Series Solutions — Question 9

Question 9 – Solution

Strategy. Convert the residual into a forced error equation, keeping track of both initial errors.

Step 1: Construct the solution truncation. Matching powers gives $(n+2)(n+1)a_{n+2}+a_n=\delta_{n1}$, with $a_0=a_1=0$. Thus all even coefficients vanish and $$\boxed{P_7(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}.}$$

Step 2: Derive the error equation. Direct differentiation yields $\mathcal R_{P_7}=x^7/5040$. Since $P_7$ satisfies both initial values, $e''+e=-x^7/5040$ with $e(0)=e'(0)=0$. The zero-data solution is $$\boxed{e(x)=-\frac 1{5040}\int_0^x\sin(x-t)t^7\,dt.}$$ Differentiating this integral twice verifies the equation and the two initial conditions, so uniqueness proves the formula.

Step 3: Bound the signed integral. For $0\le t\le x\le 1$, $0\le\sin(x-t)\le x-t$. Therefore $$-\frac 1{5040}\int_0^x(x-t)t^7dt =-\frac{x^9}{362880}\le e(x)\le 0.$$ The exact solution is $\boxed{y=x-\sin x}$, as substitution and initial values verify. Its convergent sine expansion has first omitted term $-x^9/9!$, so the alternating estimate agrees with the integral certificate. For $x>0$ the integral is strictly positive before its minus sign, giving $e(x)<0$.

Step 4: Expose the role of initial data. Let $p_M=P_7+M\sin x$ for any real $M$. Since $(\sin x)''+\sin x=0$, every $p_M$ has exactly the same residual. But $p_M'(0)=M$, and $y(1)-p_M(1)=e(1)-M\sin 1$ is unbounded as $|M|\to\infty$. Residual size certifies error only together with suitable initial or boundary errors and a stability estimate.