Wikipedia:Reference desk/Archives/Mathematics/2024 July 18

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July 18

Functions whose every derivative is positive growing slower than exponential

Is there any smooth function with the following two properties:

$f^{(n)}(x)>0$, i.e. the nth derivative of f is strictly positive for every x and n.

$\lim _{x\rightarrow \infty }{\frac {f(x)}{b^{x}}}=0$ for every b > 1. The hard case is when b is small.

Functions like $a^{x}$ (for a > 1) are the only ones I can think of with the first property, but none of them has the second property because you can always choose b < a. So I am asking whether there is any function with the first property that grows slower than exponential.

120.21.218.123 (talk) 10:09, 18 July 2024 (UTC)

Wouldn't any power series with positive coefficients that decrease compared to the coefficients of the exponential do? The exponential is $1+\sum _{k=1}^{\infty }{\frac {1}{k!}}x^{k}$, so e.g. $f(x)=1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}$ should do the trick. The next question is whether you can find a closed-form expression for this or a similar power series. --Wrongfilter (talk) 13:02, 18 July 2024 (UTC)

Good thinking. It is of course the case that the first property holds for any power series where all coefficients are positive. Plotting on a graph, I think your specific example doesn't satisfy the second property, but others where the coefficients decrease more rapidly do. 120.21.218.123 (talk) 13:26, 18 July 2024 (UTC)

$1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}>1+\sum _{k=1}^{\infty }{\frac {1}{k{+}1}}{\frac {1}{k!}}x^{k}=$ $1+\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k}=$ $1+{\frac {1}{x}}\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k{+}1}=$ $1+{\frac {1}{x}}\sum _{k=2}^{\infty }{\frac {1}{k}}x^{k}={\frac {\exp x-1}{x}}=$ $\Omega (2^{x}).$  --Lambiam 13:45, 18 July 2024 (UTC)

A half-exponential function will satisfy your requirements. Hellmuth Kneser famously defined an analytic function that is the functional square root of the exponential function.  --Lambiam 14:04, 18 July 2024 (UTC)

References

A variant of Wrongfilter's idea that I think does work:

$f(x)=\sum _{k=0}^{\infty }{\frac {x^{k}}{(k!)^{2}}}$

(taking $0^{0}$ to be $1$).

Numerical evidence suggests that $\log f(x)\sim 2{\sqrt {x}}.$ One might therefore hope that $e^{2{\sqrt {x}}}$ would also work. However, its second derivative is negative for $x<{\tfrac {1}{4}}.$  --Lambiam 22:20, 20 July 2024 (UTC)

And some higher derivatives are negative for even larger values of x. The eighth derivative is negative for $x<6.17$, for instance. 120.21.79.62 (talk) 06:48, 21 July 2024 (UTC)

The fourteenth derivative is negative for $x<20$. That's as high as WolframAlpha will let me go. 120.21.79.62 (talk) 06:54, 21 July 2024 (UTC)

Yes, shifting the graph along the x-axis by using $e^{2{\sqrt {x+c}}}$ won't help.  --Lambiam 11:46, 21 July 2024 (UTC)