D6 - Using standard Maclaurin series

D6 - Using standard Maclaurin series

Standard series turn exponentials, logarithms, trigonometric functions and powers into algebra near zero. You will recognise their patterns, substitute new expressions, multiply expansions and keep the correct validity range alongside the answer.

Five standard Maclaurin series

Why can two almost identical powers have completely different expansion ranges? Compare (1+x)2(1+x)^2 with (1+x)−2(1+x)^{-2}. One is a polynomial; the other needs an infinite expansion. First collect the standard results that make such calculations quicker.

The Maclaurin coefficient rule explains the factorials below. The derivative cycles explain why sine has odd powers and cosine has even powers. All trigonometric angles are in radians.

Exponential, logarithm, sine and cosine

ex=1+x+x22!+x33!+⋯all real xln⁡(1+x)=x−x22+x33−x44+⋯−1<x≤1sin⁡x=x−x33!+x55!−⋯all real xcos⁡x=1−x22!+x44!−⋯all real x\begin{aligned} \mathrm{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots && \text{all real }x\\ \ln(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots && -1<x\le1\\ \sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots && \text{all real }x\\ \cos x&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots && \text{all real }x \end{aligned}

These four series and their ranges are supplied in the examination booklet. The logarithm's general term is (−1)r+1xr/r(-1)^{r+1}x^r/r, starting at r=1r=1.

The logarithm has ordinary integer denominators, not factorials. Its interval includes x=1x=1, giving ln⁡2=1−1/2+1/3−⋯\ln2=1-1/2+1/3-\cdots. It excludes x=−1x=-1, where ln⁡(1+x)\ln(1+x) is not defined. Although the function exists for x>1x>1, this Maclaurin series does not represent it there.

For a general power, use the binomial coefficients you know from A-level Mathematics. Here the exponent is written nn, while rr counts the power of xx.

Binomial series

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots

For rational nn, the infinite form is valid for ∣x∣<1|x|<1. Its coefficient of xrx^r is n(n−1)⋯(n−r+1)/r!n(n-1)\cdots(n-r+1)/r!.

If nn is a positive integer, the expansion terminates and is an exact polynomial for every real xx. For n=0n=0, the expansion is the constant polynomial 1; the expression (1+x)0(1+x)^0 equals 1 when x≠−1x\ne-1, avoiding the undefined expression 000^0. For other exponents, do not include an endpoint without a separate justification. The booklet supplies the binomial formula and the condition ∣x∣<1|x|<1.

The factors in the numerator decrease by 1 each time, even when nn is negative.

Worked example 1: a negative exponent

Question. Expand (1+x)−2(1+x)^{-2} up to and including x3x^3, stating the validity range of the infinite series.

Plan. Substitute n=−2n=-2 in the binomial coefficients before simplifying signs.

(1+x)−2=1−2x+(−2)(−3)2x2+(−2)(−3)(−4)6x3+⋯=1−2x+3x2−4x3+⋯\begin{aligned} (1+x)^{-2} &=1-2x+\frac{(-2)(-3)}{2}x^2+\frac{(-2)(-3)(-4)}{6}x^3+\cdots\\ &=1-2x+3x^2-4x^3+\cdots \end{aligned}

Answer. 1−2x+3x2−4x3+⋯1-2x+3x^2-4x^3+\cdots, valid for −1<x<1-1<x<1.

Check. Differentiating (1+x)−2(1+x)^{-2} at zero gives 1,−2,6,−241,-2,6,-24; dividing by the factorials gives 1,−2,3,−41,-2,3,-4, independently confirming the coefficients.

Change only the exponent's sign. The coefficient calculation now stops because a factor becomes zero.

Worked example 2: a terminating expansion

Question. Expand (1+x)2(1+x)^2 and state where the result is valid.

Plan. Use the same coefficient rule with n=2n=2, watching for the zero factor.

(1+x)2=1+2x+2(1)2x2=1+2x+x2\begin{aligned} (1+x)^2 &=1+2x+\frac{2(1)}{2}x^2\\ &=1+2x+x^2 \end{aligned}

The cubic coefficient contains 2(1)(0)2(1)(0); all later terms vanish too.

Answer. 1+2x+x21+2x+x^2, exactly, for every real xx.

Check. Direct multiplication (1+x)(1+x)=1+x+x+x2(1+x)(1+x)=1+x+x+x^2 gives the same polynomial. At x=2x=2, outside ∣x∣<1|x|<1, it correctly gives 9.

That is the answer to the opening question: a terminating polynomial does not need the convergence restriction of a genuinely infinite binomial series.

Watch out

For 1/(1+x)m1/(1+x)^m, the exponent is −m-m. Using +m+m expands a different function. Also distinguish validity from accuracy: an exponential series is valid for every real input, but a short truncation is usually useful only near zero.

Use a new negative exponent to check both the coefficients and the condition.

Substitutions, products and validity intervals

Substituting uu for xx means changing every occurrence, including the validity condition. For example,

ln⁡(1+2x)=2x−(2x)22+(2x)33−⋯(−1<2x≤1)\ln(1+2x)=2x-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-\cdots \qquad(-1<2x\le1)

Thus the interval is −1/2<x≤1/2-1/2<x\le1/2. For ln⁡(1−2x)\ln(1-2x), the condition is −1<−2x≤1-1<-2x\le1. Dividing inequalities by −2-2 reverses their directions, giving −1/2≤x<1/2-1/2\le x<1/2. The closed endpoint has moved to the other end.

The same substitution rule gives e2x=1+2x+2x2+(4/3)x3+⋯\mathrm{e}^{2x}=1+2x+2x^2+(4/3)x^3+\cdots, valid for every real xx, and cos⁡(x2)=1−x4/2+x8/24−⋯\cos(x^2)=1-x^4/2+x^8/24-\cdots, also valid for all real xx. A substituted power must be raised as a whole.

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