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The Length of the de Rham Curve
journal of mathematical analysis and applications 223, 182–195 (1998) article no. AY985967
The Length of the de Rham Curve Serge Dubuc D´epartement de Math´ematiques et de Statistique, Universit´e de Montr´eal, C.P. 6128 Succursale Centre-ville, Montr´eal (Qu´ebec), Canada H3C 3J7
and Jean-Louis Merrien and Paul Sablonni`ere INSA de Rennes, 20 avenue des Buttes de Coesmes, 35043 Rennes Cedex, France Submitted by Joseph D. Ward Received February 4, 1997
The length L of the de Rham curve is the common limit of two monotonic sequences of lengths ln and Ln of inscribed and circumscribed polygons, respectively. Numerical computations show that their convergence is linear with the same convergence rate. This result is easy to prove for the parabola. For arbitrary de Rham curves, we prove two nearby results. First, the existence of a limit q ∈ 0; 1 of the sequence of ratios Ln+1 − L/Ln − L implies the convergence to the same limit of the two sequences ln+1 − L/ln − L and Ln+1 − ln+1 /Ln − ln . Second, the sequence Ln+1 − Ln is bounded by a convergent geometric sequence. In practice, this allows us to accelerate the convergence of both sequences by standard extrapolation algorithms. © 1998 Academic Press
1. INTRODUCTION The de Rham curve Cγ , studied in [3], is the limit of a sequence of polygons depending on a parameter γ. We are interested in the computation of the length L of this curve. This problem was already considered by other authors in the context of computer-aided geometric design, in particular for piecewise polynomial or rational curves (see, e.g., [4] and [5]). In a further paper, we shall develop applications to the problem of constructing an interpolating convex curve with prescribed length. 182 0022-247X/98 $25.00 Copyright © 1998 by Academic Press All rights of reproduction in any form reserved.
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Here is an outline of the paper: In Section 2, we recall the construction of the curve Cγ and its known properties. In Section 3, we first define upper and lower approximations of L as the lengths of two approximating polygons. They form two sequences Ln and ln which both converge monotonically to L. In Section 4, we study the convergence speed of Ln to L. Numerical computations strongly suggest that both sequences Ln and ln converge lineary to L for all γ > 1 (various examples are given in Section 6). We prove that the existence of Ln+1 − Ln ; n→+∞ Ln − Ln−1
q = lim
q ∈ 0; 1; with q 6=
γ ; γ+2
implies the existence of Ln+1 − L n→+∞ Ln − L lim
and
ln+1 − L y n→+∞ l n − L lim
moreover these limits are also q. But we did not succeed in proving the existence of this limit, except for the parabola (γ = 2). In that case, we prove in Section 5 that both limits are equal to 1/4. However, in the general case (γ > 1), we can prove that there exist constants c > 0 and 0 < κ < 1 depending on γ such that Ln+1 − Ln ≤ cκn . This shows that the convergence of Ln is at worst linear. Finally, this suggests the possibility of accelerating the convergence of the two sequences Ln and ln by the ε-algorithm or the iterated Aitken’s 12 algorithm (see e.g., [1, Chap. 2]), since they do not need the knowledge of the exact rate of convergence of these sequences.
2. CONSTRUCTION AND PROPERTIES OF THE DE RHAM CURVE Let ABC be a triangle. The curve is the limit of a sequence of polygons, P n ; n = 0; 1; 2; : : : starting with P 0 = A; B; C. Then the points dividing in three parts the sides of the polygon P n obtained at the nth step are the vertices of the next one. The three parts have lengths proportional to 1; γ; 1 successively. The number of sides of P n is 2n + 1. We denote by S0n ; S1n ; : : : ; S2nn +1 the vertices of P n . The construction of de Rham in order to get the next polygon P n+1 = S0n+1 ; S1n+1 ; : : : ; S2n+1 n+1 +1 n+1 n n n is as follows: S = 1 − βS + βS from the previous one P i 2i i+1 and n+1 n n n S2i+1 = βSi + 1 − βSi+1 for i = 0; : : : ; 2 , where β = 1/γ + 2. In Fig. 1, we show the first step in the construction of de Rham with P 0 = A; B; C and P 1 = A0 ; B0 ; C 0 ; D0 .
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FIG. 1. The first step in the construction of de Rham.
The following properties are given by de Rham. • The polygons P n are convex and the sequence P n converges to a curve Cγ which is continuous and convex. • Cγ is tangent at the midpoint of each side of P n . • If γ > 1, Cγ has a tangent at each point and the slope m is continuous. • For γ = 2, C2 is an arc of a parabola from the midpoint of AB to the midpoint of BC. For the next sections, we shall suppose γ > 1.
3. UPPER AND LOWER APPROXIMATIONS OF THE LENGTH OF THE CURVE We denote by M0n ; M1n ; : : : ; M2nn the midpoints of the sides of P n . Let Ln be the length of P n measured from the midpoint M0n of the first side to the midpoint M2nn of the last one, and let ln be the length of the polygonal line joining the midpoints: M0n M1n · · · M2nn . With this notation, Min = Sin + n+1 n /2 and M2i = Min . We write U for the euclidean norm of the vector Si+1 U. Thus, we have L0 = M00 S10 + S10 M10 = AB + BC/2; l0 = AC/2;
length of the de rham curve
185
and for all n ∈ Ln =
n 2X −1
i=0
ln =
n 2X −1
i=0
n n n Min Si+1 + Si+1 Mi+1 ;
n Min Mi+1 :
From now on, we shall omit the upper index n in Min if it is not necessary for the comprehension; similarly, we shall write Mj0 instead of Mjn+1 . Proposition 1.
For all n ∈ , the following holds: Ln+1 =
γLn + 2ln : γ+2
Proof. Let j = 2i, then for i = 0; : : : ; 2n we have, 0 0 0 0 0 0 0 + Sj+1 Mj+1 + Mj+1 Sj+2 + Sj+2 Mj+2 Mj0 Sj+1 0 0 0 0 0 = Mj0 Sj+1 + Sj+1 Sj+2 + Sj+2 Mj+2
=
2 γ γ M S + M M + S M γ + 2 i i+1 γ + 2 i i+1 γ + 2 i+1 i+1
=
2 γ Mi Si+1 + Si+1 Mi+1 + M M : γ+2 γ + 1 i i+1
Adding these equalities gives the result, since Ln+1 =
2n+1 X−1 j=0
0 0 0 Mj0 Sj+1 + Sj+1 Mj+1 :
Proposition 2. The two sequences (Ln ) and (ln ) are respectively decreasing and increasing and they converge to the same limit L, which is the length of Cγ . Proof. Again let j = 2i: 0 0 0 0 ≤ Mj0 Mj+1 + Mj+1 Mj+2 , therefore ln ≤ ln+1 . • Mi Mi+1 = Mj0 Mj+2 • Similarly,
Mi Si+1 + Si+1 Mi+1 0 0 0 0 + Sj+1 Si+1 + Si+1 Sj+2 + Sj+2 Mi+1 = Mi Sj+1 0 0 0 0 ≥ Mi Sj+1 + Sj+1 Sj+2 + Sj+2 Mi+1 0 0 0 0 0 0 0 = Mj0 Sj+1 + Sj+1 Mj+1 + Mj+1 Sj+2 + Sj+2 Mj+2 ;
therefore Ln ≥ Ln+1 .
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FIG. 2. The main parameters in the construction of de Rham.
• Now Ln ≥ AC/2 and ln ≤ AB + BC/2, so that both sequences are converging, respectively, to L and l. From the preceding proposition, we deduce L = l, which is the length of Cγ .
4. ON THE RATE OF CONVERGENCE OF THE SEQUENCE Ln We shall now study the asymptotic behavior of the sequence Ln ; we first prove that when n goes to +∞ the ratios of the lengths of any two consecutive sides of P n are uniformly bounded, then we prove that the angle between two sides tends to π. With the help of these preliminary results we shall be able to study the convergence of Ln . We denote by λ0 = S0 S1 , λ1 = S1 S2 ; : : :, λ2n = S2n S2n +1 the successive lengths of the sides of P n and by λ00 ; λ01 ; : : : ; λ02n+1 those of P n+1 . The sides of P n (resp. P n+1 ) make angles θ0 = ÜS0 S1 ; S1 S2 ; : : : ; θ2n −1 = ÜS2n −1 S2n ; S2n S2n +1 (resp. θ00 ; : : : ; θ20 n+1 −1 ). See Fig 2. We shall study the ratio Ln+1 − Ln /Ln − Ln−1 ; indeed, Brezinski and Redivo Zaglia have proved in [1] that, for a sequence un converging to u ∈ and for q ∈ with q 6= 1, un+1 − u =q n→+∞ un − u lim
if and only if
un+1 − un = q: n→+∞ un − un−1 lim
With the next proposition, we shall be able to get the rate of convergence of ln from the one of Ln . Proposition 3.
If
Ln+1 − L =q n→+∞ Ln − L lim
with q 6= 1 and q 6=
γ γ+2
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then
ln+1 − L Ln+1 − ln+1 = q: = lim n n→+∞ l − L n→+∞ Ln − ln Proof. By Proposition 1, γ + 2Ln+1 = γLn + 2ln ; then γ + 2L = γL + 2L. So that, by difference, γ + 2Ln+1 − L − γLn − L = 2ln − L and γ + 2Ln − L − γLn−1 − L = 2ln−1 − L from the preceding step. By division, we successively get lim
ln − L γ + 2Ln+1 − L − γLn − L = n−1 ; n n−1 γ + 2L − L − γL − L l −L ln − L γ + 2Ln+1 − L/Ln − L − γ = : γ + 2 − γLn−1 − L/Ln − L ln−1 − L As n tends to +∞, we get γ + 2q − γ ln+1 − L = lim n→+∞ l n − L γ + 2 − γ/q as soon as q 6= γ/γ + 2. Similarly, using γ + 2Ln+1 − L − Ln − L = 2ln − Ln and the preceding step, we easily get q−1 l n − Ln q= = lim n−1 1 − 1/q n→+∞ l − Ln−1 as soon as q 6= 1. q=
Proposition 4.
For every n ∈ we have s ! n 2X −1 4λi λi+1 cos2 θi /2 1 n+1 n L −L = λ + λi+1 1− −1 : γ + 2 i=0 i λi + λi+1 2
Proof. We recall that β = 1/γ + 2; we set α = γ/γ + 2 and again j = 2i. Then, for i = 0; 1; : : : ; 2n − 1, writing ai = βλi and bi = βλi+1 , we have λ0j = αλi ; q λ0j+1 = a2i + b2i − 2ai bi cos θi r θ = ai + bi 2 − 4ai bi cos2 i 2 r θ = β λi + λi+1 2 − 4λi λi+1 cos2 i 2 s 4λi λi+1 θ = βλi + λi+1 1 − cos2 i ; 2 λi + λi+1 2 and λ02n+1 = αλ2n .
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P2n −1 λi + λ2n /2 and a similar formula for Ln+1 , we Using Ln = λ0 /2 + i=1 n n+1 n can evaluate ε = L −L : λ λn εn = α 0 + αλ1 + · · · + αλ2n −1 + α 2 2 2 s n 2X −1 4λi λi+1 θ + βλi + λi+1 1 − cos2 i 2 λ + λ 2 i i+1 i=0 λn λ0 + λ1 + · · · + λ2n −1 + 2 − 2 2 s ! n 2X −1 α − 1λi + λi+1 4λi λi+1 θi 2 = + βλi + λi+1 1 − cos λi + λi+1 2 2 2 i=0 s ! n 2X −1 4λi λi+1 θ =β λi + λi+1 1− cos2 i − 1 : 2 λ + λ 2 i i+1 i=0 Proposition 5. Let rin = λni+1 /λni for i ∈ 0; : : : ; 2n be the ratios of two successive lengths at step n. If γ > 1, then there exist r 0 and R0 such that, for every n ∈ , for every i ∈ 0; : : : ; 2n , r 0 ≤ rin ≤ R0 : Proof. First, let us remark that the angles θin of P n are bounded away from 0 since θ00 ≤ θin ≤ π. We can suppose that there exists θ0 ∈ 0; π such that θ0 ≤ θin ≤ π: Now, omitting n and i, let r = λi+1 /λi , θ = ÜSi Si+1 ; Si+1 Si+2 and r10 =
λ0j+1 λ0j
;
r20 =
λ0j+2
λ0j+1
;
with j = 2i:
Then r10 =
1 2 r − 2r cos θ + 11/2 γ
and
r20 =
r 2
γr y − 2r cos θ + 11/2
see Fig 2. We consider the two functions f and g defined by f x =
1 2 x − 2x cos θ + 11/2 γ
and gx = γxx2 − 2x cos θ + 1−1/2 with x > 0. So we have r10 = f r and r20 = gr.
length of the de rham curve
189
Since x2 − 2x cos θ + 1 ≤ x2 + 2x + 1 = 1 + x2 ; x2 − 2x cos θ + 1 ≥ x2 − 2x cos θ0 + 1 ≥ sin2 θ0 ; we get the following bounds for f x, x ∈ + : 1+x 1 sin θ0 ≤ f x ≤ ≤ max x; ; γ γ γ−1 and similarly, we obtain minx; γ − 1 ≤ gx ≤ Now let r 0 = min
BC sin θ0 ; ;γ − 1 AB γ
and
γ : sin θ0
R0 = max
BC 1 γ ; ; : AB sin θ0 γ − 1
Using the above inequalities,we immediately get the result by induction on n ∈ . Proposition 6. There exists c ∈ and q ∈ 0; 1 such that for every n ∈ , for every i ∈ 1; : : : ; 2n , π − θin ≤ cqn . Proof. In the following computations, we are using Fig. 2 and Fig. 3. Let us consider in the polygonal line P n the triangle Si Si+1 Si+2 . We set θ = ÜSi Si+1 ; Si+1 Si+2 . If r = Si+1 Si+2 /Si Si+1 , then the triangle Si Si+1 Si+2 is similar to the triangle whose vertices are 1; 0, 0; 0, 0 0 0 ; Sj+1 Sj+2 and r cos θ; r sin θ. If j = 2i, we denote by ϕ1 = ÜSj0 Sj+1 0 0 0 0 0 0 ϕ2 = ÜSj+1 Sj+2 ; Sj+2 Sj+3 . Since the triangle Sj Si+1 Sj+2 is similar to Si Si+1 Si+2 , the vector W whose endpoints are 1; 0 and r cos θ; r sin θ makes an angle ϕ1 with the x-axis. If w = W , one has the vector identity w cos ϕ1 ; w sin ϕ1 = r cos θ − 1; r sin θ. From this, it follows that 1 1p 1 + cot2 θ: = cot θ − cot ϕ1 = cot θ − r sin θ r
FIG. 3. Computation of angles in the construction of de Rham.
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dubuc, merrien and sablonni` ere
Moreover ϕ2 = θ − π − ϕ1 , hence cot ϕ2 =
p 1 + cot θ cot ϕ1 = cot θ − r 1 + cot2 θ: cot ϕ1 − cot θ
0 , and ri = λi+1 /λi , with Setting mi = cot θi , m0j = cot θj0 , m0j+1 = cot θj+1 j = 2i, we get p q 1 + mi 2 0 0 and mj+1 = mi − ri 1 + mi 2 : mj = mi − ri q q Also, 1 + m2i ≥ 1 and 1 + m2i ≥ −mi imply
m0j+1 ≤ mi − ri ;
m0j ≤ mi − 1/ri ; m0j ≤ m0i 1 + 1/ri ;
and
m0j+1 ≤ mi 1 + ri :
Let ρ = min1/R0 ; r 0 , where R0 and r 0 are defined in the preceding proposition. By induction, there exists c1 ∈ such that, for all n ∈ and i ∈ 1; : : : ; 2n , mni ≤ c1 − nρ, so that mni tends to −∞ as n tends to +∞ and the angles θin tend to π. Similarly, there exists c2 < 0 such that, for all n sufficiently large and for i ∈ 1; : : : ; 2n , mni ≤ c2 1 + ρn . Then there exists a constant c3 > 0 such that sinπ − θin = p
1 1 + mni 2
≤ c3
1 = c3 qn 1 + ρn
with q ∈ 0; 1 and we can conclude that for some constant c4 > 0 the following holds: π − θin ≤ c4 qn :
Proposition 7. n ∈ ;
∗ There exists c ∈ + and κ ∈ 0; 1 such that, for every
Ln+1 − Ln ≤ cκn : Proof.
From Proposition 4 we know that s ! n 2X −1 2 θ /2 4λ λ cos 1 i i+1 i λ + λi+1 1− −1 : εn = Ln+1 − Ln = γ + 2 i=0 i λi + λi+1 2
length of the de rham curve √
Now as
191
1 − x ≥ 1 − x for x ∈ 0; 1, we get n
εn ≤
−1 4λi λi+1 cos2 θi /2 1 2X γ + 2 i=0 λi + λi+1 n
−1 2 2X π − θi ≤ λi + λi+1 sin2 γ + 2 i=0 2
≤
c q2n 2 2L 4 = cκn : γ+2 2
5. THE CASE OF THE PARABOLA, γ = 2 Starting with the triangle ABC, for γ = 2, we obtain an arc of parabola C2 joining the midpoint S0 of AB to the midpoint S2 of BC. Let S1 = B, 1Si = Si+1 − Si , and 12 Si = 1Si+1 − 1Si . In this particular case, we are able to evaluate the length L of C2 and to estimate the convergence rates of the sequences Ln and ln . Proposition 8.
The length L of the curve C2 is equal to 1S0 1S1 2 − 1S0 :1S1 2 1S1 :12 S0 + 1S1 12 S0 L= ln 3 1S0 :12 S0 + 1S0 12 S0 12 S0 2
+ Proof.
1S1 1S1 :12 S0 − 1S0 1S0 :12 S0 2
12 S0
:
The equation of the parabola is Mt = S0 1 − t2 + 2S1 t1 − t + S2 t 2
with t ∈ 0; 1:
Since M 0 t = 21S0 1 − t + 1S1 t, we obtain Z1 Z1 L= M 0 t dt = 2 pt1/2 dt; 0
0
2
where pt = α0 1 − t + 2α1 t1 − t + α2 t 2 , α0 = 1S0 2 , α1 = 1S0 :1S1 , and α2 = 1S1 2 . Let 1α0 = α1 − α0 = 1S0 :12 S0 ;
1α1 = α2 − α1 = 1S1 :12 S0 ;
and 12 α0 = α2 − 2α1 + α0 = 12 S0 2 : Then we have α0 α2 − α21 = 1S0 2 1S1 2 − 1S0 :1S1 2 ≥ 0 by the Schwarz inequality.
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dubuc, merrien and sablonni` ere
Now, let a2 = α0 α2 − α21 /12 α0 . Then the change of variable q 1α u = t 12 α0 + p 0 12 α0 in the integral gives L= p
Z
2 12 α0
u1 u0
p
u2 + a2 du;
p p with u0 = 1α0 / 12 α0 and u1 = 1α1 / 12 α0 , from which we immediately deduce q 2 q q 2 2 a u u1 + u1 + a u1 L= p ln + 2 u21 + a2 − 20 u20 + a2 : q a a 2 12 α0 2 u0 + u0 + a Since
q
u20 + a2 = 1S0 , p
u0 12 α0
=
q
u21 + a2 = 1S1 , and
1S0 :12 S0 ; 12 S0 2
p
u1 12 α0
=
1S1 :12 S0 ; 12 S0 2
and p
a2 12 α0
=
1S0 2 1S1 2 − 1S0 :1S1 2 ; 12 S0 3
we obtain the desired result. Proposition 9. More specifically,
The two sequences Ln and ln converge linearly to L.
1 Ln+1 − L ln+1 − L Ln+1 − ln+1 = : = lim = lim n n n n n→+∞ L − L n→+∞ l − L n→+∞ L −l 4 lim
Proof. For any function g ∈ C ∞ 0; 1, there hold, respectively, the two following asymptotic expansions for the values of the trapezoidal rule and of the midpoint rule with step length 1/2n (see, e.g., [2, pp. 189–190]): n Z1 2X −1 1 1 i 1 g0 + gt dt − n g n + g1 2 2 2 2 0 i=1 =− Z
1 0
1 g0 1 − g0 0 + O4−2n ; 12 4n
2n 1 X 2i − 1 1 g0 1 − g0 0 gt dt − n g + O4−2n : = 2 i=1 2n+1 24 4n
length of the de rham curve
193
This proves that the two sequences converge linearly to the integral with a convergence rate equal to 1/4. Applying these results to gt = M 0 t gives the two first announced limits, since Z1 M 0 tdt; L= 0
Ln = and
n 2X −1 1 1 0 M 0 i + 1 M 0 1 M 0 + 2n 2 2n 2 i=1
2n 0 2i − 1 1 X : l = n M 2 i=1 2n+1 n
It suffices to prove these equalities for n = 0, as the general formulas are summations of local formulas in each interval i/2n ; i + 1/2n , i ∈ 0; : : : ; 2n − 1. With the notation of the previous proposition, we easily compute L0 = 1S0 + 1S1
and
l0 = S0 S2 = 1S0 + 1S1 :
Since M 0 0 = 21S0 , M 0 1 = 21S1 , and M 0 1/2 = 1S0 + 1S1 , we see immediately that 1 1 0 0 0 0 L = M 0 + M 1 and l = M 0 2 2 are, respectively, the values of theR trapezoidal rule and of the midpoint 1 rule applied to the integral L = 0 M 0 t dt. Similarly, Ln and ln are, respectively, the values of the composed trapezoidal and midpoint rules, with a step of length h = 1/2n , applied to the same integral. Finally, we compute g0 0 = 2
1S0 :12 S0 = 212 S0 cos θ0 ; 1S0
g0 1 = 2
1S1 :12 S0 = 212 S0 cos θ1 ; 1S1
where θ0 = Ü1S0 ; 12 S0 and θ1 = Ü1S1 ; 12 S0 , hence g0 1 − g0 0 = 212 S0 cos θ1 − cos θ0 6= 0: Similarly, the first term of the asymptotic expansion of Ln − ln being equal to 18 g0 1 − g0 0/4n , we deduce the third limit in the same way.
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dubuc, merrien and sablonni` ere 6. EXAMPLES
In an orthonormal basis, let A = −1; 1, B = 0; 0, and C = 1; 2. The following three tables show the computed values of the successive lengths Ln and ln and the ratios of their differences for different values of γ: n
Ln
ln
0 1 2 :: :
1.825141 1.421080 1.331533 :: :
1.118034 1.264372 1.294985 :: :
n
Ln
ln
0 1 2 :: :
1.825141 1.471587 1.393683 :: :
1.118034 1.315779 1.355960 :: :
n
Ln
ln
0 1 2 :: :
1.825141 1.542298 1.484878 :: :
1.118034 1.398749 1.447380 :: :
10 11
10 11
10 11
1.3038891 1.3038888
1.3685601 1.3685599
1.4650072 1.4650070
1.3038886 1.3038887
1.3685596 1.3685597
1.4650067 1.4650069
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.221618 :: :
0.209189 :: :
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.220347 :: :
0.203198 :: :
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.203009 :: :
0.173241 :: :
0.251342 0.251460
0.250000 0.250000
0.248206 0.248208
n−1
γ = 1:5,
0.251176 0.251360
n−1
γ = 2,
0.250000 0.250000
n−1
γ = 3.
0.248205 0.248208
The last table shows the computed maximum and minimum values of λni+1 /λni at step n = 11; these extrema seem to converge to γ − 1 and 1/γ − 1, respectively, for γ ≥ 2 and to 1/γ − 1 and γ − 1 for 1 < γ ≤ 2: γ
n = 11,
1.5 1.8 2 2.5 3 5 10
max
λni+1 λni
1.986 1.251 1.001 1.500 2.000 4.000 9.000
min
λni+1 λni
0.506 0.801 0.999 0.667 0.500 0.250 0.111
length of the de rham curve
195
REFERENCES 1. C. Brezinski and M. Redivo Zaglia, “Extrapolation Methods, Theory and Practice,” NorthHolland, New York, 1991. 2. L. M. Milne-Thomson, “The Calculus of Finite Differences,” 2nd ed., Chelsea, New York, 1981. 3. G. de Rham, Sur une courbe plane, J. Math. Pures Appl. (9) 35 (1956), 25–42. 4. J. A. Roulier and B. Piper, Prescribing the length of rational B´ezier curves, Comput. Aided Geom. Design 13 (1996), 23–43. 5. J. A Roulier, Specifying the arc length of B´ezier curves, Comput. Aided Geom. Design 10 (1993), 25–56.
The Length of the de Rham Curve Serge Dubuc D´epartement de Math´ematiques et de Statistique, Universit´e de Montr´eal, C.P. 6128 Succursale Centre-ville, Montr´eal (Qu´ebec), Canada H3C 3J7
and Jean-Louis Merrien and Paul Sablonni`ere INSA de Rennes, 20 avenue des Buttes de Coesmes, 35043 Rennes Cedex, France Submitted by Joseph D. Ward Received February 4, 1997
The length L of the de Rham curve is the common limit of two monotonic sequences of lengths ln and Ln of inscribed and circumscribed polygons, respectively. Numerical computations show that their convergence is linear with the same convergence rate. This result is easy to prove for the parabola. For arbitrary de Rham curves, we prove two nearby results. First, the existence of a limit q ∈ 0; 1 of the sequence of ratios Ln+1 − L/Ln − L implies the convergence to the same limit of the two sequences ln+1 − L/ln − L and Ln+1 − ln+1 /Ln − ln . Second, the sequence Ln+1 − Ln is bounded by a convergent geometric sequence. In practice, this allows us to accelerate the convergence of both sequences by standard extrapolation algorithms. © 1998 Academic Press
1. INTRODUCTION The de Rham curve Cγ , studied in [3], is the limit of a sequence of polygons depending on a parameter γ. We are interested in the computation of the length L of this curve. This problem was already considered by other authors in the context of computer-aided geometric design, in particular for piecewise polynomial or rational curves (see, e.g., [4] and [5]). In a further paper, we shall develop applications to the problem of constructing an interpolating convex curve with prescribed length. 182 0022-247X/98 $25.00 Copyright © 1998 by Academic Press All rights of reproduction in any form reserved.
length of the de rham curve
183
Here is an outline of the paper: In Section 2, we recall the construction of the curve Cγ and its known properties. In Section 3, we first define upper and lower approximations of L as the lengths of two approximating polygons. They form two sequences Ln and ln which both converge monotonically to L. In Section 4, we study the convergence speed of Ln to L. Numerical computations strongly suggest that both sequences Ln and ln converge lineary to L for all γ > 1 (various examples are given in Section 6). We prove that the existence of Ln+1 − Ln ; n→+∞ Ln − Ln−1
q = lim
q ∈ 0; 1; with q 6=
γ ; γ+2
implies the existence of Ln+1 − L n→+∞ Ln − L lim
and
ln+1 − L y n→+∞ l n − L lim
moreover these limits are also q. But we did not succeed in proving the existence of this limit, except for the parabola (γ = 2). In that case, we prove in Section 5 that both limits are equal to 1/4. However, in the general case (γ > 1), we can prove that there exist constants c > 0 and 0 < κ < 1 depending on γ such that Ln+1 − Ln ≤ cκn . This shows that the convergence of Ln is at worst linear. Finally, this suggests the possibility of accelerating the convergence of the two sequences Ln and ln by the ε-algorithm or the iterated Aitken’s 12 algorithm (see e.g., [1, Chap. 2]), since they do not need the knowledge of the exact rate of convergence of these sequences.
2. CONSTRUCTION AND PROPERTIES OF THE DE RHAM CURVE Let ABC be a triangle. The curve is the limit of a sequence of polygons, P n ; n = 0; 1; 2; : : : starting with P 0 = A; B; C. Then the points dividing in three parts the sides of the polygon P n obtained at the nth step are the vertices of the next one. The three parts have lengths proportional to 1; γ; 1 successively. The number of sides of P n is 2n + 1. We denote by S0n ; S1n ; : : : ; S2nn +1 the vertices of P n . The construction of de Rham in order to get the next polygon P n+1 = S0n+1 ; S1n+1 ; : : : ; S2n+1 n+1 +1 n+1 n n n is as follows: S = 1 − βS + βS from the previous one P i 2i i+1 and n+1 n n n S2i+1 = βSi + 1 − βSi+1 for i = 0; : : : ; 2 , where β = 1/γ + 2. In Fig. 1, we show the first step in the construction of de Rham with P 0 = A; B; C and P 1 = A0 ; B0 ; C 0 ; D0 .
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FIG. 1. The first step in the construction of de Rham.
The following properties are given by de Rham. • The polygons P n are convex and the sequence P n converges to a curve Cγ which is continuous and convex. • Cγ is tangent at the midpoint of each side of P n . • If γ > 1, Cγ has a tangent at each point and the slope m is continuous. • For γ = 2, C2 is an arc of a parabola from the midpoint of AB to the midpoint of BC. For the next sections, we shall suppose γ > 1.
3. UPPER AND LOWER APPROXIMATIONS OF THE LENGTH OF THE CURVE We denote by M0n ; M1n ; : : : ; M2nn the midpoints of the sides of P n . Let Ln be the length of P n measured from the midpoint M0n of the first side to the midpoint M2nn of the last one, and let ln be the length of the polygonal line joining the midpoints: M0n M1n · · · M2nn . With this notation, Min = Sin + n+1 n /2 and M2i = Min . We write U for the euclidean norm of the vector Si+1 U. Thus, we have L0 = M00 S10 + S10 M10 = AB + BC/2; l0 = AC/2;
length of the de rham curve
185
and for all n ∈ Ln =
n 2X −1
i=0
ln =
n 2X −1
i=0
n n n Min Si+1 + Si+1 Mi+1 ;
n Min Mi+1 :
From now on, we shall omit the upper index n in Min if it is not necessary for the comprehension; similarly, we shall write Mj0 instead of Mjn+1 . Proposition 1.
For all n ∈ , the following holds: Ln+1 =
γLn + 2ln : γ+2
Proof. Let j = 2i, then for i = 0; : : : ; 2n we have, 0 0 0 0 0 0 0 + Sj+1 Mj+1 + Mj+1 Sj+2 + Sj+2 Mj+2 Mj0 Sj+1 0 0 0 0 0 = Mj0 Sj+1 + Sj+1 Sj+2 + Sj+2 Mj+2
=
2 γ γ M S + M M + S M γ + 2 i i+1 γ + 2 i i+1 γ + 2 i+1 i+1
=
2 γ Mi Si+1 + Si+1 Mi+1 + M M : γ+2 γ + 1 i i+1
Adding these equalities gives the result, since Ln+1 =
2n+1 X−1 j=0
0 0 0 Mj0 Sj+1 + Sj+1 Mj+1 :
Proposition 2. The two sequences (Ln ) and (ln ) are respectively decreasing and increasing and they converge to the same limit L, which is the length of Cγ . Proof. Again let j = 2i: 0 0 0 0 ≤ Mj0 Mj+1 + Mj+1 Mj+2 , therefore ln ≤ ln+1 . • Mi Mi+1 = Mj0 Mj+2 • Similarly,
Mi Si+1 + Si+1 Mi+1 0 0 0 0 + Sj+1 Si+1 + Si+1 Sj+2 + Sj+2 Mi+1 = Mi Sj+1 0 0 0 0 ≥ Mi Sj+1 + Sj+1 Sj+2 + Sj+2 Mi+1 0 0 0 0 0 0 0 = Mj0 Sj+1 + Sj+1 Mj+1 + Mj+1 Sj+2 + Sj+2 Mj+2 ;
therefore Ln ≥ Ln+1 .
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FIG. 2. The main parameters in the construction of de Rham.
• Now Ln ≥ AC/2 and ln ≤ AB + BC/2, so that both sequences are converging, respectively, to L and l. From the preceding proposition, we deduce L = l, which is the length of Cγ .
4. ON THE RATE OF CONVERGENCE OF THE SEQUENCE Ln We shall now study the asymptotic behavior of the sequence Ln ; we first prove that when n goes to +∞ the ratios of the lengths of any two consecutive sides of P n are uniformly bounded, then we prove that the angle between two sides tends to π. With the help of these preliminary results we shall be able to study the convergence of Ln . We denote by λ0 = S0 S1 , λ1 = S1 S2 ; : : :, λ2n = S2n S2n +1 the successive lengths of the sides of P n and by λ00 ; λ01 ; : : : ; λ02n+1 those of P n+1 . The sides of P n (resp. P n+1 ) make angles θ0 = ÜS0 S1 ; S1 S2 ; : : : ; θ2n −1 = ÜS2n −1 S2n ; S2n S2n +1 (resp. θ00 ; : : : ; θ20 n+1 −1 ). See Fig 2. We shall study the ratio Ln+1 − Ln /Ln − Ln−1 ; indeed, Brezinski and Redivo Zaglia have proved in [1] that, for a sequence un converging to u ∈ and for q ∈ with q 6= 1, un+1 − u =q n→+∞ un − u lim
if and only if
un+1 − un = q: n→+∞ un − un−1 lim
With the next proposition, we shall be able to get the rate of convergence of ln from the one of Ln . Proposition 3.
If
Ln+1 − L =q n→+∞ Ln − L lim
with q 6= 1 and q 6=
γ γ+2
length of the de rham curve
187
then
ln+1 − L Ln+1 − ln+1 = q: = lim n n→+∞ l − L n→+∞ Ln − ln Proof. By Proposition 1, γ + 2Ln+1 = γLn + 2ln ; then γ + 2L = γL + 2L. So that, by difference, γ + 2Ln+1 − L − γLn − L = 2ln − L and γ + 2Ln − L − γLn−1 − L = 2ln−1 − L from the preceding step. By division, we successively get lim
ln − L γ + 2Ln+1 − L − γLn − L = n−1 ; n n−1 γ + 2L − L − γL − L l −L ln − L γ + 2Ln+1 − L/Ln − L − γ = : γ + 2 − γLn−1 − L/Ln − L ln−1 − L As n tends to +∞, we get γ + 2q − γ ln+1 − L = lim n→+∞ l n − L γ + 2 − γ/q as soon as q 6= γ/γ + 2. Similarly, using γ + 2Ln+1 − L − Ln − L = 2ln − Ln and the preceding step, we easily get q−1 l n − Ln q= = lim n−1 1 − 1/q n→+∞ l − Ln−1 as soon as q 6= 1. q=
Proposition 4.
For every n ∈ we have s ! n 2X −1 4λi λi+1 cos2 θi /2 1 n+1 n L −L = λ + λi+1 1− −1 : γ + 2 i=0 i λi + λi+1 2
Proof. We recall that β = 1/γ + 2; we set α = γ/γ + 2 and again j = 2i. Then, for i = 0; 1; : : : ; 2n − 1, writing ai = βλi and bi = βλi+1 , we have λ0j = αλi ; q λ0j+1 = a2i + b2i − 2ai bi cos θi r θ = ai + bi 2 − 4ai bi cos2 i 2 r θ = β λi + λi+1 2 − 4λi λi+1 cos2 i 2 s 4λi λi+1 θ = βλi + λi+1 1 − cos2 i ; 2 λi + λi+1 2 and λ02n+1 = αλ2n .
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dubuc, merrien and sablonni` ere
P2n −1 λi + λ2n /2 and a similar formula for Ln+1 , we Using Ln = λ0 /2 + i=1 n n+1 n can evaluate ε = L −L : λ λn εn = α 0 + αλ1 + · · · + αλ2n −1 + α 2 2 2 s n 2X −1 4λi λi+1 θ + βλi + λi+1 1 − cos2 i 2 λ + λ 2 i i+1 i=0 λn λ0 + λ1 + · · · + λ2n −1 + 2 − 2 2 s ! n 2X −1 α − 1λi + λi+1 4λi λi+1 θi 2 = + βλi + λi+1 1 − cos λi + λi+1 2 2 2 i=0 s ! n 2X −1 4λi λi+1 θ =β λi + λi+1 1− cos2 i − 1 : 2 λ + λ 2 i i+1 i=0 Proposition 5. Let rin = λni+1 /λni for i ∈ 0; : : : ; 2n be the ratios of two successive lengths at step n. If γ > 1, then there exist r 0 and R0 such that, for every n ∈ , for every i ∈ 0; : : : ; 2n , r 0 ≤ rin ≤ R0 : Proof. First, let us remark that the angles θin of P n are bounded away from 0 since θ00 ≤ θin ≤ π. We can suppose that there exists θ0 ∈ 0; π such that θ0 ≤ θin ≤ π: Now, omitting n and i, let r = λi+1 /λi , θ = ÜSi Si+1 ; Si+1 Si+2 and r10 =
λ0j+1 λ0j
;
r20 =
λ0j+2
λ0j+1
;
with j = 2i:
Then r10 =
1 2 r − 2r cos θ + 11/2 γ
and
r20 =
r 2
γr y − 2r cos θ + 11/2
see Fig 2. We consider the two functions f and g defined by f x =
1 2 x − 2x cos θ + 11/2 γ
and gx = γxx2 − 2x cos θ + 1−1/2 with x > 0. So we have r10 = f r and r20 = gr.
length of the de rham curve
189
Since x2 − 2x cos θ + 1 ≤ x2 + 2x + 1 = 1 + x2 ; x2 − 2x cos θ + 1 ≥ x2 − 2x cos θ0 + 1 ≥ sin2 θ0 ; we get the following bounds for f x, x ∈ + : 1+x 1 sin θ0 ≤ f x ≤ ≤ max x; ; γ γ γ−1 and similarly, we obtain minx; γ − 1 ≤ gx ≤ Now let r 0 = min
BC sin θ0 ; ;γ − 1 AB γ
and
γ : sin θ0
R0 = max
BC 1 γ ; ; : AB sin θ0 γ − 1
Using the above inequalities,we immediately get the result by induction on n ∈ . Proposition 6. There exists c ∈ and q ∈ 0; 1 such that for every n ∈ , for every i ∈ 1; : : : ; 2n , π − θin ≤ cqn . Proof. In the following computations, we are using Fig. 2 and Fig. 3. Let us consider in the polygonal line P n the triangle Si Si+1 Si+2 . We set θ = ÜSi Si+1 ; Si+1 Si+2 . If r = Si+1 Si+2 /Si Si+1 , then the triangle Si Si+1 Si+2 is similar to the triangle whose vertices are 1; 0, 0; 0, 0 0 0 ; Sj+1 Sj+2 and r cos θ; r sin θ. If j = 2i, we denote by ϕ1 = ÜSj0 Sj+1 0 0 0 0 0 0 ϕ2 = ÜSj+1 Sj+2 ; Sj+2 Sj+3 . Since the triangle Sj Si+1 Sj+2 is similar to Si Si+1 Si+2 , the vector W whose endpoints are 1; 0 and r cos θ; r sin θ makes an angle ϕ1 with the x-axis. If w = W , one has the vector identity w cos ϕ1 ; w sin ϕ1 = r cos θ − 1; r sin θ. From this, it follows that 1 1p 1 + cot2 θ: = cot θ − cot ϕ1 = cot θ − r sin θ r
FIG. 3. Computation of angles in the construction of de Rham.
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dubuc, merrien and sablonni` ere
Moreover ϕ2 = θ − π − ϕ1 , hence cot ϕ2 =
p 1 + cot θ cot ϕ1 = cot θ − r 1 + cot2 θ: cot ϕ1 − cot θ
0 , and ri = λi+1 /λi , with Setting mi = cot θi , m0j = cot θj0 , m0j+1 = cot θj+1 j = 2i, we get p q 1 + mi 2 0 0 and mj+1 = mi − ri 1 + mi 2 : mj = mi − ri q q Also, 1 + m2i ≥ 1 and 1 + m2i ≥ −mi imply
m0j+1 ≤ mi − ri ;
m0j ≤ mi − 1/ri ; m0j ≤ m0i 1 + 1/ri ;
and
m0j+1 ≤ mi 1 + ri :
Let ρ = min1/R0 ; r 0 , where R0 and r 0 are defined in the preceding proposition. By induction, there exists c1 ∈ such that, for all n ∈ and i ∈ 1; : : : ; 2n , mni ≤ c1 − nρ, so that mni tends to −∞ as n tends to +∞ and the angles θin tend to π. Similarly, there exists c2 < 0 such that, for all n sufficiently large and for i ∈ 1; : : : ; 2n , mni ≤ c2 1 + ρn . Then there exists a constant c3 > 0 such that sinπ − θin = p
1 1 + mni 2
≤ c3
1 = c3 qn 1 + ρn
with q ∈ 0; 1 and we can conclude that for some constant c4 > 0 the following holds: π − θin ≤ c4 qn :
Proposition 7. n ∈ ;
∗ There exists c ∈ + and κ ∈ 0; 1 such that, for every
Ln+1 − Ln ≤ cκn : Proof.
From Proposition 4 we know that s ! n 2X −1 2 θ /2 4λ λ cos 1 i i+1 i λ + λi+1 1− −1 : εn = Ln+1 − Ln = γ + 2 i=0 i λi + λi+1 2
length of the de rham curve √
Now as
191
1 − x ≥ 1 − x for x ∈ 0; 1, we get n
εn ≤
−1 4λi λi+1 cos2 θi /2 1 2X γ + 2 i=0 λi + λi+1 n
−1 2 2X π − θi ≤ λi + λi+1 sin2 γ + 2 i=0 2
≤
c q2n 2 2L 4 = cκn : γ+2 2
5. THE CASE OF THE PARABOLA, γ = 2 Starting with the triangle ABC, for γ = 2, we obtain an arc of parabola C2 joining the midpoint S0 of AB to the midpoint S2 of BC. Let S1 = B, 1Si = Si+1 − Si , and 12 Si = 1Si+1 − 1Si . In this particular case, we are able to evaluate the length L of C2 and to estimate the convergence rates of the sequences Ln and ln . Proposition 8.
The length L of the curve C2 is equal to 1S0 1S1 2 − 1S0 :1S1 2 1S1 :12 S0 + 1S1 12 S0 L= ln 3 1S0 :12 S0 + 1S0 12 S0 12 S0 2
+ Proof.
1S1 1S1 :12 S0 − 1S0 1S0 :12 S0 2
12 S0
:
The equation of the parabola is Mt = S0 1 − t2 + 2S1 t1 − t + S2 t 2
with t ∈ 0; 1:
Since M 0 t = 21S0 1 − t + 1S1 t, we obtain Z1 Z1 L= M 0 t dt = 2 pt1/2 dt; 0
0
2
where pt = α0 1 − t + 2α1 t1 − t + α2 t 2 , α0 = 1S0 2 , α1 = 1S0 :1S1 , and α2 = 1S1 2 . Let 1α0 = α1 − α0 = 1S0 :12 S0 ;
1α1 = α2 − α1 = 1S1 :12 S0 ;
and 12 α0 = α2 − 2α1 + α0 = 12 S0 2 : Then we have α0 α2 − α21 = 1S0 2 1S1 2 − 1S0 :1S1 2 ≥ 0 by the Schwarz inequality.
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dubuc, merrien and sablonni` ere
Now, let a2 = α0 α2 − α21 /12 α0 . Then the change of variable q 1α u = t 12 α0 + p 0 12 α0 in the integral gives L= p
Z
2 12 α0
u1 u0
p
u2 + a2 du;
p p with u0 = 1α0 / 12 α0 and u1 = 1α1 / 12 α0 , from which we immediately deduce q 2 q q 2 2 a u u1 + u1 + a u1 L= p ln + 2 u21 + a2 − 20 u20 + a2 : q a a 2 12 α0 2 u0 + u0 + a Since
q
u20 + a2 = 1S0 , p
u0 12 α0
=
q
u21 + a2 = 1S1 , and
1S0 :12 S0 ; 12 S0 2
p
u1 12 α0
=
1S1 :12 S0 ; 12 S0 2
and p
a2 12 α0
=
1S0 2 1S1 2 − 1S0 :1S1 2 ; 12 S0 3
we obtain the desired result. Proposition 9. More specifically,
The two sequences Ln and ln converge linearly to L.
1 Ln+1 − L ln+1 − L Ln+1 − ln+1 = : = lim = lim n n n n n→+∞ L − L n→+∞ l − L n→+∞ L −l 4 lim
Proof. For any function g ∈ C ∞ 0; 1, there hold, respectively, the two following asymptotic expansions for the values of the trapezoidal rule and of the midpoint rule with step length 1/2n (see, e.g., [2, pp. 189–190]): n Z1 2X −1 1 1 i 1 g0 + gt dt − n g n + g1 2 2 2 2 0 i=1 =− Z
1 0
1 g0 1 − g0 0 + O4−2n ; 12 4n
2n 1 X 2i − 1 1 g0 1 − g0 0 gt dt − n g + O4−2n : = 2 i=1 2n+1 24 4n
length of the de rham curve
193
This proves that the two sequences converge linearly to the integral with a convergence rate equal to 1/4. Applying these results to gt = M 0 t gives the two first announced limits, since Z1 M 0 tdt; L= 0
Ln = and
n 2X −1 1 1 0 M 0 i + 1 M 0 1 M 0 + 2n 2 2n 2 i=1
2n 0 2i − 1 1 X : l = n M 2 i=1 2n+1 n
It suffices to prove these equalities for n = 0, as the general formulas are summations of local formulas in each interval i/2n ; i + 1/2n , i ∈ 0; : : : ; 2n − 1. With the notation of the previous proposition, we easily compute L0 = 1S0 + 1S1
and
l0 = S0 S2 = 1S0 + 1S1 :
Since M 0 0 = 21S0 , M 0 1 = 21S1 , and M 0 1/2 = 1S0 + 1S1 , we see immediately that 1 1 0 0 0 0 L = M 0 + M 1 and l = M 0 2 2 are, respectively, the values of theR trapezoidal rule and of the midpoint 1 rule applied to the integral L = 0 M 0 t dt. Similarly, Ln and ln are, respectively, the values of the composed trapezoidal and midpoint rules, with a step of length h = 1/2n , applied to the same integral. Finally, we compute g0 0 = 2
1S0 :12 S0 = 212 S0 cos θ0 ; 1S0
g0 1 = 2
1S1 :12 S0 = 212 S0 cos θ1 ; 1S1
where θ0 = Ü1S0 ; 12 S0 and θ1 = Ü1S1 ; 12 S0 , hence g0 1 − g0 0 = 212 S0 cos θ1 − cos θ0 6= 0: Similarly, the first term of the asymptotic expansion of Ln − ln being equal to 18 g0 1 − g0 0/4n , we deduce the third limit in the same way.
194
dubuc, merrien and sablonni` ere 6. EXAMPLES
In an orthonormal basis, let A = −1; 1, B = 0; 0, and C = 1; 2. The following three tables show the computed values of the successive lengths Ln and ln and the ratios of their differences for different values of γ: n
Ln
ln
0 1 2 :: :
1.825141 1.421080 1.331533 :: :
1.118034 1.264372 1.294985 :: :
n
Ln
ln
0 1 2 :: :
1.825141 1.471587 1.393683 :: :
1.118034 1.315779 1.355960 :: :
n
Ln
ln
0 1 2 :: :
1.825141 1.542298 1.484878 :: :
1.118034 1.398749 1.447380 :: :
10 11
10 11
10 11
1.3038891 1.3038888
1.3685601 1.3685599
1.4650072 1.4650070
1.3038886 1.3038887
1.3685596 1.3685597
1.4650067 1.4650069
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.221618 :: :
0.209189 :: :
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.220347 :: :
0.203198 :: :
Ln − Ln−1 Ln−1 − Ln−2
ln − ln−1 l − ln−2
0.203009 :: :
0.173241 :: :
0.251342 0.251460
0.250000 0.250000
0.248206 0.248208
n−1
γ = 1:5,
0.251176 0.251360
n−1
γ = 2,
0.250000 0.250000
n−1
γ = 3.
0.248205 0.248208
The last table shows the computed maximum and minimum values of λni+1 /λni at step n = 11; these extrema seem to converge to γ − 1 and 1/γ − 1, respectively, for γ ≥ 2 and to 1/γ − 1 and γ − 1 for 1 < γ ≤ 2: γ
n = 11,
1.5 1.8 2 2.5 3 5 10
max
λni+1 λni
1.986 1.251 1.001 1.500 2.000 4.000 9.000
min
λni+1 λni
0.506 0.801 0.999 0.667 0.500 0.250 0.111
length of the de rham curve
195
REFERENCES 1. C. Brezinski and M. Redivo Zaglia, “Extrapolation Methods, Theory and Practice,” NorthHolland, New York, 1991. 2. L. M. Milne-Thomson, “The Calculus of Finite Differences,” 2nd ed., Chelsea, New York, 1981. 3. G. de Rham, Sur une courbe plane, J. Math. Pures Appl. (9) 35 (1956), 25–42. 4. J. A. Roulier and B. Piper, Prescribing the length of rational B´ezier curves, Comput. Aided Geom. Design 13 (1996), 23–43. 5. J. A Roulier, Specifying the arc length of B´ezier curves, Comput. Aided Geom. Design 10 (1993), 25–56.