integral/polynomial question (1 Viewer)

[OLDMAN]

Explain why the polynomial
b_0+b_1x+...+b_nx^n has at least one real root if,
b_0/1+b_1/2+...+b_n/(n+1)=0.

 

[underthesun]

let f(x) = b_0+b_1x+...+b_nx^n

let s(x) = b_0/1+b_1/2+...+b_n/(n+1)

now s(x) = I{0-->1}f(x)dx

which means that s(x) represents the total for the area between the curve and the x-axis from x=0 to x=1. However, s(x), or total area, equals zero.

This means that there must be two segments of areas which "cancels" each other out, and hence the curve must have gone through y = 0 between x = 0 and x =1. This means that it has at least one real root.

is there a shorter way to say this?

 

[Harimau]

I suck... so im just going to try and fudge this.

Intergrate the original polynomial:

=b_0/1+b_1/2+...+b_n/(n+1)+ C, where C is some constant.

Now if b_0x/1+b_1x^2/2+...+b_nx^n+1/(n+1)=0,Then the intergral of the original polynomial would result in some constant C. Therefore, there exists at least some area either below or above the X axis

Now if b_0/1+b_1/2+...+b_n/(n+1)=0, that would mean that certain coeffiecients would be negative for the expression to be able to equal 0. This is so because if b_0, b_1.....B_n are all>0, the polynomial would be a positive definite without any roots. Therefore this polynomial has at least one real root.


EDIT: NVM, My answer is wrong. Im an idiot

 

[OLDMAN]

well done, underthesun. I wonder how much harder it would've been if I named the subject "harder polynomial" rather than "integral/polynomial" thus not giving the integral method away.

harimau : often, integral limits are simpler to work with than adding a constant.

 

[underthesun]

Originally posted by OLDMAN
I wonder how much harder it would've been if I named the subject "harder polynomial" rather than "integral/polynomial" thus not giving the integral method away.

Yeah, it was quite a hint

 

[freaking_out]

Originally posted by OLDMAN
well done, underthesun. I wonder how much harder it would've been if I named the subject "harder polynomial" rather than "integral/polynomial" thus not giving the integral method away.

harimau : often, integral limits are simpler to work with than adding a constant.

actually, the fact that the question is posted by "OLDMAN", turns me off straight away.