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Originally posted by FabianFnasLet me try to "show" there probably isn't:
I've seen people knowing how to work with calculators, producing approximations. But noone yet producing an answer. Isn't there any?
arctan(x) = 1/(2x)
tan(arctan(x) = tan(1/(2x))
x = tan(1/(2x))
From Lindemann-Weierstrass we know that the tangent of any non-zero algebraic number is transcendental, so if x was algebraic then 1/2x is algebraic and therefore tan(1/(2x))=x should be transcendental, so we get a contradiction.
x must then be transcendental and we can easily check that it's not a multiple of Pi.
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Originally posted by PalynkaGood explanation!
Let me try to "show" there probably isn't:
arctan(x) = 1/(2x)
tan(arctan(x) = tan(1/(2x))
x = tan(1/(2x))
From Lindemann-Weierstrass we know that the tangent of any non-zero algebraic number is transcendental, so if x was algebraic then 1/2x is algebraic and therefore tan(1/(2x))=x should be transcendental, so we get a contradiction.
x must then be transcendental and we can easily check that it's not a multiple of Pi.
So to find the solution of the problem, numerical methods can be our only tool? An approximation can be the only answer?
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Originally posted by FabianFnasI can't say. 😀
Good explanation!
So to find the solution of the problem, numerical methods can be our only tool? An approximation can be the only answer?
I only proved x is transcendental, but the answer could be a multiple of another transcendental number or a log or...
Although the nature of the tangent I would expect that unless it is a multiple of Pi, then probably numerical methods are the only tool.