Question

I call upon thee greatest mathematicians to help me solve "n" from this equation:
5.41*arctan(((0.03/n)+n*sin(8.87*n))/(294.4*n*(n^2-1)+(n^2+1)*cos(8.87*n)))-2=0

Written it looks like this:
http://i.imgur.com/Vu1eHEW.png

I'd be very thankful if anyone here would solve this for me. I've been struggling with it for the past few days.

Answer 1

For what course is this problem? It'd take a genius to solve this problem without some kind of math program.

Are you familiar with Newton's Method of approximating roots? That's one way to go, but again you'll need to use calculator or computer power.

Answer 2

In view of the complicated expression for the derivative, it is also possible to calculate by the bisection method that requires no evaluation of derivatives.
The two roots are well behaved and are easy to bracket.
In any case, the required accuracy cannot be substantial since the given data is given only to three or four significant figures.

Answer 3

You can use any programs you want @mathmale and @mathmate . I'd be happy if I got the solution for n, then I could calculate all my other paraemeters.

Answer 4

I have found two solutions around +1 and -1.
You only need to use the bisection method to refine it. Newton's is ok too, if you want to toil with the derivative.

Answer 5

I actually have a new equation
5.41*arctan(((0.06/n)+2*n*sin(8.87*n))/(0.267*n-(0.267/n)+(n^2+1)*cos(8.87*n)))-2=0

Written it looks like this:
http://i.imgur.com/tW7Dn95.png

I get an answer n=-234.28891252558161414819175592836

Is it normal to get such a refreactive index? Or do you get a different answer?

Answer 6

@mathmate

Answer 7

I have not heard of a negative refractive index. However, according some articles, it seems that certain materials exhibit this phenomenon. It may be that you are doing research on a special material, or you have not found the positive values of the solution, since arcTan has many possible solutions.

There are many solutions to the expression between 0 and 1 and 4 between 1 and 2.5.
see http://prnt.sc/cgxrgd

As I mentioned before, you can refine the solution using bisection method or Newton's, whichever you're more comfortable with, BUT you need to choose a range acceptable to you first, (i.e. which of the possible solutions).