Nested Radical of Period ( + + ) (++-)

Algebra Level 4

4 + 4 + 4 x = x \sqrt{4+\sqrt{4+\sqrt{4-x}}}=x What is x x

Hint: The expression can be expressed as 2 ( cos ( A ) + cos ( B ) + cos ( C ) ) 2(\cos(A)+\cos(B)+\cos(C))

Note: Please try to do this problem without resorting to factoring out the eighth degree equation, or using the calculator, or using CALCULUS


The answer is 2.50701864409.

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1 solution

Chris Lewis
Dec 18, 2019

Clearing all the square roots leads to a messy degree 8 8 polynomial, which is only solvable numerically (realistically - it does factorise into a quadratic and two cubics, but this is not obvious). It also has eight real roots, seven of which are not the solution we're looking for.

Since a numerical approach is needed anyway, a much easier method is to simply pick a sensible start point x 0 x_0 , and iterate the given function as x n + 1 = 4 + 4 + 4 x n x_{n+1}=\sqrt{4+\sqrt{4+\sqrt{4-x_n}}} for n 1 n \ge 1 . It's not guaranteed this will converge, but in this case we're lucky.

What might a sensible start value be? Well, by inspection, x x must be in the interval ( 0 , 4 ) (0,4) , so x 0 = 2 x_0=2 isn't a bad guess.

Plugging in to the above scheme, we find x 1 = 2.51532 , x 2 = 2.50687 , x 3 = 2.50702 x_1=2.51532\ldots,x_2=2.50687\ldots,x_3=2.50702\ldots , quickly converging on the solution.

No sir , The actual Problem involves to prove this in terms of values of cosines. hence brute force or newton raphson method wont be the ideal solution to take.

Sarthak Sahoo - 1 year, 5 months ago

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Not the problem you posted! That's a big clue you omitted. Incidentally, the method isn't NR at all (no calculus involved). But OK, the cosine thing is interesting. You can factorise the degree eight polynomial into a quadratic and two cubics; the solution is a root of one of the cubics, which can indeed be expressed in terms of cosines. However, this is probably not the best way to go - you would firstly need to factorise a degree eight polynomial (not easy, in general), then you'd need to check which root works in the original equation.

Better would be a geometric approach, though I can't immediately see one. If the problem you're solving gives a form to aim for, that again would be useful information.

Chris Lewis - 1 year, 5 months ago

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The values of A=4pi/19 , B=6pi/19, C=10pi/19

Sarthak Sahoo - 1 year, 5 months ago

Sure i'll edit the question to have the form as well

Sarthak Sahoo - 1 year, 5 months ago

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