of the planets, the paths of comets, and the tides of the seas first demonstrated.
Understanding these basic definition and statement~
- Let $f$ be a real-valued function defined on an interval $I$ containing the point $c$. We say $f$ is differentiable at $c$ if the limit
$$
\lim {x \rightarrow c} \frac{f(x)-f(c)}{x-c}=\lim {h \rightarrow 0} \frac{f(c+h)-f(c)}{h}
$$
exists. In this case we write $f^{\prime}(c)$ for this limit. If the function $f$ is differentiable at each point of the set $S \subseteq I$, then $f$ is said to be differentiable on $S$ and the function $f^{\prime}: S \rightarrow R$ is called the derivative of $f$ on $S$. When $f$ is differentiable at $c$ the tangent line to $f$ at $c$ is the linear function $L(x)=f(c)+f^{\prime}(c)(x-c)$.
Rolle’s Theorem: Let $f$ be a continuous function on $[a, b]$ that is differentiable on $(a, b)$ and such that $f(a)=f(b)=0$. Then there exists at least one point $c \in(a, b)$ such that $f^{\prime}(c)=0$.
Mean Value Theorem: If $f:[a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a point $c \in(a, b)$ where
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} .
$$
- If $f$ is differentiable on $(a, b)$ and $f^{\prime}$ is continuous, we say $f$ is of class $C^{1}$. If $f^{\prime}$ is differentiable, then we can get the second derivative $f^{\prime \prime} .$ If $f^{\prime \prime}$ is continuous, we say $f$ is of class $C^{2}$, and so on.
Taylor’s Theorem: Let $f:[a, b] \rightarrow \mathbb{R}$ be of class $C^{n}\left(n\right.$ is a positive integer) and let $x_{0} \in[a, b] .$ Then for each $x \in[a, b]$ with $x \neq x_{0}$ there exists a point $c$ between $x$ and $x_{0}$ such that
$f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\cdots+\frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}+\frac{f^{(n+1)}(c)}{(n+1) !}\left(x-x_{0}\right)^{n+1}$
Try the following problem to test yourself！
We say a function $f:(a, b) \rightarrow \mathbb{R}$ is uniformly differentiable if $f$ is differentiable on $(a, b)$ and for each $\varepsilon>0$ there exists a $\delta>0$ such that
$$
0<|x-y|<\delta \quad \text { and } \quad x, y \in(a, b) \quad \Rightarrow \quad\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(x)\right|<\varepsilon
$$
Prove that if $f$ is uniformly differentiable, then $f^{\prime}$ is continuous. Then give an example of a function that is differentiable but not uniformly differentiable.
Suppose $f$ is uniformly differentiable. Let $\varepsilon>0$ be given. Then there exists some $\delta>0$ such that
$$
0<|x-y|<\delta \quad \Rightarrow \quad\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(x)\right|<\frac{\varepsilon}{2} .
$$
Therefore,
$$
\left|f^{\prime}(x)-f^{\prime}(y)\right|=\left|f^{\prime}(x)-\frac{f(x)-f(y)}{x-y}\right|+\left|\frac{f(y)-f(x)}{y-x}-f^{\prime}(y)\right| \leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
$$
Hence, $f^{\prime}$ is continuous. Now consider
$$
f(x)= \begin{cases}x^{2} \sin \frac{1}{x} & \text { if } x \neq 0 \ 0 & \text { if } x=0 .\end{cases}
$$
This function is differentiable everywhere but $f^{\prime}(x)=2 x \sin \frac{1}{x}-\cos \frac{1}{x}$ for $x \neq 0$ and $\lim _{x \rightarrow 0} f^{\prime}(x)$
does not exist. Thus, $f^{\prime}$ exists but is not continuous.
Prove the Dirichlet function $f:(0,1) \rightarrow \mathbb{R}$ defined as
Let $f(x)=\sqrt{1+x^{2}}$. Show that
$$
\left(1+x^{2}\right) f^{[n+2]}(x)+(2 n+1) x f^{[n+1]}(x)+\left(n^{2}-1\right) f^{[n]}(x)=0 \text {, }
$$
for any $n \geq 1$. Use this identity to show that $f^{[2 n+1]}(0)=0$ for any $n \geq 0$. The notation $f^{[m]}(x)$ denotes the $m$ th derivative of $f$.
we showed that $\lim {x \rightarrow a} f(x)=0$ for any $a \in(0,1)$. Therefore (a) if $a \in \mathbb{R} \backslash \mathbb{Q}$, then $f(a)=0$ which implies $\lim {x \rightarrow a} f(x)=f(a)$, i.e., $f(x)$ is continuous at $a$;
(b)
Note first that
$$
f^{\prime}(x)=\frac{2 x}{2 \sqrt{1+x^{2}}}=\frac{x}{\sqrt{1+x^{2}}} .
$$
Hence $\left(1+x^{2}\right) f^{\prime}(x)=x \sqrt{1+x^{2}}=x f(x) .$ If we take the derivative of both sides of this equation, we get
$$
2 x f^{\prime}(x)+\left(1+x^{2}\right) f^{\prime \prime}(x)=f(x)+x f^{\prime}(x)
$$
or
$$
\left(1+x^{2}\right) f^{\prime \prime}(x)+x f^{\prime}(x)-f(x)=0 .
$$
Again let us take the derivative of both sides to get
$$
\left(1+x^{2}\right) f^{[3]}(x)+2 x f^{\prime \prime}(x)+x f^{\prime \prime}(x)+f^{\prime}(x)-f^{\prime}(x)=0
$$
or
$$
\left(1+x^{2}\right) f^{[3]}(x)+3 x f^{\prime \prime}(x)=0 .
$$
So the desired identity is valid when $n=1$. Assume it is still valid for $n$ and let us prove it for $n+1$. So we have
$$
\left(1+x^{2}\right) f^{[n+2]}(x)+(2 n+1) x f^{[n+1]}(x)+\left(n^{2}-1\right) f^{[n]}(x)=0 \text {. }
$$
If we take the derivative of both sides of this equation, we get
$\left(1+x^{2}\right) f^{[n+3]}(x)+2 x f^{[n+2]}(x)+(2 n+1) x f^{[n+2]}(x)+(2 n+1) f^{[n+1]}(x)+\left(n^{2}-1\right) f^{[n+1]}(x)=0$,
or
$$
\left(1+x^{2}\right) f^{[n+3]}(x)+(2 n+3) x f^{[n+2]}(x)+\left(2 n+1+n^{2}-1\right) f^{[n+1]}(x)=0 .
$$
Since $(n+1)^{2}-1=2 n+1+n^{2}-1$, we get
$$
\left(1+x^{2}\right) f^{[n+3]}(x)+(2 n+3) x f^{[n+2]}(x)+\left((n+1)^{2}-1\right) f^{[n+1]}(x)=0 \text {. }
$$
Hence by induction the desired identity is true for any $n \geq 1$. First note that $f(0)=1, f^{\prime}(0)=1$. From the identity
$$
\left(1+x^{2}\right) f^{\prime \prime}(x)+x f^{\prime}(x)-f(x)=0
$$
we get $f^{\prime \prime}(0)=1$. Assume that $f^{[2 n+1]}(0)=0$. Then from the identity
$$
\left(1+x^{2}\right) f^{[2 n+3]}(x)+(2 n+1) x f^{[2 n+2]}(x)+\left((2 n+1)^{2}-1\right) f^{[2 n+1]}(x)=0,
$$
we get
$$
f^{[2 n+3]}(0)+(2 n+1) 0 f^{[2 n+2]}(0)+\left((2 n+1)^{2}-1\right) f^{[2 n+1]}(0)=0,
$$
or $f^{[2 n+3]}(0)+\left((2 n+1)^{2}-1\right) f^{[2 n+1]}(0)=0$. Since $f^{[2 n+1]}(0)=0$, we get $f^{[2 n+3]}(0)=0$. By
induction we deduce that $f^{[2 n+1]}(0)=0$, for any $n \geq 0$.
(Implicit Function Theorem) Let $D=\left{(x, y) \in \mathbb{R}^{2}: a \leq x \leq b\right}$ and $F: D \rightarrow \mathbb{R}$ be a function where its partial derivative with respect to $y$ exists and there exist $m, M>0$ such that
$$
0<m<\frac{\partial F}{\partial y} \leq M \text { for all }(x, y) \in D
$$
Show that there exists one and only one continuous function $y(x)$ on $[a, b]$ such that
$$
F(x, y(x))=0 .
$$
Note: This means the equation $F(x, y(x))=0$ does implicitly define a unique continuous function $y$ in terms of $x$. To solve this problem consider the vector space $C[a, b]$ of all continuous realvalued functions defined on $[a, b]$ with $|f|=\max _{a \leq x \leq b}|f(x)|$ and define a map
$$
T:(C[a, b],|\cdot|) \rightarrow(C[a, b],|\cdot|)
$$
as
$$
T y(x)=y(x)-\frac{1}{M} F(x, y(x))
$$
Show $T$ is a contraction and use the Banach Contraction Mapping Theorem.
Let
$$
T:(C[a, b],|\cdot|) \rightarrow(C[a, b],|\cdot|)
$$
defined as
$$
T y(x)=y(x)-\frac{1}{M} F(x, y(x))
$$
We first claim that $T$ is a contraction on $C[a, b] .$ From the Mean Value Theorem we have
$$
F\left(x, y_{1}\right)-F\left(x, y_{2}\right)=\frac{\partial F}{\partial y}(x, c)\left(y_{1}-y_{2}\right)
$$
where $c$ is between $y_{1}$ and $y_{2}$. Hence
$$
\begin{aligned}
T_{y_{1}}(x)-T_{y_{2}}(x)=&\left[y_{1}(x)-y_{2}(x)\right]-\frac{1}{M}\left[F\left(x, y_{1}\right)-F\left(x, y_{2}\right)\right]=\left[y_{1}(x)-y_{2}(x)\right]-\frac{1}{M} \frac{\partial F}{\partial y}(x, c)\left[y_{1}(x)-y_{2}(x)\right] \
&=\left(1-\frac{1}{M} \frac{\partial F}{\partial y}(x, c)\right)\left[y_{1}(x)-y_{2}(x)\right] \leq\left(1-\frac{m}{M}\right)\left[y_{1}(x)-y_{2}(x)\right]
\end{aligned}
$$
Consider the given norm $|.|$,
$\left|T y_{1}-T y_{2}\right|=\max {a \leq x \leq b}\left|T y{1}(x)-T y_{2}(x)\right| \leq \max {a \leq x \leq b}\left(1-\frac{m}{M}\right)\left|y{1}(x)-y_{2}(x)\right| \leq\left|1-\frac{m}{M}\right|\left|y_{1}-y_{2}\right|$
Since $1-\frac{m}{M}<1$ it follows that $T$ is a contraction on $C[a, b]$, therefore by the Banach Contraction Mapping Theorem $T$ has a unique fixed point $y$. Thus for all $x \in[a, b]$,
$$
y(x)=T y(x)=y(x)-\frac{1}{M} F(x, y(x))
$$
but $M \neq 0$ thus $F(x, y(x))=0$
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Fourier analysis代写
离散数学代写
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