"If 4cos2(kπj) is the greatest root of the..." was answered by our experts

Jakayla Benton

Answered question

2022-04-22

4 \cos^{2} (\frac{k π}{j})

is the greatest root of the equation

x^{3} - 7 x^{2} + 14 x - 7 = 0

where

gcd (k, j) = 1

Evaluate

k + j

morpheus1ls1

Beginner2022-04-23Added 22 answers

Let

z = \frac{k π}{j}

. If

4 \cos^{2} z

is the root of

x^{3} - 7 x^{2} + 14 x - 7 = 0

, then

2 \cos z

is the root of

x^{6} - 7 x^{4} + 14 x^{2} - 7 = 0

Let's plug

x = 2 \cos z = 2 \frac{e^{i z} + e^{- i z}}{2}

in and see what happens.
The expression comes out to be

e^{- 6 i z} (- e^{2 i z} + e^{4 i z} - e^{6 i z} + e^{8 i z} - e^{10 i z} + e^{12 i z} + 1) = 0

Let's change the variables again,

- e^{2 i z} = t

, so

- t^{- 3} (1 + t + t^{2} + t^{3} + t^{4} + t^{5} + t^{6}) = 0

- e^{2 i z}

is the 7th-root of unity:

\frac{t^{7} - 1}{t^{3} (t - 1)} = 0

- e^{2 i z} = e^{i \frac{2 π k}{7}}, gcd (k, 7) = 1

2 i z + π i = i \frac{2 π k}{7}

z = - \frac{π}{2} + \frac{π k}{7}

Now we need

R e (t)

to be maximal, so we select k to

- \frac{π}{2} + \frac{π}{k}

be the closest to 0:

z = \pm \frac{π}{14}

The answer is either 13 or 15.

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