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 taylor polynomial for arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 +...
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.
For any positive x the error in the seventh orther Taylor polynomial is positive (that is ), and for negative x the error is -ve (look at the Lagrange form of the remainder, this may be more convieniently done with the 8th order polynomial). The observation need only be: for any positive x the error in the seventh orther Taylor polynomial is positive, since the other follows from the polynomial and the function both being odd functions.
CB
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)?