List of ways to highly accurately solving a mixed equation

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SUMMARY

The discussion focuses on solving the equation E - e * sin(E) = M with high accuracy. The Newton Method is initially employed, but participants highlight its limitations in convergence speed and accuracy. An alternative approach suggested involves using the second derivative to enhance convergence. The importance of selecting an appropriate initial guess, such as E = M, is also emphasized for improved results.

PREREQUISITES
  • Understanding of Newton's Method for root-finding
  • Familiarity with trigonometric functions and their properties
  • Basic knowledge of calculus, specifically derivatives
  • Concept of floating point precision in numerical computations
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  • Research advanced root-finding methods, such as Halley's Method
  • Explore Taylor series expansions for approximating functions
  • Learn about convergence criteria in numerical analysis
  • Investigate the impact of initial guesses on iterative methods
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Mathematicians, engineers, and computer scientists involved in numerical analysis and algorithm development, particularly those focused on solving nonlinear equations with high precision.

Philosophaie
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I would like a list of ways highly accurate to solve the equation:

E - e * sin(E) = M

I can solve this with the Newton Method:

M = 2*pi/3
e = 0.002
E = pi
d = 0.01
Do While SQRT(d^2) > 0.000001
d= (E - e * sin(E) - M) / (1 - e * cos(E))
E = E + d
Loop

This is not very accurate.

Is there some sort of trigonometric expansion or infinite series that would be more accurate?
 
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This is very puzzling. Are you saying that Newton's method is not very accurate, or that it doesn't converge fast enough for you? Either way, your equation for implementing it doesn't look correct to me. Shouldn't it be E = E - d, not E = E + d? Also, your initial guess for these parameter values would be better chosen to be E = M.

Chet
 
There are methods that converge faster than Newton - you can take the second derivative into account, for example. But they are not more accurate - the limit is always exactly the solution to the equation (or the limits of your floating point precision).
 
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