已知 y=cot⁡(x+a)y = \cot(x + a)，其中 aa 为常数。

  1. (i)

    证明 dydx=−(1+y2)\dfrac{dy}{dx} = -\bigl(1 + y^2\bigr)，并由此求出用 yy 表示的 d2ydx2\dfrac{d^2y}{dx^2} 和 d3ydx3\dfrac{d^3y}{dx^3}。

  2. (ii)

    解释为什么当 a=0a = 0 时，无法求出 yy 的麦克劳林级数。

  3. (iii)

    当 a=π2a = \dfrac{\pi}{2} 时，利用你在 (i) 中的结论求出 yy 关于 xx 的升幂级数展开式，直到包含 x3x^3 的项。

  4. (iv)

    利用你在 (iii) 中的结论和公式表（MF26）中的标准级数，求 $\dfrac{y}{x+2}关于关于x的升幂级数展开式，直到包含的升幂级数展开式，直到包含x^3$ 的项。

前几项公式推导标准级数展开近似概念隐函数与参数方程求导基本初等函数图像特征English
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完整解法

2(i)

1. Differentiate $y = \cot(x+a)$ to show the first derivative in terms of $y$.

y=cot⁡(x+a)y = \cot(x+a)
dydx=−csc⁡2(x+a)\frac{dy}{dx} = -\csc^2(x+a)
Using the trigonometric identity csc⁡2θ=1+cot⁡2θ\csc^2 \theta = 1 + \cot^2 \theta:
dydx=−(1+cot⁡2(x+a))\frac{dy}{dx} = -\bigl(1 + \cot^2(x+a)\bigr)
dydx=−(1+y2)\frac{dy}{dx} = -\bigl(1 + y^2\bigr)

2. Differentiate implicitly to find the second derivative in terms of $y$.

d2ydx2=ddx[−(1+y2)]\frac{d^2y}{dx^2} = \frac{d}{dx} \bigl[-(1 + y^2)\bigr]
Applying the chain rule:
d2ydx2=−2ydydx\frac{d^2y}{dx^2} = -2y \frac{dy}{dx}
Substitute dydx=−(1+y2)\frac{dy}{dx} = -\bigl(1 + y^2\bigr):
d2ydx2=−2y[−(1+y2)]\frac{d^2y}{dx^2} = -2y \bigl[-(1 + y^2)\bigr]
d2ydx2=2y(1+y2)=2y+2y3\frac{d^2y}{dx^2} = 2y(1 + y^2) = 2y + 2y^3

3. Differentiate implicitly again to find the third derivative in terms of $y$.

d3ydx3=ddx[2y+2y3]\frac{d^3y}{dx^3} = \frac{d}{dx} \bigl[2y + 2y^3\bigr]
d3ydx3=(2+6y2)dydx\frac{d^3y}{dx^3} = (2 + 6y^2) \frac{dy}{dx}
Substitute dydx=−(1+y2)\frac{dy}{dx} = -\bigl(1 + y^2\bigr):
d3ydx3=(2+6y2)[−(1+y2)]\frac{d^3y}{dx^3} = (2 + 6y^2) \bigl[-(1 + y^2)\bigr]
d3ydx3=−(2+2y2+6y2+6y4)\frac{d^3y}{dx^3} = -\bigl(2 + 2y^2 + 6y^2 + 6y^4\bigr)
d3ydx3=−2−8y2−6y4\frac{d^3y}{dx^3} = -2 - 8y^2 - 6y^4

答案

d2ydx2=2y(1+y2)\dfrac{d^2y}{dx^2} = 2y(1+y^2) and d3ydx3=−2−8y2−6y4\dfrac{d^3y}{dx^3} = -2 - 8y^2 - 6y^4

2(ii)

1. Explain that the function is undefined at $x = 0$

When a=0a = 0, the function is y=cot⁡xy = \cot x.
At x=0x = 0, y=cot⁡0y = \cot 0, which is undefined.

2. Relate the undefined value to the Maclaurin series

Because yy (and its derivatives) are undefined at x=0x = 0, the coefficients of the Maclaurin series cannot be evaluated.

答案

When a=0a = 0, y=cot⁡xy = \cot x. At x=0x = 0, y=cot⁡0y = \cot 0, which is undefined. Hence, the Maclaurin series cannot be found.

2(iii)

1. Determine the expressions for the second and third derivatives in terms of $y$ (from part i) and evaluate them, along with $y$, at $x=0$.

From (i), we are given dydx=−(1+y2)\frac{dy}{dx} = -(1+y^2).
Differentiating implicitly with respect to xx:
d2ydx2=−2ydydx\frac{d^2y}{dx^2} = -2y\frac{dy}{dx}
d3ydx3=−2(dydx)2−2yd2ydx2\frac{d^3y}{dx^3} = -2\left(\frac{dy}{dx}\right)^2 - 2y\frac{d^2y}{dx^2}

When x=0x=0, y=cot⁡(0+π2)=0y = \cot\left(0 + \frac{\pi}{2}\right) = 0.
Substituting y=0y=0 into the derivatives:
dydx=−(1+02)=−1\frac{dy}{dx} = -(1+0^2) = -1
d2ydx2=−2(0)(−1)=0\frac{d^2y}{dx^2} = -2(0)(-1) = 0
d3ydx3=−2(−1)2−2(0)(0)=−2\frac{d^3y}{dx^3} = -2(-1)^2 - 2(0)(0) = -2

2. Apply the Maclaurin series formula to construct the expansion up to the $x^3$ term.

Using the Maclaurin series formula 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+(−1)x+02!x2+−23!x3y \approx 0 + (-1)x + \frac{0}{2!}x^2 + \frac{-2}{3!}x^3
y≈−x−26x3y \approx -x - \frac{2}{6}x^3
y≈−x−13x3y \approx -x - \frac{1}{3}x^3

答案

−x−13x3-x - \frac{1}{3}x^3

2(iv)

1. Rewrite $\frac{1}{x+2}$ to use the standard binomial series and expand it up to the term in $x^2$.

1x+2=(2+x)−1=2−1(1+x2)−1=12(1+x2)−1\frac{1}{x+2} = (2+x)^{-1} = 2^{-1}\left(1+\frac{x}{2}\right)^{-1} = \frac{1}{2}\left(1+\frac{x}{2}\right)^{-1}
Using the standard binomial series expansion (1+x)n=1+nx+n(n−1)2!x2+…(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots from the List of Formulae:
12(1+x2)−1=12[1+(−1)(x2)+(−1)(−2)2!(x2)2+… ]\frac{1}{2}\left(1+\frac{x}{2}\right)^{-1} = \frac{1}{2}\left[ 1 + (-1)\left(\frac{x}{2}\right) + \frac{(-1)(-2)}{2!}\left(\frac{x}{2}\right)^2 + \dots \right]
≈12−14x+18x2\approx \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2

2. Multiply the series for $y$ from part (iii) with the expansion of $\frac{1}{x+2}$ to find the final series up to $x^3$.

From part (iii), the series expansion of yy is y≈−x−13x3y \approx -x - \frac{1}{3}x^3.
Substituting both expansions:
yx+2≈(−x−13x3)(12−14x+18x2)\frac{y}{x+2} \approx \left(-x - \frac{1}{3}x^3\right) \left(\frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2\right)
Expanding and ignoring terms with powers higher than x3x^3:
yx+2≈−x(12−14x+18x2)−13x3(12)\frac{y}{x+2} \approx -x\left(\frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2\right) - \frac{1}{3}x^3\left(\frac{1}{2}\right)
=−12x+14x2−18x3−16x3= -\frac{1}{2}x + \frac{1}{4}x^2 - \frac{1}{8}x^3 - \frac{1}{6}x^3
=−12x+14x2−724x3= -\frac{1}{2}x + \frac{1}{4}x^2 - \frac{7}{24}x^3

答案

−12x+14x2−724x3-\frac{1}{2}x + \frac{1}{4}x^2 - \frac{7}{24}x^3

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