# How is the Bessel function approximated by a ln function

## Homework Statement

It is stated in "Mathematical methods of Physics" by J. Mathews, 2nd ed, p274, that the Bessel function of the second kind and of order zero, i.e. $Y_0(x)$ can be approximated by $\frac{2}{\pi}\ln(x)$+constant as $x \to 0$, but no more details are given in the same text.

## Homework Equations

Could you show me how that approximation comes about?

## The Attempt at a Solution

Visually I see the resemblance between the functions, but I cannot derive the approximation myself.

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## Answers and Replies

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gabbagabbahey
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## Homework Statement

It is stated in "Mathematical methods of Physics" by J. Mathews, 2nd ed, p274, that the Bessel function of the first kind and of order zero, i.e. $Y_0(x)$ can be approximated by $\frac{2}{\pi}\ln(x)$+constant as $x \to 0$, but no more details are given in the same text.

## Homework Equations

Could you show me how that approximation comes about?

## The Attempt at a Solution

Visually I see the resemblance between the functions, but I cannot derive the approximation myself.
Hi sinbaski, welcome to PF!

Do you mean the Bessel function of the second kind? (Bessel functions of the first kind are finite at the origin. and $\frac{2}{\pi}\ln(x)$ is not)

Yes, gabbagabbahey, I meant the second kind. Sorry, that was a careless typo! :|

gabbagabbahey
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Okay, well there are a few different ways of defining the Bessel functions, so what definition are you using in your course?

In the same book $Y_m$ is defined as

$Y_m(x) = \frac{J_m(x) \cos m\pi - J_{-m}(x)}{\sin m\pi}$

while $J_m(x)$ is defined as
$J_m(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r! \Gamma(r+m+1)}\left(\frac{x}{2}\right)^{2r+m}$

vela
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I think the answer will pop out if you go back to solving the Bessel differential equation for m=0. The method of Frobenius will yield one solution J0(x), and now you need to construct a second, independent one.

I have to admit I can't solve by hand the Bessel equation
$x^2 y'' + x y' + x^2 y = 0$

Matlab gives the solution to the equation as
$A J_0(x) + B Y_0(x)$

Still I don't see how $\frac{2}{\pi}\ln(x)$ emerges from the equations :(

vela
Staff Emeritus
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You want to learn about the Frobenius method. Depending on the roots so the indicial equation, sometimes it'll yield two independent solutions, but in this case, it doesn't. Wikipedia just tells you the form of the second solution.

http://en.wikipedia.org/wiki/Frobenius_method#Double_roots

If you look on page 16 of Mathews and Walker, there's a discussion of how to solve Bessel's equation using Frobenius, including how to find the second solution.

vela
Staff Emeritus
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I did a bit more reading, and it looks like to get the constant factor of ##2/\pi##, you're going to have to look at the definition of Ym(x). It's an indeterminate form when m is an integer, so you want to evaluate it using L'Hopital's rule.

gabbagabbahey
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In the same book $Y_m$ is defined as

$Y_m(x) = \frac{J_m(x) \cos m\pi - J_{-m}(x)}{\sin m\pi}$
For integer $m$, this should really be

$$Y_m(x) = \lim_{\alpha \to m} \frac{J_{\alpha}(x) \cos \alpha \pi - J_{-\alpha}(x)}{\sin \alpha \pi}$$

while $J_m(x)$ is defined as
$J_m(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r! \Gamma(r+m+1)}\left(\frac{x}{2}\right)^{2r+m}$
Okay, so use this definition, with the (corrected) one above to find an expression for $Y_0(x)$.

Note: There's no need for you to actually solve Bessel's differential equation for this problem (the above equations are the two independent solutions), but it is a good exercise and one you may want to do after this problem (or before)

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Thank you for your kind reply! The wikipage is very comprehensive compared with the book. Now I think I have got something, following your advice:
$\lim_{m\to 0} \frac{\cos m\pi J_m(x) - J_{-m}(x)}{\sin m\pi}$
$= \lim_{m\to 0}\frac{ -\pi\sin m\pi J_m(x) + \cos m\pi \frac{\partial}{\partial m}J_m(x) + \frac{\partial}{\partial (-m)}J_{-m}(x) }{\pi \cos m\pi}$
$= \frac{2}{\pi} \lim_{m\to 0} \frac{\partial}{\partial m}J_m(x)$

Then
$\frac{\partial}{\partial m}J_m(x) = \frac{\partial}{\partial m}\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r}$
$= \ln(x/2)\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r} + \sum_{r=0}^{\infty} \left(\frac{x}{2}\right)^{m+2r} \frac{\partial}{\partial m}\frac{(-1)^r}{r!\Gamma(m+r+1)}$

Taking the limit $x \to 0, m\to 0$, the first sum reduces to $\ln(x/2)$ while the second becomes a constant. Thus we get the desired result of $\frac{2}{\pi}\ln(x)+C$

Do you think this argument is correct?

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gabbagabbahey
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$= \ln(x/2)\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r} + \sum_{r=0}^{\infty} \left(\frac{x}{2}\right)^{m+2r} \frac{\partial}{\partial m}\frac{(-1)^r}{r!\Gamma(m+r+1)}$

Taking the limit $x \to 0, m\to 0$, the first sum reduces to $\ln(x/2)$ while the second becomes a constant. Thus we get the desired result of $\frac{2}{\pi}\ln(x)+C$

Do you think this argument is correct?
I'm not sure it makes any sense to say that taking the limit $x \to 0$ and $m \to 0$ of $\ln(x/2)\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r}$ results in $\ln(x/2)$. I think you should take the limit $m \to 0$ of the whole expression (obviously you will need to first calculate $\frac{\partial}{\partial m}\frac{(-1)^r}{r!\Gamma(m+r+1)}$ ), simplify it as much as possible, and then do a Taylor Series approximation for small $x$.