# Math Help - Taylor's Theorem & Approximation (Spivak)

1. ## Taylor's Theorem & Approximation (Spivak)

I am supposed to use Taylor's theorem to approximate the solutions to

$2x^2=x sin(x)+cos(x)$ and give a bound on the error

I computed the 4th degree Taylor polynomial, $T_4$ of the right hand side at x=0. I reasoned that the intersection must occur when $|x|<1$, so I bounded the error between the Taylor polynomial and the function within $\pm \frac{1}{24}$.

Then I computed the intersection between $T_4-\frac{1}{24}$ and $2x^2$ and between $T_4+\frac{1}{24}$ and $2x^2$. So the solution to the original problem must be between these values. So I took the average of the values as the approximation and half the difference as the bounds for the error...

How does this look? I just ask because my answer had a square root within a square root, and this feels like it is as difficult to approximate as the answer to the original question.

Thanks.

2. Originally Posted by billa
I am supposed to use Taylor's theorem to approximate the solutions to

$2x^2=x sin(x)+cos(x)$ and give a bound on the error

I computed the 4th degree Taylor polynomial, $T_4$ of the right hand side at x=0. I reasoned that the intersection must occur when $|x|<1$, so I bounded the error between the Taylor polynomial and the function within $\pm \frac{1}{24}$.

Then I computed the intersection between $T_4-\frac{1}{24}$ and $2x^2$ and between $T_4+\frac{1}{24}$ and $2x^2$. So the solution to the original problem must be between these values. So I took the average of the values as the approximation and half the difference as the bounds for the error...

How does this look? I just ask because my answer had a square root within a square root, and this feels like it is as difficult to approximate as the answer to the original question.

Thanks.
You are looking for a root of:

$f(x)=x \sin(x)+\cos(x)-2x$

Expand $f(x)$ as a Taylor series about $0$ keeping the first non-constant term and an upper bound on the error of the approximation on $[-1,1]$

Solve the resulting equation when this is set to zero. Your approximate root will be the value of this solution when you set the remainder to zero, and you determine the size of the error using the upper bound on the error.

CB