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Functions 2

  1. Mar 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Let f(x) be a pllynomial of degree n, an odd positive integer, and has monotonic behaviour then the number of real roots of the equation
    f(x)+ f(2x) + f(3x)...+ f(nx)= n(n+1)/2

    2. Relevant equations



    3. The attempt at a solution

    This seems like the summation of 1+2+3+...+n, and that would be a root of the equation, but I dont know if thats the only one.
     
  2. jcsd
  3. Mar 1, 2007 #2

    AKG

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    You didn't finish asking a question. Do you mean that you have to show that the number of solutions of

    f(x) + ... + f(nx) = 0

    is n(n+1)/2, or are you asked to find the number of solutions of the equation

    f(x)+ f(2x) + f(3x)...+ f(nx)= n(n+1)/2?
     
  4. Mar 1, 2007 #3

    HallsofIvy

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    Not to mention: what do you mean by "has monotonic behavior"? Do you mean that it is monotonic for all x? If so then f(x)= 0 has exactly one root.
     
  5. Mar 4, 2007 #4
    You have to find the number of roots of the equation. And it is monotonic for all x. How did you get f(x)=0 has only one root?
     
  6. Mar 5, 2007 #5

    AKG

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    So you have to find the number of x-values such that

    f(x) + ... + f(nx) = n(n+1)/2

    Right? f(x)=0 has only one root because f is monotonic, so once it crosses the x-axis once, it will never cross again. Of course, it must cross the x-axis at least once because f is an odd-degree polynomial.
     
  7. Mar 5, 2007 #6

    HallsofIvy

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    I would recommend looking at simple cases first. It is easy to show that if f(x)= ax+ b, there is exactly one root to that equation, but what if f(x)= x3 or variations on that?
     
  8. Mar 12, 2007 #7
    Since f(x) is monotonic for all x, then it has only one solution. I.e, it only cuts the line y=0 at one point. Let f(x) be a function of the type [tex]f(x)=a_nx^n+a_{n-1}x^{n-1}...[/tex]
    Then f(x)+f(2x)+f(3x)+...+f(nx) will also be a function of the type [tex]F(x)=A_nx^n+A_{n-1}x^{n-1}...[/tex] which will also be monotonic, and hence have only one solution.

    Is that right?
     
  9. Mar 12, 2007 #8

    AKG

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    [itex]x \mapsto f(x)[/itex] is either monotonic increasing or monotonic decreasing. Without loss of generality, suppose it is monotonic increasing. Then what can you say about the functions [itex]x \mapsto f(kx)[/itex] where k is some positive integer?
     
  10. Mar 13, 2007 #9
    f(kx) is also monotonically increasing or decreasing. Graphically, if k<1, then the graph expands, and if k>1, then the graph contracts, but the graph remains basically the same.
     
  11. Mar 13, 2007 #10

    AKG

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    Right, and it is monotonic in the same way, i.e. if f(x) is increasing, then f(kx) is increasing (for k > 0) and if f(x) is decreasing then f(kx) is decreasing (for k > 0). Now you know that f(x), f(2x), ..., f(nx) are all monotonic, and either ALL strictly increasing or ALL strictly decreasing. What can you say about the sum f(x) + ... + f(nx)?
     
  12. Mar 14, 2007 #11
    f(x)+...+f(nx) is also monotonic. Thats why it can only intersect the x axis at one point. Hence, f(x)+...+f(nx) has only one root.
     
  13. Mar 14, 2007 #12

    AKG

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    Okay, but you're not done yet. You actually want to find the number of roots of f(x) + ... + f(nx) - n(n+1)/2. f(x) + ... + f(nx) is monotonic, and -n(n-1)/2 is just a constant, so what can you say about their sum? Also, ex is monotonic, but it has no zeroes. Both ex and any odd-degree polynomial, so what more do you need to say about f(x), and then about f(x) + ... + f(nx), and then about f(x) + ... + f(nx) - n(n-1)/2 in order to count the roots?
     
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