2 Attachment(s)

Zeros of function with binomial coefficients

For $\displaystyle n = 1,2,...$, I want to prove that

$\displaystyle f_n(x)=-(2x-1)^n+2nx^{2n-1}(1-x)\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{(-1)^i}{3+2i}\left(\frac{1-x}{x}\right)^{2i}$

has exactly one zero for $\displaystyle x\in\left[\frac{1}{2},1\right]$.

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

The plot in Attachment 22222 indicates the truth of the statement for $\displaystyle n = 1, 4, 7, ..., 20$. Other values for $\displaystyle n$ 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 $\displaystyle f''_n(x)$ is negative for $\displaystyle x\in\left[\frac{1}{2},1\right]$. Unfortunately, $\displaystyle f''_n(x)$ is a very complex expression. I can post this expression if someone wants to.

Could anyone help me? Many thanks!

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 $\displaystyle f_n(1)=-1<0$).

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 $\displaystyle f_n$ 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.

1 Attachment(s)

Re: Zeros of function with binomial coefficients

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

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

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

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

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

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

btw: I have corrected the $\displaystyle f_n(1)$-term in my previous post.

Re: Zeros of function with binomial coefficients

An idea: put $\displaystyle t=\frac{1-x}x$ for $\displaystyle \frac 12\leq x\leq 1 ($then $\displaystyle 0\leq t\leq 1$) and work with the function $\displaystyle h_n(t):=(-1)^{n+1}t^n(t+1)^n+2n\sum_{i=0}^{n-1}\binom{n-1}i\frac{(-1)^i}{3+2i}t^{2i+3}$.

Re: Zeros of function with binomial coefficients

- It turns out that I should put $\displaystyle u:=u(x)=\frac{1-x}{x}$ then $\displaystyle 0\leq u\leq 1$ for $\displaystyle x\in\left[\frac{1}{2},1\right]$. This is a bijection, because $\displaystyle u'(x)=-\frac{1}{x^2}<0$. Define

$\displaystyle g_n ( u ): = f_n(x) = - \left( {\frac{{1 - u}}{{1 + u}}} \right)^n + 2n\left( {u + 1} \right)^{ - 2n} \sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - 1} \right)^i }}{{3 + 2i}}u^{2i + 1} }$

Since $\displaystyle u(x)$ is a bijection, the number of zeros of $\displaystyle f_n$ on $\displaystyle \left[\frac{1}{2},1\right]$ equals the number of zeros of $\displaystyle g_n$ on $\displaystyle \left[0,1\right]$.

Notice $\displaystyle g_n(0)<0$ and $\displaystyle g_n(1)>0$ (use this thread.), so it suffices to consider the number of zeros on $\displaystyle \left(0,1\right)$. Now define

$\displaystyle h_n \left( u \right): = u^2 \left( {1 + u} \right)^{2n} g_n \left( u \right) = -u^2\left( {1 - u^2 } \right)^n + 2n\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - 1} \right)^i }}{{3 + 2i}}u^{2i + 3} }$

which has the same sign as $\displaystyle g_n(u)$ and so the same number of zeros on $\displaystyle \left(0,1\right)$.

It is straightforward to show that

$\displaystyle h'_n \left( u \right) = -2u\left( {1 - u^2 } \right)^n + 2nu^3\left( {1 - u^2 } \right)^{n-1} + 2n\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}{\left( { - 1} \right)^i }u^{2i + 2} } = 2u\left( {1 - u^2 } \right)^{n - 1} \left( {u^2 (n + 1) + nu - 1)} \right)$

This function has zeros at $\displaystyle -1,0,\frac{1}{n+1},1$. Therefore, the continuous function $\displaystyle g_n$ declines on $\displaystyle \left(0,\frac{1}{n+1}\right)$ and increases on $\displaystyle \left(\frac{1}{n+1},1\right)$. Combining this with $\displaystyle g_n(0)<0$ and $\displaystyle g_n(1)>0$, it follows that $\displaystyle g_n$ has one zero, which is on $\displaystyle \left(\frac{1}{n+1},1\right]$. Many thanks! - Although I am currently satisfied, I am wondering if an exact solution for the zero is available. Define $\displaystyle k_n\left( u \right): = \frac{\left( {1 + u} \right)^{2n}}{u} g_n \left( u \right) = -\frac{\left( {1 - u^2} \right)^n}{u} + 2n\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\frac{{\left( { - u^2} \right)^i }}{{3 + 2i}}}$

$\displaystyle = -\frac{\left( {1 - u^2} \right)^n}{u} + 2n\int_0^1\sum\limits_{i = 0}^{n - 1} {\binom{n-1}{i}\left( - u^2\right)^i }x^{2+2i}dx$

$\displaystyle =-\frac{\left( {1 - u^2} \right)^n}{u} + 2n\int_0^1x^2\left(1-(xu)^2\right)^{n-1}dx$

Because $\displaystyle g_n$ and $\displaystyle k_n$ have the same zeros on $\displaystyle (0,1)$, we know from the previous discussion that

$\displaystyle \frac{\left( {1 - u^2} \right)^n}{u} = 2n\int_0^1x^2\left(1-(xu)^2\right)^{n-1}dx$

has a unique solution on $\displaystyle (0,1)$. 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. $\displaystyle k_n(u)=0$ on $\displaystyle (0,1)$?