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Is the function $\hat{i}_0(x) = e^{-|x|} \sqrt{\frac{\pi}{2x}} I_{\frac{1}{2}}(x)$ positive or negative for negative $x$?

$I_{\alpha}(x)$ above is a modified Bessel function.

Here are my arguments. Considering that $I_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sinh(x)$, $\hat{i}_0(x)$ can be represented as follows:

$$ \hat{i}_0(x) = e^{-|x|} \sqrt{\frac{\pi}{2x}} \sqrt{\frac{2}{\pi x}} \sinh(x)$$

$$ \hat{i}_0(x) = \frac{e^{-|x|} \sinh(x)}{(\sqrt{x})^2}$$

Using the convention that $\sqrt{x} = i\sqrt{-x}$ for negative $x$, we get

$$ \hat{i}_0(x) = \frac{e^{-|x|} \sinh(x)}{x}$$

which is positive for negative $x$:

plot of \hat{i}_0(x)

However, using the original formula, Wolfram Alpha says that the function is negative for negative $x$:

plot of \hat{i}_0(x)

Am I missing something?

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Ignoring the scaling for the moment (it's always positive, so we can ignore it for now), you have the modified spherical Bessel function of the first kind $\mathrm{i}_0(x)=\frac{\sinh\,x}{x}$, which is always nonnegative... –  J. M. Jul 12 '12 at 17:43
    
@J.M.: Thanks for the pointer. Any ideas why Wolfram Alpha thinks differently? –  vitaut Jul 12 '12 at 17:47
    
I gave an explanation in the comment to Robert's answer; note that Mathematica is the thing that's running behind Wolfram Alpha. –  J. M. Jul 12 '12 at 17:51
    
You could give the following function to Wolfram Alpha, though: Exp[-Abs[x]] SphericalBesselJ[0, I x]. –  J. M. Jul 12 '12 at 17:57
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1 Answer

up vote 1 down vote accepted

What you're missing is that $\sqrt{1/x}$ is not the same as $1/\sqrt{x}$ when $x$ is negative. Indeed, $\sqrt{1/(-1)} = \sqrt{-1} = i$ but $1/\sqrt{-1} = 1/i = -i$. You might try asking Wolfram Alpha for $e^{-|x|} \dfrac{\sqrt{\pi}}{\sqrt{2x}} I_{1/2}(x)$.

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Mathematica, being the careful sort, evaluates the OP's function (omitting the scaling factor) as (Sqrt[1/x] Sinh[x])/Sqrt[x]; since the Sqrt[1/x]/Sqrt[x] is always positive, Sinh[x] is the only way the function can be negative and indeed it does become negative for negative argument. –  J. M. Jul 12 '12 at 17:48
    
@Robert Israel: You are right. Thank you for pointing this out. –  vitaut Jul 12 '12 at 18:04
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