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avatar+72 

If f(x) = (x+1)(x2+1)(x4+1)(x8+1)(x16+1)(x32​+1), then in simplified form f(2) = 2n-1. Determine the value of n.

MapleTheory  Dec 29, 2017
 #1
avatar+89729 
+2

If f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32​+1), then in simplified form f(2) = 2n-1. Determine the value of n.

 

f(2)  = 

 

(2 + 1) (2^2 + 1) (2^4 + 1)(2^8 + 1) (2^16 + 1)(2^32 + 1)  =

 

(3)(5)(17)(257)(65537)(4294967297)  =

 

18446744073709551615  = 2^n  - 1       add 1 to both sides 

 

18446744073709551616  =  2^n        take the log of both sides        

 

log (18446744073709551615)  =  log (2)^n     and we can write

 

log (18446744073709551615)  = n* log (2)    divide both sides by log 2

 

log (18446744073709551615) / log (2)  =  n  =  64 

 

 

cool cool cool

CPhill  Dec 29, 2017
 #2
avatar+502 
+2

I got the same answer on a rough page but looking at the answer I was like 'w*f is this'

Rauhan  Dec 29, 2017
 #3
avatar+72 
+1

Wow thanks!!

MapleTheory  Dec 29, 2017
 #4
avatar+89729 
0

I actually think there is a more efficient way to do this........maybe someone else knows how

 

 

cool cool angel

CPhill  Dec 29, 2017

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