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 35766 by abdo mathsup 649 cc last updated on 23/May/18

let f(x)=arctan(1+x)  and  f_n (x)= e^(−nx)   1) define fof_n (x) and f_n of(x)  2) study the nature of the series Σ_(n=0) ^(+∞) fof_n (x)  and Σ_(n=0) ^∞  f_n of(x) .

$${let}\:{f}\left({x}\right)={arctan}\left(\mathrm{1}+{x}\right)\:\:{and}\:\:{f}_{{n}} \left({x}\right)=\:{e}^{−{nx}} \\ $$$$\left.\mathrm{1}\right)\:{define}\:{fof}_{{n}} \left({x}\right)\:{and}\:{f}_{{n}} {of}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{nature}\:{of}\:{the}\:{series}\:\sum_{{n}=\mathrm{0}} ^{+\infty} {fof}_{{n}} \left({x}\right) \\ $$$${and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} {of}\left({x}\right)\:. \\ $$

Commented by maxmathsup by imad last updated on 02/Sep/18

1) we have fof_n (x)=f(f_n (x))=arctan(1+f_n (x))=arctan(1+e^(−nx) )  f_n 0f(x) = f_n (f(x))=e^(−nf(x)) =e^(−narctan(1+x))   2)Σ_(n=0) ^∞  fof_n (x) =Σ_(n=0) ^∞  arctan(1+e^(−nx) ) =Σ u_n (x)  if x<0  lim_(n→+∞) arctan(1+e^(−nx) ) ≠0  so Σ u_n (x)diverges  if x>0  u_n (x)∼ e^(−nx)    (n→+∞)  but Σ e^(−nx)  converges so the convergence  simple of Σ u_n (x) is assured.also  Σ f_n of  = Σ_n  e^(−n arctan(1+x))    =Σ_n     (e^(−arctan(1+x)) )^n  ⇒  if arctan(1+x)>0  the serie converges  if arctan(1+x)<0 the serie diverges.

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{fof}_{{n}} \left({x}\right)={f}\left({f}_{{n}} \left({x}\right)\right)={arctan}\left(\mathrm{1}+{f}_{{n}} \left({x}\right)\right)={arctan}\left(\mathrm{1}+{e}^{−{nx}} \right) \\ $$$${f}_{{n}} \mathrm{0}{f}\left({x}\right)\:=\:{f}_{{n}} \left({f}\left({x}\right)\right)={e}^{−{nf}\left({x}\right)} ={e}^{−{narctan}\left(\mathrm{1}+{x}\right)} \\ $$$$\left.\mathrm{2}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:{fof}_{{n}} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{arctan}\left(\mathrm{1}+{e}^{−{nx}} \right)\:=\Sigma\:{u}_{{n}} \left({x}\right) \\ $$$${if}\:{x}<\mathrm{0}\:\:{lim}_{{n}\rightarrow+\infty} {arctan}\left(\mathrm{1}+{e}^{−{nx}} \right)\:\neq\mathrm{0}\:\:{so}\:\Sigma\:{u}_{{n}} \left({x}\right){diverges} \\ $$$${if}\:{x}>\mathrm{0}\:\:{u}_{{n}} \left({x}\right)\sim\:{e}^{−{nx}} \:\:\:\left({n}\rightarrow+\infty\right)\:\:{but}\:\Sigma\:{e}^{−{nx}} \:{converges}\:{so}\:{the}\:{convergence} \\ $$$${simple}\:{of}\:\Sigma\:{u}_{{n}} \left({x}\right)\:{is}\:{assured}.{also} \\ $$$$\Sigma\:{f}_{{n}} {of}\:\:=\:\sum_{{n}} \:{e}^{−{n}\:{arctan}\left(\mathrm{1}+{x}\right)} \:\:\:=\sum_{{n}} \:\:\:\:\left({e}^{−{arctan}\left(\mathrm{1}+{x}\right)} \right)^{{n}} \:\Rightarrow \\ $$$${if}\:{arctan}\left(\mathrm{1}+{x}\right)>\mathrm{0}\:\:{the}\:{serie}\:{converges} \\ $$$${if}\:{arctan}\left(\mathrm{1}+{x}\right)<\mathrm{0}\:{the}\:{serie}\:{diverges}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com