# question on taylor polynomials

• Jul 19th 2009, 11:44 PM
oblixps
question on taylor polynomials
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.
• Jul 20th 2009, 03:32 AM
HallsofIvy
Quote:

Originally Posted by oblixps
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.

What is the Taylor's polynomial for arctan(x)? In particular is the term involving $x^8$, the first term not in the 7 th order polynomial, have positive or negative coefficient?
• Jul 21st 2009, 12:08 AM
oblixps
Quote:

Originally Posted by HallsofIvy
What is the Taylor's polynomial for arctan(x)? In particular is the term involving $x^8$, the first term not in the 7 th order polynomial, have positive or negative coefficient?

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.
• Jul 21st 2009, 01:09 AM
CaptainBlack
Quote:

Originally Posted by oblixps
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 $f(x)>P_7(x)$), 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
• Jul 23rd 2009, 07:11 PM
oblixps
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)?
• Jul 23rd 2009, 11:48 PM
CaptainBlack
Quote:

Originally Posted by oblixps
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)?

You have to subtract a positive quantity (add a negative) from $p_7(x)$ to get $f(x)$; hence f(x) is less than $p_7(x)$.

CB