# roll theorem question..

• February 2nd 2009, 11:08 AM
transgalactic
roll theorem question..
i am given that
$
\frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
$

prove that there is a solution for 0<x<1
using roll theorem
$
a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
$
• February 2nd 2009, 12:12 PM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
i am given that
$
\frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
$

prove that there is a solution for 0<x<1
using roll theorem
$
a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
$

Consider the function $f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x$.
Notice that $f(0) = f(1) = 0$.
Now apply Rolle's theorem and get $f'(x_0) = 0$ for some $0 < x_0 < 1$.
• February 4th 2009, 01:28 AM
transgalactic
if i put x=1 on your function
f(1) doesnt equal zero

how you get f(1)=0
??
• February 4th 2009, 04:17 AM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
how you get f(1)=0
??

That is the hypothesis you were given.
• February 5th 2009, 09:43 PM
transgalactic
i cant see the equation that if i will put x=1
i will get 0

?
• February 8th 2009, 08:53 AM
transgalactic
there is no such hypothesis in the posted question
??
• February 8th 2009, 11:20 AM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
there is no such hypothesis in the posted question
??

$f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0$
That is the hypothesis.
• February 8th 2009, 12:44 PM
transgalactic
Quote:

Originally Posted by ThePerfectHacker
$f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0$
That is the hypothesis.

so its only an assumption
its not a fact??
• February 8th 2009, 04:53 PM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
so its only an assumption
its not a fact??

You are given this assumption part of the problem.
• February 8th 2009, 08:21 PM
transgalactic
ok if f(0)=f(1) then there is a point c for which
f'(c)=0
0<c<1
but its not a solution for f(x)

its an extreme point
??
• February 8th 2009, 08:24 PM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
ok if f(0)=f(1) then there is a point c for which
f'(c)=0
0<c<1
but its not a solution for f(x)

its an extreme point

It is a solution to $f'(x) = 0$ (with $x=c$).
But $f'(x) = a_0x^n + ... + a_n$.
Thus, you found a solution to the polynomial equation $a_0x^n + ... + a_n = 0$.
• February 8th 2009, 11:48 PM
transgalactic
so you take a function
$
f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x\\
$

which is an anti derivative of my given equation that i need find solution to.
$
f'(x)=a_0 x^n + a_1 x^{n - 1} + .. + a_n
$

so the extreme point of f(x) is the solution of f'(x)

thanks:)
• February 9th 2009, 05:58 PM
ThePerfectHacker
Yes, finally you understand. (Clapping)