Omega's Pet Function

Algebra Level pending

F ( 1 ) = ω F ( 2 ) = ω 2 F ( 3 ) = 1 F(1) = \omega \qquad F(2) = \omega^2 \qquad F(3) = 1 Let F ( x ) F(x) be a quadratic polynomial satisfying the equations above, where ω \omega is a non-real cube root of unity . Find F ( 4 ) F(4) .

1 1 6 + 4 ω 6 + 4\omega 2 + 3 ω 2 + 3 \omega ω + 1 \omega+1

A P by A P , 19, USA

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2 solutions

Otto Bretscher
May 28, 2016

As with any quadratic polynomial, we have F ( 4 ) = F ( 1 ) − 3 F ( 2 ) + 3 F ( 3 ) F(4)=F(1)-3F(2)+3F(3) , which is ω − 3 ω 2 + 3 = 6 + 4 ω \omega-3\omega^2+3=\boxed{6+4\omega} since 1 + ω + ω 2 = 0 1+\omega+\omega^2=0 .

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展豪 張
May 28, 2016

By interpolation, F ( x ) = ω ( x − 2 ) ( x − 3 ) ( 1 − 2 ) ( 1 − 3 ) + ω 2 ( x − 1 ) ( x − 3 ) ( 2 − 1 ) ( 2 − 3 ) + ( x − 1 ) ( x − 2 ) ( 3 − 1 ) ( 3 − 2 ) F(x)=\omega\dfrac{(x-2)(x-3)}{(1-2)(1-3)}+\omega^2\dfrac{(x-1)(x-3)}{(2-1)(2-3)}+\dfrac{(x-1)(x-2)}{(3-1)(3-2)}
ω 2 + ω + 1 = 0 \omega^2+\omega+1=0
F ( 4 ) = 4 ω + 6 F(4)=4\omega+6

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