Deriving a formula for the coefficients of a power series.

Question

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Consider the two truncated power series $u(t)=u_0+u_1x+u_2x^2+u_3x^3+u_4x^4+\cdots+u_nx^n$ and $d(t)=d+d_1x+d_2x^2+d_3x^3+d_4x^4+\cdots+dx^n$. Show how to derive the formula \begin{align*} q_i=\frac{\sum_{k=0}^{i-1}q_kd_{i-k}}{d_0} \end{align*} where the $q_i$ are the coefficients of the truncated power series for $q(t)=\frac{u(t)}{d(t)}$.

I have tried rearranging and rewriting and many other things. Help would be greatly appreciated!

Answer 1

Hint: The statement of this problem shows some serious flaws.

• $u(t)=u_0+u_1x+u_2x^2+u_3x^3+u_4x^4+\cdots+u_nx^n$: The left-hand side indicates a function in $t$ while the right-hand side indicates a function in $x$.

• $d(t)=d+d_1x+d_2x^2+d_3x^3+d_4x^4+\cdots+dx^n$: The same flaw as above. Additionally, the first and last coefficient are $d$ but should presumably be $d\color{blue}{_0}$ and $d\color{blue}{_n}$ which are different values in general.

• $q_i=\frac{\sum_{k=0}^{i-1}q_kd_{i-k}}{d_0}$: The range of $i$ is missing here. The formula for $q_i$ seems to be incorrect.

Usually we would make the approach

\begin{align*} q(x)=q_0+q_1x+q_2x^2\cdots+q_nx^n=\frac{u(x)}{d(x)} \end{align*} from which we derive \begin{align*} &u_0+u_1x+u_2x^2+\cdots+u_nx^n\\ &\qquad=(q_0+q_1x+q_2x^2+\cdots+q_nx^n)(d_0+d_1x+d_2x^2+\cdots d_nx^n) \end{align*} We now compare terms with equal powers, analyse \begin{align*} u_0&=q_0d_0\\ u_1&=q_0d_1+q_1d_0\\ u_2&=q_0d_2+q_1d_1+q_2d_0\\ &\vdots\\ u_n&=q_0d_n+q_1d_{n-1}+\cdots+q_nd_0 \end{align*} from which another formula follows.