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.
On
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
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
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 $0$.
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.