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A sufficient condition for the convexity of the area of an isoptic curve of an oval

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A Sufficient Condition for the Convexity of the Area of an Isoptic Curve of an Oval.

M. MICHALSKA(*)

ABSTRACT - The problem of convexity of the area function of ana-isoptic of the convex curve C is considered.

1. Introduction.

An a-isoptic Ca of a plane, closed, convex curve C is a set of those points in the plane from which the curve C is seen under a fixed angle p2a. In [1] the authors discussed the problem of convexity of a-isoptic curves. We will study the problem of convexity of the area function.

Letp(t),t[ 0 , 2p] be a support function of the curveC. It is known [1], [2] that the area A(a) of a-isoptic of C is given by formula

A(a)4

0 2p 1

sin2a[p2(t)2p(t1a)(p(t) cosa1p.

(t) sina) ]dt. (1.1)

Letp(t)4a01

!

n41 Q

(ancosnt1bnsinnt) be the Fourier series of the sup- port function. We will assume that p(t)C2[ 0 , 2p] and p(t),p.

(t) L2[ 0 , 2p]. Moreover the curvature radiusR(t)4p(t)1p..

(t)D0 and the Fourier series of R(t)4a01

!

n41 Q

(12n2)(ancosnt1bnsinnt) is convergent.

(*) Indirizzo dell’A.: Institute of Mathematics, Maria Curie-SklCodowska Uni- versity, pl. M. Curie-SklCodowska 1, 20-031 Lublin, Poland; e-mail: daisyHgo- lem.umcs.lublin.pl

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Let f(x)An

!

4 2Q Q

cneinx, g(x)An

!

4 2Q Q

dneinxL2 where c2n4cn are complex numbers then we have ([5], p. 37, theorem 1.12)

THEOREM 1.1. Suppose that f and g are in L2and have coefficients cn and dn respectively. Then

1 2p

0 2p

f(x2t)g(t)dt4n

!

4 2Q Q

cndneinx (1.2)

for all x, and the series on the right converges absolutely and uniformly.

In particular 1 2p

0 2p

f(x1t) f(t)dt4n

!

4 2Q Q

NcnN2einx, (1.3)

1 2p

0 2p

f(t)g(t)dt4n

!

4 2Q Q

cnd2n, (1.4)

1 2p

0 2p

Nf(t)N2dt4n4 2Q

!

Q NcnN2. (1.5)

If f(t)4a01

!

n41 Q

(ancosnt1bnsinnt) is a real valued function and g4f8 then we have for all xR

0 2p

f2(t)dt42pa021p

!

n41 Q

(an21bn2) , (1.6)

0 2p

f(t1x) f(t)dt42pa021p

!

n41 Q

(an21bn2) cosnx, (1.7)

0 2p

f(t1x) f8(t)dt4p

!

n41 Q

n(an21bn2) sinnx (1.8)

which, with formula (1.1), give the following formula for the area of the a-isoptic

(1.9) A(a)4 4 1

sin2a

k

2pa021pn

!

Q41(an21bn2)2cosa

g

2pa021pn

!

4Q1(an21bn2) cosna

h

2

(3)

2sinaQp

!

n41 Q

n(an21bn2) sinna

l

4

4 p

sin2a

k

2a02( 12cosa)1n

!

4Q1(an21bn2)(12cosnacosa2nsinasinna)

l

4

4 p

sin2a

k

2a02(12cosa)1n

!

40 Q

(an112 1bn112 )

g

11n2 cos (n12)a2n12

2 cosna

hl

.

2. The main result.

THEOREM 2.1. Let p(t)4a01

!

n41 Q

(ancosnt1bnsinnt)be the Fouri- er series of the support function of C.If Fourier coefficients of p(t)are such that the inequality holds

a02F

!

n40 Q

(n(n12 ) )2(an112 1bn112 ) (2.1)

then A(a) is a convex function.

PROOF. We will show thatA9(t)F0 . From the above considerations we get

A8(a)

p 42a02 ( 12cosa)2 sin3a 1 1

!

n40 Q

(an2111bn211) 1

sin3a

y

n(n212 )sina( sinna2sin (n12 )a)2

22 cosa

g

11 n2 cos (n12 )a2n12

2 cosna

hl

and by some calculations (2.2) A9(a)4 p

sin4a

m

2a02( 12cosa)2( 22cosa)1

1

!

n40 Q

(an12 1

1bn12 1)[n3sin3asin (n11 )a13n2sin2acosna2 26nsinacosasinna12( 112 cos2a)( 12cosna) ]

n

.

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If the condition (2.1) holds, then A9(a)

p F 1

sin4a

!

n40 Q

(an2111bn211)]2n2(n12 )2( 12cosa)2( 22cosa)1 1n3sin3asin (n11 )a13n2sin2acosna26nsinacosasinna1 12( 112 cos2a)( 12cosna)(

4 1 sin4a

!

n40 Q

(an211

1bn211)Mn(a), where

(2.3) Mn(a)42n2(n12 )2( 12cosa)2( 22cosa)1n3sin3asin (n11 )a1 13n2sin2acosna26nsinacosasinna12( 112 cos2a)( 12cosna) for abbreviation. One can see that Mn( 0 )40 for nN andM0(a)40 . We will show that Mn8(a)F0 . We have

Mn8(a)42(n414n314n2) sina( 12cosa)( 523 cosa)1

1n4sin3acos (n11 )a14n3sin3acos (n11 )a1 18nsin2asinna28 sinacosa( 12cosna) . Ifa[ 0 , p) then sinaF0 andNsinnaN Gnsinaandn414n3F5n2for any positive integer n. Moreover, we have

Mn8(a)Fn4sina( 12cosa)[ 2( 523 cosa)2( 11cosa) ]1

14n3sina( 12cosa)[ 2( 523 cosa)2( 11cosa) ]1 18n2sina( 12cosa)( 523 cosa)2

28n2sin3a28 sinacosaQ2 sin2na 2 F F(n414n3) sina( 12cosa)[ 927 cosa]1

18n2sina( 12cosa)[ 523 cosa212cosa]2 216n2sinacosasin2 a

2 F5n2Q2 sinasin2 a

2( 927 cosa)216n2sinacosasin2 a 2 4 42n2sinasin2 a

2[ 45235 cosa28 cosa]4

(5)

42n2sinasin2 a

2( 45243 cosa)F0 .

The functionMn(a) is not decreasing for eacha[ 0 ,p) and Mn( 0 )40 . This means thatMn(a)F0 for eacha[ 0 ,p) and each positive integer n and M0(a)40 . This completes the proof of the inequality

A9(a)

p F 1

sin4a

!

n40 Q

(an2111bn211)Mn(a)F0 for each a[ 0 ,p). r

EXAMPLE 1. Let p(t)4a1bcos 3 t , a,bD0 then the curvature radiusR(t)4a28bcos 3t. If aD8b then the curve C is convex. From (2.1) the area function A(a) is convex if a2F( 8b)2.

EXAMPLE 2. Let p(t)4a1bcos 3t1ccos 5t then from (2.1) a2F F( 8b)21( 24c)2andR(t)4a28bcos 3t224ccos 5tD0 foraD8NbN 1 124NcN. We have

a2D( 8NbN 124NcN)2D( 8b)21( 24c)2.

For any convex curve C with support function p(t)4a1bcos 3t1 1ccos 5t the area function A(a) is convex.

EXAMPLE 3. Consider a support function p(t)4 1

12asint, t[ 0 , 2p). Then the curvature radius of the corresponding oval is R(t)4123asint12a

2

( 12asint)3 F0 for 0E NaN E12. We will find the Fourier series of p(t)4a01

!

n41 Q

(ancosnt1bnsinnt).

1

12asint4

!

n40 Q

(asint)n, where

sinmt4

. ` / `

´

(21 )k 22k

y

2

!

j40 k21

g

2jk

h

(21 )jcos 2(k2j)t1(21 )k

g

2kk

h z

for m42k

(21 )k 22k

y !

j40

k

g

2k1j 1

h

(21 )jsin ( 2k22j11 )t

z

for m42k11 ,

(6)

hence we have 1

12asint 4

!

k40 Q

a2k

g

2kk

h

22k 1

!

k40 Q

(21 )k11cos 2(k11 )t

!

j4k Q a2j12

22j11

g

2jj21k2

h

1

!

k40 Q

(21 )ksin ( 2k11 )t

!

j4k Q a2j11

22j

g

2jj21k1

h

.

The Fourier coefficients of p(t) are equal to

an4

. ` / `

´

k

!

40 Q

a2k

g

2kk

h

22k for n40 (21 )k11

!

j4k Q a2j12

22j11

g

2jj21k2

h

for n42k12

0 for n42k11 ,

bn4

. /

´

0 for n42k (21 )k

!

j4k Q a2j11

22j

g

2jj21k1

h

for n42k11 .

Using the inequalities

k2pn

g

ne

h

ne12n11 1En!Ek2pn

g

ne

h

ne12n1 ,

g

11n1

h

nEeE

g

11 n1

h

n112,

we have

g

2jj1112

h

22j11 E 2

kpkj11 G 2 kpkk11 ,

g

2jj1111

h

22j E 2

kpkj G 2

kpkk , for j4k,k11 ,R

(7)

and

Na2k12N 4

!

j4k Q a2j12

22j11

g

2jj21k2

h

Gj

!

4k Q a2j12

22j11

g

2jj1112

h

Gj4k

!

Q 2a2j12

k

p(k11 )

4 2a2k12

k

p(k11 ) 1 12a2,

Nb2k11N 4

!

j4k Q a2j11

22j

g

2jj21k1

h

Gj4k

!

Q a2j11

22j

g

2jj1111

h

G

G

!

j4k Q 2a2j11

kpk 4 2a2k11 kpk

1 12a2 . We can estimate the right hand side in the formula (2.1)

n

!

40 Q

(n(n12 ) )2(an211

1bn211

)4

4

!

k40 Q

( 2k11 )2( 2k13 )2a2k2121

!

k40 Q

( 2k)2( 2k12 )2b2k211E

E

!

k40 Q

( 2k11 )2( 2k13 )2

u k

2p(ka2k12

11 ) 1 12a2

v

21

1

!

k40 Q

( 2k)2( 2k12 )2

g

2ak2kpk11 1 12a2

h

2G

G 1

p( 12a2)2

!

k40 Q

[

64k(k11 )2a4k1218( 2k11 )( 2k13 )2a4k14

]

4

4 1

p( 12a2)2

!

k40 Q

[ (a4k15)R24(a4k14)9 23(a4k13)8 23a4k121

1(a4k17)R24(a4k16)9 23(a4k15)8 23a4k14]4

8a4( 924a21a4) p( 12a2)6 .

(8)

On the other hand

a024

u !

k40 Q

a2k

g

2kk

h

22k

v

24

u

11

!

k41 Q

a2k

g

2kk

h

22k

v

2D1 .

Comparing both sides we obtain that if 0E NaN E

o

2kp2215p18p17223p then the condition (2.1) holds.

PROPOSITION 2.1. Let p(t)be the support function of the curve C.If there existseD0such that fora[ 0 ,e]Fourier coefficients of p(t)sat- isfy inequalities

(2.4)

!

n40 Q

(an112 1bn112 )[n3sin3asin (n11 )a13n2sin2acosna2 26nsinacosasinna12( 112 cos2a)( 12cosna) ]F0 and

a02FC(e)

!

n40 Q

(n313n216n)(an2111bn211) , (2.5)

where C(e)42( 1 1

2cose)2( 22cose) then A(a) is a convex function.

PROOF. Ifa[0,e] then from the assumption (2.4) we haveA9(a)F0.

For a(e,p) we have

n3sin3asin (n11 )a13n2sin2acosna2

26nsinacosasinna12( 112 cos2a)( 12cosna)F 2n323n226n and if a0 satisfy (2.5) then we obtain

A9(a)

p 42a02( 12cosa)2( 22cosa)

sin4a 1

1 1 sin4a

!

n40 Q

(an211

1bn211)[n3sin3asin (n11 )a1

13n2sin2acosna26nsinacosasinna12( 112 cos2a)( 12cosna) ] F 1

sin4a

y

C(e)

!

n40 Q

(n313n216n)(an211

1bn211

) 1 C(e)2 2

!

n40 Q

(an12 1

1bn12 1)(n313n216n)

l

F0 . r

(9)

In particular, if the Fourier coefficients of p(t) are such that )n0N (nDn0 an4bn40

then p(t) satisfy (2.4) for some e0D0 . In fact if we put Nn11(a)4 1

sin4a[n3sin3asin (n11 )a1

13n2sin2acosna26nsinacosasinna12( 112 cos2a)( 12cosna) ] in (2.2) we have lim

aK01

Nn11(a)4n

2(n12 )2

4 D0 for each 0EnGn0 and N1f0 .Nn11(a) is a continuous function and therefore existsenD0 such that Nn11(a)D0 fora[ 0 ,en] we definee04min]e1,e2,R,en0(. If for e0 conditon (2.5) holds then A(a) is a convex function.

The author would like to thank the Referee of this paper for his re- marks which improve this paper.

R E F E R E N C E S

[1] W. CIES´LAK- A. MIERNOWSKI - W. MOZGAWA, Isoptics of a Closed Strictly Convex Curve, Lect. Notes in Math., 1481(1991), pp. 28-35.

[2] W. CIES´LAK- A. MIERNOWSKI - W. MOZGAWA, Isoptics of a Closed Strictly Convex Curve II, Rend. Sem. Mat. Univ. Padova,96 (1996), pp. 37-49.

[3] H. GROEMER, Geometric applications of Fourier series and spherical har- monics, Encyclopedia of Mathematics and its Applications, Cambridge Univ.

Press, Cambridge, 1996.

[4] M. A. HURWITZ,Sur quelques applications géométriques des séries de Fouri- er, Ann. Sci. Ecole Norm. Sup., 19 (1902), pp. 357-408.

[5] A. ZYGMUND, Trigonometric series, Cambridge Univ. Press, vol. I, II, 1968.

Manoscritto pervenuto in redazione il 9 gennaio 2003.

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