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Original question & working

Check an approximation by its error function · source reference

[2018/DHS/Prelim/I/2] · Question pp. 7 · Working pp. 7

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Printed question

(i) Using appropriate expansions from the List of Formulae (MF26), show that the series expansion of ln⁡(1+e2x)\ln(1+e^{2x}) in ascending powers of xx, up to and including the term in x2x^2, is ln⁡2+x+x22\ln2+x+\frac{x^2}2. [3]

(ii) Find the set of values of xx such that the value for the above expansion for ln⁡(1+e2x)\ln(1+e^{2x}) is within ±0.3\pm0.3 of the value of ln⁡(1+e2x)\ln(1+e^{2x}). [2]

(iii) Deduce an expansion for 21+e−2x\frac2{1+e^{-2x}} up to and including the term in xx. [2]

[2018/DHS/Prelim/I/2]

Printed working

ln⁡(1+e2x)=ln⁡[1+1+2x+(2x)2/2!+⋯ ]=ln⁡(2+2x+2x2+⋯ )=ln⁡2+ln⁡[1+(x+x2+⋯ )]=ln⁡2+(x+x2+⋯ )−12(x+x2+⋯ )2+⋯=ln⁡2+x+x2/2+⋯ .\begin{aligned}\ln(1+e^{2x})&=\ln[1+1+2x+(2x)^2/2!+\cdots]\\&=\ln(2+2x+2x^2+\cdots)=\ln2+\ln[1+(x+x^2+\cdots)]\\&=\ln2+(x+x^2+\cdots)-\tfrac12(x+x^2+\cdots)^2+\cdots\\&=\ln2+x+x^2/2+\cdots.\end{aligned}

−0.3<ln⁡(1+e2x)−[ln⁡2+x+x2/2]<0.3-0.3<\ln(1+e^{2x})-[\ln2+x+x^2/2]<0.3. Printed GC interval −1.55<x<1.55-1.55<x<1.55 (rounded).

Differentiate: 2e2x/(1+e2x)=1+x+⋯2e^{2x}/(1+e^{2x})=1+x+\cdots. Divide numerator/denominator by e2xe^{2x}: 2/(1+e−2x)=1+x+⋯2/(1+e^{-2x})=1+x+\cdots.

Teacher note 1

For e2xe^{2x}, use MF27: ex=1+x+x2/2!+x3/3!+⋯e^x=1+x+x^2/2!+x^3/3!+\cdots, where xx is replaced by 2x2x. Do not expand as ln⁡(1+e2x)=ln⁡[1+1+2x+(2x)2/2!+⋯ ]=(1+2x+⋯ )−(1+2x+⋯ )2/2+⋯\ln(1+e^{2x})=\ln[1+1+2x+(2x)^2/2!+\cdots]=(1+2x+\cdots)-(1+2x+\cdots)^2/2+\cdots, as each bracketed term contains a constant and terms in x,x2,x3,…x,x^2,x^3,\ldots.

Diagram 09-08-error

Error in the quadratic logarithm approximation

Mathematical review

1+e2x=2(1+x+x2+O(x3))1+e^{2x}=2(1+x+x^2+O(x^3)), so the log series gives ln⁡2+x+x2/2+O(x3)\ln2+x+x^2/2+O(x^3). More precisely, ln⁡(1+e2x)=ln⁡2+x+ln⁡cosh⁡x\ln(1+e^{2x})=\ln2+x+\ln\cosh x, hence error E(x)=ln⁡cosh⁡x−x2/2E(x)=\ln\cosh x-x^2/2 is even and nonpositive. For x>0x>0, E′(x)=tanh⁡x−x<0E'(x)=\tanh x-x<0. The unique positive root of E(x)=−0.3E(x)=-0.3 is about 1.5494761.549476; thus −1.549476<x<1.549476-1.549476<x<1.549476 (about ±1.55\pm1.55 at 3 s.f.).

Differentiating the analytic series gives 2e2x/(1+e2x)=1+x+O(x3)2e^{2x}/(1+e^{2x})=1+x+O(x^3), equivalently 2/(1+e−2x)=1+x+O(x3)2/(1+e^{-2x})=1+x+O(x^3).

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 7. Working page(s): 7. The two Series source files are byte-identical; their shared visible content is expressed once and mapped to both page sets. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.