How do you get x_n>|x|/2xn>|x|2 from 0<|x_n - x|<|x|/20<|xnx|<|x|2 ?

1 Answer
Oct 21, 2017

See below.

Explanation:

From 0<|x_n - x|<|x|/20<|xnx|<|x|2 we have equivalently

epsilon_1^2=sqrt((x_n-x)^2)ε21=(xnx)2
sqrt((x_n-x)^2)+epsilon_2^2 = sqrt(((x)/2)^2)(xnx)2+ε22=(x2)2

with {epsilon_1,epsilon_2} ne 0{ε1,ε2}0

then

(sqrt((x_n-x)^2)+epsilon_2^2)^2 = ((x)/2)^2((xnx)2+ε22)2=(x2)2 or

(x_n-x)^2+2epsilon_2^2 sqrt((x_n-x)^2)+epsilon_2^4=x^2/4(xnx)2+2ε22(xnx)2+ε42=x24

and again

(2epsilon_2^2 sqrt((x_n-x)^2))^2=(x^2/4-(x_n-x)^2-epsilon_2^4)^2(2ε22(xnx)2)2=(x24(xnx)2ε42)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)=0116(2ε22+x2xn)(2ε22+3x2xn)(2ε223x+2xn)(2ε22x+2xn)=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.