0
$\begingroup$

I am trying to find a single (i.e. non-piecewise) analytic function $f(x)$ with the following features:

$f(x), f'(x)$ defined $ \forall x \in \mathbb R$
$f'(x) > 0, \forall x$
$lim_{x \to - \infty} f(x) = - \infty$
$lim_{x \to + \infty} f(x) = + \infty$
$lim_{x \to - \infty} f'(x) = m, m \in \mathbb R^+$
$lim_{x \to + \infty} f'(x) = m, m \in \mathbb R^+$
$f'(\frac 1 2) = n, n \in \mathbb R^+, n \le \frac m {10}$ # no longer applicable, see edit below #

So a monotonically increasing curve that approximates a straight line of slope $m$ almost everywhere, except for a section with much lower slope $n$ around $x = \frac 1 2$, say for $0.1 < x < 0.9$.

How would you go about putting together such a function?

EDIT

I had forgotten to include a condition:

$f'(x) \ge n, \forall x$

Also, I realized the function I am looking for becomes symmetric with respect to the origin if I translate $x$ by $- \frac 1 2$, so in fact the previously mentioned condition on the derivative becomes:

$f'(0) = n, n \in \mathbb R^+, n \le \frac m {10}$

$\endgroup$
2
$\begingroup$

EDIT: Taking into account the OP's edits, we have to modify the previous function slightly. I'm merging it with the original answer for clarity.

Assume $m,n,a > 0, m>n$. Then, consider the function $$f(x) = \int_0^x \left(m-(m-n)e^{-\frac{u^2}{a^2}}\right) du = mx-(m-n)a\frac{\sqrt{\pi}}{2} \ \text{erf}\left(\frac{x}{a}\right)$$

Thus, we get for $f(x)$

  1. As $\text{erf}(-\infty)$ is finite, $\lim_{x\to-\infty} f(x) = -\infty$
  2. As $\text{erf}(\infty)$ is finite, $\lim_{x \to \infty} f(x) = \infty$

Also, we have $f'(x) = m-(m-n)e^{-\frac{x^2}{a^2}}$. This gives for $f'(x)$

  1. For all $x$, as $e^{-\frac{x^2}{a^2}} \le 1$, we get \begin{align} f'(x) &= m-(m-n)e^{-\frac{x^2}{a^2}} \\ &\ge m-(m-n)\\ &=n \end{align}
  2. $f'(0) = m-(m-n)e^{-\frac{0^2}{a^2}}= m-(m-n) =n \ \ \forall \ x$
  3. $\lim_{x\to-\infty} f'(x) = \lim_{x\to-\infty} \left(m-(m-n)e^{-\frac{x^2}{a^2}}\right) = m$
  4. $\lim_{x\to\infty} f'(x) = \lim_{x\to\infty} \left(m-(m-n)e^{-\frac{x^2}{a^2}}\right) = m$

Note that $m, n, a$ here are completely general. So, we can now impose any other conditions, like $n\le \frac{m}{10}$ trivially.

$\endgroup$
4
  • $\begingroup$ That's brilliant, thank you very much! I realize now that I omitted an important condition: $f'(x) \ge n, \forall x$. And in fact I would need $m$ to be a parameter in the equation. I might try to adapt your equation to achieve this, but if you have any suggestions, as you were so quick finding what I was looking for... $\endgroup$ Jan 22 at 18:12
  • $\begingroup$ Could you write it down in your main question? It'll both help me and others to answer your question better. $\endgroup$
    – Ishan Deo
    Jan 22 at 19:29
  • $\begingroup$ Sure, thanks. Done. $\endgroup$ Jan 22 at 19:42
  • $\begingroup$ There. I've changed it. $\endgroup$
    – Ishan Deo
    Jan 22 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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