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Calculus - Maclaurin series – Extension of Binomial Theorem (HL)

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Maclaurin series is a Taylor series expansion of a function f(x)f(x) about x=0x = 0. It is given by f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n.

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The General Binomial Theorem extends the binomial expansion (a+b)n(a+b)^n to cases where nn is a negative integer or a fraction (rational number).

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The expansion of (1+x)n(1+x)^n for n∈Qn \in \mathbb{Q} is valid if and only if ∣x∣<1|x| < 1. If n∈Z+n \in \mathbb{Z}^+, the series terminates and is valid for all xx.

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The binomial coefficient for non-integer nn is defined as (nr)=n(n−1)(n−2)…(n−r+1)r!\binom{n}{r} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}.

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To expand (a+x)n(a+x)^n where a≠1a \neq 1, first factor out aa: an(1+xa)na^n (1 + \frac{x}{a})^n. The condition for convergence then becomes ∣xa∣<1|\frac{x}{a}| < 1 or ∣x∣<∣a∣|x| < |a|.

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Maclaurin series for composite functions can often be found by substituting a simpler series into another, or by multiplying/dividing known series.

📐Formulae

f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯+f(n)(0)n!xn+…f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots + \frac{f^{(n)}(0)}{n!}x^n + \dots

(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 + \dots

General Term: Tr+1=n(n−1)…(n−r+1)r!xr\text{General Term: } T_{r+1} = \frac{n(n-1)\dots(n-r+1)}{r!}x^r

Condition for convergence: ∣x∣<1 for n∉Z+\text{Condition for convergence: } |x| < 1 \text{ for } n \notin \mathbb{Z}^+

💡Examples

Problem 1:

Find the first four terms of the Maclaurin series for f(x)=1−2xf(x) = \sqrt{1-2x} and state the range of values of xx for which the expansion is valid.

Solution:

We write f(x)=(1−2x)1/2f(x) = (1 - 2x)^{1/2}. Here, n=12n = \frac{1}{2} and the 'x' in the formula is replaced by −2x-2x. Using the formula (1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3:

  1. First term: 11
  2. Second term: n(−2x)=12(−2x)=−xn(-2x) = \frac{1}{2}(-2x) = -x
  3. Third term: 12(12−1)2!(−2x)2=12(−12)2(4x2)=−18(4x2)=−12x2\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2x)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2}(4x^2) = -\frac{1}{8}(4x^2) = -\frac{1}{2}x^2
  4. Fourth term: 12(−12)(−32)3×2×1(−2x)3=386(−8x3)=116(−8x3)=−12x3\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{3 \times 2 \times 1}(-2x)^3 = \frac{\frac{3}{8}}{6}(-8x^3) = \frac{1}{16}(-8x^3) = -\frac{1}{2}x^3 So, 1−2x≈1−x−12x2−12x3\sqrt{1-2x} \approx 1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3. The expansion is valid for ∣−2x∣<1|-2x| < 1, which simplifies to ∣x∣<12|x| < \frac{1}{2}.

Explanation:

We applied the extended binomial theorem substituting u=−2xu = -2x and n=12n = \frac{1}{2}. The validity is determined by the condition ∣u∣<1|u| < 1.

Problem 2:

Use the Maclaurin series for exe^x and sin⁡x\sin x to find the first three non-zero terms of exsin⁡xe^x \sin x.

Solution:

The Maclaurin series are: ex=1+x+x22!+x33!+…e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots sin⁡x=x−x33!+x55!−…\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots Multiply the series: exsin⁡x=(1+x+x22+… )(x−x36+… )e^x \sin x = (1 + x + \frac{x^2}{2} + \dots)(x - \frac{x^3}{6} + \dots) Expand the product: =1(x−x36)+x(x−x36)+x22(x−x36)+…= 1(x - \frac{x^3}{6}) + x(x - \frac{x^3}{6}) + \frac{x^2}{2}(x - \frac{x^3}{6}) + \dots =x−x36+x2−x46+x32−x512+…= x - \frac{x^3}{6} + x^2 - \frac{x^4}{6} + \frac{x^3}{2} - \frac{x^5}{12} + \dots Group powers of xx: =x+x2+(−16+12)x3+…= x + x^2 + (-\frac{1}{6} + \frac{1}{2})x^3 + \dots =x+x2+13x3+…= x + x^2 + \frac{1}{3}x^3 + \dots

Explanation:

By multiplying the known series for exe^x and sin⁡x\sin x, we can find the terms of the product. We ignore terms with powers higher than x3x^3 as we only need the first three non-zero terms.

Problem 3:

Expand f(x)=12+xf(x) = \frac{1}{2+x} as a power series in xx up to the term in x2x^2.

Solution:

Rewrite f(x)f(x) in the form (1+u)n(1+u)^n: f(x)=(2+x)−1=2−1(1+x2)−1=12(1+x2)−1f(x) = (2+x)^{-1} = 2^{-1}(1 + \frac{x}{2})^{-1} = \frac{1}{2}(1 + \frac{x}{2})^{-1} Using the binomial expansion for n=−1n = -1: (1+x2)−1=1+(−1)(x2)+(−1)(−2)2!(x2)2+…(1 + \frac{x}{2})^{-1} = 1 + (-1)(\frac{x}{2}) + \frac{(-1)(-2)}{2!}(\frac{x}{2})^2 + \dots =1−x2+x24−…= 1 - \frac{x}{2} + \frac{x^2}{4} - \dots Multiply by the constant 12\frac{1}{2}: 12(1−x2+x24)=12−x4+x28\frac{1}{2}(1 - \frac{x}{2} + \frac{x^2}{4}) = \frac{1}{2} - \frac{x}{4} + \frac{x^2}{8}

Explanation:

To use the binomial expansion (1+x)n(1+x)^n, the constant term inside the bracket must be 1. We factor out 2 and then apply the expansion formula for n=−1n=-1.