<s>[INST] Find any rational approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}}$ I need to find any ratinal approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}} $ with accuracy $d = \frac{1}{500}$.
I generally know how to do these type of tasks (using Taylor expansion), but I got problem with this one. 
I defined function $f(x) = \begin{cases} 1 \ \mbox{dla} \ x=0 \\ \frac{\sin x}{x} \ \mbox{dla} \ x \neq 0 \end{cases}$, to handle problem with dividng by zero in $f(0)$.
I calculate first and second derivative:
$f'(x) = \frac{x^2 \cdot \cos x - \sin x}{x^2}$
$f''(x) = \frac{x^3 \cdot \sin x - 2x^2 \cdot \cos x + \sin x }{x^4}$
But well, what now? I have no idea how to use the Lagrange remainder theorem here.
 [/INST] $\text{sinc}(x)$ is an entire function fulfilling
$$ \text{sinc}(x) = \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!} \tag{1}$$
hence
$$ \text{sinc}(\sqrt{2})=\sum_{n\geq 0}\frac{(-1)^n 2^n}{(2n+1)!}\tag{2}$$
The terms of the last series are decreasing in absolute value, hence
$$ \left|\,\text{sinc}(\sqrt{2})-\sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!}\right|\leq \frac{2^3}{(2\cdot 3+1)!}=\frac{1}{630}\tag{3} $$
and
$$ \sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!} = \color{red}{\frac{44}{63}}\tag{4} $$
is an approximation fulfilling the given constraints.
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