Estimation of Taylor Series

Suppose

$$f(x) = \sum_{n=0}^{\oo } \frac{f^{(n)}(a)}{n!}(x-a)^n.$$

Write

$$R_N(x) := f(x) - \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!} (x-a)^n$$

We call

$$T_N(x) = \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!} (x-a)^n$$

the $N$th degree Taylor polynomial. Notice that

$$\lim_{N\to\infty} T_N(x) = f(x)$$

if and only if

$$\lim_{N\to\infty} R_N(x) = 0.$$

We would like to estimate $f(x)$ with $T_N(x)$. We need an estimate for $R_N(x)$.

Theorem 6.7.1 (Taylor's theorem)   If $\vert f^{(N+1)}(x)\vert \leq M$ for $\vert x-a\vert\leq d$, then

$$\vert R_N(x)\vert \leq \frac{M}{(N+1)!} \vert x-a\vert^{N+1}$$   for $|x-a|&le#leq;d$.

For example, if $N=0$, this says that

$$\vert R(x)\vert = \vert f(x)-f(a)\vert \leq M \vert x-a\vert,$$

i.e.,

$$\left\vert\frac{f(x) - f(a)}{x-a}\right\vert \leq M,$$

which should look familiar from a previous class (Mean Value Theorem).

Applications:

1. One can use Theorem  to prove that functions converge to their Taylor series.

2. Returning to the relativity example above, we apply Taylor's theorem with $N=1$ and $a=0$. With $x=-v^2/c^2$ and $M$ any number such that $\vert f''(x)\vert\leq M$, we have
   $$\vert R_1(x)\vert \leq \frac{M}{2}x^2.$$
   For example, if we assume that $\vert v\vert\leq 100m/s$ we use
   $$\vert f''(x)\vert \leq \frac{3}{2}(1-100^2/c^2)^{-5/2} = M.$$

   Using $c=3\times 10^8 m/s$, we get

   $$\vert R_1(x)\vert \leq 4.17 \cdot 10^{-10} \cdot m_0.$$
   Thus for $v\leq 100m/s \sim 225$   mph, then the error in throwing away relativistic factors is $10^{-10} m_0$. This is like 200 feet out of the distance to the sun (93 million miles). So relativistic and Newtonian kinetic energies are almost the same for reasonable speeds.