150 Friends Problem

Algebra Level 5

f ( x ) = x 150 + x 2 7 x + 1 \large f(x)= x^{150} + x^{2} -7x + 1

Let x 1 x_{1} , x 2 x_{2} , x 3 x_{3} \ldots x 150 x_{150} be the roots of the above equation. Define

S = i = 1 150 1 ( 1 + x i ) 2 \large S = \huge \sum_{i=1}^{150}\dfrac{1}{(1+x_{i})^{2}}

Find S |S| .


The answer is 1982.39.

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3 solutions

Surya Prakash
Aug 29, 2015

Since, x 1 , x 2 , x 150 x_{1}, x_{2}, \ldots x_{150} are the roots of f ( x ) f(x) .

So, f ( x ) = ( x x 1 ) ( x x 2 ) ( x x 150 ) f(x) = (x-x_{1})(x-x_{2}) \ldots (x-x_{150}) .

Differentiating it w.r.t x x and dividing it with f ( x ) f(x) . We get

f ( x ) f ( x ) = i = 1 150 1 x x i \dfrac{f'(x)}{f(x)} = \sum_{i=1}^{150} \dfrac{1}{x-x_{i}}

Differentiating again w.r.t x x . We get,

( f ( x ) ) 2 f ( x ) f ( x ) f 2 ( x ) = i = 1 150 1 ( x x i ) 2 \dfrac{(f'(x))^{2} - f(x)f''(x)}{f^2 (x)} = - \sum_{i=1}^{150}\dfrac{1}{\left(x -x_{i}\right)^{2}}

Taking x = 1 x = -1 , we get, ( f ( 1 ) ) 2 f ( 1 ) f ( 1 ) f 2 ( 1 ) = i = 1 150 1 ( 1 x i ) 2 \dfrac{(f'(-1))^{2} - f(-1)f''(-1)}{f^2 (-1)} = - \sum_{i=1}^{150}\dfrac{1}{\left(-1 -x_{i}\right)^{2}}

But we have f ( 1 ) = 10 f(-1) = 10 , f ( 1 ) = 159 f'(-1) = -159 and f ( 1 ) = 22352 f''(-1) = 22352 . Substituting and solving them we get,

S = i = 1 150 1 ( 1 + x i ) 2 = 1982.39 S = \sum_{i=1}^{150}\dfrac{1}{\left(1 +x_{i}\right)^{2}} = -1982.39

So, S = 1982.39 |S| = \boxed{1982.39} .

Moderator note:

Since you tagged this problem with algebra, what would be an algebraic approach?

Bingo!! Same method! Amazing problem and congrats on 150 followers!

Aditya Kumar - 5 years, 9 months ago

Good. It took me a few seconds only to crack the idea of this problem as I had previously posted a similar problem: An Intriguing Polynomial Problem

Also Shivamani had added a similar problem being inspired my my problem: 400 Followers Problem

I've also linked your problem with mine as a Sister Problem! :D

Satyajit Mohanty - 5 years, 9 months ago

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but it's little different than yours. :)

Surya Prakash - 5 years, 9 months ago

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Similar = A little different :P

Satyajit Mohanty - 5 years, 9 months ago

I think even transformation of the polynomial helps...

Abhishek Bakshi - 5 years, 9 months ago

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Yaah! Nice idea +1. but it makes the solution little lengthy

Surya Prakash - 5 years, 9 months ago

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that's true but its a pass for someone who doesn't want to use or doesn't know calculus..

Abhishek Bakshi - 5 years, 9 months ago

U really back?? No studies?? :P

Aditya Kumar - 5 years, 9 months ago

Vieta is all you need.

Billy Sugiarto - 5 years, 9 months ago

Since you tagged this problem with algebra (as opposed to calculus), what would be an algebraic approach?

Calvin Lin Staff - 5 years, 9 months ago

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An algebraic approach would be using the transformations.

Surya Prakash - 5 years, 9 months ago

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Great, could you add that solution too?

Calvin Lin Staff - 5 years, 9 months ago

Awesome!Brilliant!

Adarsh Kumar - 5 years, 7 months ago

i solved it using algebra and nice problem!!

Dev Sharma - 5 years, 5 months ago

This is knowledge.

Lu Chee Ket - 5 years, 8 months ago
Lu Chee Ket
Oct 11, 2015

My failure in first attempt was caused by a careless mistake of 7 being taken as 1.

Lu Chee Ket - 5 years, 8 months ago

i did same

Dev Sharma - 5 years, 5 months ago
Madhav Srirangan
Sep 3, 2015

let t=1/1+x =>x=(1-t)/t sub x=(1-t)/t in f(x)=o we get, (1-t)^{150}+t^{148}*(1-t)^{2}-7t^{149}(1-t)+t^{150}=0 we get binomial expansion given the pic.

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