Why don't you try bashing this

Algebra Level 3

2 2015 = x 100 + 3 x 99 + 5 x 98 + 7 x 97 + + 195 x 3 + 197 x 2 + 199 x \large 2^{2015} =x^{100} + 3x^{99} + 5x^{98} + 7x^{97} +\ldots+ 195x^{3} + 197x^{2} + 199x

In the equation above, all coefficients in the right side of the equation follows an arithmetic progression. How many roots of x x occur along the open interval ( 17 , 18 ) (17,18) ?

0 2 Cannot be determined 6 17 1 3

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2 solutions

Otto Bretscher
Oct 24, 2015

The LHS is 2 2015 > ( 2 10 ) 200 > 1 0 600 2^{2015}>(2^{10})^{200}>10^{600} while the RHS is < 100 200 2 0 100 < 100 1000 10 0 100 < 1 0 300 <100*200*20^{100}<100*1000*100^{100}<10^{300} . Thus there are 0 \boxed{0} solutions.

Efren Medallo
Oct 24, 2015

We will investigate this by using Budan's Theorem.

Budan's Theorem states that the number of existing roots of a polynomial function P ( x ) P(x) within a given interval ( a , b ) (a,b) cannot be greater than the difference in the number of sign variations in the coefficients of P ( x + a ) P(x+a) and P ( x + b ) P(x+b) . That is,

n s P ( x + a ) s P ( x + b ) \large n \leq s_{P(x+a)} - s_{P(x+b)}

where n n is the number of possible roots in the given interval; s s is the number of sign variations in the coefficients of a given polynomial

This is all provided that the following conditions are met:

P ( x + a ) P ( x + b ) P(x+a) \geq P(x+b) , and

P ( b ) 0 P(b) \neq 0 .

So we check.

In both polynomials, that is, P ( x + 17 ) P(x+17) and P ( x + 18 ) P(x+18) , we are assured that there will be no negative coefficients occurring anywhere in the function where x is multiplied. Thus, there is only one sign variation which both occurs at the end of each function. That is, at the coefficient in x and the constant. So, s P ( x + a ) = s P ( x + b ) = 1 s_{P(x+a)} = s_{P(x+b)} = 1

n s P ( x + a ) s P ( x + b ) \large n \leq s_{P(x+a)} - s_{P(x+b)}

n = 0 \large n = 0

Which means that there are no real roots in the said interval. :)

There are other means to solve this problem. I just wanted to use this theorem. Don't forget to reshare!

Thank you. I have learned something new!

Pi Han Goh - 5 years, 7 months ago

Wait...Do you expect me to expand f ( x + 17 ) f(x+17) and \f(x+18) ) for f ( x ) = x 100 + 3 x 99 + 5 x 98 + 7 x 97 + + 195 x 3 + 197 x 2 + 199 x 2 2015 f(x) = x^{100} + 3x^{99} + 5x^{98} + 7x^{97} +\ldots+ 195x^{3} + 197x^{2} + 199x -2^{2015} then find the change of signs of coefficients?

If yes, how do you know the coefficient of constants in f ( x + 17 ) f(x+17) and f ( x + 18 ) f(x+18) are positive without any calculation?

Pi Han Goh - 5 years, 7 months ago

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since ( x + a ) n (x+a)^{n} would not yield any negative terms for any positive value of n n , I guess it will be alright anymore not to solve for their values; just knowing that they are positive is already enough. :)

Efren Medallo - 5 years, 7 months ago

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Also, the sums of the constants 1 7 100 + 3 1 7 99 . . . 17^{100} + 3•17^{99} ... will yield a sum greatly less than 2 2015 2^{2015} . This is since ( 17 / 2 ) 100 < 2 1915 (17/2)^{100} < 2^{1915}

Efren Medallo - 5 years, 7 months ago

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@Efren Medallo So you're saying the constant is a negative value right? so p 17 = 1 p_{17} = 1 ? How about the value of p 18 p_{18} ?

Pi Han Goh - 5 years, 7 months ago

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@Pi Han Goh Yes, that's right! :) Same thing for p 18 p_{18}

Efren Medallo - 5 years, 7 months ago

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