数理解析研究所講究録
915 巻 1995 年 96-106

96

ON FAIR PARAMETRIC RATIONAL CURVES

ノ裟頭型神$\Gamma\oplus*^{r}\grave{\mathrm{E}}$(Manabu Sakai )1. Introduction.
For given data $(x_{i}^{(k)}, y_{i})(k)(i=0,1;k=0,1)$ , we consider the problem of finding afair parametric rational curve with a parameter $p(>-1)$ :$x(t)=a_{0}t+a_{1}u+a_{2}t^{3}/(1+pt)+a_{3}u^{3}/(1+pu)$(1)$(0\leq t\leq 1, u=1-t)$$y(t)=b_{0}t+b_{1^{\mathcal{U}}}+b_{2}t^{3}/(1+pt)+b_{3}u^{3}/(1+pu)$or$x(t)=a_{0}t+a_{1}u+a_{2}t^{3}/(1+pu)+a_{3}u^{3}/(1+pt)$(1)$(0\leq t\leq 1, u=1-t)$$y(t)=b_{0}t+b_{1}u+b_{2}t^{3}/(1+pu)+b_{3^{\mathcal{U}^{3}}}/(1+pt)$

so that$(i=0,1;k=0,1)$ .
It is well-known that a drawback of a parametnic cubic curve $(p=0)$ is indicated by thefact that unwanted inflection points or singularities may occur on its segment. In whatfolows, the adjectiveon the curve is usually suppressed since all the curvesin this paper are parametric ones and afair curve means a one Ree of unwanted inflection
points and singularities. Techniques for eliminating the unwanted ones have beendeveloped ([1], [2], [5], [6]). One of them is to use the rational curve of the fonn (1) or adifferent looking rational curve of the form (1). Note there is no difference between useof the two rational ones since letting $1/p+1/q=-1(parrow-1+\Leftrightarrow qarrow\infty$ $-1<p<0$, then $t^{3}/(1+pt)=(1+q)t^{3}/(1+qu)$ and $u^{3}/(1+pu)=(1+q)u^{3}/(1+qt)$ .
Hence, all the subsequent results for the curve of the forn (1) are still valid for the one of
the form (1) with $p(>-1)$ replaced by $-p/(1+p)(=q)$ . The first object of Section
2 is to show that we can find a fair curve segment of the forn (1) interpolating to (2) if(2)$(x^{(k)}(i), \mathcal{Y}(k)(i))=(x_{i}^{(k)}, y^{()}i)k$’$|paramet^{\gamma}ic$”$,$$\Leftrightarrow q>0)$$\lambda(=Q\sqrt D),$$\mu(=-C_{1}/D)\geq(1+p)/(3+2p)$ where $\Delta x=x_{10}-x,$ $\Delta y=y_{1}-\mathcal{Y}0$,$C_{i}=x_{i}\Delta y-\mathcal{Y}i\Delta x||(i=0,1)$

and$D=x_{0}|y_{1}|-X_{1}||\mathcal{Y}_{0}$. Note that we can find a fair curve of

the form (1) if $\mu\geq 1/(3+p)$ . That is, if
is in the interior of the first quadrant
plane, a suitable choice of the parameter gives a fair curve of the form (1)of theinterpolating to (2), strictly speaking, according to $p(= \min(\lambda, \mu))\geq 1/3$ and $0<p<$$1/3$
we can take $p=0$ and $(3p-1)/(1-2\rho)$ , respectively since a smaller value ofwould make the truncation
be smaller provided that the data arise from a function.
$(\lambda, \mu)$$\lambda,$$(\lambda, \mu)$$p$$\mathrm{M}$$\mathrm{e}\mathrm{I}\mathrm{r}\mathrm{o}\mathrm{r}$97

For the curve of the form (1), we choose $p= \max(\mathrm{O}, 1/\rho-3)$ . The second object ofSection 2 is to show that the region for a fair curve also contains the whole third quadrant
in addition to the whole first quadrant (theoretically for sufficiently close to-l).
fair interpolatory rational curve to data $S=\{(x_{i}, y_{i}),$
Accordingly, we can find aby suitable choice of the parameter if$p$$0\leq i$$C^{2}$$\leq n\}$

$p$

$a$

$x[t_{i}, t_{i+1}]_{\mathcal{Y}}[fj, t_{j+}1, fj+2]-_{\mathcal{Y}}[t_{i}, f_{i1}+]x[(tj, rj+1, t_{j+2}]>0(j=i-1, i)$where $x[t_{i}, t_{i+1}](X[t_{i,i+1,i+2}tt])$ is the first (second) divided difference and $r_{0}=0$ ,$t_{i+1}=t_{i}+h_{i}(=\sqrt{(x_{i+1i}-X)2+(y_{i+}1-y_{i})2}(0\leq i\leq n-1)$ . In Section 3, we derivetheorem conceming the distribution of inflection points and singularities for the cubic(not) faircurve segment $(p=0)$ which has given another technique for finding
cubic curve interpolating to $S([6])$ . In Section 3, some numerical examples are given.
$a$

$a$

$C^{2}$$C^{1}$

2. On segments of parametric rational curves.
We shall show that the curve segment (1) is fair for $\mu\geq(1+p)/(3+2p)$ inpl .
the first quadrant of theInflection points. First we obtain sufficient conditions for the curve segment (1) not to
contain an inflection point. Defining $\phi(t)=t^{3}/(1+pt)-t/(1+p)$ , the segment (1)to (2) is expressed by equations$\lambda,$$(\lambda, \mu)$$a\mathrm{n}\mathrm{e}$$\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}_{\Psi^{\mathrm{O}}}1a\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}$(3)$x(t)=x_{1}t+x0u+c_{1}\phi(f)+d_{1}\phi(u),$ $y(f)=y_{1}t+y_{0}u+c_{2}\phi(f)+d_{2}\phi(\mathcal{U})$in which$(3+2p)/(1+p)^{2}(C_{1}, d_{1})=((1+p)x_{0}|+(2+p)x_{1}’-(3+2p)\Delta x$ ,(4)$-(2+p)x_{0}-|(1+p)x_{1}|+(3+2p)\Delta x)$

is given by (4) with replacing .
andInflection points of the curve (1) are determined by the equation:$(0<t<1)$ .
(5). Then the equation (5) can be
We assume for the moment
equivalently rewritten as$(c_{2}, d_{2})$

$x$

$y$

$x^{\mathrm{t}1}(t)_{\mathcal{Y}’}’(t)-X(|t)y’(t)=0$$D(=x_{0\mathcal{Y}1}^{\mathrm{t}1||}-x1y_{0})\neq 0$”(6) $w(t)(=\lambda\alpha t)+\mu\theta(u)+(1+p)^{2}\dagger\psi’(t)\psi^{\mathrm{t}}’(u)+\phi$ $(t)\psi^{\mathrm{t}}(u)1)=0(0<t<1)$

where
(7)

”

”

$\alpha t)=(2+p)\phi$$(u)-(1+p)\phi(|’ t)-(1+p)2\mathrm{t}\psi’(t)\emptyset \mathrm{t}\mathrm{t}(u)+\emptyset$$(t)\psi’(u)\}$.

By a direct calculation, $\theta(t)>0(0<t<1)$ since(8)$\theta(1)=0$,$\theta’(t)=(1+p)^{2}\psi^{(3})(u)\{\emptyset’(t)-(2+p)/(1+p)^{2}\}$$-(1+p)^{2}\phi^{()}3(t)\{\psi’(u)+1/(1+p)\}<0$”$(t)=2f(3+3pt+pt)22/(1+pt)^{3}$
where $\phi’(t)=(3t^{2}+2pt^{3})/(1+pt)^{2_{-1}}/(1+p),$
and $\phi^{(3)}(t)=6/(1+pt)^{4}$ . Therefore, if $\mu\geq(1+p)/(3+2p)$ , then no inflection point
occurs on the curve segment (1) since$\phi$$\lambda,$(9) $w(t)\geq_{\mathrm{t}}(1+p)2/(3+2p)\}[\phi"(t)\{\emptyset’(u)+1/(1+p)\}+\phi"(u)\{\emptyset’(t)+1/(1+p)\}]>0$ .

98Singularities. Next we shall obtain sufficient condition for the curve segment (1) not
to contain singularity. To this end, we have to get an equation of the image of the curve(1) by eliminating the parameter from (1) (or (3)). From (3),
$a$

$a$

$t$

.

(10)

The

$c_{1}\phi(t)+d_{1}\mu \mathcal{U})=X-x0u-x_{12}f,$$C\mu_{t})+d_{2}\mu u)=y-y0u-y_{1^{t}}$$\mathrm{c}a\mathrm{s}\mathrm{e}\Delta\neq 0$

is considered since the other case$\Delta=0$

is easily treated where$\Delta(=c_{1}d2-c2d_{1})=\{(1+p)4/(3+2p)1(D-\mathrm{G})+c1)(=\mathrm{t}(1+p)4/(3+2p)\mathrm{I}(1-\lambda-$$\mu)D)$. A combination of the two equations in (10) gives(11) $t^{3}/(1+pt)=(d_{2}X-d_{1\mathcal{Y})/}\Delta-\alpha_{1}u-\beta_{1}t, u/3(1+pu)=(c_{1}y-C_{2}X)/\Delta-\alpha_{2}u-\beta 2^{t}$where$\alpha_{1}-\beta_{1}=\frac{\{(3+2p)G1^{-}(1+p)D\}}{(1+p)^{2}(Q_{1^{-C_{1}}}-D)}=\frac{\{(3+2p)\lambda-(1+p)\}}{(1+p)^{2}(\lambda+\mu-1)}$(12)$-( \alpha_{2}-\beta 2)=\frac{\{(3+2p)(-c_{)-}1(1+p)D1}{(1+p)2(\mathrm{c}_{)}-C1-D)}=\frac{\{(3+2p)\mu-(1+p)\}}{(1+p)^{2}(\lambda+\mu-1)}$.

Further we rewrite (11) as(13)$t^{3}=p(\alpha_{1^{-}}\beta_{1})f2+q_{1}t+r_{1},$$u^{3}=p(\beta_{2^{-}}\alpha_{2})u2+q_{2}u+r_{2}$in which(i)$r_{1}=(d_{2^{X-}}d_{1\mathcal{Y}^{)/\Delta-}}\alpha 1, r_{2}=(c_{1}y-C_{2}x)/\Delta-\beta 2$(ii)$q_{1}=pr_{11}+\alpha-\beta 1,$,(14)$q2=P^{\gamma_{2}}-\alpha_{2^{+}}\beta_{2}$.

Defining(15)$k=3-p( \alpha_{1}-\beta 1)+p(\alpha 2-\beta 2)=\frac{(3+3p+p)(2\mathrm{G}_{1}-c1)-(1+p)(3+p)D}{(1+p)^{2}(Q_{)}-C_{1}-D)}$$= \frac{(3+3p+p^{2})(\lambda+\mu)-(1+p)(3+p)}{(1+p)^{2}(\lambda+\mu-1)}(\neq 0)$

for$\lambda,$$\mu\geq(1+p)/(3+2p)$ ,the summation of the two equations in (13) gives(16)$kt^{2}-\{3+q_{1}-q_{2}+2p(\alpha_{2}-\beta_{2})\}f+\{1-q2^{-}r1-r_{2}+p(\alpha_{2^{-}}\beta_{2})\}=0$

Rewriting quadratic equation (16) as$a$$(t+\alpha)^{2}=\beta$.

, easy calculation gives(i)$2\alpha k=-3-p(\gamma 1-\gamma 2)-(\alpha_{1}-\beta_{1})-(1+2p)(\alpha_{2\beta_{2}}-)$(ii)$\beta k=-1+r_{1}+(1+p)r_{2}-(1+p)(\alpha 2-\beta_{2})+\alpha^{2}k$

(17)

.

Here we consider(18)$(\alpha, \beta)$$\kappa\langle\alpha_{r_{1}},$

as functions of$\alpha_{\gamma_{\sim}},)=(-p/2,p/2),$$(r_{1}, \Gamma_{2})$instead of $(x, y)$ , and then$\kappa(\rho_{r_{1\beta_{\Gamma})(\alpha,1p\alpha)}},2=1-p+p+$

Use change of variable $t=t^{*}-\alpha$ and eliminate the parameter $t^{*}((t^{*})^{2}=\beta)$
from the first equation in (13) to give the required equation $\psi(x, y)=0$ of the image ofthe curve (1):
$a$

(19)

$\psi(x, y)=\{\beta+3\alpha 2+(2\psi-1)(\alpha 1-\beta 1)-pr1\}^{2}\beta$$-[\alpha^{3}+3\alpha\beta+\{p(\alpha^{2}+\beta)-\alpha 1(\alpha 1-^{\rho_{1}})+(1-p\alpha)\gamma_{1}]^{2}$.

instead of $(x, y)$ . Then, theFor simplicity, we can consider as a function of
singularity of is determined by the system of equations$\psi$$\psi$$(r_{1}, r_{2})$

99

(20)

$\psi(r_{1}, \gamma_{2})=\psi_{r_{1}}(r_{1,2}r)=\psi_{\gamma_{2}}(r1, \gamma_{2})=0$

Writing$\psi$as$A$ $2\beta-C^{2},$$\psi_{r_{1}}(\Gamma_{1}, r_{2})=\psi_{r_{2}}(r_{1}, \gamma_{2})=0$.

are equivalent to$2 \{-p-\frac{p}{2k}A\alpha+\frac{1-p\alpha}{k}A_{\beta}\}A\beta+\frac{1-p\alpha}{k}A^{2}=2\{1-p\alpha-\frac{p}{2k}C_{\alpha}+\frac{1-p\alpha}{k}C_{\beta}\}C$

(21)

2

$\{\frac{p}{2k}A_{\alpha}+\frac{1+p+p\alpha}{k}A_{\beta}\}A\beta+\frac{1+p+p\alpha}{k}A^{2}=2\{\frac{p}{2k}C_{\alpha}+\frac{1+p+p\alpha}{k}C_{\beta}\}c$

On eliminating A 2 from (21), we obtain(22)$\{p(1+p+p\alpha)+_{\mathrm{t}p}/k)(1+p/2)A\alpha\}A\beta$$=\{-(1-p\alpha)(1+p+p\alpha)+(p/k)(1+p/2)c\alpha\}c$

.

Since

$=|$

(23)

by (23) and the similar relations for$|$ ,$\frac{1p\alpha}{k}\frac{1-p\alpha}{+p+k}$$\frac{-p}{\frac{2kp}{2k}}$

and$\beta$$C$

we get$(1+p+p\alpha)A_{r}1-(1-p\alpha)A\gamma_{2}=-(2p+p^{2})/(2k)A_{\alpha}$(24)$(1+p+p\alpha)\beta_{r_{1}}-(1-p\alpha)\beta_{r_{2}}=0$$(1+p+p\alpha)C_{r1^{-}}(1-p\alpha)cr2=-(2p+p^{2})/(2k)C_{a}$Making linear combination of, fromto (22):$a$$(1-p\alpha)\psi r2(\gamma 1, r_{2})$(25)$\psi_{\gamma_{1}}(\gamma 1, r_{2})$and$\psi_{\gamma_{2}}(\gamma 1, r_{2})$$\psi_{r1}(r_{1,2}r)=\psi_{\gamma_{2}}(r1, r2)=0$$A_{\alpha}A\beta=C_{\alpha}C$, i.e., $(1+p+p\alpha)_{\psi r}1(\gamma_{1}, \gamma_{2})-$

we get another equation in addition.

From (22) and (25),(26)$C\{-(1-p\alpha)(1+p+p\alpha)+(p/k)(1+p/2)c\alpha\}A_{\alpha}A\beta$.

$=C_{\alpha}C\{p(1+p+p\alpha)+\mathrm{t}p/k)(1+p/2)A\alpha\}A\beta$

Two cases $C\neq 0$ and $C=0$ , will be considered separately. $(a)C\neq 0$ : Then, $A,$
since A $2\beta=C^{2}$ . From (26) we have

{

(27)

In this$(1+p+p\alpha)(1+p\alpha)$ $pC_{\alpha^{-}}(p$a$-1)A_{\alpha}$$\beta\neq 0$} $=0$ .

, we show that the curve segment (1) (or (19)) does not have a singularity. If$\mathrm{c}a\mathrm{s}\mathrm{e}$there were any singularity (loop or cusp), then we note that two values of the parameter
defined by the quadratic equation $(t+\alpha)^{2}=\beta$ must belong to $(0,1)$ , i.e.,
$t$

(28)

$0<t(=t^{*}-\alpha=\pm\sqrt{\beta}-\alpha)<1$

or $-1<\alpha<0,0\leq\beta<\alpha^{2},$where $\beta>0$ and $\beta=0$ correspond to loop and cusp, respectively.
Since $1+p+p\alpha,$ $1+p\alpha>0$ for $p>-1,$ (27) giveswith the aid of (25). Hence we have$\beta<(1+\alpha)^{2}$$a$$pC_{\alpha}=(p\alpha-1)A_{\alpha}$from which$A{}_{\alpha}C_{\alpha}\{p\beta A-(\beta\alpha-1)C\}=0$(29)$p\beta A-(p\alpha-1)C=0$, then we obtain $p\beta A-(p\alpha-1)C=0$ from (22) and (27).
orwhere if
Combining (29) with A $2\beta=C^{2}$ , we have
$A_{\alpha}$$C_{\alpha}=0$

100

(30)

$\beta=(\alpha-1/p)^{2}$

which can not satisfy the required inequalities in (28), i.e., $\beta<\alpha^{2}(-1/2\leq\alpha<0)$ or$\beta<(1+\alpha)^{2}(-1<\alpha<-1/2)$ . That is, if $C\neq 0$ , the curve segment (1) (or (19)) does
not have a singularity. (b) $C=0$ : From (24) we get $\beta=0$ or $A=0$ . Since $\beta=0$ gives$=0$ , we have only to consider the case when $A=C=0$ . Then
$A$

(i)

$\beta+3\alpha^{2}+(2p\alpha-1)(\alpha_{1}-\beta_{1})=pr_{1}$(ii)$2\alpha^{3}-2\alpha\beta+p(\alpha^{2}-\beta)(\alpha_{1^{-}}\beta 1)=r_{1}$(31)

In addition, eliminating(32)$r_{2}$.

from (17),$k\{p\alpha^{2}+2\alpha(1+p)-p\beta\}+3+2p+(1+p)(\alpha_{1}-\beta 1)+(1+p)^{2}(\alpha_{2}-\beta_{2})$$=-(2p+p^{2})r_{1}$EhIninate

$r1$

(i)

ffom 31.

$(\mathrm{i})-(\mathrm{i}\mathrm{i})$and 31 (1) $-(32)$ to give two equations, respectively$\alpha_{1}-\beta_{1}=\frac{3\alpha^{2}-2p\alpha^{3}+\beta(1+2p\alpha)}{(1-p\alpha)2_{-}\beta p^{2}}$

(33)

(ii)

$-( \alpha_{2^{-}}\beta_{2})=\frac{3(1+\alpha)^{2}+2p(1+\alpha)3+\beta \mathrm{t}1-2p\langle 1+\alpha)\}}{(1+p+p\alpha)^{2_{-}}\beta p^{2}}$.

Note that for $p\neq 0$, it is be difficult to find the solutionof (33) (the singul point$(x, y)$ of (19) by. Therefore we show that the existence of the solution
satisfying the required inequalities (28) brings $\mu<(1+p)/(3+2p)$ , as implies that the
curve segment (19) is free of singularity for $\mu\geq(1+_{F)/(3}+2p)$ . Since the both$-1/2\leq\alpha<0$ andright hand sides of (33) are monotone increasing in$(\alpha, \beta)$$a\mathrm{r}$$14(\mathrm{i})-(17))$$(\alpha, \beta)$$\lambda,$$\lambda,$$a$$\beta(0\leq\beta<\alpha^{2},$$0\leq\beta<(1+\alpha)^{2},$(34)$-1<\alpha<-1/2)$ ,$0< \alpha_{1}-\beta 1<\frac{4\alpha^{2}}{1-2p\alpha},$$0<-( \alpha_{2}-\beta 2)<\frac{4\alpha^{2}+6\alpha+(3+2p)/(1+p)}{1+p+2p\alpha}$$(-1/2\leq\alpha<0)$

or
(35)

$0< \alpha_{1}-\beta 1<\frac{4\alpha^{2}+2\alpha+1/(1+p)}{1-p-2p\alpha},$$0<-( \alpha_{2}-\beta 2)<\frac{4(1+\alpha)^{2}}{1+2p+2p\alpha}$$(-1<\alpha<-1/2)$ .

In addition, use (34) $-(35)$ to obtain(36)$( \alpha_{1}-^{\rho_{)}}1-(\alpha_{2}-\rho_{2})<\frac{3+2p}{(1+p)^{2}}$which is easily checked since the above inequality (36) is equivalent to(37)$\alpha\{2(2+2p+p^{2})\alpha+(3+3p+p^{2})\}<0$$(-1/2\leq\alpha<0)$

or
(38)

$(1+\alpha)\{2(2+2p+p^{2})\alpha+(1+p+p^{2})\}<0$$(-1<\alpha<-1/2)$ .

Combining (12) with the first inequalities of (34) $-(35)$ and (36), the necessary conditions
for the curve segment (19) to contain singularity are given by$a$$\{(3+2p)\lambda-(1+p)\}/(\lambda+\mu-1)>0,$ $\{(3+2p)\mu-(1+p)\}/(\lambda+\mu-1)>0$

101

(39)
$\{(3+2p)(\lambda+\mu)-2(1+p)\}/(\lambda+\mu-1)<(3+2p)$(i.e., $\lambda+\mu<1$ )from which follow(40)$\lambda,$

Therefore, if$\lambda$$\mu<(1+p)/(3+2p)$ .

, $\mu\geq(1+p)/(3+2p)$ , then the curve segment (19) has no singulanity. In

summary, theorem conceming the inflection points and singularity is obtained:$a$

Theorem 1. The curve segment of the form (1) interpolating to (2) is fair forplane.
$(1+p)/(3+2p)$ in the fnst quadrant of the$\lambda,$$\mu\geq$$(\lambda, \mu)$Remark. For the case $D=0$ in which the tangent vectors at the two end points areparallel provided that $(x_{i})^{2}|+(y_{i})^{2}|\neq 0(i=0,1)$ . The curve segment (1) does notcontain an inflection point if $Q_{1}(-c_{1})>0$ , i.e.,(41)$(X_{0}\Delta_{\mathcal{Y}}-_{\mathcal{Y})(}||0\Delta XX1\Delta y-y_{1}||\Delta_{X})<0$

since for $D=0,$ $w(t)=Q_{1}\alpha t)+(-C_{1})\alpha u)$ . In addition, the curve segment (1) is free of
singularity since $\alpha_{1}-\beta_{1}-(\alpha_{2^{-}}h^{)}=(3+2p)/(1+p)^{2}$ from (12), i.e., the necessinequalities (36) for the existence of the singularities does not hold where note that all the.
forargument in case I is still valid for $D=0$ with (12) and (15) having G),
Therefore, the curve segment (1) is fair for )$(-c_{1})>0$ .
$a$$a\mathrm{r}\mathrm{y}$$C_{1}$$\lambda,$$\mu$$\mathrm{c}$Fig. 1. Distribution of inflection points and singulanities $(p=-\mathrm{O}. 5)$

102
Fig. 1 gives the numerically determined distribution of inflection points and singularities
for $p=-0.5$ where $N_{i}(i=0,1,2)$ or (shown by dots) represent the regions for the
curve segment (1) to contain inflection points and no singularity or loop and noinflection point. The upper boundof represents the region for the curve segment(1) to have cusp and passes throughwith $\lambda=\mu=(1+p)/(6+6p+p^{2})$ , sinceis the solution of (33). As decreases to-l, the region for athenfair curve rapidly increases to the whole first and third quadrants; refer to Figs 1-2 and 4.
By means of Theorems 1, gene.rally speaking, we see that if the curve segment $(p=0)$
contains loop, according as decreases to-l, it contains loop and no inflection.point
inflection points and no singularity
a cusp and no inflectionone
inflection point and no singulanity, by tums,
inflection point and noprovided that $(Q\sqrt D, -c1/D)$ is in the fnst quadrant. In numerical deternination of the
distribution, we repeat the process that first give
, and then count the number of theof (33) satisfies (28).
roots of (6) belonging to $(-1,0)$ and check if the solution
At the end of this section, we remark that if the interv of the parameter isinstead of $[0,1]$ in the end conditions (2), i.e.,
$L$

$i$

$a$

$L$

$a\mathrm{r}\mathrm{y}C$$(\lambda, \mu)$$a$$(\alpha, \beta)=(-1/2, \mathrm{o})$

$p$

$a$

$a$

$p$

$\Rightarrow$$\Rightarrow$$\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\Rightarrow \mathrm{t}\mathrm{w}\mathrm{o}$$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}a\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}\Rightarrow \mathrm{n}\mathrm{o}$$(\lambda, \mu)$$(\alpha, \beta)$$a1$(2)\dagger$(x^{(k)}(\alpha), y(k)(\alpha))=(xy^{()}0’ 0)(k)k,$$t$$[\alpha, \beta]$$(X^{(k)}(\beta), \mathcal{Y}(k)(\beta))=(x^{(}, y_{1}1)k)(k)(k=0,1)$,

then we use the parametric rational curve of the form (1) with $t^{3}/(1+pt)$ and $u^{3}/(1+pu)$
replaced by $s^{3}/(1+ps)$ and $r^{3}/(1+pr)$ with $s=(t-\alpha)/(\beta-\alpha)$ allel $r=1-s$ . In this, we can findcase, lettingand$-C_{1}/D\geq(1+p)/(3+2p)$ . With help of this remark, we can easily
fair curve ifof the above
andcheck that the curve $(x(f), \mathcal{Y}(f))\equiv\{(x_{i}(f), yi(t)),$ $0\leq i\leq n-1$ :interpolating to the data $S=\{(X_{i,y)}i$ ,replacing
modified form with
$0\leq i\leq n\}$ is fair on the same assumption given in [6] where he gave
numerical example
to demonstrate the automatic removal of spurious singularities (loops or cusps) without
theoretic analysis which would be required since inflection points and singularities
is given byoccur under different conditions. The curve$C_{i}=y[\alpha, \beta]_{X_{i}}-X[\alpha, \beta]_{\mathcal{Y}_{i}}||(i=0,1)$$D=x_{0\mathcal{Y}_{1^{-x}}\mathcal{Y}0}|||1|$$a$$\mathrm{G}\sqrt D,$$x_{i(t)}$$(t_{i}, t_{i+1})$$y_{i}(t)$$(\alpha, \beta)\}$

$a$

$a1$

$(x(t), \mathcal{Y}(f))$$x(t):=x_{i}(t)=x_{i+1}s+x_{i}r+h_{i}\{C_{1()}i\psi(s)+d_{1(}i)\phi(\gamma)\}$(3)$(t_{i}\leq t\leq f_{i+1})$$y(t)$ $:=yi(t)=\mathcal{Y}i+1^{S+y+}i\Gamma hi\{C_{2(}i)\psi(s)+d_{2}(i)\psi(\Gamma)\}$

where
(4)

$(3+2p)/(1+p)^{2}(C1(i), d_{1}(i))=((1+p)x_{i}|+(2+p)x_{i}|+1^{-}(3+2p)x[ti, ti+1]$$-(2+p)x_{i}-|(1+p)x_{i1}|++(3+2p)X[fi, ti+1])$

and $(c_{2}(i), d_{2}(i))$ is given by (4) with replacing .
are determined by the consistency relation for $z\in C^{2}[f0, t_{n}]$ :
Here
$y$

$x$

$z_{i}|(=x_{i’ \mathcal{Y}_{i}}||)$(42)$\frac{z_{i+1}1}{h_{i}}+\frac{(2+p)}{(1+p)}(\frac{1}{h_{i}}+\frac{1}{h_{i-1}})_{Z}i+\frac{z_{i-1}1}{h_{i-1}}|=\frac{(3+2p)}{(1+p)}\{\frac{z[t_{i},t_{i1}+]}{h_{i}}+\frac{z[ti-1fi]}{h_{i-1}},\}(1\leq i\leq n-1)$

and the bound$a\mathrm{r}\mathrm{y}$conditions $z_{0}’|=z_{n}||=0$ which are equivalent to103(43) $(2+p)z^{\mathrm{t}}0+(1+p)z’ 1=(3+2p)_{Z[}t0,$ $t1],$ $(2+p)z_{n}|+(1+p)_{Z}n-1|=(3+2p)Z[tn-1, fn]$ .

It was proved in [1] or from (42) $-(43)$ that on letting $parrow-1$$z_{i}|arrow(h_{i-1}Z[t_{j,1}ti+]+hiZ[t_{i}-1, t_{i}])/(h_{i}+h_{i-1})(1\leq i\leq n-1)$(44)$z_{0}’arrow z[t0, t1],$$z_{n}|arrow z[t_{n-1}, t_{n}]$.

Use (44) to obtain on $[t_{i}, t_{i+1}](1\leq i\leq n-2)$G) $(=y[t_{i}, t_{i}+1]X_{i}-X[ti, t_{i}|+1]\mathcal{Y}_{i})|$$arrow h_{i}\{x[t_{i}, ti+1]_{\mathcal{Y}[f}i-1, t_{i}, t_{i+1}]-y[ti, ti+1]x[ti-1, t_{i,i1}t+]\}$(45)$-c1(=-y[ti, fi+1]_{X_{i+1}+}X1[t_{i}, ti+1]_{\mathcal{Y}}i+1)|$$arrow h_{i}\{x[ti, t_{i+1}]y[ti, t_{i}+1, t_{i2}+]-y[ti, t_{i1}+]X[t_{i}, ti+1, ti+2]\}$.

Therefore, if $x[t_{i}, ti+1]_{\mathcal{Y}}[t_{j}, t_{j+1}, t_{j2}+]-y[t_{i,i1}t+]x[(tj, fj+1, tj+2]>0(j=i-1, i)$ or $<0$$(j=i-1, i)$ , then $(\lambda, \mu)(=(\mathrm{G}lD, -C_{1}/D))$ is in the first or third quadrants foris fair for sufficiently
sufficiently close to–l, i.e., the curve segment, from (43) $-(44)(2+p)Q)=$
close to-l on $[t_{i}, t_{i+}1](1\leq i\leq n-2)$ . On$(1+p)(-C_{1})$ and $-C_{1}arrow h_{0}\{x[f_{0}, t_{1}]y[t0, t1, f2]-\mathcal{Y}[t_{0,1}t]x[r0, f1, t2]\}$ . Therefore, if, then the curve segment $(x_{0}(t), y0(t))$ isisalso fair for sufficiently close to-l. Similarly the curve segment
close
fair if $x[t_{n-1}, f_{n}]_{\mathcal{Y}}[fn-2, tn-1, tn]-y[tn-1, fn]x[t_{n2}-, t_{n-1}, tn]\neq 0$ for$p$$(x_{i}(t), \mathcal{Y}i(t))$$p$$[t_{0}, t_{1}]$$x[t_{0}, t_{1}]\mathcal{Y}[t_{0}, t1, t2]-\mathcal{Y}[t0, t1]X[t0, t1, t_{2}]\neq 0$$(x_{n-1}(t), \mathcal{Y}n-1(t))$$p$$p\mathrm{s}\mathrm{u}\mathrm{f}\mathrm{f}_{1}^{*}\mathrm{C}\mathrm{i}\mathrm{e}\mathrm{n}\iota 1\mathrm{y}$to–l.
Suppose that the tangent directions are fixed at the two end points of a segment, andandonly the magnitudes of the tangents are allowed to be varied in scalar multiples$(\eta , \kappa>0)$ , respectively. Then
, i.e.,,),)$\eta$$\mathrm{G}$$\muarrow\mu/\eta$. Therefore, if$\mathrm{G}\mathit{1}D,$$0<\kappa\leq 3\lambda(=3\mathrm{G}\sqrt D),$

$C_{1}arrow\kappa C_{1},$$\lambdaarrow\chi_{\mathrm{K}}$$Darrow\eta\kappa D$$-C1/D>0$ , then Theorem 1 enables us to findsegment (1) by suitable choice of$a$$arrow\eta Q$$\eta$

and$\kappa$, strictly speaking, if$\lambda/\kappa,$$0<\eta\leq 3\mu(=-3C_{1}/D)$ where for another$\kappa$$a$fair curve$\mu/\eta\geq 1/3$

, i.e.,

proof, see [6, .
$\mathrm{p}$54].

3. Numerical examples.
In this section, we consider two numerical examples. First we consider the different
shapes of the curve segments with different values of the parameter ; see Fig. 2 where$(8, -4)$ , i.e.,$(5,6),$
the data are given by$p$$(x_{0}^{(k)}, \mathcal{Y}_{0})(k)=(0,1),$$(x_{1}^{(k)}, \mathcal{Y}_{1})(k)=(1,1),$$=(2/17,1/17)$ and the values of the parameter $p$ are $0,$ $-0.5$ and-0.83 ($(\lambda, \mu)$$=\dagger’$anthe proposed one in thisapproximate value when a cusp occurs\dagger ’) and –14/15 (paper\dagger ’). Point (2/17, 1/17) is denoted by solid circles in Fig 1, as implies that the curve$=\dagger 1$

104segment with $p=-0.5$ contains loop. Numerical determination of the distribution
assured that (2/17, 1/17) denoted by solid circle is nearly on the boundary of with$=-0.83$ in Fig. 3, i.e., then the curve segment contains a cusp. Next $S=\{(0,0),$ $(2,4)$ ,$(4, 0)$ , $(6,2),$ $(8,10),$ $(10,2),$The signs of )( $-^{c)}1<0$ on$[t_{i}, t_{i1}+](i=3,4)$ change
decreases to-l, and so one inflection point
which does not occur with $p=0$ appears on each $[t_{i}, t_{i}+1](i=3,4)$ when letting $parrow-$, i.e., in thisin order to eliminate aloop on
no loop would be moredesirable than two inflection points. Generally speaking, a loop on two or moreconsecutive curve segments could beby letting $parrow-1$ since then each curve
segment reduces to a straight line one by (4) or (4) . Note that all the discussion in
Sections 2-3 is concemed with a singularity (loop and cusp) on a single segment. In Fig
4, we give a graph of the ration curves $(p=0, -0.3, -\mathrm{o}.5)$ on
of interest. The
algorithm proposed by Clements [1] is sufficient in practical computation of the curve
segment of the form (1) or (1) interpolating to (2).
$a$

$C$

$a$

$(10.5,2)\}(\mathrm{C}\mathrm{l}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}[1])$.

$\mathrm{G}$$\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}+\mathrm{t}\mathrm{o}-\mathrm{a}\mathrm{s}p$$1$$[t_{3}, t_{5}]$$\mathrm{c}a\mathrm{s}\mathrm{e}$$\mathrm{e}\mathrm{l}\mathrm{i}\mathrm{m}\dot{\mathrm{n}}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}$$a1$$[t_{4}, t_{5}]$Fig. 2

$L$

$p$

105

Fig. 3

106

References

1. Clements, J. C.: A convexity-preserving parametric rational interpolation, Numer.
Math. 63, 165-171, 1992.
2. M. Sakai and R,. Usmani: Shape preserving approximation by rational splines,
WSSIAA 2, 345-354, World Scientific, Singapore, 1993.
3. M. Sakai and J. W. Schmidt: Positive interpolation with rational splines, BIT 29,
140-147, 1989.
4. M. Sakai and R,. Usmani: On orders of approximation of plane curves by rational
splines, BIT 30, 735-741, 1990.
-Algorithmen, R. Oldenbourg Verlag
5. H. Sp\"ath: Eindimensionale
Munchen, 1990.
6. Su, B-O and Liu, D-Y: Computational Geometry-Curve and Surface Modeling,
Academic Press, New York, 1989.
$C^{2}$$\mathrm{S}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{e}- \mathrm{I}\mathrm{n}\mathrm{t}\mathrm{e}\varphi \mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}$