Degrees of taylor polynomials

[ptolema]

Homework Statement

use the third degree Taylor polynomial of cos at 0 to show that the solutions of x²=cos x are approx. $$\pm$$$$\sqrt{2/3}$$, and find bounds on the error.

Homework Equations

P_2n,0(x) = 1-x²/2!+x⁴/4!+...+(-1)ⁿx²ⁿ/(2n)!

The Attempt at a Solution

when it says "third degree" for cos at 0, does it mean that n=3, so P_6,0 is what is needed? or does it mean that 2n=3, so i should use P_3,0?
because P_6,0(x) = 1-x²/2!+x⁴/4!-x⁶/6! is very different from P_3,0(x) = 1-x²/2!
i'm also not so hot on the finding the error, but the degree thing is most of the problem

 

[Mark44]

Homework Statement

use the third degree Taylor polynomial of cos at 0 to show that the solutions of x²=cos x are approx. $$\pm$$$$\sqrt{2/3}$$, and find bounds on the error.

Homework Equations

P_2n,0(x) = 1-x²/2!+x⁴/4!+...+(-1)ⁿx²ⁿ/(2n)!

The Attempt at a Solution

when it says "third degree" for cos at 0, does it mean that n=3, so P_6,0 is what is needed? or does it mean that 2n=3, so i should use P_3,0?
because P_6,0(x) = 1-x²/2!+x⁴/4!-x⁶/6! is very different from P_3,0(x) = 1-x²/2!
i'm also not so hot on the finding the error, but the degree thing is most of the problem

By third degree, it means the polynomial of degree 3, so you can ignore all terms of higher degree. Since there is no term of degree 3 in the Maclaurin series for cos(x), your approximation is 1 - x²/2!