Polynomial with known root cosine
a x 4 − b x 3 − c x 2 + d x + 1
You are given that the polynomial above has a root of cos ( 1 5 2 π ) for positive integers a , b , c and d . Find the value of a + b + c + d .
The answer is 48.
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2 solutions
This method is nice! Thanks!
Is there a reason why cyclotomic polynomial works here? (pretend I'm not familiar with cyclotomic polynomials).
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The cyclotomic polynomial Φ n ( x ) is the minimal polynomial of the primitive n th roots of unity. Dividing by x ϕ ( n ) / 2 and using the tableau method, we can find the minimal polynomial of cos ( 2 π / n ) .
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Oh I see. What is a Tableau method? Is it applicable here ? Because I can't seem to apply it.
Let θ = 1 5 2 π . Then cos 5 θ = − 2 1 . This means that T 5 ( x ) = 1 6 x 5 − 2 0 x 3 + 5 x = − 2 1 where x = cos θ . Now 3 2 x 5 − 4 0 x 3 + 1 0 x + 1 = 0 . Note that x = cos 1 5 1 0 π = cos 3 2 π = − 2 1 is a solution to the equation and 3 2 x 5 − 4 0 x 3 + 1 0 x + 1 = ( 2 x + 1 ) ( 1 6 x 4 − 8 x 3 − 1 6 x 2 + 8 x + 1 ) .
Now it is clear that cos 1 5 2 π is a root of 1 6 x 4 − 8 x 3 − 1 6 x 2 + 8 x + 1 and hence a + b + c + d = 4 8 .
Starting with the cyclotomic polynomial Φ 1 5 ( x ) = x 8 − x 7 + x 5 − x 4 + x 3 − x + 1 , dividing by x 4 and applying the tableau method to x + x 1 = 2 cos ( 2 π / 1 5 ) , we find that cos ( 2 π / 1 5 ) is a root of 1 6 x 4 − 8 x 3 − 1 6 x 2 + 8 x + 1 , so, the required sum is 4 8 .