Question

if y = 8x^2 and you know the value of x with an error no more +-E then the relative error in using this approx. of x to approximate the value of y does not depend on x.

true or false?

thanks!

Answer 1

y(x+E) = 8 (x+E)^2 = 8 (x^2 + 2 x E +E^2)= 8x^2 + 16 x E + 8 E^2)
see that 16 x E term ?? It includes x.