david/A bound on the radius of absolute monotonicity of a polynomial in terms of the ratio of its first two coefficients
Let \(\psi(z) = \sum_{j=0}^s a_j z^j\) with \(a_1\ge0\) and let \(R(\psi)\) denote the radius of absolute monotonicity of \(\psi\). Then \[R \le s \frac{a_0}{a_1},\] with equality only if \[\psi(z) = a_0 \left(1+z\frac{a_1}{sa_0}\right)^s.\]
Proof. We can write \[\psi(z) = \sum_{j=0}^s \gamma_j \left(1+\frac{z}{R}\right)^j, \ \ \ \ \ \gamma_j\ge0.\] Equating coefficients, we have \[\sum_{j=0}^s \gamma_j = a_0 \] \[\sum_{j=0}^s j \gamma_j = R a_1 \] Subtracting \(s\) times the first equation from the second gives \[\sum_{j=0}^s (j-s) \gamma_j = Ra_1 - s a_0.\] The left hand side of this equality is non-positive, so we have \[R a_1 - s a_0 \le 0\] and the result is immediate.
For the uniqueness result, observe that \(R=s\frac{a_0}{a_1}\) implies \[\sum_{j=0}^s (j-s) \gamma_j = 0,\] which can hold only if \(\gamma_j=0\) for \(j\ne s\), in which case \(\gamma_0=a_0\).
Of course, a special case is the following: let \(a_0=1\) and choose \(a_1\ge 1/2\). Then \(R\le 2s\). |
This might be a step in proving the 2s conjecture, if we could also prove the following two results:
Let \(R(A,b)\) denote the radius of absolute monotonicity of a Runge–Kutta scheme with coefficients \((A,b)\), and let \(\psi(z)= \det(I-z(A-eb^T))\) denote the numerator of its stability function. Then the threshold factor \(R(\psi)\) satisfies \(R(\psi)\le R(A,b)\).
Let \(\psi(z)=a_0 + a_1 z + \cdots\) be the numerator of the stability function of a second order accurate Runge–Kutta method with coefficients \((A,b)\), with \(A\ge0, b\ge0\). Then \(a_1\ge 1/2\).
The first result might be approached using the formulas given for \(\psi\) in terms of the Shu-Osher coefficients.
The second result may follow from the order conditions (written as conditions on \(\gamma_j\) as in the proof above) and the trace inequality derived by Lajos.