Using this we can find g(ε) with a lot of combinatorics of the Taylor series, or we can
recall the trigonometric identity cos(2x) = 2cos2(x) - 1 and note |
g(ε) = 2cos(√ε) |
satisfies g(4ε) = g(ε)2 - 2: |
g(4ε) = 2cos(√(4ε)) = 2cos(2√ε) =
2(2cos2(√ε) - 1) = g(ε)2 - 2 |
Here is an anmiation illustrating the convergence of gn(ε) to g(ε). |
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Click to repeat. |
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