Taylor Polynomial Error

I was told to find the "bounds on the error" in a problem in Spivak's Calculus, but I have no idea what that means.

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

I recognize that this reduces to simply $1 - x^2/2! + R(x) = x^2$ and that $R(x)$ is the error. So, um, if we don't have the $R(x)$, since it's probably small, we get the needed approximate solution. Am I on the right track? What do I need to do to make this formal?

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Go back a couple of pages. Spivak gives you an expression for R(x). You can use that and bound it, given it your approximate solution. – Simon S Jan 18 '12 at 3:36
@SimonS I think I am having a hard time understanding the concept of "bounding" something... but R(x) =< x^4/(4!). I am not sure what I am supposed to do with this though... – Don Jan 18 '12 at 4:09

The error in the Taylor expansion is $\dfrac{x^4}{4!}\cos(\theta_x)$ where $\theta_x$ is between $0$ and $\pi/2$, and therefore $$1-\frac{x^2}{2}< \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}.$$

The solution near $\sqrt{2/3}$ of the equation $$\frac{x^4}{24}-\frac{3x^2}{2}+1=0$$ will then be larger than the true (positive) solution of $\cos x=x^2$.

The above equation is quadratic in $x^2$. The relevant solution turns out to be $\sqrt{18-10\sqrt{3}}$.

Calculate. We have $\sqrt{2/3}\approx 0.8164966$, and $\sqrt{18-10\sqrt{3}}\approx 0.824313$. We conclude that the positive solution of $\cos x=x^2$ is larger than $\sqrt{2/3}$, but by less than $0.0078165$.

By the way, Wolfram Alpha gives the approximation $0.8241323$ for the positive solution of $\cos x=x^2$. It is not a surprise that this is quite a bit closer to our upper bound than it is to $\sqrt{2/3}$.

By symmetry, $-\sqrt{2/3}$ is larger than the negative solution of $\cos x=x^2$, but by less than $0.0078165$.

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That $\sqrt{3/2}$ is a typo for $\sqrt{2/3}$? – Gerry Myerson Jan 18 '12 at 8:19
@Gerry Myerson: Thanks, yes it was. – André Nicolas Jan 18 '12 at 13:59

You're being told that one solution is $x=\sqrt{2/3}+\epsilon$, and you're being asked to find some positive real number $r$ for which you can guarantee $|\epsilon|\lt r$. You should be able to express $\epsilon$ in terms of $R(x)$ and use that expression to find $r$.

Oh, then you still have to do the $-\sqrt{2/3}$ thing, but if you can do the one, I'm sure you can do the other.

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