- #1
ReyChiquito
- 120
- 1
Hello guys, i had a little chat with a teacher of mine and he asked me how can someone plot the zero order Bessel function. Here is what I've done..
using the integral expresion for [tex]J_{0}(r)[/tex]
[tex]J_{0}(r)=\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)d\theta[/tex]
i can calculate the first order derivative with respect to r
[tex]\frac {\partial}{\partial r}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \sin(r\cos\theta)\cos\theta d\theta[/tex]
wich when evaluated in r=0 is 0. For the second derivative
[tex]\frac {\partial^2}{\partial r^2}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)\cos^{2}\theta d\theta[/tex]
wich evaluated in r=0 is equal to -1/2.
The idea is to construct the taylor series around r=0. And given the fact that
[tex]|J^{(n)}(r)|\leq\frac{1}{\pi}[/tex]
i can easily bound the error, ie, if i only take two terms of the series
[tex]J_{0}(r)=1-\frac{r^2}{4}+E(r^4)[/tex]
where
[tex]|E(r)|\leq \frac{r^4}{4!\pi}[/tex]
so, if, for instance, i want to know where is the first zero of the function, given the first approximation, i can say that is on 2 with an error of 0.21...
given the next term
[tex]J_{0}(r)=1-\frac{r^2}{4}+\frac{r^4}{64}-E(r^6)[/tex]
where
[tex]|E(r)|\leq \frac{r^6}{6!\pi}[/tex]
tells me that the zero is in 2^(3/2) with an error of 0.23
and so on...
do you guys think this is a correct procedure?
is there any other way i can construct the plot?
i really want to impress my teacher, so any help would be well received.
Thx.
using the integral expresion for [tex]J_{0}(r)[/tex]
[tex]J_{0}(r)=\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)d\theta[/tex]
i can calculate the first order derivative with respect to r
[tex]\frac {\partial}{\partial r}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \sin(r\cos\theta)\cos\theta d\theta[/tex]
wich when evaluated in r=0 is 0. For the second derivative
[tex]\frac {\partial^2}{\partial r^2}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)\cos^{2}\theta d\theta[/tex]
wich evaluated in r=0 is equal to -1/2.
The idea is to construct the taylor series around r=0. And given the fact that
[tex]|J^{(n)}(r)|\leq\frac{1}{\pi}[/tex]
i can easily bound the error, ie, if i only take two terms of the series
[tex]J_{0}(r)=1-\frac{r^2}{4}+E(r^4)[/tex]
where
[tex]|E(r)|\leq \frac{r^4}{4!\pi}[/tex]
so, if, for instance, i want to know where is the first zero of the function, given the first approximation, i can say that is on 2 with an error of 0.21...
given the next term
[tex]J_{0}(r)=1-\frac{r^2}{4}+\frac{r^4}{64}-E(r^6)[/tex]
where
[tex]|E(r)|\leq \frac{r^6}{6!\pi}[/tex]
tells me that the zero is in 2^(3/2) with an error of 0.23
and so on...
do you guys think this is a correct procedure?
is there any other way i can construct the plot?
i really want to impress my teacher, so any help would be well received.
Thx.
Last edited: