2 2 0 1 5 = x 1 0 0 + 3 x 9 9 + 5 x 9 8 + 7 x 9 7 + … + 1 9 5 x 3 + 1 9 7 x 2 + 1 9 9 x
In the equation above, all coefficients in the right side of the equation follows an arithmetic progression. How many roots of x occur along the open interval ( 1 7 , 1 8 ) ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
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 ) within a given interval ( a , b ) cannot be greater than the difference in the number of sign variations in the coefficients of P ( x + a ) and P ( x + b ) . That is,
n ≤ s P ( x + a ) − s P ( x + b )
where n is the number of possible roots in the given interval; 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 ) , and
P ( b ) = 0 .
So we check.
In both polynomials, that is, P ( x + 1 7 ) and P ( x + 1 8 ) , 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
n ≤ s P ( x + a ) − s P ( x + b )
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!
Wait...Do you expect me to expand f ( x + 1 7 ) and \f(x+18) ) for f ( x ) = x 1 0 0 + 3 x 9 9 + 5 x 9 8 + 7 x 9 7 + … + 1 9 5 x 3 + 1 9 7 x 2 + 1 9 9 x − 2 2 0 1 5 then find the change of signs of coefficients?
If yes, how do you know the coefficient of constants in f ( x + 1 7 ) and f ( x + 1 8 ) are positive without any calculation?
Log in to reply
since ( x + a ) n would not yield any negative terms for any positive value of n , I guess it will be alright anymore not to solve for their values; just knowing that they are positive is already enough. :)
Log in to reply
Also, the sums of the constants 1 7 1 0 0 + 3 • 1 7 9 9 . . . will yield a sum greatly less than 2 2 0 1 5 . This is since ( 1 7 / 2 ) 1 0 0 < 2 1 9 1 5
Log in to reply
@Efren Medallo – So you're saying the constant is a negative value right? so p 1 7 = 1 ? How about the value of p 1 8 ?
Log in to reply
@Pi Han Goh – Yes, that's right! :) Same thing for p 1 8
Problem Loading...
Note Loading...
Set Loading...
The LHS is 2 2 0 1 5 > ( 2 1 0 ) 2 0 0 > 1 0 6 0 0 while the RHS is < 1 0 0 ∗ 2 0 0 ∗ 2 0 1 0 0 < 1 0 0 ∗ 1 0 0 0 ∗ 1 0 0 1 0 0 < 1 0 3 0 0 . Thus there are 0 solutions.