Polynomials and infinite sum

Algebra Level 2

The polynomial p ( x ) = x 2 − a x + 1 4 p(x) = x^2 - ax + \frac14 , with a a a positive real, has two real roots (not necessarily distinct) r 1 r_1 and r 2 r_2 such that ∣ r 1 ∣ ≤ 1 2 \left| r_1 \right| \le \frac12 and ∣ r 2 ∣ ≤ 1 2 \left| r_2 \right| \le \frac12 . Find the value of: r 1 + r 2 + r 1 2 + r 2 2 + r 1 3 + r 2 3 + ⋯ = ∑ k = 1 ∞ ( r 1 k + r 2 k ) r_1 + r_2 + {r_1}^2 + {r_2}^2 + {r_1}^3 + {r_2}^3 + \dots = \sum_{k=1}^\infty \left({r_1}^k + {r_2}^k\right)


The answer is 2.

João Baptista by João Baptista , 24, Brazil

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1 solution

Arjen Vreugdenhil
Oct 11, 2015

The polynomial is equal to x 2 − a x + 1 4 = ( x − r 1 ) ( x − r 2 ) = x 2 − ( r 1 + r 2 ) x + r 1 r 2 ; x^2-ax+\tfrac14 = (x-r_1)(x-r_2) = x^2 - (r_1+r_2)x + r_1r_2; the combination of r 1 r 2 = 1 4 r_1r_2=\tfrac14 and ∣ r 1 ∣ , ∣ r 2 ∣ ≤ 1 2 |r_1|, |r_2| \leq \tfrac12 implies that r 1 = r 2 = ± 1 2 r_1=r_2=\pm\tfrac12 .

Since a a is positive, it follows that r 1 = r 2 = + 1 2 r_1 = r_2 = +\tfrac12 .

Now we must add twice the sum ∑ k = 1 ∞ ( 1 2 ) k = 1 2 1 − 1 2 = 1 , \sum_{k=1}^\infty (\tfrac12)^k = \frac{\tfrac12}{1-\tfrac12} = 1, so that the answer is equal to 2 \boxed{2} .

Note: if we allow a a to be negative, another solution is r 1 = r 2 = − 1 2 r_1 = r_2 = -\tfrac12 ; then the answer is − 2 3 -\tfrac23 .

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