Series Solutions — Question 7

Question 7 – Solution

Strategy. Nonlinearity couples coefficients by convolution; an exact solution then validates convergence and reveals blow-up.

Step 1: Convolve rather than square termwise. The constant forcing contributes only at $n=0$: $$\boxed{(n+1)a_{n+1}=\delta_{n0}+\sum_{k=0}^n a_ka_{n-k},\qquad a_0=0,}$$ where $\delta_{n0}=1$ for $n=0$ and $0$ otherwise. This gives $$\boxed{P_7(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315},}$$ with $a_2=a_4=a_6=0$.

Step 2: Establish parity and coefficient signs. If $y$ solves the IVP, $-y(-x)$ does too; local uniqueness gives oddness. The recurrence also proves by induction that all even coefficients vanish and all odd coefficients are positive: each required convolution contains positive odd-index products. For example, the coefficient of $x^4$ in $y^2$ is $2a_1a_3$, not $a_2^2$. Cross terms are essential.

Step 3: Identify the function and its radius. Separation yields $\arctan y=x$ near $0$, so $\boxed{y=\tan x}$. It satisfies the equation and initial condition on $(-\pi/2,\pi/2)$ and blows up at both endpoints, making this the maximal real interval through $0$. Since $\sin(\pm\pi/2)\ne 0$, the nearest zeros of cosine are nonremovable poles of the quotient. The supplied facts give $\boxed{R=\pi/2}$. Within this radius, the analytic solution’s coefficients obey the recurrence, which uniquely determines them; hence it is the constructed series. At $x=\pi/2$ the positive-term series diverges: a finite endpoint sum would bound $\tan x$ below that endpoint, contradicting blow-up. Oddness gives divergence at $-\pi/2$ as well.

Step 4: Interpret truncation and blow-up. For $0<x<\pi/2$, every omitted odd term is positive, so $0<P_7(x)<\tan x$. But $P_7$ stays bounded as $x\to(\pi/2)^-$ while $\tan x\to+\infty$. Their difference is unbounded on that interval. Every fixed Taylor truncation has this defect; convergence on smaller compact intervals does not imply uniform approximation up to a blow-up boundary.