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Intervals-Word Problem

  1. Jun 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that [itex] \frac{a_0}{1} + \frac{a_1}{2} + ... \frac{(a_n)}{(n+1)} = 0 [/itex]

    then [itex] a_0 + a_1x + ... + a_nx^n [/itex] = 0

    for some x in the interval [0, 1].

    2. Relevant equations



    3. The attempt at a solution

    I thought at first the easiest value to find for a, would be 0, but it is not included on the interval and the question wants a more theoretical approach, rather than solving for a itself. I think I would be able to use limits in order to solve this question, but the most puzzling part is how to set it up (what would the function be on which the limit is being tested)? Is it the general form: [itex] \frac{a_n}{(n+1)} [/itex] and [itex] a_nx^n [/itex] ? Thank you.
     
  2. jcsd
  3. Jun 15, 2013 #2

    LCKurtz

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    Hint: Let$$
    F(x) = \int_0^x a_0+a_1t+...+a_nt^n\, dt$$and think about Rolle's theorem with ##F(x)##.
     
    Last edited: Jun 15, 2013
  4. Jun 15, 2013 #3
    When I integrate the above function I get:

    [itex] a_0x + \frac{a_1x^2}{2} + ... + \frac{a_nx^(n+1)}{n+1} [/itex]

    Rolle's theorem guarantees that there will be some c on the closed interval (a,b) such that f'(c)=0.

    I remember reading somewhere in the textbook that [itex] \frac{1}{b-a} ∫_a^b f(x) dx [/itex] (the average value) can be found using Rolle's Theorem?
     
  5. Jun 15, 2013 #4

    LCKurtz

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    Rolle's theorem has hypotheses and conclusions. State the theorem completely. Then try applying it to F(x) and show what you get. Skip the guessing about extraneous non-related issues.
     
  6. Jun 15, 2013 #5
    I was mentioning it because I thought it could be useful in this problem, but I guess not.

    The derivative of f(x) is [itex] a_0 + a_1x ... + a_nx^n [/itex]
    f'(0)= [itex] a_0 [/itex]
    So f'(x) is never equal to zero.
    By Rolle's Theorem, this means that this equation cannot have 2 or more solutions.
     
  7. Jun 15, 2013 #6

    LCKurtz

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    I asked you to state Rolle's theorem completely. Then try applying it to F(x) and show what you get. If you are going to ignore my suggestions, how do you expect me to help you? And please use consistent notation. Don't use f(x) if you mean F(x).
     
  8. Jun 15, 2013 #7
    I amn't ignoring your suggestions Professor. I thought that by Rolle's Theorem, I should find the critical points through the first derivative, and that is why I did that.

    I will use F(x) from now on to indicate the function in this problem.

    The Rolle's Theorem states exactly this: "Let a < b. If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)= f(b) then there is a c in (a,b) with f'(c)=0. That is, under these hypotheses F has a horizontal tangent somewhere between a and b".
     
  9. Jun 15, 2013 #8

    LCKurtz

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    OK, that's better but you are still mixing notation. Use either F or f, but not both in the statement of the theorem. Now you know what the theorem says.

    Now, what does it say about the ##F(x)## I suggested, and which you calculated correctly in post #3. What are a and b in your problem and does ##F(x)## satisfy the hypotheses (show this)? If so, what does the conclusion tell you?
     
  10. Jun 15, 2013 #9
    A and B are 0 and 1 in this problem. A < B and f'(x)=0 because the problem states that some x on the interval of [a,b] gives a y-value of 0. Rolle's Theorem tells me that f(a) will have to equal f(b) since A < B, and f'(c)= 0?
     
  11. Jun 15, 2013 #10

    LCKurtz

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    Yes. We are dealing with the interval [0,1]

    I thought we were talking about your problem with ##F(x)##. Rolle's theorem has hypotheses which the function ##F(x)## must satisfy before you can draw the conclusion.

    1. Is F(x) continuous on the closed interval?
    2. Is it differentiable on the open interval?
    3 Does F(0) = F(1)?

    These are the hypotheses, and you need to verify them or at least state why they are true. When you have done that, then what does the conclusion of the theorem tell you for our F(x)?
     
  12. Jun 15, 2013 #11
    I don't think I know that f(0) is equal to f(1). I do know that the function is continuous on the interval, and differentiable. How can you come to such a conclusion? Thank you.
     
  13. Jun 15, 2013 #12

    LCKurtz

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    PLEASE use F(x) and not f(x) if you are talking about the function I suggested. I gave you a hint for F(x) and you have a formula for F(x) that you figured out in post #3. You don't have to guess whether F(0) = F(1); use the formula and check it.
     
  14. Jun 15, 2013 #13
    I must be doing something wrong because I get F(0) = 0 and F(1)≠ 0.
     
  15. Jun 15, 2013 #14

    LCKurtz

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    Show your work. What do you get for F(1)? And at this point it might be a good idea for you to read the original problem again so you haven't forgotten what you are given and what you are trying to prove.
     
  16. Jun 15, 2013 #15
    [itex] a_0x + \frac{a_1x^2}{2} + ... + \frac{a_n(x^(n+1))}{n+1} [/itex]

    Using this formula I substituted in 0 and 1.

    F(0)= [itex] a_0(0) + \frac{a_1*(0^2)}{2} + ... + \frac{a_n(0^(n+1))}{n+1} [/itex] = 0

    F(1)= [itex] a_0(1) + \frac{a_1*(1^2)}{2} + ... + \frac{a_n(1^(n+1))}{n+1} = a_0 + \frac{a_1}{2} + ... + \frac{a_n*(1^(n+1)}{n+1} [/itex]
     
  17. Jun 15, 2013 #16

    LCKurtz

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    Since ##1^{n+1}=1## you hardly need to leave it in there. Note the correction I edited in post #2 to your statement of the problem. Does that help you with F(1)?
     
  18. Jun 15, 2013 #17
    I would think that means the last quantity is negligible so F(1)= [itex] a_0 + \frac {a_1}{2} [/itex] ? But I still don't see how they're equal. I'm sorry I'm so slow to catch on.
     
  19. Jun 15, 2013 #18

    LCKurtz

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    Why would you think that? You calculated F(1) and it isn't just two terms. What does the original problem say about that quantity? You did read my post #2 showing the IF that you left out, right?
     
  20. Jun 15, 2013 #19
    Yes, I did read that post (it was a slip-up). The original problem states that the second equation is contingent upon the truth of the first equation.
     
  21. Jun 15, 2013 #20

    LCKurtz

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    But the problem is giving you the first equation by saying IF it is true then prove....

    So what is the value of F(1)?

    Also, to satisfy my curiosity, what grade level are you in and what math courses have you completed?
     
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