About this experiment
A calculator cannot know sin 2 directly, but it can add and multiply. A Taylor polynomial copies a function using only those operations: it matches the value, the slope, the curvature and every higher derivative at one point. Add terms one at a time and watch the polynomial hug the curve over a wider and wider stretch, see where it refuses to converge however many terms you add, and compare the actual error with the Lagrange bound from calculus.
Learning goals
- Build the Taylor polynomial of a function from its derivatives at a point
- Recognise the Maclaurin series of sin x, cos x, eˣ, ln(1 + x), 1/(1 − x) and √(1 + x)
- Explain why the approximation is best near the centre and gets worse further away
- Find the interval of convergence and see that outside it more terms make things worse
- Use the Lagrange remainder to bound the error of an approximation
How to use it
- 1Choose sin x with centre 0. Add terms one at a time. Why does every second step leave the graph unchanged?
- 2With degree 3, move the check point from 0 to 3. How does the error grow? How many terms do you need for an error below 0.001 at x = 2?
- 3Compare the actual error with the Lagrange bound for several degrees. Is the bound ever smaller than the error?
- 4Choose ln(1 + x). Set the check point at x = 1.5 and add terms. What happens, and how does the green band explain it?
- 5Choose 1/(1 − x) and move the centre to −2. How does the interval of convergence change? Where is the nearest problem point?
- 6Choose eˣ and centre 1. Write down the first three terms from the formula and check them against the screen.
The model
Assumptions
- Coefficients are computed from exact formulas for the derivatives, not from numerical differentiation, up to degree 20.
- The Lagrange bound readout uses M = 1, which is exact for sin x and cos x. For the other functions M depends on the interval, so no bound is shown.
- The radius of convergence about a is the distance from a to the nearest point where the function or its extension to complex numbers breaks down, such as x = −1 for ln(1 + x).
- With centre 0 the formula is shown with exact fractions, otherwise with coefficients rounded to four significant figures.
Try this in class
- Q1Write the Maclaurin polynomial of degree 4 for cos x and use it to estimate cos 0.5.
- Q2How many terms of the series for eˣ are needed to find e to four decimal places?
- Q3Use the Lagrange bound to show that x − x³/6 approximates sin x to within 0.0003 for |x| ≤ 0.5.
- Q4Why does the series for 1/(1 − x) about 0 fail at x = 2 even though the function itself is defined there?
- Q5Find the Taylor polynomial of degree 2 for √(1 + x) about 0 and use it to estimate √1.1.
Frequently asked questions
Do calculators really use Taylor series?
They use the same idea, refined. Most reduce the angle into a small range first, where a short polynomial is very accurate, and many use CORDIC or polynomials tuned to spread the error evenly instead of plain Taylor terms.
Why do some series converge everywhere and others only nearby?
The radius of convergence is set by the nearest point, possibly a complex one, where the function blows up or is undefined. sin x and eˣ have no such point, so their series work everywhere. ln(1 + x) breaks at x = −1, so its series about 0 only works within a distance of 1.
Is the small angle approximation sin x ≈ x a Taylor polynomial?
Yes, it is the degree 1 (and degree 2) Taylor polynomial of sin x about 0. The next term, −x³/6, shows the error: about 0.2 % at 0.1 rad and about 4 % at 0.5 rad.
