I need to find the zeros (roots?) of a function e.g. $$x-\sin(2x)$$ in the interval $[0,\pi]$ so that would make $$\sin(2x) = x$$
I've already found $0$ but I can't find a way to determine the other one
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Can you use numerical methods? – Archis Welankar Jul 19 '19 at 13:41
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no, just plain discrete math – Wouter Vandenputte Jul 19 '19 at 13:44
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1What do you mean with "find the zeros"? Find some interval that contains exactly one zero, compute an approximating sequence, or something else? – Lutz Lehmann Jul 19 '19 at 14:06
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finding all the solutions in the interval $[0,\pi]$ for which $x - sin(2x) = 0$ in this case – Wouter Vandenputte Jul 19 '19 at 14:09
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1@Wouter, what do you mean by "discrete mathematics" here? This is very much a continuous problem, a trascendental equation that has no solution in terms of elementary functions. – vonbrand Jul 19 '19 at 14:12
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As $\sin 2x\approx 2x $ for $x\approx 0$ you know that $$ \sin(2x)>x $$ on some interval right of $x=0$. More precisely, by alternating series properties $$\sin (2x)\ge 2x-\frac{(2x)^3}6=x\left(2-\frac{4x^2}3\right)$$ for $|x|<1$, so that you can take $x_1=\frac34$.
For $x>1$ you know that $$\sin(2x)\le 1<x.$$
Combined this means that there is a root of the equation in $$ x\in\left[\frac34,1\right]. $$ Now check the derivatives over this interval to conclude that there is exactly one root because of monotonicity.
Lutz Lehmann
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