테일러 다항식과 오차

A Taylor polynomial gives a polynomial approximation for a function $s(t)$. Since $s(t)$ is itself a polynomial, any $n$th order Taylor polynomial, with $n$ greater than or equal to the degree of $s(t)$, will simply give $s(t)$. So, in our case, $T_{100}(t)=s(t)=4.9t^2-t+10$.

If that isn't satisfying, you can go through the normal method of calculating $T_{100}(t)$.

$$\begin{aligned} s(t)&=4.9t^2-t+10 & s(5)&=4.9(25)-5+10=127.5\\ s'(t)&=9.8t-1 & s'(5)&=9.8(5)-1=48\\ s''(t)&=9.8 & s''(5)&=9.8 \end{aligned}$$

The rest of the derivatives of $s(t)$ are identically zero, so they are (in particular) zero when $t=5$. Therefore,

$$\begin{aligned} T_{100}(t)&=127.5+48(t-5)+\frac{1}{2}9.8(t-5)^2\\ &=127.5+48(t-5)+4.9(t-5)^2 \end{aligned}$$

We can now check that $T_{100}(t)$ really is the same as $s(t)$.

$$\begin{aligned} T_{100}(t)&=127.5+48(t-5)+4.9(t-5)^2\\ &=127.5+48(t-5)+4.9(t^2-10t+25)\\ &=[127.5+48(-5)+4.9(25)]+[48-4.9(10)]t+4.9t^2\\ &=10-t+4.9t^2=s(t) \end{aligned}$$

as expected.