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Approximation of values from non-closed form equation.

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Legaldose
#1
Mar7-14, 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-(c-b)^{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 5-6 digits of accuracy. I just need tips to head in the right direction. Anything will be useful. Thank you!
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scurty
#2
Mar7-14, 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.
JJacquelin
#3
Mar8-14, 01:36 AM
P: 761
[tex]1 = cos^{2}(b)[1-(c-b)^{2}][/tex]
[tex]1-cos^{2}(b) = cos^{2}(b)[-(c-b)^{2}][/tex]
[tex]sin^{2}(b) = cos^{2}(b)[-(c-b)^{2}][/tex]
[tex]tan^{2}(b) = -(c-b)^{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]

Legaldose
#4
Mar8-14, 11:44 AM
P: 72
Approximation of values from non-closed form equation.

Oh okay, thanks JJacquelin, I didn't even think to do this.


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