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I need to approximate this expression in order to sum it. Asymptotically I obtain $\frac1{\sqrt{\pi}x}+\frac1{2\sqrt{\pi} x^2} + O\left(\frac1{x^3}\right)$. Although this looks fine there is the following problem:

If $x=a(1-\frac{k}{b})$ with $ a<<b $ and $\frac{b}{2} \leq k \leq b-1, \ x$ turns out to be greater than 1 for $k < k^{\ast}$ for some $k^{\ast}$ and less than 1 for $k \geq k^{\ast}$. This makes the asymptotic expansion somewhat tricky: for fixed $a,b$ the terms are either growing or contracting and then the whole approximation is essentially wrong. Is there any way to make this asymptotic expansion more exact?

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"More exact"? That's a bit like saying "more pregnant"... anyway I seem to be getting a different asymptotic expansion in Mathematica; it does look to me you can't get something free of exponential factors. –  J. M. Aug 21 '11 at 23:41
    
corrected the function –  sigma.z.1980 Aug 21 '11 at 23:47
    
Have you tried constructing a Padé approximant out of your asymptotic series? –  J. M. Aug 21 '11 at 23:50
    
Maple agrees with your asymptotic expression. But I don't understand what you're trying to do with $a$, $b$ and $k$. If $a$ and $b$ are fixed and $b/2 \le k \le b-1$, $x$ is not going to $\infty$ and so asymptotics are irrelevant. What is it that you're really trying to do? –  Robert Israel Aug 21 '11 at 23:57
    
I want to approximate the function in order to sum the approximation. This is where I see the problem: if $x=a(1-\frac{k}{b}$ and set e.g. $a=10,\ b=100$ then for $k <90 \ x $ is greater than 1, for $k \geq 90 \ x$ is greater than 1, and then terms in the asymptotic expansion become larger. Since I need only 1 term (the largest, of course), the problem becomes unbounded. –  sigma.z.1980 Aug 22 '11 at 0:07
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Maple (with MultiSeries) reports this asymptotic for $\mathrm{erfi}(x)$ ... $$ \Biggl(\frac{1}{\sqrt{\pi} x} + \frac{1}{2 \sqrt{\pi} x^{3}} + \frac{3}{4 \sqrt{\pi} x^{5}} + \frac{15}{8 \sqrt{\pi} x^{7}} + O \Bigl(x^{(-8)}\Bigr)\Biggr) \operatorname{e} ^{x^{2}} $$ Maple's definition of erfi is equivalent to: $$ \mathrm{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} \operatorname{e} ^{s^{2}} d s $$

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I use Matlab symbolic toolbox, but the results are the same, of course. The problem is that (cancelling out $exp(x^2)$) $x$ can be both more and less than 1, thus the terms in the series either grow or contract. –  sigma.z.1980 Aug 21 '11 at 23:58
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If $x$ is less than 1, then you shouldn't be using an asymptotic series; asymptotic series are intended to be used for $x$ of large magnitude... –  J. M. Aug 22 '11 at 2:58
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