I am having some trouble with this problem. So far, I am thinking to use a similar approach to finding the approximation of sin(x) per guidance of my textbook.
This would get me something along the lines of: $\sum_{k=1}^\infty = \frac{x^{(2^k-1)}(-1)^{k+1}}{(2k-1)!}$
and then $$x- \frac {x^3}6 + .. $$
How would I proceed?
Respectfully
On
write the remainder term in the Taylor series set it less than $10^{-4}$ you will get a value of n. now write $T_n$ and put the value $x=2$
$$f(x) = \sum_{k=0}^n \dfrac{f^{(k)}(c)}{k!} (x - c)^k + \dfrac{f^{(n+1)}(z)}{(n + 1)!} (x - c)^{n+1},$$
remainder term $$R_n(x; c) = \dfrac{f^{(n+1)}(z)}{(n + 1)!} (x - c)^{n+1}.$$
$z$ is between $x$ and $c$. hope it helps
The terms
$$2,-\frac86,\frac{32}{120},-\frac{128}{5040},\frac{512}{362880},-\frac{2048}{39916800}$$
will suffice, as the last one is below $10^{-4}$ (and the series alternates).
Hence
$$\frac{141782}{155925}\approx0.9093$$
Note that you will get a better convergence by evaluating
$$\cos\left(2-\frac\pi2\right)$$ with the Taylor series
$$1-\frac{x^2}2+\frac{x^4}{4!}-\cdots$$
because the argument is smaller.