Taylor polynomials

What is it about?

The Taylor polynomial $T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$ shares the value and the first $n$ derivatives with $f$ at $a$ : $T_0$ is the horizontal line $y = f(a)$ , $T_1$ the tangent, $T_2$ the osculating parabola. Differentiating $(x-a)^k$ $k$ times gives $k!$ – that is why $c_k = \frac{f^{(k)}(a)}{k!}$ .

For sin, cos and $e^x$ the Lagrange bound $|R_n(x)| \le M \frac{|x-a|^{n+1}}{(n+1)!}$ estimates the error. For $\ln(1+x)$ and $\frac{1}{1-x}$ the polynomials only converge within the interval of convergence $|x - a| < r$ ; outside they blow up and the error grows with $n$ .

Try it

sin x about 0: press “Increase degree” several times. Why does every second step change nothing?
Matching derivatives: which ticks are green? Tap the rows to compare $f''$ and $T_2''$ .
Radius of convergence: run through all degrees – what happens to the right of $x = 1$ ?
Calculator: how many digits of $e$ are correct after each term?
Pendulum: up to which angle is $\sin\varphi \approx \varphi$ good enough?

Tasks

Task 1: Computing e

Load task

Find $T_4(1)$ for $e^x$ about $0$ , the error and the Lagrange bound. How many terms give 6 correct decimals?

Show solution

$1 + 1 + \frac12 + \frac16 + \frac1{24} = \frac{65}{24} \approx 2.708333$ , error $\approx 0.009948$ ; bound $\frac{e}{5!} \approx 0.022652$ . Six decimals need $|R_n(1)| < 5 \cdot 10^{-7}$ : degree $n = 9$ , i.e. $10$ terms, $T_9(1) \approx 2.718282$ . Show

Task 2: sin 0.5

Load task

Use $T_3(x) = x - \frac{x^3}{6}$ to approximate $\sin 0.5$ and bound the error.

Show solution

$T_3(0.5) \approx 0.479167$ , exact $0.479426$ . Lagrange: $\frac{0.5^4}{4!} \approx 0.002604$ ; since $T_3 = T_4$ even $\frac{0.5^5}{5!} \approx 0.000260$ . Show

Task 3: How far does “sin φ ≈ φ” go?

Load task

For which $\varphi$ is $|\sin\varphi - \varphi| < 0.01$ ?

Show solution

Estimate $\frac{\varphi^3}{6} < 0.01 \Rightarrow \varphi < 0.39$ ; exactly $|\varphi| < 0.392493$ , about $22.5°$ . A relative error below $1\,\%$ only holds up to about $14.0°$ . Show

Task 4: ln 2 from the series

Load task

Compute $T_5(1)$ for $\ln(1+x)$ about $0$ and compare with $\ln 2$ .

Show solution

$1 - \frac12 + \frac13 - \frac14 + \frac15 \approx 0.783333$ , while $\ln 2 \approx 0.693147$ : at the boundary point $x = 1$ the series converges only slowly. For $x = 1.5$ it diverges. Show

Ideas for the classroom

Start with the question “How does a calculator compute $\sin 0.5$ ?”; continue from the tangent to $T_2$ with the matching-derivatives view; link to the pendulum.
Common stumbling blocks: forgetting $k!$ ; writing $x^k$ instead of $(x-a)^k$ ; expecting more terms to help outside the interval of convergence; confusing the Lagrange bound with the true error.