Find the value of ∑r=1412−ar where ak(k=0,1,2,3,4,5) are fifth roots of unity.

Another proof:P(x)=x5−1=profr=04(x−αr)HenceP′(x)P(x)=5x4x5−1=∑r=041x−αrNow put x=2 in this formula.

Since a0,a1,a2,…,a4 are roots of the equationx5−1=0You can apply Transformation of Roots to find a equation whose roots are12−α0,12−α1,…12−α4Let P(y) represent the polynomial whose roots are 12−aky=12−ak=12−x⇒x=2y−1yPut in (1)(2y−1y)5−1=0⇒(2y−1)5−y5=0Use Binomial Theorem to find coefficient of y5 and y4.You will get sum of the roots using Vieta's Formulas.

If a5=1,12−α=c0+c1α+...+c4α4 where 2c0−c4=1 2c0−c4=1 2c1−c0=0 2c2−c2=0 2c3−c2=0 2c4−c3=0 The solution of this is c0=16/31, c1=8/31, c=4/31, c3=2/31, c4=1/31 Then ∑r=1412αr=1631∑r=141+831∑r=14ar+431∑r=14a2r+231∑r=14a3r+131∑r=14a4r =1631⋅4−831−431−231−131=4931

Another proof:P(x)=x5−1=profr=04(x−αr)HenceP′(x)P(x)=5x4x5−1=∑r=041x−αrNow put x=2 in this formula.

Since a0,a1,a2,…,a4 are roots of the equationx5−1=0You can apply Transformation of Roots to find a equation whose roots are12−α0,12−α1,…12−α4Let P(y) represent the polynomial whose roots are 12−aky=12−ak=12−x⇒x=2y−1yPut in (1)(2y−1y)5−1=0⇒(2y−1)5−y5=0Use Binomial Theorem to find coefficient of y5 and y4.You will get sum of the roots using Vieta's Formulas.

If a5=1,12−α=c0+c1α+...+c4α4 where 2c0−c4=1 2c0−c4=1 2c1−c0=0 2c2−c2=0 2c3−c2=0 2c4−c3=0 The solution of this is c0=16/31, c1=8/31, c=4/31, c3=2/31, c4=1/31 Then ∑r=1412αr=1631∑r=141+831∑r=14ar+431∑r=14a2r+231∑r=14a3r+131∑r=14a4r =1631⋅4−831−431−231−131=4931

"Find the value of ∑r=1412−ar where ak(k=0,1,2,3,4,5) are..." was answered by our community

Dayton Burnett

Answered question

2022-01-21

Find the value of

\sum_{r = 1}^{4} \frac{1}{2 - a^{r}}

where

a^{k} (k = 0, 1, 2, 3, 4, 5)

are fifth roots of unity.

Answer & Explanation

cevarnamvu

Beginner2022-01-22Added 12 answers

Another proof:

P (x) = x^{5} - 1 = p r o f_{r = 0}^{4} (x - α^{r})

Hence

\frac{P^{'} (x)}{P (x)} = \frac{5 x^{4}}{x^{5} - 1} = \sum_{r = 0}^{4} \frac{1}{x - α^{r}}

Now put x=2 in this formula.

eris0cg

Beginner2022-01-23Added 13 answers

Since

a_{0}, a_{1}, a_{2}, \dots, a_{4}

are roots of the equation

x^{5} - 1 = 0

You can apply Transformation of Roots to find a equation whose roots are

\frac{1}{2 - α_{0}}, \frac{1}{2 - α_{1}}, \dots \frac{1}{2 - α_{4}}

Let

P (y)

represent the polynomial whose roots are

\frac{1}{2 - a_{k}}

y = \frac{1}{2 - a_{k}} = \frac{1}{2 - x} \Rightarrow x = \frac{2 y - 1}{y}

Put in (1)

{(\frac{2 y - 1}{y})}^{5} - 1 = 0 \Rightarrow {(2 y - 1)}^{5} - y^{5} = 0

Use Binomial Theorem to find coefficient of

y^{5}

and

y^{4}

.You will get sum of the roots using Vieta's Formulas.

RizerMix

Expert2022-01-27Added 656 answers

a^{5} = 1, \frac{1}{2 - α} = c_{0} + c_{1} α + . . . + c_{4} α^{4}

where

2 c_{0} - c_{4} = 1

2 c_{0} - c_{4} = 1

2 c_{1} - c_{0} = 0

2 c_{2} - c_{2} = 0

2 c_{3} - c_{2} = 0

2 c_{4} - c_{3} = 0

The solution of this is

c_{0} = 16 / 31, c_{1} = 8 / 31, c = 4 / 31, c_{3} = 2 / 31, c_{4} = 1 / 31

Then

\sum_{r = 1}^{4} \frac{1}{2 α^{r}} = \frac{16}{31} \sum_{r = 1}^{4} 1 + \frac{8}{31} \sum_{r = 1}^{4} a^{r} + \frac{4}{31} \sum_{r = 1}^{4} a^{2 r} + \frac{2}{31} \sum_{r = 1}^{4} a^{3 r} + \frac{1}{31} \sum_{r = 1}^{4} a^{4 r}

= \frac{16}{31} \cdot 4 - \frac{8}{31} - \frac{4}{31} - \frac{2}{31} - \frac{1}{31} = \frac{49}{31}

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