Approximation of values from non-closed form equation.

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Discussion Overview

The discussion revolves around approximating the value of b from the equation 1 = cos²(b)[1-(c-b)²], where c is greater than π. Participants explore methods for finding an approximate solution for b, given the belief that a closed-form solution may not be possible.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses the need to approximate b for a given c, suggesting that a closed-form solution is not feasible.
  • Another participant introduces Newton's method as a potential approach for finding an approximate solution, providing a brief explanation of the iterative formula.
  • A different participant derives an alternative perspective by manipulating the original equation, leading to the conclusion that for real solutions, c must equal b, suggesting c = b = nπ as a possible solution.
  • A later reply acknowledges the contribution of the previous participant, indicating that the approach was not previously considered.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method for approximation, as different approaches are proposed and explored, indicating multiple competing views on how to tackle the problem.

Contextual Notes

The discussion includes assumptions about the nature of the solutions and the conditions under which they hold, particularly regarding the relationship between c and b.

Legaldose
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Hello everyone, I'm working on a problem and it turns out that this equation crops up:

1 = cos^{2}(b)[1-(c-b)^{2}]

where

c > \pi

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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1 = cos^{2}(b)[1-(c-b)^{2}]
1-cos^{2}(b) = cos^{2}(b)[-(c-b)^{2}]
sin^{2}(b) = cos^{2}(b)[-(c-b)^{2}]
tan^{2}(b) = -(c-b)^{2}
For real solution, positive term = negative term is only possible if they are =0.
Hence the solution is : c=b=n\pi
 
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Oh okay, thanks JJacquelin, I didn't even think to do this.
 

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