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Taylor Polynomials Hug a Curve

A Taylor polynomial matches a function’s value, slope, bend and every higher derivative at one center point. Add terms one at a time and watch the polynomial wrap itself around the curve, then see where it can never follow: outside the interval of convergence.

Calculus II Lessons 5.1 to 5.5 · Calculus Lesson 9.4 · about 20 minutes

Click or tap the graph to choose the x where the error is measured. With the graph focused, the left and right arrow keys move it (hold Shift for small steps), and + and − add or remove a term.

The polynomial

Show each term’s derivative and coefficient
kf(k)(a)ck = f(k)(a)/k!
Function

Center and degree

0
Degree n 1

Interval of convergence

Error at a chosen x

1
f(x)
Pn(x)
Error |f − Pn|
Bound

Building the polynomial

The degree-n Taylor polynomial of f centered at a is

Pn(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)2/2! + … + f(n)(a)(x − a)n/n!

Each coefficient is chosen so that Pn and f have the same value and the same first n derivatives at a. Degree 0 is a flat line at the right height, degree 1 is the tangent line, degree 2 also matches the bending, and so on. When a = 0 it is called a Maclaurin polynomial.

Some terms are zero. Around a = 0, sin x is odd, so all its even-power terms vanish: P3 and P4 are the same polynomial. Press + and watch nothing change on those steps.

The interval of convergence

Letting n → ∞ gives the Taylor series. For sin x, cos x and ex it converges to the function for every x. The other three only converge within a distance R, the radius of convergence, of the center:

  • 1/(1 − x) blows up at x = 1, so R = |1 − a|. Neither endpoint is included.
  • ln(1 + x) breaks down at x = −1, so R = 1 + a. The right endpoint is included (there the series is an alternating harmonic-type series); the left one, at −1, is not.
  • arctan x looks perfectly smooth on the number line, yet R = √(1 + a2). The trouble is hidden off the line: its derivative 1/(1 + x2) blows up at the complex numbers ±i, and R is the distance from a to them. Both endpoints are included.

Outside the interval, adding terms makes things worse, not better. Try 1/(1 − x) at x = −1.5 and press + a few times.

How big is the error?

Taylor’s theorem (Lagrange form) says the remainder Rn(x) = f(x) − Pn(x) equals f(n+1)(z)(x − a)n+1/(n + 1)! for some z between a and x. If |f(n+1)| ≤ M there, then

|Rn(x)| ≤ M|x − a|n+1/(n + 1)!

For sin and cos every derivative is at most 1, so M = 1. For ex, M is e raised to the larger of a and x. The page shows that bound next to the true error so you can see that the bound is always at least as big.

For ln(1 + x) to the right of the center, and for arctan x centered at 0 with |x| ≤ 1, the series alternates with shrinking terms, so the alternating series bound applies: the error is at most the size of the first term left out. For 1/(1 − x) the page shows the exact remainder, because the series is geometric.

Check yourself

  1. Write the Maclaurin polynomial P5 for sin x and use it to estimate sin 1.
    Answer

    P5(x) = x − x3/3! + x5/5!, so sin 1 ≈ 1 − 1/6 + 1/120 ≈ 0.841667. The true value is 0.841471, an error of about 0.000196, under the Lagrange bound 1/6! ≈ 0.00139.

  2. What is the interval of convergence of the Taylor series for 1/(1 − x) centered at a = −1?
    Answer

    The distance from −1 to the blow-up at 1 is R = 2, so the series converges on (−3, 1). It is the geometric series Σ (x + 1)k/2k+1, and at both endpoints its terms do not shrink to 0.

  3. Why can no number of terms of the Maclaurin series for ln(1 + x) approximate ln 4?
    Answer

    ln 4 = ln(1 + 3) needs x = 3, which is outside the interval (−1, 1]. There the terms (−1)k−13k/k grow without bound, so the partial sums swing wider and wider. Move the center to a = 2 (then R = 3) and x = 3 lies inside.

A bit of history: Brook Taylor published the general series in his Methodus incrementorum (1715); James Gregory had found series like the one for arctan x decades earlier, and Colin Maclaurin made the case a = 0 popular in 1742. Joseph-Louis Lagrange gave the remainder formula used above in 1797.