let $$f_{1}(x)=1, f_{2}(x)=1+x, f_{n}(x)=f_{n-1}(x)+xf_{n-2}(x)$$
show that $$f_{2m+1}(x)+xf_{2m-1}(x)=0,m\in N$$only have real roots.
Thank you everyone. This problem is my frend ask me.I don't known this problem come from which books,Thank you
It is known that Hermite polynomials and Legendre polynomials only have real numbers,see: http://en.wikipedia.org/wiki/Hermite_polynomials and http://en.wikipedia.org/wiki/Legendre_polynomials But this I cann't prove it.
We first rephrase the problem using a simpler formulation. Observe that we are only concerned with $f_n(x)$ for odd $n$. We have $f_1(x)=1, f_2(x)=1+x, f_3(x)=(1+x)+x(1)=1+2x$.
We have $$f_{2m+1}(x)=f_{2m}(x)+xf_{2m-1}(x)$$ Thus $$f_{2m}(x)=f_{2m+1}(x)-xf_{2m-1}(x), \; f_{2m-2}(x)=f_{2m-1}(x)-xf_{2m-3}(x)$$ Substituting this into $$f_{2m}(x)=f_{2m-1}(x)+xf_{2m-2}(x)$$ , we get $$f_{2m+1}(x)-xf_{2m-1}(x)=f_{2m-1}(x)+x(f_{2m-1}(x)-xf_{2m-3}(x))$$ $$f_{2m+1}(x)=(1+2x)f_{2m-1}(x)-x^2f_{2m-3}(x)$$
We can thus define $g_m(x)=f_{2m+1}(x)$, so that $$g_0(x)=f_1(x)=1, g_1(x)=f_3(x)=1+2x, g_m(x)=(1+2x)g_{m-1}(x)-x^2g_{m-2}(x)$$ The problem reduces to showing that $g_m(x)+xg_{m-1}(x)=0$ has only real roots.
It is easy to see (and easily proven by induction) that $f_n(x)$ has degree $\lfloor \frac{n}{2} \rfloor$, has positive leading coefficient, and has constant term $1$, so $g_m(x)$ has degree $m$, has positive leading coefficient, and has constant term $1$. In particular $g_m(x)+xg_{m-1}(x)$ has degree $m$, so it suffices to show that it has $m$ distinct real roots.
We now proceed by induction on $m \geq 1$ to prove the following statements:
When $m=1$, we have that $g_0(x)=1$, which clearly has $0$ distinct real roots. For $-\infty=\alpha_{0, 1}<x<\alpha_{0, 0}=+\infty$, we clearly have $sgn(g_{0}(x))=sgn(1)=(-1)^0$. Also, $sgn(g_1(\alpha_{0, 0}))=1, sgn(g_1(\alpha_{0, 1}))=-1$. Thus the statements indeed hold in the base case $m=1$.
Suppose that the $3$ statements hold for $m=k$. Consider $m=k+1$. We have by the induction hypothesis (statement $3$) that $sgn(g_k(\alpha_{k-1, i}))=(-1)^i, 0 \leq i \leq k$. By the intermediate value theorem (since $g_k(x)$ is a polynomial, it is continuous), $g_k(x)$ has $k$ distinct real roots $\alpha_{k, i}, 1 \leq i \leq k$, satisfying $g_{k-1, i}<g_{k, i}<g_{k-1, i-1}$. (We have shown statement $1$) Since $g_k(x)$ has degree $k$, these are all its roots.
Note that we have $\alpha_{k, k}<\alpha_{k-1, k-1}<\alpha_{k, k-1}<\alpha_{k-1, k-2}< \ldots <\alpha_{k-1, 1}<\alpha_{k, 1}$. Consider the interval $(\alpha_{k, i+1}, \alpha_{k, i})$ for $1 \leq i \leq k-1$. Note that $g_k(x)$ has no roots in this open interval, so $sgn(g_k(x))$ is constant over this interval. Since $\alpha_{k-1, i}$ is in this interval, we have for $\alpha_{k, i+1}<x<\alpha_{k, i}$, $sgn(g_k(x))=sgn(g_k(\alpha_{k-1, i}))=(-1)^i$.
Now consider the open intervals $(-\infty, \alpha_{k, k})$ and $(\alpha_{k, 1}, +\infty)$. Again, $g_k(x)$ has no roots in this interval. Therefore $sgn(g_k(x))$ is constant over this interval. For $-\infty=\alpha_{k-1, k}=\alpha_{k, k+1}<x<\alpha_{k, k}$, we have $sgn(g_k(x))=sgn(g_k(\alpha_{k-1, k}))=(-1)^k$. For $\alpha_{k, 1}<x<\alpha_{k, 0}=\alpha_{k-1, 0}=+\infty$, we have $sgn(g_k(x))=sgn(g_k(\alpha_{k-1, 0}))=(-1)^0$. This, combined with the above paragraph, shows statement $2$.
Now since $g_{k+1}(x)=(1+2x)g_k(x)-x^2g_{k-1}(x)$, we have for $1 \leq i \leq k$ that \begin{align} sgn(g_{k+1}(\alpha_{k, i}))=sgn((1+2\alpha_{k, i})g_k(\alpha_{k, i})-\alpha_{k, i}^2g_{k-1}(\alpha_{k, i}))&=sgn(-\alpha_{k, i}^2g_{k-1}(\alpha_{k, i})) \\ &=-sgn(g_{k-1}(\alpha_{k, i})) \end{align} (where the last equality follows from the fact that $g_k(0)=1$, so $\alpha_{k, i} \not =0$.)
Now since $\alpha_{k-1, i}<\alpha_{k, i}<\alpha_{k-1, i-1}$, we have by the induction hypothesis (statement $2$), that $sgn(g_{k-1}(\alpha_{k, i}))=(-1)^{i-1}$. Thus $sgn(g_{k+1}(\alpha_{k, i}))=-sgn(g_{k-1}(\alpha_{k, i}))=(-1)^i$.
Finally since $g_{k+1}(x)$ has degree $k+1$ and has positive leading coefficient, we have $sgn(g_{k+1}(\alpha_{k, 0}))=sgn(g_{k+1}(+\infty))=1$ and $sgn(g_{k+1}(\alpha_{k, k+1}))=sgn(g_{k+1}(-\infty))=(-1)^{k+1}$. This, combined with the above, proves statement $3$.
Therefore we are done by induction, so the above $3$ statements hold for all $m \in \mathbb{Z}^+$.
We now have $sgn(g_k(\alpha_{k-1, i})+\alpha_{k-1, i}g_{k-1}(\alpha_{k-1, i}))=sgn(g_k(\alpha_{k-1, i}))=(-1)^i, 1 \leq i \leq k-1$. Also, since $g_k(x)+xg_{k-1}(x)$ has degree $k$ and positive leading coefficient, $sgn(g_k(\alpha_{k-1, 0})+\alpha_{k-1, 0}g_{k-1}(\alpha_{k-1, 0}))=1$ and $sgn(g_k(\alpha_{k-1, k})+\alpha_{k-1, k}g_{k-1}(\alpha_{k-1, k}))=(-1)^k$.
By the intermediate value theorem (since $g_k(x)+xg_{k-1}(x)$ is a polynomial, it is continuous), $g_k(x)+xg_{k-1}(x)$ has $k$ distinct real roots. Since it has degree $k$, we are done.
Note that the exact same argument shows that the polynomial equation $f_{2m+1}(x)+cxf_{2m-1}(x)=0$ has only real roots, where $c \geq 0$ is a constant.
Fisk - Polynomials, Roots, & Interlacing - Page 49 - Corollary 1.95
Theorem. Suppose $P_1$ and $P_2$ have real interlacing zeros, then the recursion formula $$P_n=P_{n-1}+x\cdot P_{n-2}$$ defines a sequence of polynomials with only real zeros.
Note: I am stating a weaker paraphrased version to the theorem in the book.
Define $$g_m(x)=f_{m+1}(x)+x\cdot f_{m-1}(x).$$ Notice that the sequence $\{g_m\}$ includes, as a subset, the set of polynomials that you want to show have only real zeros. With this easier formula we arrive at the simple recursion: $$g_m(x)=g_{m-1}+x\cdot g_{m-2}(x).$$ Therefore, we only need to show that $\{g_m\}$ has two initial polynomials with real interlacing zeros. We calculate the first few polynomials, $$g_2(x)=3x+1 \\ g_3(x)=2x^2+4x+1 \\ g_4(x)=5x^2+5x+1 \\ g_5(x)=2x^3+9x^2+6x+1 \\ g_6(x)=7x^3+14x^2+7x+1 \\ \vdots $$ and calculate the roots, $$Roots(g_2)=\{-0.3333\} \\ Roots(g_3)=\{-0.2929,-1.7071\} \\ Roots(g_4)=\{-0.2764,-0.7236,\} \\ Roots(g_5)=\{-0.2670,-0.5000,-3.7321\} \\ Roots(g_6)=\{-0.2630,-0.4090,-1.3290\} \\ \vdots $$ We see that $g_2$ and $g_3$ have interlacing zeros, in fact this list is consecutively interlacing. Note that the consecutive interlacing was not guaranteed by the theorem; the only thing the theorem guarantees is that the remaining polynomials will all have only real zeros. Although, now that we know all the polynomials have only real zeros, there is another theorem of Fisk that tells us they will continue to consecutively interlace.
Definition. Interlacing means that the zeros of two polynomials fit together; for example $$a_1< b_1 < a_2 < \cdots a_n < b_n.$$ Just to be clear two polynomials can only interlace if they are within a degree of each other.
