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I want to construct a non-negative function $f:\mathbb R\to\mathbb R$ such that $f$ smoothly vanishes on the interval $[M,\infty)$. This means $\lim_{x\to M-}=0, f(x)=0$ for all $x\geq M$, and also for any $k\geq 1$, $f^{(k)}(x)=0$ for all $x\geq M$ and $\lim_{x\to M-} f^{(k)}(x)=0$ where $f^{(k)}$ means $k$'th derivative of $f$.

If $M=\infty$ instead of a real number, $f(x)=\dfrac{e^x}{(1+e^x)^2}$ works. I want a kind of a Gaussian like function that becomes zero at a finite $M$ but smoothly.

Any help is appreciated.

Landon Carter
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2 Answers2

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Let $f$ defined as follows :

$$ f : \Bbb R \to \Bbb R \\ f(x) = \cases{x < M, \hspace{0.5 cm} e^{\frac{1}{x-M}}\\x \geq M, \hspace{0.5cm} 0} $$ this function is continuous and continuously differentiable at all points of $\Bbb R$ except possibly at $M$. So we have to show that is continuously differentiable at $M$.

If we try to compute the derivatives of all orders near and at $M$, we would notice that basically, we should compute limits of this form :

$$ \lim_{x \to 0_{-}} \frac{1}{x^{k}}e^{\frac{1}{x}}.$$

with the substitution $x = X-M$ and $k >=1$. But we have :

$$ \lim_{x \to 0_{-}} \frac{e^{\frac{1}{x}}}{x^k} = \lim_{ y \to \infty} \frac{e^{-y}}{y^{k}}=0. $$

Amirhossein
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Consider a Gaussian and bring infinity to $M$.

$$e^{-1/(x-M)^2}.$$