ReyChiquito
- 119
- 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 J_{0}(r)
J_{0}(r)=\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)d\theta
i can calculate the first order derivative with respect to r
\frac {\partial}{\partial r}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \sin(r\cos\theta)\cos\theta d\theta
wich when evaluated in r=0 is 0. For the second derivative
\frac {\partial^2}{\partial r^2}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)\cos^{2}\theta d\theta
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
|J^{(n)}(r)|\leq\frac{1}{\pi}
i can easily bound the error, ie, if i only take two terms of the series
J_{0}(r)=1-\frac{r^2}{4}+E(r^4)
where
|E(r)|\leq \frac{r^4}{4!\pi}
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
J_{0}(r)=1-\frac{r^2}{4}+\frac{r^4}{64}-E(r^6)
where
|E(r)|\leq \frac{r^6}{6!\pi}
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 J_{0}(r)
J_{0}(r)=\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)d\theta
i can calculate the first order derivative with respect to r
\frac {\partial}{\partial r}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \sin(r\cos\theta)\cos\theta d\theta
wich when evaluated in r=0 is 0. For the second derivative
\frac {\partial^2}{\partial r^2}J_{0}(r)=-\frac {1}{\pi}\int_0^\pi \cos(r\cos\theta)\cos^{2}\theta d\theta
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
|J^{(n)}(r)|\leq\frac{1}{\pi}
i can easily bound the error, ie, if i only take two terms of the series
J_{0}(r)=1-\frac{r^2}{4}+E(r^4)
where
|E(r)|\leq \frac{r^4}{4!\pi}
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
J_{0}(r)=1-\frac{r^2}{4}+\frac{r^4}{64}-E(r^6)
where
|E(r)|\leq \frac{r^6}{6!\pi}
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.
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