13.Let ? be a fixed real number ?0
a) prove that for any positive integer n,sin?= 2^n sin ?/2^n ? from k=1 to n?cos ?/2^k ?
The solution I came up with is the following:
sin?=2sin ?/2 cos ?/2,so by induction,for n=1
sin?=2^n sin ?/2^n ? from k=1 to n?cos ?/2^k ?=2sin ?/2 cos ?/2=sin?
assume for n:
sin?= 2^n sin ?/2^n ? from k=1 to n (cos ?/2^k),then prove true for n+1:
sin?= 2^(n+1) sin ?/2^(n+1) ?from k=1 to n+1?cos ?/2^k ?=2(2^n sin ?/2^(n+1) cos ?/2 cos ?/4 cos ?/8…cos ?/2^n cos ?/2^(n+1) )=
2^n sin ?/2^n cos ?/2 cos ?/4 cos ?/8…cos ?/2^n (because 2sin ?/2^(n+1) cos ?/2^(n+1) =sin ?/2^n
after doing this simplification n times,we are left with 2 sin ?/2 cos ?/2=sin?
so by induction, sin?= 2^n sin ?/2^n ? from k =1 to n?cos ?/2^k ? is true for all positive integers n
b) deduce that sin?/?=? k = 1 to ??cos ?/2^k ?= lim?(n??)? k=1 to n?cos ?/2^k ?
from part a,we know that sin?= 2^n sin ?/2^n ? k=1 to n?cos ?/2^k ?,so (sin?)/?=(sin ?/2^n )/(?/2^n ) ? k=1 to n?cos ?/2^k ?
since this is true for any n,it is true as n approaches infinity,and lim?(n??) (sin ?/2^n )/(?/2^n )=lim?(n??) (sin(n))/n
=1
so (sin?)/?=(sin ?/2^n )/(?/2^n ) ? k=1 to n?cos ?/2^k ?=lim?(n??) ? k=1 to n?cos ?/2^k ?= ? k=1 to ??cos ?/2^k ?
these I have checked and have gotten the correct answers for. The following, however, I am stuck on. Could anyone show me a solution for the rest?
c) deduce the value of ? k=2 to ? ?cos ?/2^k ?.
d) prove that cos ?/2^k =1/2 ?(2+?(2+?(2+??2 )) ) (how many?),for integers k?2
e) Deduce another expression for the value obtained in c)
f) Assume that-?/2 ?? ? ?/2, ??0. Let a_1=cos?, b_1=1, a_(n+1)=(a_n+b_n )/2, and
b_(n+1)=?(b_n * a_(n+1) ) for n?1. Find lim?(n??)?a_n ? and lim?(n??)?b_n ?
a) prove that for any positive integer n,sin?= 2^n sin ?/2^n ? from k=1 to n?cos ?/2^k ?
The solution I came up with is the following:
sin?=2sin ?/2 cos ?/2,so by induction,for n=1
sin?=2^n sin ?/2^n ? from k=1 to n?cos ?/2^k ?=2sin ?/2 cos ?/2=sin?
assume for n:
sin?= 2^n sin ?/2^n ? from k=1 to n (cos ?/2^k),then prove true for n+1:
sin?= 2^(n+1) sin ?/2^(n+1) ?from k=1 to n+1?cos ?/2^k ?=2(2^n sin ?/2^(n+1) cos ?/2 cos ?/4 cos ?/8…cos ?/2^n cos ?/2^(n+1) )=
2^n sin ?/2^n cos ?/2 cos ?/4 cos ?/8…cos ?/2^n (because 2sin ?/2^(n+1) cos ?/2^(n+1) =sin ?/2^n
after doing this simplification n times,we are left with 2 sin ?/2 cos ?/2=sin?
so by induction, sin?= 2^n sin ?/2^n ? from k =1 to n?cos ?/2^k ? is true for all positive integers n
b) deduce that sin?/?=? k = 1 to ??cos ?/2^k ?= lim?(n??)? k=1 to n?cos ?/2^k ?
from part a,we know that sin?= 2^n sin ?/2^n ? k=1 to n?cos ?/2^k ?,so (sin?)/?=(sin ?/2^n )/(?/2^n ) ? k=1 to n?cos ?/2^k ?
since this is true for any n,it is true as n approaches infinity,and lim?(n??) (sin ?/2^n )/(?/2^n )=lim?(n??) (sin(n))/n
=1
so (sin?)/?=(sin ?/2^n )/(?/2^n ) ? k=1 to n?cos ?/2^k ?=lim?(n??) ? k=1 to n?cos ?/2^k ?= ? k=1 to ??cos ?/2^k ?
these I have checked and have gotten the correct answers for. The following, however, I am stuck on. Could anyone show me a solution for the rest?
c) deduce the value of ? k=2 to ? ?cos ?/2^k ?.
d) prove that cos ?/2^k =1/2 ?(2+?(2+?(2+??2 )) ) (how many?),for integers k?2
e) Deduce another expression for the value obtained in c)
f) Assume that-?/2 ?? ? ?/2, ??0. Let a_1=cos?, b_1=1, a_(n+1)=(a_n+b_n )/2, and
b_(n+1)=?(b_n * a_(n+1) ) for n?1. Find lim?(n??)?a_n ? and lim?(n??)?b_n ?