Review : Power Series — Question 7

Question 7 – Solution

Strategy. Work in $h=x-2$ and remember that multiplication by $x$ means multiplication by $h+2$, not merely by $h$.

Step 1: Reindex before multiplying. Inside the radius, $y'=\sum_{n=0}^{\infty}(n+1)a_{n+1}h^n$. Now $hy'=\sum_{n=1}^{\infty}na_nh^n$; its constant coefficient is zero. Adding $2y'$ and $y$ yields $$\boxed{H=\sum_{n=0}^{\infty}\bigl((n+1)a_n+2(n+1)a_{n+1}\bigr)h^n.}$$ The formula also works for $n=0$, where the constant term of $xy'$ comes entirely from $2y'$.

Step 2: Display the first four coefficients. The requested coefficients are $$\boxed{a_0+2a_1,\quad 2a_1+4a_2,\quad 3a_2+6a_3,\quad 4a_3+8a_4.}$$ As a direct constant-term check, $H(2)=2y'(2)+y(2)=2a_1+a_0$. Keeping the same summation index through a differentiation would miss the shift from $a_n$ to $a_{n+1}$.

Step 3: Determine exactly when the shortcut works. The claimed series actually represents $hy'+y$. Its difference from $H$ is $2y'$. The identity therefore holds throughout the interval exactly when $y'\equiv 0$, or equivalently $\boxed{a_n=0\text{ for every }n\ge 1}$. One justification of the equivalence is uniqueness of power-series coefficients, obtained by successive differentiation at the center. Any constant $a_0$ is allowed. Equality at a single isolated point would not imply this identity.

Step 4: Check against a geometric closed form. For $a_n=3^{-n-1}$, $y=(1/3)\sum(h/3)^n=1/(3-h)=1/(5-x)$ when $|h|<3$. Hence $$\boxed{H=\frac{x}{(5-x)^2}+\frac 1{5-x}=\frac 5{(5-x)^2}.}$$ The coefficient formula gives $(n+1)/3^{n+1}+2(n+1)/3^{n+2}=5(n+1)/3^{n+2}$. This agrees with the differentiated geometric identity $5/(3-h)^2=\sum_{n\ge 0}5(n+1)h^n/3^{n+2}$. The radius is $\boxed{3}$; the factors growing linearly in $n$ do not change it.