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Roll theorem prove question

  1. Mar 9, 2009 #1
    question :
    suppose that to the quadratic equation x^2+px+q=0
    has two roots [tex]x_1<x_2[/tex]

    prove that for
    [tex]
    \frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0
    [/tex]
    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 )

    ??
     
  2. jcsd
  3. Mar 9, 2009 #2

    Mark44

    Staff: Mentor

    I'm guessing that no one has responded because you haven't given enough information.
    That's "proof." A proof is the mathematical statements you write to show that some conclusion is reached when you start with a certain hypothesis. To prove something, you come up with those statements. A proof is a thing; proving it is an action.

    Anyway, Rolle's Theorem does guarantee that there is a number c in (x1, x2) for which f'(c) = 0. I'm assuming that f(x) = x^2 + px + q.

    Another solution to what? If f(x) = x^2 + px + q, there is only one solution to f'(x) = 0. You need to be specific about what equation g is a solution to.

    Many of your posts come from questions you are asking as you read someone else's proof. If you don't include all of the preceding steps, it is difficult or impossible for us to explain how the author of the proof can make some statement.

    Since the statement you're trying to prove (or whose proof you're trying to understand - this is not clear) involves the n-th derivative of a function raised to the n-th power, the most natural approach would seem to be a proof by induction, but there is no mention in your post that this is what is happening.
     
  4. Mar 10, 2009 #3
    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: Mar 10, 2009
  5. Mar 10, 2009 #4

    Mark44

    Staff: Mentor

    No, not in general. Here's a counterexample.
    f(x) = x^2 - 1
    f'(x) = 2x
    f''(x) = 2
    f(-1) = 0 and f'(0) = 0 so a = -1 and b = 0 and the interval is [-1, 0].
    There is no number in the interval [-1, 0] for which f''(x) = 0. In fact, for this function, there is no number anywhere for which f''(x) = 0.
     
  6. Mar 10, 2009 #5
    they say it in the proof .
    they say that we got a point f'(c)=0 so we check [x1,c]
    and between them there is another point "g" for which f''(g)=0
    and they say that we keep doing then till the n'th derivative

    you said that its not working for all functions
    what conditions do i need for this to work??
    i thought that for this stuff to work
    for example:
    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
    and if f'(a)=0
    is the a point "c" between [a,b] for which f''(c)=0

    but i cant see that they mention that each new solution is a solution to its anti derivative too??
     
  7. Mar 10, 2009 #6

    Mark44

    Staff: Mentor

    Probably the stated conditions of the problem.
    My counterexample shows that this is not true for all functions.
    "each new solution is a solution to its antiderivate" - this doesn't make any sense. Also, I asked in post 2 what you meant by "each new solution." Whenever you talk about a solution, it's always relative to some equation or inequality or a system of equations or inequalities. If I tell you that x = 6 is a solution, I'm not telling you anything meaningful if I don't tell you what this is a solution to.

    You said in your first post that they say "we can get another solution "g" on the interval of (x1,c) using rolls theorem." What equation is g a solution to?
     
  8. Mar 10, 2009 #7
    "You said in your first post that they say "we can get another solution "g" on the interval of (x1,c) using rolls theorem." What equation is g a solution to?"

    i ment f''(g)=0

    g is solution for the second derivative
     
  9. Mar 10, 2009 #8
    the trick that they are doing is taking a solution of a function on n'th derivative
    and a solution of the same function on the n+1 derivative

    and they say that using rolls theorem we have a solution on n+2 derivative of this function on the interval of the n'th and n+1 x value solution

    i cant understand how it could happen
    ??
     
  10. Mar 10, 2009 #9

    Mark44

    Staff: Mentor

    Here's what I would do:
    You're given that x^2 + px + q = 0 has two roots, x1 and x2, where x1 < x2.

    Show that the equation d/dx[(x^2 + px + q)^1] = 0 has 1 root in the interval (x1, x2).
    Show that the equation d^2/dx^2[(x^2 + px + q)^1] = 0 has 2 different roots in the interval (x1, x2).

    If you can follow the logic of the proof you're reading for n = 1 and n = 2, that would be a good start at following what they're doing for an arbitrary n.
     
  11. Mar 10, 2009 #10
    ok if i am given that x^2 + px + q = 0
    and it has two roots f(x1)=f(x2)=0

    so between them we have point 'c' for which f'(c)=0
    of f'(x) i have only point c
    there is no way to get a solution on f''(x)

    ??

    for what condition x1, and x2 are solution on every derivative
    ??
     
  12. Mar 10, 2009 #11
    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??
     
  13. Mar 10, 2009 #12

    Mark44

    Staff: Mentor

    OK, so you have verified that the original statement is true for n = 1, namely that the equation d/dx[(x^2 + px + q)^1] = 0 has 1 root in the interval (x1, x2). You're done with that part.
     
  14. Mar 10, 2009 #13

    Mark44

    Staff: Mentor

    Don't use f again, since you have already defined it above as f(x) = x^2 + px + q. You will confuse any readers, and yourself if you come along and redefine it as some different function. Use another letter, say g. (But don't also use g as a number in an interval.)
    So x1 and x2 are solutions of g'(x) = 0. (I changed the name of the function.)
    Did you not notice that there is another solution to the equation g'(x) = 0? It's also possible to see exactly where this solution is in relation to x1 and x2.
     
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