AN EXCEPTIONAL EXPONENTIAL FUNCTION

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1 AN EXCEPTIONAL EXPONENTIAL FUNCTION BRANKO ĆURGUS Thr ar at last two surprising rsults in this articl. Th first is that thr is actually a link btwn th standard calculus problm of finding th bst viw of a painting and graphs of xponntial functions. Th scond is that th xponntial function with th bst viw is not th on with th bas. As a bonus w gt an intrsting apparanc of th Lambrt W function. Th painting problm On of th standard calculus problms is this: A painting hangs on a wall in an art gallry. Its top dg is at distanc p and its bottom dg at distanc q abov th y lvl of an art lovr. How far from th wall should th art lovr stand to gt th bst viw? Hr th bst viw mans th maximum viwing angl; s Figur.) This statmnt of th painting problm is similar to Exrcis 44 in Stwart [9, p. 34]. In Problm 23 of Hughs-Halltt, t al. [6, p. 203], a painting is rplacd by th Statu of Librty and th valus of p and q ar spcific numbrs. In this stting th rquirmnt that th bottom of th obsrvd objct b abov th y lvl of an obsrvr sms mor natural. W lav th dtails of th solution of th painting problm to th radr. Bfor procding, howvr, w mak an obsrvation. For ach distanc from th gallry wall, thr is a corrsponding viwing angl. Morovr, for ach distanc v othr than qp from th wall thr is a scond distanc qp/v with th sam viwing angl. Figur givs a gomtric proof of this fact. Ray OD in th figur rprsnts y lvl. Lt OA = OA = p, OB = OB = q, OC = OC = v, and OD = qp/v. Right triangls OAC and OA C ar congrunt sinc thir lgs ar congrunt. Similarly, right triangls OBC and OB C ar congrunt. Hnc th markd angls at C and C ar congrunt. Furthrmor, right triangls OA C

2 2 BRANKO ĆURGUS A p D qp v θ qp C v B q θ O C Figur. A painting on a wall. θ B A and ODB ar similar sinc thir lgs ar proportional: OD = qp/v OA p Thrfor lins C A and = q v, OB OC = q v. DB ar paralll. In th sam way, triangls OB C and ODA ar similar, which implis that lins C B and DA ar also paralll. Hnc th markd angls at C and D ar congrunt. Consquntly, th viwing positions C and D provid th sam viwing angl θ. If w assum that th solution of th painting problm is uniqu that is, if w assum that thr xists a viwing position from which th viwing angl is largr than that from any othr position) thn, by what w just provd, th distanc qp is th only possibl candidat for th solution. Using ithr calculus combind with trigonomtry, or plan gomtry, it can b shown that qp is indd th uniqu solution. Th calculus solution is standard. Th plan gomtry solution and much mor on th painting problm can b found in [3]. W now procd to a smingly unrlatd topic of two spcial tangnt lins to an xponntial function. But, th painting problm is coming back. Stay alrt! An xcptional xponntial function Lt a >. Considr th xponntial function with bas a, f a x) = a x, and its graph in th xy-plan. Thr ar two distinguishd points in this stting: th origin and th y-intrcpt. Thrfor two notworthy tangnt lins to th graph of f a ar th on that passs through th origin and th on at th y-intrcpt. Thir quations ar l 0,a x) = ln a) x and l,a x) = lna) x +.

3 AN EXCEPTIONAL EXPONENTIAL FUNCTION 3 Sinc a /ln a) = ln a) /ln a) =, th tangnt lin l 0,a touchs th graph of f a at th point /lna), ). Notic that th scond coordinat of this point dos not dpnd on a. This may not b so surprising to th radrs of this Journal, sinc this proprty of xponntial functions appard in [8]. For larg a, th lins l 0,a and l,a ar almost vrtical, so th angl btwn thm is small. For a clos to, ths lins ar almost horizontal, so again th angl btwn thm is small. This suggsts th xistnc of a spcial bas β that maximizs th angl btwn th two lins. What is this bas? Th lins l 0,a and l,a intrsct at th point V a = ) lna, ). ) Surprisingly, th scond coordinat of V a dos not dpnd on a. This fact provids th link btwn th family of xponntial functions and th painting problm that w considrd arlir. W can think of V a as th y of an art lovr looking at th painting that is positiond btwn th points 0, 0) and 0, ). Th lin y = / ) rprsnts y lvl. Th angl btwn th tangnt lins l 0,a and l,a is th viwing angl. This is somwhat upsid-down viwing, but it is nvrthlss usful for obtaining th maximum angl. S Figur 2.) In th notation of th painting problm, w hav p = and q =. 2) Using th fact that th bst viw is at a distanc of qp, w find that th maximum angl btwn our two lins is attaind whn thy intrsct at th point, ). Comparing this with ), w conclud that lnβ = /. Hnc th spcial bas is β = / This is th bas for th xcptional xponntial function in th titl. Its graph is th havy curv in Figur 2. Th connction with th painting problm is th rason why w call y = x/ th bst-viw xponntial function. Nxt w us our rsult for th painting problm that says that th distancs v and qp/v from th wall provid th sam viwing angl, to

4 4 BRANKO ĆURGUS x x/ x/ 2 V Figur 2. Th xcptional xponntial function: x/. x obtain th analogous rsult for xponntial functions. Lt v = ) lna b th x-coordinat of V a ; s ). With p and q from 2), w find that qp v = lna. Thrfor th angl btwn th tangnt lins intrscting at th point ) ln a, is qual to th angl btwn th tangnt lins intrscting at th point V a. Comparing this with ), w find that th corrsponding bas is / ln a). Consquntly, th angl btwn th notworthy tangnt lins to th graph of y = a x is qual to th angl btwn thos tangnt lins to th graph of y = x/ lna). Figur 2 shows this whn a =. Th plot of th radian masur of th angl btwn th lins l 0,a and l,a as a function of th bas a is shown in Figur 3. Th markd points rprsnt th xponntial functions plottd in Figur 2. Notic

5 AN EXCEPTIONAL EXPONENTIAL FUNCTION 5 5π 24 π 6 π 8 π 2 π Figur 3.Th angl btwn tangnt lins as a function of bas. that th maximum angl which corrsponds to th bas / ) is ) arctan radians ) Th angl that corrsponds to th bas is ) arctan radians Compositions of xponntial functions and a linar function In th prcding sction w discovrd th link btwn th family of xponntial functions and th painting problm with p = / ) and q = / ). Our nxt goal is to xtnd this rsult to familis of functions obtaind by composing xponntial functions with a fixd linar function. W bgin with a dfinition. For b > and c > 0, lt F b,c b th family of functions g a x) = c a x + b ) for a >. For xampl, th family F 0, is th family of xponntial functions studid in th prcding sction. Notic that g a = ϕ f a, whr f a x) = a x and ϕx) = cx + b). Nxt w study two notworthy tangnt lins to th graphs of functions in F b,c. It is important to notic that th functions in F b,c hav th sam y-intrcpt 0, c + b) ) and that c + b) > 0. On notworthy tangnt lin is asy to find: Th tangnt lin to th graph of g a at

6 6 BRANKO ĆURGUS th y-intrcpt is l,a x) = xc ln a + c + b). Th xistnc and uniqunss of th othr tangnt lin is not obvious; that is stablishd by th following thorm. Thorm. Each function g a in F b,c has xactly on tangnt lin l 0,a that passs through th origin and is tangnt to th graph of g a in th first quadrant. In th proof of this thorm w ncountr an quation that dos not sm to b solvabl symbolically. Howvr, with th hlp of Mathmatica th author ralizd that th solution can b obtaind in trms of th Lambrt W function. In Mathmatica it is th ProductLog function. For mor about this function, which has attractd considrabl attntion rcntly, s for xampl [, 2, 5, 7]. Bfor w bgin th proof, w giv dfinitions of W and its clos rlativ Y, both of which play an important rol in this articl. Th functions W and Y and th proof Th art of solving quations is practicd with a larg toolbox of invrs functions. Whn w ncountr a function that is not on-toon, w wisly rstrict its domain, so that th rsulting nw function has an invrs. That is what is don hr with th function x x x. Although, this is a vry lgant function, it is not on-to-on. But, a simpl xrcis in calculus shows that its rstriction Tx) = x x, x, is strictly incrasing. Thrfor, T is a bijction from [, + ) to [ /, + ). W dfin W to b th invrs of T. Hnc, W is a strictly incrasing bijction from [ /, + ) to [, + ). Th graphs of W and T ar shown in Figur 4. Th function Y is dfind by Y z) = xp + Wz/) ), z. It is a composition of two strictly incrasing bijctions. Thrfor Y is a strictly incrasing bijction from [, + ) to [, + ). An asy calculation shows th invrs of Y to b Y x) = x ln x ), x. Figur 5 shows Y and its invrs. Not that sinc W0) = 0, Y 0) =.

7 AN EXCEPTIONAL EXPONENTIAL FUNCTION T = W 3 2 W Figur 4. W and its invrs Y 4 2 Y Figur 5. Y and its invrs. 2 Th following rlation btwn W and Y is a consqunc of th dfinitions. For z, Wz/)Y z) = Wz/) xp + Wz/) ) = Wz/) xp Wz/) ) = T Wz/) ) Thrfor, = z. + Wz/) = + z Y z) = Y z) + z, for z. 4) Y z) Now w hav all th tools w nd to prov th thorm. Proof. Lt a > and lt g a x) = ca x + b) b a function in F b,c. Furthr, lt x 0, g a x 0 ) ) b an arbitrary point on th graph of g a in th first quadrant. Th quation of th tangnt lin at this point is lx) = a x 0 c lna)x x 0 ) + c a x 0 + b ). 5) A littl computation shows that this lin passs through th origin if and only if x0 ln a ) x 0 ln a = b ; that is, if and only if T x 0 lna ) = b. Sinc b >, th last quation has a uniqu solution. This is whr th Lambrt W function coms to th rscu. From th dfinition of

8 8 BRANKO ĆURGUS W and 4), w find that th only positiv solution of th last quation is x 0 = + W b/ ) = Y b) + b ln a Y b) ln a. Substituting this into 5) givs th tangnt lin that passs through th origin. Sinc a x 0 = x 0 ln a = +Wb/) = Y b), th quation of this lin is l 0,a x) = xcy b) ln a, which complts th proof of th thorm. Th painting problm appars again W ar now rady to look for a link btwn th painting problm and th graphs of th functions g a in F b,c. Straightforward calculation shows that th intrsction of l,a and l 0,a is + b Y b) ) ln a, + b) Y b) c Y b) ). 6) Not that onc again th scond coordinat of this point dos not dpnd on g a. As bfor, w can think of this point as th y of an art lovr looking at th painting positiond btwn th points 0, 0) and 0, c + b) ). This is th link btwn th family Fb,c and th painting problm with p = c + b) c + b) Y b) and q = Y b) Y b). 7) W shall, of cours, illustrat this link with a pictur Figur 6), but bfor that w solv th invrs problm. Not that 2) is th spcial cas of 7) with b = 0 and c =.) Th invrs problm In th prvious sction w saw that ach family F b,c is linkd to a painting problm. Now w pos th invrs problm: Givn a painting problm with paramtrs p and q, find th corrsponding family F b,c. To solv this problm w nd to find b and c for which 7) holds. Our first attmpt was to us Mathmatica s Solv command to solv 7) as a systm of quations with unknowns b and c. Surprisingly, Mathmatica 4. did not solv this systm. Nvrthlss, th following simpl rasoning lads to th solution. Lt F = { b, c) : b >, c > 0 } and G = { p, q) : p > q > 0 }.

9 AN EXCEPTIONAL EXPONENTIAL FUNCTION 9 First notic that th mapping from F to G givn by 7) can b rprsntd as th composition of th following thr mappings: b, c) b, c + b) ) = s, t), s, t) Y s), t ) = x, y), ) xy x, y) x, y = p, q). x Th radr can vrify that th first mapping is a bijction with both domain and rang F. Th scond mapping is a bijction btwn F and th st { x, y) : x >, y > 0 }. Th third is a bijction btwn this st and G. Thus, th mapping givn by 7), bing th composition of thr bijctions, is a bijction from F to G. Th invrss of th thr bijctions displayd abov, listd in th rvrs ordr, ar ) p p, q) q, p q = x, y), x, y) Y x), y ) = s, t), ) t s, t) s, = b, c). + s Composing ths, starting with th first, w gt b = Y p/q ) = p ) p q ln and q c = p q + b. 8) This is th solution of th invrs problm: For a painting problm with paramtrs p and q, th family F b,c with b and c givn by 8) is linkd to th painting problm. For xampl, w usd spcific p and q from Figur and formulas 8) to calculat th corrsponding family F b,c linkd to th painting problm in Figur : g a x) = p q + Y p/q ) a x + p )) p q ln, x R, a >. q Th bas a C that corrsponds to th point C in Figur is found to b, using 6) and 8), )) v q + p lnp/q) a C = xp p q Th bas a D that corrsponds to th point D is calculatd similarly. This is illustratd in Figur 6. Notic that th main objcts of Figur appar upsid down in Figur 6. Th points A and B from Figur ar at th origin and th y-intrcpt in Figur 6. Notic that, for th.

10 0 BRANKO ĆURGUS viwing position C and th corrsponding function g ac, th viw lins through C ar th notworthy tangnt lins to th graph of g ac. Th havy curv in Figur 6 is th graph of th bst-viw function. W calculat this function in th nxt sction. p q θ B C qp θ D A Figur 6. Figur with th corrsponding family F b,c. It is asir to rcogniz Figur if th journal is turnd upsid down. A family of xcptional functions In ach family F b,c thr is a spcial function g β for which th notworthy tangnt lins form th maximum angl. Bcaus of th connction with th painting problm, w call g β th bst-viw function. Onc mor, rcall that in th painting problm th bst viw is from distanc qp. Thrfor, using 7), w find that th angl btwn th notworthy tangnt lins is maximum whn thy intrsct at th point ) c + b) + b Y b), Y b) Y b) Y b) c. Comparing this point to th on in 6), w conclud that β = /c Y b)). Consquntly, th bst-viw function in F b,c is g β x) = c x/c ) Y b)) + b. 9)

11 AN EXCEPTIONAL EXPONENTIAL FUNCTION For b > and c > 0, this is th family of xcptional functions in th titl of this sction. For b = 0 and c =, th xcptional xponntial function is y = x/. Whr to hang a painting? Assum that a painting of hight is to b placd in a gallry as dscribd in th painting problm. How high should it b placd? W giv two ways to answr this qustion. It is common to associat th goldn ratio φ with asthtically plasing balanc btwn two lngths. Thrfor, w first suggst choosing p and q = p so that both p/ and /q qual th goldn ratio. That is, q = φ = and p = φ = q Figur 7. Th goldn ratio. With this choic, th bst viw of th painting is at distanc qp = from th wall. Th corrsponding viwing angl is ) arctan radians Anothr way to answr th qustion is to lt th xponntial functions dcid and so us th valus for p and q = p suggstd by th family F 0, : q = and p =.582. q Figur 8. Th goldn xponntial ratio. Ths numbrs ar rmarkably clos to th numbrs suggstd by th goldn ratio, but thy provid a slightly largr viwing angl of about ; s 3). Th prcntag diffrnc in p is only 2.228% and th prcntag diffrnc in th angl is 3.609%.) Can this b just a coincidnc? W invit th radr to us Figurs 7 and 8 to mak an sthtic comparison btwn th goldn ratio and th goldn xponntial ratio.

12 2 BRANKO ĆURGUS Closing commnts. Lt g a F b,c and l 0,a b as in th thorm. It follows from th proof that thir graphs mt at th point Y b) + b Y b) lna, Y b) + b ) ) c, th scond coordinat of which dos not dpnd on th function itslf. In th spcial cas whr b = 0 and c =, this proprty was proposd in [8] as a discovr- activity in a calculus class. For b > and c > 0 in gnral, this bcoms a discovr-w activity in a class that has bn introducd to W as a nw lmntary function, as suggstd in []. Inspird by [8], in [4] th authors gav a mthod for producing familis of curvs whos tangnt lins at a fixd y-coordinat go through th origin. Th family F b,c, howvr, cannot b obtaind by this mthod. 2. At th bginning of this articl w xprssd our surpris that th xcptional xponntial function was not th on with th bas. This xprssd our particular affinity for th bas. Sinc w also hav an affinity for th numbr, w pos th following problm: Find an xcptional function g in 9) whos y-intrcpt is 0, ) and which is a composition of th xponntial function x and a linar function. 3. As notd abov, ach function in F b,c is th composition of an xponntial function and a fixd linar function. This motivats th following problm: Find a non-linar function h for which th family F h = { h g : g F 0, } has proprtis similar to Fb,c. Som obvious candidats for h ar quadratic functions and xponntial functions. 4. For a > w considrd graphs of xponntial functions f a x) = a x and thir tangnt lins l 0,a at th origin and l,a at th y-intrcpt. Ths ar spcial cass of th following problm. Problm. Givn an arbitrary point 0, u) on th y-axis, find th quation of th tangnt lin to th graph of f a that passs through 0, u). Although this problm looks lik a simpl xrcis in calculus, its solution rquirs th us of th Lambrt W function. In fact, th solution xists if and only if u. If a solution dos xist, thn th only two cass for u that can b found without th us of th Lambrt function ar u = 0 and u =. This is anothr rason why l 0,a and l,a ar notworthy. Acknowldgmnt. Many thanks to Árpád Bényi, Vslin Jungić, and Kati Stabls for hlpful suggstions.

13 AN EXCEPTIONAL EXPONENTIAL FUNCTION 3 Rfrncs [], Tim for a Nw Elmntary Function? FOCUS ) 2. [2] R. M. Corlss, G. H. Gonnt, D. E. G. Har, D. J. Jffry, and D. E. Knuth, On th Lambrt W function, Adv. Comput. Math ) [3] J. A. Frohligr and B. Hahn, Hony, whr should w sit? Math. Mag ) [4] R. A. Gordon and B. C. Ditl, Off on a tangnt, Collg Math. J ) [5] B. Hays, Why W? Amrican Scintist ) [6] D. Hughs-Halltt, t al., Calculus 3rd d.), Wily, [7] E. W. Packl and D. S. Yun, Projctil motion with rsistanc and th Lambrt W function, Collg Math. J ) [8] H. Skala, A discovr-, Collg Math. J ) [9] J. Stwart, Calculus. Concpts & Contxt 3rd d.), Thomson Brooks/Col, 2005.

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