$$.

Given two power series with radii of convergence $$R_1$$ and $$R_2$$, i.e.
\begin{align*}
\displaystyle f_1(x) = \sum_{n=0}^\infty a_n x^n, \ \vert x \vert < R_1 \\ \displaystyle f_2(x) = \sum_{n=0}^\infty b_n x^n, \ \vert x \vert < R_2 \end{align*} The sum of the power series \begin{align*} \displaystyle f_1(x) + f_2(x) &= \sum_{n=0}^\infty a_n x^n + \sum_{n=0}^\infty b_n x^n \\ &=\sum_{n=0}^\infty (a_n + b_n) x^n \end{align*} and its Cauchy product:
\begin{align*}
\displaystyle f_1(x) \cdot f_2(x) &= \left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{n=0}^\infty b_n x^n \right) \\
&=\sum_{n=0}^\infty \left( \sum_{l=0}^n a_l b_{n-l}\right) x^n
\end{align*}
both have radii of convergence greater than or equal to $$\min \{R_1,R_2\}$$.