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Question Number 207085 by necx122 last updated on 06/May/24

Let f be a function with the following  properties: (i) f(1) =1 (ii) f(2n)=n.f(n) for  any positive integer n. Find the value  of f(2^(10) )  a)1 b) 2^(10 ) c) 2^(35)  d) 2^(45)

Letfbeafunctionwiththefollowingproperties:(i)f(1)=1(ii)f(2n)=n.f(n)foranypositiveintegern.Findthevalueoff(210)a)1b)210c)235d)245

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)

f(2n)=nf(n)f(n)=n2f(n2)=n2×n4f(n4)=nq21+2+3+...+qf(n2q)[byinduction]f(210)=(210)1021+2+3+...+10f(210210)=2100255=245

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)

f(210)=f(2×29)=29×f(29)=29×[28×f(28)]=29+8×27f(27)=29+8+7+...+3×22f(22)=29+8+7+...+2×2f(2)=245

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

Thankyousir.Iknowyoutriedyourbesttoexpainyet,Istilldontunderstand

Commented by necx122 last updated on 06/May/24

This now, is very clear. Thank you so  much great teacher.

Thisnow,isveryclear.Thankyousomuchgreatteacher.

Answered by A5T last updated on 06/May/24

f(2^(10) )=2^9 f(2^9 )=2^(9+8+7+6+5+4+3+2+1) f(2)=2^(45)

f(210)=29f(29)=29+8+7+6+5+4+3+2+1f(2)=245

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