IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

If x₁, x₂, x₃, ...,x_n are the roots of the polynomial P(x) = xⁿ+x^n-1+..+1 then prove that 1/1-x₁ + 1/1-x₂+...+1/1-x_n = n/2.

neeraj_agarwal_1990

(x-x1)(x-x2)......(x-xn) = x^n+x^(n-1)+......+1

take log...

log(x-x1) + log (x-x2) +.....log(x-xn) = log(x^n+x^(n-1)+...+1)

now differentiate both sides...

1/x-x1 + 1/x-x2 +..+1/x-xn = nx^(n-1)+(n-1)x^(n-2)+...+1 /(x^n+x^(n-1) +...+1)

put x=1...

required value = (1+ 2 +...+n )/ (1+1+....+1) --->n+1 times

so required value = n/2 (using 1+2+....+n = n(n+1)/2)

hsbhatt

yup. nice soln by neeraj. but I will wait for other approaches. And for goodness sake pls don't post variations of this same approach (if any). you're gonna get boinked for that!

neeraj_agarwal_1990

i think its a common approach...there are some similar questions in complex numbers too....

hsbhatt

@Neeraj: I was going to say this when my connection timed out-

The approach is good, only you have pretty confidently taken log on both sides without heed to whether the arguments are +ve or not. And not knowing complex analysis and logs of -ve numbers in complex form.

That is why I am waiting for a watertight approach.

So guys/gals dont lose heart! There still are points for the taking

neeraj_agarwal_1990

itni tension lenge to question kaise solve hoga!

neeraj_agarwal_1990

anyways....lets see wat others have to say...

hsbhatt

nahi neeraj bhai, there are finicky mathematicians who will take you to task for such oversights (our friend Konichiwa has caught me n times being careless). Math = Rigour in arguments.
That's why I think this business of having JEE in objective format is not a good idea.

neeraj_agarwal_1990

objective karte karte aisi cheezon pe dhyaan kam jaata hai....
it has become a habit.......

rohith291991

well majority of us are aiming for iit to become engineers!! not mathematicians.... so mathematical rigour of this degree isnt gonna be needed that much...but i am speaking for the masses....in my opinion u are VERY right hsbhatt....

rohith291991

and anyway even if both sides are negative ...u will just have to modify and write log|LHS| + ipi/2 = log|RHS| + ipi/2 the ipi/2 terms get cancelled....

konichiwa2x

Well this is not a very different method, but anyway...

In general if

, are the roots of a polynomial

of degree n, then

this is very easy to see: let

then

now differentiate with respect to x to get the result. so, for example, if

,
then

, and

thus

ps. you can do this by another method, but its pretty tedious. You can look at the case e.g n=3 or n=4 and then generalize.Clear the denominators, simplify, and group on the left. using that sums of roots, products of pairs of roots, etc are -1,+1, -1 , + 1 etc from coefficients of P(x) combined with P(1) = n+1, you can prove it.

hsbhatt

Exactly what I was lookin' for!

You can avoid the log business simply by differentiating by parts.

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