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Question Number 39291 by math khazana by abdo last updated on 04/Jul/18
$${let}\:{f}\left({x}\right)=\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 05/Jul/18
$${we}\:{have}\:{e}^{−{x}^{\mathrm{2}} } =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−{x}^{\mathrm{2}} \right)^{{n}} }{{n}!}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} }{{n}!} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)=\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{\mathrm{2}{n}} \right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{c}_{{n}} \:{x}^{\mathrm{2}{n}} \:\:\:{with}\:{c}_{{n}} =\sum_{{i}+{j}={n}} \:\:\:{a}_{{i}} {b}_{{j}} \\ $$$$=\sum_{{i}+{j}={n}} \:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{{i}!}\:\left(−\mathrm{1}\right)^{{j}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{{i}!}\:\left(−\mathrm{1}\right)^{{n}−{i}} \:=\:\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{i}!}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left(\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{i}!}\right)\:{x}^{{n}} \:\:. \\ $$$$\Sigma \\ $$
Commented by abdo mathsup 649 cc last updated on 05/Jul/18
$${the}\:{radius}\:{of}\:{convergence}\:{is}\:{R}=\mathrm{1} \\ $$