let f(x)=ln(1+ e^(−x) ) developp f at integr serie .


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 38725 by maxmathsup by imad last updated on 28/Jun/18

$l e t f (x) = l n (1 + e^{- x}) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by abdo.msup.com last updated on 29/Jun/18
$we have ln(1+u) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ ln(1+e^(−x) ) =Σ_(n=1) ^∞ (((−1)^n )/n) e^(−nx) =Σ_(n=1) ^∞ (((−1)^n )/(n!)) Σ_(p=0) ^∞ (((−nx)^p )/(p!)) =Σ_(n=1) ^∞ (((−1)^n )/(n!)){Σ_(p=0) ^∞ (((−n)^p )/(p!))x^p }.$
$w e h a v e l n (1 + u) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow$ $l n (1 + e^{- x}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} e^{- n x}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n!} \sum_{p = 0}^{\infty} \frac{{(- n x)}^{p}}{p!}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n!} {\sum_{p = 0}^{\infty} \frac{{(- n)}^{p}}{p!} x^{p}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum