Why does $(1+x)^np(x)$ has at least $n+1$ terms?

Question

$p(x)$ is a polynomial. Assume $p(x)\ne0$. How to prove that $(1+x)^np(x)$ has at least $n+1$ terms?

Answer 1

Hint: Induction with differentiation to get to smaller step.

Answer 2

For a polynomial $f\in\mathbb C[X]$, $f(x)=\sum_ka_kX^k$ let $$w(f)=|\{\,k\in\Bbb N:a_k\ne 0\,\}|$$ denote the number of nonzero terms occurring in $f$ and let $$v(f)=\begin{cases}\max\{\,k\in\Bbb N:(1+X)^k\mid f\,\}&\text{if $f\ne 0$}\\\infty&\text{if $f=0$}\end{cases}$$ denote the multiplicity of $-1$ as a root of $f$. Observe that $$\tag1w(Xf)=w(f)$$ and that $$\tag2w(f')=\begin{cases} w(f)&\text{if $X\mid f$}\\w(f)-1&\text{otherwise.}\end{cases}$$ Also, $$\tag3v(Xf)=v(f) $$ and as the derivative of $(1+X)^kq(X)$ with $q(-1)\ne 0$ is $(1+X)^{k-1}r(X)$ with $r(X):=kq(X)+(1+X)q'(X)$ and $r(-1)=kq(-1)$ we have $$\tag4v(f')=v(f)-1\qquad\text{if $0<v(f)<\infty$} $$

The problem statement is equivalent to saying that $w(f)>v(f)$ for all nonzero $f$. This follows readily by induction, the base step being constant polynomials where $w(f)=1>0=v(f)$, and the induction step using $(1)$ and $(3)$ if $X\mid f$ and $(2)$ and $(4)$ if $X\nmid f$.

Answer 3

Let's notice first an equivalent statement: for $\alpha \ne 0$ and $p\not \equiv 0$, the polynomial

$$q(x)\colon = (x-\alpha)^n p(x)$$

has at least $n+1$ non-zero coefficients. It is more convenient to consider $\alpha = 1$.

Now, the polynomial $q(x)$ above has $1$ root with multiplicity at least $n$, so we have

$$q(1) = q'(1) = \cdots = q^{(n-1)}(1)=0$$

Let $q(x) = c_0 x^0+ c_1 x + \cdots c_k x^k$. The above system of equalities is equivalent to \begin{eqnarray} &c_0 + c_1 + c_2 + \cdots + c_N &= &0 \\ &0^1 c_0 + 1^1 c_1 + 2^1 c_2 + \cdots + k c_k&= &0 \\ \ldots \ldots \ldots \\ &0^{n-1} c_0 + 1^{n-1} c_1 + \cdots + k^{n-1} c_k &=& 0 \end{eqnarray}

Now let's assume that the non-zero coefficients of $q(x)$ are at position $m_1$, $m_2$, $\ldots$, $m_n$. From the above we get a system of linear equations for $c_{m_1}$, $\ldots$, $c_{m_n}$. The matrix of coefficients is a Vandermonde matrix corresponding to $(m_1, \ldots, m_n)$, so of non-zero determinant. But that implies that $m_1 = \ldots = m_n = 0$, contradiction.

Answer 4

If p(x) is non-zero then it contains at least one term in it. How many terms are in $(1+x)^n$? The binomial expansion will help here.