- #1
lavster
- 217
- 0
Hi,
as part of my maths course i am learning about bessel functions. But this is something that I am not fully comfortable with - there seems to be a lot of tricks.
There is a statement in my notes that when [tex]\alpha_n>>1[/tex],
[tex]\frac{J_0(i^{\frac{3}{2}}\alpha_n\frac{r}{a})}{J_0(i^{\frac{3}{2}}\alpha_n)} = (\frac{r}{a})^\frac{1}{2}exp[-\sqrt{i}(1-\frac{r}{a}\alpha_n][/tex]
i know that [tex]\sqrt{i}=\pm\frac{1}{\sqrt{2}}(1+i)[/tex], which i think will help show this statement. I've also noticed that the argument in exp[] is the bit in the brackets of the bessl function in the denominator minus the bit in the brackets of the bessel function on the numerator. and the multiplying factor of exp in the RHS sems to me the bits in the brackets of the bessel function deivided by each other...
can anyone tell me if this is merely coincidence or whether there is a mathematical trick or identity that helps to show this?
sorry if this is confusing...
thanks, lav
as part of my maths course i am learning about bessel functions. But this is something that I am not fully comfortable with - there seems to be a lot of tricks.
There is a statement in my notes that when [tex]\alpha_n>>1[/tex],
[tex]\frac{J_0(i^{\frac{3}{2}}\alpha_n\frac{r}{a})}{J_0(i^{\frac{3}{2}}\alpha_n)} = (\frac{r}{a})^\frac{1}{2}exp[-\sqrt{i}(1-\frac{r}{a}\alpha_n][/tex]
i know that [tex]\sqrt{i}=\pm\frac{1}{\sqrt{2}}(1+i)[/tex], which i think will help show this statement. I've also noticed that the argument in exp[] is the bit in the brackets of the bessl function in the denominator minus the bit in the brackets of the bessel function on the numerator. and the multiplying factor of exp in the RHS sems to me the bits in the brackets of the bessel function deivided by each other...
can anyone tell me if this is merely coincidence or whether there is a mathematical trick or identity that helps to show this?
sorry if this is confusing...
thanks, lav