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\begin{document}
\centerline{Math 372 lecture for Friday, Week 4}
\bigskip

\textbf{Definition.} A sequence of real numbers~$a_{0},\ldots,a_{n}$ is
\emph{logarithmically concave} or \emph{log-concave} is
\[
  a_{i}^{2}\geq a_{i-1}a_{i+1}
\]
for $1\leq i\leq n-1$.  We say~$a_{0},\ldots,a_{n}$ is \emph{strongly
log-concave} if
\[
  b_{i}^{2}\geq b_{i-1}b_{i+1}
\]
for $1\leq i\leq n-1$ where~$b_{i}:=a_{i}/\binom{n}{i}$.
\medskip

The name \emph{log-concave} comes from the fact that $a_{i}^{2}\geq
a_{i-1}a_{i+1}$ is a multiplicative version of the inequality~$a_{i}\geq
\frac{a_{i-1}+a_{i+1}}{2}$.  An easy calculation shows that strong log-concavity
is equivalent to 
\[
  a_{i}^{2}\geq\left(1+\frac{1}{i}\right)\left(1+\frac{1}{n-i}\right)a^{i-1}a^{i+1},
\]
from which it follows that strong log-concavity implies log-concavity.
\medskip

It is not always true that log-concavity implies unimodality.
Consider~$1,0,0,1$, for instance. A sequence, $a_{0},a_{1},\ldots,a_{n}$ has
\emph{no internal zeros} if whenever~$i<j<k$ and both $a_{i}$ and~$a_{k}$ are
nonzero, then~$a_{j}$ is also nonzero.  
\medskip

\textbf{Proposition 5.11} Let $\alpha=(a_{0},\ldots,a_{n})$ be a sequence of
nonnegative real numbers with no internal zeros.  If~$\alpha$ is log-concave,
the~$\alpha$ is unimodal.
\begin{proof}
If at most two values of~$\alpha$ are nonzero, then~$a_{i-1}a_{i-1}=0$ for all
$1\leq i\leq n-1$, and the result holds.  Otherwise, assume $\alpha$ is not
unimodal.  Then there exists $1\leq i\leq n-1$ such that~$a_{i-1}>a_{i}<a_{i+1}$.
Using the fact that the elements of~$\alpha$ are nonnegative, it follows
that~$a_{i}^{2}<a_{i-1}a_{i+1}$.
\end{proof}

\textbf{Example.}  The $n$-th row of Pascal's triangle
\[
  \binom{n}{0}, \binom{n}{1},\ldots,\binom{n}{n}
\]
is trivially strongly log-concave with no internal zeros.  Hence, it is
unimodal.  Here is a combinatorial proof of log-concavity of the binomial
coefficients from Richard Stanley's paper
\href{http://dedekind.mit.edu/~rstan/pubs/pubfiles/72.pdf}{Log-concave and
unimodal sequences in algebra, combinatorics, and geometry}, 1989 (see the
``Final word of warning'' at the end of that paper).  Let $\binom{[n]}{i}$ denote the
collection of subsets of~$[n]$ of size~$i$.  We describe an injection
\[
  \phi\colon\binom{[n]}{i-1}\times \binom{[n]}{i+1}\to \binom{[n]}{i}\times \binom{[n]}{i}
\]
Log-concavity then follows:
\[
  a_{i-1}a_{i+1}=\left|\binom{[n]}{i-1}\right|\left|\binom{[n]}{i+1}\right|
  \leq\left|\binom{[n]}{i}\right|^{2}=a_{i}^{2}.
\]
For any~$X\subseteq [n]$ and~$j=0,\ldots,n$, define
\[
  X_{j}=X\cap [j]=\left\{x\in X: x\leq j\right\}.
\]
Given~$(A,B)\in \binom{[n]}{i-1}\times \binom{[n]}{i+1}$, let $j$ be the
smallest nonnegative integer such that $|A_{j}|=|B_{j}|-1$.  This integer must
exist since $|A_{0}|=|B_{0}|=0$, and $|A_{n}|=|B_{n}|-2$, and the sizes of $A_{j}$
and $B_{j}$ grow in size by at most one each time $j$ increases by one.
Next, define
\[
  C:= A_{j}\cup(B\setminus B_{j})\quad\text{and}\quad D:=B_{j}\cup(A\setminus
  A_{j}).
\]
Note that these are disjoint unions since, for instance~$A_{j}\subseteq[j]$ and
$B\setminus B_{j}\subset [n]\setminus[j]$.  We also have that~$|C|=|D|=i$:
instance, 
\[
  |C|=|A_{j}|+|B|-|B_{j}|= |A_{j}|+(i+1)-(|A_{j}|+1)=i,
\]
and similarly for $|D|$. Define~$\phi(A,B)=(C,D)$.  For injectivity, note
that
\[
  A_{k}=C_{k}\quad\text{and}\quad B_{k}=D_{k}
\]
for $k=0,\ldots,j$, while
\[
  A_{j+1}\neq C_{j+1}\quad\text{and}\quad B_{j+1}\neq D_{j+1}.
\]
Thus, we can recover~$j$, $A_{j}$, and~$B_{j}$ from~$(C,D)$.  But from that
information, we can recover~$(A,B)$.  For instance,
\[
  A = C_{j}\cup (D\setminus B_{j}).
\]
That completes the proof.
\medskip

\textbf{Example.} Let~$A=\{1,3,4,6\}$.  Then~$A_{0}=\emptyset$,
$A_{1}=A_{2}=\{1\}$, $A_{3}=\{1,3\}$, $A_{4}=A_{5}=\{1,3,4\}$, and $A_{6}=A$.
\medskip

There is a surprising connection between log-concavity and polynomials with real
zeros:
\medskip

\textbf{Theorem 5.12} (I.~Newton) Let
\[
  P(x)=\sum_{i=0}^{n} b_{i}x^{i}=\sum_{i=0}^{n} \binom{n}{i}a_{i}x^{i}
\]
be a real polynomial all of whose zeros are real numbers.  Then the
sequence $b_{0},\ldots,b_{n}$ is strongly log-concave, or equivalently, the
sequence $a_{0},\ldots,a_{n}$ is log-concave. Moreover, if each $b_{i}\geq0$ (so
the zeros of $P(x)$ are non-positive [why?]) then the sequence
$b_{0},\ldots,b_{n}$ has no internal zeros.
\medskip

Note that Stanley's statement here reverses the notation introduced earlier: in
the definition of strong log-concavity, we had $b_{i}=a_{i}/\binom{n}{i}$, and
were talking about strong log-concavity of the sequence $(a_{i})$ and
here we have $a_{i}=b_{i}/\binom{n}{i}$ and are talking about strong
log-concavity of the sequence~$(b_{i})$.
\medskip

Before proving the theorem, we first prove some preliminary results.
\medskip

\textbf{Proposition.} Let~$F(x)=\sum_{i=0}^{m} a_{i}x^{i}$ be a real polynomial
of degree~$m$ with only real zeros.  Then its derivative, $F'(x)$, has only real
zeros.
\begin{proof}
 By long division, $\alpha$ is a zero of $F$ of multiplicity $r\geq1$ if and only if
 there exists a polynomial~$G$ such that
 \[
   F(x) = (x-\alpha)^{r}G(x)
 \]
 where~$G(\alpha)\neq0$.  So suppose~$\alpha$ is a zero of $F$ of
 multiplicity~$r$
 and choose $G$ as above.  Then,
 \[
   F'(x)=r(x-\alpha)^{r-1}G(x)+(x-\alpha)^{r}G'(x)=(x-\alpha)^{r-1}(G(x)+(x-\alpha)G'(x)).
 \]
 Since $G(x)+(x-\alpha)G'(x)$ evaluated at $\alpha$ is~$G(\alpha)\neq 0$, we
 conclude that~$\alpha$ is a zero of $F'$ of multiplicity~$r-1$.

 Say the zeros of $F$ are
 $\alpha_{1}<\ldots<\alpha_{k}$ with respective multiplicities
 $r_{1},\ldots,r_{k}$.  By Rolle's theorem, for each~$1\leq i<k$,
 since~$F(\alpha_{i})=F(\alpha_{i+1})=0$, there is zero~$\beta_{i}$ of~$F'(x)$
 between $\alpha_{i}$ and $\alpha_{i+1}$.  Thus, we have zeros for~$F'(x)$:
 \[
   \alpha_{1}<\beta_{1}<\alpha_{2}<\beta_{2}<\cdots<\alpha_{k-1}<\beta_{k-1}<\alpha_{k}.
 \]
 where $\alpha_{i}$ has multiplicity~$r_{i}-1$.  Counting multiplicities,
 we have found $(\sum_{i=1}^{k} r_{i})-1=\deg(F)-1=m-1$ roots of~$F'$.  Since
 $\deg(F')=m-1$, we have found all of the roots of~$F'$, and they are all
 real.
\end{proof}

\textbf{Proposition.} Let~$F(x)=\sum_{i=0}^{m} a_{i}x^{i}$ be a real polynomial
of degree~$m$ with only real zeros.  Then the polynomial $\widetilde{F}(x):=x^{m}F(1/x)$ has
only real zeros.
\begin{proof}
 Let~$k\geq0$ be the multiplicity of~$0$ as a zero of~$F$.  We can then
 write $F(x)=x^{k}G(x)$ where~$G(0)\neq 0$.  The nonzero zeros of $F$ and $G$
 are the same, with the same multiplicities.  We have
 \[
   \widetilde{F}(x)=x^{m-k}G(1/x).
 \]
 So the nonzero zeros of~$\widetilde{F}$ are the reciprocals of those of~$F$.
 Thus, all of the zeros of~$\widetilde{F}$ are real.
\end{proof}

The above two propositions say that the property of having all real zeros is
preserved under taking derivatives and under a certain type of ``inversion''.
Here is an example illustrating how these operations are useful in relating the
property of having all real zeros with strong log-concavity.  Say
\[
  P(x) = b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+b_{5}x^{5}x^{5}+b_{6}x^{6}+b_{7}x^{7}
\]
has all real zeros.  We can show that strong log-concavity holds for, say,
$b_{2},b_{3},b_{4}$ by using the two operations from the propositions to
eliminate all the terms in~$P$ except the three involving~$b_{2},b_{3}$
and~$b_{4}$.  First take two derivatives to eliminate~$b_{0}$ and~$b_{1}$:
\[
  P''(x)=(2\cdot1)\,b_{2}+(3\cdot 2)\, b_{3}x +(4\cdot3)\, b_{4}x^{2}+(5\cdot
  4)\, b_{5}x^{3}+(6\cdot5)\,b_{6}x^{4}+(7\cdot6)\,b_{7}x^{5}.
\]
Now invert to define a polynomial~$Q$:
\[
  Q(x)=x^{5}P''(1/x)
=(2\cdot1)\,b_{2}x^{5}+(3\cdot 2)\, b_{3}x^{4} +(4\cdot3)\, b_{4}x^{3}+(5\cdot
4)\, b_{5}x^{2}+(6\cdot5)\,b_{6}x+(7\cdot6)\,b_{7}.
\]
Take $3$ derivatives to eliminate the terms involving~$b_{5}, b_{6}$, and
$b_{7}$:
\[
  Q^{(3)}
=(2\cdot1)(5\cdot4\cdot3)\,b_{2}x^{2}+(3\cdot
2)(4\cdot3\cdot2)\, b_{3}x
+(4\cdot3)(3\cdot2\cdot1)\, b_{4}.
\]
To prove strong log-concavity, define $a_{i}=b_{i}/\binom{7}{i}$ and substitute:
\begin{align*}
  Q^{(3)}
&=(2\cdot1)(5\cdot4\cdot3)\,\binom{7}{2}a_{2}x^{2}+(3\cdot
2)(4\cdot3\cdot2)\, \binom{7}{3}a_{3}x
+(4\cdot3)(3\cdot2\cdot1)\, \binom{7}{4}a_{4}\\[8pt]
&=(2\cdot1)(5\cdot4\cdot3)\,\frac{7!}{2!5!}a_{2}x^{2}+(3\cdot
2)(4\cdot3\cdot2)\, \frac{7!}{3!4!}a_{3}x
+(4\cdot3)(3\cdot2\cdot1)\, \frac{7!}{4!3!}a_{4}\\[8pt]
&=\frac{7!}{2}(a_{2}x^{2}+2a_{3}x +a_{4})
\end{align*}
Since~$Q^{(3)}$ has only real zeros, it must be that the quadratic
\[
  a_{2}x^{2}+2a_{3}x +a_{4}
\]
has only real zeros.  By the quadratic equation, this happens if and only if the
discriminant is nonnegative:
\[
  4a_{3}^{2}-4a_{2}a_{4}=4(a_{3}^{2}-a_{2}a_{4})\geq0.
\]
That gives log-concavity for $a_{2},a_{3},a_{4}$, and hence strong log-concavity
for $b_{2},b_{3},b_{4}$.
\medskip

\begin{proof}[Proof of Theorem 5.12]
  Given $P$ as in the statement of the theorem, suppose that $\deg(P)=m$
  (allowing for the fact that $b_{m+1},\ldots,b_{n}$ could all be~$0$).
  Fix~$1< i< m$, and take derivatives and inversions to eliminate all
  terms in $P$ except that involving $b_{i-1},b_{i}$, and~$b_{i+1}$.  The result
  is the following polynomial which also has only real zeros:
  \[
    \frac{m!}{2}(a_{i-1}x^{2}+2a_{i}x+a_{i+1}).
  \]
  Therefore, the sequence of $b_{i}$ is strongly log-concave.

  Next suppose that each~$b_{i}\geq0$.  For the sake of contradict, suppose
  there are internal zeros. Then there exists $i+1<k$ where $b_{i}$ and $b_{j}$
  are positive and $b_{j}=0$ for all $i<j<k$.  By taking derivatives and inverting, as above,
  eliminate the terms of~$P$ having degree less then $i$ or larger then~$k$.  We
  are left with a polynomial having two terms (since $b_{j}=0$ for all~$i<j<k$):
  \[
    c+dx^{k-i}
  \]
  for some positive $c,d$, and with $k-i\geq2$. Further, this polynomial has
  only real zeros.  To find them we solve 
  \[
    x^{k-i} = -\frac{c}{d}.
  \]
  This equation has~$k-i$ solutions over the complex numbers, with at most one
  real solution (when~$k-i$ is odd, then~$-\sqrt[k-i]{c/d}$ is a solution).
  Thus, we reach a contradiction.
\end{proof}

Answer to the ``why?'' in Theorem 5.12: If~$P$ is constant, then $0$ is the only
possible zero of $P$.  Otherwise, since the~$b_{i}\geq0$, the derivative~$P'(x)$
is positive for $x>0$. Since~$P(0)=b_{0}\geq0$, it follows that~$P(x)>0$
for~$x>0$.
\end{document}


