PII. S0161171201004173 http://ijmms.hindawi.com © Hindawi Publishing Corp.
ILLUMINATION BY TAYLOR POLYNOMIALS
ALAN HORWITZ
(Received 22 November 1999)
Abstract.Letf (x)be a differentiable function on the real lineR, and letP be a point not on the graph off (x). Define the illumination index ofPto be the number of distinct tangents to the graph off which pass throughP. We prove that iffis continuous and nonnegative onR,f≥m >0 outside a closed interval ofR, andf has finitely many zeros onR, then any pointPbelowthe graph offhas illumination index 2. This result fails in general iffis not bounded away from 0 onR. Also, iffhas finitely many zeros andf
is not nonnegative onR, then some point below the graph has illumination index not equal to 2. Finally, we generalize our results to illumination by odd order Taylor polynomials.
2000 Mathematics Subject Classification. 26A06.
1. Introduction. The central problem in differential calculus is to find the tangent line to a given curvey=f (x)at a given point(c, f (c))on the graph off. A somewhat more complicated problem is: given a pointP=(s, t)not on the graph off, find all values ofcso that the tangent line to the graph offat(c, f (c))passes throughP. If such acexists, we say that the point(c, f (c))illuminatesP. A typical example is: find all tangents toy=x2which pass through the point(2,3). In this case, each of the points(1,1)and(3,9)would illuminateP. Of course, it is certainly possible thatno
tangent line at all passes through the given point(s, t)—for example, ify=x2and
P=(1,3). A simple, but interesting exercise is: letPbe any point below the graph of y=x2. Prove that there are exactly two tangents to the graph which pass throughP. In considering this type of problem, the following question naturally arises: given f (x), for which pointsP=(s, t)is there a tangent line tofwhich passes throughP? Also, how many tangents pass throughP?
The questions above lead to some potentially interesting ideas for research. For example, suppose thatf is convex onR, and letP be any point below the graph of y=f (x). Are there always exactly two tangents to the graph which pass throughP? What if one assumes thatf(x) >0 onR? We give the answers inSection 2(see, in particular,Theorem 2.6).
InSection 3, we prove a converse result toTheorem 2.6. It is also natural to try to extend our results to illumination byhigherorder Taylor polynomials. InSection 4, we prove results similar toTheorem 2.6for illumination byoddorder Taylor polynomials. Most of the proofs extend verbatim, but some results from [3] are needed.
2. Illumination by tangent lines
Definition2.1. Letf (x) be a differentiable function on the real line, and letP
there arekdistinct tangents to the graph off which pass throughP. We include the possibility thatk= ∞.
Remark2.2. Call a tangent lineTmultiple ifTis tangent to the graph offat more
than one point. If only one tangent lineTpasses throughP, butTis a multiple tangent, we still define the illumination index ofP to be one. One could, of course, define an illumination index which takes into account the number of points of tangency of each tangent line.
As noted earlier, any point below the graph ofy =x2 has illumination index 2. We now generalize this to convexC2functions in general, with the added condition that f is bounded below by a positive number outside some closed interval (see Theorem 2.6below). First we prove a couple of lemmas.
Lemma2.3. Letf (x)∈C2(R), and suppose that there existsT >0such thatf(x)≥ m >0on|x|> T. LetTc(x)be the tangent line tof at(c, f (c)). Then for any fixeds, lim|c|→∞Tc(s)= −∞.
Proof. For fixeds, letg(c)=Tc(s), which implies thatg(c)=(s−c)f(c). Let
U=max(s, T ),u=min(s,−T ). It follows that
g(c)is ≤
0, c > U
≥0, c < u (2.1)
which implies that
g(c)is
decreasing on(U ,∞)
increasing on(−∞, u). (2.2)
Also, sincef(x)≥m >0 on|x|> T,
lim c→∞g
(c)= −∞, lim c→−∞g
(c)= ∞. (2.3)
Partition[U ,∞)into infinitely many subintervals,[ck−1, ck], of constant widthh > 0. By (2.3), givenM >0, there exists C >0 such that g(c)≤ −M forc ≥C. Now g(ck)=g(ck−1)+
ck ck−1g
(t)dt≤g(U )−Mhifck
−1≥C. Since this inequality holds for anyM >0,g(ck)→ −∞. Also, since this inequality holds for any increasing sequence
{ck} → ∞, withck−ck−1constant, limc→∞g(c)= −∞. A similar argument shows that limc→−∞g(c)= −∞.
Remark2.4. Lemma 2.3is a little easier to prove under the stronger assumption
thatf(x)is positive and bounded away from 0 on the real line. One can then just examine the errorEc(x)=f (x)−Tc(x)and use Taylor’s remainder formula.
Lemma2.5. Suppose thatf(x)is continuous, nonnegative, and has finitely many
zeros inR. Then at most two distinct tangent lines tof can pass through any given pointP in the plane.
Proof. Suppose that three distinct tangents,T1,T2, andT3pass throughP, and
generality, that x1< x2< x3. Since f is convex on any open interval, each pair of tangents has a unique point of intersection. LetI1=intersection point ofT1andT2, and letI2=intersection point ofT2andT3. Since all three tangents pass throughP,
I1=I2=P. IfI1=(s1, t1)andI2=(s2, t2), then, again, sincefis convex on any open interval,x1< s1< x2andx2< s2< x3, which implies thats1< s2, which contradicts the fact thatI1=I2.
Theorem2.6. Suppose thatf(x)is continuous, nonnegative, and has finitely many
zeros inR. Assume also that there existsT >0such thatf(x)≥m >0on|x|> T. Let
P=(s, t)witht < f (s). Then there are exactly two distinct tangent lines to the graph off which pass throughP.
Proof. Sincet < f (s), forcsufficiently close tos,Tc(s)=f (c)+f(c)(s−c) > t.
ByLemma 2.3, lim|c|→∞Tc(s)= −∞. Hence, for|c|sufficiently large,Tc(s) < t. By the Intermediate Value Theorem,Tc(s)=tfor at least two values ofc. Note also that for a convex function,c1≠c2implies thatTc1≠Tc2. Hence the illumination index ofPis at least two. ByLemma 2.5, the illumination index ofPis at most two. This proves the theorem.
The following example shows thatTheorem 2.6doesnothold in general for func-tions that only satisfyf(x) >0 onR.
Example2.7. Letf (x)=0x(0te−u2du)dt=(1/2)erf(x)√π x+(1/2)e−x2−1/2,
where erf(x)=0xe−t 2
dt. Sincef(x)=e−x2
, limx→±∞f(x)=0. We now show that
no tangent tof passes through the point (0, t) when t <−1/2< f (0)=0. If the tangent lineTc tof passes through(s, t), thenf (c)+f(c)(s−c)=t. So consider the functionh(x)=f (x)+f(x)(s−x)−t=(1/2)e−x2
−1/2+(1/2)erf(x)√π s−t. Ifs=0, thenh(x)=(1/2)e−x2−1/2−t⇒h(x)= −xe−x2, which implies thath(x) is increasing forx <0 and decreasing forx >0. Since limx→±∞((1/2)e−x2−1/2−t)=
−1/2−t, ift <−1/2 thenhis always positive and thus has no real zeros.
Our definition of the illumination indexk includes the possibility thatk= ∞. Of course, for polynomials the illumination index is always finite (indeed, it is bounded above by the degree of the polynomial). The following example shows that there are entire functions, however, wherealmost every point not on the graph has infinite illumination index.
Example2.8. Let f (x)=sinx, and letP =(s, t)be any point not on the graph
off, witht≠±1. The tangent line at(c, f (c))passes throughPif and only iff (c)+ f(c)(s−c)=t, that is, wheng(c)=sinc+(s−c)cosc−t=0. For nsufficiently large and even,g(nπ )=(−1)n(s−nπ )−t <0, while fornsufficiently large and odd,
Remark 2.9. Given f, one may define, for each nonnegative integer k, the set
Dk, equal to the set of points in the plane with illumination indexk. TheDk form a partition ofR2−G, where G is the graph of f. For example, if f (x)=x3, it is not hard to show that D3= {(s, t):s >0, 0< t < s3} ∪ {(s, t):s <0, s3< t <0},
D2= {(s, t):s≠0, t=0},D1=G−(D2∪D3), andDk= ∅fork=0 ork >3.
3. A converse result. Suppose thatf (x)is not convex onR. Is it possible for every point below the graph off to have illumination index 2? The answer is no, and thus we have the following partial converse ofTheorem 2.6.
Theorem3.1. Letf∈C3(R)and suppose thatf(x)has finitely many zeros inR.
Iff(x)is not nonnegative onR, then there is a pointP below the graph off with illumination index not equal to2.
Proof. If f(x)≤0 on R, then clearly any point P below the graph of f has
illumination index 0. Hence we may assume that there are real numberssandusuch thatf(s) >0,f(u)=0, andf(x)changes sign atx=u, withf≥0 betweens andu. We consider the cases < u, the other case being similar. LetP=(s, t), witht to be chosen shortly. Now the tangent line at(c, f (c))passes throughPif and only if f (c)−cf(c)+f(c)s=t, which holds if and only ifh(c)=0, where
h(c)=f (c)−cf(c)+f(c)s,
h(c)=(s−c)f(c), h(c)=(s−c)f(c)−f(c). (3.1)
Note that h(s)=f (s), h(s)=0, and h(s)= −f(s) <0, so thath(s) is a local maximum ofh(c). Sinceh(c)≤0 on(s, u)andh(c)≥0 on(u, u+),h(u)is a local minimum ofh(c). Note thath(u) < h(s). LetT be the liney=h(u), the tangent to hat(u, h(u)).
Case1. T only intersects the graph ofhat(u, h(u)). Then lett=h(u).
Case 2. T intersects the graph of h at some pointQ≠(u, h(u)). If a Qexists
such thath−T changes sign atQ, theny=h(u)+intersects the graph ofhin at least three points for some >0. If no suchQexists, thenhmust have another local minimum atQ. Theny=h(u)+intersects the graph ofhin at least four points for some >0. In either case, lett=h(u)+, withchosen sufficiently small so that h(u)+ < h(s). Since the zeros ofhcorrespond to values ofcsuch that the tangent line at(c, f (c))passes throughP, for case two there are at least three such values ofc. However, it is possible that some of the corresponding tangents could be multiple. It was shown in [1], however, thatfcan have onlyfinitely manymultiple tangent lines in any bounded interval. Also, since each tangent is tangent at only finitely many points, we can also choosesufficiently small so that none of the tangents corresponding to the zeros ofhis multiple. Thus at least threedistincttangents pass throughP.
In each case covered, P lies below the graph off sinceh(s)=f (s). Hence the illumination index ofPis either one or greater than or equal to three, and thus cannot equal two.
ways, the odd order Taylor polynomialsPc(x)behave like tangent lines. Suppose that
f∈Cr+1(−∞,∞), and letPc(x)denote the Taylor polynomial tofof orderratx=c. In [2] it was proved that iff(r+1)(x)≠0 on[a, b], then there is a uniqueu,a < u < b, such thatPa(u)=Pb(u). This defines a meanm(a, b)≡u. We shall prove a slightly stronger version of this result. The method of the proof is very similar to that used in [3], where further results and generalizations of the meansm(a, b)were proved.
For the rest of this section we assume thatr is anoddpositive integer. LetEc(x)=f (x)−Pc(x). By the integral form of the remainder, we have
Ec(x)= 1 r!
x
c
f(r+1)(t)(x−t)rdt. (4.1)
Lemma 4.1. Suppose that f(r+1)(x) is continuous, nonnegative, and has finitely
many zeros in[a, b]. ThenPb−Pahas precisely one real zeroc,a < c < b.
Proof. By (4.1),
Ea(x)= 1 r!
x
a
f(r+1)(t)(x−t)rdt,
Eb(x)= 1 r!
b
xf
(r+1)(t)(t−x)rdt,
Ea(x)= 1 (r−1)!
x
a
f(r+1)(t)(x−t)r−1dt.
(4.2)
This implies that
Ea(x) <0 forx < a, Ea(x) >0 forx > a. (4.3)
HenceEa(x)is strictly increasing on(a, b). Similarly,Eb(x)is strictly decreasing on
(a, b). SinceEa(a)=0 andEb(b)=0, there is a uniquec,a < c < b, such thatEb(c)−
Ea(c)=0. This implies thatPb(c)−Pa(c)=0. Now
Eb−Ea(x)= − b
a
f(r+1)(t)(x−t)rdt (4.4)
which implies that
Eb−Ea(x)= −r b
af
(r+1)(t)(x−t)r−1dt (4.5)
which is less than or equal 0 onR. Sincef(r+1) has finitely many zeros, this implies that Eb−Ea is strictly decreasing onR. Hence Eb−Ea has precisely one real zero, which implies thatPb−Pahas precisely one real zeroc,a < c < b.
Lemma 4.2. Suppose that f(r+1)(x) is continuous, nonnegative, and has finitely
many zeros in[a, b]. LetP be any point in thexy-plane. Then at most two distinct Taylor polynomials of orderratx=c,a≤c≤b, can pass throughP.
Proof. Suppose that three distinct Taylor polynomials of orderr, Pc
loss of generality, assume thata≤c1< c2< c3≤b. ByLemma 4.1,c1< s < c2and
c2< s < c3, which is a contradiction. Hence at most two distinct Taylor polynomials of orderr can pass throughP.
Lemma4.3. Letf (x)∈Cr+1(R). Suppose that there existsT >0such thatf(r+1)(x)≥ m >0on|x|> T. Then for any fixeds,lim|c|→∞Pc(s)= −∞.
Proof. The proof is almost identical to that ofLemma 2.3, and we omit it.
Theorem4.4. Suppose thatf(r+1)(x)is continuous, nonnegative, and has finitely
many zeros inR. In addition, assume that there existsT >0such thatf(r+1)(x)≥m >0
on |x|> T. LetP =(s, t) witht < f (s). Then there are exactly two distinct Taylor polynomials of orderr to the graph offwhich pass throughP.
Proof. Sincet < f (s), forcsufficiently close tos,Pc(s)=f (s)+rk=1(f(k)(c)/k!)(s−
c)k > t. By Lemma 4.3, lim
|c|→∞Pc(s) = −∞, and hence, for |c| sufficiently large,
Pc(s) < t. By the Intermediate Value Theorem,Pc(s)=tfor at least two values ofc. Also, it is not hard to show that if f(r+1)(x) >0 on R, thenc1≠c2 implies that
Pc1≠Pc2. Hence the illumination index ofPis at least two. ByLemma 4.2, it isat most two. This completes the proof.
Example4.5. Letf (x)=ex+x4,P=(0,0), andr=3. ThenTheorem 4.4applies,
and the illumination index ofPequals 2. We now verify this by estimating the actual values ofc. The third order Taylor polynomial tof at(c, f (c))is
Pc(x)= 3
k=0
ec(x−c)k
k! +c
4+4c3(x−c)+6c2(x−c)2+4c(x−c)3, (4.6)
and so we get
Pc(0)=ec−ecc+ 1 2e
cc2−1 6e
cc3−c4=0. (4.7)
Numerical estimates give the solutionsc1≈ −0.9953 andc2≈0.9782.
Note that iff (x)=exinstead, then the illumination index ofPequals 1. This does not contradictTheorem 4.4sincef(iv)(x)→0 asx→ −∞.
References
[1] A. Horwitz,Reconstructing a function from its set of tangent lines, Amer. Math. Monthly
96(1989), no. 9, 807–813.MR 91a:26003. Zbl 718.41050.
[2] ,Means and Taylor polynomials, J. Math. Anal. Appl.149(1990), no. 1, 220–235. MR 91e:26013. Zbl 706.26019.
[3] ,Means and averages of Taylor polynomials, J. Math. Anal. Appl.176(1993), no. 2, 404–412.MR 94g:41053. Zbl 807.41017.
Alan Horwitz: Penn State University,25Yearsley Mill Road, Media, PA19063, USA
