How can I approximate the root of $\cos x = x$ with a Taylor series?

Question

I want an expression to approximate the dottie number. The reason I want to use a Taylor series rather than, say, Newton's Method is because I want to guarantee that the approximation is within a certain percentage of error which is something that is easy to do with Taylor polynomials.

The problem is that the only way I seem to be able to do this is by setting the Taylor series for $\cos x$ equal to $x$. This leaves me with the following expression:

$$-x+\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}=0$$ Expanded this looks like: $$1-x-\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots=0$$

I can solve for $x$ at $n=1,2$ but of course when $n>2$ the degree of the taylor polynomial is higher than a quartic and I can't solve the equation in general (Abel's Impossibility Theorem). To approximate those roots I would have to use something like Newton's method and we are back at square one.

Is it even possible to approximate the Dottie number to an arbitrary precision? Am I missing something?

Answer 1

You can find a zero of $f(x)=\cos(x)-x$ to arbitrary precision, and have an upper bound of your error estimate, using the bisection method.

We know $f(0)=1>0$, $f(1)<0$, so we evaluate $f(1/2)$. We find $f(1/2)>0$, so we change our interval of interest from $[0,1]$ to $[1/2,1]$ and continue by evaluating $f(3/4)$. It turns out $f(3/4)<0$, so we are now in $[1/2,3/4]$.

Answer 2

First of all if you plan to use Taylor series you need to modify your equation to $$\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}= 1-\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots=x$$ Which is not easy to solve for x.

You may try other methods such as bisection method.