research33 min studyLesson 940 in Mathematics

Positivity, real-rootedness and log-concavity

Separate basis-dependent positivity from root geometry and derive coefficient-shape consequences under sufficient hypotheses.

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What you’ll understand

  • Distinguish coefficient positivity, log-concavity, unimodality and real-rootedness.
  • Apply Newton inequalities with the correct degree normalization.
  • Use counterexamples to reject unsupported converse implications.

A generating polynomial can have positive coefficients, a single peak, log-concave coefficients, or only real zeros. These are different properties, and the logical arrows between them need hypotheses. In algebraic combinatorics there is a second ambiguity: positivity depends on the chosen basis. Schur-positive, monomial-positive and elementary-positive symmetric functions are distinct notions. We first examine finite univariate polynomials over R, then connect the resulting cautions back to representation multiplicities and face-ring Hilbert numerators.

Nonnegative coefficients do not control roots

For a finite sequence a0 through ad, log-concavity means ai squared is at least ai−1 ai+1 at every interior index. Assume the entries are nonnegative and there are no internal zeros, meaning the positive indices form a single interval. Then successive positive ratios are nonincreasing, so the sequence can change from rising to falling at most once; it is unimodal, allowing a plateau. Without the support assumption, the locally checked inequalities can miss a long gap between separated positive entries.

Nonnegative coefficients alone imply neither log-concavity nor real-rootedness. Even a positive log-concave sequence can have nonreal roots: the polynomial 1+t+t² has coefficients (1,1,1) satisfying equality in its interior adjacent-product comparison, but discriminant minus three. Conversely, a polynomial with mixed-sign coefficients may have only real roots without giving the desired nonnegative counting sequence. Root statements and coefficient-shape statements must therefore be compared within an explicitly declared nonnegative class.

ai2≥ai−1ai+1(1≤i<d),{i:ai>0} an interval⟹(a0,…,ad) is unimodala_i^2\ge a_{i-1}a_{i+1}\quad(1\le i<d),\qquad \{i:a_i>0\}\text{ an interval}\quad\Longrightarrow\quad(a_0,\ldots,a_d)\text{ is unimodal}
The implication assumes all coefficients are nonnegative. An interval of positive support rules out internal gaps.↔ Scroll wide equations horizontally.

Newton inequalities give a stronger normalized conclusion

Suppose A(t) has degree d, nonnegative coefficients and only real roots. A positive real argument cannot be a root because all summands are nonnegative and at least one is positive. Hence all nonzero roots are negative; any zero roots contribute an initial power of t. Factoring the remaining polynomial into positive multiples of terms t+r with r>0 shows that its positive coefficients occupy a consecutive interval. Thus the support condition needed for unimodality follows from these hypotheses.

Newton inequalities assert log-concavity after dividing coefficient ai by the binomial coefficient choose(d,i). This is stronger than ordinary log-concavity because the binomial coefficients themselves are strictly log-concave in the interior. One route identifies normalized coefficients with normalized elementary symmetric functions of nonnegative root parameters. We use the theorem rather than prove its full inequality here. Use the actual degree d for this sharp normalization; padding the coefficient list with trailing zeros changes the binomial factors and must not be confused with the same normalized sequence.

A(t)=∑i=0daiti real-rooted, ai≥0⟹(ai(di))2≥ai−1(di−1)ai+1(di+1)A(t)=\sum_{i=0}^{d}a_it^i\text{ real-rooted},\ a_i\ge0\quad\Longrightarrow\quad\left(\frac{a_i}{\binom di}\right)^2\ge\frac{a_{i-1}}{\binom d{i-1}}\frac{a_{i+1}}{\binom d{i+1}}
The degree is exactly d and the polynomial is nonzero. These Newton inequalities imply ordinary log-concavity, with no internal gaps under the stated root and sign assumptions.↔ Scroll wide equations horizontally.

A positive basis does not automatically give a root theorem

Littlewood–Richardson coefficients are nonnegative because they count tableaux or representation multiplicities. That does not force every polynomial formed from them to have real roots. Likewise a Cohen–Macaulay complex has a nonnegative h-vector, but this alone provides no universal unimodality or real-rootedness theorem. The triangle boundary already has h(t)=1+t+t² and nonreal roots. Geometry or representation theory can supply stronger conditions in particular families, but each conclusion needs the relevant additional theorem.

Positivity also changes under a change of basis. For example, p2=m2 is monomial-positive, yet p2=s2−s11 is not Schur-positive. Schur-positivity does imply monomial-positivity because Kostka coefficients are nonnegative. Reversing that implication fails in this example. A productive research question therefore names the objects, the coefficient basis or generating statistic, and the exact strength of the desired property before looking for a combinatorial interpretation or a real-rootedness-preserving recurrence.

Check real roots and normalized log-concavity

For A(t)=(1+t)(1+2t)(1+3t), compute coefficients, roots and Newton-normalized coefficients.

  1. Multiplying the first two factors gives 1+3t+2t². Multiplying by 1+3t gives A(t)=1+6t+11t²+6t³.

  2. The factors give roots −1,−1/2,−1/3, all real and negative. Ordinary checks give 6²=36≥11 and 11²=121≥36, so the coefficient sequence is log-concave.

  3. Divide by the degree-three binomial sequence (1,3,3,1), giving (1,2,11/3,6). Newton checks are 4≥11/3 and 121/9≥12. Both hold, establishing the stronger normalized inequalities in this example.

The result

The polynomial is positive-coefficient and real-rooted; its normalized sequence passes Newton inequalities and its ordinary sequence is unimodal with its peak at coefficient 11.

Explore the model

Interactive lab

Interactive lab · probability

A real-rooted binomial family

Pause, reset, then set p=0.5 and n=4. Predict probabilities (1,4,6,4,1)/16, mean 2 and variance 1. Their generating polynomial is ((1+t)/2)⁴ with the repeated real root −1. Divide the displayed probabilities by choose(4,i): every normalized coefficient is 1/16, so Newton inequalities hold with equality. The lab covers a binomial family, not arbitrary Schur-positive or h-polynomials.

A real-rooted binomial familyThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forReal-rootedness can explain log-concavity in a specified family, but coefficient positivity by itself cannot supply it.

Check your understanding

Which implication is valid with the listed hypotheses?

Try it yourself

Compare coefficient log-concavity and real-rootedness for B(t)=1+3t+3t² and C(t)=1+t³. Explain the role of internal zeros.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the quadratic discriminant for B. For C, test its coefficient inequalities at both interior positions before deciding whether it is unimodal.

Compare with the solution

B has coefficients (1,3,3), with 3²≥1·3, so it is log-concave and unimodal. Its discriminant is 9−12=−3, so it is not real-rooted. C has coefficients (1,0,0,1); at both interior positions the log-concavity comparison is 0≥0, so these local inequalities hold, but the sequence falls from 1 to 0 and rises back to 1 and is not unimodal. The two internal zeros show why the support hypothesis is required. Its roots also include a nonreal pair, so neither example contradicts Newton’s sufficient conditions.

Take it with you

  • Positivity requires a named coefficient basis and is weaker than root control.
  • Real-rootedness with nonnegative coefficients implies normalized Newton log-concavity.
  • Log-concavity implies unimodality only with nonnegative interval support.

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