Alugaritm agaman

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Iga Alugaritm agaman nɣ Alugaritm anipiri (s tanglizt : Natural logarithm) yat tsɣnt bahra ittyusann ɣ tusnakt. Ar tsnfal tasɣnt ad afaris s tmrnit zun d tisɣnin ilugaritmiyn yaḍnin. Ar tt nttara s mk ad : $\ln()$ .

Ar nttini is iga alugaritm agaman s uzadur n $e$ acku $ln(e)=1$ . Iga alugaritm agaman tamnzut n tsɣnt : $x\longmapsto 1/x$ ɣ uzilal $]0;+\infty [$ .

Amzruy

Alugaritm ad dars yan yism yaḍn ad t igan d "anipiri". Ism ad ikka d John Napir, yan umusnak askatlandi, amskar amzwaru n tflwit talugaritmiyt (tmzaray f talli nssn).

Tettyawskar tazmilt n ulugaritm zɣ dar Grégoire de Saint-Vincent d Alphonse Antonio de Sarasa ɣ usggʷas n 1649.

Isnmal

Ad tili $a\in ]0;+\infty [$ , nzḍar ad nini mas d asnml n tsɣnt talugaritmiyt iga tt unrar lli illan ɣ izddar n tsɣnt n $x\mapsto 1/x$ gr $a$ d 1. Ar tt nttara s tɣarast ad:

$\ln a=\int _{1}^{a}{\frac {1}{x}}\,dx.$

Iɣ darnɣ $a<1$ rad nini mas d anrar lli darnɣ illan iga uzdir.

Tṭṭaf tasɣnt talugaritmiyt aydatn kullu tn n tsɣnin tilugaritmin.^[1]

${\textstyle \ln(ab)=\ln a+\ln b.}$

Nzḍar ad nml ayad s tibḍit n uɣrd lli f nsawl ɣ umzwaru s snat tfulin ɣ akud ann nɣrd s usnfl n umutti $x=ta$ ɣ tfult tiss snat, zun d mk ad:

${\begin{aligned}\ln(ab)=\int _{1}^{ab}{\frac {1}{x}}\,dx&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{a}^{ab}{\frac {1}{x}}\,dx\\[5pt]&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{at}}\,d(at)\\[5pt]&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{t}}\,dt\\[5pt]&=\ln a+\ln b.\end{aligned}}$

Aydatn

Taɣlalt d tsbk n tasɣnt n ulugaritm agaman

Zɣ aylli izrin ar nttafa mas d tasɣnt $x\mapsto ln(x)$ tlla ɣar ɣ uzilal $]0;+\infty [$ , nzḍar ad tt nzllm ɣ ammas n uzilal ad nit d:

$\forall x\in \mathbb {R} _{+}^{*},\ \ln '(x)={\frac {1}{x}}$

S ɣikAd tilia $ln$ tasɣnt tamaɣlalt ɣ uzilal $]0;+\infty [$ , acku tazllumt nns tga tumnigt rad nini ma sd $ln$ tga tasɣnt igmmn ɣ ammas uzilal $]0;+\infty [$

Timhal f tasɣnt n ulugaritm agaman

Ad tg $f$ yat tsɣnt $\ f(x)=\ln(ax)$ s $a$ d $x$ sin imḍann umnign. Tazllumt nns tga tazllumt nit n ulugaritm agaman, ilmma:

$\ f(x)=\ln(x)+k\ /\ k\in \ R$

Acku ${\textstyle f(1)=k}$ rad nini mas d $ln(a)=k$ , s yat tɣarast igan tamatayt:

$\forall (a;b)\in ]0:+\infty [^{2},\ \ln(ab)=\ln(a)+\ln(b)$

Zɣ uyda yad rad naf aydatn ddaw as:

$\forall (a;b)\in ]0:+\infty [^{2},\ \ln \left({\frac {a}{b}}\right)=\ln(a)-\ln(b)$

Iɣ iga $n$ amḍan waxiḍ:

$\forall a\in ]0;+\infty [,\ \forall n\in \mathbb {Z} ,\ \ln(a^{n})=n\ln(a)$
$\forall a\in ]0;+\infty [,\ \forall r\in \mathbb {Q} ,\ \ln(a^{r})=r\ln(a)$

Iɣ iga $n$ amḍan ana:

$\forall a\in \mathbb {R} ^{*},\ \forall \;n\in \mathbb {Z} ^{*},\ \ln(a^{n})=2n\ln |a|$
$\forall (a_{1},a_{2},....,a_{k})\in ]0:+\infty [^{k},\ \sum _{n=1}^{k}\ln(a_{n})=\ln(\prod _{n=1}^{k}a_{n})$

$\prod$ ad igan tamatart n ufaris.

Aydatn yaḍnin

$\ln 1=0$
$\ln e=1$
$\ln(xy)=\ln x+\ln y\quad {\text{i }}\;x>0\;{\text{d }}\;y>0$
$\ln(x^{y})=y\ln x\quad {\text{i }}\;x>0$
$\ln x<\ln y\quad {\text{i }}\;0<x<y$
$\lim _{x\to 0}{\frac {\ln(1+x)}{x}}=1$
$\lim _{\alpha \to 0}{\frac {x^{\alpha }-1}{\alpha }}=\ln x\quad {\text{i }}\;x>0$
${\frac {x-1}{x}}\leq \ln x\leq x-1\quad {\text{i}}\quad x>0$
$\ln {(1+x^{\alpha })}\leq \alpha x\quad {\text{i}}\quad x\geq 0\;{\text{d }}\;\alpha \geq 1$