问:
Convergence of $p$-series.
I've been given a few problems of the form:

Find all values of $p$ such that the series converges:

$\sum_{k=1}^{\infty} \frac{\ln(k)}{k^p}$

$\sum_{k=1}^{\infty} \frac{1}{(c+k)^p}$ with $c > 0$

Naively, I'd assume that both series converge for $p > 1$ (following from that any $p$-series converges for $p > 1$). However, it's obviously (?) not that simple.

How does one approach such a problem?

答:
Hint:

For the first series (which is a Bertrand's series), use the integral test.

For the second series, use asymptotic equivalence:
$$(c+k)^p\sim_{k\to\infty}k^p, \quad\text{ whence }\quad\frac1{(c+k)^p}\sim_{k\to\infty}\frac1{k^p}.$$

Title: Convergence of $p$-series.


问:
Uniform convergence of $\frac{n^3x}{1+n^4x^2}$
I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.

So far i have attempted to prove pointwise converge as such;

Let $ N = 1/\epsilon x  $

For every $n \ge N$ we have $   \vert\frac{n^3x}{1+n^4x^2} - 0 \vert =   \frac{n^3x}{1+n^4x^2} < \frac{1}{Nx} = \frac{1}{(1/\epsilon x)x} = \epsilon  $

Is this enough of a proof for pointwise convergence? and if so, how do i prove that the convergence is uniform (if it is in this case)?

答:
Let $ n>0$.

For  $ x=0 $, $ f_n(0)=0$, so
$$\lim_{n\to+\infty}f_n(0)=0=f(0)$$

For $ x\in(0,1]$,

$$f_n(x)\sim \frac{n^3x}{n^4x^2}\sim \frac{1}{nx}$$
So,
$$\lim_{n\to+\infty}f_n(x)=\lim_{n\to+\infty}\frac{1}{nx}=0=f(x)$$

The convergence is pointwise to zero function at $ [0,1]$.

For the uniform convergence, let
$$g_n(x)=|f_n(x)-0|=\frac{n^3x}{1+n^4x^2}$$

the numerator of the derivative $ g'_n(x)$ is
$$n^3(1+n^4x^2)-2n^4x.n^3x=$$
$$n^3(1-n^4x^2)$$
Thus

$$M_n=\sup_{x\in[0,1]}g_n(x)=g_n(\frac{1}{n^2})=\frac n2$$

But $\lim_{n\to+\infty}M_n=+\infty$, then the convergence is Not Uniform at $[0,1]$.

ContentLicense: CC BY-SA 4.0


问:
If possible, write each of the vectors X_1, X_2 and X_3 as a linear combination of vectors in T
Let S={X_1,X_2,X_3} be a linearly independent a set of vectors in R^3 and T={Y_1,Y_2,Y_3} be a set of vectors in R^3, where Y_1=X_1+〖2X〗_2-2X_3, Y_2=-X_1-X_2+3X_3 and Y_3=-2X_1-〖3X〗_2+4X_3.

```<code>If possible, write each of the vectors X_1, X_2 and X_3 as a linear combination of vectors in T.
</code>```

After reducing the matrix, i get all zeros.
I would appreciate some help please
Thankyou

答:
Hints: Write the vectors as column vectors. Then you have:
$$
(X_1,X_2,X_3)\cdot\begin{pmatrix}1 &-1&-2\\2&-1&-3\\-2&3&4\end{pmatrix}=(Y_1,Y_2,Y_3).
$$
Let $A$ be the second matrix on the RHS above. You can check $\det(A)=-1$ so $A$ is invertible so, yes, you can  definitely write $X_1,X_2$, and $X_3$ as linear combinations of vectors in $T$. It remains to solve for nine numbers:
$$
\begin{pmatrix}1 &-1&-2\\2&-1&-3\\-2&3&4\end{pmatrix}\begin{pmatrix}b_1 &b_2&b_3\\b_4&b_5&b_6\\b_7&b_8&b_9\end{pmatrix}=I_3.
$$
Once you do that, then, for example, $X_1=b_1Y_1+b_4Y_2+b_7Y_3$. To get the $b_\ldots$, use the rule of matrix multiplication to obtain a 3-equation system for $\{b_1,b_4,b_7\}$, another 3-equation system for $\{b_2,b_5,b_8\}$, etc.


问:
Question about diagonalization of matrix
The following question is taken from Pg. no. 156 of "Linear Algebra Done Right (3rd Edition)" by Sheldon Axler.

Define $T\in L(R^2)$ by
\begin{equation}
T(x,y)= (41x + 7y, -20x + 74y)
\end{equation}
The matrix of $T$ with respect to the standard basis of $R^2$ is
\begin{equation}
\begin{pmatrix} 41 & 7 \\ -20 & 74 \end{pmatrix}
\end{equation}
which is not a diagonal matrix. However, $T$ is diagonalizable, because the matrix of $T$ with respect to the basis $(1, 4), (7, 5)$ is
\begin{equation}
\begin{pmatrix} 69 & 0 \\ 0 & 46 \end{pmatrix}
\end{equation}

Now I understand the first part that the matrix formed by standard basis is
\begin{equation}
\begin{pmatrix} 41 & 7 \\ -20 & 74 \end{pmatrix}
\end{equation}
since for $(1,0)$ we have $(41+0, -20+0)$ and for $(0,1)$ we have $(0+7, 0+74)$ and so the matrix is of the form given above.

However, in the second part for the basis $(1, 4), (7, 5)$, should it not be something like this
\begin{equation}
\begin{pmatrix} 41(1)+7(4) &  41(7)+7(5) \\ -20(1)+74(4) & -20(7)+74(5) \end{pmatrix}
\end{equation}

Also, how can we guess a basis for which a matrix is diagonalizable?

答:
You have $T(1,4) = (69,276) = 69(1,4)$, and similarly $T(7,5) = (322,230) = 46(7,5)$, which is why the resulting matrix is diagonal.

Also, how can we guess a basis for which a matrix is diagonalizable?

You don't have to guess: the basis consists of eigenvectors for the linear transformation.


问:
If $f$ measurable and $f=g$ a.e. implies $g$ measurable, then $\mu$ is complete
This is the Proposition 2.11 of the book Real Analysis from Folland.

The following implications are valid if and only if the measure is complete:

(a) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.

(b) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

The complete solution to this question can be found in this answer. But note that it was considered that $f,g:(X,\mathcal{M})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, other solutions like [1] or others that I found on the internet also assume that the image is $\mathbb{R}$ or $\overline{\mathbb{R}}$.

I'm thinking about the general case, where, $f:(X,\mathcal{M})\to(Y,\mathcal{N})$.
In this answer, the proposition $\mu$ is complete $\Rightarrow$ (a) is demonstrated in the general case.

I tried to show the opposite direction but I couldn't, in the general case is this proposition true? How to demonstrate this?

答:
If you only ask that there is some sigma algebra $(Y, \mathcal N)$ such that "for all $f , g : X \to Y$ such that $f$ measurable and $f = g$ a.e., we have $g$ measurable", in general you can't conclude that the measure on $X$ is complete. For example, when $Y$ is a singleton or another set with the coarse sigma-algebra, any $f, g : X \to Y$ are measurable, and the assumption gives no information.

As soon as $Y$ contains a nontrivial measurable set, the implication is true. The argument when $Y = \mathbb R$ still works: let $\mathbf 0$ be any nontrivial measurable subset of $Y$ with complement $\mathbf 1$, and take $0 \in \mathbf 0$ and $1 \in \mathbf 1$. When $A \subset X$ is contained in a set of measure $0$, take $f$ to be the constant function $0$ and $g$ to be $1$ on $A$ and $0$ on $X - A$. Then $f$ is constant, hence measurable, and $f = g$ a.e. so that $g^{-1}(\mathbf 1) = A$ is measurable.


问:
Least squares: How does the projection theorem lead to the normal equations?
I understand why $\vec{{P}}$ (the orthogonal projection of $\vec{{b}}$ on w)

is the closest vector to $\vec{{b}}$ (The vector that does not give us a solution to the
equation and that's why we need to do the process)

But I do not understand the equation:

$$A^TA\vec{{x}}=A^T \vec{{b}}$$
Why does multiplying each side of the original equation $A\vec{{x}}=\vec{{b}}$ (We know she has no answer)
by $A^T$ give us the closest answer?

Thank you

答:
Given vector $\vec b$ and matrix $A$, the goal is to find $\vec x$ so that $A\vec x$ is as close to $\vec b$ as possible.

Let $W$ be the subspace of all vectors of the form $A\vec x$. Observe that $W$ is the set of all linear combinations of columns of $A$.

We are looking for the vector in $W$ that is closest to $\vec b$. Write $A\hat x$ for this vector. By the projection theorem, $A\hat x$ is the orthogonal projection of $\vec b$ onto $W$. Equivalently, $A\hat x-\vec b$ is orthogonal to every vector in $W$. As a special case, we conclude:

$$
\text{$A\hat x-\vec b$ is orthogonal to every column of matrix $A$}.
\tag{*}$$

Constraint (*) can be written in matrix form as:
$$ A^T(A\hat x-b) = 0,$$ which leads to the equation $A^TA\hat x=A^T\vec b$.


问:
Finding PDF of sum of 2 uniform random variables
I need help understanding the question and its solution below.

Suppose that X is uniformly distributed in [0,a] and Y is uniformly
  distributed in [0,b], $0 < a \le b$, and that X and Y are independent.
  Find the PDF of $Z=X+Y$.

The solution is as follows.

For simplicity, define the unit square pulse

$$q(t) = u(t) - u(t-1)$$

where u(t) is the unit step function. Then the PDFs of X and Y are

$$f_X(x) = \frac{1}{a} q(\frac{x}{a})$$
$$f_Y(y) = \frac{1}{b} q(\frac{x}{b})$$

The PDF of Z is the convolution of $f_x$ and $f_Y$. i.e.

$$f_Z(z) = \int_{-\infty}^{\infty}\frac{1}{ab}q(\frac{\tau}{a})q(\frac{z-\tau}{b})d\tau$$

Therefore,

\begin{equation}
  f_Z(z) = \frac{1}{ab}\begin{cases}
    z, & \text{if $0<z\le a$}.\\
    a, & \text{if $a<z\le b$}.\\
    a+b-z, & \text{if $b<z\le a+b$}.\\
    0, & \text{elsewhere}.\\
  \end{cases}
\end{equation}

I have a few questions.

(1) How is the $f_X(x)$ and $f_Y(x)$ obtained? Why involve the unit step function instead of just taking $f_X(x) = \frac{1}{a}$ and $f_Y(x) = \frac{1}{b}$?

(2) I don't know how to get $f_Z$. I sketched out the 3 ranges for z.

I think my sketches must be wrong. My understanding of the convolution is that it is the resulting overlapping region between $f_X(x)$ and $f_Y(z-b)$ (the corresponding q functions seem to be a factor of $f_Y$ and $f_X$?), so I don't know how the overlapping region can be more than the red squares, bounded by $x = a$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
By definition, the answer is given by:
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{a}{1 \over a}
{\bracks{0 < z - x < b} \over b}\,\dd x
\,\right\vert_{\ 0\ <\ a\ \leq\ b}}
\\[5mm] = &\
{1 \over ab}\int_{0}^{a}
\bracks{z - b < x < z}\,\dd x
\\[5mm] = &\
{\bracks{z - b < 0} \over ab}
\bracks{0 < z < a}\int_{0}^{z}\dd x
\\[2mm] + &\
{\bracks{z - b < 0} \over ab}
\bracks{z > a}\int_{0}^{a}\dd x
\\[2mm] + &\
{\bracks{0 < z - b  a}\int_{z - b}^{a}\dd x
\\[5mm] = &\
{\bracks{0 < z < a} \over ab}\,z +
{\bracks{a < z < b} \over ab}\,a
\\[2mm] + &\
{\bracks{b < z < a + b} \over ab}
\,\pars{a + b - z}
\\[5mm] \substack{0\ <\ a\ \leq\ b
\,\,\,\,\,\,\,\,\,\, \\[1mm]
{\LARGE =}} &\
\!\!\!\!\!
\left\{\begin{array}{lcl}
\ds{z \over ab} & \mbox{if} & \ds{z \leq a}
\\[2mm]
\ds{1 \over b} & \mbox{if} & \ds{a < z \leq b}
\\[2mm]
\ds{{1 \over b} + {1 \over a} - {z \over ab}} & \mbox{if} & \ds{b < z < a + b}
\\[2mm]
\ds{0} && \mbox{otherwise}
\end{array}\right.
\end{align}
The following picture is a particular example with $\ds{\color{red}{a = 1}}$ and $\ds{\color{red}{b = 2}}$:

ContentLicense: CC BY-SA 3.0

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
By definition, the answer is given by:
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{a}{1 \over a}
{\bracks{0 < z - x < b} \over b}\,\dd x
\,\right\vert_{\ 0\ <\ a\ \leq\ b}}
\\[5mm] = &\
{1 \over ab}\int_{0}^{a}
\bracks{z - b < x < z}\,\dd x
\\[5mm] = &\
{\bracks{z - b < 0} \over ab}
\bracks{0 < z < a}\int_{0}^{z}\dd x
\\[2mm] + &\
{\bracks{z - b < 0} \over ab}
\bracks{z > a}\int_{0}^{a}\dd x
\\[2mm] + &\
{\bracks{0 < z - b < a} \over ab}
\bracks{z > a}\int_{z - b}^{a}\dd x
\\[5mm] = &\
{\bracks{0 < z < a} \over ab}\,z +
{\bracks{a < z < b} \over ab}\,a
\\[2mm] + &\
{\bracks{b < z < a + b} \over ab}
\,\pars{a + b - z}
\\[5mm] \substack{0\ <\ a\ \leq\ b
\,\,\,\,\,\,\,\,\,\, \\[1mm]
{\LARGE =}} &\
\!\!\!\!\!
\left\{\begin{array}{lcl}
\ds{z \over ab} & \mbox{if} & \ds{z \leq a}
\\[2mm]
\ds{1 \over b} & \mbox{if} & \ds{a < z \leq b}
\\[2mm]
\ds{{1 \over b} + {1 \over a} - {z \over ab}} & \mbox{if} & \ds{b < z < a + b}
\\[2mm]
\ds{0} && \mbox{otherwise}
\end{array}\right.
\end{align}
The following picture is a particular example with $\ds{\color{red}{a = 1}}$ and $\ds{\color{red}{b = 2}}$:


问:
What is the norm on the functional space used in defining the generator of a homogeneous Markov process?
From Wikipedia:

Given a strongly continuous semigroup $T : \mathbb{R}_+ \to L(B)$ on a Banach space $B$, its infinitesimal generator $A$ of a strongly continuous semigroup $T$ is defined as a mapping $B \to B$ as
  $$
    A\,x = \lim_{t\downarrow0} \frac1t\,(T(t)- I)\,x , \forall x \in B
$$
  whenever the limit exists wrt the norm on the Banach space $B$.

For a homogeneous Markov process $X_t, t \geq 0$ with state space $\mathbb R^n$, we can define a Banach space $B$ as the set of some measurable functions from $\mathbb R^n$ to $\mathbb R$ equipped with some norm (?), and a semigroup $T : \mathbb{R}_+ \to L(B)$ as
$$
[[T(t)](f)](x) := \mathbf{E}^{x} [f(X_{t})], \forall f \in B, \forall x \in \mathbb R^n
$$

From Wikipedia

The infinitesimal generator of the process $X$ is the operator $A$, which is defined to act on suitable functions $f : \mathbb R^n → \mathbb R$ by
  $$
    A f (x) = \lim_{t \downarrow 0} \frac{\mathbf{E}^{x} [f(X_{t})] - f(x)}{t}, \forall x \in \mathbb R^n $$

Compared to the generator for a general semigroup, we can try to rewrite the definition of $A$ as
$$
    A f  = \lim_{t \downarrow 0} \frac{ F(t,f) - f}{t},  $$
where $F(t,f)(x) := \mathbf{E}^{x} [f(X_{t})]$.
Now I was wondering what is the norm on $B$, the set of some measurable functions from $\mathbb R^n$ to $\mathbb R$, wrt which the above limit is taken?

Thanks and regards!

答:
In the most general definition, you can take $T_t:B\to B$ to be defined by

$$(T_tf)(x) = E^xf(X_t),$$

where $B$ is the space of all bounded and Borel measurable mappings from $\mathbb{R}^n$ to $\mathbb{R}$, equipped with the uniform norm. The uniform norm is almost invariably used, as this makes the space complete. In general, however, this space is much too large, and the generator will only exist on a much smaller set. The most useful theory is obtained by considering the restriction of $T_t$ to $C_0(\mathbb{R}^p)$, the subspace of $B$ of continuous mappings which vanish at infinity. When $T_t$ maps $C_0(\mathbb{R}^p)$ into itself and $t\mapsto T_tf$ is continuous at zero for all $f\in C_0(\mathbb{R}^p)$ the semigroup is said to be Feller. In this case, the generator exists on a dense subspace of $C_0(\mathbb{R}^p)$.

For more on this, see "Markov processes" by Ethier & Kurtz, or "Functional analysis for Probability and Stochastic Processes" by Bobrowski.


问:
If f has closed graph and is bounded, f is continuous and the domain is closed
Let $\mathbb{R}^n$ has Euclidean distance, $\emptyset \neq S \subset \mathbb{R}^n$ and function $f:S \to \mathbb{R}^m$.

I want to prove this.

If graph of $f$ (i.e. $\{(x,f(x)) \subset \mathbb{R}^{n+m} \mid x \in S \}$) is closed in $\mathbb{R}^{n+m}$ and $f$ is bounded, then $S$ is closed in $\mathbb{R}^n$ and $f$ is continuous.

Could anyone give me some ideas ?

I proved that if $f$ is continuous and $S$ is closed, then graph of $f$ is closed, but I do not have ideas about the converse proposition.

答:
Let $G$ be the graph of $f$. If $S$ is not closed, let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $S$ converging to some point $x\in\Bbb R^n\setminus S$. The sequence $\langle f(x_n):n\in\Bbb N\rangle$ is bounded in $\Bbb R^m$, so it has a convergent subsequence, and without loss of generality we may assume that it actually converges, say to $y$. Then $\big\langle\langle x_n,f(x_n)\rangle:n\in\Bbb N\big\rangle$ in $G$ of $f$ converges to $\langle x,y\rangle$, and since $G$ is closed, $\langle x,y\rangle\in G$, and therefore $x\in S$. This contradiction shows that $S$ must be closed.

Now let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a convergent sequence in $S$ with limit $x\in S$. We’d like to show that $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(x)$, as that would show that $f$ is continuous. As a first step we’ll show that $\sigma$ has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $\langle f(x_{n_k}):k\in\Bbb N\rangle$ converges to $f(x)$.

The sequence $\big\langle\langle x_n,f(x_n)\rangle:n\in\Bbb N\big\rangle$ in $G$ is bounded, so it has a convergent subsequence $\big\langle\langle x_{n_k},f(x_{n_k})\rangle:k\in\Bbb N\big\rangle$, and since $G$ is closed, this subsequence must converge to some $\langle x,y\rangle\in G$. Then $y=f(x)$, since $\langle x,y\rangle\in G$, so $\langle f(x_{n_k}):k\in\Bbb N\rangle$ converges to $f(x)$. This shows that every sequence in $S$ converging to $x$ has a subsequence whose image under $f$ converges to $f(x)$.

If $\sigma$ does not converge to $f(x)$, there are an open nbhd $U$ of $f(x)$ and a subsequence $\langle f(x_{m_k}):k\in\Bbb N\rangle$ of $\sigma$ such that $f(x_{m_k})\notin U$ for each $k\in\Bbb N$. But then $\langle x_{m_k}:k\in\Bbb N\rangle$ is a sequence in $S$ converging to $x$ that cannot have a subsequence whose image under $f$ converges to $f(x)$, which is impossible. Thus, $\sigma$ converges to $f(x)$, and $f$ is continuous.


答:
One strategy to check for uniform convergence is to find the maximum $M_n$ of $f_n(x) = \frac{n^3x}{1 + n^4 x^2}$ over $[0,1]$ using calculus, then check whether $M_n \to 0$ (keeping in mind that $0 < f_n(x) \leq M_n$ for all $x$).

That said, the same idea works with any upper bound for $f_n(x)$ that approaches $0$ as $n\to \infty$ and does not depend on $x$. For instance, one upper bound can be attained by combining an upper bound over $[0,1/n^2]$ and another over $[1/n^2,1]$.

With the first approach we find that
$$
f_n'(x) = \frac{n^3(1 + n^4x^2) - n^3x(2n^4x)}{[1 + n^4 x^2]} = 0 \implies\\
n^3(1 + n^4 x^2)  = 2 n^3 \cdot n^4 x^2 \implies\\
(1 + n^4 x^2)  = 2 n^4 x^2 \implies\\
1 = n^4 x^2 \implies\\
x = \pm 1/n^2.
$$
We can see that the critical point at $x = 1/n^2$ is indeed a maximum since $f_n'(0) > 0$ and $f_n'(1) < 0$. Thus, our maximum is
$$
M_n = f_n(1/n^2) = \frac{n^3 \cdot \frac 1{n^2}}{1 + n^4 \cdot \frac 1{n^4}} = \frac n2.
$$
So, we find in fact that $M_n$ does not converge to $0$, which means that $f_n$ fails to converge uniformly.

Title: Uniform convergence of $\frac{n^3x}{1+n^4x^2}$

Body:
I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.

So far i have attempted to prove pointwise converge as such;

Let $ N = 1/\epsilon x  $

For every $n \ge N$ we have $   \vert\frac{n^3x}{1+n^4x^2} - 0 \vert =   \frac{n^3x}{1+n^4x^2} < \frac{1}{Nx} = \frac{1}{(1/\epsilon x)x} = \epsilon  $

Is this enough of a proof for pointwise convergence? and if so, how do i prove that the convergence is uniform (if it is in this case)?


问:
Polya's Urn: Need Help Calculating Expected Value of Concentration of Red Balls at Stage n
An urn U contains $r_0$ red balls and $g_0$ green balls. At each stage a ball is selected at random from the urn, we observe its color, we return it to the urn and then we add another ball of the same color. We denote by $b_n$ the total number of balls in U at stage n, by $R_n$ the number of red balls and by $G_n$ the number of green balls at stage n. Finally, we denote by $C_n$ the "concentration" of red balls at stage n,

$C_n$ = $\frac{R_n}{b_n}$ = $\frac{R_n}{R_n + G_n}$

Show that E[$C_n$] = $\frac{r_0}{r_0 + g_0}$ for all n ∈ N

I'm not sure where to begin with this problem. If anyone could walk me through it, or at least give me a place to start, that would be greatly appreciated.

If it helps, the first part of this problem (this is the second part) showed that E[$C_{n+1}|R_n = i$] = $\frac{i}{b_n}$. I already solved that part of the problem.

答:
HINT:

We are adding a ball of the same color based on their initial probability of being picked up.

Take an example,

Concentration of red balls to start with, $C_0 = \displaystyle \frac{r_0}{r_0 + g_0}$

After first draw and addition of an extra ball, expected number of red balls

$ E_1 = \displaystyle \frac{r_0}{r_0 + g_0} (r_0 + 1) + \frac{g_0}{r_0 + g_0} (r_0) = \frac{r_0(r_0 + g_0 + 1)}{r_0 + g_0}$

(If we draw red with the probability of $r_0 / (r_0 + g_0)$, we have $r_0 + 1$ red balls otherwise if we draw green, we continue to have $r_0$ red balls.

Total number of balls after first draw $ = r_0 + g_0 + 1$

So, $C_1 = \displaystyle \frac{r_0}{r_0 + g_0}$ which is same as we started.


答:
Since $1/n \in [0,1]$,

$$\sup_{x \in [0,1]}\left|\frac{n^3x}{1+n^4x^2} \right| \geqslant \frac{n^3(1/n)}{1+n^4(1/n)^2} = \frac{n^2}{1+n^2} \underset{n \to \infty}{\longrightarrow}  1 \neq 0,$$

the convergence is not uniform.


答:
Here are some pointers.

For the first problem, you can immediately observe that $p > 1$ for the series to converge, but is this tight? To answer this question, you may want to proceed by contradiction and ask yourself the question "if for some $ p \geq 1 + \epsilon > 1$, the series did not converge, what would this imply about the $p$-series with $p = 1 + \epsilon / 2 > 0$"? I'll flesh this out with an example: suppose, say this series did not converge for $p = 1.01$. Then for sufficiently large $k$, say $k \geq k_0$, we have that $k^{0.005} > \ln k$, so
$$\sum_{k = k_0}^\infty \frac{k^{0.005}}{k^{1.01}} = \sum_{k = k_0}^\infty \frac{1}{p^{1.005}} > \sum_{k = k_0}^\infty \frac{\ln k}{k^{1.01}} = \infty$$
so this would imply that the $p$ series with $p = 1.005$ does not converge, which we know is not true. The key idea is that $\ln k$ grows slower than any polynomial in $k$. I'll let you fill in the details for the general case.

For the second problem, we can immediately see that it converges for $p > 1$ by the direct comparison test, but now we have to ask if it could converge for $p \leq 1$. We claim that it does not converge for $p = 1$, and as a result does not converge for $p \leq 1$. To see this, note that every term of the series is at least $1 / (\lceil c \rceil + k)$. From here, can you see that this series is always divergent? For an explicit example, consider $c = 5.073$. Then the series
$$\sum_{k = 1}^{\infty} \frac{1}{c + k} \geq \sum_{k = 1}^{\infty} \frac{1}{6 + k} = \frac 1 7 + \frac 1 8 + \frac 1 9 + \cdots$$
Is it clear why this series must be divergent?

Title: Convergence of $p$-series.

Body:
I've been given a few problems of the form:

Find all values of $p$ such that the series converges:

$\sum_{k=1}^{\infty} \frac{\ln(k)}{k^p}$

$\sum_{k=1}^{\infty} \frac{1}{(c+k)^p}$ with $c > 0$

Naively, I'd assume that both series converge for $p > 1$ (following from that any $p$-series converges for $p > 1$). However, it's obviously (?) not that simple.

How does one approach such a problem?


问:
Prove that $AM = MF$ where $M$ is the intersection between the perpendicular of circle $ω$ and line $AF$

Given is circle $ω$ and line $k$ disjoint with circle $ω$. Point $M$ is
the foot of the perpendicular from the center of the circle onto line $k$. A line
intersects line $k$ at point $A$ and circle $ω$ at points $B$ and $C$. Lines $\overline{BM}$ and
$\overline{CM}$ intersect circle $ω$ for the second time at points $D$ and $E$. Line $\overline{DE}$ meets
line $k$ at point $F$. Prove that $\overline{AM} = \overline{MF}$.

Once again, I've been assigned another set of geometry problems, and this is one of the harder ones that I haven't been able to figure out yet. I've asked my instructor what I could use for this problem, and he told me that Power of a Point would be useful here. But so far, I've tried applying Power of a Point, but I'm not getting anywhere with the few points that currently exist. Can anyone help out?

答:
Menelaus for triangle $AFP$ and transversal $M-E-C$: $${AM\over MF}\cdot {EF\over EP}\cdot {CP\over CA}=1$$ and Menelaus for triangle $AFP$ and transversal $M-B-D$: $${AM\over MF}\cdot {DF\over DP}\cdot {BP\over BA}=1$$

IF we multiply these equation we get:
$$\Big({AM\over MF}\Big)^2\cdot {DF\cdot EF\over \color{red}{EP\cdot DP}}\cdot {\color{red}{CP\cdot BP}\over CA\cdot BA}=1$$

By the power of the point $P$ with respect to circle $\omega $ we have:

$$\color{red}{ BP\cdot CP  = DP\cdot EP}$$ so we have
$$\Big({AM\over MF}\Big)^2\cdot {DF\cdot EF\over CA\cdot BA}=1$$

By the power of the points $A$ and $F$ we have: $$CA \cdot BA = AO^2-r^2 = x^2+d^2-r^2$$
and $$DF\cdot EF = FO^2-r^2 = y^2+d^2-r^2$$ so we have now $${x^2\over y^2}\cdot {y^2+d^2-r^2\over x^2+d^2-r^2} = 1\implies x^2= y^2$$
and we are done.


问:
The error in Taylor polynomial of $\sin x$ is $10^{-6}$...

If we want to expand $f(x)=\sin x$ with Taylor polynomial centered at
$x=0$. and error of that for$x=\frac{\pi}{2}$ be at most $10^{-6}$.
then what order of polynomial should be at? ($n=?$)

As a hint our lecturer said we should use $c=\frac{\pi}{2}$ at formula of remainder of Taylor series.

Here is my work:

The reminder of Taylor series for the function $f(x)$ where $x_0=0$ is :

$$R_n(x)=\cfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$

And I estimate $\frac{\pi}{2}$ to closest integer which is $2$ for the value of $x$.

To find $n$ I should solve inequality:

$$|R_n(x)|\le 10^{-6}\quad\rightarrow\quad|\cfrac{f^{(n+1)}(\frac{\pi}{2})}{(n+1)!}\times2^{n+1}|\le 10^{-6}$$

$(n+1)$'st derivative of $f(x)=\sin x$ can be one of the $\pm\sin x$ ,$\pm\cos x$. So $|f^{(n+1)}(\frac{\pi}{2})|\in\{0,1\}$. If it is equal to zero then we have $0\le 10^{-6}$ and it is absolutely wrong so the best I can do is using $|f^{(n+1)}(\frac{\pi}{2})|=1$. hence the inequality is $\left|\cfrac{2^{n+1}}{(n+1)!}\right|\le 10^{-6}$ . After plugging in some $n$s I got $n\ge13$. so the order of the Taylor polynomial should be $13$.

Is my approach right ? as you can see in my answer I had some difficulties to find $|f^{(n+1)}(\frac{\pi}{2})|$ did I right at this point?

答:
As $c$ can be anywhere in $[0,\frac\pi2]$, the best you can say is

$$\left|f^{(n+1)}(c)\right|\le1.$$

Now we can use the exact value of $\frac\pi2$ (unless a calculator is forbidden), and the possible remainders are bounded by the inverses of

$$8.11E-01,
3.94E+00,
4.79E+01,
1.09E+03,
3.97E+04,
2.12E+06
$$

and the order $11$ suffices.

Notice that as this is an alternating series, you need not precompute the number of terms, and you can stop when the current term becomes smaller than the requested error.


问:
Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?
Suppose for all real $x$ that $g^{(n)}(x) = g^{(n-1)}(x)$ and $g^{(n-1)}(0) = -1$ where $g^{n}(x)$ is the nth derivative. Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?

Isn't this answer just the form $g(x) = -e^x$? and Hence $\lim_{x \to \infty} \frac{g(x)}{e^x} = -1$? How could I more rigorously show this?

答:
The taylor series of a function is given as:
$$g(x)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n$$
so in your case:
$$g(x)=-\sum_{n=0}^\infty\frac{x^n}{n!}=-e^x$$
so:
$$\frac{g(x)}{e^x}=-1$$
which is not dependent on $x$ and so converges for all $x$


问:
Applications of design theory
I have recently started reading up on design theory, with the ultimate purpose of doing some original research in that area. I understand the mathematics fairly well, but am not understanding the application. Can someone here explain that? Alternatively are there some articles, web-sites, papers explaining the application of design theory in agriculture, biological experiments etc.

For example after establishing the existence of/constructing a certain type of design mathematically, how do we use that knowledge in practice?

答:
EDIT: replaced dead link.

See this site; "The extract deals with her interest in how a simple mathematical model applies to the weaving process and the use of combinatorial designs for agricultural experiments."

Title: Applications of design theory


问:
A category which direct limits but no general colimits
I am looking for a (at best, real life) category that has direct limits, but no general small colimits, or a category that has inverse limits, but no general small limits. Are there any interesting examples that are not too obviously made to be an example for this?

I am asking this because I wonder why many lectures include the construction of direct/inverse limits as an exercise, instead of general small (co)limits. In particular, I wonder this for topology and algebra classes.

答:
1- Consider a first order functional (meaning it only has function symbols) language $L$, and a first order theory $T$.

If $T$ is especially simple, e.g. it's an equational theory, then the category of models of $T$ has all colimits, it's even presentable.

If we make $T$ more complicated, it might not have all colimits; but if we keep it simple enough, it'll still have filtered colimits. Alex Kruckman's example of fields is a particular example of this situation, where the "extra" axiom (on top of the equational ones) is (for instance) $\forall x, \exists y,( xy = 1 \lor x= 0)$

Now let's take this example and generalize it to find other examples. Suppose you have  formula $\varphi(x_1,...,x_n,y_1,...,y_m)$ built from atomic formulas using only $\land, \lor$. Then the models of $\forall x_1,...,\forall x_n, \exists y_1,...\exists y_m, \varphi(x_1,...,y_m)$ has all filtered colimits, and they're computed as in sets. This is quite easy to prove: just prove by structural induction on the formula that it is satisfied in the filtered colimit structure for a particular tuple $(a_1,...,a_n,b_1,...,b_m)$ if and only if it is at some finite stage.

But they might not have arbitrary colimits.
For instance one can encode a cardinality bound with such a formula ($\forall x_1,...,x_n, \bigvee_{i,j}x_i= x_j$ encodes a bound of cardinality $<n$). So "groups of cardinal $\leq n$" for a fixed $n$ are an instance which of course doesn't have all colimits.

Another somewhat less silly example is the category of groups of exponent "either $2$ or $3$", which you can encode with the formula $\forall x, \forall y (x^2 =1 \lor y^3 = 1)$. Now this doesn't have coproducts (can you figure out why ?).

2- Another example I like is an example which relates the question of having (co)limits and preserving (co)limits for a functor. There are nice examples of functors that commute with filtered colimits but not all colimits (e.g. taking fixed points of a $G$-action in $G$-sets, or taking global sections of a sheaf, etc.). I claim that these examples provide examples of categories which don't have all colimits, but that have filtered ones.

Indeed let $F:C\to D$ be a functor, and let $E$ be the category whose objects are $Ob(C)\coprod Ob(D)$, where an arrow between objects of $C$ is an arrow in $C$, an arrow between objects in $D$ is an arrow in $D$, and an arrow $c\to d$ is an arrow $F(c)\to d$ (and there are no arrows from an object in $D$ to an object in $C$).

Let $K: I\to C$ be a diagram, and assume it has a colimit in $C$. Then $F$ preserves that colimit if and only if the obvious diagram $I\to C\to E$ has a colimit. Indeed, let $Q, j_i: K(i)\to Q$ be a colimit for $D$ in $C$; and assume $F(Q),F(j_i)$ forms a colimit in $D$. Then $Q, j_i$ forms a colimit in $E$. Indeed the universal property is obviously satisfied for objects of $C$, and for objects of $D$ by preservation.

Conversely, assume $I\to C\to E$ has a colimit. Then, since it maps to the objects in the diagram,it must be in $C$. Then by analyzing the universal property, one can easily see that it has to be a colimit in $C$, and that it must be preserved by $F$.

In particular, suppose $C,D$ are cocomplete, then $F:C\to D$ preserves (filtered) colimits if and only if $E$ has them. So if $F$ preserves filtered colimits but not general ones, as in the examples I mentioned (let me mention some others: the functor taking a category to its core groupoid, the functor taking a ring $R$ to its $K_0$, most forgetful functors from algebra to sets,...), $E$ has filtered colimits, but not general ones.

As Kevin Arlin points out in the comment below, it's extremely frequent for right adjoints between presentable categories to preserve filtered colimits (it's equivalent to the left adjoint preserving compact objects), but much less frequent for them to preserve arbitrary colimits.


答:
@nosuchthingasmagic gives a link to a question that gives a calculus-based derivation. Here's another one.

The standard notation is $\vec y = X \vec {\beta}$. Then we note that this has no exact solution, so really $X \vec {\beta} = \vec y + \vec {\epsilon}$. If $\vec \beta$ is optimal, then $X^T \vec {\epsilon}=\vec 0$. So when we multiply both sides by $X$, $\epsilon$ disappears and we're left with $X^TX \vec {\beta} = X^T\vec y$.

Why $X^T \vec {\epsilon}=\vec 0$? Well, that's saying that every row of $X^T$ is orthogonal to $\vec {\epsilon}$. And the rows of $X^T$ represent all the observations of a particular feature. So this is saying that for each feature, the observations of that feature are uncorrelated with the errors. If there were a correlation, we could adjust the corresponding $\beta_i$ to get a better fit.

Title: Least squares: How does the projection theorem lead to the normal equations?

Body:
I understand why $\vec{{P}}$ (the orthogonal projection of $\vec{{b}}$ on w)

is the closest vector to $\vec{{b}}$ (The vector that does not give us a solution to the
equation and that's why we need to do the process)

But I do not understand the equation:

$$A^TA\vec{{x}}=A^T \vec{{b}}$$
Why does multiplying each side of the original equation $A\vec{{x}}=\vec{{b}}$ (We know she has no answer)
by $A^T$ give us the closest answer?

Thank you


答: Just note that for $(x_n)=1/n,\ \frac{n^3x_n}{1+n^4x_n^2}=\frac{n^2}{1+n^2}\to 1\neq 0$ as $n\to \infty$ so the convergence is not uniform.


问:
$ \cos x\geq 1-\frac{x^2}{2} $
Prove that for $x\in\mathbb{R}$
$$
\cos x\geq 1-\frac{x^2}{2}.
$$

My try:

Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$
If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.

But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$

答:
Note
$$\frac{x^2}2-1 + \cos x =\int_0^x (t - \sin t)dt  = \int_0^x dt\int_0^t (1-\cos s)ds \ge 0$$
Thus
$$ \cos x\geq 1-\frac{x^2}{2} $$

Title: $ \cos x\geq 1-\frac{x^2}{2} $


问:
accumulation point of recursive sequence
Given is a sequence with:

$(a_0)=1$, $(a_1=1)$, $a_{n+2}=\frac{1+a_{n+1}}{a_n}$

I now have to show what the accumulation points are:

I guess that the sequence is jumping from number to number like this:
1->1->2->3->2->1->1->2..

So the acc.points should indeed be "1, 2 and 3".

Is this correct? If yes, how can I 'show' this?
Furthermore: Is there any way to build some subsequence of $a_n$ that converges against "1, 2 and 3" ?

Thank you :)

答:
Prove by induction that for $n=4,9,14,\dots$, $a_n=3$, that for $n=0,1,5,6,10,11,\dots$, $a_n=1$ and that for the remaining indices, $a_n=2$. That is all you want to do.

As per the accumulation points, recall that a number $\alpha$ is an accumulation point or limit point of a sequence $\langle a_n\rangle$ if there exists a subsequence of $\langle a_n\rangle$ that converges to $\alpha$. Thus, $1,2,3$ are accumulation points of $\langle a_n\rangle$.


问:
Astroid area with Green Theorem
I know that the area of an astroid can be calculated by Green's theorem through the integral: $$ A =  \frac{3a^2}{2}\int_{0}^{2\pi}[\cos^4t \sin^2t+\sin^4t\cos^2t]dt$$.

However, I found the following expression in the class notes of a former teacher of mine: $$ A =  \frac{3a^2}{2}\int_{0}^{\frac{\pi}{2}}[1-\cos^2(2t)]dt$$

This integral results in the appropriate area: $\frac{3a^2 \pi}{8}$, as well as the first integral. However, I could not understand if this second integral does in fact make sense. Can you help me understand?

答:
There is a match and both the answers are correct since
$$
{
\cos^4 x\sin^2x+\sin^4 x\cos^2x=
\\\cos^2 x\sin^2x(\cos^2 x+\sin^2x)=
\\\cos^2 x\sin^2x=
{1\over 4}\sin^2 2x=\\
{1\over 4}(1-\cos^2 2x)
}
$$
Note that the integral bounds have changed to lead to a same answer.

Title: Astroid area with Green Theorem


问:
Slope of tangent line in a function

Find the slope of the tangent line for the point $(\frac{1}{4},
> g(\frac{1}{4}))$ where $g(x) = \frac{x^2e^{4x}}{4}$.

I "did" this in two ways: by definition and by deriving. I can derive $g$ and check the derivative on $x=\frac{1}{4}$. But I'm stuck doing it by the definition.

This is my attempt so far:

$$\lim_{x \rightarrow \frac{1}{4}}  \frac{g(x) - g(\frac{1}{4})}{x-\frac{1}{4}} = \lim_{x \rightarrow \frac{1}{4}} \frac{ \frac{x^2e^{4x}}{4} - \frac{e}{4 \times 16}}{x-\frac{1}{4}} = \lim_{x \rightarrow \frac{1}{4}} \frac{ x^2e^{4x} - \frac{e}{16}}{4x-1} = \lim_{x \rightarrow \frac{1}{4}} \frac{ 16x^2e^{4x} - e}{16(4x-1)} = \lim_{x \rightarrow \frac{1}{4}} \frac{ (16x^2e^{4x-1} - 1)e}{16(4x-1)}$$

答:
Hint

You can either use the L^Hopital's rule or the following approximation:
$$
e^u\approx 1+u\quad,\quad |u|<<1
$$
where $u=4x-1$.

Title: Slope of tangent line in a function


答:
With the substitution $$u=t-\pi$$

$\int_0^{2\pi}f(t)dt$ becomes $\int_{-\pi}^{\pi}f(u)du$

the integrand is an even function, so, it gives
$$2\int_0^{\pi}f(u)du$$

A new substitution $ v=u-\frac{\pi}{2}$

gives $$2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f(v)dv$$
and finally

$$4\int_0^{\frac{\pi}{2}}f(v)dv$$

Title: Astroid area with Green Theorem

Body:
I know that the area of an astroid can be calculated by Green's theorem through the integral: $$ A =  \frac{3a^2}{2}\int_{0}^{2\pi}[\cos^4t \sin^2t+\sin^4t\cos^2t]dt$$.

However, I found the following expression in the class notes of a former teacher of mine: $$ A =  \frac{3a^2}{2}\int_{0}^{\frac{\pi}{2}}[1-\cos^2(2t)]dt$$

This integral results in the appropriate area: $\frac{3a^2 \pi}{8}$, as well as the first integral. However, I could not understand if this second integral does in fact make sense. Can you help me understand?


问:
complex holomorphic function which only has finite roots
Suppose that D is a bounded region, f $\in$ H(D)$\bigcap$C($\bar D$).Prove that f has only finite roots if f$\neq$0 on $\partial D$.

答: Assume $f$ would have an infinite number of distinct roots. Let $(z_i)$ be a sequence of distinct roots. Since $\bar D$ is compact there exists a sub sequence of $(z_i)$ with a limit $\zeta$. Since $f$ is continuous on $\bar D$, we have $f(\zeta)=0$. Hence lies in the interiour of $D$. A holomorphic function on a region with an accumulation point of zeros inside that region is constant. Hence $f=0$. End of proof. </p>


问:
Prove $2^n = O(n!)$
I proved by induction that for all $n \leq 4$,
$2^n+1 \lt (n+1)!$, therefore $2^n \lt n!$, but I just don't know how to prove the big $O$.

I know the definition that there exists $c$ for which $n! \leq c \cdot 2n$, but I just don't understand how to find $c$.

答:
Hint:

To go from $2^n$ to $2^{n+1}$, you multiply by $2$.

To go from $n!$ to $(n+1)!$, you multiply by $n+1$.

For almost all $n$, $n+1\geq 2$.

Title: Prove $2^n = O(n!)$


答:
You got it backwards, and wrote $2n$ instead of $2^n$.  So you want $2^n\le c\cdot n!$.  But if you write it out, you have $2\cdot2\dots2/(n\cdot (n-1)\dots2\cdot1)=2/n\cdot2/(n-1)\cdot\dots2/2\cdot2/1\le2$.  So we are done.

Actually, we also have that, since $n!$ increases faster than $2^n$, $2^n=o(n!)$.  But when one function is little-oh of another function, that implies that it is also big-oh of it.

Title: Prove $2^n = O(n!)$

Body:
I proved by induction that for all $n \leq 4$,
$2^n+1 \lt (n+1)!$, therefore $2^n \lt n!$, but I just don't know how to prove the big $O$.

I know the definition that there exists $c$ for which $n! \leq c \cdot 2n$, but I just don't understand how to find $c$.


问:
Quadratic equation where constant term is an integral function
Consider a quadratic equation ${x^2} + 2x = k + \int\limits_0^1 {\left| {t + k} \right|dt}$. Then choose the correct option(s),

(A) Roots are Real

(B) Roots are Imaginary

(C) Roots are Distinct

(D) Roots are complex number

My approach is as follow

Let $-c=k + \int\limits_0^1 {\left| {t + k} \right|dt}$

$a=1$ & $b=2$

$b^2-4ac$

$4-4(k + \int\limits_0^1 {\left| {t + k} \right|dt})$

How do I check the nature of the roots

答:
As pointed out in comments and in John Omielan's answer, to show that there are always two roots, it suffices to show that $\newcommand{\abs}[1]{|#1|}\newcommand{\dd}{\; \mathrm{d}} $
$$1+k+\int_0^1 \abs{t+k} \dd t \ge 0$$ for every $k\in\mathbb R$.
From triangle inequality (or reverse triangle inequality) we get that $\abs{t+k}\ge \abs{k}-\abs{t}$. Since we are looking only at $t$ between $0$ and $1$, we can actually use $\abs{t+k}\ge \abs k - t$  to get
\begin{align*}
1+k+\int_0^1 \abs{t+k} \dd t \\
&\ge  1+k+\int_0^1 (\abs{k}-t) \dd t \\
&= 1+k+\int_0^1 \abs{k} \dd t  - \int_0^1 t \dd t \\
&= 1+k+\abs{k}  - \frac12\\
&= \frac12+k+\abs{k}  \\
&\ge \frac12
\end{align*}
(We have used $x+\abs x\ge 0$.)

I wouldn't be trying to find some kind of proof which does not need to be split in the cases if I hadn't seen in the other answer that this always has two distinct real roots. So the credit should go there.

As a side note, I will also mention that there is this older questions: Comment upon the nature of roots of the quadratic equation $x^2+2x=k+ \int_0^1 |t+k| \, dt$ depending on the value of $k \in \mathbb{R}$.
However, it is now deleted, so it's visible only to 10k+ users. One interesting thing I see there is that $\int |x| \dd x = \frac{x\cdot|x|}2$ is used there to evaluate the integral.


问:
If 9.999... = 10, then is there a general proof for any number that has infinite trailing 9s?
I've read about $9.999...=10$, and I would say that I understand it. However, I am looking to apply that proof to all real numbers with trailing 9s. For example:

$72.999...=73$

I have the following potential proof for this particular scenario:

$73/3=x$

$x=24.333...$

$3x=72.999...$

$72.999...=73$

This works, but how would I create a general proof for all numbers that have an infinite trail of 9s?

答:
Let $x = k.999999\ldots $ for any integer $k$.

Compute $10 x - x$.


问:
Identity regarding Gamma function (kind of)
During some work (i.e. trying to prove series expansion of a certain function) I stumbled upon this identity, which I can't however prove:

$$
\sum\limits_{i=0}^k \frac{(-1)^i}{(n+2i)}\cdot \frac{1}{i!(k-i)!} = \frac{2^k}{n(n+2)(n+4)\ldots(n+2k)}.
$$

Here $n$ is just a fixed positive integer (a dimension of a space, but it's not important). It works in low cases, but all I can do is simply rearrange it term by term, which obviously isn't optimal for general case. I tried induction and some combinatorics tricks, but I'm not good enough at using them.

The reason for Gamma function in the title is that the Maple simply evaluates both sides as
$$
\frac{\Gamma\left(\frac{n}{2}\right)}{2 \cdot \Gamma \left(\frac{n}{2}+k+1\right)}.
$$
I can see it at the right hand side, but the relation of the sum on the LHS to the Gamma function seems out of the blue to me. Does anybody have any hints or ideas for that identity?

答:
You want to multiply both sides by $k!$. Then the LHS becomes

$$\sum_{i=0}^k \frac{(-1)^i}{n + 2i} {k \choose i}$$

which is a bit easier to understand. Starting from the binomial expansion $(1 - x)^k = \sum_{i=0}^k (-1)^i {k \choose i} x^i$ we see that the LHS is

$$I(n, k) = \int_0^1 x^{n-1} (1 - x^2)^k \, dx.$$

This is a variant of the beta function integral. It can be converted into a beta function integral on the nose using the substitution $u = x^2$, which gives

$$I(n, k) = \frac{1}{2} \int_0^1 u^{ \frac{n}{2} - 1 } (1 - u)^k \, du = \frac{1}{2} B \left( \frac{n}{2}, k+1 \right) = \frac{1}{2} \frac{\Gamma \left( \frac{n}{2} \right) \Gamma(k+1)}{\Gamma \left( \frac{n}{2} + k + 1 \right)}$$

and after dividing by $k! = \Gamma(k+1)$ we get Maple's answer.

Edit: For a direct proof starting from the RHS, replace $n$ with a complex variable $z$. Then the LHS is the partial fraction decomposition of the RHS; that is, we in fact have

$$\sum_{i=0}^k \frac{(-1)^i}{i! (k-i)! (z + 2i)} = \frac{2^k}{z(z + 2) \dots (z + 2k)}$$

for complex $z \neq -2i$. To see this it suffices to compute

$$\lim_{z \to -2i} \frac{2^k (z + 2i)}{z(z + 2) \dots (z + 2k)} = \frac{2^k}{(-2i)(-2i + 2) \dots (2k - 2i)} = \frac{(-1)^i}{i! (k-i)!}.$$

Edit #2: Another direct proof, this time starting from the LHS. If $a_n$ is a sequence, write $(\Delta a)_n = a_{n+1} - a_n$ for its forward finite difference. Then we have the general identity

$$(\Delta^k a)_0 = \sum_{i=0}^k (-1)^{k-i} {k \choose i} a_i.$$

Setting $a_i = \frac{1}{n + 2i}$ gives the LHS times $(-1)^k k!$. Now we can argue by induction on $k$: we have

$$(\Delta a)_i = \frac{1}{n + 2i + 2} - \frac{1}{n + 2i} = \frac{-2}{(n + 2i)(n + 2i + 2)}$$
$$(\Delta^2 a)_i = \frac{-2}{(n + 2i + 2)(n + 2i + 4)} - \frac{-2}{(n + 2i)(n + 2i + 2)} = \frac{2^2 2!}{(n + 2i)(n + 2i + 2)(n + 2i + 4)}$$

and in general by induction

$$(\Delta^k a)_i = \frac{(-1)^k 2^k k!}{(n + 2i)(n + 2i + 2) \dots (n + 2i + 2k)}.$$

Setting $i = 0$ then gives the desired identity. As in the previous argument $n$ can be a complex number $z \neq -2i$.

The $2$s are in some sense a red herring. Really this identity is a mild variant of the identity

$$\sum_{i=0}^k \frac{(-1)^i}{z + i} {k \choose i} = \frac{k!}{z(z + 1) \dots (z + k)}.$$

which can be proven using any of the above approaches. Then substitute $z \mapsto \frac{z}{2}$ and clear denominators appropriately. This identity is probably quite classical but I don't know if it has a name; it's related to the Norlund-Rice integral. After substituting $z \mapsto - \frac{1}{z}$ and clearing denominators appropriately, it can be thought of in terms of one of the generating functions of the Stirling numbers of the second kind.


答:
First of all, assume that $x\in(0,1)$. Then $x=0.d_1d_2\ldots d_n99999\ldots$, with $d_n\ne9$. Therefore $10^nx=a+0.999\ldots$, where $a=10^{n-1}a_1+10^{n-2}a_2+\cdots+a_0$. So, $10^nx=a+1$, and therefore $x=0.d_1d_2\ldots d_{n-1}(d_n+1)$.

In the general case, $x$ can be written as $m+x'$, with $m\in\Bbb Z$ and where $x'\in(0,1)$ has trailing $9$'s. So, the previous argument aples to $x'$.

Title: If 9.999... = 10, then is there a general proof for any number that has infinite trailing 9s?

Body:
I've read about $9.999...=10$, and I would say that I understand it. However, I am looking to apply that proof to all real numbers with trailing 9s. For example:

$72.999...=73$

I have the following potential proof for this particular scenario:

$73/3=x$

$x=24.333...$

$3x=72.999...$

$72.999...=73$

This works, but how would I create a general proof for all numbers that have an infinite trail of 9s?


问:
In "$\,f_w(x) = \sum_{i}\frac{1}{2}\|wx_i + c - Y\|_2^2\,$", what do the stacked "$2$"s at the end mean?
I got an equation and for me it is not clear for what the last stacked 2's is for?

$$f_w(x) = \sum_{i}\frac{1}{2}\|wx_i + c - Y\|_2^2$$

答: That’s the square of the two-norm (aka the Euclidean norm).</p>


问:
The weights of $n$ items are $X_1, \ldots , X_n$, assumed independent ${\cal U}(0, 1)$. Compute PDFs and expectations.
The weights of n items are $X_1, \ldots, X_n$, assumed independent uniform distribution ${\cal U}(0, 1)$. Mary and John each have a bin (or suitcase) which can carry total weight 1. Mary likes to pack in her bin only the heaviest item. John likes to pack the items in order $1, 2, \ldots, n$, packing each item if it can fit in the space remaining. Denote by $W_M$ the weight of Mary's suitcase and by $W_J$ the weight of John's suitcase.

Find the pdf-s of $W_M$ and $W_J$ and then compute the expectations of these random variables.

Hint. You need to know that, for any $c > 0$ and any positive integer $k$ we have $\int\limits_{x_1,...,x_k \geq 0, x_1+...+x_k \leq c} \frac{c^k}{k!}$.

I am not sure where to begin for this problem, so if anyone could walk me through it or give me a place to start that would be greatly appreciated.

答:
Mary's bag carries only the heaviest item. This is often called the $n$th order statistic. Check out: https://en.wikipedia.org/wiki/Order_statistic

One way to get the PDF for it, is by first getting the CDF: $P(\max(X_1,\ldots,X_n)\leq x) = P(X_1\leq x, \ldots, X_n \leq x) = P(X_1\leq x)\cdots P(X_n\leq x) = x^n.$ Can you get the PDF from here?

For John's bag, this is the sum of the first $k$ weights such that $\sum_{i=1}^k X_i \leq 1 < \sum_{i=1}^{k+1} X_i$. For this problem, I'd take some time to think about how it is related to this problem: Summing (0,1) uniform random variables up to 1 and go from there to get the PDF.


答:
The subscript is relative to the norm being used. The "2" stands for the euclidean norm. The superscript is a power.
$$
\|X\|_2^2 = \left(\left(\sum_{i=1}^n X_i^2 \right)^{1/2}\right)^2 = \sum_{i=1}^n X_i^2
$$

Other examples of norms could be

$$
\|X\|_1 = \sum_{i=1}^n |X_i|, \quad \|X\|_{\infty}= \max_{i=1,\cdots, n}|X_i|.
$$

Title: In "$\,f_w(x) = \sum_{i}\frac{1}{2}\|wx_i + c - Y\|_2^2\,$", what do the stacked "$2$"s at the end mean?

Body:
I got an equation and for me it is not clear for what the last stacked 2's is for?

$$f_w(x) = \sum_{i}\frac{1}{2}\|wx_i + c - Y\|_2^2$$


问:
Problem arithmetic $ b=qa+r $ $\Rightarrow$ $a\wedge b=a\wedge r$
I want to Prove that:

$ b=qa+r   $$\Rightarrow$$a\wedge b=a\wedge r$

I tried to prove this, but my results doesn't match the correct result

My attempt :

$b=aq+r$$\Rightarrow $$r=b-qa$ $(r≠0)$

Suppose $(*) a\wedge b=k$. $k\in  \mathbb{Z^{*}} $ and $k≠a$

(  note :$a\wedge b$=$gcd(a, b) $)

$a\wedge b=k$$\Rightarrow $ $a=a'k  $ et $b=b'k $

$r=b-qa$$\Rightarrow $$ k(b'-qa') =r$

$\Rightarrow $$ b'-qa'=\frac{r} {k} $

$b'-qa'\in \mathbb{Z} $$\Rightarrow $$ \frac{r} {k} \in \mathbb{Z} $

$\Rightarrow $$r=dk$

$\Rightarrow $$r\wedge k=k(**) $

So after$ (*, **) $ we can see $r\wedge k=a\wedge b $

$(k≠a) $

Where's my mistake?

答:
Let the expresion
$$b=aq+r$$
Suppose $d=gcd(a,b)$ by hypotesis
$d|a$ and $d|b$, notice that $r=b-aq$
and since $d|ax+by$ for $x,y\in \mathbb{Z}$ in particular for $x=q$ and $y=-1$ we get $d|r$ and $d|a$ then $d$ is a divisor of $r$ and $a$.
Now we prove that $d$ is the maxime, consider $d^{\prime}$ and divisior of $r$ and $a$ we should prove that $d|d^{\prime}$.

Since $d^{\prime}|a$ and $d^{\prime}|r$, notice that $b=aq+r$
and $d^{\prime}|ax+ry$ in particular $d^{\prime}|b$
now  notice tahat
$d^{\prime}|b$ and $d^{\prime}|a$ then $d^{\prime}$ is a divisor of $a$ and $b$
by hipotesis since $d$ is the $gcd(a,b)$ then $d|d^{\prime}$
And therfore $d=gcd(a,r)=gcd(a,b)$


答: Since $$0.999\ldots =1$$ we add the integer $n$ to both sides to get $$n.999\ldots =n+1.$$


问:
Confluent hypergeometric integrand over finite domain
Tricomi's (confluent hypergeometric) function is $$U(a,b,z) = \frac{1}{\Gamma(a)}\int_0^\infty  e^{-zt}t^{a-1}(1+t)^{b-a-1}dt$$ for $\Re(a)>0$.

Is there any analytical expression/special function for the integral over a finite domain $$\int_x^y  e^{-zt}t^{a-1}(1+t)^{b-a-1}dt,$$ where $x,y\in\mathbb{C}$ are constants?

答:
The answer was given by @Maxim in the comments. The special function is called Humbert series. With the integral representation here, a substitution $t=ux$ immediately implies

$$\int_0^x (1+t)^{b-a-1}t^{a-1}e^{-zt} \mathrm{d}t = \frac{x^a}{a} \Phi_1(a,a-b+1,a+1;-x,-zx)$$

because $\frac{\Gamma(a)\Gamma(1)}{\Gamma(a+1)}=\frac{1}{a}$.


问:
Proof of a fact regarding uniformly convergent functions
Denote the pointwise limit of $f_{n}(x)$ as $f(x)$. Define $M_n = sup_{x \ \in E} |f_{n}(x) - f(x)|$. Show that $f_n \to f$ uniformly iff $M_n \to 0$ as $n \to \infty$

Not sure how to start. Any tips would be appreciated!I know I have to show both directions since it is an iff statement.

答:
hint

In the initial definition, You just need replace the condition

$$(\forall x\in E)\;|f_n(x)-f(x)|<\epsilon$$
by its equivalent
$$\sup_{x\in E}\{|f_n(x)-f(x)|\}<\epsilon$$


问:
Why is the divergence of curl expected to be zero?
I was reading the proof of Helmholtz decomposition theorem where I found the relation between the rotational and the irrotational fields are not symmetric. And by that I mean if the divergence of the gradient field is supposed to be zero except where the particle exist aka:

$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$

Then why is the divergence of the curl always zero?

$$\nabla\cdot\nabla\times \bf A = 0$$

and not:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

where $\bf c $ is a vector with speed of light magnitude.

Edit: This question is about asymmetry of electromagnetic potential fields and I have no idea why it was migrated to math forum. Anyways, if mathematicians can see this question my question is:

Suppose there is a vector field $ F=\nabla(1/r)+\nabla \times \bf A $ made out of a scalar potential $1/r$ and a vector potential $\bf A$ where these relations hold:
$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$
and:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

So both potential fields have critical points, considering $\bf F$ should have been sufficiently smooth, can we still apply Helmholtz decomposition theorem?

答:
That the divergence of a curl is zero, and that the curl of a gradient is zero are exact mathematical identities, which can be easily proven by writing these operations explicitly in terms of components and derivatives.

On the other hand, a Laplacian (divergence of gradient) of a function is not necessarily zero. Equating it to a charge or another source is a matter of specific physical interpretation, which goes beyond pure math.

OwnerDisplayName: Martin Spinoza

OwnerDisplayName: Vadim


答:
This is a good question and the answer lies in the misuse of the notation $\nabla$, and it is also why I like to write $\operatorname{\mathbf{grad}}, \operatorname{\mathbf{curl}}, \operatorname{div}$ instead of $\nabla, \nabla \times, \nabla \cdot$, resp. That both $\operatorname{div} \operatorname{\mathbf{curl}} \mathbf{v}=0$ and $\operatorname{\mathbf{curl}}\operatorname{\mathbf{grad}} f=0$ hold for arbitrary $f, \mathbf{v}$ scalar or vector fields, resp.,  is nothing but the equality of mixed partial derivatives $\frac{\partial^2}{\partial x \partial y}=\frac{\partial^2}{\partial y \partial x} $as it is easily seen by writing them out in components. But this means neither $\nabla \cdot \nabla \equiv  0$ nor $\nabla \times \nabla \equiv  \bf{0}$ for such repeated operator is not defined; $\nabla$ is not a vector and it is neither perpendicular to nor parallel with itself in any sense.

There is a formulation of vector analysis using (exterior) differential forms where a differentiation operator is introduced that does act as you would like $\nabla$ behave. It can replace $\operatorname{\mathbf{grad}}, \operatorname{\bf{curl}}, \operatorname{div}$ with a single operator and can also repeatedly act on its argument with the result that you would expect. Differential forms are the natural generalization of vector analysis (including the vector product) for higher than 3 dimensional space, and there are attempts to replace and teach all of vector analysis in that way including the 3D case. You can find much written on this subject here in phys.stack.

Title: Why is the divergence of curl expected to be zero?

Body:
I was reading the proof of Helmholtz decomposition theorem where I found the relation between the rotational and the irrotational fields are not symmetric. And by that I mean if the divergence of the gradient field is supposed to be zero except where the particle exist aka:

$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$

Then why is the divergence of the curl always zero?

$$\nabla\cdot\nabla\times \bf A = 0$$

and not:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

where $\bf c $ is a vector with speed of light magnitude.

Edit: This question is about asymmetry of electromagnetic potential fields and I have no idea why it was migrated to math forum. Anyways, if mathematicians can see this question my question is:

Suppose there is a vector field $ F=\nabla(1/r)+\nabla \times \bf A $ made out of a scalar potential $1/r$ and a vector potential $\bf A$ where these relations hold:
$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$
and:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

So both potential fields have critical points, considering $\bf F$ should have been sufficiently smooth, can we still apply Helmholtz decomposition theorem?

OwnerDisplayName: hyportnex


问:
Prove $0$ is the only eigenvalue of $A$.
Let $A$ be an $n$ x $n$ real matrix. Prove $0$ is
eigen value of $A$.

PROOF:

Let $m∈ℤ^{+}$

so $A^{m}$x$=λ^{m}$x

there exists$m≥1$ s.t $A^{m}=0$.

$0=λ^{m}$x, but $x≠0$, so $λ^{m}=0$

$m>0$ , so $λ^{m}=0$ IFF so $λ=0$.

so all $λ=0$

答:
Here is a rewrite of your proof that I think is an improvement.

Proof: We are given that $A$ is $n \times n$, and that $A$ is nilpotent. Let $m$ denote the smallest positive integer for which $A^m = 0$.

First, we show that $A$ must have $0$ as an eigenvalue. Let $\mathbf x \in \Bbb R^n$ be any non-zero vector. Let $j_0$ denote the smallest positive integer $j$ for which $A^j\mathbf x = 0$; we note that such an integer exists because taking $j = m$ yields $A^j\mathbf x = 0 \mathbf x = 0$.  We can see that $A(A^{j_0 - 1} \mathbf x) = 0 \cdot \mathbf x$, and $A^{j_0 - 1} \mathbf x \neq \mathbf 0$. So, $A^{j_0 - 1}\mathbf x$ is an eigenvector of $A$ associated with $\lambda = 0$.

Now, suppose that $\lambda$ is an eigenvector of $A$. Let $\mathbf x \in \Bbb R^n$ be an associated eigenvector so that $\mathbf x \neq \mathbf 0$ and $A \mathbf x  = \lambda \mathbf x$.

Claim: $A^k \mathbf x = \lambda^k \mathbf x$.

Proof of claim: We proceed by induction

Base case: $A^1 \mathbf x= \lambda^1 \mathbf x$.

Inductive step: If $A^n \mathbf x = \lambda^n \mathbf x$, then
$$
A^{n+1} \mathbf x = A (A^n \mathbf x) = A \lambda^n \mathbf x =
\lambda^n A \mathbf x = \lambda^n \cdot \lambda \mathbf x = \lambda^{n+1} \mathbf x. \quad \square
$$

We know that $A^m = 0$. By the claim, it follows that
$$
\mathbf 0 = 0 \cdot \mathbf x = A^m \mathbf x = \lambda^m \mathbf x.
$$
If $\lambda^m \mathbf x = 0$, then we must have $\lambda^m = 0$. However, $m$ is a positive integer, so $\lambda^m = 0$ implies that $\lambda = 0$.

That is, if $\lambda$ is an eigenvalue of $A$, then $\lambda = 0$. So, $\lambda = 0$ is the only eigenvalue of $A$.


问:
Möbius function - average
I had to solve an exercise comprised of 3 parts:

a) For q, a integers, $q\geq 2$ and $\gcd(a,q)=1$ to show that
$$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv a\pmod q}}\mkern-18mu \mu (n) = 0$$

b) For $m \in \mathbb{Z}_{>0}$, to show

$$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv  a\pmod q ,\\ \gcd(m,n)=1}}\mkern-18mu\mu (n) = 0$$

c) For $q, a$ integers, $q\geq 2$ and $\gcd(a,q)=d>1$, to show that
$$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv \text{a(mod q)}}}\mkern-18mu \mu (n) = 0$$

Out of this I solved a), b) using ideas from Dirichlet characters, Tauberian Theorem and some other tools from complex analysis, but for c) I cannot seem to get anywhere. As a small hint, I need to embed b) but not sure how this would work. I would appreciate any suggestion in the right direction.

答:
If $\gcd(a,q)=d$ then the change of variables $n=kd$ yields
$$
\sum_{\substack{n\le x \\ n\equiv a\pmod q}} \mu(n) = \sum_{\substack{k\le x/d \\ k\equiv a/d\pmod{n/d}}} \mu(kd) = \mu(d) \sum_{\substack{k\le x/d \\ k\equiv a/d\pmod{n/d} \\ (k,d)=1}} \mu(k).
$$

Title: Möbius function - average


问:
In how many ways can we arrange 4 letters of the word "ENGINE"?
I need to know a combinatoric solution to this problem, with Generating functions in book, gives us 102.

It might be a very simple problem, but Im very confused with this.
 Would be very nice of you to give me some help.
Im really grateful.

答:
By grouping the similar things we got,

G, I, (E, E), (N, N)

4 letters are different, then we can select =
4C4, arrange = 4C4 × 4!

2 E and 2 letters are different, then we can select =    3C2, arrange = 3C2 × $\frac{4!}{2!       }$

2 N and 2 letters are different, then we can select =    3C2, arrange = 3C2 × $\frac{4!}{2!}$

Taking 2 from same group and 2 from other same group, this can be
selected in only 1 way - (for example, 2 E, 2 N), we can select = 1,
arrange = 1 × $\frac{4!}{2! 2!      }$

By adding,

Total number of selection = 8

Total number of arrangement = 102


问:
$30$ red balls and $20$ black balls are being distributed to $5$ kids. Each kid gets at least one red ball. In how many ways can we distribute balls?
$30$ red balls and $20$ black balls are being distributed to $5$ kids, so that each kid gets at least one red ball. In how many ways can we distribute balls?

Circle the correct answers:

a) $\binom{29}{4}$ $\binom{24}{20}$

b) $\binom{29}{5}$ $\binom{24}{5}$

c)$|Sur(N_{30},N_{5})|S(20,5)$ Note: $|Sur(N_{30},N_{5})|$ is the number of surjections, and $S(20,5)$ is a Stirling number of the second kind

d) None of $3$ previous answers are correct

My approach:

First I gave each of $5$ kids one red ball, which leaves me with $25$ red balls. Now I used the stars and bars method to distribute the balls I am left with.

Red balls: $x_1+x_2+x_3+x_4+x_5=25,x_i\geq 0, i=1,..,5$. This equation has $\binom{5+25-1}{25}=\binom{29}{25}=\binom{29}{4}$.

Black balls: $x_1+x_2+x_3+x_4+x_5=20,x_i\geq 0, i=1,..,5$. This equation has $\binom{5+20-1}{20}=\binom{24}{20}=\binom{24}{4}$.

So I would say a) is the correct answer.. am I right?

答:
I don't know what the "stars and bars" method is, and I despise permutations, combinations, etc. and all of those combinatoric black boxes. So this is how I would do it:

Just place all of the balls on a straight line (going from left to right) in front of you. Choose 4 random spots between the balls. You have $\binom{29}{4}$ ways of doing so.

Give the first kid the balls before the first spot, the second kit the balls between the first and second spot, ... the last kid the balls after the last spot. Thus you have $\binom{29}{4}$ ways of distributing the red balls.

For the black balls, first go to the store and buy 5 new black balls. But then also add the condition of giving each kid at least one black ball. The same argument yields $\binom{24}{4}$ ways of distributing them. After the distribution is done, get one black ball back from each kid, so you have the same distribution as if you didn't get those extra 5 balls (and did not care to give at least one black ball to each kid).

So indeed the answer seems to be $\binom{29}{4}\binom{24}{4}$ ways of distributing the red and black balls.


问:
Work out the percentage of a single unit in a scale
1st question on this forum, and I know it will be an easy-peasy one for someone out there!!  I'm not even sure if the title of the question is accurate for the question I am asking.....I'm also not sure what tags to put it under!

Anyway, here goes....

I have a line (l) which is of an arbitrary length, which represents an arbitrary range of minutes (r).

For example, the line is 900 wide and represents 720 minutes.

How do I work out, as a percentage of l, what one minute is?

Thanks,

ETFairfax

答:
Algebraic Mastmin is correct in that the percentage is independent of the length, and that each minute is $\frac{100}{720} = \frac{5}{36}$ percent of the line. But I thought you might also be interested in how long a minute is (in length) on the line.

If there are 720 minutes on the line, one minut is $\frac{1}{720}$ of this line. To get a better understanding of why this is correct, you can try adding them together again:
$$ \underbrace{\frac{1}{720} + \frac{1}{720} + \cdots + \frac{1}{720}}_{\text{720 times}} = 720 \cdot \frac{1}{720} = \frac{720}{720} = 1$$
We see that we get the whole line. Now we need to find out the length of $\frac{1}{720}$ of that line. The length of the whole line is 900. We want $\frac{1}{720}$ of that.
$$\frac{900}{720} = \frac{5}{4}$$
You can try for yourself to multiply this by 720 and see that you get the length of the line. We see that each minute is $\frac{5}{4}$ long.


问:
What wrong with pointwise convergent
It is true that the uniform limit of continuous functions is continuous as it has proved as a theorem, but what is wrong with pointwise limit? I mean why this theorem it doesn't work for if the convergence is pointwise?

Can any one clarify this to me may be using epsilon definition if possible?

Thank you.

答:
I think this is a reasonable question. In the standard $\epsilon/3$ proof of the theorem, you have (with $f_n\to f$)
$$|f(x)-f(a)| \le |f(x)-f_n(x)| + |f_n(x)-f_n(a)| + |f_n(a)-f(a)|\,,$$
and, using uniform convergence, you pick $N$ so that $|f_n(z)-f(z)|<\epsilon/3$ for all $n\ge N$. If you have only pointwise convergence, you choose an $N$ that works for $a$ but then may well find that you need increasing values of $N$ (going to $\infty$) for different values of $x$ as $x\to a$. You need a fixed such $N$ to get a $\delta$ from the limit definition for $f_N$ to make the middle term $<\epsilon/3$.

If you try to implement the proof with the standard example @Ittay Weiss gave, or many others, you will see precisely what I'm saying here.

Title: What wrong with pointwise convergent


问:
Homeomorphic metric spaces where the identity map is not a homeomorphism
In our Topology lectures, we were told that there are examples where two metric spaces $(M, d)$ and $(M, d')$ are homomeomorphic (i.e. there exists some homeomorphism between them, which is not necessarily the identity map) but the identity map $\operatorname{Id}\colon(M, d) \rightarrow (M, d')$ is not a homeomorphism. I was hoping someone could provide an example of this.

答: Take any metric space $(X,d)$ for which there is a bijective discontinuous function $f\colon M\longrightarrow M$ and define $d'(x,y)=d\bigl(f(x),f(y)\bigr)$. Then $f$ is a homeomorphism (actually, it's an isometry) from $(M,d')$ into $(M,d)$. However, the identity is not a homeomorphism from $(M,d')$ into $(M,d)$.


答:
Let $M=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, and let $d$ be the usual metric on $M$. Define a new metric $d'$ on $M$ as follows:

$$d'(x,y)=\begin{cases}
|x-y|,&\text{if }0<x,y<1\text{ or }x=y\text{ or }\{x,y\}=\{0,1\}\\
x,&\text{if }0<x<y=1\\
y,&\text{if }0<y<x=1\\
1-x,&\text{if }0=y<x<1\\
1-y,&\text{if }0=x<y<1
\end{cases}$$

This simply interchanges the rôles of $0$ and $1$: the map

$$h:M\to M:x\mapsto\begin{cases}
x,&\text{if }0<x<1\\
1,&\text{if }x=0\\
0,&\text{if }x=1
\end{cases}$$

is a homeomorphism between $\langle M,d\rangle$ and $\langle M,d'\rangle$ in either direction, but the identity map is not: $0$ is the only non-isolated point with respect to $d$, while $1$ is the only non-isolated point with respect to $d'$.

Title: Homeomorphic metric spaces where the identity map is not a homeomorphism

Body: In our Topology lectures, we were told that there are examples where two metric spaces $(M, d)$ and $(M, d')$ are homomeomorphic (i.e. there exists some homeomorphism between them, which is not necessarily the identity map) but the identity map $\operatorname{Id}\colon(M, d) \rightarrow (M, d')$ is not a homeomorphism. I was hoping someone could provide an example of this.


问:
Solve $(x^2-y^2)dx+2xy\ dy=0$
I'm at the beggining of a differential equations course, and I'm stuck solving this equation:
$$(x^2-y^2)dx+2xy\ dy=0$$
I'm asked to solve it using 2 different methods. I proved I can find integrating factors of type $\mu_1(x)$ and $\mu_2(y/x)$. If I'm not wrong, these two integrating factors are:
$$\mu_1(x)=x^{-2} \ \ , \ \ \mu_2(y/x)=\left(1+\frac{y^2}{x^2}\right)^{-2}.$$
Then, I've used $\mu_1(x)$ to find the function $\psi(x,y)$ (I think it's called "potential function") so that
$$\frac{\partial \psi}{\partial x}=x^{-2}(x^2-y^2) \ \ , \ \ \frac{\partial \psi}{\partial y}=2xy.$$
Then, a solution to my equation is $\psi(x,y)=C$, being $C$ a constant.
I've got as solution:
$$\boxed{\psi(x,y)=x+\frac{y^2}{x}=C.}$$
I want to know first if this solution is correct. Then, I know that given two different integrating factors of different type, in my case $\mu_1(x)$ and $\mu_2(y/x)$, then
$$\frac{\mu_1(x)}{\mu_2(y/x)=K} \ \ , \ \ \text{$K$ constant}$$
defines also the solution to the equation. But what I get is
$$\boxed{\frac{x^{-2}}{\left(1+\frac{y^2}{x^2}\right)^{-2}}=K.}$$
I've graphed both solutions and they are not the same (correct me if I'm wrong). What of these is the right solution? Did I did something wrong along the way? Are both solutions the same? Any help will be appreciated.

PD: I know it can also be solved as an homogeneus one, but I'm interested in using these two methods and finding what I did wrong along the way.

EDIT: I realised I got wrong the integrating factor $\mu_2(y/x)$. Instead it it
$$\mu_2(y/x)=\left(1+\frac{y^2}{x^2}\right)^2.$$
Now, I can state as solution
$$\frac{\mu_2}{\mu_1}=K \ \implies \ \boxed{\left(x+\frac{y}{x}\right)^2=K}$$
which is equivalent to my first solution.

答:
$$(x^2-y^2)dx+2xy\ dy=0$$
$$x^2dx-y^2 dx+x dy^2=0$$
$$dx+\dfrac {x dy^2-y^2dx}{x^2}=0$$
$$dx+d \left (\dfrac {y^2 }{x} \right)=0$$
$$x+\dfrac {y^2 }{x}=C$$
It seems to me that there is a sign mistake somewhere.

Title: Solve $(x^2-y^2)dx+2xy\ dy=0$


问:
Proving Sequent using Natural Deduction
RTP: $P \to Q, R \to \neg Q, (S\to \neg P)\to R \vdash (\neg T ∨ P)\to(T \to S)$ using primitive rules of natural deduction. I've attempted this question multiple times but keep getting stuck on trying to get $S$ so I can use a couple arrow introductions to get the conclusion.

Attempt:

1 (1) $P \to Q$ Assumption

2 (2) $R \to \neg Q$ Assumption

3 (3) $(S\to \neg P)\to R$ Assumption

4 (4) $(\neg T ∨ P)$ Assumption

5 (5) $T$ Assumption

4,5 (6) $P$ $4,5∨$E

1,4,5 (7) $Q$ $1,6\to$E

8 (8) $S$ Assumption

9 (9) $S\to \neg P$ Assumption

8,9 (10) $\neg P$ Assumption

4,5,9 (11) $\neg T$ 6,10RAA(8)

12 (12) R Assumption

2,12 (13) $\neg Q$ $2,12\to$E

1,2,4,5 (14) $~R$ 7,13RAA(12)

I realise now that I've eliminated almost all the necessary assumptions with only line 9 left over (those at lines 4 and 5 which will be eliminated via arrow introduction for the conclusion once I have $S$), but I don't know how I can proceed from here. Any help would be greatly appreciated.

答:
$\def\fitch#1#2{~~\begin{array}{|l|}\hline #1\\\hline#2\\\hline\end{array}~~}$

You seek to derive $S$ under the assumption of $T$, $\lnot T\vee P$, and the premises.

You can immediately derive $P$ and $Q$ using disjunctive syllogism and modus ponens.

Likewise you could there after derive $\neg R$ and $\neg(S\to\neg P)$ using modus tollens.

However, I'd rather use reduction to absurdity to derive $S$.

Discharge the necessary contradictions and you are done.

$$\fitch{~~1.~P\to Q\\~~2.~R\to\lnot Q\\~~3.~(S\to\lnot P)\to R}{\fitch{~~4.~\lnot T\lor P}{\fitch{~~5.~T}{~~6.~P\\~~7.~Q\\\fitch{~~8.~\lnot S}{\vdots\\14.~\lnot Q}\\15.~S}\\16.~T\to S}\\17.~(\lnot T\lor P)\to(T\to S)}$$

Using logic.tamu.edu/daemon.html

```<code>P->Q, R->~Q, (S->~P)->R |- (~TvP)->(T->S)
OK    1          (1)   P->Q           A
OK    2          (2)   R->~Q          A
OK    3          (3)   (S->~P)->R     A
OK    4          (4)   ~TvP           A
OK    5          (5)   T              A
OK    4,5        (6)   P              4,5vE
OK    1,4,5      (7)   Q              1,6->E
OK    8          (8)   ~S             A
:     :          :     :              :
OK    2,3,8      (14)  ~Q             ..,..->E
OK    1,2,3,4,5  (15)  S              7,14RAA(8)
OK    1,2,3,4    (16)  T->S           15->I(5)
OK    1,2,3      (17)  (~TvP)->(T->S) 16->I(4)
</code>```

Title: Proving Sequent using Natural Deduction

Body:
$\def\fitch#1#2{~~\begin{array}{|l|}\hline #1\\\hline#2\\\hline\end{array}~~}$

You seek to derive $S$ under the assumption of $T$, $\lnot T\vee P$, and the premises.

You can immediately derive $P$ and $Q$ using disjunctive syllogism and modus ponens.

Likewise you could there after derive $\neg R$ and $\neg(S\to\neg P)$ using modus tollens.

However, I'd rather use reduction to absurdity to derive $S$.

Discharge the necessary contradictions and you are done.

$$\fitch{~~1.~P\to Q\\~~2.~R\to\lnot Q\\~~3.~(S\to\lnot P)\to R}{\fitch{~~4.~\lnot T\lor P}{\fitch{~~5.~T}{~~6.~P\\~~7.~Q\\\fitch{~~8.~\lnot S}{\vdots\\14.~\lnot Q}\\15.~S}\\16.~T\to S}\\17.~(\lnot T\lor P)\to(T\to S)}$$

Using logic.tamu.edu/daemon.html

P->Q, R->~Q, (S->~P)->R |- (~TvP)->(T->S)
OK    1          (1)   P->Q           A
OK    2          (2)   R->~Q          A
OK    3          (3)   (S->~P)->R     A
OK    4          (4)   ~TvP           A
OK    5          (5)   T              A
OK    4,5        (6)   P              4,5vE
OK    1,4,5      (7)   Q              1,6->E
OK    8          (8)   ~S             A
:     :          :     :              :
OK    2,3,8      (14)  ~Q             ..,..->E
OK    1,2,3,4,5  (15)  S              7,14RAA(8)
OK    1,2,3,4    (16)  T->S           15->I(5)
OK    1,2,3      (17)  (~TvP)->(T->S) 16->I(4)


问:
Question about if $g\circ f$ is surjective then $g$ is surjective.
Suppose we have $f:A\rightarrow B$ and $g:B\rightarrow C$

if $g\circ f$ is surjective then $g$ is surjective.

This seems one directional as in $\Rightarrow$, but my Professor said if it is a definition then it is always an $\iff$ no matter how the statement is said (or that is how I am interpreting his words, correct me if I am wrong!)

Proof:

Let $c\in C$ be arbritrary. Given that $g\circ f$ is surjective, there exists an $a\in A$ such that $g\circ f(a)= g(f(a))=c$. This implies we have some $b=f(a)\in B$ such that $g(b)= c$. Hence we have shown $g$ is surjective.

But I am wondering is it necessarily true if we try to prove the backwards direction? I seem to run into the problem that I am uncertain if $B$ in $f$ is the same $B$ in $g$.

答:
The backwards direction isn't true in general. Here's a counter example.

let $A=B=C=\mathbb{R}$

let $f(x)=1$.

let $g(x)=x$.

$g(x)$ is surjective.

$f(g(x)) = 1$ which is not surjective

There must have been something he said wrong, or you misunderstood.

EDIT: I agree with peek-a-boo. Is it possible he was describing 'if and only if"? "if and only if" does correspond to definitions.


问:
Combinatorics: Group voting amongst themselves
5 members of a group vote for each other, how many ways can they do so such that they cannot caste a vote for themselves?
I believe it is $4^{5}$ since each member has 4 different possibilities. Am I correct in saying so?

答: Yes this is correct. Person 1 has 4 ways of voting. Person 2 has 4 ways and so on. So you get $4*4*4*4*4$ = $4^{5}$


答:
Your proof is correct.  The converse is not true.  For example, $g: \{1,2\} \rightarrow \{1,2\}, 1 \mapsto 1, 2\mapsto 2$ is surjective, but if $f: \{1\} \rightarrow \{1\}$ is defined by $1 \mapsto 1$ then $g \circ f$ is not surjective.

Probably what your professor meant was that when we define something, we sometimes use "if" and let "only if" be implied. e.g.,

We say that $f:A\rightarrow B$ is surjective if for all $b \in B$ there exists some $a \in A$ with $f(a) = b$.

This is a definition, so really "if" means "if and only if".  The reason is that a definition must be an "if and only if" condition to be sensible; otherwise it is ill-defined.  Context clues tell you it is a definition: the use of "we say that" and italics around the term surjective (and perhaps that the term has not been used before in the text.)  In your case, the statement is not a definition, so the converse is not implied.

Title: Question about if $g\circ f$ is surjective then $g$ is surjective.

Body:
Suppose we have $f:A\rightarrow B$ and $g:B\rightarrow C$

if $g\circ f$ is surjective then $g$ is surjective.

This seems one directional as in $\Rightarrow$, but my Professor said if it is a definition then it is always an $\iff$ no matter how the statement is said (or that is how I am interpreting his words, correct me if I am wrong!)

Proof:

Let $c\in C$ be arbritrary. Given that $g\circ f$ is surjective, there exists an $a\in A$ such that $g\circ f(a)= g(f(a))=c$. This implies we have some $b=f(a)\in B$ such that $g(b)= c$. Hence we have shown $g$ is surjective.

But I am wondering is it necessarily true if we try to prove the backwards direction? I seem to run into the problem that I am uncertain if $B$ in $f$ is the same $B$ in $g$.


问:
The half-life $\lambda$ of a radioactive substance decaying according to the law $y = y_oe^{kt}$ is defined to be the time

The half-life $\lambda$ of a radioactive substance decaying according to the law $y = y_oe^{kt}$ is
defined to be the time it takes the amount to decrease to $1/2$ of the initial amount %y_o%.

a) Express the half-life $\lambda$ in terms of k.

b) Show using your expression for $\lambda$ that if at time $t_1$ the amount is $y_1$, then at time
$t_1 + \lambda$ it will be $y_1/2$, no matter what $t_1$ is.

I was able to do (a), and got the final expression:

$\lambda = \frac{-\ln{2}}{k}$

But I'm not sure how to do (b).

if at time $t_1$ the amount is $y_1$

$y_1=y_0e^{kt_1}$

then at time $t_1 + \lambda$ it will be $y_1/2$

What I did is the following:

$y_1=y_0e^{k(t_1+\lambda)}$

I replaced $\lambda = \frac{-\ln{2}}{k}$

$y_1=y_0e^{k(t_1-\frac{\ln{2}}{k})}$

$y_1=y_0e^{kt_1}e^{-\ln{2}}$

But the hypothesis was that $y_1=y_0e^{kt_1}$, so

$y_1 = y_1e^{-\ln{2}}$

Finally

$1 = \frac{1}{2}$

Which is absurd.

What am I doing wrong?

答:
Your part (a) is correct; note that $k$ is negative for an exponential decay model, thus $\lambda$ will be positive.

For part (b), if $y_1$ is the amount at $t_1$, then you correctly have $$y_1 = y_0 e^{k t_1}.$$  Now at time $t = t_1 + \lambda$, the right-hand side of the exponential decay equation becomes $$y_0 e^{k(t_1 + \lambda)} = y_0 e^{kt_1 + k\lambda} = y_0 e^{kt_1} e^{k\lambda}.$$  But we already showed $y_0 e^{kt_1} = y_1$, so substituting back gives us $$y = y_1 e^{k\lambda}$$ at time $t_1 + \lambda$.  And since $$\lambda = -\frac{\ln 2}{k},$$ we substitute this too:  $$y = y_1 e^{k (- \ln 2)/k} = y_1 e^{-\ln 2} = y_1 e^{\ln (1/2)} = \frac{y_1}{2},$$ as claimed.

Your confusion stems from setting the left-hand side of the exponential decay equation to $y_1$ at time $t_1 + \lambda$.  That's not what you should do.  The amount at this time is not $y_1$; it is some amount smaller than $y_1$, for which you need to substitute the appropriate values into the right-hand side to simplify, as I showed above.


问:
Basis of the space of vector fields on smooth manifolds
Let $M$ be a smooth ($C^\infty$) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$. $\mathfrak{X}(M)$ is a real vector space and it is also a module over $\mathfrak{F}(M)$.

We know that partial derivatives constitute a basis for tangent space at any point $p\in M$.
Is there some sort of basis for $\mathfrak{X}(M)$ (as a vector space or as a module)? Do partial derivatives constitute a basis here as well?

I think this question Basis of vector fields on manifold is similar to mine, but because of the way it's written, I'm not really sure I understand the question and I'm not sure we're asking about the same thing.

答:
The set of vector fields of a manifold $M$ is infinite dimensional. For example, over $\mathbb{R}$, it is isomorphic to $\mathcal{C}^{\infty}(\mathbb{R})$ because the tangent bundle is trivial, and the datum of a vector field $(x,V_x)$ over $\mathbb{R}$ is the same as the datum of the function $x \mapsto \langle V_x,e_1\rangle$ where $e_1$ is the constant unit vector field.

In fact, as soon as you have a non-zero vector field $X$ over $M$, the familly $\left\{fX~|~ f \in \mathcal{C}^{\infty}(M) \right\}$ spans an infinite dimensional linear subspace of $\mathfrak{X}(M)$. So one cannot hope to find a basis of it in the general case.

In addition, $\mathfrak{X}(M)$ is a $\mathcal{C}^{\infty}(M)$-module of finite rank. But this does not mean that there is a basis of $\mathfrak{X}(M)$ as a module. This can happen only if the tangent bundle is trivial, that is $TM \simeq M\times \mathbb{R}^n$. This is because if $(X_1,\ldots,X_m)$ is a basis of $\mathfrak{X}(M)$ as a module, then the map
\begin{align}
TM& \longrightarrow M \times \mathbb{R}^m \\
(p,v) &\longmapsto (x,v_1,\ldots, v_m)
\end{align}
where $v=\sum v_i X_i(p)$, would be surjective and injective, and linear in the fibers. Thus, it would be a trivilaization of $TM$ and $M$ will be a parallelizable manifold (and moreover, $m = \dim M$).


问:
There are no sections of blowup $Bl_0(\mathbb{A}^2)$ of affine plane at the origin
Let $p:Bl_0(\mathbb{A}^2) \to \mathbb{A}^2$ be blowup of plane
at the origin. Is there a geometric reason why there
are no sections $s: \mathbb{A^2} \to Bl_0(\mathbb{A^2})$, that is
no maps to vareties with $p \circ s= id$ on affine plane $\mathbb{A^2}$.

Where I'm trying to get to is: in this comment by Zhen Lin used this
recognizing the total space of the tautological line bundle $O(-1)$
on $\mathbb{P}^1$ to be identical to the blowup of affine plane at the origin.

答: Here's one simple geometric reason: Consider any two distinct lines $L_1, L_2$ through the origin in $\newcommand{\AA}{\mathbb{A}}\AA^2$. The proper transforms of these two lines under the blowup at the origin are two non-intersecting lines in $\newcommand{\Bl}{\operatorname{Bl}}\Bl_0(\AA^2)$. If there were a section $s\colon \AA^2 \to \Bl_0(\AA^2)$ of the blowup map, then it would have to map each of these lines to its proper transform, because the image of a connected space under a continuous map is connected, and regular maps are continuous. But this is impossible, because this forces $s(0, 0)$ to have two different values.


问:
For a random iid sample of size n, probability of some $X_j$ of being minimum
Suppose we have an identically and independently distributed sample of continuous random variables $X_1, ..., X_n$.

How do I calculate, using integration areas, the probability that for some $j$, $X_j = \underset{(X_i)}{min} X_i $ ?

I know that there's a logic that goes like this:  since they are identical and independent, each has the same probability of being the minimum. And, since one of them must be the minimum, we have that each random variable has probability $1/n$.

But I tried explicitly calculating this for a $U(0,1)$ pdf (which should arguably be the easiest) and I kept getting stuck. For two RVs, $X$ and $Y$, calculating $Pr(X \leq Y)$ is straightforward, since we define the areas: $x, y \in [0,1]$ and $x \leq y$ resulting in:

$$Pr(X \leq Y) = \int_0^1 \int_0^y f(x,y) dxdy = \frac{x^2}{2}\vert_0^1 = \frac{1}{2}$$

Now, adding another random variable $Z$, and taking: $Pr(X = min\{X,Y,Z\}) = Pr(X \leq Y, X \leq Z)$, and having $x, y, z \in [0,1]$, $x \leq y$, $x \leq z$, and taking integrals as above, results in $1/6$ (instead of the actual answer, $1/3$). I suspect this has to do with splitting between two cases: $y \leq z$ and $y \geq z$, but I don't see how to make such observation fit in the areas of integration.

Body:
The event

$X=min(X,Y,Z)$ can be written as

$$(0<X<Y<Z<1)\cup (0<X<Z<Y<1)=A\cup B$$

1.

$$\mathbb{P}[A]=\int_0^1  dx\int_x^1dy\int_y^1dz=\frac{1}{6}$$

2.

$$\mathbb{P}[B]=\int_0^1  dx\int_x^1dz\int_z^1dy=\frac{1}{6}$$

Thus

$$\mathbb{P}[X=min(X,Y,Z)]=\frac{1}{6}+ \frac{1}{6}= \frac{1}{3}  $$


问:
Is $T: \mathbb{R} \rightarrow \mathbb{R}:(x,y,z) \rightarrow (y+z, x+z, x+y)$ an isomorphism?
Intuitively I want to say yes. I figured that the best way to go about this problem was to check if $T$ is bijective.

Injective:

$$(0,0,0) = (y+z, x+z, x+y)$$
We can rewrite that as matrix
$$\
\left[
\begin{array}{ccc|c}
0 & 1 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
\end{array}
\right]
$$
After evaluating you will find that this only has a trivial solution. Thus $ker(T) = \{0\} \implies$ Injective.

Surjective:

For this I would need to show that $im(T) = \mathbb{R}^3$, but I'm not sure how I would do this. I Think you should show that for all $\mathbf{v} \in \mathbb{R}^3$ you get $\mathbf{v} \in im(T)$ and vice versa. But I'm not sure how to show this.

答:
Either you use the rank-nullity theorem as suggested in the comments, or you formally invert the matrix, or after some simple manipulation you find directly that
$$
T(x,y,z)=(\alpha,\beta,\gamma)
$$
with
$$
x=\frac{-\alpha+\beta+\gamma}2,\quad
y=\frac{\alpha-\beta+\gamma}2,\quad
z=\frac{\alpha+\beta-\gamma}2.
$$


问:
Quick homework question help (triple Venn diagram questions with maybe)

I have already filled out all the necessary information on the Venn diagram. But some follow up questions are confusing such

find (A and B and maybe C) or

find (A and B but maybe C)

Not sure what these two questions are asking. I have provided an example diagram if it helps.

Thank you.

答:
English is ambiguous, but a plausible interpretation is the following:

$$A \& B \& (\text{ maybe } C) = A\texttt{ AND }B.$$

$$(A \& B )\text{ but maybe } C = (A \texttt{ AND }B) \texttt{ XOR } C.$$


问:
Name for collection of all derivatives (and/or integrals) for a given function?
Is there a name for the set of functions that make up a function's derivatives? Integrals?

For example's sake, let's say there is a fourth degree polynomial function $f(x) = 1 + x^2 + x^3 + x^4$. What would the collection $f(x)$, $f'(x)$, $f''(x)$, ... be called? Can this be extended to integration?

I tried searching for something along the lines of generalizing the set of derivatives or a generalization of gradient, but found generalizations in the multivariate or different coordinate senses, not to n'th derivatives.

答:
The answer to this question really depends on your motivation for it. If you just want a name for the sequence of pointwise values of a function and its derivatives at a point, then "the sequence of pointwise values of a function and its derivatives at a point" is an answer.

However, you probably want to do something with this sequence, thus you need a definition that hints at some more structure on it. In that case, the closest notion I can think of is the one of "Jet" ; https://en.wikipedia.org/wiki/Jet_(mathematics). It arises in the study of symmetries of differential equations, because it allows to rewrite such equations as algebraic ones; see Why is a PDE a submanifold (and not just a subset)?.

(On these things, I like the book of Peter Olver "Applications of Lie groups to differential equations").


问:
Problem involving Cauchy's inequality and inequality between quadratic and arithmetic means
I am working on a problem in an elementary mathematics book. I have come up with a solution, but it doesn't seem to work. At this point I'm thinking there may be an error in the book, but maybe I'm missing something.

The problem goes :
Let $a_{1},a_{2},a_{3},a_{4}$ be positive reals. Prove that :
\begin{equation}
\sum_{1 \le i < j < k \leq 4} \frac{a_{i}^{2} + a_{j}^{2} + a_{k}^{2}}{a_{i}+a_{j}+a_{k}} \geq a_{1} + a_{2} + a_{3} + a_{4}
\end{equation}
with equality iff $a_{1} = a_{2} = a_{3} = a_{4}$.

I think we are supposed to use this inequality to derive the solution :
\begin{equation}
\sqrt{ \frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2}}{n} } \geq \frac{a_{1} + a_{2} + \dots + a_{n}}{n}
\end{equation}
I have derived that :
\begin{align}
\require{cancel}
\sqrt{ \frac{a_{i}^{2} + a_{j}^{2} + a_{k}^{2}}{3} } & \geq \frac{a_{i} + a_{j} + a_{k}}{3}\\
\sqrt{a_{i}^{2}+a_{j}^{2}+a_{k}^{2}} & \geq \frac{\sqrt{3}(a_{i}+a_{j}+a_{k})}{3} \\
a_{i}^{2} + a_{j}^{2} + a_{k}^{2} & \geq \frac{\cancel{3}(a_{i}+a_{j}+a_{k})^{2}}{\cancel{9}}\\
\frac{a_{i}^{2}+a_{j}^{2}+a_{k}^{2}}{(a_{i}+a_{j}+a_{k})} & \geq \frac{(a_{i}+a_{j}+a_{k})}{3}
\end{align}
So :
\begin{align}
\frac{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}{a_{1}+a_{2}+a_{3}} & \geq \frac{a_{1}+a_{2}+a_{3}}{3} \\
\frac{a_{2}^{2}+a_{3}^{2}+a_{4}^{2}}{a_{2}+a_{3}+a_{4}} & \geq \frac{a_{2}+a_{3}+a_{4}}{3}
\end{align}
So :
\begin{equation}
\sum_{1\leq i < j < k \leq 4} \frac{a_{i}^{2}+a_{j}^{2}+a_{k}^{2}}{a_{i}+a_{j}+a_{k}} \geq \frac{a_{1}+a_{2}+a_{3}+a_{2}+a_{3}+a_{4}}{3}
\end{equation}
But this doesn't prove the needed result. Can someone help with this ?

答:
The error is that $1\leq i\lt j\lt k\leq 4$ contains not just $(1,2,3)$ and $(2,3,4)$

It contains $(1,2,4)$ and $(1,3,4)$ as well.

There are $\binom 43=4$ such tuples $(i,j,k)$ and indeed that gives you $$\geq\frac{\sum\limits_{1\leq i\lt j\lt k\leq 4}a_i+a_j+a_k}3\\ =\frac{(a_1+a_2+a_3)+(a_1+a_2+a_4)+(a_1+a_3+a_4)+(a_2+a_3+a_4)}3$$

which simplifies to $a_1+a_2+a_3+a_4$

In general, we have,

$$\sum_{1\leq i_1\lt i_2\lt\dots\lt i_k\leq n}\sum_{m=1}^k a_{i_m}=\binom{n-1}{k-1}\sum_1^n a_i$$

which can be proven by a simple counting argument. Try it out!

Following which, here is a more general version of the inequality given in your problem:

$$\sum_{1 \le i_1\lt i_2\lt\dots\lt i_k\leq n}\frac{\sum\limits_{m=1}^k a_{i_m}^2}{\sum\limits_{m=1}^k a_{i_m}} \geq \frac{\binom{n-1}{k-1}}k\sum_1^n a_i$$ with equality iff all $a_i$'s are equal.

which admits the same proof as yours.

The inequality in your question is the special case of the result above with $n=4$ and $k=3$


问:
Find expected number of draws from first principles
This is a variation of a common probability problem: Given a bag of N blue marbles and M red marbles what's the number of expected draws until you get a blue marble?

With $p_R = \frac{M}{M+N}$ and $p_B = \frac{N}{M+N}$, the expected number of trials before the first success is a discrete RV distributed according to a Geometric distribution. This distribution's expected value is known and in this case is $\mathbb{E}[X]= \frac{1}{p_{B}}=\frac{M+N}{N}$.

Given that this is correct (which I think it is), is there a way to derive this result from first principles, e.g. if we didn't know about the Geometric distribution?

答:
Let's say $D(N,M)$ is the expected number of draws until the first blue when starting from $N$ blue marbles and $M$ red marbles.  You start by drawing one marble and if it is blue you stop, while if you draw a red marble, with probability $\frac{M}{N+M}$ you draw again.

With replacement this means $D(N,M) = 1+ \frac{M}{N+M} D(N,M)$ which is easily solved to give (as you say) $$D(N,M) = \frac{N+M}{N}$$

Without replacement this means $D(N,M) = 1+ \frac{M}{N+M} D(N,M-1)$ starting with  $D(N,0)=1$ as you must draw blue if there is no red.  So $D(N,1)=1+\frac{1}{N+1}\times 1=\frac{N+2}{N+1}$ and then $D(N,2)=1+\frac{2}{N+2}\times \frac{N+2}{N+1} = \frac{N+3}{N+1}$ and so on.  It is an easy induction to show $$D(N,M) = \frac{N+M+1}{N+1}$$


问:
A probability problem involving probability density function
I am going through a probability book where the following has been claimed :

If $$f_{X}(x) = \int_{-\infty}^{\infty}f_{U}(x-y)f_{V}(y)dy$$

then $f_{X}=f_{U}*f_{V}$

where $X, U, V$ are continuous random variables where $U$ and $V$ are independent.

I don't understand this argument.

答: It's just the probability density of $X=U+V$, which is given by the convolution of the PDFs of $U$ and $V$. So it's just the definition of convolution, for which the symbol '$*$' is used. Addition of independent RVs corresponds to the convolution of their PDFs.</p>


问:
Maximum value at an endpoint
Let $f: R \to R$ be a continuous function that satisfies $\forall x \in R$
$$f(x) \leq \frac{f(x - h) + f(x + h)}{2}$$

$\forall h > 0$. Show that the maximum value of f on any bounded closed interval $[a,b]$ is attained at one of the endpoints. That is, either $f(a)$ or $f(b)$ is the maximum value of f on the interval [a,b].

I'm not actually convinced the maximum is at an endpoint. I'm honestly not sure how to analyze this function. Can anyone give me some insight here? Thank you!

Note: This is an an analysis class where we have not covered differentiation. So we cannot use that.

答:
If the maximum is not attained at the endpoints, then it is attained at some $c \in (a,b)$, with $f(a) < f(c)$ and $f(b) < f(c)$. Assume without loss of generality that $c$ is no closer to $b$ than it is to $a$, and let $h = c-a$. Then $f(c-h) = f(a) < f(c)$, and since $b$ is no closer to $c$ than it is to $a$, we have $c+h \in [a,b]$, which implies that $f(c+h) \leq f(c)$. Putting these together,

$$f(c) \leq \frac{f(c-h)+f(c+h)}2 < \frac{f(c)+f(c)}2 = f(c),$$

a contradiction.

Title: Maximum value at an endpoint


答: Suppose the maximum value $M$ of $f$on $[a,b]$ is attained at a point $c$. If $c=1$ or $c=b$ we are done. Suppose $a<c<b$. Let $h$ be the minimum of the numbers $c-a$ and $b-c$. Then $M=f(c)\leq \frac {f(c-h)+f(c+h)} 2\leq \frac {M+M} 2=M$. This forces $f(c-h)=f(c+h)=M$ I will leave it to you to check that $\{c-h,c+h\}$ contains one of the end points $a,b$.

Title: Maximum value at an endpoint

Body:
Let $f: R \to R$ be a continuous function that satisfies $\forall x \in R$
$$f(x) \leq \frac{f(x - h) + f(x + h)}{2}$$

$\forall h > 0$. Show that the maximum value of f on any bounded closed interval $[a,b]$ is attained at one of the endpoints. That is, either $f(a)$ or $f(b)$ is the maximum value of f on the interval [a,b].

I'm not actually convinced the maximum is at an endpoint. I'm honestly not sure how to analyze this function. Can anyone give me some insight here? Thank you!

Note: This is an an analysis class where we have not covered differentiation. So we cannot use that.


问:
Proving that a function is not holomorphic on the disk
Let $f \in C(\overline{D(0,1)})$ and $f(\frac{1}{n})=\frac{1}{n+1}$ for $n = 2,3 ...$ Show that $f \notin H(D(0,1))$

Attempted Solution: If I understand the problem correctly, I should show that $f$ is not differentiable on the unit disc centered at the origin. Hence, the limit:

$$ \lim_{z\rightarrow 0} \frac{f(z) -f(0)}{z-0}         $$

does not exist. I'm guessing that I should approach $0$ through the sequence $\frac{1}{n}$ and that should somehow enable me to show that the limit does not exist. But, I don't quite understand how I should go about doing that. Can anyone give me some clues?

答: Zeros of $f(z)-\frac z {z+1}$ have a limit point so the Identity Theorem shows $f(z)=\frac z {z+1}$  in the open disk.   But $\frac z {z+1}$ is not a continuous function on the boundary. [The limit of this function as $z \to -1$ does not exist].


问:
Probability of receiving a job offer on 3rd application
Suppose that when you submit a job application you have a probability of $0.1$ to receive an interview, and a job interview results in a job offer with probability $0.2$. Also assume you only submit one application at a time (i.e. you wait to know if you have been rejected from the job, either at the application stage or the interview stage before applying to another job).

What is the probability of getting your first job offer after submitting at most 3 job applications?

Attempt: Either you get a job offer after your first, second, or third application.

The probability of getting an offer after your 1st application is $(0.1)(0.2)=0.02$.

The probability of getting an offer after your 2nd application is $(1-0.02)(0.1)(0.2)=0.0196$

The probability of getting an offer after your 3rd application is $(1-0.02)(1-0.02)(0.1)(0.2)=0.0192$

Thus, $0.02+0.0196+0.0192=0.0588$

I don't think my approach is correct, it doesn't make sense to me that you have  a higher probability of getting an offer on your first application than you do on your 2nd or 3rd. I'm stumped as to what the mistake is that I am making.

答: The probability you labeled as that of getting an offer after your 2nd application is the probability that you didn't get an offer after your first application and you got an offer after your second...so your actual probability to get a job after the second application is 0.02+0.0192....


问:
Is my proof for Surjective functions correct?
The definition was: $f: A\rightarrow B$ is surjective if and only if $\forall C\subseteq B, f^{-1}(C)\not=\emptyset$

My proof:

Showing the forward direction we are given $f$ is surjective implying $C\not=\emptyset$. Then $\exists c\in C$ and $\exists a\in A$ such that $f(a)=c\in C$, by definition of surjectivity. Now, by definition of inverse images it is implied that $a\in f^{-1}(C)$, hence $f^{-1}(C)\not=\emptyset$. Now showing the backwards direction we assume $f^{-1}(C)\not=\emptyset$ with the assumption that $C$ is nonempty. Suppose there is a $b\in B$ then $b\in \{b\}$ which implies $\{b\}\subseteq B$. Since $f^{-1}(C)$ is nonempty we have an $a\in A$ such that $f(a)\in \{b\}$, but this implies there is a $a\in A$ such that $f(a)=b$. So for each element in the codomain $B$, there exists an element in $A$ that can be mapped back to the element in $B$. Hence $f$ is surjective.

I am 95% confident my proof is correct, but since this is a proof-'writing' class I wanted to get feedback on how I can improve wording, syntax, and anything else I might not be considering.

答:
Edited Proof:

Showing the forward direction we are given $f$ is surjective and $C\not=\emptyset$. Then let $c\in C$ be arbritary then there exists $a\in A$ such that $f(a)=c\in C$, by definition of surjectivity. Now, by definition of inverse images it is implied that $a\in f^{-1}(C)$, hence $f^{-1}(C)\not=\emptyset$. Now showing the backwards direction we assume $f^{-1}(C)\not=\emptyset$ with the assumption that $C$ is nonempty. Suppose there is a $b\in B$ then $b\in \{b\}$ which implies $\{b\}\subseteq B$. Since $f^{-1}(C)$ is nonempty we have an $a\in A$ such that $f(a)\in \{b\}$. So for each element in the codomain $B$, there exists an element in $A$ that can be mapped back to the element in $B$. Hence $f$ is surjective.

But now my issue is my conclusion feels wonky because I have been talking about the subsets but in my conclusion I am referring to the elements. Is that alright or is there a more concise way to finish it off with how I have written the proof.


问:
How do I solve the D'Alembert equation: $x(\frac{dy}{dx})=y+\frac{1}{4}(\frac{dy}{dx})^4$?
$$x\frac{dy}{dx}=y+\frac{1}{4}\left(\frac{dy}{dx}\right)^4$$

I know it's the D'Alembert equation. How can I solve it?

答:
Let $p=y'$. Then the equation becomes
$$ xp=y+\frac14 p^4. $$
Differentiating both sides w.r.t $p$ gives
$$ \frac{dx}{dp}p+x=\frac{dy}{dp}+p^3. $$
Using
$$ \frac{dy}{dp}=\frac{dy}{dx}\frac{dx}{dp}=p\frac{dx}{dp} $$
one has
$$ \frac{dx}{dp}p+x=p\frac{dx}{dp}+p^3 $$
or
$$ x=p^3. $$
So
$$ y'=p=\sqrt[3]{x} $$
and hence
$$ y=\frac{3}{4}x^{\frac{4}{3}}+C. $$