1The idea in one paragraph

Polynomials are the easiest functions there are: you can evaluate, differentiate and integrate them with nothing but arithmetic. A Maclaurin series writes another function as a polynomial with infinitely many terms, a₀ + a₁x + a₂x² + a₃x³ + …, chosen so that at x = 0 it has the same value as the function, the same gradient, the same second derivative, and so on for ever. Matching the derivatives fixes every coefficient: aₙ = f⁽ⁿ⁾(0) ÷ n!. Cut the series off after a few terms and you have a polynomial that is an excellent approximation near 0, getting worse further away. From five standard series you can build the series of many other functions by substituting, multiplying, differentiating and integrating, and you can even find the series of a function you only know through a differential equation.

2Where the coefficients come from

Suppose f(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + …. Put x = 0 and every term but the first vanishes, so a₀ = f(0). Differentiate once and put x = 0: a₁ = f′(0). Differentiate again: f″(x) = 2a₂ + 3 × 2a₃x + …, so f″(0) = 2a₂. Once more: f‴(0) = 3 × 2 × 1 × a₃ = 3! a₃. The pattern is aₙ = f⁽ⁿ⁾(0)/n!, which gives the formula in the booklet:

Maclaurin series: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + …, with the n-th term (xⁿ/n!) f⁽ⁿ⁾(0).

Example 1: eˣ. Every derivative of eˣ is eˣ, and e⁰ = 1. So every f⁽ⁿ⁾(0) is 1, and

Figure 1 shows what the partial sums do. The line 1 + x matches the value and the slope at 0: it is the tangent. Adding x²/2 matches the bending as well, and adding x³/6 matches the rate at which the bending changes. Each term makes the polynomial cling to eˣ over a wider stretch.

A quick use: e^0.1 ≈ 1 + 0.1 + 0.005 + 0.000167 = 1.105167, and the true value is 1.105171. Four terms give five correct decimal places, because the powers of 0.1 shrink so fast.

Example 2: sin x. The derivatives cycle: sin x, cos x, −sin x, −cos x, sin x, …. At 0 these are 0, 1, 0, −1, 0, 1, …. The even powers vanish and the odd ones alternate in sign.

An odd function has only odd powers, which is a good check. Figure 2 shows the polynomials of degree 1, 3, 5 and 7 each staying on the sine curve for longer. The series for cos x follows the same way (derivatives at 0 are 1, 0, −1, 0, …), or faster by differentiating the sine series (section 5).

Example 3: ln(1 + x). Here ln x itself has no Maclaurin series, because ln 0 does not exist; that is why the standard series is for ln(1 + x).

The factorials cancel to leave the tidy pattern ± xⁿ/n. Lay out a derivation like this, a column of derivatives and a column of values at 0, so each mark can be seen.

3The standard series, and where each one works

The last column matters when you use a series to approximate. The guide does not ask you to prove where a series works, but it helps to know. Figure 3 shows the problem with ln(1 + x): inside −1 < x ≤ 1 the partial sums close in on the curve, but beyond x = 1 each extra term makes things worse, because the powers xⁿ grow faster than the n in the denominator.

The binomial series. For (1 + x)ᵖ with p a fraction or a negative number, the Maclaurin method gives exactly the extended binomial expansion of 1.10: f(0) = 1, f′(0) = p, f″(0) = p(p − 1), and so on. For example, √(1 + x) = (1 + x)^1/2 = 1 + x/2 − x²/8 + x³/16 − …. With x = 0.1 this gives √1.1 ≈ 1 + 0.05 − 0.00125 + 0.0000625 = 1.0488125, against the true 1.0488088.

Accuracy against simplicity. More terms mean more accuracy, but the right choice of series can matter more. Five terms of the ln(1 + x) series at x = 1 give ln 2 ≈ 0.783, which is poor, because x = 1 is at the edge of where the series works. Subtracting the series for ln(1 − x) from that for ln(1 + x) gives ln((1 + x)/(1 − x)) = 2(x + x³/3 + x⁵/5 + …), and x = 1/3 makes (1 + x)/(1 − x) = 2. Three terms then give ln 2 ≈ 0.69300, against the true 0.69315. Mathematicians of the Kerala school in India, from Madhava in the fourteenth century onwards, had series for sine, cosine and arctangent and used them for exactly this kind of calculation, centuries before Taylor and Maclaurin.

4New series by substitution and by multiplication

You rarely need to differentiate from scratch. Figure 4 shows the four ways to make a new series from a standard one.

Substitution. Replace x throughout by something else: x², −x, 2x, 3x². The guide's own example is e^x²:

The same move gives cos 2x = 1 − (2x)²/2! + (2x)⁴/4! − … = 1 − 2x² + (2/3)x⁴ − …, where the brackets matter: (2x)² is 4x², not 2x². It also gives ln(1 − x) = −x − x²/2 − x³/3 − …, by putting −x for x, and 1/(1 + x²) = 1 − x² + x⁴ − x⁶ + …, from (1 + u)⁻¹ with u = x². A substitution also changes where the series works: ln(1 + 3x) needs −1 < 3x ≤ 1.

Multiplication. Multiply two series as you would two long brackets, but keep only the terms up to the power you are asked for. The guide's example is eˣ sin x, up to x⁴.

Collecting by powers, as here, keeps you from multiplying out terms you will throw away. Differentiating eˣ sin x four times would give the same coefficients, much more slowly.

5New series by differentiation and integration

A series can be differentiated or integrated term by term, like any polynomial.

Differentiation. d/dx of the sine series is 1 − 3x²/3! + 5x⁴/5! − …, which simplifies to 1 − x²/2! + x⁴/4! − …: the cosine series, as it should be. From 1/(1 − x) = 1 + x + x² + x³ + …, differentiating gives 1/(1 − x)² = 1 + 2x + 3x² + 4x³ + ….

Integration: where arctan x comes from. The direct method is messy for arctan x, because its derivatives get complicated fast. Integration is quicker. The derivative of arctan x is 1/(1 + x²), whose series you found by substitution:

When you integrate an indefinite series, work out the constant: here it is 0 because arctan 0 = 0. Forgetting to check it is the usual error.

Integrals no formula can do. The function e^−x² has no antiderivative made of standard functions, yet its area matters (it is the shape of the normal distribution in 4.9). Its series integrates easily. Figure 5 shows the area from 0 to 0.5.

Limits. A series also settles limits of the form 0/0 (5.13) without l'Hôpital's rule:

6Series from a differential equation

Sometimes you do not know f(x) at all, only a differential equation it satisfies and one value. The Maclaurin formula needs only the derivatives at 0, and the differential equation supplies them: it gives y′ directly, and differentiating it gives y″, y‴ and so on.

Example 4. The function y satisfies dy/dx = x + y², with y = 1 when x = 0. Find the Maclaurin series for y up to the term in x³.

None of the exact methods of 5.18 solves this equation: it is not separable, not homogeneous, and not linear, because of the y². The series is what you can get. Figure 6 compares the cubic with the true solution, computed numerically: they agree to 0.003 at x = 0.2, then separate.

Example 5: when you know the function but it is awkward. Find the series for y = e^{sin x} up to x⁴. Differentiating e^{sin x} four times directly is heavy. Instead, notice that y′ = cos x × e^{sin x} = y cos x, a differential equation, and differentiate that.

The product rule, applied each time to terms like y′ cos x, produces the next line. Keep y, y′, y″ as symbols until the end and substitute the numbers at 0 only then.