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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 &= 1024n^2 - 4096/3n + 320\\ b &= 12288n^3 - 14336n^2 + 4864/3n + 256. \end{align*} Then \begin{equation} V(x)=(1-x)^6H(x)+(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(ax + b(1-x)) \end{equation} for a polynomial $H(x)$.

We claim for all $n\ge6$ that $H(x)$ is positive for all $x>0$ for the simple reason that all coefficients of $H(x)$ are positive. As $a(n)$ and $b(n)$ is positive for all $n\ge1$, the claim follows.

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

There is a finite set of pairs $(i,j)$ of nonnegative integers, independent of $n$, such that the exponents of $W(x)$ have the form $ni+j$. Let $a_{i,j}$ be the corresponding coefficient. It is a polynomial in $n$. It turns out that $0\le i\le 24$ for even $i$, and $-4\le j\le 6$.

We have \begin{equation} W = \sum_{i,j}a_{i,j}(n)x^{ni+j}. \end{equation}

If $n\ge6$, then $ni+j>ni'+j'$ if and only if $i>i'$ or $i=i'$ and $j>j'$. This gives a total order on the pairs $(i,j)$. Every exponent of $x$ in $W(x)/(1-x)^6$ which matters has the form $ni+j+y$ where $0\le y<n(i'-i)+j'-j$, where $(i',j')$ is the successor of $(i,j)$ for our order.

The coefficient of $x^{ni+j+y}$ in $W(x)\frac{1}{(1-x)^6}$ is the sum of all $a_{i_1,j_1}(n)\cdot\binom{ni+j-ni_1-j_1+y+5}{5}$ for all $(i_1,j_1)\le(i,j)$.

Thus we obtain finitely many (actually $86$) explicit polynomials $f(n,y)$ in $n$ and $y$ (of small degrees in $n$ and $y$) for which we have to prove that they are positive for all $n\ge6$ and all $y$ in the specified range.

We distinguish two cases: If $(n'i+j')-(ni+j)=1$ (recall that $(i',j')$ is the successor of $(i,j)$ again), then only $y=0$ is possible. In all these cases, we see that $f(n+6,0)$ has nonnegative coefficients. Thus $f(n,0)\ge0$ whenever $n\ge6$.

Now assume that $(n'i+j')-(ni+j)>1$. In these cases it turns out that $i'=i+2$, so $0\le y< 2n-e$ for some integer $e$. Thus $y=2n-e-1-o$ for some non-negative $o$. Thus $f(n,o)=f((y+e+1+o)/2,o)=g(y,o)$. This latter polynomial, in $y$ and $o$, has to be non-negative for all non-negative $o$ and $y$. It turns out that except for a single example actually all the coefficients are nonnegative. This single exception however, when multiplied with a suitable polynomial with positive coefficients, has positive coefficients too. (See the end of the Sage code.)

Our argument used $n\ge6$. The cases $2\le n\le5$ are checked directly for $V(x)$.

# 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 = V - (2*N + 1)^2 *(N + 1)^2*X^(24*N-1)*(1-X)^4*(a*X + b*(1-X))

# Compute the pairs (i,j) for the exponents ni+j of W, and the
# corresponding coefficients a_ij. 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()]
aij = {}
K.<n> = QQ[]
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)), ZZ(e(N=0))  # Clumsy method to compute pairs (i,j)
    aij[i,j] = aij[i,j] + c if (i,j) in aij else c

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

# compute the coefficients of W(X)/(1-X)^6
K = K.extend_variables(('y', 'o'))
K.inject_variables()
trouble = []
aijk = aij.keys()
aijk.sort()
for k in range(len(aijk)-1):
    (i,j) = aijk[k]
    # f(n,y) is the coefficient of X^(ni+j+y) for 0 <= y < yr
    f = sum(aij[z]*prod(n*i+j+y-n*z[0]-z[1]+5-zz for zz in range(5))/120
            for z in aijk[:k+1])
    (ii,jj) = aijk[k+1] # successor of (i,j)
    yr = n*(ii-i)+(jj-j)
    # check non-negativity of the polynomial f
    if yr == 1:
        # Checks if the coefficient of X^(ni+j), which is a
        # polynomial in n, has nonnegative coefficients upon replacing
        # n with n+6
        f = f(n=n+6,y=0)
        if min(f.coefficients()+[0]) < 0:
            print "False"
            break
    else:
        # The coefficient of X^(ni+j+y), which is a polynomial in n and y,
        # has to be nonnegative for 0 <= y < yr, so y = yr-1-o for
        # non-negative o. Here yr has the form 2n-e for an integer e. Thus
        # y = 2n-e-1-o. So upon replacing n with (y+1+o+e)/2,
        # we get a polynomial in y and o which has to be nonnegative
        # for all y and o. In most cases, this holds because the
        # coeffcients are non-negative.
        if ii-i != 2:
            print "False"
        e = j-jj
        f = f(n=(y+1+o+e)/2)
        if min(f.coefficients()+[0]) < 0:
            trouble.append(f)

# Finally, take care of the only case which requires an ad hoc argument
print len(trouble) == 1

f = trouble[0]
c = o^2 + 93*y^2 + 36*o + 570*y + 715
print min(c.coefficients()+(c*f).coefficients()) >= 0

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}

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.

# 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

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 $x^{\deg V}V(1/(1+x))$ 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.

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.

Remark: The following is a somewhat simplified solution (with updated accompanying Sage code).

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 &= 1024n^2 - 4096/3n + 320\\ b &= 12288n^3 - 14336n^2 + 4864/3n + 256 \end{align*} Then \begin{equation} V(x)=(1-x)^6H(x)+(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(ax + b(1-x)) \end{equation} for a polynomial $H(x)$.

Here $a,b$ are positive for $n\ge1$. We show that all coefficients of $H(x)$ are positive whenever $n\ge6$.

Now assume that $(n'i+j')-(ni+j)>1$. In these cases it turns out that $i'=i+2$, so $0\le y< 2n-e$ for some integer $e$. Thus $y=2n-e-1-o$ for some non-negative $o$. Thus $f(n,o)=f((y+e+1+o)/2,o)=g(y,o)$. This latter polynomial, in $y$ and $o$, has to be non-negative for all non-negative $o$ and $y$. It turns out that except for a single example actually all the coefficients are nonnegative. This single exception can be written in the form $o((o^2-d(y))^2+h(y))+r(y)$, where $h(y)$ and $r(y)$ have nonnegative coefficients. (See the end of the Sage code.)

By the way, $V$ actually seems to be a nonnegative linear combination of $x^i(1-x)^{\deg V-i}$. This can be obtained from the recursion (fixed $n$) \begin{equation} V_0=V,\;\; V_{i+1}=\frac{V_i-V_i(1)x^{\deg V_i}}{1-x}. \end{equation} I have an explicit expression of the coefficients $a_i$ in $V=\sum a_ix^i(1-x)^{\deg V-i}$. But I don't see an easy argument showing positivity.

The scrollable field where I had put the Sage code cuts off part of the program. (An MO bug?) Instead you find it here now.

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 &= 1024n^2 - 4096/3n + 320\\ b &= 12288n^3 - 14336n^2 + 4864/3n + 256. \end{align*} Then \begin{equation} V(x)=(1-x)^6H(x)+(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(ax + b(1-x)) \end{equation} for a polynomial $H(x)$.

We claim for all $n\ge6$ that $H(x)$ is positive for all $x>0$ for the simple reason that all coefficients of $H(x)$ are positive. As $a(n)$ and $b(n)$ is positive for all $n\ge1$, the claim follows.

Now assume that $(n'i+j')-(ni+j)>1$. In these cases it turns out that $i'=i+2$, so $0\le y< 2n-e$ for some integer $e$. Thus $y=2n-e-1-o$ for some non-negative $o$. Thus $f(n,o)=f((y+e+1+o)/2,o)=g(y,o)$. This latter polynomial, in $y$ and $o$, has to be non-negative for all non-negative $o$ and $y$. It turns out that except for a single example actually all the coefficients are nonnegative. This single exception however, when multiplied with a suitable polynomial with positive coefficients, has positive coefficients too. (See the end of the Sage code.)

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 $x^{\deg V}V(1/(1+x))$ 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 = V - (2*N + 1)^2 *(N + 1)^2*X^(24*N-1)*(1-X)^4*(a*X + b*(1-X))

# Compute the pairs (i,j) for the exponents ni+j of W, and the
# corresponding coefficients a_ij. 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()]
aij = {}
K.<n> = QQ[]
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)), ZZ(e(N=0))  # Clumsy method to compute pairs (i,j)
    aij[i,j] = aij[i,j] + c if (i,j) in aij else c

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

# compute the coefficients of W(X)/(1-X)^6
K = K.extend_variables(('y', 'o'))
K.inject_variables()
trouble = []
aijk = aij.keys()
aijk.sort()
for k in range(len(aijk)-1):
    (i,j) = aijk[k]
    # f(n,y) is the coefficient of X^(ni+j+y) for 0 <= y < yr
    f = sum(aij[z]*prod(n*i+j+y-n*z[0]-z[1]+5-zz for zz in range(5))/120
            for z in aijk[:k+1])
    (ii,jj) = aijk[k+1] # successor of (i,j)
    yr = n*(ii-i)+(jj-j)
    # check non-negativity of the polynomial f
    if yr == 1:
        # Checks if the coefficient of X^(ni+j), which is a
        # polynomial in n, has nonnegative coefficients upon replacing
        # n with n+6
        f = f(n=n+6,y=0)
        if min(f.coefficients()+[0]) < 0:
            print "False"
            break
    else:
        # The coefficient of X^(ni+j+y), which is a polynomial in n and y,
        # has to be nonnegative for 0 <= y < yr, so y = yr-1-o for
        # non-negative o. Here yr has the form 2n-e for an integer e. Thus
        # y = 2n-e-1-o. So upon replacing n with (y+1+o+e)/2,
        # we get a polynomial in y and o which has to be nonnegative
        # for all y and o. In most cases, this holds because the
        # coeffcients are non-negative.
        if ii-i != 2:
            print "False"
        e = j-jj
        f = f(n=(y+1+o+e)/2)
        if min(f.coefficients()+[0]) < 0:
            trouble.append(f)

# Finally, take care of the only case which requires an ad hoc argument
print len(trouble) == 1

f = trouble[0]
c = o^2 + 93*y^2 + 36*o + 570*y + 715
print min(c.coefficients()+(c*f).coefficients()) >= 0