Expanding denominator in a power series, mismatch of the expansion Below is a snippet from the book Ralston:First course in numerical analysis but it seems to me that something is wrong with $(10.2-11):$ the denominator divided by $a_1\lambda_1^m$ starts with $1$ not with $\lambda_1$. Anyway, I cannot expand and obtain $\lambda_1+\beta_i(\frac{\lambda_2}{\lambda_1})^m+[\text{terms in }(\frac{\lambda_3}{\lambda_1})^m,...,(\frac{\lambda_n}{\lambda_1})^m \text{ and higher powers}].$
Moreover it seems to me that $(\frac{\lambda_2}{\lambda_1})^m$ is of the same order as
$(\frac{\lambda_3}{\lambda_1})^m,...$

 A: The series expansion is sound. We obtain
\begin{align*}
&\frac{a_1\lambda_1^{m+1}+\sum_{j=2}^{n}a_j\lambda_j^{m+1}}{a_1\lambda_1^m+\sum_{j=2}^na_j\lambda_j^m}
=\frac{a_1\lambda_1^{m}\left(\lambda_1^{m}+\sum_{j=2}^{n}\frac{a_j\lambda_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^{m}\right)}{a_1\lambda_1^m\left(1+\sum_{j=2}^n\frac{a_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^m\right)}\\
&\quad=\frac{\lambda_1^{m}+\sum_{j=2}^{n}\frac{a_j\lambda_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^{m}}{1+\sum_{j=2}^n\frac{a_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^m}\\
&\quad=\left(\color{blue}{\lambda_1^{m}+\sum_{j=2}^{n}\frac{a_j\lambda_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^{m}}\right)\\
&\quad\quad\quad\cdot\left(\color{blue}{1} - \left(\sum_{j=2}^n\frac{a_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^m\right)+
\left(\sum_{j=2}^n\frac{a_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^m\right)^2-\cdots\right)\tag{1}\\
&\quad=\lambda_1^{m}+\sum_{j=2}^{n}\frac{a_j\lambda_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^{m}+
\mathrm{terms\ of\ higher\ order}\tag{2}\\
\end{align*}
We see in (1) $\beta_i:=\frac{a_2\lambda_2}{a_1}$ the factor in the sum with $j=2$. The author considers also all terms of the sum $\sum_{j=2}^n\frac{a_j}{a_1}\left(\frac{\lambda_j}{\lambda_1}\right)^m$ to be of the same order.
Comment:

*

*In (1) we make a geometric series expansion $\frac{1}{1+z}=1-z+z^2-z^3+\cdots$.


*In (2) we see the explicitly stated terms are derived from the multiplication of the blue terms in the line above. Multiplication with other terms which are not marked as blue produce higher order terms.
