If r 1 , r 2 , r 3 , r 4 , r 5 are the complex roots of the equation x 5 − 3 x 4 − 1 = 0 . Find the value of
r 1 9 1 + r 2 9 1 + r 3 9 1 + r 4 9 1 + r 5 9 1
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I am speechless.
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thanks. try this
We can solve this using Newton's Sums with a twist.
Let x 5 − 3 x 4 − 1 = x 5 − a 4 x 4 + a 3 x 3 − a 2 x 2 + a 1 x − a 0 = 0
⇒ a 4 = 3 , a 3 = 0 , a 2 = 0 , a 1 = 0 , a 0 = 1 .
Also let b k = r k 1 , where k = 1 , 2 , 3 , 4 , 5 , then we have:
\(\begin{array} {} S_1 = \displaystyle \sum_{k=1}^5 {b_k} = \dfrac {a_1}{a_0} = 0 & S_2 = \displaystyle \sum_{cycl}^5 {b_1b_2} = \dfrac {a_2}{a_0} = 0 \\ S_3 = \displaystyle \sum_{cycl}^5 {b_1b_2b_3} = \dfrac {a_3}{a_0} = 0 & S_4 = \displaystyle \sum_{cycl}^5 {b_1b_2b_3b_4} = \dfrac {a_4}{a_0} = 3 \\ S_5 = \displaystyle \sum_{cycl}^5 {b_1b_2b_3b_4b_5} = \dfrac {1}{a_0} = 1 \end{array} \)
Now let P n = b 1 n + b 2 n + b 3 n + b 4 n + b 5 n , where n = 1 , 2 , 3 . . .
Then, we have:
P 9 = b 1 9 + b 2 9 + b 3 9 + b 4 9 + b 5 9 = r 1 9 1 + r 2 9 1 + r 3 9 1 + r 4 9 1 + r 5 9 1 = S 1 P 8 − S 2 P 7 + S 3 P 6 − S 4 P 5 + S 5 P 4 = − 3 P 5 + P 4 = − 3 [ − 3 P 1 + 5 ( 1 ) ] − 4 ( 3 ) = − 3 [ 0 + 5 ] − 1 2 = − 1 5 − 1 2 = − 2 7
( x − 3 ) = x 4 1
x 5 x − 3 = x 9 1
x = r 1 ∑ r 5 ( x 4 1 − x 5 3 ) = x = r 1 ∑ r 5 x 9 1
we require x = r 1 ∑ r 5 x 9 1
Also r i 5 = 1 as they are five roots of unity.
x = r 1 ∑ r 5 x 5 x − 3 × 5
x = r 1 ∑ r 5 x − 3 × 5
after this r 1 + r 2 + r 3 + r 4 + r 5 = 0
so i got my answer as − 1 5
Plz reply @Chew-Seong Cheong sir, plz tell where i was wrong
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oh damn i realised they are complex roots not five roots of unity. sorry for troubling u
First of all, there are only 4 complex roots of this equation and one real root. I don't think this problem was well specified and I don't agree with your answer. one real root is 3.015 and the complex roots are -.5321+-4962i and .526098+-.591794i so just to be clear there are not 5 complex roots to this equation.
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You are right that Newton Sums and Vieta Formulas work on all the complex roots of an equation and that include real roots. Real roots like imaginary roots are all complex roots. I have used Wolfram Alpha to check the roots and then calculated r k 1 and added them up, the result was (-26.99999101+0i)
\(\begin{array} {} \dfrac{1}{(3.01215+0i)^9} & = 4.89903613938914\times 10^{-05}+0i\\ \dfrac{1}{(-0.532172-0.496268i)^9} & = -15.5778+7.93898i\\ \dfrac{1}{(-0.532172+0.496268i)^9} & = -15.5778+-7.93898i\\ \dfrac{1}{(0.526098-0.591794i)^9} & = 2.07778+-7.90286i\\ \dfrac{1}{(0.526098+0.591794i)^9} & = 2.07778+7.90286i \\ \dfrac {1}{r_1^9} + \dfrac {1}{r_2^9} + \dfrac {1}{r_3^9} + \dfrac {1}{r_4^9} + \dfrac {1}{r_5^9} & = -26.99999101+0i \end{array} \)
can you please tell me how S4 =3 and S1 = 0, actually S4= 0 and S1=3 . Please recheck and reply ...
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let x = y 1 ,then y 5 1 − 3 y 4 1 − 1 = 0 ⟹ y 5 + 3 y − 1 = 0 by vietas ∑ ( x ) = 3 ∑ ( y ) = 0 so 1 y 5 = − 3 y + 1 − − ( i ) y 4 = − 3 + y 1 − − ( i i ) ( i ) ∗ ( i i ) ⟹ y 9 = ( − 3 y + 1 ) ( − 3 + y 1 ) = 9 y − 6 + x − − ( i i i ) ∑ ( i i i ) ⟹ ∑ ( y 9 ) = 9 ∑ ( y ) − ∑ ( 6 ) + ∑ ( x ) = 0 − 3 0 + 3 = − 2 7