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Math Help - roll theorem prove question..

  1. #1
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    roll theorem prove question..

    question :
    suppose that to the quadratic equation x^2+px+q=0
    has two roots x_1<x_2

    prove that for
    <br />
\frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0<br />
    there are n different roots on the interval (x1,x2)
    ??

    the prove:
    by rolls theorem we have a point "c" on the interval of (x1,x2) for which

    f'(c)=0

    then they say
    "that we can get another solution "g" on the interval of (x1,c) using rolls theorem"

    but its not rolls theorem
    we dont have two points for which there is a point "j" for which f'(j)=0

    we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )

    ??
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  2. #2
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    is it ok to use rolls theorem here??
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  3. #3
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    ok
    my problem is very simple

    if i have a function f(x)
    and i have a point "a" for which f(a)=0
    and i have point "b" for which f'(b)=0
    is the a point "c" between [a,b] for which f''(c)=0
    ??

    (i got one oint of f(x) and the other on f'(x) they are not on the same level)
    Last edited by transgalactic; March 10th 2009 at 03:05 AM.
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    question :
    suppose that to the quadratic equation x^2+px+q=0
    has two roots x_1<x_2
    so we can write x^2+ px+ q= (x- x_1)(x- x_2).

    prove that for
    <br />
\frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0<br />
    there are n different roots on the interval (x1,x2)
    ??
    Write that as \frac{d^n (x-x_1)^n(x-x_2)^n}{dx^n}

    the prove:
    by rolls theorem we have a point "c" on the interval of (x1,x2) for which

    f'(c)=0

    then they say
    "that we can get another solution "g" on the interval of (x1,c) using rolls theorem"

    but its not rolls theorem
    we dont have two points for which there is a point "j" for which f'(j)=0we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )

    ??
    They aren't referring to f now, they are applying Rolle's theorem to f'.
    It should be obvious that for f(x)= (x^2+ px+ q)^n= (x-x_1)^n(x-x_2)^n, f' is also 0 at x_1 and x_2. In fact that is true for all deritives up to the n-1 derivative.
    Last edited by HallsofIvy; March 10th 2009 at 04:37 AM.
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    ok
    my problem is very simple

    if i have a function f(x)
    and i have a point "a" for which f(a)=0
    and i have point "b" for which f'(b)=0
    is the a point "c" between [a,b] for which f''(c)=0
    ??

    (i got one oint of f(x) and the other on f'(x) they are not on the same level)
    Do you mean "is there a point "c" between a and b for which f"(c)= 0"?
    Not necessarily. It depends on whether f'(a)= 0 also which is true for the polynomial in your first post.

    For example if f(x)= (x- 1)(x- 2)= x^2- 3x+ 2, f(1)= 0 and f(2)= 0 so there must exist "b" between 1 and 2 such that f'(b)= 2x- 3= 0. It is obvious that that point is b= 3/2 but f"(x)= 2 which is never 0.

    But if f(x)= (x^2- 3x+ 2)^2= (x-1)^2(x-2)^2, then f'(x)= 2(x-1)(x-2)^2+ 2(x-1)^2(x-2) which is 0 at x= 1, x= 2 and at x= 3/2. Now applying Rolle's theorem to f', on the intervals from 1 to 3/2 and from 3/2 to 2 there must be a number b between 1 and 3/2 such that f"(b)= 0 and and a number c between 3/2 and 2 such that f"(c)= 0. That is, there are two values between 1 and 2 that make the second derivative 0 as the theorem says.
    Last edited by HallsofIvy; March 10th 2009 at 12:12 PM.
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  6. #6
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    when you say
    "on the intervals from 1 to 3/2 and from 3/2 to 1"
    this is the same solution
    you cant take the same interval but on the opposite direction
    and say that its a different solution (its the same solution for the second derivative)

    and when you say:
    "there must be a number b between 0 and 3/2 such that f"(b)= 0 "

    b need to be between the interval(between 1 to 3/2) not
    "between 0 and 3/2"
    ??
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  7. #7
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    if you want to apply rolls theorem on f'(x)
    then we got a one solution for f''(x) between 1 and 3/2 and onther solution for f''(x) between 3/2 and 2

    ??
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  8. #8
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    i cant understand what is the general rule.

    in my solution they say "we have one point on f(a)=0
    the other on f'(b) so there is a point c where f''(c)=0

    that is not rolls theorem
    ??
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  9. #9
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    Quote Originally Posted by transgalactic View Post
    when you say
    "on the intervals from 1 to 3/2 and from 3/2 to 1"
    Sorry, that was a typo. I meant "1 to 3/2 and 3/2 to 2".
    I have edited my previous post.

    this is the same solution
    you cant take the same interval but on the opposite direction
    and say that its a different solution (its the same solution for the second derivative)

    and when you say:
    "there must be a number b between 0 and 3/2 such that f"(b)= 0 "

    b need to be between the interval(between 1 to 3/2) not
    "between 0 and 3/2"
    ??
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    i cant understand what is the general rule.

    in my solution they say "we have one point on f(a)=0
    the other on f'(b) so there is a point c where f''(c)=0

    that is not rolls theorem
    ??
    Are you certain that was what they said? it should be "we have one point on f'(a)=0 the other on f'(b) so there is a point c where f''(c)=0" which is Rolle's theorem applied to f' instead of f.
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  11. #11
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    i am given that f(x)=x^2 + px + q
    so i can present f(x) as f(x)=(x-x1)(x-x2)
    so if we look at n=2 we get
    f(x)=(x-x1)^2(x-x2)^2

    f'(x)=2(x-x1)(x-x2)^2 + 2(x-x1)^2(x-x2)=2(x-x1)(x-x2)[x-x2+x-x1]=
    =2(x-x1)(x-x2)[2x-x1-x2]

    so x1 and x2 are solution on f'(x)
    but this is a practical observation.
    what do i need to say so for g(x)=[f(x)]^n
    then x1,x2 are the roots till the n'th derivative??
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  12. #12
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    Quote Originally Posted by transgalactic View Post
    i am given that f(x)=x^2 + px + q
    so i can present f(x) as f(x)=(x-x1)(x-x2)
    so if we look at n=2 we get
    f(x)=(x-x1)^2(x-x2)^2

    f'(x)=2(x-x1)(x-x2)^2 + 2(x-x1)^2(x-x2)=2(x-x1)(x-x2)[x-x2+x-x1]=
    =2(x-x1)(x-x2)[2x-x1-x2]

    so x1 and x2 are solution on f'(x)
    but this is a practical observation.
    what do i need to say so for g(x)=[f(x)]^n
    then x1,x2 are the roots till the n'th derivative??
    If f(x)= (x-x1)^n(x-x2)^n then f'(x= n(x-x1)^(n-1)(x-x2)^n+ n(x-x1)^n(x-x2)^(n-1). f"= n(n-1)(x-x1)^(n-2)(x-x2)^n+ 2n^2(x-x1)^(n-1)(x-x2)^(n-1)+ n(n-1)(x-x1)^n(x-x2)^(n-2), etc. That is, the ith derivative is a sum of terms each including factors of (x-x1) and (x-x2) to at least the n-i power. As long as i< n, each term will include at least one factor of each of (x-x1) and (x-x2) and so will be 0 at x= x1 and x= x2.
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  13. #13
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    so in our example
    x1 and x2 are solutions till the n-1
    int the nth derivative x1 and x2 are not a solution
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