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Let us consider the following equation:
Prove that
If the answer to the last question is positive, this would mean that log(Pi) is NOT a transcendental number, but rather, an algebraic number. A remarkable result in itself!
Source for picture: algebraic numbers
Solution and related problem
Any real number larger or equal to 1 is a solution, so there is nothing particular with log(Pi). A more subtle version of this problem is to ask the student to solve the following equation:
We know from the previous problem that if x^5 - x^2 - 1 = x^2 - 1, the equality holds. Thus to find a solution, we just need to solve x^5 - x^2 - 1 = x^2 - 1. The cubic root of 2 is a solution.
More generally, let's define
Then the (unique) real-valued solution to the equation f(x) = 0 is given by
In particular, if p = 3, then x = 2. If p = 2 + log(2) / log(3), then x = 3. Note that the function f is monotonic, and thus invertible. What is the inverse of f?
Another curious problem
Find an exact,non-trivial solution to the following equation:
sin x + sin(2x sin x) = sin 3x.
This is part of a general type of equations, namely f(x) + f(x f(x)) = 2 f(1), where one of the solutions must of course also be solution of f(x) = 1. Here, f(x) = 2 sin x, and thus x = Pi / 6 is a solution. See plot of sin x + sin(2x sin x) - sin 3x, below.
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Comment
A reader posted the following comment:
I think this is a really neat example, a great problem to give to high school students. To solve it, notice that the first two square roots look like sqrt(x + z) - sqrt(x - z). Call this y. Simple algebra gives y = +-sqrt(2x +- 2). Plug in a value, say x = 2, and the answer must be sqrt(2x - 2) as claimed.
Going deeper, let's turn this into a polynomial system. Let z = sqrt(x^{2} - 1), w = sqrt(2x - 2), v = sqrt(x - z), u = sqrt(x + z). Then we have a system of five equations
(Set each to 0). Use the Dixon resultant to eliminate u, v, w, z. The answer is that x = 1; that is, x must be 1. But this is false! Why? Because the solution is not 0-dimensional. Dixon doesn't have to work then. Nice example.
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