Question Number 207085 by necx122 last updated on 06/May/24
$${Let}\:{f}\:{be}\:{a}\:{function}\:{with}\:{the}\:{following} \\ $$$${properties}:\:\left({i}\right)\:{f}\left(\mathrm{1}\right)\:=\mathrm{1}\:\left({ii}\right)\:{f}\left(\mathrm{2}{n}\right)={n}.{f}\left({n}\right)\:{for} \\ $$$${any}\:{positive}\:{integer}\:{n}.\:{Find}\:{the}\:{value} \\ $$$${of}\:{f}\left(\mathrm{2}^{\mathrm{10}} \right) \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:{b}\right)\:\mathrm{2}^{\mathrm{10}\:} {c}\right)\:\mathrm{2}^{\mathrm{35}} \:{d}\right)\:\mathrm{2}^{\mathrm{45}} \\ $$
Answered by A5T last updated on 06/May/24
![f(2n)=nf(n) f(n)=(n/2)f((n/2))=(n/2)×(n/4)f((n/4)) =(n^q /2^(1+2+3+...+q) )f((n/2^q )) [by induction] ⇒f(2^(10) )=(((2^(10) )^(10) )/2^(1+2+3+...+10) )f((2^(10) /2^(10) ))=(2^(100) /2^(55) )=2^(45)](https://www.tinkutara.com/question/Q207086.png)
$${f}\left(\mathrm{2}{n}\right)={nf}\left({n}\right) \\ $$$${f}\left({n}\right)=\frac{{n}}{\mathrm{2}}{f}\left(\frac{{n}}{\mathrm{2}}\right)=\frac{{n}}{\mathrm{2}}×\frac{{n}}{\mathrm{4}}{f}\left(\frac{{n}}{\mathrm{4}}\right) \\ $$$$=\frac{{n}^{{q}} }{\mathrm{2}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{q}} }{f}\left(\frac{{n}}{\mathrm{2}^{{q}} }\right)\:\:\left[{by}\:{induction}\right] \\ $$$$\Rightarrow{f}\left(\mathrm{2}^{\mathrm{10}} \right)=\frac{\left(\mathrm{2}^{\mathrm{10}} \right)^{\mathrm{10}} }{\mathrm{2}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{10}} }{f}\left(\frac{\mathrm{2}^{\mathrm{10}} }{\mathrm{2}^{\mathrm{10}} }\right)=\frac{\mathrm{2}^{\mathrm{100}} }{\mathrm{2}^{\mathrm{55}} }=\mathrm{2}^{\mathrm{45}} \\ $$
Commented by A5T last updated on 06/May/24
![f(2^(10) )=f(2×2^9 )=2^9 ×f(2^9 )=2^9 ×[2^8 ×f(2^8 )] =2^(9+8) ×2^7 f(2^7 )=2^(9+8+7+...+3) ×2^2 f(2^2 ) =2^(9+8+7+...+2) ×2f(2)=2^(45)](https://www.tinkutara.com/question/Q207090.png)
$${f}\left(\mathrm{2}^{\mathrm{10}} \right)={f}\left(\mathrm{2}×\mathrm{2}^{\mathrm{9}} \right)=\mathrm{2}^{\mathrm{9}} ×{f}\left(\mathrm{2}^{\mathrm{9}} \right)=\mathrm{2}^{\mathrm{9}} ×\left[\mathrm{2}^{\mathrm{8}} ×{f}\left(\mathrm{2}^{\mathrm{8}} \right)\right] \\ $$$$=\mathrm{2}^{\mathrm{9}+\mathrm{8}} ×\mathrm{2}^{\mathrm{7}} {f}\left(\mathrm{2}^{\mathrm{7}} \right)=\mathrm{2}^{\mathrm{9}+\mathrm{8}+\mathrm{7}+…+\mathrm{3}} ×\mathrm{2}^{\mathrm{2}} {f}\left(\mathrm{2}^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}^{\mathrm{9}+\mathrm{8}+\mathrm{7}+…+\mathrm{2}} ×\mathrm{2}{f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{45}} \\ $$
Commented by necx122 last updated on 06/May/24
$${Thank}\:{you}\:{sir}.\:{I}\:{know}\:{you}\:{tried}\:{your} \\ $$$${best}\:{to}\:{expain}\:{yet},\:{I}\:{still}\:{dont}\:{understand} \\ $$
Commented by necx122 last updated on 06/May/24
$${This}\:{now},\:{is}\:{very}\:{clear}.\:{Thank}\:{you}\:{so} \\ $$$${much}\:{great}\:{teacher}. \\ $$
Answered by A5T last updated on 06/May/24
$${f}\left(\mathrm{2}^{\mathrm{10}} \right)=\mathrm{2}^{\mathrm{9}} {f}\left(\mathrm{2}^{\mathrm{9}} \right)=\mathrm{2}^{\mathrm{9}+\mathrm{8}+\mathrm{7}+\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}} {f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{45}} \\ $$