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..(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Ploting zero order Bessel function

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