Let f(x) be the Taylor Polynomial P_7(x) of order 7 of arctan(x) about x=0. Then it follows that, if -1.5<x<1.5,
A) f(x) = arctan(x)
B) f(x) </= arctan(x)
C) f(x) >/= arctan(x)
D) f(x) > arctan(x) if x < 0, but f(x) < arctan(x) if x > 0
E) f(x) < arctan(x) if x < 0, but f(x) > arctan(x) if x > 0
my book says the answer is D) but i have no idea how they figured out whether or not f(x) is greater than arctan(x). please help.
the term involving x^8 is 0 but the coefficient of the term involving x^9 is positive. but i'm still unclear about how to know whether or not f(x) is greater than or less than arctan(x) and why D is the correct answer.
the solution in my book reads:
Note that the Taylor seires for arctan(x) satisfies the Alternating Series Test and that f(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 = P_7(x). If x<0, then the first omitted term, (x^9)/9 is negative. Hence P_7(x) exceeds arctan(x).
But that doesn't make sense since if the error of the seventh order polynomial is negative, how can P_7(x) be greater than arctan(x)?