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Maclaurin series and approximation · Track orders and small angles

Keep enough terms after cancellation

Work backwards from the requested final power before truncating a series. Exact rewrites, division and outer factors can change how many terms you need.

  1. Rewrite before approximating

    Identify the small variable and the requested highest power. Use exact identities to expose standard series in that variable.

    sin⁡(x+π/6)\sin(x+\pi/6)

    =12cos⁡x+32sin⁡x.=\frac12\cos x+\frac{\sqrt3}{2}\sin x.

    Small variableA shifted angle need not be small

    When x is small, x+π/6x+\pi/6 is near π/6\pi/6, not zero. Use angle addition before applying the small-angle series. Angles must be in radians.

    Standard formExpose the actual argument

    Factor constants and rewrite a reciprocal as a negative power when useful. For example, e1−2x=e e−2xe^{1-2x}=e\,e^{-2x}; substitute −2x-2x into the exponential series, including every power.

    This is a strategy for combining known expansions. A question asking for derivation from derivatives or a differential relation still needs that requested reasoning.

  2. Count powers in the final expression

    Track the lowest surviving powers after cancellation, multiplication and division. The requested final order is not always the order needed inside a bracket.

    xr×xk=xr+k.x^r\times x^k=x^{r+k}.

    xk/xm=xk−m(x≠0).x^k/x^m=x^{k-m}\quad(x\ne0).

    DivisionExpand further when powers are lowered

    To find (sin⁡x−x)/x3(\sin x-x)/x^3 through x2x^2, keep the sine term in x5x^5. The linear terms cancel and division lowers the remaining powers by three.

    Outer factorA factor can reduce the bracket order

    In θ(1+θ)−1\theta(1+\theta)^{-1} through θ2\theta^2, the reciprocal bracket is needed only through its linear term. In a product, retain every cross-product that contributes to a requested power.

    A request for the first few nonzero terms differs from a request through a specified power. Check cancellations before deciding that enough terms have been found.

  3. Check validity and what the approximation claims

    Use the condition belonging to each chosen series, transformed by its substituted argument. Keep original domain exclusions even after simplifying.

    (1+u)−1=1−u+u2−⋯(1+u)^{-1}=1-u+u^2-\cdots

    ∣u∣<1.|u|<1.

    ConvergenceThere is no universal interval

    The reciprocal geometric series above requires ∣u∣<1|u|<1. Other standard series have their own intervals and endpoint behaviour. Positive-integer binomial expansions are finite polynomials and need no infinite-series convergence restriction.

    AccuracyA valid series does not certify a decimal error

    Use ≈ for a finite approximation. A truncation through a stated power does not by itself guarantee a stated number of correct decimal places; justify any error bound separately.

    Simplification may give a finite limit at a point where the original quotient is undefined. State the limit separately rather than assigning the original expression a value there.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Division: plan three powers further

Find (sin⁡x−x)/x3(\sin x-x)/x^3 through x2x^2 for small nonzero x. Predict which sine term will supply the final quadratic term.

Show reasoning

  1. 1. Keep the term that will become quadratic

    The linear terms cancel. To reach power two after dividing by a cubic, keep sine through power five.

    sin⁡x=x−x36+x5120+⋯ .\begin{aligned}\sin x&=x-\frac{x^3}{6}\\&\quad+\frac{x^5}{120}+\cdots.\end{aligned}

  2. 2. Divide and preserve the excluded point

    The quotient is undefined at zero, although its limit is −1/6. Keeping sine only through its cubic term would miss the requested quadratic term.

    sin⁡x−xx3=−16+x2120+⋯(x≠0).\begin{aligned}&\frac{\sin x-x}{x^3}\\&=-\frac16+\frac{x^2}{120}+\cdots\\&\qquad(x\ne0).\end{aligned}

Example 11: the outer angle supplies a power

After the exact geometry in the linked question, the area is a2tan⁡θ/(1+tan⁡θ)a^2\tan\theta/(1+\tan\theta). For small positive θ in radians, retain terms through θ2\theta^2.

Show reasoning

  1. 1. Check the tangent truncation

    The next tangent term is cubic, so it cannot change the constant, linear or quadratic coefficients of this expression near zero. The denominator is nonzero there.

    tan⁡θ=θ+θ33+⋯ .\tan\theta=\theta+\frac{\theta^3}{3}+\cdots.

  2. 2. Expand the reciprocal only as far as needed

    The outer θ means only the constant and linear bracket terms contribute through θ². The geometric reciprocal series uses |θ|<1; the source geometry also restricts the original angle.

    (1+θ)−1=1−θ+θ2−⋯ .\begin{aligned}&(1+\theta)^{-1}\\&=1-\theta+\theta^2-\cdots.\end{aligned}

  3. 3. Collect the requested powers

    This gives the source’s area approximation in square centimetres. Continue with the lesson question for the geometry and the exact area identity.

    area⁡(ACE)≈a2θ(1−θ).\begin{aligned}&\operatorname{area}(ACE)\\&\approx a^2\theta(1-\theta).\end{aligned}
Continue with the lesson question