Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was trying to construct some functions, but I don´t know how I can do it Dx. The first it´s a continuous function, that has exactly the set $$ \left\{ 0 \right\} \cup \left\{ {\frac{1} {n}} \right\} $$ as local strictly maximum points. My problem is with the point 0. Dx

The second is construct ( if it exist ) an infinite differentiable function, such that $$ \eqalign{ & f^{\left( k \right)} \left( 0 \right) = 1\,\,\forall k \in {\Bbb N} \cr & f\left( 0 \right) = 1 \cr} $$ and it´s different from the exponential.

here it´s obvious that if exist such function exist, and it´s different from the exponencial, could not be analytic, otherwise, i´ll be the exponencial, but I don´t have some example If someone can help me with that )=

share|cite|improve this question
For the second question, take $e^x+f(x)$ where $f(x)$ is the standard example of a nonzero smooth function whose derivatives at $0$ all vanish. – Henning Makholm Oct 12 '11 at 21:09
up vote 3 down vote accepted

We will produce a function $g$ such that $0$ together with the set $\{\frac{1}{n}\}$, where $n$ ranges over $\mathbb{N}$ is the set of local strictly minimum points. If we set $f(x)=-g(x)$, then $f$ has the property that was asked for.

Let $g(0)=0$, and for any positive integer $n$, let $g(1/n)=\frac{1}{n}$. It remains to define $g(x)$ for $x$ in the intervals $\left(\frac{1}{n+1},\frac{1}{n}\right)$, and for $x>1$.

For any positive integer $k$, let $P_k$ be the point $(1/k,1/k)$. Let $m_n$ be the number midway between $1/(n+1)$ and $1/n$. Let $g(m_n)=2/n$, and let $M_n=(m_n,2/n)$.

For $x$ between $m_n$ and $1/n$, define $g(x)$ by specifying that $(x,g(x))$ lies on the straight line segment that joins $M_n$ and $P_n$. Similarly, for $x$ between $1/(n+1)$ and $m_n$, define $g(x)$ by specifying that $(x,g(x))$ lies on the straight line segment that joins $P_{n+1}$ and $M_n$. Finally, for $x>1$, let $g(x)=1+x$.

This $g(x)$ does the job. Continuity at $0$ is ensured because near $0$, our function does not dip much below $0$. If we also want things to work for negative $n$, reflect the curve we have described across the $y$-axis.

The function $g$ has many points of non-differentiability. It is not hard to smooth it out, so that it is everywhere infinitely differentiable.

For the second question, @Henning Makholm's comment is a full answer.

share|cite|improve this answer
@Henning Makholm: Definitely not on purpose! Thanks for pointing out the typo. – André Nicolas Oct 13 '11 at 2:33
$@$André, so I thought. I had tried correcting it myself, but my edit must have been lost in crosstalk. – Henning Makholm Oct 13 '11 at 2:43
@Henning Makholm: Thanks for trying to fix it. Last I looked, it is fixed. Am not sure which one of us did it. – André Nicolas Oct 13 '11 at 2:50

Your Answer


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.