Algebraic phase unwrapping along the real axis: extensions and stabilizations

Abstract

The unwrapped phase of a complex function is defined with a line integral of the gradient of the arctangent of the ratio of the real and imaginary parts of the function. The phase unwrapping, which is a problem to reconstruct the unwrapped phase of an unknown complex function from its finite observed samples, has been a key for estimating useful physical quantity in many signal and image processing applications. In the light of the functional data analysis, it is natural to estimate first the unknown complex function by a certain piecewise complex polynomial and then to compute the exact unwrapped phase of the piecewise complex polynomial with the algebraic phase unwrapping algorithms (Yamada et al. in IEEE Trans Signal Process 46(6), 1639–1664, 1998; Yamada and Bose in IEEE Trans Circuits Syst I Fundam Theory Appl 49(3), 298–304, 2002; Yamada and Oguchi in Multidimens Syst Signal Process 22(1–3), 191–211, 2011). In this paper, we propose several useful extensions and numerical stabilizations of the algebraic phase unwrapping along the real axis which was established originally in Yamada and Oguchi (Multidimens Syst Signal Process 22(1–3), 191–211, 2011). The proposed extensions include (i) removal of a certain critical assumption premised in the original algebraic phase unwrapping, and (ii) algebraic phase unwrapping for a pair of bivariate polynomials. Moreover, in order to resolve certain numerical instabilities caused by the coefficient growth in an inductive step in the original algorithm, we propose to compute directly a certain subresultant sequence without passing through the inductive step. The extensive numerical experiments exemplify the notable improvement, in the performance of the algebraic phase unwrapping, made by the proposed numerical stabilization.

Appendices

Appendices

1.1 Appendix 1: On the expression and the integrability of $\theta _{C}$

Without loss of generality, we can assume $G(t):=\mathrm{GCD}(C(t),\overline{C}(t))\in \mathbb{R }[t]$ and $B(t)\ne 0$ for all $t\in \mathbb{R }$. Let $\mathcal{Z }_{C}:=\{t\in \mathbb{R }\mid C(t)=0\}$. Then by $C(t)=G(t)B(t)$, it follows that

$$\begin{aligned} \frac{C^{\prime }(t)}{C(t)}=\frac{G^{\prime }(t)}{G(t)}+\frac{B^{\prime }(t)}{B(t)}\quad \text{ for } t\in \mathbb{R }\setminus \mathcal{Z }_{C}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \mathfrak{I }\left\{ \frac{C^{\prime }(t)}{C(t)}\right\} =\mathfrak{I } \left\{ \frac{B^{\prime }(t)}{B(t)}\right\} \quad \text{ for } \text{ all } t\in \mathbb{R }\setminus \mathcal{Z }_{C}, \end{aligned}$$

which, together with $|\mathcal{Z }_{C}|<\infty$, ensures (3). Furthermore, by the continuity of

$$\begin{aligned} H(t):=\mathfrak{I }\left\{ \frac{B^{\prime }(t)}{B(t)}\right\} =\frac{B^{\prime }_{(1)}(t)B_{(0)}(t)-B_{(1)}(t)B^{\prime }_{(0)}(t)}{\{B_{(0)}(t)\}^{2}+\{B_{(1)}(t)\}^{2}}, \end{aligned}$$

the integral in (4) is well-defined. $\square$

1.2 Appendix 2: Proof of Proposition 1

By

$$\begin{aligned} (\arctan \{\mathcal{Q }_{A}(t)\})^{\prime }=\frac{A^{\prime }_{(1)}(t)A_{(0)}(t)-A_{(1)}(t)A^{\prime }_{(0)}(t)}{\{A_{(0)}(t)\}^{2}+\{A_{(1)}(t)\}^{2}} \end{aligned}$$

for all $t\in (a,b)\setminus \mathcal Z _{A_{(0)}}$ and $|\mathcal{Z }_{A_{(0)}}|<\infty$, we can express the $\theta _{A}(t^{*})$ in (5) in terms of $\arctan \{\mathcal{Q }_{A}(t)\}$ as follows.

• (I) If $\mathcal{Z }_{A_{(0)}}=\varnothing$ or $t^{*}\le \mu _{1}$, we have

  $$\begin{aligned} \theta _{A}(t^{*})&= \theta _{A}(a)+\int \nolimits _{ a}^{{ t}^{*}} \left( \arctan \left\{ \mathcal{Q }_{A}(t)\right\} \right) ^{\prime }dt\\&= \theta _{A}(a)-\lim _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}+\lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}. \end{aligned}$$

  and $\varLambda (t^{*})=\sum \nolimits _{\mu _{i}\in (a,t^{*})}\mathcal{X }(\mu _{i})=0$ in (7).

• (II) If $\mathcal{Z }_{A_{(0)}}\ne \varnothing$ and $t^{*}>\mu _{1}$, by letting $\mu _{k}:=\max (\{\mu _{1},\mu _{2},\ldots ,\mu _{z}\}\cap [a,t^{*}))$, we have

  $$\begin{aligned} \theta _{A}(t^{*})&= \theta _{A}(a)+\int \nolimits _{a}^{{t}^{*}} \left( \arctan \left\{ \mathcal{Q }_{A}(t)\right\} \right) ^{\prime }dt \nonumber \\&= \theta _{A}(a)+\int \nolimits _{a}^{\mu _{1}}\left( \arctan \left\{ \mathcal{Q }_{A}(t)\right\} \right) ^{\prime }dt \nonumber \\&\quad +\sum _{i=1}^{k-1}\int \nolimits _{\mu _{i}}^{\mu _{i+1}} \left( \arctan \left\{ \mathcal{Q }_{A}(t)\right\} \right) ^{\prime }dt+\int \nolimits _{\mu _{k}}^{t^{*}} \left( \arctan \left\{ \mathcal{Q }_{A}(t)\right\} \right) ^{\prime }dt \nonumber \\&= \theta _{A}(a)-\lim _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}+\lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}\nonumber \\&\quad +\sum _{i=1}^{k}\lim _{{\begin{array}{c} {\tau _{1}\rightarrow \mu _{i}-0}\\ {\tau _{2} \rightarrow \mu _{i}+0}\end{array}}}(\arctan \{\mathcal{Q }_{A}(\tau _{1})\}-\arctan \{\mathcal{Q }_{A}(\tau _{2})\}). \end{aligned}$$

  (19)

  Furthermore, for $\mu _{i}$ $(i=1,2,\ldots ,k)$ and sufficiently small $\varepsilon >0$, we have the following relations.

  • (i) If $A_{(0)}(t)A_{(1)}(t)>0$ for $t\in (\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)<0$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, then

    $$\begin{aligned} \left\{ \begin{array}{l} \lim \nolimits _{t\rightarrow \mu _{i}-0}\arctan \{\mathcal{Q }_{A}(t)\}=\pi /2,\\ \lim \nolimits _{t\rightarrow \mu _{i}+0}\arctan \{\mathcal{Q }_{A}(t)\}=-\pi /2, \end{array}\right. \, \text{ and } \,\mathcal{X }(\mu _{i})=1. \end{aligned}$$

  • (ii) If $A_{(0)}(t)A_{(1)}(t)<0$ for $t\in (\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)>0$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, then

    $$\begin{aligned} \left\{ \begin{array}{l} \lim \nolimits _{t\rightarrow \mu _{i}-0}\arctan \{\mathcal{Q }_{A}(t)\}=-\pi /2,\\ \lim \nolimits _{t\rightarrow \mu _{i}+0}\arctan \{\mathcal{Q }_{A}(t)\}=\pi /2, \end{array}\right. \ \text{ and } \ \mathcal{X }(\mu _{i})=-1. \end{aligned}$$

  • (iii) Otherwise,

    $$\begin{aligned} \lim _{t\rightarrow \mu _{i}-0}\arctan \{\mathcal{Q }_{A}(t)\}=\displaystyle \lim _{t\rightarrow \mu _{i}+0}\arctan \{\mathcal{Q }_{A}(t)\}=\pm \pi /2, \ \text{ and } \ \mathcal{X }(\mu _{i})=0. \end{aligned}$$

  From (i), (ii) and (iii), we have

  $$\begin{aligned} \lim _{{\begin{array}{c}{\tau _{1} \rightarrow \mu _{i}-0}\\ {\tau _{2} \rightarrow \mu _{i}+0}\end{array}}}(\arctan \{\mathcal{Q }_{A}(\tau _{1})\}-\arctan \{\mathcal{Q }_{A}(\tau _{2})\})={\mathcal{X }}(\mu _{i})\pi . \end{aligned}$$

  (20)

  Finally, (19) and (20) yield (7). $\square$

1.3 Appendix 3: Proof of Lemma 1

For the readers’ convenience, we present proofs of all statements.

• (A) Proof of (a): Assume ${\Psi }_{q}(t^{*})=0$ at some $t^{*}\in [a,b]$. Since $\Psi _{q}(t)$ is $\mathrm{GCD}(\Psi _{0},\Psi _{1})$, we have $\Psi _{0}(t^{*})=\Psi _{1}(t^{*})=0$. Moreover, $\Psi _{0}(t):= {\displaystyle \frac{A_{(0)}(t)}{(t-a)^{e_{0}}}}$ and $\Psi _{1}(t):={\displaystyle \frac{A_{(1)}(t)}{(t-a)^{e_{1}}}}$ imply $\Psi _{0}(a)\ne 0$ and $\Psi _{1}(a)\ne 0$, and hence, $A_{(0)}(t^{*})=\Psi _{0}(t^{*})=\Psi _{1}(t^{*})=A_{(1)}(t^{*})=0$ at some $t^{*}\in (a,b]$. This contradicts $A(t)=A_{(0)}(t)+jA_{(1)}(t)\ne 0$ for all $t\in [a,b]$. As a result, $\Psi _{q}(t)\ne 0$ for all $t\in [a,b]$.

• (B) Proof of (b): Assume $\Psi _{k}(t^{*})=\Psi _{k+1}(t^{*})=0$ at some $t^{*}\in [a,b]$. Then $\mathrm{GCD}(\Psi _{0},\Psi _{1})\equiv \mathrm{GCD}(\Psi _{k},\Psi _{k+1})$ implies $\Psi _{0}(t^{*})=\Psi _{1}(t^{*})=0$, which contradicts $A(t)=A_{(0)}(t)+jA_{(1)}(t)\ne 0$ for all $t\in [a,b]$.

• (C) Proof of (c): Suppose $\Psi _{k}(t^{*})=0$. Then from (b), i.e., $\Psi _{k-1}(t^{*})\ne 0$ and $\Psi _{k+1}(t^{*})\ne 0$, and $\Psi _{k+1}(t^{*})=-\Psi _{k-1}(t^{*})+H_{k}(t^{*})\Psi _{k}(t^{*})=- \Psi _{k-1}(t^{*})$, we have $\Psi _{k+1}(t^{*})\Psi _{k-1}(t^{*})<0$.

• (D) Proof of (d): Since $(t-a)^{e_{i}}>0$ $(i=0,1)$ for $a<t\le b$, the proof is obvious.

• (E) Proof of (e): From $\Psi _{i}(a)\ne 0$ and the continuity of $\Psi _{i}(t)\,(i=0,1)$, we have

  $$\begin{aligned} \lim _{t \rightarrow a+0}\mathrm{sgn}(A_{(i)}(t))=\lim _{t \rightarrow a+0}\mathrm{sgn}\left( \frac{A_{(i)}(t)}{(t-a)^{e_{i}}}\right) =\lim _{t \rightarrow a+0}\mathrm{sgn}(\Psi _{i}(t))=\mathrm{sgn}(\Psi _{i}(a))\ne 0. \end{aligned}$$

  $\square$

1.4 Appendix 4: Proof of Theorem 1

We derive computable expressions for

$$\begin{aligned} \lim \nolimits _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}\quad \text{ and } \quad \lim \nolimits _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi \end{aligned}$$

in (7) as follows.

• (A) Computable expression for $\lim \nolimits _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}$ in (7):

  • (I) If $A_{(0)}(a)\ne 0$, we have  ${\lim \nolimits _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}}=\arctan \{\mathcal{Q }_{A}(a)\}$.

  • (II) If $A_{(0)}(a)=0$, then

    $$\begin{aligned} \lim _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}=\left\{ \begin{array}{ll} \pi /2&{}\ \text{ if } \,{\lim \nolimits _{t \rightarrow a+0}}\mathrm{sgn}(A_{(0)}(t)A_{(1)}(t))=1,\\ -\pi /2&{}\ \text{ if } \,{\lim \nolimits _{t \rightarrow a+0}}\mathrm{sgn}(A_{(0)}(t)A_{(1)}(t))=-1. \end{array}\right. \end{aligned}$$

    (21)

    From Lemma 1(e) and (21), $\lim \nolimits _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}$ in (7) can be expressed as

    $$\begin{aligned} \lim _{t \rightarrow a+0}\arctan \{\mathcal{Q }_{A}(t)\}=\mathrm{sgn}(\Psi _{0}(a)\Psi _{1}(a))\pi /2. \end{aligned}$$

• (B) Computable expression for $\lim \nolimits _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi$ in (7): To derive the relation between $V\{\Psi (t)\}$ and $\mathcal{X }(\mu _{i})$ $(i=1,2,\ldots ,z)$, we have to know the behavior of $V\{\Psi (t)\}$. Since the real polynomials $\Psi _{k}(t)$ $(0\le k\le q)$ are all continuous, any point where $V\{\Psi (t)\}$ changes must be in the neighborhood of a zero of some $\Psi _{k}(t)$ $(0\le k\le q)$. Let us observe the behavior of $V\{\Psi (t)\}$ in the neighborhood of a zero of $\Psi _{k}(t)$ for $0<k<q$. Suppose

  $$\begin{aligned} \Psi _{k}(\eta )=0\quad \text{ for }\,\eta \in [a,b]. \end{aligned}$$

  From Lemma 1(c) and the continuity of $\Psi _{k-1}(t)$ and $\Psi _{k+1}(t)$, there exists a sufficiently small $\varepsilon >0$ such that

  $$\begin{aligned} \Psi _{k-1}(t)\Psi _{k+1}(t)<0\quad \text{ for } \text{ all }\;t\in (\eta -\varepsilon ,\eta +\varepsilon )\cap [a,b]. \end{aligned}$$

  (22)

  From (22), all the possibilities of the sign of $(\Psi _{k-1}(t),\Psi _{k}(t),\Psi _{k+1}(t))$ in $t\in (\eta -\varepsilon ,\eta +\varepsilon )\cap [a,b]$ are $(+,\pm ,-)$, $(-,\pm ,+)$, $(+,0,-)$ or $(-,0,+)$. In all cases, the number of sign changes among $(\Psi _{k-1}(t),\Psi _{k}(t),\Psi _{k+1}(t))$ is 1. Therefore $V\{\Psi (t)\}$ does not change in the neighborhood of a zero of $\Psi _{k}(t)$ for $0<k<q$. Moreover from Lemma 1(a), any change of $V\{\Psi (t)\}$ is caused only by the sign changes of $(\Psi _{0}(t),\Psi _{1}(t))$ in the neighborhood of a zero of $\Psi _{0}(t)$. To wrap up, for any point $\xi _{i}$ $(i=0,1,\ldots ,z)$ such that

  $$\begin{aligned} a\le \xi _{0}<\mu _{1}<\xi _{1}<\mu _{2}<\xi _{2}<\cdots <\mu _{z}<\xi _{z}<b, \end{aligned}$$

  we have

  $$\begin{aligned} V\{\Psi (t)\}=\left\{ \begin{array}{ll} V\{\Psi (\xi _{0})\}&{}\ \text{ if }\,a\le t<\mu _{1},\\ V\{\Psi (\xi _{1})\}&{}\ \text{ if }\,\mu _{1}< t<\mu _{2},\\ &{}\vdots \\ V\{\Psi (\xi _{z})\}&{}\ \text{ if } \mu _{z}< t<b.\\ \end{array}\right. \end{aligned}$$

  (23)

  • (I) If $\mathcal{Z }_{A_{(0)}}=\varnothing$ or $t^{*}\in (a,\mu _{1})$, we have ${{\lim }_{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}}=\arctan \{\mathcal{Q }_{A}(t^{*})\}$ and $[V\{\Psi (t^{*})\}-V\{\Psi (a)\}]\pi =[V\{\Psi (\xi _{0})\}- V\{\Psi (\xi _{0})\}]\pi =0=\sum \nolimits _{\mu _{i}\in (a,t^{*})}\mathcal{X }(\mu _{i})$.

  • (II) If $\mathcal{Z }_{A_{(0)}}\ne \varnothing ,\mu _{1}<t^{*}<b$ and $t^{*}\ne \mu _{i}$ $(i=1,2,\ldots ,z)$, Lemma 1(b) and the continuity of $\Psi _{1}(t)$ ensure the existence of a sufficiently small $\varepsilon >0$ for $\mu _{i}$ such that

    $$\begin{aligned} \left. \begin{array}{ll} [\mu _{i}-\varepsilon ,\mu _{i}+\varepsilon ]\subset (a,b)\\ \Psi _{1}(t)\ne 0\ \text{ for } \text{ all } t\in (\mu _{i}-\varepsilon ,\mu _{i}+\varepsilon )\\ \Psi _{0}(t)\Psi _{1}(t)\ne 0\ \text{ for } \text{ all } t\in (\mu _{i}-\varepsilon ,\mu _{i})\cup (\mu _{i},\mu _{i}+\varepsilon ) \end{array}\right\} . \end{aligned}$$

    (24)

    We fix arbitrarily $\xi _{i-1}\in (\mu _{i}-\varepsilon ,\mu _{i})$ and $\xi ^{\prime }_{i}\in (\mu _{i},\mu _{i}+\varepsilon )$.

    • (i) If $A_{(0)}(t)A_{(1)}(t)\!>\!0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}\!{\Psi }_{0}(t){\Psi }_{1}(t) \!>\!0\bigr )$ for $t\!\in \!(\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)$ $<0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}{\Psi }_{0}(t){\Psi }_{1}(t)<0\big )$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, we have $\mathrm{sgn}(\Psi _{0}(\xi _{i-1}))=\mathrm{sgn}(\Psi _{1}(\xi _{i-1}))$ and $\mathrm{sgn}(\Psi _{0}(\xi ^{\prime }_{i}))=-\mathrm{sgn}(\Psi _{1}(\xi ^{\prime }_{i}))$. Then the number of sign changes of $(\Psi _{0}(\xi _{i-1}),\Psi _{1}(\xi _{i-1}))$ is $0$, and that of $(\Psi _{0}(\xi ^{\prime }_{i}),\Psi _{1}(\xi ^{\prime }_{i}))$ is $1$. Moreover by (23), we have $V\{\Psi (\xi ^{\prime }_{i})\}=V\{{\Psi }(\xi _{i})\}$. Hence, we have

      $$\begin{aligned} V\{\Psi (\xi _{i})\}-V\{{\Psi }(\xi _{i-1})\}=V\{\Psi (\xi ^{\prime }_{i})\}-V\{{\Psi }(\xi _{i-1})\}=1 \ \text{ and } \ \mathcal{X }(\mu _{i})=1. \end{aligned}$$

    • (ii) If $A_{(0)}(t)A_{(1)}(t)\!<\!0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}\!{\Psi }_{0}(t){\Psi }_{1}(t)\!<\!0\bigr )$ for $t\!\in \!(\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)$ $>0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }} {\Psi }_{0}(t){\Psi }_{1}(t)>0\big )$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, we have $\mathrm{sgn}(\Psi _{0}(\xi _{i-1}))=-\mathrm{sgn}(\Psi _{1}(\xi _{i-1}))$ and $\mathrm{sgn}(\Psi _{0}(\xi ^{\prime }_{i}))=\mathrm{sgn}(\Psi _{1}(\xi ^{\prime }_{i}))$. Then the number of sign changes of $(\Psi _{0}(\xi _{i-1}),\Psi _{1}(\xi _{i-1}))$ is $1$, and that of $(\Psi _{0}(\xi ^{\prime }_{i}),\Psi _{1}(\xi ^{\prime }_{i}))$ is $0$. Moreover by (23), we have $V\{\Psi (\xi ^{\prime }_{i})\}=V\{{\Psi }(\xi _{i})\}$. Hence, we have

      $$\begin{aligned} V\{\Psi (\xi _{i})\}-V\{\Psi (\xi _{i-1})\}\!=\!V\{\Psi (\xi ^{\prime }_{i})\}-V\{{\Psi }(\xi _{i-1})\}\!=\!-1 \quad \text{ and } \,\mathcal{X }(\mu _{i})\!=\!-1. \end{aligned}$$

    • (iii) If $A_{(0)}(t)A_{(1)}(t)\!>\!0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}\!{\Psi }_{0}(t){\Psi }_{1}(t)\!>\!0\bigr )$ for $t\!\in \!(\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)$ $>0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}{\Psi }_{0}(t){\Psi }_{1}(t)>0\big )$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, we have $\mathrm{sgn}(\Psi _{0}(\xi _{i-1}))=\mathrm{sgn}(\Psi _{1}(\xi _{i-1}))$ and $\mathrm{sgn}(\Psi _{0}(\xi ^{\prime }_{i}))=\mathrm{sgn}(\Psi _{1}(\xi ^{\prime }_{i}))$. Then the number of sign changes of $(\Psi _{0}(\xi _{i-1}),\Psi _{1}(\xi _{i-1}))$ is $0$, and that of $(\Psi _{0}(\xi ^{\prime }_{i}),\Psi _{1}(\xi ^{\prime }_{i}))$ is $0$. Moreover by (23), we have $V\{\Psi (\xi ^{\prime }_{i})\}=V\{{\Psi }(\xi _{i})\}$. Hence, we have

      $$\begin{aligned} V\{\Psi (\xi _{i})\}-V\{\Psi (\xi _{i-1})\}=V\{\Psi (\xi ^{\prime }_{i})\}-V\{{\Psi }(\xi _{i-1})\}=0 \ \text{ and } \ \mathcal{X }(\mu _{i})=0. \end{aligned}$$

    • (iv) If $A_{(0)}(t)A_{(1)}(t)\!<\!0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}\!{\Psi }_{0}(t){\Psi }_{1}(t)\!<\!0\bigr )$ for $t\!\in \!(\mu _{i}-\varepsilon ,\mu _{i})$ and $A_{(0)}(t)A_{(1)}(t)$ $<0\bigl ({\mathop {\Longleftrightarrow }\limits ^\mathrm{Lemma~1(d) }}{\Psi }_{0}(t){\Psi }_{1}(t)<0\big )$ for $t\in (\mu _{i},\mu _{i}+\varepsilon )$, we have $\mathrm{sgn}(\Psi _{0}(\xi _{i-1}))=-\mathrm{sgn}(\Psi _{1}(\xi _{i-1}))$ and $\mathrm{sgn}(\Psi _{0}(\xi ^{\prime }_{i}))=-\mathrm{sgn}(\Psi _{1}(\xi ^{\prime }_{i}))$. Then the number of sign changes of $(\Psi _{0}(\xi _{i-1}),\Psi _{1}(\xi _{i-1}))$ is $1$, and that of $(\Psi _{0}(\xi ^{\prime }_{i}),\Psi _{1}(\xi ^{\prime }_{i}))$ is $1$. Moreover by (23), we have $V\{\Psi (\xi ^{\prime }_{i})\}=V\{{\Psi }(\xi _{i})\}$. Hence, we have

      $$\begin{aligned} V\{\Psi (\xi _{i})\}-V\{\Psi (\xi _{i-1})\}= V\{\Psi (\xi ^{\prime }_{i})\}-V\{{\Psi }(\xi _{i-1})\}=0 \, \text{ and } \, \mathcal{X }(\mu _{i})=0. \end{aligned}$$

    As a result, in all cases (i), (ii), (iii) and (iv), we have

    $$\begin{aligned} V\{\Psi (\xi _{i})\}-V\{\Psi (\xi _{i-1})\}=\mathcal{X }(\mu _{i}). \end{aligned}$$

    Finally, by letting $\mu _{k}:=\max (\{\mu _{1},\mu _{2},\ldots ,\mu _{z}\}\cap [a,t^{*}))$ in the definition of $\varLambda (t^{*})$, we have, from $A_{(0)}(t^{*})\ne 0$ and (23),

    $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi&\!= \arctan \{\mathcal{Q }_{A}(t^{*})\}+\sum _{i=1}^{k}\mathcal{X }(\mu _{i})\pi \\&\!=\!&\arctan \{\mathcal{Q }_{A}(t^{*})\}\!+\!\sum _{i=1}^{k}[V\{\Psi (\xi _{i})\}\!-\! V\{\Psi (\xi _{i-1})\}]\pi \\&\!=\!&\arctan \{\mathcal{Q }_{A}(t^{*})\}+ [V\{\Psi (\xi _{k})\}-V\{\Psi (\xi _{0})\}]\pi \\&\!=\!&\arctan \{\mathcal{Q }_{A}(t^{*})\} +[V\{\Psi (t^{*})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

  • (III) If $\mathcal{Z }_{A_{(0)}}\!\ne \!\varnothing$ and $t^{*}=\mu _{k}$ for some $k\in \{1,2,\ldots ,z\}$, Lemma 1(b) and (d) ensure $\mathrm{sgn}({\Psi }_{1}(\mu _{k}))$ $=\mathrm{sgn}(A_{(1)}(\mu _{k}))\ne 0$, and we have

    $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}=\left\{ \begin{array}{ll} \pi /2&{}\ \text{ if } { \lim \nolimits _{t \rightarrow \mu _{k}-0}}\mathrm{sgn}(A_{(0)}(t)A_{(1)}(\mu _{k}))=1,\\ -\pi /2&{}\ \text{ if } {\lim \nolimits _{t \rightarrow \mu _{k}-0}}\mathrm{sgn}(A_{(0)}(t)A_{(1)}(\mu _{k}))=-1. \end{array}\right. \end{aligned}$$

    (25)

    Fix $\xi _{i-1}$ and $\xi ^{\prime }_{i}$ $(i=1,2,\ldots ,k)$ in exactly same way as shown in the beginning of (II). Then we have $V\{\Psi (\xi _{i})\}-V\{\Psi (\xi _{i-1})\}=\mathcal{X }(\mu _{i})$ $(i=1,2,\ldots ,k-1)$.

    • (i) If $\lim \nolimits _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}=\pi /2$, Lemma 1(d) and (25) ensure $\mathrm{sgn}(\Psi _{0}(\xi _{k-1}))=\mathrm{sgn}(\Psi _{1}(\xi _{k-1}))\ne 0$ and $\mathrm{sgn}(\Psi _{0}(\mu _{k}))=0$, which imply that both of the numbers of sign changes in $(\Psi _{0}(\xi _{k-1}),\Psi _{1}(\xi _{k-1}))$ and in $(\Psi _{0}(\mu _{k}),\Psi _{1}(\mu _{k}))$ are $0$. Hence, we have

      $$\begin{aligned} V\{\Psi (\mu _{k})\}=V\{\Psi (\xi _{k-1})\}. \end{aligned}$$

      As a result, we have

      $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi&= \pi /2+\sum _{i=1}^{k-1}\mathcal{X }(\mu _{i})\pi \nonumber \\&= \pi /2+\displaystyle \sum _{i=1}^{k-1}[V\{\Psi (\xi _{i})\}- V\{\Psi (\xi _{i-1})\}]\pi \nonumber \\&= \pi /2+[V\{\Psi (\xi _{k-1})\}-V\{\Psi (\xi _{0})\}]\pi \nonumber \\&= \pi /2+[V\{\Psi (\mu _{k})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

      (26)

    • (ii) If $\lim \nolimits _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}\!=\!-\pi /2$, Lemma 1(d) and (25) ensure $\mathrm{sgn}(\Psi _{0}(\xi _{k-1}))=-\mathrm{sgn}(\Psi _{1}(\xi _{k-1}))\!\ne \!0$ and $\mathrm{sgn}(\Psi _{0}(\mu _{k}))\!=\!0$, which imply that the number of sign changes in $(\Psi _{0}(\xi _{k-1}),\Psi _{1}(\xi _{k-1}))$ is $1$ while the number of sign changes in $(\Psi _{0}(\mu _{k}),\Psi _{1}(\mu _{k}))$ is $0$. Hence, we have

      $$\begin{aligned} V\{\Psi (\mu _{k})\}=V\{\Psi (\xi _{k-1})\}-1. \end{aligned}$$

      As a result, we have

      $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi&= -\pi /2+\sum _{i=1}^{k-1}\mathcal{X }(\mu _{i})\pi \nonumber \\&= -\pi /2+\sum _{i=1}^{k-1}[V\{\Psi (\xi _{i})\}-V \{\Psi (\xi _{i-1})\}]\pi \nonumber \\&= -\pi /2+[V\{\Psi (\xi _{k-1})\}-V\{\Psi (\xi _{0})\}]\pi \nonumber \\&= -\pi /2+[V\{\Psi (\mu _{k})\}+1-V\{\Psi (a)\}]\pi \nonumber \\&= \pi /2+[V\{\Psi (\mu _{k})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

      (27)

    From (26) and (27), for both cases (i) and (ii), we have

    $$\begin{aligned} {\lim }_{t \rightarrow t^{*}-0}{\arctan }\{\mathcal{Q }_{A}(t)\}+\varLambda (t^{*})\pi =\pi /2+[V\{\Psi (t^{*})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

  • (IV) Suppose that $\mathcal{Z }_{A_{(0)}}\ne \varnothing$ and $t^{*}=b$.

    • (i) If $A_{(0)}(b)\ne 0$, i.e., $V\{{\Psi }(\xi _{z})\}=V\{{\Psi }(b)\}$, we can deduce, in almost same way as in proof for (II),

      $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}\!+\!\varLambda (t^{*})\pi \!=\!\arctan \{\mathcal{Q }_{A}(t^{*})\}\!+\![V\{\Psi (t^{*})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

    • (ii) If $A_{(0)}(b)=0$, by imposing additionally, to the conditions of $\varepsilon >0$ in (24),

      $$\begin{aligned} \left. \begin{array}{ll} b-\varepsilon >a\\ \Psi _{1}(t)\ne 0\, \text{ for } \text{ all } t\in (b-\varepsilon , b]\\ \Psi _{0}(t)\Psi _{1}(t)\ne 0\, \text{ for } \text{ all } t\in (b-\varepsilon ,b) \end{array}\right\} \end{aligned}$$

      and fixing arbitrarily $\xi _{z}\in (b-\varepsilon ,b)$, we can deduce, in almost same way as in proof for (III),

      $$\begin{aligned} \lim _{t \rightarrow t^{*}-0}\arctan \{\mathcal{Q }_{A}(t)\}+\varLambda ({t}^{*})\pi =\pi /2+ [V\{\Psi (t^{*})\}-V\{\Psi (a)\}]\pi . \end{aligned}$$

From (A) and (B), we obtain (9) for all $t^{*}\in (a,b]$. $\square$

1.5 Appendix 5: Proof of Theorem 2

• (A) Proof of (a): From

  $$\begin{aligned} \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial y}(x,y)+j\frac{\partial f_{(1)}}{\partial y}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} =\frac{\left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}, \end{aligned}$$

  the denominator of $\frac{\partial }{\partial x}\left( \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial y}(x,y)+j\frac{\partial f_{(1)}}{\partial y}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} \right)$ is $\left[ \{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}\right] ^{2}$, and the numerator is

  $$\begin{aligned}&\left[ \left( \frac{\partial ^{2}f_{(1)}}{\partial x\partial y}(x,y)\right) f_{(0)}(x,y)-f_{1} (x,y)\left( \frac{\partial ^{2}f_{(0)}}{\partial x\partial y}(x,y)\right) \right] \left[ \{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}\right] \\&\quad -\left[ \left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial x}(x,y)\right) \!+\!\left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) \right] \left[ \{f_{(0)}(x,y)\}^{2}\!-\!\{f_{(1)}(x,y)\}^{2}\right] \\&\quad -2\left[ \left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) \left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) -\left( \frac{\partial f_{(0)}}{\partial x}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) \right] f_{(0)}(x,y)f_{(1)}(x,y). \end{aligned}$$

  Similarly, from

  $$\begin{aligned} \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial x}(x,y)+j\frac{\partial f_{(1)}}{\partial x}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} =\frac{\left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial x}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}, \end{aligned}$$

  the denominator of $\frac{\partial }{\partial y}\left( \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial x}(x,y)+j\frac{\partial f_{(1)}}{\partial x}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} \right)$ is $\left[ \{f_{(0)}(x,y)\}^{2} +\{f_{(1)}(x,y)\}^{2}\right] ^{2}$, and the numerator is

  $$\begin{aligned}&{\left[ \left( \frac{\partial ^{2}f_{(1)}}{\partial y\partial x}(x,y)\right) f_{(0)}(x,y)\!-\!f_{1}(x,y) \left( \frac{\partial ^{2}f_{(0)}}{\partial y\partial x}(x,y)\right) \right] \left[ \{f_{(0)}(x,y)\}^{2}\!+\!\{f_{(1)}(x,y)\}^{2}\right] }\\&-\left[ \left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) +\left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial x}(x,y)\right) \right] \left[ \{f_{(0)}(x,y)\}^{2}-\{f_{(1)}(x,y)\}^{2}\right] \\&-2\left[ \left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) \left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) -\left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) \left( \frac{\partial f_{(0)}}{\partial x}(x,y)\right) \right] f_{(0)}(x,y)f_{(1)}(x,y). \end{aligned}$$

  Then, since $f_{(i)}\in C^{2}(D)\,(i=0,1)$ ensure $\frac{\partial ^{2} f_{(i)}}{\partial x \partial y}(x,y)=\frac{\partial ^{2} f_{(i)}}{\partial y \partial x}(x,y)$ for all $(x,y)\in D$, we have

  $$\begin{aligned} \frac{\partial }{\partial x}\left[ \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial y}(x,y)+j\frac{\partial f_{(1)}}{\partial y}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} \right] =\frac{\partial }{\partial y}\left[ \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial x}(x,y)+j\frac{\partial f_{(1)}}{\partial x}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} \right] \end{aligned}$$

  for all $(x,y)\in D$.

• (B) Proof of (b): Define

  $$\begin{aligned} P(x,y)&:= \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial x}(x,y)+j\frac{\partial f_{(1)}}{\partial x}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} \text{ and } \nonumber \\ Q(x,y)&:= \mathfrak I \left\{ \frac{\frac{\partial f_{(0)}}{\partial y}(x,y)+j\frac{\partial f_{(1)}}{\partial y}(x,y)}{f_{(0)}(x,y)+jf_{(1)}(x,y)}\right\} . \end{aligned}$$

  Then, since $f_{(i)}\,(i=0,1)$ are $C^{2}(D)$ functions, $P$ and $Q$ are $C^{1}(D)$ functions. Moreover, from (a), $P$ and $Q$ satisfy $\frac{\partial P}{\partial y}(x,y)=\frac{\partial Q}{\partial x}(x,y)$ for all $(x,y)\in D$. Hence, from Poincar$\acute{\mathrm{e}}$’s lemma (Fact 1(b)), there exists a function $\theta _{f}\in C^{2}(D)$ satisfying

  $$\begin{aligned} \frac{\partial \theta _{f}}{\partial x}(x,y)=P(x,y)\, \text{ and } \, \frac{\partial \theta _{f}}{\partial y}(x,y)=Q(x,y)\, \text{ for } \text{ all } \,(x,y)\in D, \end{aligned}$$

  (28)

  and the function $\theta _{f}$ is the scalar potential of the vector field $(P(x,y),Q(x,y))$ over $D$. Eq. (28) implies that the function $\theta _{f}$ is determined as

  $$\begin{aligned} \theta _{f}(x,y)=\int \left[ P(x,y)dx+Q(x,y)dy\right] \end{aligned}$$

  uniquely if we impose additionally the condition $\theta _{f}(x_{0},y_{0})=\theta _{0}$.

• (C) Proof of (c): Define $P(x,y)$ and $Q(x,y)$ as in (B). From (a), i.e., $\frac{\partial P}{\partial y}(x,y)=\frac{\partial Q}{\partial x}(x,y)$ for all $(x,y)\in D$, and Green’s theorem (Fact 1(a)), we have

  $$\begin{aligned} \oint \nolimits _{\partial \Omega }\left[ P(x,y)dx+Q(x,y)dy\right] =\int \int \nolimits _\Omega \left( \frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y)\right) dxdy=0. \end{aligned}$$

  In particular, if $\gamma ^\mathrm{I}$ and $\gamma ^\mathrm{I\!I}$ are piecewise $C^{1}$ paths in $D$ with the same initial and final points, by letting $\partial \Omega :=\gamma ^\mathrm{I}-\gamma ^\mathrm{I\!I}$, we have

  $$\begin{aligned} \oint \nolimits _{\gamma ^\mathrm{I}-\gamma ^\mathrm{I\!I}}\left[ P(x,y)dx+Q(x,y)dy\right] =0, \end{aligned}$$

  which implies

  $$\begin{aligned} \int \nolimits _{\gamma ^\mathrm{I}}\left[ P(x,y)dx+Q(x,y)dy\right] =\int \nolimits _{\gamma ^\mathrm{I\!I}}\left[ P(x,y)dx+Q(x,y)dy\right] . \end{aligned}$$

• (D) Proof of (d): By using the parameterizations $\gamma ^\mathrm{I}(t):=(x_\mathrm{I}(t),y_\mathrm{I}(t))$ and $\gamma ^\mathrm{I\!I}(\tau ):=(x_\mathrm{I\!I}(\tau ),y_\mathrm{I\!I}(\tau ))$, we deduce, from (c),

  $$\begin{aligned}&\int \nolimits _{a}^{b}\mathfrak I \left\{ \frac{\left( f_{(0)}(\gamma ^\mathrm{I}(t))\right) ^{\prime }+j\left( f_{(1)}(\gamma ^\mathrm{I}(t))\right) ^{\prime }}{f_{(0)}(\gamma ^\mathrm{I}(t))+jf_{(1)}(\gamma ^\mathrm{I}(t))}\right\} dt\\&\quad \!=\!\!\int \nolimits _{a}^{b}\frac{\frac{d}{dt}\!\left( f_{(1)}(x_\mathrm{I}(t),y_\mathrm{I}(t))\right) f_{(0)}(x_\mathrm{I}(t),y_\mathrm{I}(t))\!-\!f_{(1)}(x_\mathrm{I}(t),y_\mathrm{I}(t))\frac{d}{dt}\left( f_{(0)}(x_\mathrm{I}(t),y_\mathrm{I}(t)\right) }{\{f_{(0)}(x_\mathrm{I}(t),y_\mathrm{I}(t))\}^{2}\!+\!\{f_{(1)}(x_\mathrm{I}(t),y_\mathrm{I}(t))\}^{2}}dt\\&\quad \!=\!\int \nolimits _{\gamma ^\mathrm{I}(a)}^{\gamma ^\mathrm{I}(b)}\left[ \frac{\left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial x}{(0)}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}dx\right. \\&\qquad \!+\!\left. \frac{\left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}dy\right] \\&\quad \!=\!\int \nolimits _{\gamma ^\mathrm{I\!I}(c)}^{\gamma ^\mathrm{I\!I}(d)}\left[ \frac{\left( \frac{\partial f_{(1)}}{\partial x}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial x}{(0)}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}dx\right. \\&\quad \quad \!+\!\left. \frac{\left( \frac{\partial f_{(1)}}{\partial y}(x,y)\right) f_{(0)}(x,y)-f_{(1)}(x,y)\left( \frac{\partial f_{(0)}}{\partial y}(x,y)\right) }{\{f_{(0)}(x,y)\}^{2}+\{f_{(1)}(x,y)\}^{2}}dy\right] \\&\quad \!=\!\int \nolimits _{c}^{d}\frac{\frac{d}{d\tau }\left( f_{(1)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I}(\tau ))\right) f_{(0)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I\!I}(\tau ))\!-\!f_{(1)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I}(\tau ))\frac{d}{d\tau }\left( f_{(0)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I\!I}(\tau )\right) }{\{f_{(0)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I\!I}(\tau ))\}^{2}+\{f_{(1)}(x_\mathrm{I\!I}(\tau ),y_\mathrm{I\!I}(\tau ))\}^{2}}d\tau \\&\quad \!=\!\int \nolimits _{c}^{d}\mathfrak I \left\{ \frac{\left( f_{(0)}(\gamma ^\mathrm{I\!I}(\tau ))\right) ^{\prime }+j\left( f_{(1)}(\gamma ^\mathrm{I\!I}(\tau ))\right) ^{\prime }}{f_{(0)}(\gamma ^\mathrm{I\!I}(\tau ))+jf_{(1)}(\gamma ^\mathrm{I\!I}(\tau ))}\right\} d\tau .\\ \end{aligned}$$

  $\square$

1.6 Appendix 6: Proof of Proposition 4

• (A) Proof of (a):

  • (I) If $k=1$, $\mathrm{Sres}_{i}(P_{0},P_{1},t)$ $(i\in [\deg (P_{2}),\deg (P_{1})-1])$ can be expressed as a constant multiple of $P_{2}(t)$ as one of the first three expressions in Fact 2. Clearly, these are special cases of (10).

  • (II) If $2\le k\le q-1$, i.e., $\deg (P_{k+1})\le i\le \deg (P_{k})-1\le \deg (P_{k-1})-2\le \deg (P_{k-2})-3 \le \cdots \le \deg (P_{2})-(k-1)$, by using the forth expression in Fact 2 repeatedly, we deduce

    $$\begin{aligned}&\mathrm{Sres}_{i}(P_{0},P_{1},t)\\&=(-1)^{(\deg (P_{0})-\deg (P_{1})+1)(\deg (P_{1})-i)}(\mathrm{lc}(P_{1})) ^{\deg (P_{0})-\deg (P_{2})}\\&\quad \quad \times \,\mathrm{Sres}_{i}(P_{1},P_{2},t)\\&=(-1)^{(\deg (P_{0})-\deg (P_{1})+1)(\deg (P_{1})-i)} (\mathrm{lc}(P_{1}))^{\deg (P_{0})-\deg (P_{2})}\\&\quad \quad \times (-1)^{(\deg (P_{1})-\deg (P_{2})+1) (\deg (P_{2})-i)}(\mathrm{lc}(P_{2}))^{\deg (P_{1})-\deg (P_{3})}\\&\quad \quad \times \,\mathrm{Sres}_{i}(P_{2},P_{3},t)\\&=\cdots \\&=\prod _{n=0}^{k-2}(-1)^{(\deg (P_{n})-\deg (P_{n+1})+1) (\deg (P_{n+1})-i)}(\mathrm{lc}(P_{n+1}))^{\deg (P_{n})-\deg (P_{n+2})}\\&\quad \quad \times \,\mathrm{Sres}_{i}(P_{k-1},P_{k},t)\\&=\prod _{n=0}^{k-2}(-1)^{(\deg (P_{n})-\deg (P_{n+1})+1) (\deg (P_{n+1})-i)}(\mathrm{lc}(P_{n+1}))^{\deg (P_{n})-\deg (P_{n+2})}\\&\quad \times \,\left\{ \begin{array}{ll} (-1)^{\deg (P_{k-1})-\deg (P_{k})+1}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1}) -\deg (P_{k})+1}P_{k+1}(t)\\ \quad \quad \quad \quad \text{ for } i\!=\!\deg (P_{k})-1,\\ 0\quad \quad \quad \,\,\text{ for } i\!\in \![\deg (P_{k+1})\!+\!1,\deg (P_{k})\!-\!2]\,\,(\text{ if } \deg (P_{k+1})\!<\!\deg (P_{k})\!-\!2),\\ (-1)^{(\deg (P_{k-1})-\deg (P_{k})+1)(\deg (P_{k})-\deg (P_{k+1}))}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-\deg (P_{k+1})}\\ \quad \times (\mathrm{lc}(P_{k+1}))^{\deg (P_{k})-\deg (P_{k+1})-1}P_{k+1}(t)\\ \quad \quad \quad \quad \text{ for } i=\deg (P_{k}+1), \end{array}\right. \\&=\left\{ \begin{array}{l} \lambda _{\deg (P_{k})-1}P_{k+1}(t)\\ \quad \quad \quad \quad \quad \text{ for } i\!=\!\deg (P_{k})-1,\\ 0\quad \quad \quad \quad \,\,\text{ for } i\!\in \![\deg (P_{k+1})\!+\!1,\deg (P_{k})\!-\!2]\,\,\,(\text{ if } \deg (P_{k+1})\!<\!\deg (P_{k})\!-\!2),\\ \lambda _{\deg (P_{k+1})}(\mathrm{lc}(P_{k+1}))^{\deg (P_{k}) -\deg (P_{k+1})-1}P_{k+1}(t)\\ \quad \quad \quad \quad \quad \text{ for } i=\deg (P_{k+1}). \end{array}\right. \end{aligned}$$

• (B) Proof of (b): If $\det (M_{i}(P_{0},P_{1}))\ne 0$ for all $i\in [0,\deg (P_{1})-1]$, from (6), we have $\det (M_{i}(P_{0},P_{1}))$ $=\mathrm{lc}(\mathrm{Sres}_{i}(P_{0},P_{1},t))$ for all $i\in [0,\deg (P_{1})-1]$. Hence $\deg (\mathrm{Sres}_{i}(P_{0},P_{1},t))=i$ for all $i\in [0,$ $\deg (P_{1})-1]$. Assume that there exists some $k\in [1,q-1]$ s.t. $\deg (P_{k+1})<\deg (P_{k})-1$. Then, from (a), we have $\deg (\mathrm{Sres}_{\deg (P_{k})-1}(P_{0},P_{1},t))=\deg (P_{k+1})<\deg (P_{k})-1$, which contradicts $\deg (\mathrm{Sres}_{i}(P_{0},$ $P_{1},t))$ $=i$ for all $i\in [0,\deg (P_{1})-1]$. Therefore, we have $\deg (P_{k+1})=\deg (P_{k})-1=\deg (P_{1})-k$ for all $k\in [1,q-1]$. Since $\deg (P_{k+1})=\deg (P_{k})-1=\deg (P_{1})-k$ for all $k\in [1,q-1]$ ensures $\deg (P_{k})-\deg (P_{k+1})+1=2$ and $\deg (P_{k})-\deg (P_{k+2})=2$ for all $k\in [1,q-2]$, we have

  $$\begin{aligned} \lambda _{\deg (P_{k+1})}&= (-1)^{(\deg (P_{0})-\deg (P_{1})+1) (\deg (P_{1})-\deg (P_{k+1}))}(\mathrm{lc}(P_{1}))^{\deg (P_{0})-\deg (P_{2})}\\&\quad \times \prod _{n=1}^{k-2}(-1)^{2(\deg (P_{n+1})-\deg (P_{k+1}))} (\mathrm{lc}(P_{n+1}))^{2}\\&\quad \times (-1)^{2}(\mathrm{lc}(P_{k}))^{2}\\&= \left( (-1)^{k}\mathrm{lc}(P_{1})\right) ^{\deg (P_{0})-\deg (P_{1})+1} \prod _{n=2}^{k}\left( \mathrm{lc}(P_{n})\right) ^{2}. \end{aligned}$$

  $\square$

1.7 Appendix 7: Proof of Proposition 5

• (A) Proof of (a): We derive computable expressions for

  $$\begin{aligned} \deg (P_{l+1}),\mathrm{lc}(P_{l+1}) \text{ and } \mathrm{sgn}(\mathrm{lc}(P_{l+1})). \end{aligned}$$

  • (I) Computable expression for $\deg (P_{l+1})$: From Proposition 4(a), for $i\in [\deg (P_{l+1}),\deg (P_{l})-1], \mathrm{Sres}_{i}(P_{0},P_{1},t)$ can be expressed as

    $$\begin{aligned} \mathrm{Sres}_{i}(P_{0},P_{1},t)=\left\{ \begin{array}{l} \lambda _{\deg (P_{l})-1}P_{l+1}(t)\\ \quad \quad \quad \quad \quad \quad \text{ for } i=\deg (P_{l})-1,\\ 0\quad \quad \quad \quad \quad \,\,\text{ for } i\in [\deg (P_{l+1})+1,\deg (P_{l})-2]\\ \quad \quad \quad \quad \quad \quad (\text{ if }\deg (P_{l+1})<\deg (P_{l})-2),\\ \lambda _{\deg (P_{l+1})}(\mathrm{lc}(P_{l+1}))^{\deg (P_{l})- \deg (P_{l+1})-1}P_{l+1}(t)\\ \quad \quad \quad \quad \quad \quad \text{ for } i=\deg (P_{l+1}). \end{array}\right. \end{aligned}$$

    (29)

    • (i) If $\deg (P_{l+1})=\deg (P_{l})-1$, we have, from the third expression in (29), $\deg (\mathrm{Sres}_{\deg (P_{l})-1}\,(P_{0},P_{1},t))= \deg (\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},t))=\deg (P_{l+1})=\deg (P_{l})-1$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-1}(P_{0},P_{1}))= \mathrm{lc}(\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},t))\ne 0$. Hence, we have

      $$\begin{aligned} \deg (P_{l+1})=\deg (P_{l})-\min \{s\in \mathbb N ^{*}\mid \det (M_{\deg (P_{l})-s}(P_{0},P_{1}))\ne 0\}. \end{aligned}$$

    • (ii) If $\deg (P_{l+1})=\deg (P_{l})-2$, let us examine first the case $s=1$. Then we have, from the first expression in (29), $\deg (\mathrm{Sres}_{\deg (P_{l})-1}$ $(P_{0},P_{1},t))=\deg (P_{l+1})<\deg (P_{l})-1$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-1}(P_{0},P_{1}))=0$. Next, let us examine the case $s=2$. Then we have, from the third expression in (29), $\deg$ $(\mathrm{Sres}_{\deg (P_{l})-2}(P_{0},P_{1},t))\!=\!\deg (\mathrm{Sres}_{\deg (P_{l+1})} (P_{0},P_{1},t))\!=\!\deg (P_{l+1})\!=\!\deg (P_{l})\!-2$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-2}(P_{0},P_{1}))\!=\! \mathrm{lc}(\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},$ $P_{1},t))$ $\ne 0$. Hence, we have

      $$\begin{aligned} \deg (P_{l+1})=\deg (P_{l})-\min \{s\in \mathbb N ^{*}\mid \det (M_{\deg (P_{l})-s}(P_{0},P_{1}))\ne 0\}. \end{aligned}$$

    • (iii) If $\deg (P_{l+1})\!\le \!\deg (P_{l})-3$, let us examine first the case $s=1$. Then we have, from the first expression in (29), $\deg (\mathrm{Sres}_{\deg (P_{l})-1}(P_{0},P_{1},t))=\deg (P_{l+1})<\deg (P_{l})-1$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-1}(P_{0},P_{1}))=0$. Next let us examine the cases $s=\{2,3,\ldots ,\deg (P_{l})-\deg (P_{l+1})-1\}$, we have, from the second expression in (29), $\deg (\mathrm{Sres}_{\deg (P_{l})-s}(P_{0},P_{1},t))=\deg (0)=-\infty <\deg (P_{l})-s$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-2}$ $(P_{0},P_{1}))=0$. Third let us examine the case $s=\deg (P_{l})-\deg (P_{l+1})$, we have, from the third expression in (29), $\deg (\mathrm{Sres}_{\deg (P_{l})-s}(P_{0},P_{1},t))=\deg (\mathrm{Sres}_{\deg (P_{l+1})} (P_{0},P_{1},t))=\deg (P_{l+1})=\deg (P_{l})-s$. Moreover, from (6), we have $\det (M_{\deg (P_{l})-s}(P_{0},P_{1}))=\mathrm{lc}(\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},$ $P_{1},t))\ne 0$. Hence, we have

      $$\begin{aligned} \deg (P_{l+1})=\deg (P_{l})-\min \{s\in \mathbb N ^{*}\mid \det (M_{\deg (P_{l})-s}(P_{0},P_{1}))\ne 0\}. \end{aligned}$$

  • (II) Computable expression for $\mathrm{lc}(P_{l+1})$ and $\mathrm{sgn}(\mathrm{lc}(P_{l+1}))$:

    • (i) If $(\deg (P_{l})-\deg (P_{l+1}))$ is odd, from $\deg (\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},t))=\deg (P_{l+1})$, we have, from (6)

      $$\begin{aligned} \det (M_{\deg (P_{l+1})}(P_{0},P_{1}))&= \mathrm{lc}(\mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},t))\nonumber \\&= \lambda _{\deg (P_{l+1})}(\mathrm{lc}(P_{l+1}))^{\deg (P_{l}) -\deg (P_{l+1})-1}\times \mathrm{lc}(P_{l+1})\nonumber \\&= \lambda _{\deg (P_{l+1})} (\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})}. \end{aligned}$$

      (30)

      Hence, we deduce

      $$\begin{aligned} \mathrm{lc}(P_{l+1})=\root \deg (P_{l})-\deg (P_{l+1}) \of {{\frac{\det (M_{\deg (P_{l+1})}(P_{0},P_{1}))}{\lambda _{\deg (P_{l+1})}}}}. \end{aligned}$$

      Moreover, from (30), we have

      $$\begin{aligned}&\mathrm{sgn}\left( \lambda _{\deg (P_{l+1})}\det (M_{\deg (P_{l+1})}(P_{0},P_{1}))\right) \nonumber \\&\quad =\mathrm{sgn}\left( \lambda ^{2}_{\deg (P_{l+1})}(\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})}\right) \\&\quad = \mathrm{sgn}\left( (\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})}\right) =\mathrm{sgn}(\mathrm{lc}(P_{l+1})). \end{aligned}$$

    • (ii) If $(\deg (P_{l})-\deg (P_{l+1}))$ is even, i.e., $(\deg (P_{l})-\deg (P_{l+1})-1)$ is odd, we have, from the first and third expressions of (29), for any $\tau \in \mathbb R$,

      $$\begin{aligned} \left. \begin{array}{l} \mathrm{Sres}_{\deg (P_{l})-1}(P_{0},P_{1},\tau )=\lambda _{\deg (P_{l})-1}P_{l+1}(\tau )\\ \mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},\tau )=\lambda _{\deg (P_{l+1})}(\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})-1}P_{l+1}(\tau ) \end{array}\right\} .\nonumber \\ \end{aligned}$$

      (31)

      Therefore, by using any $\tau \in \mathbb R$, we deduce

      $$\begin{aligned} \mathrm{lc}(P_{l+1})=\root \deg (P_{l})-\deg (P_{l+1})-1 \of {{\frac{\lambda _{\deg (P_{l})-1} \mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},\tau )}{\lambda _{\deg (P_{l+1})}\mathrm{Sres}_{\deg (P_{l})-1}(P_{0},P_{1},\tau )}}}. \end{aligned}$$

      Moreover, from (31), we have

      $$\begin{aligned}&\mathrm{sgn}\left( \lambda _{\deg (P_{l})-1}\lambda _{\deg (P_{l+1})}\mathrm{Sres}_{\deg (P_{l})-1}(P_{0},P_{1},\tau ) \mathrm{Sres}_{\deg (P_{l+1})}(P_{0},P_{1},\tau )\right) \\&\quad =\mathrm{sgn}\left( \lambda ^{2}_{\deg (P_{l})-1}\lambda ^{2}_{\deg (P_{l+1})} (\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})-1}(P_{l+1}(\tau ))^{2}\right) \\&\quad =\mathrm{sgn}\left( (\mathrm{lc}(P_{l+1}))^{\deg (P_{l})-\deg (P_{l+1})-1}\right) = \mathrm{sgn}(\mathrm{lc}(P_{l+1})). \end{aligned}$$

• (B) Proof of (b): If $\det (M_{i}(P_{0},P_{1}))\ne 0$ for all $i\in [0,\deg (P_{1})-1]$, we can regard Eq.  (12) as a special case of (A)-(II)-(i), and hence obtain, for all $k\in [1,q-1]$,

  $$\begin{aligned} \mathrm{lc}(P_{k+1})=\frac{\det (M_{\deg (P_{1})-k}(P_{0},P_{1}))}{\left( (-1)^{k}\mathrm{lc}(P_{1})\right) ^{\deg (P_{0})-\deg (P_{1})+1} \prod \limits _{n=2}^{k}\left( \mathrm{lc}(P_{n})\right) ^{2}}. \end{aligned}$$

  Moreover, since Eq. (12) ensures $\mathrm{sgn}(\lambda _{\deg (P_{k\!+\!1})})\!=\!\! \mathrm{sgn}\left( \left( (\!-1)^{k}\mathrm{lc}(P_{1})\right) ^{(\deg (P_{0})\!-\!\deg (P_{1})\!+\!1)}\right)$, we have, for all $k\in [1,q-1]$,

  $$\begin{aligned} \mathrm{sgn}(\mathrm{lc}(P_{k+1}))&= \mathrm{sgn}\left( \lambda _{\deg (P_{k+1})} \det (M_{\deg (P_{k+1})}(P_{0},P_{1}))\right) \\&= \mathrm{sgn}\left( \left( (-1)^{k}\mathrm{lc}(P_{1})\right) ^{\deg (P_{0})- \deg (P_{1})+1}\det (M_{\deg (P_{1})-k}(P_{0},P_{1}))\right) . \end{aligned}$$

  $\square$

1.8 Appendix 8: Proof of Theorem 3

• (A) Proof of (a): If $\deg (\Psi _{0})\ge \deg (\Psi _{1})$ and $q\ge 2$, from Proposition 4(a) and (8), we have

  $$\begin{aligned} \mathrm{sgn}({\Psi }_{k}(t^{*}))&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}(P_{k}(t^{*}))\\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}\left( \frac{\mathrm{Sres}_{\deg (P_{k})} (P_{0},P_{1},t^{*})}{\lambda _{\deg (P_{k})}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-\deg (P_{k})-1}}\right) \\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}\left( \frac{\mathrm{Sres}_{\deg (P_{k})} (P_{0},P_{1},t^{*})}{\lambda _{\deg (P_{k})}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-\deg (P_{k})-1}}\right) \\&\quad \times \,\mathrm{sgn}\left( \lambda ^{2}_{\deg (P_{k})}(\mathrm{lc}(P_{k})) ^{2(\deg (P_{k-1})-\deg (P_{k})-1)}\right) \\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}(\lambda _{\deg (P_{k})}) \bigl (\mathrm{sgn}(\mathrm{lc}(P_{k}))\bigr )^{\deg (P_{k-1})-\deg (P_{k})-1}\\&\quad \times \,\mathrm{sgn}\left( \mathrm{Sres}_{\deg (P_{k})}(P_{0},P_{1},t^{*})\right) \\&= (-1)^{\frac{(k-1)k}{2}}\kappa ^{\langle 0\rangle }_{\deg (\Psi _{k})}\bigl (\mathrm{sgn}(\mathrm{lc}(P_{k}))\bigr ) ^{\deg (\Psi _{k-1})-\deg (\Psi _{k})-1}\\&\quad \times \,\mathrm{sgn}\left( \mathrm{Sres}_{\deg ({\Psi }_{k})}({\Psi }_{0}, {\Psi }_{1},t^{*})\right) , \end{aligned}$$

  where we used ${\Psi }_{0}(t)=P_{0}(t)$, ${\Psi }_{1}(t)=P_{1}(t)$, $\deg ({\Psi }_{k})=\deg (P_{k})$ and $\mathrm{sgn}(\lambda _{\deg (P_{k})})=\kappa ^{\langle 0\rangle }_{\deg (\Psi _{k})}$ by (11). In particular, if $\det (M_{i}({\Psi }_{0},{\Psi }_{1}))\ne 0$ for all $i\in [0,\deg ({\Psi }_{1})-1]$, we have, from (12), $\deg ({\Psi }_{q})=\deg ({\Psi }_{1})-(q-1)=0$, and hence $q=\deg ({\Psi }_{1})+1$. Moreover we have, from $\deg (\Psi _{k-1})-\deg ({\Psi }_{k})-1=0$ and (12), $\kappa ^{\langle 0\rangle }_{\deg (\Psi _{k})}\bigl (\mathrm{sgn}(\mathrm{lc}(P_{k}))\bigr )^{\deg (\Psi _{k-1})-\deg (\Psi _{k})-1}=\kappa ^{\langle 0\rangle }_{\deg (\Psi _{k})}=\mathrm{sgn}(\lambda _{\deg (P_{k})})=(-1)^{(k-1)(\deg ({\Psi }_{0})-\deg (\Psi _{1})-1)}\bigl (\mathrm{sgn}(\mathrm{lc}(\Psi _{1}))\bigr )^{\deg (\Psi _{0})-\deg (\Psi _{1})+1}$. As a result, we have (14).

• (B) Proof of (b): If $\deg (\Psi _{0})<\deg (\Psi _{1})$, i.e., $\deg (P_{0})<\deg (P_{1})$, and $q\ge 3, P_{2}(t)=P_{0}(t)-0\times P_{1}(t)=P_{0}(t)$ and we have $\deg (P_{1})>\deg (P_{2})>\cdots >\deg (P_{q})$. Then by replacing $P_{0}(t)$ and $P_{1}(t)$ in Proposition 4(a) with $P_{1}(t)$ and $P_{2}(t)$, for any $k=\{2,3,\ldots ,q-1\}$, $\mathrm{Sres}_{i}(P_{1},P_{0},t)$ $(i\in [\deg (P_{k+1}),\deg (P_{k})-1])$ can be expressed as

  $$\begin{aligned} \mathrm{Sres}_{i}(P_{1},P_{0},t)&\!= \mathrm{Sres}_{i}(P_{1},P_{2},t)\\&\!=\!&\left\{ \begin{array}{l} \lambda ^{\langle 1\rangle }_{\deg (P_{k})\!-\!1}P_{k+1}(t)\\ \quad \quad \quad \quad \quad \quad \text{ for } i\!=\!\deg (P_{k})\!-\!1,\\ 0\quad \quad \quad \quad \quad \,\,\text{ for } i\!\in \![\deg (P_{k+1})\!+\!1,\deg (P_{k})\!-\!2]\,\,(\text{ if } \deg (P_{k+1})\!<\!\deg (P_{k})\!-\!2),\\ \lambda ^{\langle 1\rangle }_{\deg (P_{k+1})}(\mathrm{lc}(P_{k+1}))^ {\deg (P_{k})\!-\!\deg (P_{k+1})\!-\!1}P_{k+1}(t)\\ \quad \quad \quad \quad \quad \quad \text{ for }\,i=\deg (P_{k+1}), \end{array}\right. \end{aligned}$$

  where, for $i=\deg (P_{k})-1, \deg (P_{k+1})$,

  $$\begin{aligned} \lambda ^{\langle 1 \rangle }_{i}&:= \prod _{n=1}^{k-2}(-1)^{(\deg (P_{n})-\deg (P_{n+1})+1)(\deg (P_{n+1})-i)} (\mathrm{lc}(P_{n+1}))^{\deg (P_{n})-\deg (P_{n+2})}\\&\quad \times (-1)^{(\deg (P_{k-1})-\deg (P_{k})+1) (\deg (P_{k})-i)}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-i}. \end{aligned}$$

  As a result, we have,

  $$\begin{aligned} \mathrm{sgn}({\Psi }_{k}(t^{*}))&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}(P_{k}(t^{*}))\\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}\left( \frac{\mathrm{Sres}_{\deg (P_{k})}(P_{1},P_{0},t^{*})}{\lambda ^{\langle 1\rangle }_{\deg (P_{k})}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-\deg (P_{k})-1}}\right) \\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}\left( \frac{\mathrm{Sres}_{\deg (P_{k})}(P_{1},P_{0},t^{*})}{\lambda ^{\langle 1\rangle }_{\deg (P_{k})}(\mathrm{lc}(P_{k}))^{\deg (P_{k-1})-\deg (P_{k})-1}}\right) \\&\quad \times \,\mathrm{sgn}\left( \left( \lambda ^{\langle 1\rangle }_{\deg (P_{k})}\right) ^{2}(\mathrm{lc}(P_{k}))^{2(\deg (P_{k-1})-\deg (P_{k})-1)}\right) \\&= (-1)^{\frac{(k-1)k}{2}}\mathrm{sgn}(\lambda ^{\langle 1\rangle }_{\deg (P_{k})})\bigl (\mathrm{sgn}(\mathrm{lc}(P_{k})) \bigr )^{\deg (P_{k-1})-\deg (P_{k})-1}\\&\quad \times \,\mathrm{sgn}\left( \mathrm{Sres}_{\deg (P_{k})}(P_{1},P_{0},t^{*})\right) \\&= (-1)^{\frac{(k-1)k}{2}}\kappa ^{\langle 1\rangle }_{\deg (\Psi _{k})}\bigl (\mathrm{sgn}(\mathrm{lc}(P_{k}))\bigr )^{\deg (\Psi _{k-1})-\deg (\Psi _{k})-1}\\&\quad \times \,\mathrm{sgn}\left( \mathrm{Sres}_{\deg ({\Psi }_{k})}({\Psi }_{1}, {\Psi }_{0},t^{*})\right) , \end{aligned}$$

  where we used ${\Psi }_{0}(t)=P_{0}(t)$, ${\Psi }_{1}(t)=P_{1}(t)$, $\deg ({\Psi }_{k})=\deg (P_{k})$ and $\mathrm{sgn}(\lambda ^{\langle 1\rangle }_{\deg (P_{k})})=\kappa ^{\langle 1\rangle }_{\deg (\Psi _{k})}$. In particular, if $\det (M_{i}({\Psi }_{1},{\Psi }_{0}))\ne 0$ for all $i\in [0,\deg ({\Psi }_{0})-1]$, in almost same way as in proof for (A), we have $q=\deg ({\Psi }_{0})+2$ and (16). $\square$