Estimation of Taylor Series


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


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

We call

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

the $ N$th degree Taylor polynomial. Notice that

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

if and only if

$\displaystyle \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

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

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

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


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

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


  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

    $\displaystyle \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

    $\displaystyle \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

    $\displaystyle \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.

William Stein 2006-03-15