Please show all steps Thanks For approximately what values o

Please show all steps! Thanks! For approximately what values of x can you replace sin x by x - x3/6 with an error of magnitude no greater than 3 times 10-4? |x|

Solution

Taylors theorem with Lagrange-error term for f(x) = sin(x) up to order 5 around a=0 is f(x) = x - x^3/6 + f^(5)(t)*x^5/120 where f^(5) is the 5th derivative of f(x), thus f^(5)(x) = cos(x) and t is some point in the interval (0,x) (if x>0) or (x,0) (if x<0). The magnitude of the error is |f^(5)(t)*x^5/120| = |cos(t)|*|x|^5/120. Now you don\'t know at which point t you will get the exact error, but you know that |cos(t)| <= 1 for all t, so by using that you can know for sure that the magnitude of the error is no greater than |x|^5/120. Now solving for x in |x|^5/120 = 0.0003 gives |x|^5 = 0.036 which is |x| = (0.036)^(1/5) = 0.51... Now to make sure we don\'t exceed the error we have to round down, thus -0.51 < x < 0.51. But a equally good answer would have been -0.51 < x < 0.51 etc.