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