Zeros of function with binomial coefficients

**sander** · September 10th 2011, 12:24 AM

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!

**InvisibleMan** · September 10th 2011, 04:24 AM

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.

**sander** · September 10th 2011, 06:16 AM

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.

**girdav** · September 10th 2011, 06:49 AM

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 $h_n(t):=(-1)^{n+1}t^n(t+1)^n+2n\sum_{i=0}^{n-1}\binom{n-1}i\frac{(-1)^i}{3+2i}t^{2i+3}$ .

**sander** · September 11th 2011, 03:21 AM

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!
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 $(0,1)$ ?