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Nvmervs e et exponentialis natvralis

Svmmarivm

Haec lectio nvmervm ee constrvit vt svpremvm seqventiae (1+1n)n\left( 1 + \frac{1}{n} \right)^n, qvae ex vsvra composita oritvr: binomio Newtoniano demonstrat seqventiam crescere nec valorem 33 attingere, et svpremvm eivs inter ipsam et seqventiam decrescentem (1+1n)n+1\left( 1 + \frac{1}{n} \right)^{n + 1} inclvdit. Deinde demonstrat ee etiam svpremvm esse svmmarvm inversorvm factorialivm, errorem harvm svmmarvm circvmscribit, qvo fit vt ee qvavis praecisione compvtari possit, et ostendit ee irrationalem esse. Postremo exponentialem natvralem definit, cvm capitalizatione continva vsvrae conivngit et inaeqvalitatem ex≥1+xe^x \geq 1 + x demonstrat, qvae dicit rectam y=1+xy = 1 + x svb imagine exponentialis iacere.

Proposita discendi

  1. Demonstrare seqventiam (1+1n)n\left( 1 + \frac{1}{n} \right)^n crescentem et circvmscriptam esse, et nvmervm ee vt eivs svpremvm definire.
  2. Valores propinqvos nvmeri ee per svmmam inversorvm factorialivm compvtare, cvm margine erroris rigoroso, et demonstrare ee irrationalem esse.
  3. Exponentialem natvralem cvm vsvra composita et cvm eivs capitalizatione continva conivngere, et effectvm capitalizationis diversa freqventia factae comparare.
  4. Inaeqvalitatem ex≥1+xe^x \geq 1 + x demonstrare et adhibere, eamqve interpretari vt sitvm rectae y=1+xy = 1 + x respectv imaginis fvnctionis y=exy = e^x.

Seqventia vsvrae compositae

Sors 11 per vnvm annvm deponitvr vsvra centvm centesimarvm in annvm. Si vsvra semel tantvm solvitvr, anno exevnte, sors dvplicatvr et valet 22. Si bis solvitvr, singvlis semestribvs cvm dimidio modo vsvrae, vsvra primi semestris in secvndo ipsa vsvram parit: sors bis per 1+121 + \frac{1}{2} mvltiplicatvr et valet 94\frac{9}{4}. Qvattvor solvtionibvs trimestribvs valet (1+14)4=625256\left( 1 + \frac{1}{4} \right)^4 = \frac{625}{256}, pavlo plvs qvam 2,44, et dvodecim solvtionibvs menstrvis pavlo plvs qvam 2,61. Generaliter, si annvs in nn tempora aeqvalia dividitvr et in singvlis fractio 1n\frac{1}{n} modi vsvrae solvitvr, sors nn vicibvs per 1+1n1 + \frac{1}{n} mvltiplicatvr, et anno exevnte valet

an=(1+1n)na_n = \left( 1 + \frac{1}{n} \right)^n

Crescente nn saepivs mvltiplicatvr, sed per factorem propiorem nvmero 11, nec evidens est vter effectvs praevaleat: potentiae nvmeri 11 omnes valent 11, potentiae avtem basis fixae maioris qvam 11 sine margine crescvnt. Haec sectio demonstrat seqventiam crescere, terminos avtem eivs valorem 33 non attingere; svpremvm eivs est nvmervs ee. Demonstrationes nitvntvr binomio Newtoniano lectionis de combinatoria et binomio Newtoniano, inaeqvalitate Bernovlliana, (1+x)n≥1+nx(1 + x)^n \geq 1 + nx pro x≥−1x \geq -1, in lectione de axiomatibvs Peani demonstrata, et completvdine nvmerorvm realivm, in lectione de generibvs nvmerorvm envntiata.

Incrementvm seqventiae

Per binomivm Newtonianvm ana_n est svmma n+1n + 1 terminorvm. Terminvs indicis kk est (nk)1nk\binom{n}{k} \frac{1}{n^k}, et scribitvr vt prodvctvm ex 1k!\frac{1}{k!} et k−1k - 1 factoribvs positivis minoribvs qvam 11; convento vt prodvctvm sine factoribvs valeat 11, haec scriptvra valet etiam pro k=0k = 0 et k=1k = 1:

an=∑k=0n1k!∏j=1k−1(1−jn)=1+1+12!(1−1n)+13!(1−1n)(1−2n)+⋯a_n = \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right) = 1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) + \cdots

In hac scriptvra nn apparet tantvm in factoribvs 1−jn1 - \frac{j}{n}, qvi cvm nn avgentvr, et in nvmero terminorvm, qvi etiam avgetvr. Ambo effectvs eodem vergvnt; haec est sententia theorematis qvod infra ponitvr.

TheoremaIncrementvm seqventiae

Pro omni nvmero natvrali nn valet an<an+1a_n < a_{n + 1}, id est seqventia (1+1n)n\left( 1 + \frac{1}{n} \right)^n stricte crescens est.

Evolvtiones binomiales ipsorvm ana_n et an+1a_{n + 1} termino cvm termino comparantvr. Figvra pro n=3n = 3 qvemqve terminvm vt massvlam colvmnae repraesentat qvae a 22 incipit, qvia dvo primi termini omni casv valent 11.

Demonstratio

  1. (nk)1nk=n(n−1)⋯(n−k+1)k! nk=1k!⋅nn⋅n−1n⋯n−k+1n\binom{n}{k} \frac{1}{n^k} = \frac{\resaltar{n(n - 1) \cdots (n - k + 1)}}{k! \, n^k} = \resaltar{\frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n - 1}{n} \cdots \frac{n - k + 1}{n}}

    Per binomivm Newtonianvm, cvm x=1x = 1 et y=1ny = \frac{1}{n}, terminvs indicis kk ipsivs ana_n est (nk)1nk\binom{n}{k} \frac{1}{n^k}. Coefficiens binomialis est prodvctvm kk factorvm divisvm per k!k!, et qvisqve factor per vnvm ex kk factoribvs nn denominatoris dividitvr.

  2. an=∑k=0n1k!∏j=1k−1(1−jn)a_n = \dato{an}{\sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)}

    Primvs qvotiens valet 11, et qvisqve eorvm qvi postea venivnt est n−jn=1−jn\frac{n - j}{n} = 1 - \frac{j}{n}, pro jj a 11 vsqve ad k−1k - 1. Terminis additis scriptvra ipsivs ana_n obtinetvr.

  3. an+1=∑k=0n1k!∏j=1k−1(1−jn+1)+1(n+1)!∏j=1n(1−jn+1)a_{n + 1} = \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n + 1} \right) + \resaltar{\frac{1}{(n + 1)!} \prod_{j=1}^{n} \left( 1 - \frac{j}{n + 1} \right)}

    Eadem scriptvra, cvm n+1n + 1 loco nn, vnvm terminvm plvs habet, illvm indicis k=n+1k = n + 1, qvi positivvs est qvia omnes eivs factores positivi svnt.

  4. 0<1−jn<1−jn+1\dato{f}{0 < 1 - \resaltar{\frac{j}{n}} < 1 - \resaltar{\frac{j}{n + 1}}}

    Pro 1≤j≤n−11 \leq j \leq n - 1, avcto denominatore, jn\frac{j}{n} minvitvr, et factor avgetvr; praeterea omnes factores positivi svnt.

  5. 1k!∏j=1k−1(1−jn)<1k!∏j=1k−1(1−jn+1)\frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)} < \frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n + 1} \right)}

    Ob monotoniam prodvcti nvmerorvm positivorvm, qvisqve terminvs ipsivs ana_n indicis k≥2k \geq 2, si qvis est, minor est qvam terminvs eivsdem indicis ipsivs an+1a_{n + 1}; termini indicvm 00 et 11 in vtraqve svmma valent 11.

  6. a1=2<a2=94<a3=6427<a4=625256a_1 = 2 < a_2 = \frac{9}{4} < a_3 = \frac{64}{27} < a_4 = \frac{625}{256}

    Svmmando, ana_n non excedit svmmam n+1n + 1 primorvm terminorvm ipsivs an+1a_{n + 1}, cvi adhvc deest terminvs positivvs indicis n+1n + 1. Ergo an<an+1a_n < a_{n + 1}. In figvra termini seqventiae ascendvnt.

Margo svperior

Incrementvm non svfficit vt seqventia svpremvm habeat: etiam nvmeri natvrales crescvnt. Margine svperiore opvs est, et ad evm obtinendvm ana_n cvm svmma inversorvm factorialivm comparatvr,

sn=∑k=0n1k!=1+1+12+16+⋯+1n!s_n = \sum_{k=0}^{n} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots + \frac{1}{n!}

qvae oritvr si in scriptvra ipsivs ana_n omnes factores 1−jn1 - \frac{j}{n} tollvntvr. Hae svmmae per se momentvm habent, et sectio qvae postea venit demonstrabit eas ad evndem nvmervm atqve ana_n accedere, et mvlto celerivs. Problema propositvm 5 lectionis de combinatoria postvlabat vt demonstraretvr an<3a_n < 3; secvndvm ex lemmatibvs qvae infra ponvntvr hvnc marginem continet, cvm inaeqvalitate media an≤sna_n \leq s_n, qvae postea adhibebitvr. Primvm factorialia cvm potentiis nvmeri 22 comparat, et id est qvod efficit vt sns_n per progressionem geometricam circvmscribatvr.

LemmaFactorialia et potentiae nvmeri dvo

Pro omni nvmero natvrali kk valet k!≥2k−1k! \geq 2^{k - 1}, aeqvalitate tantvm si k=1k = 1 vel k=2k = 2.

Ratiocinatio fit per indvctionem in kk, principio lectionis de axiomatibvs Peani: transevndo a kk ad k+1k + 1, factoriale per k+1k + 1 mvltiplicatvr, qvi saltem 22 est, potentia avtem tantvm dvplicatvr. Figvra pro kk a 11 vsqve ad 44 factoriale cvm potentia comparat, vt dvas colvmnas.

Demonstratio

  1. 1!=1=201! = 1 = 2^{0}

    Casvs basis: pro k=1k = 1 ambo membra valent 11.

  2. k!≥2k−1\dato{h}{k! \geq 2^{k - 1}}

    Hypothesis indvctionis: ponamvs inaeqvalitatem valere pro aliqvo nvmero natvrali kk.

  3. (k+1)!=(k+1)⋅k!≥2⋅2k−1=2k(k + 1)! = \resaltar{(k + 1)} \cdot k! \geq \resaltar{2} \cdot 2^{k - 1} = 2^{k}

    Factoriale ipsivs k+1k + 1 est illvd ipsivs kk mvltiplicatvm per k+1k + 1, qvi saltem 22 est; cvm omnes factores positivi sint, hypothesis indvctionis inaeqvalitatem pro k+1k + 1 dat.

  4. k!≥2k−1,3!=6>4=22k! \geq 2^{k - 1}, \qquad 3! = 6 > 4 = 2^2

    Per principivm indvctionis inaeqvalitas pro omni nvmero natvrali kk valet. Si k≥2k \geq 2, factor k+1k + 1 saltem 33 est, et inaeqvalitas pro k+1k + 1 stricta est: aeqvalitas fit in k=1k = 1 et in k=2k = 2, et in his solis.

LemmaMargo seqventiae

Pro omni nvmero natvrali nn valet an≤sn<3a_n \leq s_n < 3; prima inaeqvalitas stricta est pro n≥2n \geq 2.

Qvisqve terminvs evolvtionis ipsivs ana_n per terminvm respondentem ipsivs sns_n circvmscribitvr, et qvisqve terminvs ipsivs sns_n per terminvm progressionis geometricae rationis 12\frac{1}{2}. Figvra pro n=4n = 4 tres colvmnas svper basim commvnem 22 comparat.

Demonstratio

  1. 1k!∏j=1k−1(1−jn)≤1k!⋅1\frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)} \leq \frac{1}{k!} \cdot \resaltar{1}

    Qvisqve factor 1−jn1 - \frac{j}{n} inter 00 et 11 iacet; itaqve prodvctvm eorvm non excedit 11, et qvisqve terminvs ipsivs ana_n non excedit terminvm 1k!\frac{1}{k!} ipsivs sns_n.

  2. an≤sn\dato{as}{a_n \leq s_n}

    Svmmando pro kk a 00 vsqve ad nn prima inaeqvalitas obtinetvr. Si n≥2n \geq 2, terminvs indicis 22 factorem 1−1n<11 - \frac{1}{n} < 1 habet, et inaeqvalitas stricta est.

  3. 1k!≤12k−1,k≥1\dato{f}{\frac{1}{k!} \leq \frac{1}{2^{k - 1}}}, \quad k \geq 1

    Per lemma svpra positvm k!≥2k−1k! \geq 2^{k - 1} pro omni nvmero natvrali kk; inversis dvobvs nvmeris positivis, inaeqvalitas sensvm mvtat.

  4. sn=1+∑k=1n1k!≤1+∑k=1n12k−1s_n = 1 + \resaltar{\sum_{k=1}^{n} \frac{1}{k!}} \leq 1 + \resaltar{\sum_{k=1}^{n} \frac{1}{2^{k - 1}}}

    Terminvm indicis 00 separamvs et ceteros circvmscribimvs.

  5. 1+∑k=1n12k−1=1+2(1−12n)=3−12n−1<31 + \resaltar{\sum_{k=1}^{n} \frac{1}{2^{k - 1}}} = 1 + \resaltar{2 \left( 1 - \frac{1}{2^n} \right)} = 3 - \frac{1}{2^{n - 1}} < 3

    Qvod illvstratvm est, svmma est nn terminorvm progressionis geometricae primi termini 11 et rationis 12\frac{1}{2}, in lectione de svmmationibvs et progressionibvs compvtata.

  6. an≤sn≤3−12n−1<3a_n \leq s_n \leq 3 - \frac{1}{2^{n - 1}} < 3

    Ambabvs inaeqvalitatibvs conexis lemma obtinetvr. In figvra, cvm n=4n = 4, colvmna geometrica spatio 18\frac{1}{8} infra 33 consistit, et dvae ceterae infra eam.

Seqventia decrescens qvae eam comitatvr

Si basis 1+1n1 + \frac{1}{n} ad exponentem vno maiorem elevatvr, altera seqventia obtinetvr, cvivs termini maiores svnt qvam primae, qvia per factorem maiorem qvam 11 mvltiplicantvr:

bn=(1+1n)n+1=an(1+1n)>anb_n = \left( 1 + \frac{1}{n} \right)^{n + 1} = a_n \left( 1 + \frac{1}{n} \right) > a_n

Primi eivs termini svnt b1=4b_1 = 4, b2=278b_2 = \frac{27}{8} et b3=25681b_3 = \frac{256}{81}: descendvnt. Theorema qvod infra ponitvr hoc generaliter demonstrat; consilivm est qvotientem dvorvm terminorvm continvorvm considerare, instrvmentvm vero inaeqvalitas Bernovlliana.

TheoremaDecrementvm seqventiae comitis

Pro omni nvmero natvrali nn valet bn>bn+1b_n > b_{n + 1}, id est seqventia (1+1n)n+1\left( 1 + \frac{1}{n} \right)^{n + 1} stricte decrescens est.

Satis est ostendere qvotientem bnbn+1\frac{b_n}{b_{n + 1}} maiorem esse qvam 11, qvia termini positivi svnt. Figvra primos terminos vtrivsqve seqventiae ostendit.

Demonstratio

  1. bnbn+1=(n+1n)n+1(n+2n+1)n+2=((n+1)2n(n+2))n+1⋅n+1n+2\frac{b_n}{b_{n + 1}} = \frac{\left( \frac{n + 1}{n} \right)^{n + 1}}{\left( \frac{n + 2}{n + 1} \right)^{n + 2}} = \dato{c}{\left( \frac{(n + 1)^2}{n(n + 2)} \right)^{n + 1} \cdot \frac{n + 1}{n + 2}}

    Qvamqve basim vt qvotientem scribimvs, factorem n+1n+2\frac{n + 1}{n + 2} a potentia denominatoris separamvs et dvas potentias exponentis n+1n + 1 in vnam conivngimvs.

  2. (n+1)2n(n+2)=n(n+2)+1n(n+2)=1+1n(n+2)\frac{\resaltar{(n + 1)^2}}{n(n + 2)} = \frac{\resaltar{n(n + 2) + 1}}{n(n + 2)} = \dato{q}{1 + \frac{1}{n(n + 2)}}

    Nvmerator basis denominatorem excedit nvmero 11, qvia (n+1)2=n2+2n+1=n(n+2)+1(n + 1)^2 = n^2 + 2n + 1 = n(n + 2) + 1.

  3. (1+1n(n+2))n+1≥1+n+1n(n+2)\dato{B}{\left( 1 + \frac{1}{n(n + 2)} \right)^{n + 1} \geq 1 + \frac{n + 1}{n(n + 2)}}

    Per inaeqvalitatem Bernovllianam, cvm x=1n(n+2)x = \frac{1}{n(n + 2)} et exponente n+1n + 1, potentia hvivs basis saltem 1+(n+1)x1 + (n + 1)x est.

  4. bnbn+1≥(1+n+1n(n+2))n+1n+2=(n2+3n+1)(n+1)n(n+2)2\frac{b_n}{b_{n + 1}} \geq \resaltar{\left( 1 + \frac{n + 1}{n(n + 2)} \right)} \frac{n + 1}{n + 2} = \resaltar{\frac{(n^2 + 3n + 1)(n + 1)}{n(n + 2)^2}}

    In qvotientem svbstitvimvs et ad commvnem denominatorem redvcimvs; svmma n(n+2)+(n+1)n(n + 2) + (n + 1) valet n2+3n+1n^2 + 3n + 1.

  5. (n2+3n+1)(n+1)n(n+2)2=n3+4n2+4n+1n3+4n2+4n>1\frac{\resaltar{(n^2 + 3n + 1)(n + 1)}}{\resaltar{n(n + 2)^2}} = \frac{\resaltar{n^3 + 4n^2 + 4n + 1}}{\resaltar{n^3 + 4n^2 + 4n}} > 1

    Nvmeratorem et denominatorem evolvimvs: nvmerator denominatorem excedit nvmero 11.

  6. b1=4>b2=278>b3=25681>b4=31251024b_1 = 4 > b_2 = \frac{27}{8} > b_3 = \frac{256}{81} > b_4 = \frac{3125}{1024}

    Ergo bn>bn+1b_n > b_{n + 1}. In figvra termini bnb_n descendvnt dvm termini ana_n ascendvnt, et qvisqve bnb_n svpra qvemqve ana_n manet.

Nvmervs e

Dvae seqventiae inter se accedvnt, sed non se secant: omnis terminvs primae minor est qvam omnis terminvs secvndae, etiamsi indices eorvm diversi sint.

LemmaDvae seqventiae non se secant

Pro omnibvs nvmeris natvralibvs nn et mm valet an<bma_n < b_m.

Transitvr per maiorem dvorvm indicvm: prima seqventia crescit, secvnda decrescit, et in qvoqve indice secvnda svpra primam est. Figvra casvm n=2n = 2 et m=5m = 5 perseqvitvr.

Demonstratio

  1. bp=ap(1+1p)>ap\dato{p}{b_p = a_p \left( 1 + \frac{1}{p} \right) > a_p}

    In eodem indice pp, terminvs bpb_p est apa_p mvltiplicatvs per 1+1p1 + \frac{1}{p}, qvi maior est qvam 11, et apa_p positivvs est.

  2. n≤p→an≤ap\dato{a}{n \leq p \rightarrow a_n \leq a_p}

    Sit pp maior nvmerorvm natvralivm nn et mm. Cvm n≤pn \leq p et seqventia ana_n crescens sit, terminvs eivs indicis nn non excedit terminvm indicis pp.

  3. m≤p→bp≤bm\dato{b}{m \leq p \rightarrow b_p \leq b_m}

    Cvm m≤pm \leq p et seqventia bnb_n decrescens sit, terminvs eivs indicis pp non excedit terminvm indicis mm.

  4. an≤ap<bp≤bma_n \leq a_p < b_p \leq b_m

    Tribvs inaeqvalitatibvs conexis, media stricta, lemma obtinetvr. In figvra, cvm n=2n = 2 et m=5m = 5, index maior est p=5p = 5.

Vnde efficitvr copiam terminorvm ana_n elementa habere et svpra circvmscriptam esse, nvmero 33 vel qvolibet bmb_m. Ob completvdinem nvmerorvm realivm svpremvm habet, qvod littera ee designatvr.

Definitio. Nvmervs ee est svpremvm copiae terminorvm seqventiae (1+1n)n\left( 1 + \frac{1}{n} \right)^n, pro nn natvrali.
TheoremaNvmervs e inter dvas seqventias

Pro omni nvmero natvrali nn valet an<e<bna_n < e < b_n.

Adhibentvr definitio svpremi vt minimi marginis svperioris et monotonia vtrivsqve seqventiae. Figvra dvas seqventias ostendit, qvae ad ee ex lateribvs oppositis accedvnt.

Demonstratio

  1. an<bm,n,m∈N\dato{ab}{a_n < b_m}, \quad n, m \in \mathbb{N}

    Per lemma svpra positvm omnis terminvs primae seqventiae minor est qvam omnis terminvs secvndae.

  2. (∀m∈N)(e≤bm)\dato{eb}{(\forall m \in \mathbb{N})(e \leq b_m)}

    Qvisqve bmb_m igitvr margo svperior est copiae terminorvm ana_n, et svpremvm, qvod minimvs marginvm svperiorvm est, nvllvm excedit.

  3. an<an+1≤e\dato{ae}{a_n < \resaltar{a_{n + 1}} \leq \resaltar{e}}

    Cvm ee margo svperior sit et seqventia ana_n stricte crescens, qvisqve terminvs minor est qvam is qvi post evm venit, qvi ee non excedit.

  4. e≤bn+1<bn\dato{be}{e \leq \resaltar{b_{n + 1}} < \resaltar{b_n}}

    Eodem modo, cvm seqventia bnb_n stricte decrescens sit, ee non excedit terminvm qvi post venit, qvi minor est qvam bnb_n.

  5. an<e<bna_n < e < b_n

    Ambabvs inaeqvalitatibvs conexis theorema obtinetvr: ee est vnicvs nvmervs qvi inter omnes terminos ana_n et omnes terminos bnb_n iacet.

  6. a74<e<b17a_{74} < e < b_{17}

    Machinvla calcvlatoria invenitvr a74a_{74} svperare 2,7 et b17b_{17} minorem esse qvam 2,8; itaqve ee inter 2,7 et 2,8 est. Inclvsio lenta est: distantia inter dvas seqventias est bn−an=annb_n - a_n = \frac{a_n}{n}, qvae vt inversvm ipsivs nn minvitvr.

Nvmervs ee fere valet 2,718281828. Iacobvs Bernovlli evm anno 1683 invenit dvm vsvram compositam investigat, et Leonhardvs Evler, qvi mvltas eivs proprietates detexit, evm littera ee designavit. Vna cvm π\pi est vna ex dvabvs constantibvs fvndamentalibvs analyseos.