# Upper and lower bounds for the smallest zero of a function

The function $G_m(x)$ is what I encountered during my search for approximates of Riemann $\zeta$ function:

$$f_n(x)=n^2 x\left(2\pi n^2 x-3 \right)\exp\left(-\pi n^2 x\right)\text{, }x\ge1;n=1,2,3,\cdots,\tag{1}$$ $$F_m(x)=\sum_{n=1}^{m}f_n(x)\text{, }\tag{2}$$

$$G_m(x)=F_m(x)+F_m(1/x)\text{, }\tag{3}$$

Numerical results showed that $G_m(x)$ is zero near $m+1.2$ for $m=1,2,...,8$.

Please refer to fig 1 below for the plot of $\log|G_m(x)|$ v.s. $x$ for $m=1,2,...,8$

Let us denote these zeros by $x_0(m)$. I am interested if it can be proved that

(A) $x_0(m)$ is the smallest zero of $G_m(x)$ for $x\ge1$

(B) there exist bounds $\mu(m),\nu(m)$ such that $0<\mu(m)\le x_0(m)\le \nu(m)$;$\mu(m),\nu(m)\to\infty$, when $m\to\infty$.

Here are the things I tried.

Because $G_m(1)>0$ and $G_m(x)\to F_m(1/x)<0$ when $x\to\infty$, so there exist a zero for $G_m(x)$ between $x=1$ and $x=\infty$.

But I was not able to find the bounds for this zero.

It is tempting to speculate that $x_0(m)$ is the only zero for $G_m(x)$ and $m+1<x_0(m)<m+2$.

The values for $x_0(m), (m=1,2,...,10)$ are given by:

$x_0(1)$=2.24203, $x_0(2)$=3.21971, $x_0(3)$=4.21913, $x_0(4)$=5.22283, $x_0(5)$=6.22764, $x_0(6)$=7.23268, $x_0(7)$=8.23764, $x_0(8)$=9.24241, $x_0(9)$=10.2469, $x_0(10)$=11.2512.

• Since you said that $G_m(1) > 0$ and $G_m(x) \to F_m(\frac{1}{x}) < 0$ when $x \to \infty$, then not only does this imply that there is a zero between the two values but it also implies that it is the only one, right? Unless you need a proof of that, you should be fine off saying that $x_0(m)$ is the only zero. – Daccache Aug 19 '14 at 19:11
• thanks for the comment! – mike Aug 19 '14 at 20:20
• do you mean $G_m(x)=F_m(x)+F_{\color{#C00000}{m}}(1/x)$ in $(3)$? – robjohn Aug 22 '14 at 23:58
• @robjohn Yes. I corrected it. Thanks! – mike Aug 23 '14 at 0:10

While this is just a partial answer, I hope this serves at least as a step in the right direction for proving what you need to.

First, to work with something more concrete, I substituted the expressions for $f_n(x)$ into $F_m(x)$ and then that into $G_m(x)$ in order to get an explicit set of functions:
$$F_m(x) = \sum\limits_{n=1}^mn^2x(2\pi n^2x - 3)\exp(-n^2\pi x)$$ $$G_m(x) = \sum\limits_{n=1}^mn^2x(2\pi n^2x - 3)\exp(-n^2\pi x) + \sum\limits_{n=1}^m\frac{n^2}{x}\left(\frac{2\pi n^2}{x} - 3\right)\exp\left(-\frac{n^2\pi}{x}\right)$$ While this class of functions is particularly nasty, we can make some ways with proving the proposed bounds $m + 1 > x_0(m) > m + 2$ hold. (By the way, the inequalities are reversed here than in your question, since the function $G_m(x)$ is positive and decreases in $x_0(m)$'s neighborhood, so $G_m(m + 1) > G_m(m + 2)$, thus making the former the upper bound and the latter the lower bound, not vice versa.)

So basically what we need to show is that $G_m(m + 1)$ is always positive for all $m$, and $G_m(m + 2)$ is always negative, thus by the intermediate value theorem (since the function in question is continuous), $G_m(x)$ has a root in between $m + 1$ and $m + 2$.

Substituting in $x = m + 1$ for the function, we get:
$$G_m(m + 1) = \sum\limits_{n=1}^mn^2(m + 1)(2\pi n^2m + 2\pi n^2 - 3)\exp(-n^2\pi (m + 1)) + \sum\limits_{n=1}^m\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$ If we can show that each factor in each summation is always positive, then the whole summation is positive and so the functions are always positive at that point. (Actually, that would be the best case scenario of it satisfying a sufficient but not necessary condition; it can have some negative sums as long as the total value of the positive terms is larger than the negative ones.)

From the first summation, factor by factor:

1st summation, 1st factor: $n^2$
Since the square of any number is always positive, $n^2$ is positive.

1st summation, 2nd factor: $m + 1$
Obviously $m$ is a positive number by definition, and so $m + 1$ is also always positive.

1st summation, 3rd product: $(2\pi n^2m + 2\pi n^2 - 3)$
$2\pi n^2m + 2\pi n^2$ must be greater than 3 for this factor to be positive. Taking the 'lowest' case of $n = m = 1$, we get $2\pi + 2\pi = 4\pi,$ and $4\pi > 3$, so this factor will always be positive.

1st summation, 4th product: $\exp(-n^2\pi (m + 1))$
An exponential term is never negative or zero.

Now, we go on to the second summation:
2nd summation, 1st product: $\frac{n^2}{m + 1}$
$n^2$ and $m + 1$ are always positive, so their quotient is too.

2nd summation, 2nd product: $\frac{2\pi n^2}{m + 1} - 3$
Alas, here we run into trouble. Taking the case $n = 1, m = 2$, we see that the resulting term is negative. When will it be negative? Like the corresponding factor in the first summation, the fraction must be greater than $3$:

$\frac{2\pi n^2}{m + 1} > 3 \implies 2\pi n^2 > 3(m + 1) \implies n > \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}$
So, for the first few terms in the sum, the product will be negative, but once n is sufficiently large to satisfy the inequality, the product will become positive.
2nd summation, 3rd product: $\exp\left(-\frac{n^2\pi}{m + 1}\right)$
Again, an exponential term is never negative.

So, what does this all mean since not all terms are positive? All it means is that we need to prove that the first summation is larger than the second one (which I couldn't do), so that their difference is still positive:
$$\sum\limits_{n=1}^mn^2(m + 1)(2\pi n^2m + 2\pi n^2 - 3)\exp(-n^2\pi (m + 1)) > \sum\limits_{n=1}^m\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$
Or, if we strip out the positive terms from the second summation, (I'm calling the terms $a_n$ and $b_n$ for the 1st and 2nd summations respectively)
$$\left[\sum\limits_{n=1}^ma_n + \sum\limits_{n > \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}}^mb_n\right] > \sum\limits_{n < \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}}^mb_n$$
If we prove either inequality (the first one is stronger than the second), we deduce that the function is always positive at the point $m + 1$. We can use an extremely similar argument for the point $m + 2$ to prove it is negative, and thus we will have proved the bounds. About the first question (whether $x_0(m)$ is the smallest zero of $G_m(x)$), if we take that the function is decreasing from $G_m(1)$ to $x_0(m)$, we can prove it to be the smallest zero by contradiction (and if we also accept that $\lim\limits_{x \to \infty} G_m(x) = 0$, then we can prove that it is the only zero.) For suppose that there exists other zeros smaller than $x_0(m)$; that is, in the interval $[1, x_0(m)]$. Since the function is continuous, the only way for it to have a smaller zero is if the function dips below zero and back up again (since it needs to pass through zero at $x_0(m)$). But since the function is always decreasing on that interval, then we reach a contradiction, since after the first 'smaller' zero the function would be negative and would need to be increasing to cross the x-axis again. I know this is far from rigorous, but either way proving the bounds would also prove this.

I apologize for the (extremely!) long answer, but I found out a lot about this function and didn't want anything to go to waste. Cheers!

@GeorgeDaccache Nice results!

I just need the space and easy of use in this answer area to convey some thoughts that I have.

First let me define two integers $n_0(m)$ and $n_1(m)$ as

$$n_0(m) =\lfloor{\sqrt{3/(2\pi)}\sqrt{m + 1}}\rfloor$$ $$n_1(m) =\lceil{\sqrt{3/(2\pi)}\sqrt{m + 1}}\text{ }\rceil$$

For convenience we also define $a_n(m)$ and $b_n(m)$ as: $$a_n(m)=n^2(m + 1)(2\pi n^2(m + 1) - 3)>0$$ $$b_n(m)=\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)$$

So that $$b_n(m)\gt 0, m\ge n\gt n_1(m)$$ $$b_n(m)\lt 0, 1\le n\lt n_0(m)$$

Then what we want to prove is the following:

$$\sum\limits_{n=1}^m a_n(m)\exp(-n^2\pi (m + 1)) + \sum\limits_{n=n_1(m)}^m b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right) > \sum\limits_{n=1}^{n_0(m)} (-b_n(m))\exp\left(-\frac{n^2\pi}{m + 1}\right)\tag{1}$$

Since for $A>0$,

$$\exp(-A n^2)>\exp(-A m^2)$$ $$\exp(-A n^2)<\exp(-A * 1^2)=\exp(-A)$$

We can replace the exponential terms by their corresponding limit terms, factor them out of the summation and thus only need to prove the following:

$$\exp(-(n_0(m))^2\pi (m + 1))\sum\limits_{n=1}^{n_0(m)} a_n(m)+\exp(-m^2\pi (m + 1))\sum\limits_{n=n_1(m)}^m a_n(m) + \exp\left(-\frac{m^2\pi}{m + 1}\right)\sum\limits_{n=n_1(m)}^m b_n(m) > \exp\left(-\frac{1^2\pi}{m + 1}\right)\sum\limits_{n=1}^{n_0(m)} (-b_n(m))\tag{2}$$

The nice thing about (2) is that we can now complete the summation of $n$ in (2).

@GeorgeDaccache, can you continue to work along this direction? Since you already spent so much effort on this problem. I feel that you might just be one step away from providing a complete answer.

Best regards- mike

EDIT: Numerical results for $m=10$ showed that the terms associated with $a_n$ are quite tiny, so we can now focus on proving that:

$$\sum\limits_{n=n_1(m)}^m b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right) > \sum\limits_{n=1}^{n_0(m)} (-b_n(m))\exp\left(-\frac{n^2\pi}{m + 1}\right)\tag{3}$$

Well, I was able to prove $(3)$ in your answer above, meaning the bounds are proven! The 'proof' though is still much less formal than I'd like it to be, so it might need a little refinement. Here goes:

Intuitively, the LHS side of the inequality is the sum of all the positive terms in $b_n(m)$, and the RHS is the sum of all the negative terms. If the LHS is greater than the RHS, then the total value of $b_n(m)$ will be positive, so the original function $G_m(x)$ will be positive at $x = m + 1$ for all $m$, proving our first bound. What I thought of was that if the value of the RHS at each case is less than the value on the LHS, then the total value is positive. It is natural, then, to consider the number of summands on each summation. Once we prove that the number of summands on the LHS is larger than the RHS (lemma 1), then we find the total value of each side. (Actually, we just find directly which one is larger; it's a bit hard to explain until that point, so just wait till the end to see how.)

Lemma 1: $|\{1, ..., n_0(m)\}| \leq |\{n_1(m), ..., m\}|$
Consider the following table of values:

The first row is the given value of $m$, the second is the number of negative terms, the third is the starting point for where the terms become positive, ($m - n_0(m)$ is the number of positive terms), and the fourth is included for reference as the number of negative terms in the summation if it were equally distributed into half positive and half negative terms. What we want to deduce from this table is that there is always an equal or greater number of positive terms than negative ones, that is the majority of terms are positive. How? Well, if we compare the number of negative terms there would be had the number of terms been equally positive and negative (4th row) to the number of negative terms there actually are (2nd row), we see that the inequality holds, with the number of negative terms being equal in the cases $m = 1, 2, 3$. How are we supposed to prove that the $n_0(m)$ is always smaller than $\left \lfloor{m/2}\right \rfloor$? I reasoned that in the 'base' cases $m = 1, 2, 3$, they are equal, and $n_0(m)$ behaves like $\sqrt{m}$ asymptotically, and $\left \lfloor{m/2}\right \rfloor$ behaves like $m$ asymptotically, and the former grows faster than the latter, so $\left \lfloor{m/2}\right \rfloor$ will never become less than $n_0(m)$. So, at any value of m, the number of terms which are negative in $b_n(m)$ is always less than or equal to the number of positive terms.

Define $N_0 = \{1, ..., n_0(m)\}$, and $N_1 = \{n_1(m), ..., m\}$. We already showed that $|N_0| \leq |N_1|$, and now we shall show that every element in $N_0$ is smaller in value than in $N_1$.

Lemma 2: Every element in $N_0$ is smaller than every element in $N_1$.
This is since the two sets $N_0$ and $N_1$, being subsets of the natural numbers, are well-ordered, and thus have a greatest element and least element. In this case, if the greatest element in $N_0$ is less than the least element in $N_1$, then the elements of $N_0$ are all smaller than the elements of $N_1$. Noting that $\sup(N_0) = n_0(m)$, and $\inf(N_1) = n_1(m)$, and since $n_1(m) = n_0(m) + 1$ (by definition of floor and ceiling), then every element in $N_0$ is smaller than every element in $N_1$.

Finally, we consider the actual values of each summation. For any given $m$, the value of each summation only depends on $n$'s values. Consider $(3)$:
$$\sum\limits_{n=n_1(m)}^m b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right) > \sum\limits_{n=1}^{n_0(m)} (-b_n(m))\exp\left(-\frac{n^2\pi}{m + 1}\right)$$ $$\sum\limits_{n=n_1(m)}^m \frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right) > \sum\limits_{n=1}^{n_0(m)} -\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$ $$\sum\limits_{n=n_1(m)}^m \left(\frac{2\pi n^4 - 3n^2(m + 1)}{(m + 1)^2}\right)\exp\left(-\frac{n^2\pi}{m + 1}\right) > \sum\limits_{n=1}^{n_0(m)} -\left(\frac{2\pi n^4 - 3n^2(m + 1)}{(m + 1)^2}\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$
Thinking of the sets of functions (one for each $m$) as a set of mappings from $N_1$ to the natural numbers, we just need to prove that the values $N_1$ produces (LHS) are all larger than the values $N_0$ produces (RHS).

The original functions we considered are all increasing for $n \geq n_1(m)$, and decreasing before, so it maps the set $N_0$ onto even smaller numbers, and the larger set $N_1$ onto even larger numbers. Thus, the RHS sums a lower amount of increasingly smaller numbers, and the LHS sums a higher amount of increasingly larger numbers, so we conclude the RHS is smaller than the LHS, proving $(3)$ and thus the bound. Remember though that this whole effort is only for the point $m + 1$, and very similarly $m + 2$ can be proven as the other bound.

Please don't hesitate to point out any errors, since there very well might be some. I haven't had time to look over it again since I've been getting rather busy lately.

Cheers!

• Thanks a lot for the answer. But I think that you did not take into account the exponential part when you compare term by term in the summation. I did numerical calculation and found out an interesting pattern. Let us define $A(n)= b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right)$. Then for the LHS, we have $A(n_1(m))>A(n_1(m)+1)>...>A(m)>0$. For the RHS, we have $-A(n_0(m))>-A(n_0(m)-1)>...>-A(1)>0$. – mike Aug 24 '14 at 8:44
• Ah, my bad! It completely slipped my mind with the exponentials, but I'm afraid this is all I can do at the moment. Indeed your observation is interesting, and might very well prove to be the fruitful in proving the bounds. Good luck though! – Daccache Aug 24 '14 at 16:50