已知 y=ln⁡(2−e−2x)y = \ln(2 - \text{e}^{-2x})。

  1. (a)

    证明 dydx=4e−y−2\frac{\text{d}y}{\text{d}x} = 4\text{e}^{-y} - 2。

  2. (b)

    由此求出 yy 的麦克劳林级数，直到包含 x2x^2 的项。

  3. (c)

    利用公式表（MF26）中的标准级数，将 ln⁡(2−e−2x)\ln(2 - \text{e}^{-2x}) 展开至 x2x^2 项，并利用此展开式检验（b）部分中所求级数的正确性。

前几项公式推导标准级数展开函数概念隐函数与参数方程求导English
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完整解法

9(a)

1. Rewrite the equation in exponential form to prepare for implicit differentiation without relying on fractions.

y=ln⁡(2−e−2x)y = \ln(2 - \text{e}^{-2x})
ey=2−e−2x\text{e}^y = 2 - \text{e}^{-2x}

2. Differentiate both sides implicitly with respect to $x$.

eydydx=−(−2)e−2x\text{e}^y \frac{\text{d}y}{\text{d}x} = -(-2)\text{e}^{-2x}
eydydx=2e−2x\text{e}^y \frac{\text{d}y}{\text{d}x} = 2\text{e}^{-2x}

3. Substitute $\text{e}^{-2x}$ in terms of $y$ and rearrange to obtain the required expression.

From the first step, we know that e−2x=2−ey\text{e}^{-2x} = 2 - \text{e}^y. Substituting this in:
eydydx=2(2−ey)\text{e}^y \frac{\text{d}y}{\text{d}x} = 2(2 - \text{e}^y)
eydydx=4−2ey\text{e}^y \frac{\text{d}y}{\text{d}x} = 4 - 2\text{e}^y
Divide both sides by ey\text{e}^y:
dydx=4ey−2eyey\frac{\text{d}y}{\text{d}x} = \frac{4}{\text{e}^y} - \frac{2\text{e}^y}{\text{e}^y}
dydx=4e−y−2(Shown)\frac{\text{d}y}{\text{d}x} = 4\text{e}^{-y} - 2 \quad \text{(Shown)}

答案

dydx=4e−y−2\frac{\text{d}y}{\text{d}x} = 4\text{e}^{-y} - 2 (Shown)

9(b)

1. Evaluate $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x = 0$.

Substitute x=0x = 0 into the original equation:
y=ln⁡(2−e0)=ln⁡(1)=0y = \ln(2 - \mathrm{e}^{0}) = \ln(1) = 0

Substitute y=0y = 0 into the result from part (a):
dydx=4e0−2=2\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{0} - 2 = 2

2. Differentiate $\frac{\mathrm{d}y}{\mathrm{d}x}$ implicitly to find the second derivative.

d2ydx2=ddx(4e−y−2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x} \left( 4\mathrm{e}^{-y} - 2 \right)
d2ydx2=−4e−ydydx\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4\mathrm{e}^{-y} \frac{\mathrm{d}y}{\mathrm{d}x}

3. Evaluate the second derivative at $x = 0$ and construct the Maclaurin series.

At x=0,y=0,dydx=2x = 0, y = 0, \frac{\mathrm{d}y}{\mathrm{d}x} = 2:
d2ydx2=−4(1)(2)=−8\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4(1)(2) = -8

The Maclaurin series is given by:
y=y(0)+y′(0)x+y′′(0)2!x2+…y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \ldots
y=0+2x+−82x2y = 0 + 2x + \frac{-8}{2}x^2
y=2x−4x2y = 2x - 4x^2

答案

2x−4x22x - 4x^2

9(c)

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

e−2x=1+(−2x)+(−2x)22!+⋯≈1−2x+2x2\text{e}^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \dots \approx 1 - 2x + 2x^2

2. Substitute the expansion of $\text{e}^{-2x}$ into $2 - \text{e}^{-2x}$.

2−e−2x≈2−(1−2x+2x2)=1+2x−2x22 - \text{e}^{-2x} \approx 2 - (1 - 2x + 2x^2) = 1 + 2x - 2x^2

3. Expand $\ln(1 + u)$ where $u = 2x - 2x^2$, up to the term in $x^2$.

ln⁡(2−e−2x)≈ln⁡(1+(2x−2x2))\ln(2 - \text{e}^{-2x}) \approx \ln(1 + (2x - 2x^2))

Using the standard series ln⁡(1+u)=u−u22+…\ln(1+u) = u - \frac{u^2}{2} + \dots:
=(2x−2x2)−12(2x−2x2)2= (2x - 2x^2) - \frac{1}{2}(2x - 2x^2)^2
=2x−2x2−12(4x2−8x3+… )= 2x - 2x^2 - \frac{1}{2}(4x^2 - 8x^3 + \dots)
Discarding terms with powers higher than x2x^2:
=2x−2x2−2x2= 2x - 2x^2 - 2x^2
=2x−4x2= 2x - 4x^2

4. Compare the obtained expansion with the series found in part (b) to verify correctness.

In part (b), evaluating yy, y′y', and y′′y'' at x=0x=0 yields 00, 22, and −8-8 respectively, giving the Maclaurin series y≈0+2x+−82x2=2x−4x2y \approx 0 + 2x + \frac{-8}{2}x^2 = 2x - 4x^2.
Since the expansion using standard series is also 2x−4x22x - 4x^2, it confirms the correctness of the series found in part (b).

答案

ln⁡(2−e−2x)≈2x−4x2\ln(2 - \text{e}^{-2x}) \approx 2x - 4x^2, which matches the series from part (b).

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