TMJC JC1 Promo 9758/2019/Q6

  1. (a)

    已知 y=ln⁡(2−e2x)y = \ln(2 - \mathrm{e}^{2x})，其中 x<12ln⁡2x < \frac{1}{2}\ln 2，证明
    ey[d2ydx2+(dydx)2]=−4e2x.\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -4\mathrm{e}^{2x}.

    由此，求 yy 展开至包含 x3x^3 项的麦克劳林级数。

  2. (b)

    求出 1−2x2+3x\frac{\sqrt{1-2x}}{2+3x} 按 xx 的升幂展开至包含 x2x^2 项的展开式。写出使此展开式有效的 xx 的值域范围。

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收敛域English
先不看答案，让 AI 诊断你卡在哪里？

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

10(a)

1. Express the equation in implicit form to avoid differentiating a natural logarithm.

Given y=ln⁡(2−e2x)y = \ln(2 - \mathrm{e}^{2x})
ey=2−e2x\mathrm{e}^y = 2 - \mathrm{e}^{2x}

2. Differentiate implicitly with respect to $x$ once.

eydydx=−2e2x\mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{2x}

3. Differentiate implicitly a second time to show the required result.

Using the product rule on the left hand side:
eyd2ydx2+(eydydx)dydx=−4e2x\mathrm{e}^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x}\right) \frac{\mathrm{d}y}{\mathrm{d}x} = -4\mathrm{e}^{2x}
eyd2ydx2+ey(dydx)2=−4e2x\mathrm{e}^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \mathrm{e}^y \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -4\mathrm{e}^{2x}
ey[d2ydx2+(dydx)2]=−4e2x\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -4\mathrm{e}^{2x} (Shown)

4. Evaluate $y$, $\frac{\mathrm{d}y}{\mathrm{d}x}$, and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ at $x=0$.

When x=0x = 0:
y=ln⁡(2−e0)=ln⁡(1)=0y = \ln(2 - \mathrm{e}^0) = \ln(1) = 0

Substitute into the first derivative equation:
e0dydx=−2e0  ⟹  dydx=−2\mathrm{e}^0 \frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -2

Substitute into the second derivative equation:
e0[d2ydx2+(−2)2]=−4e0\mathrm{e}^0 \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (-2)^2 \right] = -4\mathrm{e}^0
1⋅[d2ydx2+4]=−4  ⟹  d2ydx2=−81 \cdot \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4 \right] = -4 \implies \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -8

5. Differentiate the second derivative equation to find the third derivative at $x=0$.

Differentiate ey[d2ydx2+(dydx)2]=−4e2x\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -4\mathrm{e}^{2x} with respect to xx:
ey[d3ydx3+2dydxd2ydx2]+eydydx[d2ydx2+(dydx)2]=−8e2x\mathrm{e}^y \left[ \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} \right] + \mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -8\mathrm{e}^{2x}

Substitute x=0,y=0,dydx=−2,d2ydx2=−8x=0, y=0, \frac{\mathrm{d}y}{\mathrm{d}x}=-2, \frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-8:
1⋅[d3ydx3+2(−2)(−8)]+1⋅(−2)[−8+(−2)2]=−81 \cdot \left[ \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2(-2)(-8) \right] + 1 \cdot (-2) \left[ -8 + (-2)^2 \right] = -8
d3ydx3+32−2(−4)=−8\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 32 - 2(-4) = -8
d3ydx3+40=−8  ⟹  d3ydx3=−48\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 40 = -8 \implies \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -48

6. Construct the Maclaurin series using the evaluated values.

y=y(0)+y′(0)x+y′′(0)2!x2+y′′′(0)3!x3+…y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots
y=0+(−2)x+−82x2+−486x3+…y = 0 + (-2)x + \frac{-8}{2}x^2 + \frac{-48}{6}x^3 + \dots
y=−2x−4x2−8x3+…y = -2x - 4x^2 - 8x^3 + \dots

答案

y=−2x−4x2−8x3+…y = -2x - 4x^2 - 8x^3 + \dots

10(b)

1. Apply the binomial expansion to $(1-2x)^{\frac{1}{2}}$ up to the term in $x^2$.

(1−2x)12=1+(12)(−2x)+(12)(−12)2!(−2x)2+…=1−x−12x2+…\begin{aligned} (1-2x)^{\frac{1}{2}} &= 1 + \left(\frac{1}{2}\right)(-2x) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(-2x)^2 + \dots \\ &= 1 - x - \frac{1}{2}x^2 + \dots \end{aligned}

2. Factorize to obtain the $(1+ax)^n$ form for $(2+3x)^{-1}$ and expand up to the term in $x^2$.

(2+3x)−1=[2(1+32x)]−1=12(1+32x)−1=12[1+(−1)(32x)+(−1)(−2)2!(32x)2+… ]=12(1−32x+94x2+… )=12−34x+98x2+…\begin{aligned} (2+3x)^{-1} &= \left[ 2\left(1 + \frac{3}{2}x\right) \right]^{-1} \\ &= \frac{1}{2}\left(1 + \frac{3}{2}x\right)^{-1} \\ &= \frac{1}{2} \left[ 1 + (-1)\left(\frac{3}{2}x\right) + \frac{(-1)(-2)}{2!}\left(\frac{3}{2}x\right)^2 + \dots \right] \\ &= \frac{1}{2} \left( 1 - \frac{3}{2}x + \frac{9}{4}x^2 + \dots \right) \\ &= \frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2 + \dots \end{aligned}

3. Multiply the two series together, ignoring terms with powers higher than $x^2$.

1−2x2+3x=(1−2x)12(2+3x)−1=(1−x−12x2+… )(12−34x+98x2+… )=1(12)+1(−34x)+1(98x2)−x(12)−x(−34x)−12x2(12)+…=12−34x+98x2−12x+34x2−14x2+…=12−54x+138x2+…\begin{aligned} \frac{\sqrt{1-2x}}{2+3x} &= (1-2x)^{\frac{1}{2}}(2+3x)^{-1} \\ &= \left( 1 - x - \frac{1}{2}x^2 + \dots \right) \left( \frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2 + \dots \right) \\ &= 1\left(\frac{1}{2}\right) + 1\left(-\frac{3}{4}x\right) + 1\left(\frac{9}{8}x^2\right) - x\left(\frac{1}{2}\right) - x\left(-\frac{3}{4}x\right) - \frac{1}{2}x^2\left(\frac{1}{2}\right) + \dots \\ &= \frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{1}{4}x^2 + \dots \\ &= \frac{1}{2} - \frac{5}{4}x + \frac{13}{8}x^2 + \dots \end{aligned}

4. Determine the range of values of $x$ for which the expansion is valid.

The expansion for (1−2x)12(1-2x)^{\frac{1}{2}} is valid when:
∣−2x∣<1  ⟹  ∣x∣<12|-2x| < 1 \implies |x| < \frac{1}{2}

The expansion for (1+32x)−1\left(1 + \frac{3}{2}x\right)^{-1} is valid when:
∣32x∣<1  ⟹  ∣x∣<23\left|\frac{3}{2}x\right| < 1 \implies |x| < \frac{2}{3}

For the product expansion to be valid, both conditions must be satisfied. We take the intersection (the stricter condition):
∣x∣<12|x| < \frac{1}{2}
−12<x<12-\frac{1}{2} < x < \frac{1}{2}

答案

12−54x+138x2\frac{1}{2} - \frac{5}{4}x + \frac{13}{8}x^2
Valid for −12<x<12-\frac{1}{2} < x < \frac{1}{2}

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。

同一知识点的其他题

  1. 1已知 y=1+tan⁡−12xy = \sqrt{1 + \tan^{-1} 2x}，其中 −12<x<12-\frac{1}{2} < x < \frac{1}{2}。证明 yd2ydx2+(dydx)2+8x(1+4x2)2=0.y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0.…
  2. 2已知 y=tan⁡−115xy = \tan^{-1} \dfrac{1}{5} x，证明 (25+x2)d2ydx2+2xdydx=0\left(25 + x^2\right) \dfrac{d^2 y}{dx^2} + 2x \dfrac{dy}{dx} = 0。 通过对此结果的进一步求导，求…
  3. 3已知 ey=2+ex\text{e}^y = 2 + \text{e}^x。 证明 d2ydx2+(dydx)2=dydx\frac{\text{d}^2y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \frac{\text{d}y}{\text{d}x}…
  4. 4已知 y=1+tan⁡xy = \sqrt{1+\tan x}。 证明 2y dydx=1+(y2−1)22y\,\dfrac{dy}{dx} = 1 + (y^{2}-1)^{2} 且 y d2ydx2+(dydx)2=2y(y2−1)(dydx)y\,\dfrac{d^{2}y}{dx^{2}} + \left(\dfrac{dy}{dx}\right)^{2} = 2y(y^{2}-1)\left(\dfrac{dy}{dx}\right)…
  5. 5已知 y=sin⁡(ex−1)y = \sin\left(e^x - 1\right)。 证明 d2ydx2=dydx−ye2x\dfrac{d^2 y}{dx^2} = \dfrac{dy}{dx} - ye^{2x}。 由此求出 yy 的麦克劳林级数展开式，保留至包含 x2x^2…