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Math Help - Zeros of function with binomial coefficients

  1. #1
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    Zeros of function with binomial coefficients

    For n = 1,2,..., I want to prove that
    f_n(x)=-(2x-1)^n+2nx^{2n-1}(1-x)\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}
    has exactly one zero for x\in\left[\frac{1}{2},1\right].

    It is easy to see that f_n\left(\frac{1}{2}\right)>0 (use the thread "Inequality with binomial coefficients"), f_n\left(1\right) = -1 <0, and f_n(x) is continuous. This implies a positive number of zeros on the interval \left[\frac{1}{2},1\right]. However, I have not been able to prove uniqueness.

    The plot in Zeros of function with binomial coefficients-eq.jpg indicates the truth of the statement for n = 1, 4, 7, ..., 20. Other values for n can be easily constructed using the Matlab code in eq.txt. This code also confirms that the number of zeros equals one, which may not be very obvious from the plot.

    A potential way to prove the result is to show that f''_n(x) is negative for x\in\left[\frac{1}{2},1\right]. Unfortunately, f''_n(x) is a very complex expression. I can post this expression if someone wants to.
    Could anyone help me? Many thanks!
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    Last edited by sander; September 10th 2011 at 06:04 AM.
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  2. #2
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    Re: Zeros of function with binomial coefficients

    Well, why not just calculate its first derivative and equate it to zero.
    if there are no stationary points inside (1/2,1) then f_n is strictly decreasing (btw, you meant f_n(1)=-1<0).

    Yes this maybe also tough to calculate, you can let Maple,Matlab,Mathematica to calculate this. (I prefer Maple, it seems more quick than Mathematica).

    Have you tried equating f' to zero, and see if there are no solutions?
    Even if there is a solution, you need to check that f_n at these stationary points the following cases are met:
    1. we don't have one stationary point which its f_n value is negative and another two stationary points which f_n value is positive cause in this case we have two zeros(actually 3).
    and other options, which I can't account for now cause I haven't calculated its first derivative, but the principle in this way to look for stationary and excluding any possibility for more than one zero.
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  3. #3
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    Re: Zeros of function with binomial coefficients

    It is clear from my numerical results from Matlab that f'_n(x)=0 has one solution for n=2,3,.... Of course this would be a sufficient condition for my proposition that f_n(x)=0 has exactly one solution on \left[\frac{1}{2},1\right]. However, I want an analytical proof. In fact, the first derivative of f_n(x) is already quite complicated:

    \frac{f'_n(x)}{2n} = (2n-1)x^{2n-2}(1-x)\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}-x^{2n-1}\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}-x^{2n-2}\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{2i(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}-(2x-1)^{n-1}
    =(2n-1-2nx)x^{2n-2}\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}-x^{2n-2}\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{2i(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}-(2x-1)^{n-1}

    I don't see a way how to derive the zeros of f'_n(x) from this expression. Additionally, Maple provides a polynomial of degree 2n-1 which is intuitive. See the Maple code in eqMaple.txt.

    Maple indicates that f_n(x) has for n=2,3,... only 2 real roots, one of them on \left[\frac{1}{2},1\right]. See the Maple code in eqMaple.txt. Again, this is not an analytical proof for the general case of an arbitrary integer n.

    Can anybody help me with an analytical proof that f_n(x)=0 has exactly one solution on \left[\frac{1}{2},1\right]?

    btw: I have corrected the f_n(1)-term in my previous post.
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  4. #4
    Super Member girdav's Avatar
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    Re: Zeros of function with binomial coefficients

    An idea: put t=\frac{1-x}x for \frac 12\leq x\leq 1 (then 0\leq t\leq 1) and work with the function h_n(t):=(-1)^{n+1}t^n(t+1)^n+2n\sum_{i=0}^{n-1}\binom{n-1}i\frac{(-1)^i}{3+2i}t^{2i+3}.
    Last edited by girdav; September 11th 2011 at 06:03 AM.
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  5. #5
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    Re: Zeros of function with binomial coefficients

    1. It turns out that I should put u:=u(x)=\frac{1-x}{x} then 0\leq u\leq 1 for x\in\left[\frac{1}{2},1\right]. This is a bijection, because u'(x)=-\frac{1}{x^2}<0. Define
       g_n ( u ): = f_n(x) = - \left( {\frac{{1 - u}}{{1 + u}}} \right)^n  + 2n\left( {u + 1} \right)^{ - 2n} \sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - 1} \right)^i }}{{3 + 2i}}u^{2i + 1} }
      Since u(x) is a bijection, the number of zeros of f_n on \left[\frac{1}{2},1\right] equals the number of zeros of g_n on \left[0,1\right].
      Notice g_n(0)<0 and g_n(1)>0 (use this thread.), so it suffices to consider the number of zeros on \left(0,1\right). Now define
      h_n \left( u \right): = u^2 \left( {1 + u} \right)^{2n} g_n \left( u \right) =  -u^2\left( {1 - u^2 } \right)^n  + 2n\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - 1} \right)^i }}{{3 + 2i}}u^{2i + 3} }
      which has the same sign as g_n(u) and so the same number of zeros on \left(0,1\right).
      It is straightforward to show that
       h'_n \left( u \right) = -2u\left( {1 - u^2 } \right)^n + 2nu^3\left( {1 - u^2 } \right)^{n-1} + 2n\sum\limits_{i = 0}^{n - 1}  {\binom{n-1}{i}{\left( { - 1} \right)^i }u^{2i + 2} } = 2u\left( {1 - u^2 } \right)^{n - 1} \left( {u^2 (n + 1) + nu - 1)} \right)
      This function has zeros at -1,0,\frac{1}{n+1},1. Therefore, the continuous function g_n declines on \left(0,\frac{1}{n+1}\right) and increases on \left(\frac{1}{n+1},1\right). Combining this with g_n(0)<0 and g_n(1)>0, it follows that g_n has one zero, which is on \left(\frac{1}{n+1},1\right]. Many thanks!
    2. Although I am currently satisfied, I am wondering if an exact solution for the zero is available. Define k_n\left( u \right): = \frac{\left( {1 + u} \right)^{2n}}{u} g_n \left( u \right) =  -\frac{\left( {1 - u^2} \right)^n}{u} + 2n\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - u^2} \right)^i }}{{3 + 2i}}}
       =  -\frac{\left( {1 - u^2} \right)^n}{u} + 2n\int_0^1\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\left( - u^2\right)^i }x^{2+2i}dx
       =-\frac{\left( {1 - u^2} \right)^n}{u} + 2n\int_0^1x^2\left(1-(xu)^2\right)^{n-1}dx
      Because g_n and k_n have the same zeros on (0,1), we know from the previous discussion that
      \frac{\left( {1 - u^2} \right)^n}{u} = 2n\int_0^1x^2\left(1-(xu)^2\right)^{n-1}dx
      has a unique solution on (0,1). Using the Maple code
      restart: (simplify(int(x^2*(1-(x*u)^2)^abs(n-1),x=0..1)));
      or from the Wolfram online integrator (with or without | | signs), it follows that the right-hand side has something to do with the hypergeometric distribution.

      To summarize: Does anyone know whether an analytical expression is available for the unique solution of the previous equation, i.e. k_n(u)=0 on (0,1)?
    Last edited by sander; September 12th 2011 at 12:45 AM.
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