Sum of roots of polynomial
Consider the polynomial p ( x ) defined as:- p ( x ) = x 4 + x 3 + 2 x 2 + 3 x + 1 Let r 1 , r 2 , r 3 , r 4 be the roots of p ( x ) then the value of ⎝ ⎛ i = j ∑ ( 1 − r i ) ( 1 − r j ) 1 ⎠ ⎞ ( i = 1 ∑ 4 1 − r i 1 ) can be expressed as b a where g c d ( a , b ) = 1 find a + b
Details and assumptions:- I have used the notation ∑ ( ( 1 − r i ) ( 1 − r j ) 1 to represent the sum of all possible products of the form ( 1 − r i ) ( 1 − r j ) 1 taken two at a time and i = j (i think it also called cyclic sum)
The answer is 109.
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1 solution
I also used Souryajit Roy's method.
Here are some details omitted in his reply:
p ( x ) = x 4 + x 3 + 2 x 2 + 3 x + 1
p 1 ( x ) = x 4 − 5 x 3 + 1 1 x 2 − 1 4 x + 8
p 2 ( x ) = 8 x 4 − 1 4 x 3 + 1 1 x 2 − 5 x + 1
By Vieta's Formula, result= (11/8) (14/8) = 77/32
1 0 9
There is another very simple way to solve this problem by using differential calculus and fundamantal theoram of algebra
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Can you add your solution? I'm interested in learning how FTA is used.
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My solution is not that intresting i only used FTA to express p(x) as a product of 4 linear polynomials and then used differntiation:-
By FTA we can express p(x) as product of 4 linear polynomials so:-
p ( x ) = i = 1 ∏ 4 ( x − r i ) . . . . . ( 1 ) Differentiating both sides od equation (1):-
p ′ ( x ) = i = j = k ∑ ( x − r i ) ( x − r j ) ( x − r k ) . . . . . . . ( 2 )
Differentiating equation (2)both sides:-
p ′ ′ ( x ) = 2 i = j ∑ ( x − r i ) ( x − r j ) . . . . . ( 3 )
Dividing equation (2) by (1):-
f ( x ) f ′ ( x ) = i = 1 ∑ 4 ( x − r i 1 ) . . . . ( 4 )
Dividing equation (3) by (1):-
f ( x ) f ′ ′ ( x ) = 2 i = j ∑ ( ( x − r i ) ( x − r j ) 1 ) . . . . ( 5 )
Multiplying equations (4) and (5) gives:-
2 f 2 ( x ) ( f ′ ( x ) ) ( f ′ ′ ( x ) ) = ( i = j ∑ ( ( x − r i ) ( x − r j ) 1 ) ( i = 1 ∑ 4 ( x − r i 1 ) . . . . ( 6 )
Put x = 1 ) in above equation to give required answer
Sir is there any other mathod to deal with these type of problems(except replacing x by x+1) what are the theorams and facts that i need to know in order to handle such questions smoothly @Calvin Lin sir
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@Aman Sharma – Nice! I just constructed a polynomial p ( x ) with roots ( 1 − r i ) ( 1 ≤ r ≤ 4 ) using Vieta's. I think yours is better, though!
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@Sean Ty – Yeah i also used vieta formula mathod to check the correctness of my answer
Did it the same way!!
You can find second sum by taking ln(natural logarithm) and differentiating and find value at x=1
One can just calculate the product.But there's a easier way to compute the second term in the product.One can find the polynomial whose roots are 1 − r i 1 for i = 1 , . . , 4 by transformation.(First let y = 1 − x i.e, x = 1 − y to find the polynomial p 1 ( x ) whose roots are 1 − r i and then let y = x 1 i.e x = y 1 in p 1 ( x ) to get polynomial p 2 ( x ) whose roots are 1 − r i 1 .)