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Question Number 33719 by prof Abdo imad last updated on 22/Apr/18
simplify S_n (x) =(1+x^2 )(1+x^4 )....(1+x^2^n ) 2) find lim_(n→+∞) S_n (x) if ∣x∣<1 .
$${simplify}\:{S}_{{n}} \left({x}\right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)….\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \left({x}\right)\:{if}\:\mid{x}\mid<\mathrm{1}\:. \\ $$
Commented by prof Abdo imad last updated on 23/Apr/18
we have (1+x^2 )(1+x^4 ) =1+x^4 +x^2 +x^6 =1+x^2 +x^4 +x^6 =((1−x^8 )/(1−x^2 )) if x^2 ≠1 let suppose that S_n (x) =((1−x^2^(n+1) )/(1−x^2 )) S_(n+1) (x)=(1+x^2 )(1+x^4 ).....(1+x^2^(n+1) ) = (1+x^2^(n+1) ) S_n (x)=(((1+x^2^(n+1) )(1−x^2^(n+1) ))/(1−x^2 )) =((1−(x^2^(n+1) )^2 )/(1−x^2 )) = ((1−x^2^(n+2) )/(1−x^2 )) the result is true at term n+1 and we must study tbe case x=+^− 1 2) we have proved that S_n (x) =((1 −x^2^(n+1) )/(1−x^2 )) so for ∣x∣<1 lim_(n→+∞) S_n (x)=(1/(1−x^2 ))
$${we}\:{have}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:=\mathrm{1}+{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+{x}^{\mathrm{6}} \\ $$$$=\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+{x}^{\mathrm{6}} \:=\frac{\mathrm{1}−{x}^{\mathrm{8}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\:{if}\:{x}^{\mathrm{2}} \neq\mathrm{1} \\ $$$${let}\:{suppose}\:{that}\:{S}_{{n}} \left({x}\right)\:=\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${S}_{{n}+\mathrm{1}} \left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…..\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right) \\ $$$$=\:\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)\:{S}_{{n}} \left({x}\right)=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)\left(\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left({x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\:=\:\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{2}} } }{\mathrm{1}−{x}^{\mathrm{2}} }\:{the}\:{result}\:{is}\:{true}\:{at}\:{term} \\ $$$${n}+\mathrm{1}\:\:{and}\:{we}\:{must}\:{study}\:{tbe}\:{case}\:{x}=\overset{−} {+}\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\: \\ $$$${S}_{{n}} \left({x}\right)\:=\frac{\mathrm{1}\:−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}} }\:\:{so}\:{for}\:\mid{x}\mid<\mathrm{1}\:{lim}_{{n}\rightarrow+\infty} {S}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18
(1+x^2 )(1+x^4 )=1+x^2 +x^4 +x^6 (1+x^2 )(1+x^4 )(1+x^8 )=(1+x^2 +x^4 +x^6 )(1+x^8 ) =1+x^2 +x^4 +x^6 +x^8 +x^(10) +x^(12) +x^(14) S_n (x)=(1−x^(2n) )/(1−x^2 ) so the value of S_n (x) is 1/(1−x^2 ) when limit n tends to infinity i_(x→0)
$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)=\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{8}} \right)=\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} \right)\left(\mathrm{1}+{x}^{\mathrm{8}} \right) \\ $$$$=\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} +{x}^{\mathrm{8}} +{x}^{\mathrm{10}} +{x}^{\mathrm{12}} +{x}^{\mathrm{14}} \\ $$$$ \\ $$$${S}_{{n}} \left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{2}{n}} \right)/\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$${so}\:{the}\:{value}\:{of}\:{S}_{{n}} \left({x}\right)\:{is}\:\mathrm{1}/\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:{when}\:{limit}\:{n} \\ $$$${tends}\:{to}\:{infinity} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{i}} \\ $$
Commented by prof Abdo imad last updated on 22/Apr/18
its not correct because ((1−x^(2n) )/(1−x^2 )) =((1−(x^2 )^n )/(1−x^2 )) =(((1−x^2 )(1+x^2 +x^4 +...+x^(2n−2) ))/(1−x^2 )) =1+x^2 +x^4 +...+x^(2n−2) ≠ S_n
$${its}\:{not}\:{correct}\:{because}\:\frac{\mathrm{1}−{x}^{\mathrm{2}{n}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{{n}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:=\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+…+{x}^{\mathrm{2}{n}−\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+…+{x}^{\mathrm{2}{n}−\mathrm{2}} \:\:\neq\:{S}_{{n}} \\ $$

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