Series Solutions — Question 1

Question 1 – Solution

Strategy. Keep the lowest index separate, then use the recurrence both to construct solutions and to bound their tails.

Step 1: Align the powers. The constant coefficient gives $a_2=0$. For $n\ge 1$, $(n+2)(n+1)a_{n+2}=a_{n-1}$, equivalently $$\boxed{a_{m+3}=\frac{a_m}{(m+3)(m+2)}\quad(m\ge 0).}$$ The chains start at $a_0,a_1,a_2$; the last is identically zero.

Step 2: Build the normalized series. With empty products interpreted as $1$, $$u(x)=\sum_{k\ge 0}\frac{x^{3k}}{\prod_{j=1}^k(3j)(3j-1)} =1+\frac{x^3}{6}+\frac{x^6}{180}+\frac{x^9}{12960}+\cdots,$$ $$v(x)=\sum_{k\ge 0}\frac{x^{3k+1}}{\prod_{j=1}^k(3j+1)(3j)} =x+\frac{x^4}{12}+\frac{x^7}{504}+\frac{x^{10}}{45360}+\cdots.$$

Step 3: Establish actual solutions and independence. For either series, the ratio of consecutive nonzero term magnitudes tends to zero at every fixed $x$. Thus both have infinite radius, may be differentiated termwise, and satisfy the equation and stated initial data. Their Wronskian satisfies $W'=uv''-u''v=0$ and $W(0)=1$. Hence $\boxed{y=Au+Bv}$ is the unique solution for those initial values, on $\mathbb R$.

Step 4: Bound all omitted terms. The first omitted coefficient of $u$ is $1/1710720$. On $|x|\le 1$, ratios within the remaining tail are at most $1/(15\cdot 14)=1/210$. Therefore $$\boxed{|u(x)-U_9(x)|\le\frac{|x|^{12}}{1710720}\frac 1{1-1/210} <5.88\times 10^{-7}.}$$ The plot magnifies the actual signed remainder; the geometric estimate covers both signs of $x$.