Mystery polynomial equation in 2019.

Algebra Level 5

If the number c c is the largest solution of the equation ( x − cos ⁡ π 2020 ) ( x − cos ⁡ 3 π 2020 ) ( x − cos ⁡ 5 π 2020 ) ⋯ ( x − cos ⁡ 2019 π 2020 ) = 1 2 1010 , \left(x-\cos {\frac{\pi}{2020}}\right) \left(x-\cos {\frac{3\pi}{2020}}\right) \left (x-\cos {\frac{5\pi}{2020}} \right )\cdots \left( x-\cos {\frac{2019\pi}{2020}}\right) =\frac{1}{2^{1010}}, then find the number ⌊ 1 arccos ⁡ c ⌋ . \left \lfloor{\frac{1}{\arccos{c}}}\right\rfloor.

Note: The expression ⌊ 1 arccos ⁡ c ⌋ \left \lfloor{\frac{1}{\arccos{c}}}\right\rfloor represents the floor of the number 1 arccos ⁡ c . \frac{1}{\arccos{c}}.


The answer is 964.

Arturo Presa by Arturo Presa , 62, USA

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1 solution

Arturo Presa
Jul 20, 2019

The given equation can be rewritten as 2 1009 ( x − cos ⁡ π 2020 ) ( x − cos ⁡ 3 π 2020 ) ( x − cos ⁡ 5 π 2020 ) . . . ( x − cos ⁡ 2009 π 2020 ) = 1 / 2. 2^{1009}(x-\cos\frac{\pi}{2020}) (x-\cos\frac{3\pi}{2020})(x-\cos\frac{5\pi}{2020})...(x-\cos\frac{2009\pi}{2020})=1/2. The polynomial in the left side can be expressed on the interval [ − 1 , 1 ] [-1,1] as cos ⁡ ( 1010 arccos ⁡ x ) , \cos(1010 \arccos x), that is the 1010th Chebyshev polynomial of first kind. Therefore, the given equation becomes cos ⁡ ( 1010 arccos ⁡ x ) = 1 2 . \cos ({1010 \arccos x})= \frac{1}{2} . This polynomial equation of degree 1010 has 1010 distinct solutions in the interval [ − 1 , 1 ] , [-1, 1], which are all numbers of the form cos ⁡ ( π 3030 + 2 n π 1010 ) \cos(\frac{\pi}{3030}+\frac{2n\pi}{1010}) together with the numbers of the form cos ⁡ ( 5 π 3030 + 2 n π 1010 ) , \cos(\frac{5\pi}{3030}+\frac{2n\pi}{1010}), where n n is any whole number less than or equal to 504. These two lists of numbers are decreasing and then the largest of all these values is cos ⁡ π 3030 . \cos \frac{\pi}{3030}. Then c = cos ⁡ π 3030 . c= \cos \frac{\pi}{3030}. So, the solution to this problem is ⌊ 1 arccos ⁡ c ⌋ = ⌊ 1 arccos ⁡ ( cos ⁡ π 3030 ) ⌋ = ⌊ 3030 π ⌋ = 964 . \lfloor \frac{1}{\arccos c}\rfloor=\lfloor \frac{1}{\arccos (\cos \frac{\pi}{3030})}\rfloor=\lfloor \frac{3030}{\pi}\rfloor=\boxed{964}.

2 Helpful 1 Interesting 4 Brilliant 1 Confused

How did you simplify the polynomial in the left hand side?

Mohammad Saqib - 1 year, 10 months ago

The only thing that you need to know is that the n n th Chebyshev polynomial of first kind can be expressed as cos ⁡ ( n arccos ⁡ x ) \cos (n \arccos x) over the interval [ − 1 , 1 ] [-1, 1] and also as 2 n − 1 ( x − cos ⁡ π 2 n ) ( x − cos ⁡ 3 π 2 n ) ( x − cos ⁡ 5 π 2 n ) . . . ( x − cos ⁡ ( 2 n − 1 ) π 2 n ) 2^{n-1}(x-\cos \frac{\pi}{2n})(x-\cos\frac{3\pi}{2n})(x-\cos\frac{5\pi}{2n})...(x-\cos\frac{(2n-1)\pi}{2n}) at any real value of x . x.

Arturo Presa - 1 year, 10 months ago

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