Interesting polynomial?

Calculus Level 4

Given that f ( x ) = x n + x n 1 + x n 2 + + x 2 + x + 1 f(x) = x^n + x^{n-1} + x^{n-2} + \cdots + x^2 + x+ 1 , evaluate the following:

lim n ω ( ω 1 ) f ( ω ) n + 5 \displaystyle \lim_{n\to\infty} \dfrac{\omega (\omega - 1) f'(\omega)}{n+5}

where ω \omega denotes a primitive ( n + 1 ) th (n+1)^\text{th} root of unity.


The answer is 1.

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2 solutions

Chew-Seong Cheong
Jan 26, 2016

f ( x ) = x n + x n 1 + x n 2 + . . . + x 2 + x + 1 = x n + 1 1 x 1 f ( x ) = ( n + 1 ) x n x 1 x n + 1 1 ( x 1 ) 2 = ( n + 1 ) x n ( x 1 ) x n + 1 + 1 ( x 1 ) 2 = n x n + 1 ( n + 1 ) x n + 1 ( x 1 ) 2 ω ( ω 1 ) f ( w ) n + 5 = ω ( ω 1 ) ( n ω n + 1 ( n + 1 ) ω n + 1 ) ( n + 5 ) ( ω 1 ) 2 = ω ( n ω n + 1 ( n + 1 ) ω n + 1 ) ( n + 5 ) ( ω 1 ) = n ω n + 2 ( n + 1 ) ω n + 1 + ω ( n + 5 ) ( ω 1 ) Note that ω n + 1 = 1 = n ω n 1 + ω ( n + 5 ) ( ω 1 ) = ( n + 1 ) ( ω 1 ) ( n + 5 ) ( ω 1 ) = n + 1 n + 5 lim n ω ( ω 1 ) f ( w ) n + 5 = lim n n + 1 n + 5 = lim n 1 + 1 n 1 + 5 n = 1 \begin{aligned} f(x) & = x^n + x^{n-1} + x^{n-2} + ... + x^{2} + x + 1 \\ & = \frac{x^{n+1}-1}{x-1} \\ \Rightarrow f'(x) & = \frac{(n+1)x^{n}}{x-1} - \frac{x^{n+1}-1}{(x-1)^2} \\ & = \frac{(n+1)x^{n}(x-1)-x^{n+1}+1}{(x-1)^2} \\ & = \frac{nx^{n+1} - (n+1) x^{n}+1}{(x-1)^2} \\ \Rightarrow \frac{\omega (\omega -1)f'(w)}{n+5} & = \frac{\omega (\omega -1)(n\omega^{n+1} - (n+1) \omega^{n}+1)}{(n+5)(\omega-1)^2} \\ & = \frac{\omega (n\omega^{n+1} - (n+1) \omega^{n}+1)}{(n+5)(\omega-1)} \\ & = \frac{n\omega^{n+2} - (n+1) \omega^{n+1}+\omega}{(n+5)(\omega-1)} \quad \quad \small \color{#3D99F6}{\text{Note that } \omega ^{n+1} = 1} \\ & = \frac{n\omega - n-1+\omega}{(n+5)(\omega-1)} \\ & = \frac{(n+1)(\omega -1)}{(n+5)(\omega-1)} \\ & = \frac{n+1}{n+5} \\ \Rightarrow \lim_{n \to \infty} \frac{\omega (\omega -1)f'(w)}{n+5} & = \lim_{n \to \infty} \frac{n+1}{n+5} \\ & = \lim_{n \to \infty} \frac{1+\frac{1}{n}}{1+\frac{5}{n}} \\ & = \boxed{1} \end{aligned}

Abhi Kumbale
Jan 26, 2016

just replace -(w^(n-1)+...+w+1)= w^n not -w^n, so you'll get lim((n+1)/(n+5))=1

Nikola Djuric - 5 years, 4 months ago

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