Prove that for $(1-x)^m (1+x)^n$ there is no $k$ such that the coefficients of $x^k$ and $x^{k+1}$ are both $0$

Solution 1:

Write $$f(x) = (1-x)^m (1+x)^n = \sum_{j=0}^{n+m} c_j x^j $$

Observe that (apply index re-numeration)

$$ x^{n+m} f(1/x) = (-1)^m (1-x)^m (1+x)^n = \sum_{j=0}^{n+m} c_j x^{n+m-j} = \sum_{j=0}^{n+m} c_{n+m-j} x^{j} $$

Hence $c_{n+m-j} = (-1)^m c_j$.

We discuss the effect of two consecutive zero coefficients. Let us put an upper index (m,n) at the function and coefficients to indicate which m and n were taken. For $m=n$,

$$f^{n,n}(x) = (1-x)^n (1+x)^n = (1-x^2)^n = \sum_{i=0}^n \binom{n}{i} (-1)^i x^{2i} $$

Hence we have zero coefficients $c_j$ for odd $j$ and coefficients with alternating sign for even $j$.

Let us now look at the case where $m>n$, then

$$f^{m,n}(x) = (1-x)^m (1+x)^n = (1-x^2)^n (1-x)^{m-n} = \sum_{j=1}^m c^{m,n}_j x^{j} $$

For $m-n=1$, we have $c^{m=n+1,n}_j = c^{n,n}_j - c^{n,n}_{j-1}$ so, with the described behavior of $c^{n,n}_j $, there can never be a zero $c^{m=n+1,n}_j$.

Let us generalize this recursion for arbitrary $m>n$. The idea is to go backwards to lower $m$, arriving at $m=n$ where one can compare to the known coefficient structure (see above).

Suppose $c^{m,n}_{k+1} = c^{m,n}_{k} =0$. Then we must have, with some arbitrary constants $a_i$,

$$ c^{m-1,n}_{k-1} = c^{m-1,n}_{k} = c^{m-1,n}_{k+1} = a_1\\ c^{m-2,n}_{k-2} = a_2 ; c^{m-2,n}_{k-1} = a_2+ a_1 ; c^{m-2,n}_{k} a_2 + 2 a_1; c^{m-2,n}_{k+1} = a_2 + 3 a_1 \\ c^{m-3,n}_{k-3} = a_3 ; c^{m-3,n}_{k-2} = a_3 + a_2 ; c^{m-3,n}_{k-1} = a_3 + 2 a_2+ a_1 ;\\ c^{m-3,n}_{k} = a_3 + 3 a_2 + (1+2) a_1; c^{m-3,n}_{k+1} = a_3 + 4 a_2 + (1+2+3) a_1$$

Continuing with this process for $s$ steps, we have that $c^{m-s,n}_{k}$ is a polynomial of degree $s-1$ in $k$, i.e. $c^{m-s,n}_{k} = \sum_{q=0}^{s-1} d_q k^q$ with $s$ many free constants $d_q$. This polynomial has to fit $s+2$ many coefficients $c^{m-s,n}_{k-s},\dots , c^{m-s,n}_{k+1}$.

Now let $s = m-n$, hence $m=n$. Then we have (see above) that these coefficients are alternatingly zero, positive ($\binom{n}{k}$), zero, negative (-$\binom{n}{k+2}$), etc. A polynomial fitting $s+2$ such points must have degree $s+1$ in general, unless special symmetries are graceful. This is true in two cases: a) for small $m,n$, since then the binomials themselves are low-order polynomials in $k$. However these cases can be checked manually not to contain consecutive zero coefficients; b) for the central term $\binom{m+n}{(m+n)/2}$ only, since the coefficients to the left and the right of this term are (anti-)symmetric, see above. Other than those cases, the fit is not possible since the polynomial in $k$ is only of degree $s-1$. This shows that two consecutive zero coefficients at any $m > n$ will not appear.

The case $m < n$ works analogously, which completes the proof.