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It is well-known that $f(x) = 1 - e^{-\frac{1}{x^2}}$ satisfies $f^{(i)}(0) = 1$ for all natural numbers $i.$ However, its support $\{x\in \mathbb{R}:f(x)\neq 0\} = \mathbb{R}.$

Question: Can one modify $1 - e^{-\frac{1}{x^2}}$ so that all derivatives at $x=0$ are $1$ and support is $[-1,1]?$

I tried $1 - e^{1-\frac{1}{x^2}}$ because it has value $0$ at both $x=-1$ and $x=1.$ However, it is not differentiable at those points.

Idonknow
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  • What modifications are you allowing? You can certainly define $f(x)=e^{\frac {-1}{x^2}}$ on $[-1,1]$ and $0$ elsewhere. It is discontinuous at $-1,1$ I suspect you want bump functions The first example in the article seems to be just what you want. – Ross Millikan Mar 05 '19 at 06:14
  • @RossMillikan: Any modification is allowed. The first example in the article does not satisfy $f^{(i)}(0) = 0.$ In fact, its second derivative is negative at $x=0.$ What I want is $f^{(i)}(0) = 0$ for all $i$ and its support is in $[-1,1].$ – Idonknow Mar 05 '19 at 06:23
  • @RossMillikan I apologize for typing my question wrongly. Please refer to the edited version. – Idonknow Mar 05 '19 at 07:21

3 Answers3

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How about $$ f(x) = \exp\left[-\frac{1}{x^2(1-x^2)}\right]\mathbf{1}_{(-1,1)}, $$ where $\mathbf{1}_A$ is the indicator function of the set $A$. This function is smooth, has support $[-1,1]$, and has all derivatives vanish at $x = 0$.

EDIT: For your second question, this function $$ f(x) = \frac{\exp(x)}{1+\exp\left(\frac{1}{1-x^2}-\frac{1}{x^2}\right)}\mathbf{1}_{(-1,1)} $$ is smooth, has support $[-1,1]$, and has all derivatives equal to $1$ at zero. To show this, let $g(x) = (1+\exp[(1-x^2)^{-1}-x^{-2}])^{-1}$. Note that $g(0)=1$, $g^{(n)}(0) = 0^n$, and $f(x) = e^x g(x)$. Then $$ f^{(n)}(0) = \sum_{i=0}^n\binom{n}{i}e^0g^{(i)}(0) = 1+\sum_{i=1}^n\binom{n}{i}g^{(i)}(0) = 1. $$

Note that $f$ is not analytic at $0$ here, although it's pretty close.

eyeballfrog
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    Nice. But is it clear that all derivatives of $f$ vanishes at $x=0?$ – Idonknow Mar 05 '19 at 06:50
  • I apologize for typing my question wrongly. Please refer to the edited version. In particular, I would like to have $f$ such that $f^{(i)}(0) = 1$ for all $i$ and support of $f$ is $[-1,1].$ – Idonknow Mar 05 '19 at 07:22
  • @Idonknow Yes, it is clear. Every derivative takes the form $f^{(n)}(x) = R(x)f(x)$, where $R(x)$ is a rational function. No matter what order pole $R(x)$ has at zero, it's swamped by the $\exp(-x^{-2})$ in $f(x)$. As for the modified question, that's a little easier since such an $f$ can be analytic at $0$. – eyeballfrog Mar 05 '19 at 17:34
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With the new requirements, it looks like you might want to define your function as:

  • infinitely smooth step function (this one will do) times $e^x$ on $[-1,0]$;
  • decreasing step function times $e^x$ on $[0,1]$

This way you'll have all derivatives equal to 0 at both ends, and all derivatives equal to 1 at $x=0$.

All right, if you want an explicit formula, you may have it.

$$f(x)=\begin{cases}\dfrac{e^{-{1\over1+x}}}{e^{-{1\over1+x}}+e^{{1\over x}}}\cdot e^x, & -1<x<0 \\[10pt] \dfrac{e^{-{1\over1-x}}}{e^{-{1\over x}}+e^{-{1\over1-x}}}\cdot e^x, & \phantom{-}0<x<1\end{cases}$$

Ivan Neretin
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Following the advice of Ivan Neretin and his comment on my former answer, a possible example of the function you are searching for is the following one $$ f(x)= \begin{cases} e^x \cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}} & x\in]-1,+1[\\ \\ 0 & x\notin]-1,+1[ \end{cases} $$ You have that $$ \begin{split} \frac{\mathrm{d}f}{\mathrm{d}x}&=e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}+e^x\cdot\frac{\mathrm{d}}{\mathrm{d}x} e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}\\ &=e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}- e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\frac{e^{-\frac{1}{x^2}}}{1-x^2}\\ \end{split} $$ For $x=0$, the second term of the right side of the equation is $0$ since it is a product of functions one of which is the (infinitely) flat function, while the first one is $1$: proceeding by induction you can prove that $$ f^{(i)}(0)=1\quad\forall i\in\Bbb N, $$ again by the properties of the flat function.

Daniele Tampieri
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