Does there exist a function $f\in C^n(\mathbb{R},\mathbb{R})$ for $n\ge2$ such that $f^{(n)}$ has exactly $n$ zeros, $f^{(n-1)}$ has exactly $n-1$ zeros and so on ? Where $f^{(n)}$ is the nth derivative of $f$

This question arose from lot of question asked here on MSE.

  • $\begingroup$ If u do consider on multiplicity then u can take the trivial example $x^n$. If u doesn't count multiplicity, then u can take something like this: f(x)=(x-x_1)^1+(x-x_2)^2+(x-x_3)^3+....+(x-x_n)^n , with pairwise disjoint x_j. Did u want something like that,or did i misunderstand you? $\endgroup$ – Marm Jun 26 '14 at 12:37
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    $\begingroup$ @Balla: $f(x)=x^n$ doesn't work. Then $f^{(n)}$ has no zeros (instead of $n$), etc. $\endgroup$ – Hans Lundmark Jun 26 '14 at 12:41
  • $\begingroup$ Oh, iam sorry. I was reading to fast, i think :/, nice question, btw $\endgroup$ – Marm Jun 26 '14 at 12:43
  • $\begingroup$ Did you want an explicit formula for such a function? You can definitely construct a function for any finite value of $n$, but finding an explicit function sounds rough. $\endgroup$ – cnick Jun 26 '14 at 12:49
  • $\begingroup$ @cnick This is what I would like, yes. $\endgroup$ – user146010 Jun 26 '14 at 12:54

Yes, the function $f(x)=e^{-x^2}$ has this property; the $n$th derivative is $f^{(n)}(x)=(-1)^n H_n(x) \, e^{-x^2}$, where $H_n$ is the Hermite polynomial of degree $n$, which because of orthogonality has exactly $n$ real zeros (they have to interlace with the zeros of $H_{n-1}$).

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The function $f(x)=2-\frac{1}{1+x^2}$ has the desired property for at least $0 \le n \le 7.$ The derivatives $f^{(n)}$ (for $n>0$) each have a denominator a power of $1+x^2$, and a constant in front of a polynomial $p_n(x)$ which for odd $n$ hazs a factor of $x$ and the rest is in powers of $x^2$, whereas for even $n$ there is no factor of $x$ in $p_n$ and the rest is in even powers of $x$.








That's where I stopped; to check for example that $p_6$ has $6$ zeros I rewrote $x^2$ as $t$ and looked at the cubic, which at $t=0,0.1,1,0,5,0$ has values respectively $-1,+0.7,-8,+104$ so that there are three positive zeros for the $t$ polynomial, hence $6$ zeros for $p_6$ And $p_7$ was easy sonce the cubic here factors as $(t-1)(t^2-6t+1)$, which has $3$ positive zeros, so that (recalling the factor $x$) $p_7$ ends up with seven zeros.

The pattern here makes it seem possible that this function $f(x)$ has simultaneously the desired numbers of zeros for each derivative; however I don't see any good way to show that, by induction or otherwise.

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    $\begingroup$ I would guess that there is some orthogonality which explains why these polynomials have only real zeros. $\endgroup$ – Hans Lundmark Jun 26 '14 at 14:50
  • $\begingroup$ Yes that would be nice, and finish it as in your example above. (+1 on that answer, BTW). If I see how to use orthogonal functions here I'll add to above answer. $\endgroup$ – coffeemath Jun 26 '14 at 15:13

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