tuzhong007’s blog

problem concerning euler constant

2019-12-21T00:00:00+00:00

Recently when I was browsing MSE, I came across a question about Euler constant.

The problem is:

How to show that $$ \lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots $$ No clue at all. Need help! Appreciated!

However, a certain part of the second answer confused me.

Abel’s partial summation technique:
Let $A(n) = \displaystyle \sum_{k=1}^n a(k)$ . We then have

$$ \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} $$ (The above is nothing but the discrete version of integration by parts). $$ \sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt $$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Here is the original post.

The first part is clear. But for the second part, the equation $\displaystyle \sum_{n=1}^{N} a(n) f(n) = \displaystyle \int_{1^-}^{N^+} f(t) d(A(t))$ in particular, I was suprised to see such a peculiar expression of integral which I cann’t recognize at all and that a discrete summation could be turned into an integral so easily. So I asked Mr.Zhou Ruisong for help. He deemed that $\displaystyle \int_{1^-}^{N^+} f(t) d(A(t))$ stood for $\displaystyle \lim_{a\rightarrow 1^{-}} \lim_{b \rightarrow N^{+}} \int_{a^-}^{b^+} f(t) d(A(t))$ and that $A(n)=A(\lfloor n\rfloor)$ for real $n$ , but with uncertainty. Along with his two sensible assumptions, I tried to wrote down a proof of the formular.

Proof Assume that $A:\mathbb{R}\rightarrow\mathbb{R}$ and $A(n)=A(\lfloor n\rfloor)$ for real $n$ , $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous.

Let’s consider the integer inteval $[1,N]$ .

Given $\epsilon>0$ . Let a partition $P=\{x_0,x_1,\dots,x_n\}$ s.t. $U(f,P,A)-L(f,P,A)<\epsilon$ ,
$x_{i-1}<x_{i}$ $\forall i\in [1,n]$ , $x_{0}=1-a$ where $a\in (0,1)$ , $x_{n}=N+b$ where $b\in (0,1)$ and $\{x\vert x=k-\frac{1}{2},k\in \mathbb{N}^{*}\cap [1,N+1]\}\subset P$ .

Then,

$$ \begin{align*}\Bigg \vert \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert < \epsilon \qquad \forall t_{i}\in[x_{i-1},x_{i}].\end{align*} $$

In particular, we let $t_{i}=\lfloor x_{i}\rfloor$ if $x_{i-1}<\lfloor x_{i}\rfloor < x_{i}$ . Thus,

$$ \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ]=\sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ]. $$

Hence,

$$ \begin{align*} \Bigg \vert \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert= \Bigg \vert \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert < \epsilon.\end{align*} $$

Since $\epsilon$ is arbitrary, we get

$$ \begin{align*} \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] =\int_{1-a}^{N+b} fdA.\end{align*} $$

Finally, letting $N+b \rightarrow N^{+}$ , $1-a \rightarrow 1^{-}$ , we obtain

$$ \begin{align*} \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] =\int_{1^{-}}^{N^{+}} fdA,\end{align*} $$

which is the desired result.

(Afterwards I discovered that $1^{-}$ denotes $1-\epsilon$ , where $\epsilon>0$ is arbitrarily small and $N^{+}$ denotes $N+\epsilon$ , where $\epsilon>0$ is arbitrarily small, according to the other answer of the author. The idea of the proof remains unchanged, though.)

linear transformations and derivatives(FUNCTIONS OF SERVERAL VARIABLES)

2019-11-16T00:00:00+00:00

Linear transformations A mapping $L$ from a vector space $V$ into a vector space $W$ is said to be a linear transformation if

$$ \begin{equation} \begin{aligned} L(\alpha \textbf{v}_{1}+\beta \textbf{v}_{2})= \alpha L(\textbf{v}_{1})+\beta L(\textbf{v}_{2}) \end{aligned}\end{equation} $$

for all $\textbf{v}_{1} ,\textbf{v}_{2} \in V$ and for all scalars $\alpha$ and $\beta$ .

In fact, for every linear transformation there exists a matrix corresponding to it.

Consider a linear transformation $L$ from $V$ into $W$ , where $V$ and $W$ are vector spaces of dimension $n$ and $m$ , respectively. Let $E=\{\textbf{v}_{1} ,\textbf{v}_{2}, \dots, \textbf{v}_{n}\}$ be an ordered basis for $V$ and $F=\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$ be an ordered basis for $W$ . Let $L$ be a linear transformation mapping $V$ into $W$ . If $\textbf{v}$ is any vector in $V$ , then we can express $\textbf{v}$ in terms of the basis $E$ :

$$ \begin{equation} \begin{aligned} \textbf{v}=\sum_{j=1}^{n}x_{j} \textbf{v}_{j}. \end{aligned}\end{equation} $$

For $j=1,2,\dots,n$ , let $\textbf{a}_{j}=(a_{1j},a_{2j},\dots,a_{mj})^{T}$ be the coordinate vector of $L(\textbf{v}_{j})$ with respect to $\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$ ; i.e.,

$$ \begin{equation} \begin{aligned} L(\textbf{v}_{j})=\sum_{k=1}^{m}a_{kj} \textbf{w}_{k}. \end{aligned}\end{equation} $$

Then,

$$ \begin{equation} \begin{aligned} L(\textbf{v})&=L(\sum_{j=1}^{n}x_{j} \textbf{v}_{j}) \\ &=\sum_{j=1}^{n}x_{j}L(\textbf{v}_{j}) \\ &=\sum_{j=1}^{n}x_{j}(\sum_{k=1}^{m}a_{kj} \textbf{w}_{k}) \\ &=\sum_{k=1}^{m}(\sum_{j=1}^{n}x_{j}a_{kj}) \textbf{w}_{k}. \end{aligned}\end{equation} $$

If we let $A=(a_{ij})=(\textbf{a}_{1},\textbf{a}_{2},\dots,\textbf{a}_{n})$ and $y_{i}=\sum_{j=1}^{n}x_{j}a_{kj}$ , then

$$ \begin{equation} \begin{aligned} \textbf{y}_{i}=(y_{1},y_{2},\dots,y_{m})^{T}=A\textbf{x} \end{aligned}\end{equation} $$

is the coordinate vector of $L(\textbf{v})$ with respect to $\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$ . We therefore find a corresponding matrix $A$ to $L(V,W)$ .

Derivatives

If $f$ is a real function with domain $(a,b) \subset R^{1}$ and if $x\in (a,b)$ , then $f'(x)$ is usually defined to be the real number

$$ \begin{equation} \begin{aligned} \lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}, \end{aligned}\end{equation} $$

provided, of course, that the limit exists. Thus

$$ \begin{equation} \begin{aligned} f(x+h)-f(x)=f'(x)h+r(h) \end{aligned}\end{equation} $$

where the “remainder” $r(h)$ is small, in the sense that

$$ \begin{equation} \begin{aligned} \lim_{h\rightarrow0} \frac{r(h)}{h}=0. \end{aligned}\end{equation} $$

Note that the equality above might indicate that the difference $f(x+h)-f(x)$ could be approximated by a linear function $f'(x)h$ . The point is, we could also regard the derivative of $f$ , this linear function, as a linear operator on $R^{1}$ which takes $h$ to $f'(x)h$ .

Then we can naturally extend the concept of derivative to a higher dimensional space. In this case the derivative of function $\textbf{f}$ should be a linear transformation, or, equivalently, a matrix.

9.11 Definition Suppose $E$ is an open set in $R^{n}$ , $\textbf{f}$ maps $E$ into $R^{m}$ , and $\textbf{x}\in E$ . If there exists a linear transformation $A$ of $R^{n}$ into $R^{m}$ such that

$$ \begin{equation} \begin{aligned} \lim_{\textbf{h}\rightarrow0} \frac{\vert \textbf{f}(\textbf{x}+\textbf{h})- \textbf{f}(\textbf{x})- A\textbf{h}\vert} {\vert\textbf{h}\vert}=\textbf{0}, \end{aligned}\end{equation} $$

then we say that $\textbf{f}$ is differentiable at $\textbf{x}$ , and we write

$$ \begin{equation} \begin{aligned} \textbf{f}'(\textbf{x})=A. \end{aligned}\end{equation} $$

a problem arised when using mathematical induction

2019-10-30T00:00:00+00:00

When I was tring to work out the proof of the theorem 8.3(Dominated convergence theorem), in one step I applied the mathematical induction to show that

$$ \begin{equation} \sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j} =\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

Then it occurs to me, that why can’t I just simply letting $n\rightarrow \infty$ and get the intended result? After thinking for a while(referring to 4 days), I maybe understood eventually. The reason is that when $n\rightarrow \infty$ , the series must include infinitely many terms. And since there is not an $n$ s.t. $n+1=\infty$ , the situation where the series includes infinitely many terms can’t be reached by the recursion process of the mathematical induction.

relations between some double sequences and series of functions(SEQUENCE AND SERIES OF FUNCTIONS)

2019-10-26T00:00:00+00:00

8.3 Theorem Given a double sequence $\{a_{i,j}\}$ , $i=1,2,3,\ldots, j=1,2,3,\ldots,$ suppose that

$$ \begin{equation} \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots)\end{equation} $$

and $\sum b_i$ converges. Then

$$ \begin{equation} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

Proof Let $E$ be a countable set, consisting of the points $x_0, x_1, x_2, \ldots,$ and suppose $x_{n} \rightarrow x_{0}$ as $n \rightarrow \infty$ . Define

$$ \begin{equation} \begin{aligned} f_i(x_n)&=\sum_{j=1}^{n}a_{i,j} \hspace{1cm}(i,n=1,2,3,\ldots),\\ f_i(x_0)&=\sum_{j=1}^{\infty}a_{i,j} \hspace{1cm}(i=1,2,3,\ldots),\\ g(x)&=\sum_{i=1}^{\infty}f_{i}(x) \hspace{1cm}(x\in E). \end{aligned}\end{equation} $$

And they are pretty well-defined. Since

$$ \begin{equation} \lim_{n\rightarrow\infty} f_i(x_n)=f_i(x_0)\end{equation} $$

and

$$ \begin{equation} \lim_{n \rightarrow \infty}x_n=x_0,\end{equation} $$

$f_i$ is continuous at $x_{0}$ . In addition, since

$$ \begin{equation} f_i(x_0)\leq \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots),\\ f_i(x_n)\leq \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots),\end{equation} $$

and $\sum b_i$ converges, we know that $g(x)=\sum f_i$ converges uniformly, thus $g$ is continuous at $x_{0}$ too. The continuity of $g$ implies that

$$ \begin{equation} \begin{aligned} \lim_{n \rightarrow \infty}g(x_n)&=g(x_0), \\ g(x_0)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}&=\lim_{n \rightarrow \infty}g(x_n)=\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j}. \end{aligned}\end{equation} $$

Note that

$$ \begin{equation} g(x_n)=\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j} =\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j},\end{equation} $$

which can be proved by using the mathematical induction.

Firstly, it’s obivious that the equation is correct when $n=1$ . Assume that it holds for $n=k(k\in N^{*})$ . i.e.,

$$ \begin{equation} g(x_k)=\sum_{i=1}^{\infty}\sum_{j=1}^{k}a_{i,j}= \sum_{j=1}^{k}\sum_{i=1}^{\infty}a_{i,j},\end{equation} $$

and then consider the situation where $n=k+1$ . Since

$$ \begin{equation} \begin{aligned} g(x_{k+1})&=\sum_{i=1}^{\infty}\sum_{j=1}^{k+1}a_{i,j}= \sum_{i=1}^{\infty}\sum_{j=1}^{k}a_{i,j}+\sum_{i=1}^{\infty}a_{i,k+1}\\ &=\sum_{j=1}^{k}\sum_{i=1}^{\infty}a_{i,j}+\sum_{i=1}^{\infty}a_{i,k+1}\\ &=\sum_{j=1}^{k+1}\sum_{i=1}^{\infty}a_{i,j}, \end{aligned}\end{equation} $$

hence we conclude that

$$ \begin{equation} g(x_n)=\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j} =\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

Thus,

$$ \begin{equation} \begin{aligned} g(x_0)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}&=\lim_{n \rightarrow \infty}g(x_n)=\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j}\\ &=\lim_{n \rightarrow \infty}\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j}\\ &=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}, \end{aligned}\end{equation} $$

i.e.,

$$ \begin{equation} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

Inclusion-exclusion

2019-10-06T00:00:00+00:00

The principle of inclusion-exclusion provides us with a technique for solving problems like couting. For example, how many numbers not exceeding $1000$ are divisible by $7$ or $11$ ?

solution: Let $A$ be the set of positive numbers which are not exceeding $1000$ and divisible by 7, and let $B$ be the set of positive numbers which are not exceeding $1000$ and divisible by 11. Define $\vert A\vert$ to be the number of elements in $A$ , we have

$$ \begin{equation} \begin{aligned} \vert A\cup B\vert&=\vert A\vert+\vert B\vert-\vert A\cap B\vert \\ &=\lfloor \frac{1000}{7}\rfloor+\lfloor \frac{1000}{11}\rfloor-\lfloor \frac{1000}{7\times11}\rfloor \\ &=220. \end{aligned} \end{equation} $$

Thus, the answer should be $220$ .

In fact, there’s a theorem.

Theorem(The principle of inclusion-exclusion) Let $A_1, A_2, \dots, A_n$ be finite sets. Then

$$ \begin{equation} \begin{aligned} \vert\bigcup_{i=1}^{n}A_i\vert&=\sum_{1\le i\le n}\vert A_i\vert -\sum_{1\le i<j\le n}\vert A_i\cap A_j\vert\\ &+\sum_{1\le i<j<k\le n}\vert A_i\cap A_j\cap A_k\vert-\cdots+(-1)^{n}\vert \bigcap_{i=1}^{n}A_i\vert. \end{aligned} \end{equation} $$

Sample: [codeforces 1228E]Another filling the grid

Let $S_{i,j}$ be a collection of situations with i rows and j colomns do not satisfy the condition, that is, for every grid in this i rows or j colomns, its value should be greater than 1. We could use a fomular to calculate this $\vert S_{i,j}\vert$ :

$$ \vert S_{i,j}\vert=C_{n}^{i}\times C_{n}^{j}\times (k-1)^{n(i+j)-ij}\times (k-1)^{n^2-n(i+j)+ij} $$

Now we want to obtain $\vert \bigcup_{i,j}S_{i,j}\vert$ . By observation, we know that $S_{i,j}=S_{i,0}\cap S_{0,j}$ , $S_{i,0}\subset S_{j,0}$ if $j<i$ , $S_{0,i}\subset S_{0,j}$ if $j<i$ and $S_{i,j}\cap S_{k,l}=S_{i,0}\cap S_{0,j}\cap S_{k,0}\cap S_{0,l}=S_{max(i,k),max(j,l)}$ . In additional, given the theorem, we can figure out that every $S_{i,j}$ contributes to $\vert \bigcup_{i,j}S_{i,j}\vert$ with the coefficient $(-1)^{i+j}$ .

Then apply the theorem, we get

$$ \vert \bigcup_{i,j}S_{i,j}\vert=\sum_{i=0}^{n}\sum_{j=0}^{n}(-1)^{i+j}\times C_{n}^{i}\times C_{n}^{j}\times (k-1)^{n(i+j)-ij}\times (k-1)^{n^2-n(i+j)+ij}, $$

note that $\vert S_{0,0}\vert$ , by fomular, equals to $k^n$ , which is the total number of ways to fill the grid. And therefore the answer would be this $\vert \bigcup_{i,j}S_{i,j}\vert$ . The time complexity of the algorithm should be $O(n^{2}logn)$ .

The answer is required to take mod. The law of addition and multiplication under mod operation is well known. But this time we used the formular $C_{n}^{i}=\frac{n!}{(n-i)!i!}$ to calculate combination, hence the inverse of an element is needed. There is a trick to find the inverse of an element: applying Fermat’s little theorem,

$$ a^{p-1}\equiv1(mod \hspace{0.2cm}p) $$

where $p$ is prime, $(a,p)=1$ . Thus,

$$ a\times a^{p-2}\equiv 1(mod \hspace{0.2cm}p) $$

i.e., $a^{p-2}$ is the inverse of $a$ .

my code:

#include<bits/stdc++.h>
const int mod =1e9+7;
typedef long long ll;
int n,k;
namespace NT{
	ll tm(ll x){
		return (x%mod+mod)%mod;
	}
	ll pw(ll x,ll y){
		ll res=1,bas=x;
		while(y){
			if(y&1){
				res=tm(res*bas);
			}
			bas=tm(bas*bas);
			y>>=1;
		}
		return res;
	}
	ll ginv(ll x){
		return pw(x,mod-2);
	}
	ll fac[255],Com[255];
	ll Ans;
	void getfac(){
		fac[0]=1;
		ll F=1;
		for(int i=1;i<=n;++i){
			F=tm(F*i);
			fac[i]=F;
		}
	}
	void getCom(){
		for(int i=0;i<=n;++i){
			ll cur=tm(fac[n]*ginv(fac[n-i]));
			cur=tm(cur*ginv(fac[i]));
			Com[i]=cur;
//			printf("%lld ",Com[i]);
		}
	}
	void getAns(){
		for(int i=0;i<=n;++i){
			for(int j=0;j<=n;++j){
				ll coe=1;
				if((i+j+1)%2==0)coe=-1;
				ll sij=tm(Com[i]*Com[j]);
				ll k_1p=pw(k-1,n*(i+j)-i*j);
				ll kp=pw(k,n*n+i*j-n*(i+j));
				sij=tm(sij*k_1p);
				sij=tm(sij*kp);
				Ans=tm(Ans+coe*sij);
//				printf("%lld ",sij*coe);
			}
		}
	}
	void work(){
		getfac();
		getCom();
		getAns();
		printf("%lld",Ans);
	}
}
using namespace NT;
int main(){
	scanf("%d%d",&n,&k);
	work();
	return 0;
}
 
// printf("%lld",);

norm space(SEQUENCE AND SERIES OF FUNCTIONS)

2019-09-29T00:00:00+00:00

After studying some global properties of series of functions, the concept of norm space furnished a new version to describe them.

7.14 Definition If $X$ is a metric space, $\mathscr{C}(X)$ will denote the set of all complex-valued, continuous, bounded functions with domain $X$ .

We associate with each $f\in\mathscr{C}(X)$ its supremum norm

$$ \Vert f\Vert=\sup\limits_{x\in X}\vert f(x)\vert. $$

If $h=f+g$ , then

$$ \vert h(x)\vert \leq \vert f(x)\vert+\vert g(x)\vert\leq \vert f\vert+\vert g\vert $$

for all $x\in X$ . Hence

$$ \Vert f+g\Vert \leq \Vert f\Vert+\Vert g\Vert. $$

If we define the distance between $f\in\mathscr{C}(X)$ and $g\in\mathscr{C}(X)$ to be $\Vert f-g\Vert$ , it follows that $\mathscr{C}(X)$ becomes a metric space.

A new version mentioned above, is that we can restate some previous theorems as follows:

A sequence $\{f_n\}$ converges to f with respect to the metric of $\mathscr{C}(X)$ if and only if $f_n\rightarrow f$ uniformly on $X$ .

This is an example of the supremum norm. We could also define some different norms, like:

$$ \Vert f\Vert=(\int_{0}^{1}\vert f(x)\vert^{p}dx)^{1/p}, $$

where $X=[0,1]$ .

philosophy of some proof(THE RIEMANN-STIETJES INTEGRAL)

2019-09-01T00:00:00+00:00

6.6 Theorem Suppose $\alpha$ is a monotonically increasing function on $[a,b]$ . f $\in \mathscr{R}(\alpha)$ on $[a,b]$ if and only if for every $\epsilon>0$ there exists a partition P such that

$$ \begin{equation} U(P,f,\alpha)-L(P,f,\alpha)<\epsilon. \end{equation} $$

This theorem provides a fundamental idea about solving questions like “how to check a function to be Riemann-integrable or not”. By definition, we have

$$ U(P,f,\alpha)-L(P,f,\alpha)=\sum\limits_{i=1}^{n}(M_i-m_i)\Delta\alpha_{i}, $$

where for a partition P,

$$ \begin{equation} \begin{aligned} M_i&=\sup f(x) \hspace{1cm}(x_{i-1}\leq x \leq x_{i}), \\ m_i&=\inf f(x) \hspace{1cm}(x_{i-1}\leq x \leq x_{i}), \\ \Delta\alpha_{i}&=\alpha(x_{i})-\alpha(x_{i-1}). \end{aligned} \end{equation} $$

Given $\epsilon>0$ . Naturally, we could imagine that if $f$ is Riemann-integrable, then $\sum\limits_{i=1}^{n}(M_i-m_i)\Delta\alpha_{i}$ should be small enough for some partition P. We hope to construct some P which meet the requirement.

There is an idea on constructing a suitable P. For each term, we either make $(M_i-m_i)$ or $\Delta\alpha_{i}$ small. Then this P may suffice.

The proof of the theorem below should be an example.

6.8 Theorem If f is continuous on $[a, b]$ then f $\in \mathscr{R}(\alpha)$ on $[a,b]$ .

Proof Given $\epsilon>0$ . Since $f$ is continuous, and $[a,b]$ is compact, we know $f$ is uniformly continuous, i.e., given $\epsilon>0$ , there exist a $\delta>0$ such that $\forall x,y\in [a,b]$ , if $d(x,y)<\delta$ , then $d(f(x),f(y))<\epsilon$ . Hence, if we choose a partition with $\Delta\alpha_{i}<\delta$ , then $(M_i-m_i)<\epsilon$ . We thus get

$$ \begin{equation} \begin{aligned} U(P,f,\alpha)-L(P,f,\alpha)&=\sum\limits_{i=1}^{n}(M_i-m_i)\Delta\alpha_{i} \\ &< \epsilon\sum\limits_{i=1}^{n}\Delta\alpha_{i} \\ &= \epsilon (\alpha(b)-\alpha(a)). \end{aligned} \end{equation} $$

Since $\epsilon$ is arbitrary, for each $\eta>0$ , we pick $\epsilon$ which satisfy the inequality $\epsilon (\alpha(b)-\alpha(a))<\eta$ . We finally get

$$ \begin{equation} \begin{aligned} U(P,f,\alpha)-L(P,f,\alpha)<\eta, \end{aligned} \end{equation} $$

and therefore $f\in \mathscr{R(\alpha)}$ .

test2

2019-08-11T00:00:00+00:00

I want to display diverse kinds of mathematics formular in this blog, so I will try to work out some exercises of baby rudin as samples.

Sample 1(exercise 4.2):

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$ , prove that

$$ \begin{equation} f(\overline{E})\subset \overline{f(E)} \end{equation} $$

for every set $E\subset X$ . ( $\overline{E}$ denotes the closure of $E$ .) Show, by an example, that $f(\overline{E})$ can be a proper subset of $\overline{f(E)}$ .

Solution:
Since $\overline{f(E)}$ is closed, $f^{-1}[\overline{f(E)}]$ is closed. Since $E\subset f^{-1}[\overline{f(E)}]$ , we have $\overline{E}\subset f^{-1}[\overline{f(E)}]$ , thus $f(\overline{E})\subset \overline{f(E)}$ .
Next, let $X$ and $Y$ be $R^1$ , $E=\{x|x\in N^{*}\}$ , and $f(x)=\frac{1}{x}$ . Then $f$ is continuous, and $f(\overline{E})=\{\frac{1}{x}|x\in N^{*}\}$ , while $\overline{f(E)}=\{\frac{1}{x}|x\in N^{*}\}\cup \{0\}$ . Hence $f(\overline{E})$ is a proper subset of $\overline{f(E)}$ .

Sample 2(Exercise 3.14 (a)):

If $s_n$ is a complex sequence, define its arithmetic means $\sigma_n$ by

$$ \begin{equation} \sigma _n=\frac{s_0+s_1+\cdot\cdot\cdot+s_n}{n+1}\hspace{1cm} (n=0,1,2,...). \end{equation} $$

If $\lim{s_n}=s$ , prove that $\lim{\sigma_n}=s$ .

Solution:

Given an $\epsilon_0>0$ .

$\lim{s_n}=s$ , that is to say, $\forall \epsilon_1>0$ , $\exists N_1>0$ , s.t. $\forall n\geq N_1$ , $\mid s_n-s\mid <\epsilon_1(1)$ .

We noticed that

$$ \begin{equation} |\sigma_n-s|=|\frac{\sum_{i=0}^{n}s_i-s}{n+1}|\leq \frac{\sum_{i=0}^{n}|s_i-s|}{n+1} \end{equation} $$

We now let $\epsilon_1=\frac{\epsilon_0}{2}$ , and then apply (1). Now we have

$$ \begin{equation} \begin{aligned} |\sigma_n-s|&\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|+\sum_{i=N_1}^{n}|s_i-s|}{n+1}\\ &\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|+(n-N_1+1)\epsilon_1}{n+1}\\ &\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|}{n+1}+\frac{\epsilon_0}{2} \end{aligned} \end{equation} $$

if $n\geq N_1$ .

Put $A_n=\frac{\sum_{i=0}^{N_1-1}\mid s_i-s \mid}{n+1}$ . We can see $\lim A_n=0$ , i.e., $\forall \epsilon_2>0$ , $\exists N_2>0$ , s.t. $\forall n\geq N_2$ , $A_n<\epsilon_2(2)$ . Then we let $\epsilon_2=\frac{\epsilon_0}{2}$ , and $N_0=max\{N_1,N_2\}$ . Hence, $\mid\sigma_n-s\mid \leq A_n+\frac{\epsilon_0}{2}\leq \frac{\epsilon_0}{2}+\frac{\epsilon_0}{2}=\epsilon_0$ , if $n\geq N_0$ .

Thus we obtain the conclusion that $\forall \epsilon_0>0$ , $\exists N_0>0$ , s.t. $\forall n\geq N_0$ , $\mid\sigma_n-s\mid<\epsilon_0$ , which means that $\lim{\sigma_n}=s$ .

test1

2019-08-10T15:45:59+00:00

Try to solve the equation $x+1=2$ :

$$ \begin{equation} \begin{aligned} x+1&=2 \\ x+1-1&=2-1 \\ x&=1\\ \end{aligned} \end{equation} $$

Then we get $x=1$ .