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Approximation of values from nonclosed form equation.by Legaldose
Tags: approximation, approximation method, closed, equation, form, nonclosed, solution, values 
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#1
Mar714, 07:17 PM

P: 72

Hello everyone, I'm working on a problem and it turns out that this equation crops up:
[tex]1 = cos^{2}(b)[1(cb)^{2}][/tex] where [tex]c > \pi[/tex] Now I'm pretty sure you can't solve for b in closed form (at least I can't), so what I need to do is for some value of c, approximate the value of b to about 56 digits of accuracy. I just need tips to head in the right direction. Anything will be useful. Thank you! 


#2
Mar714, 07:47 PM

P: 392

http://en.wikipedia.org/wiki/Newton's_method
Put simply, ##x_{n+1} = x_n  \frac{f(x_n)}{f'(x_n)}##. Just iterate the formula a few times to get an approximate answer. 


#3
Mar814, 01:36 AM

P: 759

[tex]1 = cos^{2}(b)[1(cb)^{2}][/tex]
[tex]1cos^{2}(b) = cos^{2}(b)[(cb)^{2}][/tex] [tex]sin^{2}(b) = cos^{2}(b)[(cb)^{2}][/tex] [tex]tan^{2}(b) = (cb)^{2}[/tex] For real solution, positive term = negative term is only possible if they are =0. Hence the solution is : [tex]c=b=n\pi[/tex] 


#4
Mar814, 11:44 AM

P: 72

Approximation of values from nonclosed form equation.
Oh okay, thanks JJacquelin, I didn't even think to do this.



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