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Fvnctiones hyperbolicae

Svmmarivm

Haec lectio cosinvm, sinvm et tangentem hyperbolicos considerat, tres fvnctiones qvae obtinentvr exponentialem natvralem exe^x cvm reciproca eivs e−xe^{-x} conivngendo. Earvm paritatem, signvm, monotoniam et copiam imaginvm determinat, et identitatem cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1 demonstrat, ex qva pvnctvm (cosh⁡t,sinh⁡t)\left( \cosh t, \sinh t \right) ramvm dextrvm hyperbolae x2−y2=1x^2 - y^2 = 1 percvrrit et per vnvmqvodqve eivs pvnctvm semel tantvm transit, vnde nomen earvm dvcitvr. Deinde formvlas additionis et argvmenti dvpli dedvcit, et fvnctiones inversas obtinet, qvae per logarithmos exprimvntvr. Postremo catenariam describit, qvae forma est catenae svspensae, et aream sectoris hyperbolici, qvae analogiam cvm fvnctionibvs circvlaribvs praenvntiat.

Proposita discendi

  1. Cosinvm, sinvm et tangentem hyperbolicos ex exponentiali natvrali definire, eorvmqve paritatem, signvm, monotoniam et margines demonstrare.
  2. Identitatem fvndamentalem demonstrare eaqve ad ramos hyperbolarvm parametris describendos vti, agnoscendo cviqve pvncto rami vnicvm parametrvm respondere.
  3. Formvlas additionis et argvmenti dvpli dedvcere, easqve ad identitates et inaeqvalitates demonstrandas adhibere.
  4. Fvnctiones hyperbolicas inversas cvm dominiis svis obtinere, aeqvationes qvadraticas in eye^y solvendo, iisqve ad aeqvationes hyperbolicas solvendas vti.

Definitio et proprietates primae

In lectione de nvmero e et exponentiali natvrali definita est fvnctio exe^x, positiva et stricte crescens in tota recta reali. Eivs reciproca e−x=1exe^{-x} = \frac{1}{e^x} etiam positiva est, sed stricte decrescens, et eivs delineatio est reflexio delineationis exe^x respectv axis yy, vt in lectione de operationibvs fvnctionvm ostensvm est. Dimidia svmma et dimidia differentia harvm dvarvm fvnctionvm, et earvm qvotiens, tam saepe in geometria, in physica et in calcvlo occvrrvnt vt nomen proprivm accipiant.

Pro omni nvmero reali xx definivntvr cosinvs hyperbolicvs, sinvs hyperbolicvs et tangens hyperbolica ipsivs xx vt dimidia svmma exe^x et e−xe^{-x}, earvm dimidia differentia et qvotiens secvndae per primam, respective.
cosh⁡x=ex+e−x2,sinh⁡x=ex−e−x2,tanh⁡x=sinh⁡xcosh⁡x=ex−e−xex+e−x\cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}

Tres fvnctiones in toto R\mathbb{R} definitae svnt: denominator tangentis est svmma dvorvm nvmerorvm positivorvm et nvmqvam evanescit. In x=0x = 0 habetvr e0=1e^0 = 1, ita vt cosh⁡0=1\cosh 0 = 1, sinh⁡0=0\sinh 0 = 0 et tanh⁡0=0\tanh 0 = 0. Adiectivvm «hyperbolicvs» in tertio capite ivstificabitvr, vbi videbitvr has fvnctiones hyperbolam describere.

LemmaParitas et decompositio exponentialis

Cosinvs hyperbolicvs est fvnctio par, sinvs avtem et tangens hyperbolica fvnctiones impares svnt. Praeterea, pro omni reali xx valent ex=cosh⁡x+sinh⁡xe^x = \cosh x + \sinh x et e−x=cosh⁡x−sinh⁡xe^{-x} = \cosh x - \sinh x.

Si pro xx ponitvr −x-x, fvnctiones exe^x et e−xe^{-x} mvnera sva permvtant. Figvra delineationes cosinvs et sinvs hyperbolicorvm ostendit cvm binis pvnctis abscissarvm oppositarvm in vnaqvaqve.

Demonstratio

  1. cosh⁡(−x)=e−x+ex2=cosh⁡x\cosh(-x) = \frac{\resaltar{e^{-x} + e^{x}}}{2} = \cosh x

    Si pro xx ponitvr −x-x, dimidia svmma non mvtatvr, qvia dvo eivs addenda inter se permvtantvr: cosinvs hyperbolicvs par est, et eivs delineatio symmetrica est respectv axis yy.

  2. sinh⁡(−x)=e−x−ex2=−sinh⁡x\sinh(-x) = \frac{\resaltar{e^{-x} - e^{x}}}{2} = -\sinh x

    Dimidia differentia contra signvm mvtat: sinvs hyperbolicvs impar est, et eivs delineatio symmetrica est respectv originis.

  3. tanh⁡(−x)=−sinh⁡xcosh⁡x=−tanh⁡x\tanh(-x) = \frac{\resaltar{-\sinh x}}{\resaltar{\cosh x}} = -\tanh x

    Tangens, qvotiens fvnctionis imparis per fvnctionem parem, impar est.

  4. cosh⁡x+sinh⁡x=ex,cosh⁡x−sinh⁡x=e−x\cosh x + \sinh x = \resaltar{e^x}, \qquad \cosh x - \sinh x = \resaltar{e^{-x}}

    Definitiones cosinvs et sinvs hyperbolicorvm addendo et svbtrahendo dvae exponentiales recvperantvr: cosh⁡x\cosh x et sinh⁡x\sinh x svnt, respective, pars par et pars impar exponentialis.

ExemplvmDelineationes cosinvs et sinvs hyperbolicorvm

Delineationes cosh⁡\cosh et sinh⁡\sinh constrvvntvr ordinatas ex2\frac{e^x}{2} et e−x2\frac{e^{-x}}{2} abscissa post abscissam addendo et svbtrahendo.

Demonstratio

  1. y=ex2,e02=12y = \frac{e^x}{2}, \quad \frac{e^0}{2} = \frac{1}{2}

    Delineatio ex2\frac{e^x}{2} ex delineatione exe^x obtinetvr eam verticaliter ad dimidivm comprimendo: axem yy in 12\frac{1}{2} secat, dextrorsvm sine margine crescit et sinistrorsvm ad axem xx accedit.

  2. y=e−x2y = \frac{e^{-x}}{2}

    Delineatio e−x2\frac{e^{-x}}{2} est reflexio primae respectv axis yy.

  3. cosh⁡0=12+12=1\cosh 0 = \resaltar{\frac{1}{2} + \frac{1}{2}} = \resaltar{1}

    In vnaqvaqve abscissa cosh⁡x\cosh x est svmma dvarvm ordinatarvm. In 00 ambae valent 12\frac{1}{2}: delineatio per (0,1)(0, 1) transit et symmetrica est respectv axis yy, qvia cosinvs hyperbolicvs par est.

  4. cosh⁡x−ex2=e−x2>0\cosh x - \frac{e^x}{2} = \frac{e^{-x}}{2} > 0

    Cosinvs hyperbolicvs svperat ex2\frac{e^x}{2} exacte qvantitate e−x2\frac{e^{-x}}{2}, qvae positiva est et tam parva fit qvam libet cvm xx crescit: dextrorsvm delineatio cosh⁡\cosh ad delineationem ex2\frac{e^x}{2} desvper accedit, et sinistrorsvm, propter symmetriam, ad delineationem e−x2\frac{e^{-x}}{2}.

  5. sinh⁡0=12−12=0\sinh 0 = \resaltar{\frac{1}{2} - \frac{1}{2}} = \resaltar{0}

    Sinvs hyperbolicvs est differentia ordinatarvm. Eivs delineatio per originem transit et symmetrica est respectv originis, qvia sinvs hyperbolicvs impar est.

  6. cosh⁡x−sinh⁡x=e−x>0\cosh x - \sinh x = e^{-x} > 0

    Dextrorsvm delineatio sinh⁡\sinh ad delineationem ex2\frac{e^x}{2} desvbter accedit, dvm delineatio cosh⁡\cosh desvper accedit: dvae inter se appropinqvant neqve vmqvam se tangvnt, qvia earvm differentia positiva est.

Signvm sinvs hyperbolici ex eivs definitione legitvr: sinh⁡x>0\sinh x > 0 aeqvivalet ex>e−xe^x > e^{-x}, et qvia exponentialis stricte crescens est, hoc fit exacte cvm x>−xx > -x, id est cvm x>0x > 0. Ergo sinvs hyperbolicvs negativvs est si x<0x < 0, nvllvs in 00 et positivvs si x>0x > 0; tangens hyperbolica, qvotiens sinvs per nvmervm positivvm, idem signvm habet. Cosinvs hyperbolicvs contra semper positivvs est, atqve etiam plvs vervm est.

LemmaMargo inferior cosinvs hyperbolici

Pro omni reali xx valet cosh⁡x−1=2sinh⁡2x2\cosh x - 1 = 2 \sinh^2 \frac{x}{2}. Itaqve cosh⁡x≥1\cosh x \geq 1, et aeqvalitas tantvm fit cvm x=0x = 0.

Consilivm est in nvmeratore cosh⁡x−1\cosh x - 1 qvadratvm differentiae agnoscere. Figvra delineationem cosinvs hyperbolici svpra rectam y=1y = 1 ostendit, differentia inter vtramqve vt segmento verticali repraesentata.

Demonstratio

  1. ex=(ex2)2,e−x=(e−x2)2,ex2e−x2=1e^x = \left( e^{\frac{x}{2}} \right)^2, \quad e^{-x} = \left( e^{-\frac{x}{2}} \right)^2, \quad \dato{p}{e^{\frac{x}{2}} e^{-\frac{x}{2}} = 1}

    Per leges exponentivm exe^x est qvadratvm ex2e^{\frac{x}{2}}, e−xe^{-x} est qvadratvm e−x2e^{-\frac{x}{2}}, et prodvctvm harvm dvarvm potentiarvm est e0=1e^0 = 1.

  2. cosh⁡x−1=ex−2+e−x2=(ex2−e−x2)22=2sinh⁡2x2\cosh x - 1 = \frac{\resaltar{e^x - 2 + e^{-x}}}{2} = \frac{\resaltar{\left( e^{\frac{x}{2}} - e^{-\frac{x}{2}} \right)^2}}{2} = \dato{c}{2 \sinh^2 \frac{x}{2}}

    Ergo nvmerator cosh⁡x−1\cosh x - 1 est qvadratvm differentiae, et ea differentia est dvplvm sinh⁡x2\sinh \frac{x}{2}.

  3. cosh⁡x−1=2sinh⁡2x2≥0\cosh x - 1 = \resaltar{2 \sinh^2 \frac{x}{2}} \geq 0

    Qvadratvm non est negativvm: cosinvs hyperbolicvs est saltem 11.

  4. cosh⁡x=1↔ex2=e−x2↔x=0\cosh x = 1 \leftrightarrow e^{\frac{x}{2}} = e^{-\frac{x}{2}} \leftrightarrow \resaltar{x = 0}

    Aeqvalitas postvlat vt basis qvadrati nvlla sit, id est ex2=e−x2e^{\frac{x}{2}} = e^{-\frac{x}{2}}; qvia exponentialis iniectiva est, hoc postvlat x2=−x2\frac{x}{2} = -\frac{x}{2}, scilicet x=0x = 0.

Inaeqvalitas cosh⁡x≥1\cosh x \geq 1 est inaeqvalitas inter mediam arithmeticam et mediam geometricam nvmerorvm positivorvm exe^x et e−xe^{-x}, qvorvm media geometrica est 11.