# 牛顿方法及其应用Newton’s Method and Other Applications-微积分辅导|PROBLEMS

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1. $18.01$ Single Variable Calculus

Fall 2006

Lecture 3 Derivatives of Products, Quotients, Sine, and Cosine

Derivative Formulas

There are two kinds of derivative formulas:

1. Specific Examples: $\frac{d}{d x} x^{n}$ or $\frac{d}{d x}\left(\frac{1}{x}\right)$
2. General Examples: $(u+v)^{\prime}=u^{\prime}+v^{\prime}$ and $(c u)=c u^{\prime}$ (where $c$ is a constant)

A notational convention we will use today is:

$$(u+v)(x)=u(x)+v(x) ; \quad u v(x)=u(x) v(x)$$

Proof of $(u+v)=u^{\prime}+v^{\prime}$. (General)

Start by using the definition of the derivative.

\begin{aligned} (u+v)^{\prime}(x) &=\lim {\Delta x \rightarrow 0} \frac{(u+v)(x+\Delta x)-(u+v)(x)}{\Delta x} \ &=\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x)+v(x+\Delta x)-u(x)-v(x)}{\Delta x} \ &=\lim _{\Delta x \rightarrow 0}\left{\frac{u(x+\Delta x)-u(x)}{\Delta x}+\frac{v(x+\Delta x)-v(x)}{\Delta x}\right} \ (u+v)^{\prime}(x) &=u^{\prime}(x)+v^{\prime}(x) \end{aligned}

Follow the same procedure to prove that $(c u)^{\prime}=c u^{\prime}$.

Derivatives of $\sin x$ and $\cos x$. (Specific)

Last time, we computed

\begin{aligned} \lim {x \rightarrow 0} \frac{\sin x}{x} &=1 \ \left.\frac{d}{d x}(\sin x)\right|{x=0} &=\lim {\Delta x \rightarrow 0} \frac{\sin (0+\Delta x)-\sin (0)}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{\sin (\Delta x)}{\Delta x}=1 \ \left.\frac{d}{d x}(\cos x)\right|{x=0} &=\lim {\Delta x \rightarrow 0} \frac{\cos (0+\Delta x)-\cos (0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\cos (\Delta x)-1}{\Delta x}=0 \end{aligned}

So, we know the value of $\frac{d}{d x} \sin x$ and of $\frac{d}{d x} \cos x$ at $x=0$. Let us find these for arbitrary $x$.

$$\frac{d}{d x} \sin x=\lim _{\Delta x \rightarrow 0} \frac{\sin (x+\Delta x)-\sin (x)}{\Delta x}$$

Recall:

$$\sin (a+b)=\sin (a) \cos (b)+\sin (b) \cos (a)$$

So,

\begin{aligned} \frac{d}{d x} \sin x &=\lim {\Delta x \rightarrow 0} \frac{\sin x \cos \Delta x+\cos x \sin \Delta x-\sin (x)}{\Delta x} \ &=\lim {\Delta x \rightarrow 0}\left[\frac{\sin x(\cos \Delta x-1)}{\Delta x}+\frac{\cos x \sin \Delta x}{\Delta x}\right] \ &=\lim {\Delta x \rightarrow 0} \sin x\left(\frac{\cos \Delta x-1}{\Delta x}\right)+\lim {\Delta x \rightarrow 0} \cos x\left(\frac{\sin \Delta x}{\Delta x}\right) \end{aligned}

Since $\frac{\cos \Delta x-1}{\Delta x} \rightarrow 0$ and that $\frac{\sin \Delta x}{\Delta x} \rightarrow 1$, the equation above simplifies to

$$\frac{d}{d x} \sin x=\cos x$$

A similar calculation gives

$$\frac{d}{d x} \cos x=-\sin x$$

2. Product formula (General)

$$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$$

Proof:

$$(u v)^{\prime}=\lim {\Delta x \rightarrow 0} \frac{(u v)(x+\Delta x)-(u v)(x)}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x+\Delta x)-u(x) v(x)}{\Delta x}$$

Now obviously,

$$u(x+\Delta x) v(x)-u(x+\Delta x) v(x)=0$$

so adding that to the numerator won’t change anything.

$$(u v)^{\prime}=\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x)-u(x) v(x)+u(x+\Delta x) v(x+\Delta x)-u(x+\Delta x) v(x)}{\Delta x}$$

We can re-arrange that expression to get

$$(u v)^{\prime}=\lim _{\Delta x \rightarrow 0}\left(\frac{u(x+\Delta x)-u(x)}{\Delta x}\right) v(x)+u(x+\Delta x)\left(\frac{v(x+\Delta x)-v(x)}{\Delta x}\right)$$

Remember, the limit of a sum is the sum of the limits.

$$\begin{gathered} {\left[\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x)-u(x)}{\Delta x}\right] v(x)+\lim {\Delta x \rightarrow 0}\left(u(x+\Delta x)\left[\frac{v(x+\Delta x)-v(x)}{\Delta x}\right]\right)} \ (u v)^{\prime}=u^{\prime}(x) v(x)+u(x) v^{\prime}(x) \end{gathered}$$

Note: we also used the fact that

$$\lim _{\Delta x \rightarrow 0} u(x+\Delta x)=u(x) \quad \text { (true because } u \text { is continuous) }$$

This proof of the product rule assumes that $u$ and $v$ have derivatives, which implies both functions are continuous.

1. A sequence of numbers $x_{1}, x_{2}, \ldots, x_{n}, \ldots$ satisfies the recurrence r tion $x_{n+1}=a x_{n}+b x_{n-1}$ for $n \geq 1$, where $a$ and $b$ are constants. Pruve that
$$\left[\begin{array}{c} x_{n+1} \ x_{n} \end{array}\right]=A\left[\begin{array}{c} x_{n} \ x_{n-1} \end{array}\right]$$
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2.4. PROBLEMS
where $A=\left[\begin{array}{ll}a & b \ 1 & 0\end{array}\right]$ and hence express $\left[\begin{array}{c}x_{n+1} \ x_{n}\end{array}\right]$ in terms of $\left[\begin{array}{l}x_{1} \ x_{0}\end{array}\right]$. If $a=4$ and $b=-3$, use the previous question to find a formula for $x_{n}$ in terms of $x_{1}$ and $x_{0}$.
$$\left.x_{n}=\frac{3^{n}-1}{2} x_{1}+\frac{3-3^{n}}{2} x_{0}\right]$$
2. Let $A=\left[\begin{array}{cc}2 a & -a^{2} \ 1 & 0\end{array}\right]$.
(a) Prove that
$$A^{n}=\left[\begin{array}{cc} (n+1) a^{n} & -n a^{n+1} \ n a^{n-1} & (1-n) a^{n} \end{array}\right] \quad \text { if } n \geq 1$$
(b) A sequence $x_{0}, x_{1}, \ldots, x_{n}, \ldots$ satisfies the recurrence relation $x_{n+1}=$ $2 a x_{n}-a^{2} x_{n-1}$ for $n \geq 1$. Use part (a) and the previous question to prove that $x_{n}=n a^{n-1} x_{1}+(1-n) a^{n} x_{0}$ for $n \geq 1$.
3. Let $A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$ and suppose that $\lambda_{1}$ and $\lambda_{2}$ are the roots of the quadratic polynomial $x^{2}-(a+d) x+a d-b c$. ( $\lambda_{1}$ and $\lambda_{2}$ may be equal.) Let $k_{n}$ be defined by $k_{0}=0, k_{1}=1$ and for $n \geq 2$
$$k_{n}=\sum_{i=1}^{n} \lambda_{1}^{n-i} \lambda_{2}^{i-1} .$$
Prove that
$$k_{n+1}=\left(\lambda_{1}+\lambda_{2}\right) k_{n}-\lambda_{1} \lambda_{2} k_{n-1},$$
if $n \geq 1$. Also prove that
$$k_{n}= \begin{cases}\left(\lambda_{1}^{n}-\lambda_{2}^{n}\right) /\left(\lambda_{1}-\lambda_{2}\right) & \text { if } \lambda_{1} \neq \lambda_{2} \ n \lambda_{1}^{n-1} & \text { if } \lambda_{1}=\lambda_{2}\end{cases}$$
Use mathematical induction to prove that if $n \geq 1$,
$$A^{n}=k_{n} A-\lambda_{1} \lambda_{2} k_{n-1} I_{2},$$
[Hint: Use the equation $A^{2}=(a+d) A-(a d-b c) I_{2}$.]