My calculus book states that for $f(x)=\sin(x)$ the Maclaurin polynomial is:
$$P_{2n-1}(x)=P_{2n}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}$$
I was just wondering: how do I get $P_0$ from this?
It seems I would get $P_0=(-1)^{-1}\frac{x^{-1}}{-1!}$ (taking $n=0$). But this doesn't make sense, does it?
In other words, why is the general formula for the Maclaurin polynomial not valid for $P_0$? Why does $n$ have to be at least $1$? And how do I get $P_0$ in such case from the general formula?
PS, of course I know that $P_0=0$, since $\sin(0)=0$
Because even exponents do not appear anyway, it makes no difference if we truncate the Mac-Laurin-Polynomial at exponent $2n-1$ or at exponent $2n$.
If you take $n=0$, you have $P_0=0$, because the constant term is $0$.
This actually coincides with $P_{-1}$ because we have the empty sum which is $0$ as well.
Another thing is $P_{2n-1}(0)=P_{2n}(0)$. If $x=0$ , we have the zero-polynomial, so $P_{2n-1}(0)=P_{2n}(0)=0$ for all $n$.