Series Solutions — Question 10

Question 10 – Solution

Strategy. Resolve the shrinking oscillation scale before judging a series truncation.

Step 1: Derive and sum the series. The recurrence is $a_{n+4}=-a_{n+2}/[\varepsilon(n+4)(n+3)]$, with $a_0=a_1=a_3=0$, $a_2=1/2$. Therefore $$\boxed{y_\varepsilon=\sum_{j=1}^\infty \frac{(-1)^{j+1}x^{2j}}{(2j)!\varepsilon^{j-1}} =\varepsilon\bigl(1-\cos(x/\sqrt\varepsilon)\bigr).}$$ The radius is infinite for each fixed positive parameter. The formula verifies all four initial derivatives and the equation directly.

Step 2: Distinguish levels of derivative convergence. The bounds $|y_\varepsilon|\leq 2\varepsilon$ and $|y_\varepsilon'|\leq\sqrt\varepsilon$ prove uniform convergence to zero on $\mathbb R$. But $y_\varepsilon''=\cos(x/\sqrt\varepsilon)$. For fixed $x\ne 0$, the sequences $\varepsilon_n=x^2/(2\pi n)^2$ and $\widetilde\varepsilon_n=x^2/((2n+1)\pi)^2$ give values $1$ and $-1$. No pointwise limit exists there. The reduced equation $y''=0$ cannot retain the original second-derivative datum.

Step 3: Test fixed truncations. The degree-two polynomial is $x^2/2$, which fails to approach zero at fixed nonzero $x$. For any fixed degree $2m\geq 4$, its highest term is a nonzero constant times $\varepsilon^{1-m}$ and dominates the lower powers as $\varepsilon\downarrow 0$. Thus the truncation diverges while the full sum tends to zero. The Taylor remainder is not uniform in this parameter at fixed $x$.

Step 4: Rescale and certify. With $y_\varepsilon/\varepsilon=1-\cos\xi$, Taylor’s theorem gives, for $|\xi|\leq R$, $$\left|1-\cos\xi-\sum_{j=1}^m\frac{(-1)^{j+1}\xi^{2j}}{(2j)!}\right| \leq\frac{R^{2m+2}}{(2m+2)!}.$$ Multiplying by $\varepsilon$ bounds the original displacement error. For the graph, $m=3$, $R=\pi$, so the profile error is at most $\pi^8/40320$.