Final Thoughts — Question 2

Question 2

The exact solution of $y'=y$, $y(0)=1$, is supplied as $y=e^x$. A polynomial approximation is $$P(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}.$$ You may use Taylor’s theorem with a fourth-derivative remainder.

Tasks

1. Calculate the residual $P'-P$ and determine whether $P$ solves the IVP exactly on any open interval containing $0$.

2. Let $E(x)=e^x-P(x)$. Prove that for $0\le x\le 1$, $$\frac{x^4}{24}\le E(x)\le\frac{e x^4}{24}.$$

3. Give exact lower and upper bounds for $E(1)$, and identify whether $P$ overestimates or underestimates the solution for $0<x\le 1$.

4. Plot the error and its two bounds in your solution. Explain why matching the initial value and several derivatives at $0$ does not make the polynomial an exact interval solution.

Question 2 – Solution

Strategy. Use a residual to test exactness and Taylor’s theorem to assess the approximation. These answer different questions.

Step 1: Exactness test. Since $P'=1+x+x^2/2$, $$\boxed{P'-P=-\frac{x^3}{6}.}$$ The residual is nonzero whenever $x\ne 0$. Although $P(0)=1$, the polynomial does not solve the equation on any open interval containing $0$.

Step 2: Error bounds. For $0<x\le 1$, Taylor’s theorem supplies a number $\xi$ with $0<\xi<x$ such that $$E(x)=\frac{e^\xi x^4}{24}.$$ Because $1\le e^\xi\le e$, the requested bounds follow; at $x=0$ all three expressions are zero.

Step 3: Size and sign. At $x=1$, $P(1)=8/3$, so $$\boxed{\frac 1{24}\le E(1)=e-\frac 83\le\frac e{24}.}$$ For $x>0$ the lower bound is positive, so $P$ underestimates the exact solution. The two endpoint bounds are in fact strict at $1$.

Step 4: Local agreement is limited evidence. The values of $P$ and its first three derivatives agree with those of $e^x$ at $0$. That is local Taylor information. Solving the ODE requires $P'(x)=P(x)$ at every point of an interval, which the nonzero residual disproves.