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I've been working on a project and proved a few relevant results, but got stuck on one tricky problem:

Conjecture. If $2\leq n\in\mathbb{N}$ and $0<x<1$ is a real number, then $$F_n(x)=\log\left(\frac{(1+x^{4n-1})(1+x^{2n})(1-x^{2n+1})}{(1+x^{2n+1})(1-x^{2n+2})}\right)$$ is a convex function of $x$.

I have a heuristic argument. Can you help with a rigorous proof or valuable tools?

Further motivation. If you succeed with this, then I'll be honored to have you as a co-author in this work. The problem itself can be found in Section 4.

Note. Write $F=\log\frac{P}Q$, then $F''>0$ amounts to the positivity of the polynomial $$V:=PQ^2P''+(PQ')^2-P^2QQ''-(P'Q)^2.$$

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  • 1
    $\begingroup$ When you take the second derivative, you get a rational function $p(x)/q(x)$ whose denominator is clearly positive on $(0,1)$. Right? So this reduces to showing that the polynomial $p(x)$ is positive on $(0,1)$. $\endgroup$ – Nate Eldredge Jan 10 '17 at 0:21
  • $\begingroup$ While for a fixed n, such problems can in theory be solved by quantifier elimination of real closed fields, I'm not sure if this is still the case when you have an integer variable in the exponent. $\endgroup$ – Fan Zheng Jan 10 '17 at 0:24
  • $\begingroup$ @NateEldredge: Yes, you get positivity of a polynomial. $\endgroup$ – T. Amdeberhan Jan 10 '17 at 0:35
  • $\begingroup$ @FanZheng: Thanks for the comment. But, we need concrete proofs if you don't mind my saying. $\endgroup$ – T. Amdeberhan Jan 10 '17 at 0:36
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    $\begingroup$ Let's recall the old version mathoverflow.net/questions/246919/… $\endgroup$ – Pietro Majer Jan 10 '17 at 22:03
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Fix $n$, and let $V(x)$ be the polynomial from the question, for which we want to show that it is positive on the open interval $(0,1)$. Set \begin{align*} a(n) &= 1024n^2 - 4096/3n + 320\\ b(n) &= 12288n^3 - 14336n^2 + 4864/3n + 256\\ W(x) &= x^4\left(V(x)-(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(a(n)x + b(n)(1-x))\right). \end{align*}

For all $n\ge12$ we claim that $(1-x)^6$ divides $W(x)$, and that all the coefficients of $W(x)/(1-x)^6$ are nonnegative. In particular, $W(x)$ is positive for $x>0$. As $a(n)$ and $b(n)$ is positive for all $n\ge1$, we see that $V(x)$ is positive for all $0<x<1$, provided that $n\ge12$. The cases $2\le n\le 11$ are easily checked directly, for instance by noting that the coefficients of $(1+x)^{\deg V}V(1/(1+x))$ are positive.

I'm not sure about the easiest kind to prove the assertion about the coefficients of $W(x)/(1-x)^6$. We compute them via \begin{equation} \frac{W(x)}{(1-x)^6}=W(x)\sum_{k\ge0}\binom{k+5}{5}x^k. \end{equation}

All what follows is assisted/confirmed by the Sage code below.

The exponents of $W(x)$ have the form $2ni+j$ for $0\le i\le 12$ and $0\le j\le 10$. Let $a_{i,j}$ be the corresponding coefficient. It is a polynomial in $n$.

We have \begin{equation} \frac{W(x)}{(1-x)^6} = \sum_{i,j}a_{i,j}(n)x^{2ni+j}\sum_k\binom{k+5}{5}x^k. \end{equation}

With $r\ge0$ and $0\le s\le 2n-1$, the coefficient of $x^{2nr+s}$ in $W(x)\frac{1}{(1-x)^6}$ is the sum of all $a_{i,j}(n)\cdot\binom{nr+s-ni-j+y}{5}$ for all pairs $(i,j)$ where $i<r$ and $0\le j\le 10$, or $i=r$ and $j\le s$.

We distinguish the cases $0\le s\le 9$ from the cases $s\ge10$. In these latter cases, we write $s=10+y$. Recall that $s\le 2n-1$, so $10+y=2n-1-o$ for a nonnegative $o$. Upon replacing $n$ with $(11+y+1)/2$, we get polynomials in $n$ and $o$.

In the former cases, the coefficients are polynomials in $n$, which have positive coefficients upon replacing $n$ with $n+12$. So these coefficients are positive for all $n\ge12$.

In the latter cases it turns out that in all but one case the coefficients are positive. So for nonnegative $y$ and $o$, the values are nonnegative.

For the single exception we see that if we multiply it with a suitable polynomial, the resulting coefficients are positive.

Remark (answering Jason's question): It is a known fact that a polynomial $V$ is positive on $(0,1)$ if and only if it is a nonnegative linear combination of polynomials $x^i(1-x)^j$. The problem is that it may involve terms where $i+j$ is bigger than $\deg V$. (This doesn't happen here, though.) I used an LP solver to play a little with $V$ and $\frac{V}{(1-x^2)^4x^{2n-2}}$. From that a pattern showed up which lead to a solution.

By the way, $V$ actually seems to be a nonnegative linear combination of $x^i(1-x)^{\deg V-i}$. This is equivalent to say that all the coefficients of $(1+x)^{\deg V}V(1/(1+x))$ are nonnegative. It is not hard to compute explicit expressions for these coefficients, but I don't see an easy arguments why they can't be negative.

# Formally compute W
var('X N')
P = (1+X^(4*N-1))*(1+X^(2*N))*(1-X^(2*N+1))
Q = (1+X^(2*N+1))*(1-X^(2*N+2))
V = P*Q^2*P.diff(X,2)+(P*Q.diff(X))^2-P^2*Q*Q.diff(X,2)-(P.diff(X)*Q)^2
a = 1024*N^2 - 4096/3*N + 320
b = 12288*N^3 - 14336*N^2 + 4864/3*N + 256
W = X^4*(V - (2*N + 1)^2 *(N + 1)^2*X^(24*N-1)*(1-X)^4*(a*X + b*(1-X)))

# Check that (1-X)^6 divides W(X)
print all(W.diff(X,i)(X=1).polynomial(QQ) == 0 for i in [0..5])

# Compute the coefficients a_ij for the exponents 2ni+j of W. Somewhat
# clumsy, as I don't know how to deal with polynomials where exponents
# are symbolic expressions.
#
# l is the list of summands of W
l = [z.canonicalize_radical() for z in W.expand().operands()]
K.<n> = QQ[]
aij = {(i,j):K(0) for i in [0..12] for j in [0..10]}
for term in l:
    c = term(X=1)                        # get coefficient of term
    e = (term.diff(X)/c)(X=1)            # get exponent of term
    c = K(c.polynomial(QQ))              # convert c to proper polynomial in n
    i, j = ZZ(e.diff(N))//2, ZZ(e(N=0))  # Clumsy method to compute pairs (i,j)
    aij[i,j] += c

# Check if coefficients aij[i,j] were correctly computed
Wnew = sum(c(n=N)*X^(2*N*i+j) for (i,j),c in aij.items())
print (W-Wnew).canonicalize_radical().is_trivial_zero()

def bino(k): # binomial coefficient binom(-6,k)=binom(k+5,5)
    return prod(k+5-z for z in range(5))/120

# compute the coefficients of W(X)/(1-X)^6
K = K.extend_variables(('y', 'o'))
K.inject_variables()
for r in [0..12]:
    for s in [0..10]:
        d = 1 if s == 10 else 0
        # compute coefficient of X^(2nr+s+d*y)
        f = sum(aij[i,j]*bino(2*n*(r-i)+s+d*y-j) for i in [0..r]
                for j in [0..s if i == r else 10])
        # check non-negativity of the polynomial f
        if d == 0:
        # Checks if the coefficient of X^(2nr+s), which is a
        # polynomial in n, has nonnegative coefficients upon replacing
        # n with n+12
            f = f(n=n+12)
            if min(f.coefficients()+[0]) < 0:
                print "False"
        else:
        # The coefficient of X^(2nr+10+y), which is a polynomial in n
        # and y, has to be nonnegative for 0 <= 10+y <= 2n-1, so 10+y
        # = 2n-1-o for non-negative o. So upon replacing n with
        # (11+y+o)/2, we get a polynomial in y and o which has to be
        # nonnegative for all nonnegative y and o. In all but one
        # case, this holds because the coeffcients are non-negative.
            f = f(n=(11+y+o)/2)
            if min(f.coefficients()+[0]) < 0:
                c = o^2 + 23*o*y + 1360*y^2 + 99*o + 340*y + 1675
                print min(c.coefficients()+(c*f).coefficients()) >= 0
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  • $\begingroup$ An interesting different approach. $\endgroup$ – T. Amdeberhan Jan 11 '17 at 21:29
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    $\begingroup$ I tried to reproduce your big equation, but instead of (1-x)^3 H(x), where H has all positive coefficients, I find (1-x)^2 H(x), where H(x) is not divisible by (1-x) and has both positive and negative coefficients. What is wrong with the equation or with the Mathematica code I'm using? $\endgroup$ – Matt F. Jan 12 '17 at 5:58
  • $\begingroup$ The code is: n = 2; P = (1 + x^(4 n - 1)) (1 + x^(2 n)) (1 - x^(2 n + 1)); Q = (1 + x^(2 n + 1)) (1 - x^(2 n + 2)); V = P Q^2 D[P, {x, 2}] + (P D[Q, x])^2 - P^2 Q D[Q, {x, 2}] - (D[P, x] Q)^2; a = 64 n^2 - 256/3 n + 20; b = 768 n^3 - 1024 n^2 + 272 n - 24; c = 14336/3 n^4 - 306304/45 n^3 + 35792/15 n^2 - 6104/15 n + 3; LHS = Simplify[Factor[ V] ((1 - x^2)^-4) (x^-(2 n - 2))]; RHS2 = (n + 1)^2 (2 n + 1)^2 x^(22 n - 4) (a x^2 + b x (1 - x) + c (1 - x)^2); RHS1 = Factor[LHS - RHS2]; {RHS1, Limit[RHS1/((1 - x)^2), x -> 1]} $\endgroup$ – Matt F. Jan 12 '17 at 6:19
  • $\begingroup$ @Matt: There is a typo in my formula, the constant term in $c$ is $32$, not $3$. Thanks for telling that something is wrong! $\endgroup$ – Peter Mueller Jan 12 '17 at 7:06
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    $\begingroup$ I think it is fascinating that all those numerous coefficients turn out to be nonnegative. This can hardly be a mere coincidence. Then, there must be some fascinating and possibly very useful general theory behind this fascinating phenomenon. $\endgroup$ – Iosif Pinelis Jan 13 '17 at 17:16
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This proof is mainly based on some of the suggestions in the early version of this answer. Also, we are building here on the comment by Peter Mueller concerning the polynomial $a$ defined below. It turns out that similar techniques can be used for all other nontrivial polynomials arising in the proof, which allowed us to replace multiple uses of Mathematica command Reduce[] in the last version of this answer by showing that the coefficients of the mentioned relevant polynomials are all nonnegative. I have still retained some use of Reduce[] -- but only to quickly make the routine check for $n\le11$.

Note that \begin{equation} G:= F_n''(x) x^{2-2 n} \left(x^{2 n}+1\right)^2 \left(x^{4 n}+x\right)^2 \left(x^{2 n+1}-1\right)^2 \left(x^{2 n+1}+1\right)^2 \left(x^{2 n+2}-1\right)^2 \end{equation} is a polynomial in $n$, $x$, and $y:=x^{2n}$, of degree $2$ in $n$: \begin{equation} G=a+bn+cn^2, \end{equation} where $a,b,c$ are certain polynomials in $x,y$.

The check of $G\ge0$ for $n=2,\dots,11$ is straightforward. So, assume $n\ge12$, whence $0<y<x^{24}<x^{12}$. We also always assume that $0<x<1$, $0<y<x^{24}$ , \begin{equation} n=\frac{\ln y}{2\ln x}; \tag{1} \end{equation} the latter relation is of course just another form of the equality $y=x^{2n}$.

Note that $0<y<1\ \&\ 0<x<1$ implies $x=1/(1+u)\ \&\ y=1/(1+v)$ for some $u,v\ge0$. As suggested by Peter Mueller, it turns out that \begin{equation} a\; (1 + u)^{10} (1 + v)^{11}|_{x=1/(1 + u), y=1/(1 + v)} \end{equation} is a polynomial in $u,v$ with all coefficients nonnegative. Details of all calculations can be seen in the Mathematica notebook and its pdf image

So, \begin{equation} a\ge0 \end{equation} for all $x,y$ in $(0,1)$.

Next, introduce \begin{equation} x_* := \frac{869}{1000},\quad x_0:=\frac{7655}{10000},\quad y_0:= x_0^{24}\approx0.00164,\quad x_1 := \frac{985}{1000},\quad y_1:= x_1^{24}\approx0.696. \end{equation}

The following cases/subcases/subsubcases are exhaustive:

Case 1: $y\le y_1$. Then the condition $0<y\le y_1\ \&\ 0<x<1\ \&\ 0<y<x^{24}$ implies $x=x_1/(1+u)\ \&\ y=x^{24}/(1+v)$ or $x=x_1+(1-x_1)/(1+u)\ \&\ y=y_1/(1+v)$ for some $u\ge0$ and $v\ge0$, depending on whether $x<x_1$ or $x\ge x_1$. It turns out that \begin{equation} c\; (1 + u)^{248} (1 + v)^{10}|_{x=x_1/(1+u)\ \&\ y=x^{24}/(1+v)} \end{equation} and \begin{equation} c\; (1 + u)^{10} (1 + v)^{10}|_{x=x_1+(1-x_1)/(1+u)\ \&\ y=y_1/(1+v)} \end{equation} are each a polynomial in $u,v$ with all coefficients nonnegative. So, \begin{equation} c\ge0 \quad\text{in Case 1}. \end{equation} So, for the derivative $G'_n$ of $G$ in $n$, we have \begin{equation}\tag{2} G'_n=b+2cn\ge b+24c=(G'_n)|_{n=12}. \end{equation}

Now we have to distinguish two subcases of Case 1, with further subsubcases of Subcase 1.1:

Subcase 1.1: $y\le y_1$ and ($y\ge y_0$ or $x\le x_*$).

Subsubcase 1.1.1: $y\le y_1$ and $y\ge y_0$ and $x\ge x_1$. Then $y\le y_1\ \&\ y\ge y_0\ \&\ x\ge x_1 \ \&\ 0<x<1\ \&\ 0<y<x^{24}$ implies $$x=x_{111}:=x_1+(1-x_1)/(1+u)\ \&\ y=y_{111}:=y_0+(y_1-y_0)/(1+v)$$ for some $u\ge0$ and $v\ge0$. It turns out that \begin{equation} (G'_n)|_{n=12}\; (1 + u)^{10} (1 + v)^{11}|_{x=x_{111}, y=y_{111}} \end{equation} and \begin{equation} G|_{n=12}\; (1 + u)^{10} (1 + v)^{11}|_{x=x_{111}, y=y_{111}} \end{equation} are each, again, a polynomial in $u,v$ with all coefficients nonnegative.

Subsubcase 1.1.2: $y\le y_1$ and $y\ge y_0$ and $x<x_1$. Then $x=x_{112}:=x_0+(x_1-x_0)/(1+u)\ \&\ y=y_{112}:=y_0+(x^{24}-y_0)/(1+v)$ for some $u\ge0$ and $v\ge0$. It turns out that \begin{equation} (G'_n)|_{n=12}\; (1 + u)^{272} (1 + v)^{11}|_{x=x_{112}, y=y_{112}} \end{equation} and \begin{equation} G|_{n=12}\; (1 + u)^{272} (1 + v)^{11}|_{x=x_{112},y=y_{112}} \end{equation} are each, again, a polynomial in $u,v$ with all coefficients nonnegative.

Subsubcase 1.1.3: $y\le y_1$ and $x\le x_*$. Then $x=x_{113}:=x_*/(1+u)\ \&\ y=y_{113}:=x^{24}/(1+v)$ for some $u\ge0$ and $v\ge0$. It turns out that \begin{equation} (G'_n)|_{n=12}\; (1 + u)^{272} (1 + v)^{11}|_{x=x_{113},y=y_{113}} \end{equation} and \begin{equation} G|_{n=12}\; (1 + u)^{272} (1 + v)^{11}|_{x=x_{113},y=y_{113}} \end{equation} are each, again, a polynomial in $u,v$ with all coefficients nonnegative.

So, $G|_{n=12}\ge0$ and $(G'_n)|_{n=12}\ge0$ in all the three subsubcases of Subcase 1.1. In view of $(2)$, we have $G\ge0$ in Subcase 1.1.

Subcase 1.2: $y\le y_1$ and $y<y_0$ and $x>x_*$. Then, of course, the condition $y\le y_1$ is redundant. Let here \begin{equation} \rho(x):=\frac7{2(1-x)}. \end{equation} The condition $x_*<x<1\ \&\ 0<y<y_0$ implies $x=x_{12}:=x_*+(1-x_*)/(1+u)$ and $y=y_{12}:=y_0/(1+v)$ for some $u\ge0$ and $v\ge0$. It turns out that \begin{equation} (G'_n)|_{n=\rho(x)/2}\; u\,(1 + u)^{10} (1 + v)^{11}|_{x=x_{12},y=y_{12}} \end{equation} and \begin{equation} G|_{n=\rho(x)/2}\; u^2\,(1 + u)^{10} (1 + v)^{11}|_{x=x_{12},y=y_{12}} \end{equation} are each, again, a polynomial in $u,v$ with all coefficients nonnegative. So, similarly to Subcase 1.1, for all $n\ge\rho(x)/2$ we get $G\ge G|_{n=\rho(x)/2}\ge0$. In Subcase 1.2, it remains to note that for our particular $n$, as in $(1)$, we indeed have $n\ge\rho(x)/2$. This follows because for $y<y_0$ and $x>x_*$ \begin{equation} \ln\frac1y>\ln\frac1{y_0}>1.08\frac7{2(1-x_*)}\,\ln\frac1{x_*} >\frac7{2(1-x)}\,\ln\frac1{x}=\rho(x)\ln\frac1{x}. \end{equation} So, $G\ge0$ in Subcase 1.2 as well.

It remains to consider

Case 2: $y>y_1$. Then condition $y>y_1\ \&\ 0<x<1\ \&\ 0<y<x^{24}[<x^{12}]$ implies $x=x_2:=x_1+(1-x_1)/(1+u),y=y_2:=y_1+(x^{12}-y_1)/(1+v)$ for some $u\ge0$ and $v\ge0$. It turns out that \begin{equation} b\;(1 + u)^{140} (1 + v)^{11}|_{x=x_{2},y=y_{2}} \end{equation} is also a polynomial in $u,v$ with all coefficients nonnegative. Hence, $b\ge0$ in Case 2. So, if $c\ge0$, then trivially $G\ge0$ (since $a\ge0$ always). So, without loss of generality, $c<0$ and hence $G$ is concave in $n$. So, it is enough to bracket the $n$ as in $(1)$ between some simple rational expressions $n_1$ and $n_2$ such that $G|_{n=n_1}\ge0$ and $G|_{n=n_2}\ge0$.

In Case 2, $y$ is "close" to $1$ and hence, in view of inequality $y<x^{24}$, so is $x$: $x>y_1^{1/12}=x_1=0.985$. So, it is a bit more convenient here to change variables $x, y$ to "small variables" $u:=1-x$ and $v:=1-y$. Then $0<v<1-y_1\approx0.304$ and $1-u=x>y^{1/24}=(1-v)^{1/24}>1-v/3$, whence $u<v/3$. Let now $t:=u/v$, so that $0<t<1/3$. It follows, in Case 2, that $t=(1/3)/(1+r)$ and $v=(1-y_1)/(1+s)$ for some $r\ge0$ and $s\ge0$.

The mentioned brackets $n_1$ and $n_2$ are respectively $\ell_1/2$ and $\ell_2/2$, where
\begin{equation} \ell_1:=\frac1t,\quad\ell_2:=\frac{(2 - v) (1 - t v)}{t (1 - v) (2 - t v)}. \end{equation} It is not hard to see (details on this are in the same Mathematica notebook) that then indeed $n_1 < n < n_2$. It remains to show that $G|_{n=n_1}\ge0$ and $G|_{n=n_2}\ge0$.

It turns out that \begin{equation} (G'_n)|_{n=\ell_1/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19} \end{equation} and \begin{equation} (G'_n)|_{n=\ell_2/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19}\; q \end{equation} are each a polynomial in $r,s$ with all coefficients nonnegative, where \begin{multline*} q:= (1 + r)^{10} (1 + s)^{19} \\ \times(11673186598630578538556565100133681446610566511878526881 \\ + 16777216000000000000000000000000000000000000000000000000 s)^2 \\ \times\big(31853088866210192846185521700044560482203522170626175627 \\ + 33554432000000000000000000000000000000000000000000000000 (r + s + r s)\big)^2. \end{multline*}

The proof is complete.

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  • $\begingroup$ Are the coefficients of $\log^2 x, \log^2t, \log x\log t$ positive individually? $\endgroup$ – T. Amdeberhan Jan 11 '17 at 1:36
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    $\begingroup$ Only the coefficient of $\ln^2 x$ is (rather, seems, and that should not be hard to check) positive on the entire big square $(0,1)^2. $\endgroup$ – Iosif Pinelis Jan 11 '17 at 2:08
  • $\begingroup$ Now a complete proof, using some of the suggestions in the previous version of this answer. $\endgroup$ – Iosif Pinelis Jan 15 '17 at 22:10
  • $\begingroup$ As an illustration, I have added the proof of $a\ge0$ using resultants. $\endgroup$ – Iosif Pinelis Jan 17 '17 at 14:42
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    $\begingroup$ @IosifPinelis: As an alternative argument to show $a(x,y)\ge0$, one can use that all the coefficients of the numerator of $a(1/(1+x),1/(1+y))$ are nonnegative. As you remarked previously (in a comment to my answer), it is really amazing that positivity of certain polynomials related to the question is a result of positive coefficients. $\endgroup$ – Peter Mueller Jan 17 '17 at 18:48
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I sketch a method that should show $F_n$ is convex for $n$ sufficiently large.

A Taylor expansion gives that $F_n(x) = x^{2n} (1-x)^2 + O(x^{4n-1})$, and so $$F_n''(x) = 2x^{2n-2}[2n^2(1-x)^2 + x^2 - n(1-x)(1+3x)] + O(x^{4n-3}).$$ This should show that $F_n''(x) > 0$ for $n$ large, provided $x < 1-\frac{C}{n}$ for some fixed $C>0$.

On the other hand, according to Mathematica, $F_n''(1) = 4n^2 - \frac{16}{3} n + \frac{5}{4}$, which has its largest root at $\frac{8 + \sqrt{19}}{12} \approx 1.02991$. (By the way, there doesn't seem to be any reason that $n$ should be restricted to be an integer, and this is in line with some plots I made indicating $F_n(x)$ is convex for $n > 1.03$.) Moreover, Mathematica gives $$\lim_{n\rightarrow \infty} n^{-2}F_n''(1-\frac{y}{n}) = \frac{16 e^{4y}}{(1+e^{4y})^2} > 0.$$

Putting these two pieces together should be able to show $F_n(x)$ is convex on $(0,1)$ for $n$ sufficiently large. Unfortunately, it might be very painful to work out all the error terms explicitly.

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  • $\begingroup$ I like this argument. on the other hand, I had some large $n$ analysis already. Please see Section 6 of math.temple.edu/~tewodros/Catalan_Convexity.pdf $\endgroup$ – T. Amdeberhan Jan 10 '17 at 18:44
  • $\begingroup$ @MattYoung : For this argument to succeed, even for large $n$, one would also need to show that the convergence of $n^{-2}F_n''(1-\frac{y}{n})$ to the positive expression $\frac{16 e^{4y}}{(1+e^{4y})^2}$ is uniform in all small enough values of $y>0$. This seems likely to be true, since $(1-y/n)^n$ converges uniformly over any bounded set of values of $y$ (as $n\to\infty$). However, Mathematica by itself would probably not be enough to show the uniformity. $\endgroup$ – Iosif Pinelis Jan 23 '17 at 14:40
  • $\begingroup$ @Iosif Pinelis Yes, that's true, one would need to prove uniform convergence. I imagined that one could show an explicit asymptotic for $n^{-2} F_n''(1-y/n)$ with an error term depending on $n$ and $y$. However, I didn't want to do it myself! $\endgroup$ – Matt Young Jan 23 '17 at 16:59
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Not an answer, just a plot of $F_n(x)$ for $n=2,\ldots,10$:


          Fn(x)
          Fn(x)2


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    $\begingroup$ It's also worth extending these graphs to the right, to show that the function is well-defined for x=1 and x>1. Furthermore, the function is not convex for x>1, e.g. not after x>1.2 if n=3. This suggests that the answer will not be easy. $\endgroup$ – Matt F. Jan 10 '17 at 2:37
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    $\begingroup$ @MattF.: Thanks for your comment. Another plot now displayed. $\endgroup$ – Joseph O'Rourke Jan 10 '17 at 2:52
  • $\begingroup$ Did anyone try to compute the power series expansion of F' and F''? $\endgroup$ – Fan Zheng Jan 10 '17 at 4:07
  • $\begingroup$ Yes, I've attempted that too,but it get messy. See page 10 and up in math.temple.edu/~tewodros/Catalan_Convexity.pdf $\endgroup$ – T. Amdeberhan Jan 10 '17 at 4:29
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This is not an answer, just a reformulation into something which looks perhaps tractable (along the lines of the comment by G. Paseman). Define, for $m\geq 1$ and $x\in (0,1)$, $$g_m(x):=\frac{d^2}{dx^2}[\log(1+x^m)]=\frac{x^{m-2}(m(m-1)-mx^m)}{(1+x^m)^2}$$ and $$h_m(x):=\frac{d^2}{dx^2}[\log(1-x^m)]=-\frac{x^{m-2}(m(m-1)+mx^m)}{(1-x^m)^2}.$$ Then, for $n\geq 2$, $$F_n''(x)=g_{4n-1}(x)+g_{2n}(x)+h_{2n+1}(x)-g_{2n+1}(x)-h_{2n+2}(x).$$

Two observations

  • The contribution of the first, second, and last terms is non-negative for all $n\geq 2$ and $x\in(0,1)$, while that of the remaining terms is non-positive.
  • It would therefore suffice to show the following: for $n$ and $x$ as above, $$g_{2n+1}(x)-h_{2n+1}(x)\leq_? g_{4n-1}(x)+g_{2n}(x)-h_{2n+2}(x).$$

Added: Since I apparently cannot add a comment, I would just note that while it is certainly (always) possible I overlook something, I have just double checked the algebra with maple, and believe that the expressions are correct (there doesn't seem to be a missing factor of m). Another note (in addition to the symmetry observed in the comment by Paseman) is the (trivial) identity $h_{2m}=g_m+h_m$, leading to another form for the desired bound: $$g_{2n+1}(x)+g_{n+1}(x)-h_{2n+1}(x)\leq_? g_{4n-1}(x)+g_{2n}(x)-h_{n+1}(x).$$

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    $\begingroup$ Do my eyes deceive me, or is it true that $g_m(-x)=h_m(x)$ for odd $m$? Gerhard "And Can We Use This?" Paseman, 2017.01.09. $\endgroup$ – Gerhard Paseman Jan 10 '17 at 5:54
  • $\begingroup$ Upon reflection, it might be better to combine the two items giving the h term and come up with a j term (2d derivative of log of ( 1 - (x^{2n+1}/simple polynomial)). Gerhard "Term Rewriting For Better Living" Paseman, 2017.01.09. $\endgroup$ – Gerhard Paseman Jan 10 '17 at 6:14
  • $\begingroup$ One can also write everything n terms of the h only: $$F_n''=\sum_{1\le i\le 4}h_{p_i} -\sum_{1\le i\le 4}h_{q_i} $$ with: $$p:=(2n+1,2n+1,4n,8n-2)$$ $$q:=(2n,2n+2,4n-1,4n+2)$$ maybe matching conveniently the good terms $(q)$ with the bad ones $(p )$ one can reduce the complexity $\endgroup$ – Pietro Majer Jan 10 '17 at 22:59
  • $\begingroup$ Yes, but working with the g's is better (assuming the domain is (0,1)) as one does not have to deal with the 1/(1-x) factor in doing comparisons. Gerhard "Let The Physicists Subtract Infinities" Paseman, 2017.01.10. $\endgroup$ – Gerhard Paseman Jan 10 '17 at 23:03
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This is only a comment to @DimaPasechnik, but I cannot put the picture in a comment. The surface to the right of the $x=y$ is a plot of Dima's function (barring mistakes); clearly not convex.

enter image description here

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    $\begingroup$ Right. I should have rotated my plot few times before posting that... :-) $\endgroup$ – Dima Pasechnik Jan 13 '17 at 21:55
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I conjecture a stronger statement: that the function $F(x,y)$ of two variables $x,y$ obtained from $F_n(x)$ by substituting $y$ for $x^{2n}$ is convex for $0<y<x<0$.

EDIT2: this conjecture is wrong as stated. See e.g. the plot in the answer by Yaakov Baruch, or rotate the plot in Sage...

While this reduces to checking positive definiteness of a matrix of bivariate polynomials of degree about 20, obtained from the Hessian of $F$ by clearing common (positive) denominator, this should be possible to make to work by standard real algebraic geometry tools. Here is an (ugly) plot of $F$:

plot on [0,1]x[0,1] obtained by exporting the plot from jmol in Sage(math)

sage: var('x y')
sage: F(x,y)=log((x+y^2)*(1+y)*(1-x*y)/(x*(1+x*y)*(1-x^2*y)))
sage: plot3d(F,[0,1],[0,1])

EDIT: comments say that as $y\to 0+$, the Hessian becomes indefinite. That is, for my original claim to hold, one need to assume $y>\epsilon>0$, with $\epsilon$ possibly depending upon $x$.

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  • $\begingroup$ And about $\partial_y F$? Don't we need it to be positive (for $0<y<x<1$) to conclude that $F(x,x^{2n})$ is convex? $\endgroup$ – Pietro Majer Jan 13 '17 at 10:56
  • $\begingroup$ A twice diff. function is convex on an open set $U$ iff its Hessian is positive semidefinite on $U$. I don't see why 1st partials matter at all.. $\endgroup$ – Dima Pasechnik Jan 13 '17 at 12:11
  • $\begingroup$ Sure, for $F(x,y)$, but I was talking about the composition, $F_n(x):=F(x,y(x))$, with $y(x):=x^{2n}$ as you said. So $F_n''(x)=D^2F[1,\dot y][1,\dot y]\ {\mathbf +}\ \partial_yF \ddot y$. $\endgroup$ – Pietro Majer Jan 13 '17 at 13:04
  • $\begingroup$ Tarski's results tell us that something like this actually follows from the original claim. The claim tells us that $F_n(x,y)''>0$ follows from $\{y > x^q \leftrightarrow q > 2n: q \in \mathbf{Q}\}$. If we take $x,y,n$ as real variables these are all statements in the theory of real-closed fields. Then we apply several properties that theory: since this is true for infinitely many real $n$, it must be true for all sufficiently large $n$, it must follow from a finite subset of the hypotheses, and it must be provable. It should be possible to adapt @PeterMueller's proof along these lines. $\endgroup$ – Matt F. Jan 13 '17 at 13:44
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    $\begingroup$ @DimaPasechnik : Alas, your function $F$ is not convex even for $y<x$, since the determinant of the Hessian matrix is $-4(1-x)^2<0$ at $y=0+$. In fact, something less than the convexity of $F$ would have sufficed (in addition to $F_y\ge0$). The subscripts here denote the differentiation in the corresponding variables. Indeed, $F''_n(x)=F_y y_{xx}+F_{xx}+F_{yx}y_x+F_{yy}y_x^2$. So, it would be enough that $F_y\ge0$, $F_{xx}\ge0$, and the discriminant $d:=F_{yx}^2-4F_{xx}F_{yy}$ be $\le0$. However, $d=4(1-x)^2>0$ if $0<x<1$ and $y=0+$. $\endgroup$ – Iosif Pinelis Jan 13 '17 at 13:59

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