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I would like to know what the value of the bessel functions of the first kind and the modified bessel of the first kind is at 0.

I think for order 0 they are 1 and for orders greater than 0 they are 0.

ie

$J_0 (0) = 1$ and $J_v(0) = 0$ for $v$ > 0

$I_0 (0) = 1$ and $I_v(0) = 0$ for $v$ > 0

Is the above correct?

evan54
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1 Answers1

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The Bessel and modified Bessel of the first kind can be expressed in series form as \begin{align} J_{\nu}(x) &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k! (\nu + k)!} \left( \frac{x}{2} \right)^{2k+\nu} \end{align} and \begin{align} I_{\nu}(x) &= \sum_{k=0}^{\infty} \frac{1}{k! (\nu + k)!} \left( \frac{x}{2} \right)^{2k+\nu}. \end{align} The first term of each is \begin{align} J_{\nu}(x) &= \left( \frac{x}{2} \right)^{\nu} + \cdots \\ I_{\nu}(x) &= \left( \frac{x}{2} \right)^{\nu} + \cdots \end{align} Since $0^{\nu} = 0$ for $\nu \geq 1$ and $0^{0}= 1$ for $\nu = 0$ then it is evident that $J_{0}(0) = 1$ and $J_{\nu}(0) = 0$ for $\nu \geq 0$. Applying the same to the modified function yields the similar result.

Leucippus
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    Should the couple of "ν ≥ 0" just be "ν > 0"? (i.e. without the "or equal to", since the ν = 0 case is separate) – tscizzle Mar 18 '23 at 23:43