Previous in Relation and Functions Next in Relation and Functions
Question Number 33293 by abdo imad last updated on 14/Apr/18
$${developp}\:{f}\left({x}\right)\:=\:\frac{{e}^{{x}} }{{x}−\mathrm{1}}\:{at}\:{integr}\:{serie} \\ $$
Commented by abdo imad last updated on 19/Apr/18
$$−{f}\left({x}\right)\:=\:\frac{{e}^{{x}} }{\mathrm{1}−{x}}\:\:{so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have}\: \\ $$$$−{f}\left({x}\right)\:=\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}!}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \right)=\left(\sum_{{n}=\mathrm{0}} ^{\infty} {a}_{{n}} {x}^{{n}} \right)\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} {b}_{{n}} {x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{c}_{{n}} \:{x}^{{n}} \:\:{with}\:{c}_{{n}} =\sum_{{i}+{j}={n}} {a}_{{i}} \:{b}_{{j}} \:=\sum_{{i}=\mathrm{0}} ^{{n}} {a}_{{i}} {b}_{{n}−{i}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{i}!}\:\Rightarrow\:{f}\left({x}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \left(\:\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{i}!}\right){x}^{{n}} \:\:. \\ $$