How do you get #x_n>|x|/2# from #0<|x_n - x|<|x|/2# ?

1 Answer
Oct 21, 2017

See below.

Explanation:

From #0<|x_n - x|<|x|/2# we have equivalently

#epsilon_1^2=sqrt((x_n-x)^2)#
#sqrt((x_n-x)^2)+epsilon_2^2 = sqrt(((x)/2)^2)#

with #{epsilon_1,epsilon_2} ne 0#

then

#(sqrt((x_n-x)^2)+epsilon_2^2)^2 = ((x)/2)^2# or

#(x_n-x)^2+2epsilon_2^2 sqrt((x_n-x)^2)+epsilon_2^4=x^2/4#

and again

#(2epsilon_2^2 sqrt((x_n-x)^2))^2=(x^2/4-(x_n-x)^2-epsilon_2^4)^2#

or factoring

#-1/16 (2 epsilon_2^2 + x - 2 x_n) (2 epsilon_2^2 + 3 x - 2 x_n) (2 epsilon_2^2 - 3 x + 2 x_n) (2 epsilon_2^2 - x + 2 x_n)=0#

or

#{(2 epsilon_2^2 + x - 2 x_n=0), (2 epsilon_2^2 + 3 x - 2 x_n=0), (2 epsilon_2^2 - 3 x + 2 x_n=0), (2 epsilon_2^2 - x + 2 x_n=0):}#

or equivalently

#{(x/2 - x_n < 0), (3/2 x - x_n < 0), ( - 3/2 x + x_n < 0), (- x/2 + x_n < 0):}#

or equivalently

#x/2< x_n# and #x_n < x/2#

#3/2x < x_n# and #x_n < 3/2 x#

Those results are contradictory.