The Abstract Of Mathematical Analysis I 于2020年11月8日2020年11月8日由Sukuna 发布
1. Limits Two important limit \[\text { e }:=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}\]
\[\lim_{x\to 0}\frac{\sin x}{x}=1\]
Definition 3. inferior limit and superior limit \[\varliminf_{k \rightarrow \infty} x_{k}:=\lim <em>{n \rightarrow \infty} \inf </em>{k \geq n} x_{k}\]
\[\varlimsup_{k \rightarrow \infty} x_{k}:=\lim <em>{n \rightarrow \infty} \sup </em>{k \geq n} x_{k}\]
Theorem 2. Stolz Let
{\displaystyle (a{n}){n\geq 1}}
and
{\displaystyle (b{n}){n\geq 1}}
be two sequences of real numbers. Assume that
{\displaystyle (b{n}){n\geq 1}}
is a strictly monotone and divergent sequence (i.e. strictly increasing and approaching
, or strictly decreasing and approaching
) and the following limit exists:
\[\lim_ {n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l\]
Then, the limit
\[\lim _{n\to \infty }{\frac {a{n}}{b{n}}}=l\]
Theorem 3. Toeplitz limit theorem Supports that
n,k\subseteq \mathbb N^{+}
,
and
\[\sum_{k=1}^{n}{t_{nk}} = 1,\quad \lim_{n \rightarrow \infty}{t_{nk}} = 0\]
if
\[\lim_{n \rightarrow \infty}{a_{n}} = a\]
, let
\[x_{n} = \sum_{k=1}^{n}{t_{nk}a_{k}}\]
, s.t.
\[\lim_{n \rightarrow \infty}{x_{n}} = a\]
By using
, we can quickly infer The Cauchy proposition theorem.
By using
t_{n k}=\frac{b_{k+1}-b_{k}}{b_{n+1}-b_{1}}
, we can quickly infer The Stolz theorem.
Stirling’s formula Specifying the constant in the
error term gives
, yielding the more precise formula:
n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}
2. Continuity Definition 0 A function
is continuous at the point
, if for any neighbourhood
of its value
at a there is a neighbourhood
of a whose image under the mapping
is contained in
.
3. Differential calculus Definition 0 The number
\[f^{\prime}(a)=\lim _{E \ni x \rightarrow a} \frac{f(x)-f(a)}{x-a}\]
is called the derivative of the function
at
.
Definition 1 A function
defined on a set
is differentiable at a point x ∈ E that is a limit point of E if
f (x + h) − f (x) = A(x)h + \alpha(x;h)
, where
is a linear function in
and
as
,
.
Definition 2 The function
of Definition 1 , which is linear in
, is called the differential of the function
at the point
and is denoted
or
. Thus,
\mathrm d f (x)(h) = A(x)h
.
We obtain
\[\frac{\mathrm{d} f(x)(h)}{\mathrm{d} x(h)}=f^{\prime}(x)\]
We denote the set of all such vectors by
or
. Similarly, we denote by
or
the set of all displacement vectors from the point
along the y-axis. It can then be seen from the definition of the differential that the mapping
\[\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(f\left(x_{0}\right)\right)\]
The derivative of an inverse function If a function
is differentiable at a point x0 and its differential
\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a
is invertible at that point, then the differential of the function
inverse to
exists at the point
and is the mapping
\[\mathrm{d} f^{-1}\left(y_{0}\right)=\left[\mathrm{d} f\left(x_{0}\right)\right]^{-1}: T \mathbb{R}\left(y_{0}\right) \rightarrow T \mathbb{R}\left(x_{0}\right)\]
inverse to
\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a
.
The derivative of some common function formula \[\left(x^{\mu}\right)^{\prime}=\mu x^{\mu-1}\]
\[(\sin x)^{\prime}=\cos x\]
\[(\cos x)^{\prime}=-\sin x\]
\[(\tan x)^{\prime}=\sec ^{2} x\]
\[(\cot x)^{\prime}=-\csc ^{2} x\]
\[(\sec x)^{\prime}=\sec x \tan x\]
\[(\csc x)^{\prime}=-\csc x \cot x\]
\[\left(a^{x}\right)^{\prime}=a^{x} \ln a \quad(a>0, a \neq 1)\]
\[\left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x}\]
\[\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(a>0, a \neq 1)\]
\[(\ln x)^{\prime}=\frac{1}{x}\]
\[(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}\]
\[(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}\]
\[(\arctan x)^{\prime}=\frac{1}{1+x^{2}}\]
\[(\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}}\]
\[(\tanh x)' =\frac{1}{\cosh ^{2} x}\]
\[(\operatorname{coth} x)' =-\frac{1}{\sinh ^{2} x}\]
\[(\operatorname{arsinh} x)'=\left(\ln \left(x+\sqrt{1+x^{2}}\right)\right)' = \frac{1}{\sqrt{1+x^{2}}}\]
\[(\operatorname{arcosh} x)'=(\ln \left(x \pm \sqrt{x^{2}-1}\right))'= \pm \frac{1}{\sqrt{x^{2}-1}}\]
\[(\operatorname{artanh} x)'=(\frac{1}{2} \ln \frac{1+x}{1-x})' = \frac{1}{1-x^{2}}\]
\[(\operatorname{arcoth} x)'=(\frac{1}{2} \ln \frac{x+1}{x-1})' = \frac{1}{x^{2}-1}\]
L’Hôpital’s rule The theorem states that for functions
and
which are differentiable on an open interval
except possibly at a point
contained in
, if
\[\lim_{x\to c}{\frac {f(x)}{g(x)}}=\lim_{x\to c}{\frac {f'(x)}{g'(x)}}\]
Taylor’s theorem Let
be an integer and let the function
be
times differentiable at the point
. Then there exists a function
R_k : \mathbb R \to\mathbb R
such that ,
\[f(a+x)=f(a)+f'(a)(x)+{\frac {f''(a)}{2!}}x^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}x^{k}+R_k(x;a)\]
and,
\[R_{k}(x;a)=\int_{a}^{a+x} \frac{f^{(k+1)}(t)}{k !}(a+x-t)^{k} \mathrm d t\]
prove: q.e.d
remainder term using little
notation,
R_{k}(x;a)=o\left(|x|^{k}\right), \quad x \rightarrow 0
(The Peano remainder term)
The Lagrange form remainder term( Mean-value forms)
\[R_{n}(x;a)=\frac{f^{(n+1)}(\theta)}{(n+1) !}x^{n+1}\quad (\theta\in(a,a+x))\]
4. Integral Antiderivative Definition In calculus, an antiderivative , inverse derivative , primitive function , primitive integral or indefinite integral of a function
is a differentiable function
whose derivative is equal to the original function
Suppose
, the notation is
\[\int f(x)\mathrm dx=F(x)\]
So all the antiderivative of
become a family set
.
also the equation below is obviously.
\[\mathrm d \int f(x) \mathrm{d} x=f(x) \mathrm{d} x, \quad \int F^{\prime}(x) \mathrm{d} x=F(x)+c\]
Theorem: Integration by parts \[\int u(x)v'(x)\,\mathrm dx=u(x)v(x)-\int u'(x)v(x)\,\mathrm dx\]
Example: Wallis product the Wallis product for
, published in 1656 by John Wallis states that
Prove: so that:
so that:
Simplify the Polynomial and Integral If
\[Q(z)=\left(z-z{1}\right)^{k{1}} \cdots\left(z-z{p}\right)^{k{p}}\]
and
is a proper fraction , there exists a unique representation of the fraction
in the form
\[\frac{P(z)}{Q(z)}=\sum_{j=1}^{p}\left(\sum_{k=1}^{k_{j}} \frac{a_{j k}}{\left(z-z_{j}\right)^{k}}\right)\]
and if
and
are polynomials with real coefficients and
\[Q(x)=\left(x-x_{1}\right)^{k_{1}} \cdots\left(x-x_{l}\right)^{k_{l}}\left(x^{2}+p_{1} x+q_{1}\right)^{m_{1}} \cdots\left(x^{2}+p_{n} x+q_{n}\right)^{m_{n}}\]
there exists a unique representation of the proper fraction
in the form
\[\frac{P(x)}{Q(x)}=\sum_{j=1}^{l}\left(\sum_{k=1}^{k_{j}} \frac{a_{j k}}{\left(x-x_{j}\right)^{k}}\right)+\sum_{j=1}^{n}\left(\sum_{k=1}^{m_{j}} \frac{b_{j k} x+c_{j k}}{\left(x^{2}+p_{j} x+q_{j}\right)^{k}}\right)\]
where
and
are real numbers.
and with these formulas below:
And from that we get the recursion:
\[\int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m+1}}=\frac{1}{2 m a^{2}} \frac{x}{\left(x^{2}+a^{2}\right)^{m}}+\frac{2 m-1}{2 m a^{2}} \int \frac{\mathrm{d} x}{\left(x^{2}+a^{2}\right)^{m}}\]
Primitives of the Form \[\int R(\cos x, \sin x)\mathrm dx\]
We make the change of variable
. Since:
\[\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}, \qquad \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\]
so that
\[\mathrm{d} t=\frac{\mathrm{d} x}{2 \cos ^{2} \frac{x}{2}} \quad \Rightarrow\quad \mathrm{d} x=\frac{2 \mathrm{d} t}{1+\tan ^{2} \frac{x}{2}}=\frac{2\,\mathrm{d} t}{1+t^{2}}\]
It follows that
\[\int R(\cos x, \sin x) \mathrm{d} x=\int R\left(\frac{1-t^{2}}{1+t^{2}}, \frac{2 t}{1+t^{2}}\right) \frac{2}{1+t^{2}} \mathrm{d} t\]
not only
can to do this, but here are a lot of formula:
\[\tan a=\frac{2 \tan \frac{a}{2}}{1-\tan ^{2} \frac{a}{2}}\]
,
\[\cot \alpha=\frac{1-\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}\]
,
\[\sec \alpha=\frac{1+\tan ^{2} \frac{\alpha}{2}}{1-\tan ^{2} \frac{\alpha}{2}}\]
,
\[\csc \alpha=\frac{1+\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}\]
Integration Riemann Sums partition A partition P of a closed interval
,
, is a finite system of points
of the interval such that
a = x_0 < x_1 <\cdots < x_n = b
.
If a function
is defined on the closed interval
and
is a partition with distinguished points on this closed interval, the sum
\[\sigma(f ; P, \xi):=\sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}\]
where
\Delta x_i = x_i − x_{i−1}
, is the Riemann sum of the function
corresponding to the partition
with distinguished points on
.
The largest of the lengths of the intervals of the partition
, denoted
, is called the mesh of the partition.
we define:
\[\int_{a}^{b} f(x) \mathrm{d} x:=\lim <em>{\lambda(P) \rightarrow 0} \sum</em>{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}\]
Integral mean value theorem If
is a continuous function on the closed, bounded interval
, then there is at least one number
in
for which
\[\int_{a}^{b} f(x) \mathrm{d} x=f(\xi)(b-a)\]
The second Integral mean value theorem If
are continuous functions on the closed, bounded interval
,
is monotonous on
, then there is at least one number
in
for which
\[\int_{a}^{b}(f \cdot g)(x) \mathrm{d} x=g(a) \int_{a}^{\xi} f(x) \mathrm{d} x+g(b) \int_{\xi}^{b} f(x) \mathrm{d} x\]
Newton-Leibniz formula Let
be a continuous real-valued function defined on a closed interval
. Let
be the function defined, s.t.
\[\frac{d}{\mathrm{d} x} \int_{a}^{x} f(t) \mathrm{d} t=f(x), \quad \forall x \in[a, b]\]
Substitution Rule For Definite Integrals Suppose
f \in C[a, b], \varphi:[\alpha, \beta] \rightarrow[a, b]
and
\varphi^{\prime} \in \mathcal{R}[\alpha, \beta],\varphi(\alpha)=a, \varphi(\beta)=b
, s.t.
\[\int_{a}^{b} f(x) \mathrm{d} x=\int_{\alpha}^{\beta} f(\varphi(t)) \varphi^{\prime}(t) \mathrm{d} t\]