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Exemplaria incrementi et decrementi exponentialis

Svmmarivm

Haec lectio fvnctiones exponentialem et logarithmicam adhibet ad magnitvdines describendas qvae periodis aeqvalibvs eadem proportione mvtantvr. Incrementvm discretvm ratione constanti et exemplar continvvm P(t)=P0ektP(t) = P_0 e^{kt} tractat, demonstrat in hoc tempvs dvplicationis et tempvs dimidiationis non pendere ab instanti qvo metivntvr, idqve applicat ad dissolvtionem radioactivam, ad datationem per carbonivm 14 et ad legem refrigerationis Newtonianam. Deinde docet qvomodo parametri exemplaris ex dvobvs datis determinentvr, demonstrat data exponentialia in delineatione semilogarithmica in rectam disponi, et conclvditvr exemplari logistico, cvivs incrementvm circvmscriptvm est et cvivs delineatio circa pvnctvm svvm medivm symmetrica est.

Proposita discendi

  1. Exemplaria incrementi et decrementi discreta et continva agnoscere, et rationem mvtationis per periodvm cvm constanti exemplaris conferre.
  2. Tempvs dvplicationis et tempvs dimidiationis compvtare et adhibere, eaqve ad dissolvtionem radioactivam, ad datationem per carbonivm 14 et ad refrigerationem corporis applicare.
  3. Parametros exemplaris exponentialis ex dvobvs datis determinare et data exponentialia in delineatione semilogarithmica agnoscere.
  4. Exemplar logisticvm investigare: eivs circvmscriptionem, eivs monotoniam et symmetriam delineationis circa pvnctvm qvo dimidivm marginis svae attingit.

Incrementvm ratione constanti

Magnitvdo dvobvs modis elementaribvs avgeri potest. Si singvlis periodis eadem qvantitas ei additvr, valores eivs progressionem arithmeticam efficivnt et delineatio eivs recta est; sic crescit sors vsvra simplici, qvae qvotannis eandem vsvram accipit. Si contra singvlis periodis per evndem factorem mvltiplicatvr, valores eivs progressionem geometricam efficivnt; sic crescit sors vsvra composita, cvivs vsvra cvivsqve anni in sortem cvmvlatam compvtatvr, et sic crescit popvlatio bacteriorvm in qva qvaeqve cellvla certo tempore dividitvr. Ambae progressiones in lectione de svmmationibvs, progressionibvs et indvctione tractatae svnt; haec lectio de altera agit et de eivs forma continva.

Sit P0>0P_0 > 0 valor initialis magnitvdinis et PnP_n eivs valor post nn periodos. Magnitvdo dicitvr ratione constanti rr per periodvm mvtari, cvm r>−1r > -1, cvm qvisqve valor ex eo qvi antecedit fit mvltiplicando per 1+r1 + r, id est cvm qvotiens dvorvm valorvm consecvtivorvm semper idem est:

Pn+1Pn=1+r\frac{P_{n + 1}}{P_n} = 1 + r

Nvmervs rr plervmqve centesimis exprimitvr: popvlatio qvae 33 per centvm qvotannis avgetvr habet r=3100r = \frac{3}{100}, et svbstantia qvae qvintam massae svae portionem singvlis horis amittit habet r=−15r = -\frac{1}{5}. Si r>0r > 0, de incremento dicitvr; si −1<r<0-1 < r < 0, de decremento.

LemmaIncrementvm ratione constanti

Si Pn+1=(1+r)PnP_{n + 1} = (1 + r) P_n pro omni nn, cvm P0>0P_0 > 0 et r>−1r > -1, tvm Pn=P0(1+r)nP_n = P_0 (1 + r)^n pro omni natvrali nn. Omnes valores positivi svnt; seqventia stricte crescens est si r>0r > 0 et stricte decrescens si −1<r<0-1 < r < 0.

Formvla est terminvs generalis progressionis geometricae rationis 1+r1 + r, et per indvctionem demonstratvr. Figvra primos valores ostendit cvm P0=1P_0 = 1, primvm cvm r=12r = \frac{1}{2}, deinde cvm r=−12r = -\frac{1}{2}.

Demonstratio

  1. P1=(1+r)P0=P0(1+r)1P_1 = (1 + r) P_0 = P_0 (1 + r)^{\resaltar{1}}

    Pro n=1n = 1 formvla est hypothesis ad P0P_0 applicata.

  2. Pn+1=(1+r)Pn=(1+r)P0(1+r)n=P0(1+r)n+1P_{n + 1} = (1 + r) \resaltar{P_n} = (1 + r) \resaltar{P_0 (1 + r)^n} = \dato{f}{P_0 (1 + r)^{n + 1}}

    Si formvla pro nn valet, hypothesis eam pro n+1n + 1 praebet; per principivm indvctionis pro omni natvrali valet.

  3. P0>0,1+r>0→Pn>0P_0 > 0, \quad 1 + r > 0 \rightarrow \dato{p}{P_n > 0}

    Cvm P0P_0 et 1+r1 + r positivi sint, omnes eorvm potentiae et prodvcta positiva svnt: omnes valores positivi svnt.

  4. Pn+1−Pn=rPnP_{n + 1} - P_n = \resaltar{r P_n}

    Differentia dvorvm valorvm consecvtivorvm est rPnr P_n, qvae signvm ipsivs rr habet: seqventia crescit si r>0r > 0 et decrescit si −1<r<0-1 < r < 0.

Exempli gratia, sors 10001000 vsvra composita 55 per centvm in annvm post tres annos fit

C3=1000(1+5100)3=92618C_3 = 1000 \left( 1 + \frac{5}{100} \right)^3 = \frac{9261}{8}

id est 1157,625, cvm vsvra simplici fieret 11501150. Differentia, initio parva, cvm nvmero annorvm sine margine crescit; problema solvtvm 2 demonstrat vsvram compositam inde ab altero anno semper simplicem svperare, et sortivm et modorvm vsvrae stvdivm ordinatvm ad cvrsvm Mathematica Nvmmaria pertinet.