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Referring to this functon: $ J_n(x)= \sum_{k=0}^{\infty}\frac{\substack{(-1)^k}} {\substack{k!\Gamma(k+n+1)}}(\frac{\substack{x}} {\substack{2}})^{2k+n} , n\geq0.$

When changing the $n$ subscript value (i.e. determining the order of Bessel function), I know that when graphing order zero, Bessel function peaks about $x=0$ and graphing order one makes it intercept the origin at $(0,0)$. However, I would like to know the reason for this.

Kevin Nguyen
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  • The reason for what? It's just what the series at $n = 0; 1$ give. – Enrico M. Apr 18 '18 at 11:35
  • Pull the $n$ from inside the sum to the outside, and you immediately get the behaviour for $x\to 0$ – gammatester Apr 18 '18 at 11:39
  • Sorry English is my second language. When graphing Bessel function of order zero, it's clear that it peaks about $x=0$. If the question asked that they wanted you to change it so that Bessel function contains a point of inflection about $(0,0)$, how would you explain that you have to change the order of the function? There must be a reason why it changes like this. – Kevin Nguyen Apr 18 '18 at 11:41
  • I'll give some context. I'm trying to graph the vibration of a circular membrane when hit in the middle. So, to do this I used Bessel function of order zero. If I were to hit it on the side, I tried to generally represent the vibration of the drum membrane as Bessel function of order one. I don't really know how to explain that I should change the order. – Kevin Nguyen Apr 18 '18 at 11:45

2 Answers2

1

Let's start with the true definition of Bessel type 1 functions:

$$J_{\nu}(x) = \left(\frac{x}{2}\right)^{\nu}\sum_{k = 0}^{+\infty} \left(\frac{x}{2}\right)^{2k} \frac{(-1)^k}{\Gamma(k+1)\Gamma(k+\nu+1)}$$

Now when $\nu = 0$ (your $n$), we just get

$$J_{0}(x) = \sum_{k = 0}^{+\infty} \left(\frac{x}{2}\right)^{2k} \frac{(-1)^k}{\Gamma^2(k+1)}$$

Which has the following plot

enter image description here

Taking the first terms of the series give you:

$$1 - \frac{x^2}{4} + \frac{x^4}{64} + \ldots$$

And as $x \to 0$ you get $1$.

When instead $\nu = 1$, the series is

$$J_{1}(x) = \left(\frac{x}{2}\right)\sum_{k = 0}^{+\infty} \left(\frac{x}{2}\right)^{2k} \frac{(-1)^k}{\Gamma(k+1)\Gamma(k+2)}$$

The plot of which is

enter image description here

Here if you take the first terms of the series you get

$$\frac{x}{2}\left(1 - \frac{x^2}{8} + \ldots\right)$$

And as $x\to 0$ you get $0$.

Enrico M.
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0

Just look at the first few terms. $$J_0(x) = 1 - \frac{1}{4}x^2 + \dots$$ and $$J_1(x) = \frac{1}{2}x - \frac{1}{16} x^3+\dots$$ As with all convergent Maclaurin series, these tell us the behavior near $x=0$.

Perhaps the $\Gamma$ function in there is making it difficult. When $n$ is an integer, those gamma functions can be written as factorials instead.

GEdgar
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