It is given that f(x)=ln⁡(1+e−x)f(x) = \ln\left(1+e^{-x}\right).

  1. ((i))

    Using standard series from the List of Formulae (MF26) or otherwise, show that the first three non-zero terms in the Maclaurin series for f(x) is ln⁡a+bx+cx2\ln a + bx + cx^2, where a, b and c are constants to be determined.

  2. ((ii))

    Hence find an expansion for 11+ex\dfrac{1}{1+e^{x}}, up to and including the term in xx.

Derivation first few termsStandard series expansions中文
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Worked solution

1((i))

1. Expand $e^{-x}$ using the standard series up to the $x^2$ term.

1+e−x=1+(1−x+(−x)22!+… )≈2−x+x221 + e^{-x} = 1 + \left(1 - x + \frac{(-x)^2}{2!} + \dots\right) \approx 2 - x + \frac{x^2}{2}

2. Substitute the expansion into the given function and factorise the argument to prepare for the logarithmic standard series.

f(x)=ln⁡(1+e−x)≈ln⁡(2−x+x22)=ln⁡[2(1+(−x2+x24))]f(x) = \ln(1+e^{-x}) \approx \ln\left(2 - x + \frac{x^2}{2}\right) = \ln\left[2\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)\right]

3. Apply the product law of logarithms to split the expression.

ln⁡[2(1+(−x2+x24))]=ln⁡2+ln⁡(1+(−x2+x24))\ln\left[2\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)\right] = \ln 2 + \ln\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)

4. Expand $\ln(1+X)$ using the standard series up to the $x^2$ term, simplify, and identify $a$, $b$, and $c$.

Using ln⁡(1+X)≈X−X22\ln(1+X) \approx X - \frac{X^2}{2}:
f(x)≈ln⁡2+(−x2+x24)−12(−x2+x24)2f(x) \approx \ln 2 + \left(-\frac{x}{2} + \frac{x^2}{4}\right) - \frac{1}{2}\left(-\frac{x}{2} + \frac{x^2}{4}\right)^2
=ln⁡2−x2+x24−12(x24)+…= \ln 2 - \frac{x}{2} + \frac{x^2}{4} - \frac{1}{2}\left(\frac{x^2}{4}\right) + \dots
=ln⁡2−12x+18x2= \ln 2 - \frac{1}{2}x + \frac{1}{8}x^2
Comparing coefficients with ln⁡a+bx+cx2\ln a + bx + cx^2, we obtain a=2a = 2, b=−12b = -\frac{1}{2}, and c=18c = \frac{1}{8}.

Answer

a=2,b=−12,c=18a = 2, b = -\frac{1}{2}, c = \frac{1}{8}

1((ii))

1. Derive the constants $a$, $b$, and $c$ for the Maclaurin series of $\mathrm{f}(x)$ as required from part (i).

From part (i), we find the derivatives at x=0x=0:
f(x)=ln⁡(1+e−x)  ⟹  f(0)=ln⁡2\mathrm{f}(x) = \ln(1+e^{-x}) \implies \mathrm{f}(0) = \ln 2
f′(x)=−e−x1+e−x=−1ex+1  ⟹  f′(0)=−12\mathrm{f}'(x) = \dfrac{-e^{-x}}{1+e^{-x}} = \dfrac{-1}{e^x+1} \implies \mathrm{f}'(0) = -\dfrac{1}{2}
f′′(x)=ddx(−(ex+1)−1)=(ex+1)−2ex  ⟹  f′′(0)=14\mathrm{f}''(x) = \dfrac{d}{dx}\left(-(e^x+1)^{-1}\right) = (e^x+1)^{-2}e^x \implies \mathrm{f}''(0) = \dfrac{1}{4}
The Maclaurin series is f(0)+f′(0)x+f′′(0)2!x2\mathrm{f}(0) + \mathrm{f}'(0)x + \dfrac{\mathrm{f}''(0)}{2!}x^2:
f(x)≈ln⁡2−12x+18x2\mathrm{f}(x) \approx \ln 2 - \dfrac{1}{2}x + \dfrac{1}{8}x^2

2. Express the target expression in terms of $\mathrm{f}'(x)$.

Observe the first derivative of f(x)\mathrm{f}(x):
f′(x)=−e−x1+e−x\mathrm{f}'(x) = \dfrac{-e^{-x}}{1+e^{-x}}
Multiplying the numerator and denominator by exe^x:
f′(x)=−1ex+1\mathrm{f}'(x) = \dfrac{-1}{e^x+1}
Rearranging this gives a direct relationship with the target expression:
11+ex=−f′(x)\dfrac{1}{1+e^x} = -\mathrm{f}'(x)

3. Differentiate the Maclaurin series of $\mathrm{f}(x)$ to find the required expansion.

Differentiating the polynomial approximation of f(x)\mathrm{f}(x) with respect to xx:
f′(x)≈ddx(ln⁡2−12x+18x2)\mathrm{f}'(x) \approx \dfrac{d}{dx}\left(\ln 2 - \dfrac{1}{2}x + \dfrac{1}{8}x^2\right)
f′(x)≈−12+14x\mathrm{f}'(x) \approx -\dfrac{1}{2} + \dfrac{1}{4}x
Therefore, substituting into 11+ex=−f′(x)\dfrac{1}{1+e^x} = -\mathrm{f}'(x):
11+ex≈−(−12+14x)=12−14x\dfrac{1}{1+e^x} \approx -\left(-\dfrac{1}{2} + \dfrac{1}{4}x\right) = \dfrac{1}{2} - \dfrac{1}{4}x

Answer

12−14x\dfrac{1}{2} - \dfrac{1}{4}x

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