Review : Systems of Equations — Question 7

Question 7 – Solution

Strategy. A model assumption converts function recovery into coefficient recovery, while extrapolation can amplify measurement errors.

Step 1: Solve the exact coefficient equations. The data give $a=1$, $a+b+c=2$ and $a+2b+4c=5$. Thus $b+c=1$ and $2b+4c=4$, yielding $c=1$ and $b=0$. Every coefficient is forced, so $$\boxed{p(t)=1+t^2,\qquad p(3)=10.}$$

Step 2: Retain the measurement perturbation. The first two equations are unchanged, while $2b+4c=4+\epsilon$. Solving gives $$a=1,\quad b=-\epsilon/2,\quad c=1+\epsilon/2, \qquad \boxed{p_\epsilon(T)-p(T)=\tfrac\epsilon 2 T(T-1).}$$ The error vanishes at both unchanged measurement times and equals $\epsilon$ at $T=2$, as required.

Step 3: Compare interpolation and extrapolation. The error is independent of $\epsilon$ exactly when $T(T-1)=0$, so the nonnegative insensitive times are $0$ and $1$. On $[0,1]$, $|T(T-1)|\le 1/4$; on $[1,2]$, $T(T-1)$ increases from $0$ to $2$. Therefore $$\boxed{\sup_{0\le T\le 2}|p_\epsilon-p|\le\eta,\qquad |p_\epsilon(10)-p(10)|\le 45\eta.}$$ Both bounds are sharp at $|\epsilon|=\eta$, the first at $T=2$. The figure shows the true curves for $\epsilon=0,\pm 0.4$.

Step 4: State the model’s limitation. For any real $K$, the smooth function $q_K(t)=1+t^2+K t(t-1)(t-2)$ matches all three original readings. For $K\ne 0$ it is not quadratic, and $q_K(3)=10+6K\ne 10$. Thus three readings uniquely determine a member of the quadratic family, not an arbitrary smooth response. Extrapolation bounds also rely on the quadratic family and on the stated location of the measurement error.