roll theorem question..

**transgalactic** · February 2nd 2009, 11:08 AM

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 \\ $

**ThePerfectHacker** · February 2nd 2009, 12:12 PM

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 $0 < x_0 < 1$ .

**transgalactic** · February 4th 2009, 01:28 AM

if i put x=1 on your function
f(1) doesnt equal zero

how you get f(1)=0
??

**ThePerfectHacker** · February 4th 2009, 04:17 AM

Originally Posted by transgalactic

how you get f(1)=0
??

That is the hypothesis you were given.

**transgalactic** · February 5th 2009, 09:43 PM

i cant see the equation that if i will put x=1
i will get 0

?

**transgalactic** · February 8th 2009, 08:53 AM

there is no such hypothesis in the posted question
??

**ThePerfectHacker** · February 8th 2009, 11:20 AM

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.

**transgalactic** · February 8th 2009, 12:44 PM

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??

**ThePerfectHacker** · February 8th 2009, 04:53 PM

Originally Posted by transgalactic

so its only an assumption
its not a fact??

You are given this assumption part of the problem.

**transgalactic** · February 8th 2009, 08:21 PM

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
??

**ThePerfectHacker** · February 8th 2009, 08:24 PM

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 $a_0x^n + ... + a_n = 0$ .

**transgalactic** · February 8th 2009, 11:48 PM

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

**ThePerfectHacker** · February 9th 2009, 05:58 PM

Yes, finally you understand.

Thread: roll theorem question..

LinkBack

Thread Tools

Display

roll theorem question..

Similar Math Help Forum Discussions

Counterfeit dice roll question

roll theorem prove question..

probability die roll

roll a die

roll of the die

Search Tags