Appendix A: Proof of Lemma 3.2
We firstly give an estimation of \(\{r_i\}_{i\ne j}\) in (3.3) through the following lemma.
Lemma A.1
Let \(\{r_i\}_{i\ne j}\) be the ones in (3.3). They satisfy
$$\begin{aligned} \begin{array}{ll} |r_i|&{}\le \left( |a_i|+|a_{i+1}z_1|+\cdots +|a_{n}z_1^{n-i}|\right) 2\mu , \qquad i=n, n-1,\ldots ,j+1, \\ |r_i|&{}\le \left( |a_i|+|a_{i-1}{z_1}^{-1}|+\cdots +|a_{0}z_1^{-i}|\right) 2\mu , \qquad i=0,1,\ldots ,j-1. \end{array} \end{aligned}$$
Proof
Let \(\{\delta _i\}_{i=j+1}^{n-1}\) and \(\{\varepsilon _i\}_{i=j+1}^{n-1}\) correspond to the round-off error in addition and multiplication respectively. According to the IEEE floating point arithmetic, we have
$$\begin{aligned} \hat{c}_{n-1}=a_n, \end{aligned}$$
and
$$\begin{aligned} \begin{aligned} \hat{c}_{n-2}&=\text {fl}\left( a_{n-1}+z_1\hat{c}_{n-1}\right) \\&=a_{n-1}\left( 1+\delta _{n-1}\right) +z_1\hat{c}_{n-1} \left( 1+\delta _{n-1}\right) \left( 1+\varepsilon _{n-1}\right) \\&=a_{n-1}+\hat{c}_{n-1}z_1-r_{n-1}, \end{aligned} \end{aligned}$$
where \(|\delta _{n-1}|\le \mu , |\varepsilon _{n-1}|\le \mu \) and
$$\begin{aligned} |r_{n-1}|=|a_{n-1}\delta _{n-1}+z_1\hat{c}_{n-1}\left( \delta _{n-1}+\varepsilon _{n-1}\right) |\le \left( |a_{n-1}|+|z_1a_n|\right) 2\mu . \end{aligned}$$
Recursively, for \(i=n-2,\ldots ,j+1,\)
$$\begin{aligned} \begin{aligned} \hat{c}_{i-1}&=\text {fl}(a_{i}+z_1\hat{c}_{i})\\&=a_i(1+\delta _{i})+z_1\hat{c}_{i}\left( 1+\delta _{i}\right) \left( 1+\varepsilon _{i}\right) \\&=a_i+z_1\hat{c}_{i}-r_i,\\ \end{aligned} \end{aligned}$$
where \(|\delta _{i}|\le \mu \) and \( |\varepsilon _{i}|\le \mu \). Expanding the expression of \(r_i\) and ignoring the high order terms of \(\mu \) give that
$$\begin{aligned} \begin{aligned} |r_i|&\le (|a_i|+|z_1\hat{c}_i|)2\mu \\&\le \left( |a_i|+|z_1\left( a_{i+1}+z_1\hat{c}_{i+1}\right) |\right) 2\mu \\&\le \cdots \\&\le \left( |a_i|+|a_{i+1}z_1|+\cdots +|a_{n}z_1^{n-i}|\right) 2\mu . \end{aligned} \end{aligned}$$
Similarly, we have
$$\begin{aligned} |r_i|\le \left( |a_i|+|a_{i-1}{z_1}^{-1}|+\cdots +|a_{0}z_1^{-i}|\right) 2\mu , \qquad i=0,1,\ldots ,j-1. \end{aligned}$$
\(\square \)
The proof of Lemma 3.2 is as follows.
Proof (Proof of Lemma 3.2)
It is obvious that
$$\begin{aligned} \begin{aligned} \frac{\sum _{i \not = j}\left| z_2^i-\left( \frac{z_2}{z_1}\right) ^jz_1^{i} \right| \times |r_i|}{\sum _i |a_i z_2^i|}&\le \frac{\sum _{i \not = j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|}+\frac{\sum _{i \not = j}\left| \left( \frac{z_2}{z_1}\right) ^jz_1^{i} \right| \times |r_i|}{\sum _i |a_i z_2^i|}.\\ \end{aligned} \end{aligned}$$
(A.1)
We firstly give an estimation of the first part in (A.1).
According to Lemma A.1,
$$\begin{aligned} \begin{aligned} \frac{\sum _{i> j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|}&\le \frac{\sum _{i> j}(|z_2/z_1|)^i\left( |a_iz_1^i|+\cdots +|a_nz_1^n|\right) }{\sum _i |a_i z_2^i|}2\mu \\&\le \max _{i>j} \left\{ \left| {z_2}/{z_1}\right| ^i\right\} \times \frac{\sum _{i > j}\left( |a_iz_1^i|+\cdots +|a_nz_1^n|\right) }{\sum _i |a_i z_2^i|}2\mu .\\ \end{aligned} \end{aligned}$$
When \(|z_2/z_1|\le 1\), we have
$$\begin{aligned} \max _{i>j} \left\{ |z_2/z_1|^i\right\} \le |z_2/z_1|^j \end{aligned}$$
and
$$\begin{aligned} \frac{\sum _{i > j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|} \le \frac{\sum _{i}|a_i z_1^i|}{|a_jz_1^j|} \times \frac{|a_jz_2^j|}{\sum _i |a_i z_2^i|}2\mu . \end{aligned}$$
When \(|z_2/z_1|\ge 1\), we have
$$\begin{aligned} \begin{aligned} \frac{\sum _{i> j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|}&\le \frac{\sum _{i> j}\left( |a_iz_2^i|+\cdots +|a_nz_2^n||z_1/z_2|^{n-i}\right) }{\sum _i |a_i z_2^i|}2\mu \\&\le \frac{\sum _{i > j}\left( |a_iz_2^i|+\cdots +|a_nz_2^n|\right) }{\sum _i |a_i z_2^i|} 2\mu \\&\le 2\mu . \end{aligned} \end{aligned}$$
Therefore,
$$\begin{aligned} \begin{aligned} \frac{\sum _{i > j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|}&\le \max \left\{ 1,\frac{\sum _{i}|a_i z_1^i|}{|a_jz_1^j|} \times \frac{|a_jz_2^j|}{\sum _i |a_i z_2^i|}\right\} 2\mu . \end{aligned} \end{aligned}$$
Similarly,
$$\begin{aligned} \frac{\sum _{i < j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|} \le \max \left\{ 1,\frac{\sum _{i}|a_i z_1^i|}{|a_jz_1^j|} \times \frac{|a_jz_2^j|}{\sum _i |a_i z_2^i|}\right\} 2\mu . \end{aligned}$$
It follows that
$$\begin{aligned} \frac{\sum _{i \ne j}\left| z_2^i\right| |r_i|}{\sum _i |a_i z_2^i|} \le \max \left\{ 1,\left( \frac{\sum _i|a_i z_1^i|}{|a_jz_1^i|} \times \frac{|a_jz_2^i|}{\sum _i |a_i z_2^i|}\right) \right\} 4\mu . \end{aligned}$$
We can similarly analyze the second part in (A.1) and get the following estimation
$$\begin{aligned} \begin{aligned} \frac{\sum _{i \ne j}\left| \left( \frac{z_2}{z_1}\right) ^jz_1^{i} \right| \times |r_i|}{\sum _i |a_i z_2^i|}&\le \max \left\{ 1,\left( \frac{\sum _i|a_i z_1^i|}{|a_jz_1^i|} \times \frac{|a_jz_2^i|}{\sum _i |a_i z_2^i|}\right) \right\} 4\mu . \end{aligned} \end{aligned}$$
Summing up those formulas above, we have
$$\begin{aligned} \frac{\sum _{i \not = j}\left| z_2^i-\left( \frac{z_2}{z_1}\right) ^jz_1^{i} \right| \times |r_i|}{\sum _i |a_i z_2^i|}\le \max \left\{ 1, \frac{\sum _i|a_i z_1^i|}{|a_jz_1^j|} \times \frac{|a_jz_2^j|}{\sum _i |a_i z_2^i|} \right\} 8 \mu . \end{aligned}$$
\(\square \)
Appendix B: Some polynomial-related estimations
Let
$$\begin{aligned} {} p(x)=\sum _{i=0}^n a_ix^i=a_n(x-x_n)(x-x_{n-1})\ldots (x-x_1). \end{aligned}$$
(B.1)
Comparing the both sides in (B.1), the coefficients of p(x) satisfy
$$\begin{aligned} |a_i|=\left| \sum _{1\le l_{n-i}<\cdots <l_{1}\le n} a_nx_{l_{1}}x_{l_{2}}\cdots x_{l_{n-i}}\right| . \end{aligned}$$
(B.2)
The following proposition estimates the polynomial coefficients.
Proposition B.1
Let \(|x_1|\le |x_2|\le \cdots \le |x_n|\) be the roots of p(x). For a large K, if \(K|x_r|\le |x_{r+1}|,\) then
$$\begin{aligned} |a_r| = \left| a_n\prod \nolimits _{r+1\le i\le n}x_i\right| (1+\delta _r), \end{aligned}$$
where \(\delta _r\) depends on K, n, r, and \(\delta _r\rightarrow 0\) as \(K\rightarrow \infty .\)
Proof
According to the expression of \(a_r\) in (B.2), we have
$$\begin{aligned} |a_r| =\left| a_n\prod \nolimits _{r+1\le i\le n}x_i\right| (1+\delta _r), \end{aligned}$$
where
$$\begin{aligned} \delta _r=\frac{\left| \sum _{1\le l_{n-r}<\cdots <l_{1}\le n} x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-r}}\right| -\left| \prod \nolimits _{r+1\le i\le n}x_i\right| }{\left| \prod \nolimits _{r+1\le i\le n}x_i\right| }. \end{aligned}$$
Since \( K|x_r|\le |x_{r+1}|,\) we have
$$\begin{aligned} \left| \prod \nolimits _{r+1\le i\le n}x_i\right| \ge K |x_r|\left| \prod \nolimits _{r+2\le i\le n}x_i\right| . \end{aligned}$$
(B.3)
So are the other terms in the expression of \(|a_r|\). Therefore
$$\begin{aligned} \begin{aligned} |\delta _r|\le \sum _{1\le l_{n-r}<\cdots <l_{1}\le n} \left| \frac{x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-r}}}{\prod \nolimits _{r+1\le i\le n}x_i} \right| -1\le \mathscr {O}\left( \frac{C_n^{n-r}}{K}\right) . \end{aligned} \end{aligned}$$
Therefore, \(\delta _r\) depends on K, n, r, and \(\delta _r\rightarrow 0\) as \(K\rightarrow \infty .\)\(\square \)
The following proposition estimates the absolute sum of a polynomial at some point.
Proposition B.2
Let \(|x_1|\le |x_2|\le \cdots \le |x_n|\) be the roots of p(x). If \(K|x_{r-1}|\le |x_r|\le |x_{r+1}|/K,\) then
$$\begin{aligned} |a_rx_r^r|=|a_{r-1}x_r^{r-1}|(1+\delta _1), \end{aligned}$$
and
$$\begin{aligned} \sum _{i=0}^n |a_ix_r^i|= (|a_rx_r^r|+|a_{r-1}x_r^{r-1}|)(1+\delta _2), \end{aligned}$$
where \(\delta _1\rightarrow 0 \) and \( \delta _2\rightarrow 0\) as \(K\rightarrow \infty .\) For \(r=1\) or \(r=n\), these still hold if \(|x_1|\le |x_2|/K\) or \(K|x_{n-1}|\le |x_n|\) respectively.
Proof
Due to Proposition B.1, since \(K|x_{r-1}|\le |x_r|\le |x_{r+1}|/K,\) we have
$$\begin{aligned} |a_r| = \left| a_n\prod \nolimits _{r+1\le i\le n}x_i\right| (1+\delta _r), \end{aligned}$$
and
$$\begin{aligned} |a_{r-1}| =\left| a_n\prod \nolimits _{r\le i\le n}x_i\right| (1+\delta _{r-1}), \end{aligned}$$
where \(\delta _r\rightarrow 0\) and \(\delta _{r-1}\rightarrow 0 \) as \(K\rightarrow \infty .\) Therefore,
$$\begin{aligned} \left| a_rx_r^r\right| =\frac{|1+\delta _{r-1}|}{|1+\delta _r|}\left| a_{r-1}x_r^{r-1}\right| = |a_{r-1}x_r^{r-1}|(1+\delta _1), \end{aligned}$$
where \(\delta _1\rightarrow 0\) as \(K\rightarrow \infty .\)
When \(i>r,\) we have
$$\begin{aligned} \begin{aligned} |a_ix_r^i|&=\left| \sum \nolimits _{1\le l_{n-i}<\cdots <l_{1}\le n} a_nx_{l_{1}}x_{l_{2}}\cdots x_{l_{n-i}}x_r^i\right| \\&\le C_{n}^{n-i} \frac{1}{K^{i-r}} \left| a_nx_{n}x_{n-1}\cdots x_{r+1}x_r^r\right| \\&\le C_{n}^{n-i} \frac{1}{K^{i-r}} |a_rx_r^r|(1+\delta _r)\\&=|a_rx_r^r|\hat{\delta }_i, \end{aligned} \end{aligned}$$
(B.4)
where \(\hat{\delta }_i \rightarrow 0\) as \(K\rightarrow \infty .\) Similarly, when \(i<r-1,\) we have \(|a_ix_r^i|= |a_rx_r^r|\hat{\delta }_i \), where \(\hat{\delta }_i \rightarrow 0\) as \(K\rightarrow \infty .\) It leads to
$$\begin{aligned} \sum _{i=0}^n |a_ix_r^i| = \left( |a_rx_r^r|+|a_{r-1}x_r^{r-1}|\right) (1+\delta _2), \end{aligned}$$
where \(\delta _2\rightarrow 0\) as \(K\rightarrow \infty .\)
We have a similar analysis for \(r=1\) or \(r=n\) and we omit the details. \(\square \)
Appendix C: Performance of composite deflation for general cases
Theorem C.1
Let \(|x_1|\le |x_2|\le \cdots \le |x_n|\) be the roots of a polynomial and \(\hat{x}_i=x_i(1+\gamma _i)\) be the approximation of \(x_i\) for all i. If \(|x_r|\ge K|x_k|\) or \(|x_r|\le |x_k|/K\) for a large K and choose \(j={\arg \max }_{i=0,1,\ldots ,n} |a_i\hat{x}_r^i|\) to deflate \(\hat{x}_r\), then for \(k>r,\)
$$\begin{aligned} \begin{aligned} \left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j \times \frac{\sum _i |a_i \hat{x}_r^i|}{\sum _i|a_i \hat{x}_k^i|} \le \gamma \beta \times \frac{\left| \sum _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-j}}\right| }{\sum _{m={j}}^{n}\left| \sum _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| },\\ \end{aligned} \end{aligned}$$
where \(\gamma \rightarrow 1\) as \(\gamma _r,\gamma _k\rightarrow 0\) and \(\beta \rightarrow n+1\) as \(K \rightarrow \infty . \)
For \(k<r,\)
$$\begin{aligned} \begin{aligned} \left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j \times \frac{\sum _i |a_i \hat{x}_r^i|}{\sum _i|a_i \hat{x}_k^i|} \le \hat{\gamma }\hat{\beta } \times \frac{\left| \sum _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-j}}\right| }{\sum _{m=0}^{j}\left| \sum _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| },\\ \end{aligned} \end{aligned}$$
where \(\hat{\gamma } \rightarrow 1\) as \(\gamma _r,\gamma _k\rightarrow 0\) and \(\hat{\beta }\rightarrow n+1\) as \(K \rightarrow \infty . \)
Proof
When \(k>r,\) since \(|x_r|\le |x_k|\), we have \(|x_r|\le |x_k|/K\). Considering that \( j={\arg \max }_{i=0,1,\ldots ,n} |a_i\hat{x}_r^i| \), when \(i<j\), we have
$$\begin{aligned} \begin{aligned} |a_i\hat{x}_k^i|&=|a_i\hat{x}_r^i||\hat{x}_k/\hat{x}_r|^i \\&\le |a_j\hat{x}_r^j||\hat{x}_k/\hat{x}_r|^i\\&= |a_j\hat{x}_r^j||\hat{x}_k/\hat{x}_r|^j|\hat{x}_k/\hat{x}_r|^{i-j}\\&\le \left| \frac{1+\gamma _k}{1+\gamma _r}\right| ^{i-j} \frac{1}{K^{j-i}}|a_j\hat{x}_k^j|. \end{aligned} \end{aligned}$$
It leads to
$$\begin{aligned} \sum _{i=0}^n |a_i\hat{x}_k^i|= (1+\delta _1)\sum _{i=j}^{n}|a_i\hat{x}_{k}^i|, \end{aligned}$$
where \(\delta _1\rightarrow 0\) as \(K\rightarrow \infty .\) Therefore,
$$\begin{aligned} \begin{aligned} \left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j \times \frac{\sum _i |a_i \hat{x}_r^i|}{\sum _i|a_i \hat{x}_k^i|}&=\left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j \times \frac{\sum _{i=0}^{n}|a_i \hat{x}_r^i|}{(1+\delta _1) \sum _{m={j}}^{n}|a_m\hat{x}_{k}^m|}\\&=\frac{1}{1+\delta _1}\left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j \times \frac{\sum _{i=0}^{n}|a_i \hat{x}_r^i|}{|a_j\hat{x}_r^j|}\times \frac{|a_j\hat{x}_r^j|}{ \sum _{m={j}}^{n}|a_m\hat{x}_{k}^m|}\\&\le \frac{n+1}{1+\delta _1} \times \left| \frac{\hat{x}_k}{\hat{x}_r}\right| ^j\times \frac{|a_j\hat{x}_r^j|}{ \sum _{m={j}}^{n}|a_m\hat{x}_{k}^m|}\\&\le \frac{n+1}{1+\delta _1} \frac{(1+|\gamma _k|)^{j}}{(1-|\gamma _r|)^{j}} \times \left| \frac{{x}_k}{{x}_r}\right| ^j\times \frac{(1+|\gamma _r|)^{n}}{(1-|\gamma _k|)^{n}}\frac{|a_jx_r^j|}{ \sum _{m={j}}^{n}|a_m{x}_{k}^m|}\\&= \gamma \beta \times \frac{\left| \sum _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-j}}\right| }{\sum _{m={j}}^{n}\left| \sum _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| },\\ \end{aligned} \end{aligned}$$
where
$$\begin{aligned} \gamma =\frac{(1+|\gamma _k|)^{j}}{(1-|\gamma _r|)^{j}}\frac{(1+|\gamma _r|)^{n}}{(1-|\gamma _k|)^{n}}\rightarrow 1 \ \text {as}\ \gamma _r,\gamma _k \rightarrow 0, \end{aligned}$$
and
$$\begin{aligned} \beta =\frac{n+1}{1+\delta _1}\rightarrow n+1 \ \text {as}\ K\rightarrow \infty . \end{aligned}$$
When \(k<r\), we can analyze it similarly and we omit the details. \(\square \)
It should be mentioned that
$$\begin{aligned} \frac{|x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-j}}|}{|x_{\hat{l}_{1}}x_{\hat{l}_{2}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}|}=\frac{|x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-m}}|}{|x_{\hat{l}_{1}}x_{\hat{l}_{2}}\cdots x_{\hat{l}_{n-m}}|}\frac{|x_{l_{n-m+1}}\cdots x_{l_{n-j}}|}{|{x}_{k}^{m-j}|} \end{aligned}$$
can be much smaller than 1. However, it is hard to precisely say how small it is in this general case. But for the following cases, we can.
-
Case 1:
\(K_1|x_p|\le |x_{p+1}|\) for a large \(K_1\).
When \(r\le p\) and \(k\ge p+1\), j can be estimated.
If \(i>p,\) similarly to (B.4), we have
$$\begin{aligned} \begin{aligned} |a_i\hat{x}_r^i|=|a_p\hat{x}_r^p|(1+\hat{\delta }_{ip}), \ \text {where} \ \hat{\delta }_{ip}\rightarrow 0 \ \text {as} \ K_1\rightarrow \infty . \end{aligned} \end{aligned}$$
Hence that
$$\begin{aligned} j={\arg \max }_{i=0,1,\ldots ,n} |a_i\hat{x}_r^i| \le p. \end{aligned}$$
Similarly to (B.3), we have
$$\begin{aligned} \left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}x_{l_{2}}\cdots x_{l_{n-j}}\right| =(1+{\tilde{\delta }}_1) \left| \sum \limits _{1\le l_{p-j}<\cdots <l_{1}\le p} x_{n}\cdots x_{p+1}x_{l_{1}}\cdots x_{l_{p-j}}\right| , \end{aligned}$$
and
$$\begin{aligned} \sum _{m=j}^{n}\left| \sum _{1\le \hat{l}_{n-m}<\cdots<\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| =(1+{\tilde{\delta }}_2)\sum _{m=p}^{n}\left| \sum _{p+1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| , \end{aligned}$$
where \({\tilde{\delta }}_1, {\tilde{\delta }}_2\rightarrow 0\) as \(K_1\rightarrow \infty .\) Then
$$\begin{aligned} \begin{aligned} \frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}\cdots x_{l_{n-j}}\right| }{\sum \limits _{m={j}}^{n}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots<\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }&=\frac{1+{\tilde{\delta }}_1}{1+{\tilde{\delta }}_2} \frac{ \left| \sum \limits _{1\le l_{p-j}<\cdots<l_{1}\le p} x_{n}\cdots x_{p+1}x_{l_{1}}\cdots x_{l_{p-j}}\right| }{\sum \limits _{m=p}^{n}\left| \sum \limits _{p+1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| } \end{aligned} \end{aligned}$$
can be much smaller than 1 since that
$$\begin{aligned} \left| \frac{x_{n}x_{n-1}\ldots x_{p+1}x_{l_{1}}x_{l_{2}}\ldots x_{l_{p-j}}}{x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}}\right| = \left| \frac{x_{n}x_{n-1}\ldots x_{p+1}}{x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}x_k^{m-p}}\right| \left| \frac{x_{l_{1}}x_{l_{2}}\ldots x_{l_{p-j}}}{x_k^{p-j}}\right| \end{aligned}$$
can be much smaller than 1. Especially, when \(\{x_i\}_{i=p+1}^n\) are clustered, it has order of
$$\begin{aligned} \left| \frac{x_{l_{1}}x_{l_{2}}\ldots x_{l_{p-j}}}{x_k^{p-j}}\right| \le \frac{1}{K_1^{p-j}} \end{aligned}$$
since \(|x_{l_{i}}|\le |x_k|/K_1,\forall i\in \{1,\ldots , p-j\} .\)
When \(|x_r|\ge |x_{p+1}|\) and \(|x_k|\le |x_{p}|\), similarly, we have \(j\ge p\), and
$$\begin{aligned} \begin{aligned}&\frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}\ldots x_{l_{n-j}}\right| }{\sum \limits _{m=0}^{j}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots<\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }\\&\quad =\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2} \frac{ \left| \sum \limits _{1\le l_{p-j}<\cdots<l_{1}\le p} x_{n}\ldots x_{p+1}x_{l_{1}}\ldots x_{l_{p-j}}\right| }{\sum \limits _{m=0}^{p}\left| \sum \limits _{1\le \hat{l}_{p-m}<\cdots<\hat{l}_{1}\le p}x_{n}\ldots x_{p+1}x_{\hat{l}_{1}}\ldots x_{\hat{l}_{p-m}}{x}_{k}^{m-j}\right| }\\&\quad =\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2}\frac{ \left| \sum \limits _{1\le l_{p-j}<\cdots<l_{1}\le p} x_{l_{1}}x_{l_{2}}\ldots x_{l_{p-j}}\right| }{\sum \limits _{m=0}^{p}\left| \sum \limits _{1\le \hat{l}_{p-m}<\cdots <\hat{l}_{1}\le p}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{p-m}}{x}_{k}^{m-j}\right| },\\ \end{aligned} \end{aligned}$$
where \(\hat{\delta }_1,\hat{\delta }_2\rightarrow 0\) as \(K_1\rightarrow \infty .\)
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Case 2:
\(K_1|x_{p_1}|\le |x_{p_1+1}|\) and \(K_2|x_{p_2}|\le |x_{p_2+1}|\) for large \(K_1\) and \(K_2\), assuming \(p_1<p_2\).
Assume \(p_1+1\le r\le p_2.\) Similarly to the analysis in case 1, we have
$$\begin{aligned} j\in \{p_1,p_1+1,\ldots ,p_2\}. \end{aligned}$$
When \(p_2+1\le k\le n\), we have
$$\begin{aligned} \begin{aligned} \frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}\ldots x_{l_{n-j}}\right| }{\sum \limits _{m={j}}^{n}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots<\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }&=\frac{1+{\tilde{\delta }}_1}{1+{\tilde{\delta }}_2} \frac{ \left| \sum \limits _{p_1+1\le l_{p_2-j}<\cdots<l_{1}\le p_2} x_{n}\cdots x_{p_2+1}x_{l_{1}}\cdots x_{l_{p_2-j}}\right| }{\sum \limits _{m=p_2}^{n}\left| \sum \limits _{p_2+1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}\cdots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }, \end{aligned} \end{aligned}$$
where \({\tilde{\delta }}_1,{\tilde{\delta }}_2\rightarrow 0\) as \(K_1,K_2\rightarrow \infty .\)
It can be much smaller than 1 since that
$$\begin{aligned} \left| \frac{x_{n}x_{n-1}\ldots x_{p_2+1}x_{l_{1}}x_{l_{2}}\ldots x_{l_{p_2-j}}}{x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}}\right| =\left| \frac{x_{n}x_{n-1}\ldots x_{p_2+1}}{x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}x_k^{m-p_2}}\right| \left| \frac{x_{l_{1}}x_{l_{2}}\ldots x_{l_{p_2-j}}}{x_k^{p_2-j}}\right| , \end{aligned}$$
where \(K_2|x|\le |x_k|\) for all \(x\in \{x_{l_1},\ldots , x_{l_{p_2-j}}\}\).
When \(1\le k\le p_1,\) we have
$$\begin{aligned} \begin{aligned}&\frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}\ldots x_{l_{n-j}}\right| }{\sum \limits _{m=0}^{j}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots<\hat{l}_{1}\le n}x_{\hat{l}_{1}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }\\&\quad =\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2} \frac{ \left| \sum \limits _{p_1+1\le l_{p-j}<\cdots<l_{1}\le p_2} x_{n}\ldots x_{p_2+1}x_{l_{1}}\ldots x_{l_{p_2-j}}\right| }{\sum \limits _{m=0}^{p_1}\left| \sum \limits _{1\le \hat{l}_{p_1-m}<\cdots<\hat{l}_{1}\le p_1}x_{n}\ldots x_{p_1+1}x_{\hat{l}_{1}}\ldots x_{\hat{l}_{p_1-m}}{x}_{k}^{m-j}\right| }\\&\quad =\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2} \frac{ \left| \sum \limits _{p_1+1\le l_{p_2-j}<\cdots<l_{1}\le p_2} x_{l_{1}}x_{l_{2}}\ldots x_{l_{p_2-j}}\right| }{\sum \limits _{m=0}^{p_1}\left| \sum \limits _{1\le \hat{l}_{p-m}<\cdots <\hat{l}_{1}\le p_1}x_{p_2}\ldots x_{p_1+1}x_{\hat{l}_{1}}\ldots x_{\hat{l}_{p_1-m}}{x}_{k}^{m-j}\right| }, \end{aligned} \end{aligned}$$
where \(\hat{\delta }_1,\hat{\delta }_2\rightarrow 0\) as \(K_1,K_2\rightarrow \infty .\) It can be much smaller than 1 since
$$\begin{aligned} \left| \frac{x_{l_{1}}x_{l_{2}}\ldots x_{l_{p_2-j}}}{x_{p_2}\ldots x_{p_1+1}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{p_1-m}}{x}_{k}^{m-j}}\right| =\left| \frac{x_k^{p_1-m}}{x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{p_1-m}}}\right| \left| \frac{x_k^{j-p_1}}{\varPi _{l\in {\{p_1+1,p_1+2,\ldots ,p_2\}\setminus \{l_1,\ldots ,l_{p_2-j}\}}}x_l}\right| , \end{aligned}$$
where \(|x_l| \ge K_1 |x_k|\) for all \({l\in {\{p_1+1,\ldots ,p_2\}\setminus \{l_1,\ldots ,l_{p_2-j}\}}}\).
We can have a similar analysis for \(r\le p_1\) and \(r\ge p_2+1\). We omit the details.
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Case 3:
\(K_1|x_{p_1}|\le |x_{p_1+1}|\), \(K_2|x_{p_2}|\le |x_{p_2+1}|\), \(\dots \), \(K_i|x_{p_i}|\le |x_{p_i+1}|\) for large \(\{K_l\}_{l=1}^{i} \), assuming \(p_1<p_2<\cdots <p_i\). In this case, when \(p_j+1\le r\le p_{j+1},\) choose
$$\begin{aligned} j\in \{p_j,p_j+1,\ldots ,p_{j+1}\} \end{aligned}$$
to deflate \(\hat{x}_r\). We can have a similar estimation to case 2 and we omit the details.
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Case 4:
\(K_1|x_{1}|\le |x_{2}|\), \(K_2|x_{2}|\le |x_{3}|\), \(\dots \), \(K_{n-1}|x_{n-1}|\le |x_{n}|\) for large \(\{K_l\}_{l=1}^{n-1} \).
For convenience, we denote \(K_0=K_n=\infty .\)
In this case, when deflating \(\hat{x}_r\), we have
$$\begin{aligned} j \in \{r-1,r\}. \end{aligned}$$
When \(k>r,\) similarly to (B.3),
$$\begin{aligned} \left| \sum \limits _{1\le l_{n-j}<\cdots <l_{1}\le n}x_{l_{1}}\ldots x_{l_{n-j}}\right| =(1+{\tilde{\delta }}_1)\left| x_{n}x_{n-1}\ldots x_{j+1}\right| , \end{aligned}$$
where \({\tilde{\delta }}_1 \rightarrow 0\) as \(K_j\rightarrow \infty .\) According to Proposition B.2,
$$\begin{aligned} \sum \limits _{m=j}^{n}\left| \sum _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| =2\left| x_{n}x_{n-1}\ldots x_{k+1}x_{k}^{k-j}\right| (1+{\tilde{\delta }}_2), \end{aligned}$$
and \({\tilde{\delta }}_2\rightarrow 0\) as \(K_{k-1},K_{k}\rightarrow \infty .\)
Then we have
$$\begin{aligned} \begin{aligned} \frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}\ldots x_{l_{n-j}}\right| }{\sum \limits _{m=j}^{n}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }&=\frac{1+{\tilde{\delta }}_1}{1+{\tilde{\delta }}_2}\frac{\left| x_{n}x_{n-1}\ldots x_{j+1}\right| }{2\left| x_{n}x_{n-1}\ldots x_{k+1}x_{k}^{k-j}\right| }\\&=\frac{1+{\tilde{\delta }}_1}{1+{\tilde{\delta }}_2}\left\{ \begin{array}{lll} \frac{1}{2} \left| \frac{x_{k-1}}{x_{k}}\right| \ldots \left| \frac{x_{r+1}}{x_{k}}\right| , &{} j=r,\\ \frac{1}{2} \left| \frac{x_{k-1}}{x_{k}}\right| \ldots \left| \frac{x_{r}}{x_{k}}\right| , &{} j=r-1, \end{array}\right. \end{aligned} \end{aligned}$$
where \({\tilde{\delta }}_1 \rightarrow 0\) as \(K_j\rightarrow \infty \) and \({\tilde{\delta }}_2\rightarrow 0\) as \(K_{k-1},K_{k}\rightarrow \infty .\)
Similarly, when \(k<r\), we have
$$\begin{aligned} \begin{aligned} \frac{\left| \sum \limits _{1\le l_{n-j}<\cdots<l_{1}\le n}x_{l_{1}}x_{l_{2}}\ldots x_{l_{n-j}}\right| }{\sum \limits _{m=0}^{j}\left| \sum \limits _{1\le \hat{l}_{n-m}<\cdots <\hat{l}_{1}\le n}x_{\hat{l}_{1}}x_{\hat{l}_{2}}\ldots x_{\hat{l}_{n-m}}{x}_{k}^{m-j}\right| }&=\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2}\frac{\left| x_{n}x_{n-1}\ldots x_{j+1}\right| }{2\left| x_{n}x_{n-1}\ldots x_{k+1}x_{k}^{k-j}\right| }\\&=\frac{1+\hat{\delta }_1}{1+\hat{\delta }_2}\left\{ \begin{array}{lll} \frac{1}{2} \left| \frac{x_{k}}{x_{r}}\right| \ldots \left| \frac{x_{k}}{x_{k+1}}\right| , &{} j=r,\\ \frac{1}{2} \left| \frac{x_{k}}{x_{r-1}}\right| \ldots \left| \frac{x_{k}}{x_{k+1}}\right| , &{} j=r-1, \end{array}\right. \end{aligned} \end{aligned}$$
where \(\hat{\delta }_1\rightarrow 0\) as \(K_j\rightarrow \infty \) and \(\hat{\delta }_2\rightarrow 0\) as \( K_{k-1},K_{k} \rightarrow \infty \).
All the cases above show that composite deflation can lead to improved the backward stability of the remaining approximate roots when they vary widely in order of magnitude.
