Funtion bessel

**babs** · April 2nd 2008, 05:31 AM

Hello to everybody!
I have a problem with Bessel funtion.
Can I say that Jo^2(x)=Jo(x)*Jo(x)

**ThePerfectHacker** · April 2nd 2008, 06:42 AM

If you asking that $J_{\mu} ^2 (x) = J_{\mu} (x) J_{\mu}(x)$ the answer is yes. The same thing with other functions too. For example, $\sin^2 x = \sin x \cdot \sin x$ .

**CaptainBlack** · April 2nd 2008, 06:56 AM

Originally Posted by ThePerfectHacker

If you asking that $J_{\mu} ^2 (x) = J_{\mu} (x) J_{\mu}(x)$ the answer is yes. The same thing with other functions too. For example, $\sin^2 x = \sin x \cdot \sin x$ .

However it's a notation to be deprecated as it allows ambiguous notation since we also have:

$J_{\mu} ^2 (x) = J_{\mu} (J_{\mu} (x))$

to be encouraged is:

$(J_{\mu} (x))^2 = J_{\mu} (x) J_{\mu}(x)$

The same goes for sin, also I don't like the usage $\sin x$ rather than $\sin (x)$

RonL

**ThePerfectHacker** · April 2nd 2008, 06:59 AM

Originally Posted by CaptainBlank

$J_{\mu} ^2 (x) = J_{\mu} (J_{\mu} (x))$

But this does not make sense. Bessel function's domain is usually positive. While the range is both positive and negative. So when you compose them you get a problem.

**CaptainBlack** · April 2nd 2008, 07:01 AM

Originally Posted by ThePerfectHacker

But this does not make sense. Bessel function's domain is usually positive.

Why don't you check that, what is usual for one person is not neccessarily so for another.

RonL

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