Polynomials $P(x)=\sum_{i=0}^{\lfloor n/2 \rfloor} (-1)^i\frac{(2i)!}{2^i\cdot i!}\binom{n}{2i}x^{n-2i}$ have $n$ real roots For any positive integer $n$, I conjecture that the polynomial

$$P(x)=\sum_{i=0}^{\lfloor n/2 \rfloor} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n}{2i}x^{n-2i}$$ 

has $n$ real roots.
I've checked some polynomials in Mathematica and I strongly believe this result is true.
Does anyone have any reference or proof for this result?
Any help would be appreciated.
 A: Let's look at a small example. The first few such polynomials are:


*

*$P_0=1$

*$P_1=x$

*$P_2=x^2-1$

*$P_3=x^3-3x$

*$P_4=x^4-6x^2+3$


We can compute the total number of real roots of $P_4$ via Sturm's theorem. The details are given at the linked Wikipedia article. The summary of how it works is that we apply the Euclidean algorithm to $P_4$ and $P_4'$, and remember the remainders we get at each step, with a suitable choice of sign. This is called the "canonical Sturm chain" for $P_4$. Then the the number of roots of $P_4$ in $(a,b)$ can be obtained by taking the difference between the number of sign changes we see in the canonical Sturm chain at $a$, and the number we see at $b$.
Explicitly, the polynomials we end up considering are:


*

*$P_4=x^4-6x^2+3$

*$P_4'=4x^3-12x=4P_3$

*$(\frac{1}{4}x)(4P_3)-P_4=x^4-3x^2-x^4+6x^2-3=3x^2-3=3P_2$

*$(\frac{4}{3}x)(3P_2)-4P_3=4x^3-4x-4x^3+12x=8x=8P_1$

*$(\frac{3}{8}x)(8P_1)-3P_2=3x^2-3x^2+3=3=3P_0$


Since the leading coefficient of each of these polynomials is positive and each has degree $1$ smaller than the one before, they alternate in sign near $-\infty$, so the sign sequence is $+-+-+$: there are $4$ changes of sign. Similarly, there are $0$ changes of sign near $\infty$ as all the polynomials are positive there. So Sturm's theorem tells us that $P_4$ has $4$ real roots.

The same argument works in general. We will establish the following two identities:
\begin{align}
P_n'&=nP_{n-1}\\
x P_{n-1}-P_n&=(n-1)P_{n-2}
\end{align}
Together, these imply that the canonical Sturm chain for $P_n$ consists of the polynomials $P_k$ for $k \leq n$, each scaled by some positive constant multiple. Since each $P_n$ is monic, this in turn means that the canonical Sturm chain has $n$ sign changes at $-\infty$ and $0$ sign changes at $\infty$, and so $P_n$ has $n$ real roots.
Each of these identities can be established by direct calculation. We adopt the convention that any coefficients which have negative factorials in the denominator are $0$; this allows us to forget about the upper limit of the sum. So:
$$P_n(x)=\sum_{i=0}^\infty (-1)^i \frac{(2i)!}{i!2^i}\binom{n}{i}x^{2n-i}=\sum_{i=0}^\infty (-1)^i \frac{n!}{(n-2i)!i!2^k}x^{2n-i}$$
With this in mind, we have:
\begin{align}
P_n'(x)&=\sum_{i=0}^\infty (-1)^i \frac{n!}{(n-2i)!i!2^i}(n-2i)x^{n-1-2i}\\
&=\sum_{i=0}^\infty (-1)^i \frac{n!}{(n-1-2i)!i!2^i}x^{n-1-2i}\\
&=n\sum_{i=0}^\infty (-1)^i \frac{(n-1)!}{(n-1-2i)!i!2^i}x^{n-1-2i}\\
&=nP_{n-1}(x)
\end{align}
and:
\begin{align}P_n(x)-xP_{n-1}(x)&=\sum_{i=0}^\infty (-1)^i \left[ \frac{n!}{(n-2i)!i!2^i}-\frac{(n-1)!}{(n-1-2i)!i!2^i}\right]x^{n-2i}\\
&=\sum_{i=0}^\infty(-1)^i\frac{(n-1)!}{(n-1-2i)!i!2^i}\left(\frac{n}{n-2i}-1\right)x^{n-2i}\\
&=\sum_{i=1}^\infty (-1)^i \frac{(n-1)!}{(n-1-2i)!i!2^i}\frac{2i}{n-2i}x^{n-2i}\\
&=\sum_{i=1}^\infty(-1)^i \frac{(n-1)!}{(n-2i)!(i-1)!2^{i-1}}x^{n-2i}\\
&=(n-1)\sum_{i=1}^\infty (-1)^i \frac{(n-2)!}{(n-2i)!(i-1)!2^{i-1}}x^{n-2i}\\
&=-(n-1)\sum_{j=0}^\infty (-1)^j \frac{(n-2)!}{(n-2-2j)!j!2^j}x^{n-2-2j}\\
&=-(n-1)P_{n-2}(x)
\end{align}
which completes the proof.

(I don't know a whole lot about this, but Wikipedia tells me that polynomials which obey the identity $P_n'=nP_{n-1}$ form an Appell sequence, which means there may be a simpler or more intuitive way of thinking about all of this in terms of umbral calculus...)
A: It will be demonstrated that
$$P_n(x)=\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n}{2i}x^{n-2i}\tag{1}$$ 
is the characteristic polynomial of the $n\times n$ dimensional matrix 
$$
M^{(n)}=\begin{pmatrix}
0&\sqrt1&0&\dots& 0&0\\
\sqrt1& 0&\sqrt2&\dots& 0&0\\
0&\sqrt2&0&\ddots& 0&0\\
\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\
0&0&0&\ddots&0&\sqrt{n-1}\\
0&0&0&\dots&\sqrt{n-1}&0
\end{pmatrix},\tag{2}
$$
which can be recognized as a finite-dimensional version of the position operator for harmonic oscillator. In short writing $M_{ij}=\sqrt{i}\delta_{i,j-1}+\sqrt{j}\delta_{i-1,j}$. 
Indeed the characteristic polynomials of the matrices $M^{(1)}$ and $M^{(2)}$ are $P_1(x)=x$ and $P_2(x)=x^2-1$, respectively. We are going to demonstrate that if expression (1) is valid for $n-1$ and $n-2$ it is valid for $n$ as well.
Let $A^{(n)}=xI^{(n)}-M^{(n)}$.
$$\begin{align}
P_n(x)&=|A^{(n)}|=x\left|A^{(n)}_{[nn]}\right|-\sqrt{n-1}\left|A^{(n)}_{[n,n-1]}\right|=x\left|A^{(n)}_{[nn]}\right|
-(n-1)\left|\left(A^{(n)}_{[n,n-1]}\right)_{[n-1,n-1]}\right|\\
&=x\left|A^{(n-1)}\right|
-(n-1)\left|A^{(n-2)}\right|=xP_{n-1}(x)-(n-1)P_{n-2}(x)\\
&\stackrel{I.H.}{=}x\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n-1}{2i}x^{n-2i-1}-(n-1)\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n-2}{2i}x^{n-2i-2}\\
&=\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n-1}{2i}x^{n-2i}+(n-1)\sum_{i=1}^{\infty} (-1)^i\dfrac{(2i-2)!}{2^{i-1}\cdot (i-1)!}\dbinom{n-2}{2i-2}x^{n-2i}\\
&=\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n-1}{2i}x^{n-2i}+\sum_{i=1}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n-1}{2i-1}x^{n-2i}\\
&=\sum_{i=0}^{\infty} (-1)^i\dfrac{(2i)!}{2^i\cdot i!}\dbinom{n}{2i}x^{n-2i}.\\
\end{align}
$$
Thus by induction the claim is proved. As matrices $M^{(n)}$ are hermitian (symmetric) all its eigenvalues and therefore the roots of the characteristic polynomials are real. 
In the computation of the determinant the Laplace expansion was used with $A_{[i,j]}$ denoting $ij$-minor of $A$.
A: As it turns out, what you have are the so-called "probabilists' Hermite polynomials" (see the DLMF as well).
That is,
$$P(x)=\operatorname{He}_n(x)=(-1)^n\exp\left(\frac{x^2}{2}\right)\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(-\frac{x^2}{2}\right)=$$
and these orthogonal polynomials have long been established to have real roots, proven through either Sturmian sequences (as in Micah's answer) or through its having a symmetric Jacobi matrix (as in user's answer). See Chihara's book for more details.
Since you mention that you are using Mathematica, here is how to use it to verify the identity for the first few members:
And @@ Table[Expand[2^(-n/2) HermiteH[n, x/Sqrt[2]] == 
                    Sum[(-1)^i Binomial[n, 2 i] (2 i - 1)!! x^(n - 2 i),
                        {i, 0, Quotient[n, 2]}]],
             {n, 0, 20}]
   True

