Hy, well I have a problem with the transcendental equation $x - c \sin(x) = 0$, where $c$ is some positive constant. I tried using Newton's method for finding the roots but it didn't work well. The problem is also the number of solutions, because as $c$ is increasing, there will be more solutions.
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1What do you mean by it didn't work well? Newton's method is the classical approach to Kepler's equation. – Jack D'Aurizio Jan 25 '18 at 18:12
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Let's look at the two functions $y=x$ and $y=c\sin(x)$ and see where they intersect.
Depending on the constant $c \gt 0$, we could have a different number of solutions. With a very large $c$, the graph will have a large amplitude and oscillate across the line many times, while a very small $c$ close to $0$ will cause $c\sin(x)$ to appear identical to the line, but not quite be the same line. To my knowledge there isn't a simple formula for the solutions using an arbitrary $c$. However, from observation we can see $x=0$ is always a solution. (Why?)
I would suggest picking a value for $c$ and then going from there.
WaveX
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I am sorry but your explanation is not easy to follow because you mix 1) Newton's method, which indeed converges to a root if one is not far from this root, and 2) considerations about the curve of $f$. – Jean Marie Jan 25 '18 at 18:57
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Hmm. How would you say I could better explain this? Should I make my explanation focus more on the considerations of the curve and remove the part on Newton's method? – WaveX Jan 25 '18 at 19:05
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The second option is better. Personnaly, I would compare the intersection of the fixed line with equation $y=x$ with varying curve $y=c \sin(x)$. – Jean Marie Jan 25 '18 at 19:07
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