i am given that

$\displaystyle

\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

$\displaystyle

a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\

$

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- Feb 2nd 2009, 11:08 AMtransgalacticroll theorem question..
i am given that

$\displaystyle

\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

$\displaystyle

a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\

$ - Feb 2nd 2009, 12:12 PMThePerfectHacker
- Feb 4th 2009, 01:28 AMtransgalactic
if i put x=1 on your function

f(1) doesnt equal zero

how you get f(1)=0

?? - Feb 4th 2009, 04:17 AMThePerfectHacker
- Feb 5th 2009, 09:43 PMtransgalactic
i cant see the equation that if i will put x=1

i will get 0

? - Feb 8th 2009, 08:53 AMtransgalactic
there is no such hypothesis in the posted question

?? - Feb 8th 2009, 11:20 AMThePerfectHacker
- Feb 8th 2009, 12:44 PMtransgalactic
- Feb 8th 2009, 04:53 PMThePerfectHacker
- Feb 8th 2009, 08:21 PMtransgalactic
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

?? - Feb 8th 2009, 08:24 PMThePerfectHacker
- Feb 8th 2009, 11:48 PMtransgalactic
so you take a function

$\displaystyle

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.

$\displaystyle

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:) - Feb 9th 2009, 05:58 PMThePerfectHacker
Yes, finally you understand. (Clapping)