Repeated Roots — Question 4

Question 4 – Solution

Strategy. Use the repeated-root sample recurrence, but retain every admissible root of the nonlinear identification equation.

Step 1: Derive the recurrence. Since $S_n=(A+Bn)z^n$, $$2zS_{n+1}-z^2S_n =\bigl(2(A+B(n+1))-(A+Bn)\bigr)z^{n+2} =S_{n+2}.$$

Step 2: Find both admissible models. The observed values imply $7/e^2=8z/e-z^2$, or $$(z-1/e)(z-7/e)=0.$$ Both roots are positive. Since $A=S_0=1$ and $B=S_1/z-A$, the candidates are $$\begin{array}{c|c|c|c} z&r&A&B\\\hline 1/e&-1&1&3\\ 7/e&\ln 7-1&1&-3/7 \end{array}$$ Each gives a different repeated-root equation $y''-2ry'+r^2y=0$, and each fits the original three observations. Selecting only one positive root of the identification quadratic would be unjustified.

Step 3: Use the fourth observation. For the first candidate, $S_3=(1+9)e^{-3}=10/e^3$, matching the new datum. The other gives $$S_3=(1-9/7)(7/e)^3=\boxed{-98/e^3},$$ so it is rejected. The identified equation and data are $$\boxed{y''+2y'+y=0},\qquad \boxed{y=(1+3t)e^{-t}},\qquad \boxed{y'(0)=2}.$$

Step 4: Verify and interpret the ambiguity. At $n=0,1,2,3$, the selected multiplier $1+3n$ gives $1,4,7,10$, with the required exponential factors. This solution tends to zero. The rejected candidate has $r=\ln 7-1>0$ and negative leading multiplier coefficient, so it eventually tends to $-\infty$.

The initial three sample constraints lead to a quadratic equation with two admissible positive multipliers, despite being exact. The fourth observation distinguishes them. Counting unknown parameters and observations alone does not prove a nonlinear inverse problem has a unique solution.