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Note:This is more a math-recreational question

Consider the series $\exp_p(x)=\sum_{k=0}^{\infty}\frac{x^k}{(k!)^p}$ which is some systematic modification of the exponential function. It's $\exp_1(x)=\exp(x) $ if p=1. I'm interested in the behaviour for $x=0\to -\infty$ depending on change of the parameter $p$.

[edit] For the plots, and also for the discussion of the oscillation I removed the constant 1 from the function. Then I edited that question for correction. But after I got some answer I better roll back the edits of the title and the formula and state that removal for the plots and the discussion of the oscillation only here explicitely instead. Please excuse that inconvenience [edit 2]

First oberservation is, that if $p>1$ the function begins to oscillate. If $p=2$ the oscillation diminuishes but for $p>2$ in general the oscillation seems to increase and even for a couple of values $p=2 + \epsilon $ which I checked manually where the (possibly) eventual increase occurs after an initial decrease.

So my questions are:

  1. is it true, that at p=2 and x from $ 0\to -\infty$ the amplitude of the oscillation diminueshes to zero?
  2. is it true, that at $p=2+\epsilon, \epsilon>0$ and x from $ 0\to -\infty$ the amplitude of the oscillation eventually increases without bound?
  3. How could I approach that question, for instance by considering the form of the power series, the analysis of the powers of the factorials. Are there any "keystones" which might be helpful?

This is a plot for $p=2$ so $f(x)=\exp_2(-(x^2)) -1 $. I used the squaring of the x to see more of the oscillation. The -1 is to locate the oscillation around the x-axis.

enter image description here

This is a plot for $p=2.02$ so $f(x)=\exp_{2.02}(-x^2) -1 $. Again I used the squaring of the x to see more of the oscillation. Bigger epsilons let the oscillation increase at an x nearer to the vertical $x=0$-line.

enter image description here

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1 Answer 1

up vote 17 down vote accepted

For p = 2, $exp_2(x) = J_0(2 \sqrt{x})$, where $J_0$ is a Bessel function. The asymptotics of this are known: as $x \to -\infty$, $exp_2(x) = \frac {\sin \left( 2\,\sqrt {x}+1/4\,\pi \right) }{\sqrt {\pi }\, x^{1/4}} + O(x^{-3/4})$.

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Hi Professor Israel! Welcome to Math.SE. I hope you'll enjoy it here. –  Willie Wong Mar 11 '11 at 19:09
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@Robert israel:thanks; that settles the question for p=2. Hmm, how are the $x^{1/4}$ in the denominator defined for the negative x? The principal branch of the log? –  Gottfried Helms Mar 11 '11 at 19:21
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@Gottfried Helms: I think there is a typo. The sum is $\exp_2 (x) = J_0(2 \sqrt{|x|})$ for $x<0$. –  Fabian Mar 11 '11 at 20:45
    
@Fabian: hmm, Pari/GP gives for $besseli(0,2*sqrt(-10))$ the value $0.228843818615 + 0*I$ (where I is the imaginary unit) and this is the same as by explicite evaluation of the series $exp_2(-10)$. If I used $besseli(0,2*sqrt(abs(-10)))$ I get something like 90.5. –  Gottfried Helms Mar 11 '11 at 22:12
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@Gottfired Helms: but remember $I_0(2 \sqrt{-10})= I_0(2i \sqrt{10}) = J_0(2 \sqrt{10})$ and $J_0$ is the Bessel function in Robert Israel's post. –  Fabian Mar 11 '11 at 22:56
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