问:
How can I find the probability of a dart landing in region S?
Someone plays darts that has a join density function based on the surface z = 6 - r where z is the likelihood of the dart landing at $(r,\theta)$. How can I find the value of C so that f is a joint density function and then find the probability that the thrown dart lands in S.

$$f(r,\theta)=
\begin{cases}
C_{1}(6-r), & \text{if} (r,\theta)∈ R\\
0, & \text{otherwise}
\end{cases}$$

Where R is the region inside of r = 2+4cosθ (which includes the little loop) and S is just the region inside the inner loop.

Right now I am thinking this might be how to solve for C if that's correct how would I actually find the probability?

Here is your integral

答:
Your calculation of $C_1$ is correct, although most people would have chosen the limits $-2\pi/3 \le \theta \le 2\pi/3$.  Yours is simply $2\pi$ added to this.  The exact value is $$C_1 = \left(12 \sqrt{3} + \frac{208\pi}{9}\right)^{-1}.$$

To get the desired probability, we have to integrate for the inner lobe, i.e. $$2 \int_{\theta = 2\pi/3}^\pi \int_{r=0}^{|2 + 4 \cos \theta|} C_1 (6-r)r \, dr \, d\theta.$$  Note the use of $2$ (to exploit symmetry) and the absolute value on the upper limit of $r$ because for $2\pi/3 < \theta \le \pi$, we have $r(\theta) < 0$.  I get $$\frac{82\pi - 135\sqrt{3}}{52\pi + 27\sqrt{3}}.$$

Since there is interest in the actual computation of the integral for the inner lobe, I will provide an outline.  You should be able to fill in the gaps using various elementary trigonometric identities.  Note that on $\theta \in [2\pi/3, \pi]$, $r \le 0$, thus $(6-|r|)r = (6+r)r$, and

$$\begin{align}
2C_1 \int_{\theta = 2\pi/3}^\pi \int_{r=0}^{2 + 4 \cos \theta} (6+r)r \, dr \, d\theta
&= 2C_1 \int_{\theta = 2\pi/3}^\pi \left[3r^2 + \frac{r^3}{3}\right]_{r=0}^{2 + 4 \cos \theta} \, d\theta \\
&= \frac{8C_1}{3} \int_{\theta = 2\pi/3}^\pi  (1 + 2 \cos \theta)^2 (11 + 4 \cos \theta) \, d\theta \\
&= \frac{8C_1}{3} \int_{\theta = 2\pi/3}^\pi 41 + 60 \cos \theta + 30 \cos 2\theta + 4 \cos 3\theta \, d\theta \\
&= \frac{8C_1}{3} \left[ 41\theta + 60 \sin \theta + 15 \sin 2\theta + \frac{4}{3} \sin 3\theta \right]_{\theta=2\pi/3}^\pi.
\end{align}$$
The rest is obvious.


问:
The Newton Algorithm - Proving Convexity of $x^r$ and finding the global minima
I am working on some exercises for the Newton Algorithm and wanted to consult you all. For part c) I would like some help.

Just based on a graphical interpretation I would assume that $r$ just needs to be even because for odd $r$ the function is not convex on the entire domain, $x \in R$

Again for the global minima, I can show that the only stationary point occurs at $x=0$, and that the function is increasing on $[0, \infty)$ and on $(- \infty, 0]$ Is this sufficient?

I am a bit confused how to show the last part and would appreciate some hints. The newton algorithm has the form

$x^{k+1} = x^k + \alpha d_k$ where $d_k = \frac{\nabla f(x_k)}{||\nabla^2 f(x_k)||_2}$

Thanks for your help!

EDIT:

Thank you to the person who replied to me below. I have tried the following, compute derivative and second derivative and by analysing the second derivative, found that only for $r$ being even do we have that $f''(x) \geq 0, \forall x$.

For the stationary points I found the point where the derivative equals 0 which is $x* = 0$. Plugging this into the second derivative is useless so what I did is argue that for even $r$ we have a symmetric function whose derivative for $x>0$ is strictly increasing and so the point must be a global minima.

For part c) using the optimal $x* = 0$, the newton method became $x_k - \frac{\alpha x^k}{r-1}$ which I substituted into the inequality and got $$| $x_k - \frac{\alpha x^k}{r-1}| \geq (1-\delta)|x_k|$$ which upon rearranging I found that $\delta \geq \frac{\alpha}{r-1}$

What do you think? Is this valid/rigorous enough?

答:
We just need $r$ to be be positive even number. In particular, we can choose $r$ to be $2$ if you wish.

Indeed we have

$$|x^k-\alpha \cdot  \frac{x^k}{r-1}| = |x^k| \left|1-\frac{\alpha}{r-1}\right|$$

We can pick $\delta$ to be $\frac{\alpha}{r-1}$, provided we pick $\alpha$ to be less than $r-1$.

ContentLicense: CC BY-SA 4.0


问:
How do we solve the integral $\int \frac{1}{x^2-y}\mathrm{d}x$?
The integral to be solved is given by:

$$ I = \int \frac{1}{x^2-y}\mathrm{d}x$$

I was wondering what integral substitution I would need to make. I looked at symbolab and it directed me to use $x = u\sqrt{y}$ as a substitution, but where does it come from?

答:
You don't have to make a substitution to solve this integral. Take a look at the following if $y>0$:
$$\frac{1}{x^2-y} = \frac{1}{(x-\sqrt y)(x+\sqrt y)} = \frac{1}{2\sqrt y} \frac{(x+\sqrt y) - (x-\sqrt y)}{(x-\sqrt y)(x+\sqrt y)}$$
$$=\frac{1}{2\sqrt y} \left(\frac{1}{x-\sqrt y} - \frac{1}{x+\sqrt y}\right)$$
I hope you can directly the integrate this now. By the way, this is popularly known as the partial fractions technique.

If $y=0$, there is nothing to talk about. If $y<0$, you can still split the integrand as:
$$\frac{1}{x^2-y} = \frac{1}{(x-i\sqrt{-y})(x+i\sqrt{-y})}=\frac{1}{2i\sqrt{-y}} \left(\frac{1}{x-i\sqrt{-y}} - \frac{1}{x+i\sqrt{-y}}\right)$$
There is nothing special about the introduction of $i = \sqrt{-1}$, you can still integrate everything in the usual way, while keeping in mind that only the real part of the expression is of interest to us. The imaginary part will anyway turn out to be zero in this case.

P.S. I assumed $y$ is a constant w.r.t. $x$ because that's what we do unless otherwise specified.


问:
A basic intuition on a probability problem
Two players take turns shooting at a target, with each shot by player $i$ hitting the target with probability $p_i$, $i=1,2$. Shooting ends when two consecutive shots hit the target. Let $\mu_i$ denote the mean number of shots taken when player $i$ shoots first, $i=1,2$.

Now, I have calculated that $$\mu_1 - \mu_2 = \frac{p_2 - p_1}{p_1+p_2-p_1p_2-2}=\frac{q_2-q_1}{1+q_1q_2}$$

where $q_i=1-p_i, i=1,2 $

i.e. if the player with higher winning probability starts the game then the mean number of shots taken is higher than the case when the the player with lower winning probability starts the game. I don't understand the intuition behind this.

答: For the intuition, take an extreme case, where the better player is almost sure to hit, and the other player is quite a bit less accurate. If the bad player goes first, there are two chances for an end by the time $3$ shots have been taken. If the good player goes first, an end by the time $3$ shots have been taken is less likely. </p>

ContentLicense: CC BY-SA 3.0


问:
Evaluate $\int_{0}^{\pi/2}\cos^{2n}(x)\text{d}x$.
I try this. Notice that,
$$
\begin{split}
\cos^{2n}x &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n} = \frac{1}{2^{2n}} \sum_{k=0}^{2n} \binom{2n}{k}e^{ikx}e^{-i(2n-k)x} \\ &= \frac{1}{2^{2n}} \sum_{k=0}^{2n} \binom{2n}{k}e^{i(2k-2n)x} \end{split}
$$
The terms with $k\ne n$ integrate to zero over $[0,\pi/2]$, and we are left with
$$
\int_0^{\pi/2}\cos^{2n}x \,dx = \int_0^{\pi/2}\frac{1}{2^{2n}}\binom{2n}{n} \,dx = \frac{\pi}{2^{2n+1}}\binom{2n}{n}
$$
I'm right or not?

答: The answer is correct, the reasoning is wrong, since $$\int_0^{\pi/2} \exp(2 i x) d x = i \neq 0.$$


问:
Group of Order $5$ Generated by $(1,2,3,4,5)$
I'm reading Algebraic Combinatorics by Richard P. Stanley and had a quick question about some verbiage:

"Let $G$ be the group of order $5$ generated by the cycle (1,2,3,4,5)"

Doesn't this cycle generate $S_5$ which has order $120$? What group would it be with order $5$? What does this group look like?

答:
There's only one group of order  $5$: the cyclic one.  One can think in terms of  $\mathbb Z_5:=\mathbb Z/5\mathbb Z $.  That's the additive group of integers $\bmod5$.

If you're thinking  of $S_5$, then the cycle $(12345) $ has order $5$.  Thus it generates a copy of the cyclic group of order  $5$.  That is $\langle (12345)\rangle \cong C_5$, as it is also denoted.

In general,  an element $a $ of order  $n $ in a group $G $ generates a cyclic group of order  $n $:  $\langle a\rangle\cong C_n\le G$.

Also there's only one group of order any given prime (up to isomorphism), and it's cyclic


答:
I like this elegant proof and you are right. However, perhaps the logic doesn't quite follow, it may just be how you've written it for brevity.

A given component $e^{i(2k-2n)x}$ does not necessarily have

$$\int_{0}^{\frac{\pi}{2}}e^{i(2k-2n)x} dx = 0$$

e.g. take $k=n+1$. However, we in fact have that

$$\int_{0}^{\frac{\pi}{2}}e^{-2kix} + e^{2kix}dx = 2\int_{0}^{\frac{\pi}{2}} \cos\left(2kx\right)dx=0$$

and this is why your proof holds and symmetry of the binomial coefficient.

Title: Evaluate $\int_{0}^{\pi/2}\cos^{2n}(x)\text{d}x$.

Body:
I try this. Notice that,
$$
\begin{split}
\cos^{2n}x &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n} = \frac{1}{2^{2n}} \sum_{k=0}^{2n} \binom{2n}{k}e^{ikx}e^{-i(2n-k)x} \\ &= \frac{1}{2^{2n}} \sum_{k=0}^{2n} \binom{2n}{k}e^{i(2k-2n)x} \end{split}
$$
The terms with $k\ne n$ integrate to zero over $[0,\pi/2]$, and we are left with
$$
\int_0^{\pi/2}\cos^{2n}x \,dx = \int_0^{\pi/2}\frac{1}{2^{2n}}\binom{2n}{n} \,dx = \frac{\pi}{2^{2n+1}}\binom{2n}{n}
$$
I'm right or not?


问:
Set of solutions for nearest point to a convex polyhedron (simplex)
I'm working on a problem that involves the sets of solutions to the problem of finding nearest point of a convex polyhedron to a given point. Let $y$ be a point in a convex polyhedron defined by a set of linear inequalities $Ay \leq b$ and $x \in \Bbb R^n$ a target point. Specifically, I'm concerned with solutions when in terms of Euclidean distance where the convex polyhedron is the interior of a simplex, i.e.,

\begin{gather*}
  \min_y \, y^\top y - 2 x^\top y \\
  \text{subject to } Ay \leq b
\end{gather*}

with

\begin{align*}
  A = \begin{bmatrix} -I_n \\ 1_n^\top \end{bmatrix} \quad \text{and} \quad b = \begin{bmatrix} 0_n \\ 1 \end{bmatrix} .
\end{align*}

Now, my question concerns the solutions as $x$ changes. In particular, I'm pretty sure that the sets of $\Bbb R^n$ that get mapped to the simplex in this way can be separated by half-planes. Intuitively, this is because for each $x$ not in the simplex, the nearest point should either be a corner or a face. Using the algorithm here, I convinced myself this was true for $\Bbb R^2$, as I could find a pattern that looked something like this:

where the orange sets are closest to a corner and the blue sets are closer to a face. I'm looking for either:

A reference to this type of result and, ideally, a way to describe those sets that get mapped to the corners and faces when the dimensionality is larger than 2.

A way to make this argument rigorous.

答:
I had written a previous answer, but I think my new answer supersedes it, so I'm replacing the old one with what follows.

I claim that if $S$ is convex polyhedron in $\mathbf{R}^n$, then:

for each $x \in \mathbf{R}^n$, there is a unique point $f(x) \in S$ closest to $x$;

for each $x \in \mathbf{R}^n$, the point $f(x)$ is the orthogonal projection of $f$ onto some face of $S$ (of dimension $0, \dots, n$);

this face can be chosen to be the smallest face of $S$ containing $f(x)$;

this choice being made, the boundaries of the sets in the corresponding partition of $\mathbf{R}^n$ are contained in the union of finitely many hyperplanes $H$, which can be determined as follows: for each face $k$-dimensional face $F$ of $S$ ($k = 1, \dots, n$) , and each $(k-1)$-dimensional subface $F'$  of $F$, let $H$ be the collection of all points whose orthogonal projection onto the subspace extending $F$ belongs to the subspace extending $F'$;

if $T \subseteq \mathbf{R}^n$ is one of the sets of the partition, and $F$ is the associated face, then $f(x)$ is the orthogonal projection of $x$ onto $F$ for all points $x$ in the closure of $T$ (not only for $x$ in $T$).

That the boundaries can be curved when $S$ is not assumed convex is shown by the following example. Let $S$ be the quadrilateral $ABCD$ with $A = (1,0)$, $B = (0,0)$, $C = (0,2)$, $D = (-1,-1)$. Then one of the boundaries is a portion of the parabola with focus $A$ and directrix $BC$.

I'll begin with some general comments about projecting onto a convex set.

If $S$ is any subset of $\mathbf{R}^n$, the function $x \mapsto d(x,S)$ is continuous on $\mathbf{R}^n$. If $S$ is closed, then by a simple compactness argument, the distance $d(x,S)$ is attained at at least one point $s \in S$.

If $S$ is closed and convex, then this point $s$ is unique, for if the minimum distance were attained in two points $s_1$ and $s_2$, the midpoint of the segment $s_1s_2$ would be still closer to $x$. Thus we can define a projection function $f \colon \mathbf{R}^n \to S$ assigning to each $x \in \mathbf{R}^n$ the unique point $f(x)$ of $S$ that is closest to it.

The projection function $f$ is in fact continuous. If $x \in \mathbf{R}^n$ and $\varepsilon > 0$, let $S' = S - B(f(x),\varepsilon)$. The distance $d(x,S')$ is attained at some point $s' \in S'$, and by the uniqueness of the projection we have $d(x,S') = d(x,s') > d(x,f(x)) = d(x,S)$. Now as $y \to x$, we have $d(y,S) \to d(x,S)$ and $d(y,S') \to d(x,S')$, hence for $y$ sufficiently close to $x$, we must have $f(y) \in B(f(x),\varepsilon)$, which shows that $f$ is continuous.

Now assume that $S$ is a convex polyhedron. For any $s \in S$, there is a face of minimum dimension $d$ (with $0 \leq d \leq n$) to which $s$ belongs. This face is contained in all other faces containing $s$, and it is also the unique face containing $s$ within its interior (where "interior" means the interior relative to the subspace extending the face). Thus we can partition space into a finite number of subsets by assigning to each $x \in \mathbf{R}^n$ the face associated in this way with $f(x)$.

If $f(x)$ belongs to the interior of a particular face $F$, it is easy to see that $f(x)$ is determined as the orthogonal projection of $x$ onto the subspace extending $F$ (since otherwise there must be nearby points of $F$ that are closer to $x$). More generally, if $f(x)$ belongs to a face $F$, though not necessarily to its interior, then $f(x)$ must be the orthogonal projection of $x$ onto the subspace extending some subface of $F$.

Now assume that $x$ is a boundary point of one of the subsets into which $\mathbf{R}^n$ is partitioned. Then $x$ is as close as we like to points associated with two distinct faces, say $F$ and $G$. By the continuity of $f$, the point $f(x)$ belongs to $F \cap G$. Thus
$f(x)$ can be obtained by orthogonal projection onto some subface of $F \cap G$. At the same time, by continuity, $f(x)$ must be the orthogonal projection of $x$ onto $F$ as well as onto $G$. The faces $F$ and $G$ being assumed distinct, in any case $f(x)$ is simultaneously the orthogonal projection of $x$ onto a face of $S$ (either $F$ or $G$) and a proper subface of that face. Since we can always replace the subface by one of dimension one less than that of the face, the result follows.


问:
A triple integral over the unit ball
I'm trying to find the following triple integral over the unit ball.

$$\iiint_{x^2 + y^2 + z^2 \leq 1} e^{(1-x^{2}-y^{2})^{3/2}} {\rm d} x \, {\rm d} y \, {\rm d} z$$

I am able to find a suitable parametrization but I can't get the bits in the exponent in terms of $r$.  Any hints would be appreciated.

答: Hint. Use cylindrical coordinates, integrating with respect to $z$ on the inside.

Title: A triple integral over the unit ball


问:
Connection between universal properties and the existence of a left adjoint
I've noticed that in many cases, whenever a functor $G:\mathscr B\to\mathscr A$ has a left adjoint, there's some kind of universal property around (I don't know how to state this more precisely other than using the word "around"). Examples (without explanation):

$G: \mathbf{Vect}_k\to \mathbf{Set}$ is the forgetful functor;

$G: \mathbf{Grp}\to \mathbf{Set}$ is the forgetful functor;

$G: \mathbf{Ab}\to \mathbf{Grp}$ is the functor that forgets that an abelian group is abelian;

$G: \mathbf{Set}\to \mathbf{Top}$ is the functor that assigns to a set the corresponding topological space with the indiscrete topology;

$G: \mathbf{Top}\to \mathbf{Set}$ is the forgetful functor.

So I was wondering whether there's some general connection between universal properties and the existence of left adjoints. This question probably is too imprecise, but here's what I got when trying to generalize the above examples:

Suppose $F: \mathscr A \to \mathscr B$, $G:\mathscr B\to \mathscr A$, and $F$ is left adjoint to $B$.  Then for all $A\in\mathscr A, B\in\mathscr B$, $$\mathscr B(F(A),B)\simeq \mathscr A(A,G(B)).$$

In this case, my conjecture is that there is the following universal property:

For any $\phi\in \mathscr A(A,G(B))$, there exists a unique $\overline \phi\in \mathscr (F(A),B)$ such that the following diagram commutes:

When $G$ is one of the functors from the above example, it's always clear what $\pi$ is, but in the general case I don't really see what the arrow $\pi$ would be. Is there a way to define it in general so that the diagram commutes? If my conjecture is true, I would call the quoted part "the general (case of) universal property" (not sure if this is standard or not). And it would follow that whenever a functor has a left adjoint, there is an associated universal property.

Apparently, the converse of my conjecture is true (if there's a way to define $\pi$ that I'm confused about) because if there's such universal property, one could define the isomorphism from the definition of adjunction by $\phi\mapsto \overline\phi$ and in the other direction by $\psi\mapsto G(\psi)\circ \pi$. (Let me know if this is not true.)

Is there more stuff that can be said about the connection of left adjoints and universal properties?

答: If you look at the Wikipedia article article on adjoint functors, one of the three equivalent definitions of an adjoint pair uses universal properties. To answer one of your specific questions: your map $\pi : A \to GF(A)$ is the unit of the adjunction evaluated at $A$ and is given explicitly by applying the hom-set adjunction to the identity map $1_{F(A)} : F(A) \to F(A)$.


问:
Why is $\sqrt{-8}\times\sqrt{-2}=-4$

Why is $\sqrt{-8}\times\sqrt{-2}=-4$

<img src="https://i.stack.imgur.com/Mbli9.png" alt="$\sqrt{-8}\times\sqrt{-2}=-4$" />

答:
As mentioned in the comments, the correct answer is $-4$.

If you are in case, wondering about why this is not equal to $\sqrt{16}=4$, its becuase

$$\sqrt{xy}=\sqrt x\sqrt y$$ if and only if at least one of $x,y$ is non negative

Title: Why is $\sqrt{-8}\times\sqrt{-2}=-4$


答:
The integral boils down to

$$\begin{aligned} I &= \iiint ^{\sqrt{1-\left( x^{2} +y^{2}\right)}}_{-\sqrt{1-\left( x^{2} +y^{2}\right)}} e^{\left( 1-x^{2} -y^{2}\right)^{\frac{3}{2}}} {\rm d} z \, {\rm d} x \, {\rm d} y\\ &= 2\iint e^{\left( 1-x^{2} -y^{2}\right)^{\frac{3}{2}}}\sqrt{1-\left( x^{2} +y^{2}\right)} {\rm d}x \, {\rm d}y \end{aligned}$$

where the double integral must be calculated over the unit circle. Substitute $x = r \cos \theta$ and $y=r\sin \theta$

$$\begin{aligned} I=\iint e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdrd\theta \\
=\int ^{1}_{0} e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdr\cdotp \int d\theta \\ \end{aligned}$$
Consider the integral

$$\begin{aligned}A=\int ^{1}_{0} e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdr\ \end{aligned}$$
Substitute$$\begin{aligned}\ 1-r^{2} =t^{2}\\
-2rdr=2tdt\\
rdr=-tdt\\ \end{aligned}$$

$$\begin{aligned}{A=\int ^{1}_{0} e^{t}}^{3}\cdotp t^{2} dt=\frac{e-1}{3}\\
So,\ I=4\pi \cdotp \frac{e-1}{3}\\ \end{aligned}$$
Does this help?

Title: A triple integral over the unit ball

Body:
I'm trying to find the following triple integral over the unit ball.

$$\iiint_{x^2 + y^2 + z^2 \leq 1} e^{(1-x^{2}-y^{2})^{3/2}} {\rm d} x \, {\rm d} y \, {\rm d} z$$

I am able to find a suitable parametrization but I can't get the bits in the exponent in terms of $r$.  Any hints would be appreciated.


答:
Here, you need to regard $y$ as a constant, at least for this integral.

The right substitution really depends on the sign of $y$. If $y > 0$, then $x = u\sqrt{y}$ is appropriate. If $y < 0$, then you want $x = u \sqrt{-y}$.

The advantage of making the substitution versus what epsilon-emperor does, is that you will need to integrate a rational function, either $\frac{1}{u^2 - 1}$ or $\frac{1}{u^2 + 1}$ that has no parameters in it, rather than $\frac{1}{x^2 - y}$, which is a pain.

The answers will look different in each case. You know that the integral of $\frac{1}{u^2 + 1}$ will be an arctangent function, whereas (at least if you want to avoid inverse hyperbolic functions), you will integrate $\frac{1}{u^2 - 1}$ by partial fractions.

Title: How do we solve the integral $\int \frac{1}{x^2-y}\mathrm{d}x$?

Body:
The integral to be solved is given by:

$$ I = \int \frac{1}{x^2-y}\mathrm{d}x$$

I was wondering what integral substitution I would need to make. I looked at symbolab and it directed me to use $x = u\sqrt{y}$ as a substitution, but where does it come from?


答:
Epsilon-Emperor's trick is very good. Here is the solution using trigonometry substitution giving the same result (assuming y>0;if y<0, use similar  $\tan \theta$ substitution instead；if y=0, then it is trival).

$I = \int \frac{1}{x^2-y}\mathrm{d}x$

${x^2-y}=$y(($\frac{x}{\sqrt y})^2-1)$

set $\frac{x}{\sqrt y}$= $\sec\theta$,$dx=\sqrt y  $ $\sec\theta$   $\tan \theta $ d$ \theta$

$y((\frac{x}{\sqrt y})^2-1)$ = $y\tan\theta ^2$,

$I = \int \frac{1}{y\tan\theta ^2}\sqrt y  $ $\sec\theta$   $\tan \theta$ d$ \theta$

=$\int \frac{1}{\sqrt y\tan\theta}  $ $\sec\theta$ d$ \theta$

=$\int \frac{1}{\sqrt y}  $ $\csc\theta$ d$ \theta$

=$\frac{1}{\sqrt y} $$ln|\csc\theta$-$\cot\theta$|+C

= $\frac{1}{2\sqrt y}$ $ln|\frac{x-\sqrt{y}}{ x+\sqrt {y}}$|+C

(last step from $\frac{x}{\sqrt y}= \sec \theta$)


问:
differences between Ideals and principal ideals in math notational descriptions
In the following post: (Confusion between principal ideal and ideal) on clarification between the concepts of ideals and principal ideals, @Yury stated: "... if $I$ is a principal ideal then every element of $I$ is a multiple of $$ (for some fixed $\in I$) (2) Every ideal contains a principal ideal but not the other way around."  I try to write out precisely what it means by not every ideal is a principal ideal in mathematical notation.

What I would like is to see if I can phrase what Yury stated in mathematical notations. I know this may sound a bit pedantic.  I just would like to make sure I am crystal clear if I cross all the Ts and dot all the Is when it comes to formulating it in terms of the correct quantifiers.

Definition of an Ideal:

An $\textbf{ideal}$ of a ring $R$ is a subring $I$ of $R$ such that for all $x\in R$ and $y\in I$, both $xy \in I$ and $yx \in I$

Definition of Principal Ideal:

Let $R$ be a commutative ring with a unit element.  An ideal $I$ of $R$ is $\textbf{principal}$ if there exists $d\in R$ such that $I=(d)=
\{rd\mid r\in R\}.$  In this case, $d$ is said to $\textbf{generate} I$

Yury's statement part (1) would mean: if we given a commutative ring $R$ with unit element, for any ideal $I$ of $R$, there exists an $a\in R$ such that $(a) \subset I$, where $(a)=\{ar: \forall r \in R\}$ contains principal ideals trivially.   For part (2) of his statement, in mathematical notation, it translate to:  There exists an ideal $I$ of $R$ such that $I \not\subset (a)$, meaning there exists an ideal $I$ of $R$ and a $x \in I$ $x \neq ar$ for all $a$, $r \in R$

I am not certain if how I put Yury's statements in mathematical notations is accurate.  IF someone can comment and point out any errors, it will be much appreciated.  Thank you in advance

答:
I'll stick to the commutative case for clarity. I'm not sure what @Yury means by "but not the other way around" but hopefully the following bullet points will clear up the confusion.

If $R$ is a commutative ring and $I$ is an ideal of $R$, then for every $a \in I$, we have that $(a) \subseteq I$. In particular since $0 \in I$, $I$ contains a principal ideal.

It is trivially true that every ideal is contained in a principal ideal, namely $I \subseteq R = (1)$.

There are some rings where every ideal is principal, these are called principal ideal rings.

There are also some rings with ideals which are not principal. The two classic examples of this are that the ideals $(2 , x) \subseteq \mathbb{Z}[x]$ and $(x , y) \subseteq k[x , y]$ (for $k$ any field) are not principal.


问:
Pointwise and uniform convergence of $f_n(x)=\sqrt[n]{nx^2+1}$
Study the pointwise and uniform convergence of $f_n(x)=\sqrt[n]{nx^2+1}$ for $x\in\mathbb{R}$.

My attempt: let $x_0 \in \mathbb{R}$ fixed, it is
$$\lim_{n\to\infty} \sqrt[n]{nx^2_0+1}=1$$
So $f_n$ converges pointwise to $f(x)=1$ for all $x\in\mathbb{R}$.

For the study of the uniform convergence, I have to evaluate
$$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}|\sqrt[n]{nx^2+1}-1|$$
The $n$-th root is an increasing function, so since $nx^2+1 \geq 1$ for all $x\in\mathbb{R}$ it is $\sqrt[n]{nx^2+1}-1 \geq 0$ and so $|\sqrt[n]{nx^2+1}-1|=(\sqrt[n]{nx^2+1}-1)$; moreover, it is
$$\frac{\text{d}}{\text{d}x}(\sqrt[n]{nx^2+1}-1)=\frac{2x}{n}(nx^2+1)^{\frac{1-n}{n}} \geq 0 \iff x \geq 0$$
So the function $f_n-f$ is increasing function of $x$ and it is an unbounded function of $x$, since
$$\lim_{x \to \infty} (f_n(x)-f(x)) = \lim_{x \to -\infty} (f_n(x)-f(x)) =\infty$$
So it is
$$\sup_{x \in \mathbb{R}}(\sqrt[n]{nx^2+1}-1)=\infty$$
So $\lim_{n \to \infty} \sup_{x \in \mathbb{R}} |\sqrt[n]{nx^2+1}-1|=\infty$, this means that the $f_n$ doesn't converge uniformly to $f(x)=1$ in $\mathbb{R}$.
However, given $a>0$ and considered the interval $[-a,a]$, by the same argument of monotonicity it is
$$\sup_{x \in [-a,a]} |\sqrt[n]{nx^2+1}-1|=\sqrt[n]{na^2+1}-1 \to 0 \ \text{as} \ n \to \infty$$
So $f_n$ converges uniformly to $f(x)=1$ in every compact subset $[-a,a]$ of $\mathbb{R}$.

My questions are the following:

When I study the positivity of the derivative, I omit the term $(nx^2+1)^{\frac{1-n}{n}}$ because I see it as an exponential and so I conclude it is always positive. I have in general problems with the $n$-th power and $n$-th roots, is this correct?

When I study the uniform converge in $\mathbb{R}$ I have to take the limit of a supremum which is already $\infty$; this never happened in my previous study, so I'm unsure about this step. It has some sense to take the limit of something that is already $\infty$?

The fact that the convergence is uniform in every compact subset of $\mathbb{R}$ but it can't be extended in $\mathbb{R}$ is because when we introduce $a>0$ for defining the interval $[-a,a]$ we are fixing $a$ and so we can conclude with the estimation on the supremum while the elements of $\mathbb{R}$ are considered elements that vary without a fixed bound (unlike $a$) and this makes the difference for the uniform convergence? Or is there a deeper reason why the uniform convergence fails in $\mathbb{R}$ but holds in every compact subset of $\mathbb{R}$?

Thank you.

答:
When you look at the limit for fixed $x_0$, why aren't you justifying the limit you write? I don't think it's dead obvious that limit is $1$.

After that, when you try to find the supremum of $|\sqrt[n]{nx^2 +1} - 1|$ for $x \in R$, what is your first impression about how big you can make this expression, if you're allowed to choose $x$ any way you like?


问:
How many sequences of length 10 with elements $\{a, b, c, d\}$ have exactly $3$ out of $4$ elements?
My logic is since $3$ out of $4$ elements are chosen, each element would appear once.
So a sequence would look like: $a\,b\,c\,x\,x\,x\,x\,x\,x\,x$

We have $7$ spots $x$ that can be whatever elements so: $3^7$

Choosing $3$ out of $4$ elements:  $4\choose 3$$ = 4$
So in total, there are $4\cdot(3^7)$ sequences

Can someone tell me if this is right or not? Thanks

答:
The Stirling number of the second kind $\left\{n \atop k\right\}$ counts the number of partitions of an $n$-set into exactly $k$ parts, so $k! \left\{n \atop k\right\}$ is the number of surjections of an $n$-set onto a $k$-set.  Your desired count is hence
$$\binom{4}{3} 3! \left\{10 \atop 3\right\} = 4 \cdot 6 \cdot 9330 = 223920.$$


答:
The argument below should also give you the intuition as to why if $p_1 > p_2$, we have $\mu_1 > \mu_2$. What the argument says is that the onus is on the shooter who shoots given that the other shooter has hit the target. We hence want this person to have a higher probability of actually hitting the target.

If player $1$ starts first, then the expected number of shots fired is given by
\begin{align}
\mu_1 & = \underbrace{2p_1p_2}_{2 \text{ shots if both shoot on their first attempt}} \\
& + \underbrace{p_1(1-p_2)(2+\mu_1)}_{\text{If $1$ hits and $2$ fails, need to restart with $1$ and $2$ shots have gone}}\\
& + \underbrace{(1-p_1)(1+\mu_2)}_{\text{If $1$ doesn't hit, then restart with $2$ and $1$ shot is already gone}}
\end{align}
Similarly, if player $2$ starts first, we then have
$$\mu_2 = 2p_2p_1 + p_2(1-p_1)(2+\mu_2) + (1-p_2)(1+\mu_1)$$
Subtract the two equations, do some nice algebra and you will be able to isolate the $\mu_1 - \mu_2$ term.

Another way would be to solve for $\mu_1$ and $\mu_2$ explicitly and compute $\mu_1 - \mu_2$.

Title: A basic intuition on a probability problem

Body:
Two players take turns shooting at a target, with each shot by player $i$ hitting the target with probability $p_i$, $i=1,2$. Shooting ends when two consecutive shots hit the target. Let $\mu_i$ denote the mean number of shots taken when player $i$ shoots first, $i=1,2$.

Now, I have calculated that $$\mu_1 - \mu_2 = \frac{p_2 - p_1}{p_1+p_2-p_1p_2-2}=\frac{q_2-q_1}{1+q_1q_2}$$

where $q_i=1-p_i, i=1,2 $

i.e. if the player with higher winning probability starts the game then the mean number of shots taken is higher than the case when the the player with lower winning probability starts the game. I don't understand the intuition behind this.


问:
If $\lim_{x \to 0}g(x)=L$. How do I prove $\lim_{h \to 0} g(f(x+h)-f(x)) = \lim_{y \to 0}g(y)$ using $\epsilon-\delta$
If I know $\lim_{x \to 0}g(x)=L$. For my particular problem $L=0$.
So I've been trying to prove: $\lim_{h \to 0} g(f(x+h)-f(x)) = \lim_{y \to 0}g(y)=0$ using $\epsilon-\delta$.

(Also I'm assuming the $g(x)$ is continuous at $0$ and $f(x)$ is continuous at $x$)

So I know that $\forall\epsilon$, there is some $\delta$ such that
$0<|k|<\delta\Rightarrow|g(k)-L|<\epsilon$    (1)

I have to prove that $\forall\epsilon_1$,$\exists\delta_1$ such that
$0<|h|<\delta_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon_1$

If $\exists\epsilon=\epsilon'$, and $\delta=\delta'$ satisfies (1), such that
$0<|k|<\delta'\Rightarrow|g(k)-L|<\epsilon'$    (2)

If I to find value for $\delta_1$, when $\epsilon_1=\epsilon'$ such that
$0<|h|<\delta_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$  (3)

I can prove $\forall\epsilon_1,\exists\delta_1$.

If I take $f(x+h)-f(x)=k$,  then RHS of (2) and (3) are same.
$$\Rightarrow0<|f(x+h)-f(x)|<\delta'\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$$
So I have to find $\delta_1$ such that
$0<|h|<\delta_1\Rightarrow0<|f(x+h)-f(x)|<\delta'$?

This seems similar to another $\epsilon-\delta$ type condition but...I'm not sure.

I know $\lim_{h\to0}f(x+h)=f(x)$, since I assumed $f(x)$ is continuous at $x$.

Since $|h|>0\Rightarrow f(x+h)\neq f(x)\Rightarrow|f(x+h)-f(x)|>0$? If this is true, I think it has been proved.

(Also please check if there are any other mistakes or misassumptions)

答:
For given $\epsilon >0 ~~~\exists~ \delta'>0$ such that $|g(x)-L|<\epsilon$  whenever $|x|<\delta'$   $\hspace{1cm}$  $\cdots$ (1).

Now since $f$ is continuous at $x$, for the above $\delta'>0~~~\exists~\delta>0$ such that $|f(x+h)-f(x)|<\delta'$ whenever $|h|<\delta$. From (1) we get $|g(f(x+h)-f(x))-L|<\epsilon$ whenever $|h|< \delta$. This proves it.


问:
$0$ element of $A\otimes \mathbb{Q}$
I am studying tensor product and I feel that it is not easy to distinguish $a\otimes b=0$ or $\neq 0$.

I'm thinking the following special case and it should be true.

Let $A$ be an abelian group, in other word a $\mathbb{Z}$ module, and  $a\otimes 1 \in A\bigotimes_{\mathbb{Z}} \mathbb{Q}$.
Then, $a\otimes 1 =0$ if and only if there exists some $0\neq n\in \mathbb{Z}$ such that $na=0$.

I think "if part" is trivial, but the other part is difficult. I want a proof of this claim or a counter examaple.

答:
When you tensor over $\mathbb{Z}$, you can pass integers over the tensor. As you seem to be alluding to, this allows you to annihilate torsion elements from $A$. Specifically, if $an = 0 \in A$, then
$$
a \otimes 1
= a \otimes \bigl( n \cdot \tfrac{1}{n} \bigr)
= (an) \otimes \tfrac{1}{n}
= 0 \otimes \tfrac{1}{n}
= 0 \otimes 0 \in A \otimes \mathbb{Q}
$$

How do you characterize a nonzero element $a \in A$ that is not torsion? If $na = 0 \in A$ then $n=0$ (linear independence of the singleton set $\{a\}$). So every pure tensor looks like
$$
a \otimes 1 = (an) \otimes \tfrac{1}{n},
$$
but $an \neq 0 \in A$ and $\tfrac{1}{n} \neq 0 \in \mathbb{Q}$, so this is not the zero element in $A \otimes \mathbb{Q}$.

Title: $0$ element of $A\otimes \mathbb{Q}$


答:
Define the abelian group $A_0$ as follows:

The underlying set is $(A \times (\mathbb{Z} \setminus \{0\})) / \sim$, where $\sim$ is the equivalence relation defined by $(a, n) \sim (b, m) \iff \exists u \in \mathbb{Z} \setminus \{0\} (u(ma - nb) = 0)$. The group operation is defined by

$$[(a,n)] + [(b,m)] = [(ma+nb,nm)]$$

(check this is well-defined and gives an abelian group!)

In other words, $A$ is the abelian group of formal quotients $a/n$ where $a \in A$ in $n \in \mathbb{Z} \setminus \{0\}$. In even more different words, we have just constructed the localization $A_0$. By basic facts of commutative algebra, $A_0 \cong A \otimes_{\mathbb{Z}} \mathbb{Z}_0 = A \otimes_{\mathbb{Z}} \mathbb{Q}$. I'll explain this isomorphism, then say how it solves the problem.

Claim $A_0 \cong A \otimes_{\mathbb{Z}} \mathbb{Q}$.

Proof Sketch. First, we define $[(a,n)] \mapsto a \otimes 1/n : A_0 \to A \otimes_{\mathbb{Z}} \mathbb{Q}$ -- check this is well-defined and a homomorphism. Next, we construct the inverse by the universal property of tensor products. Define the function $(a,p/q) \mapsto [(pa,q)] : A \times \mathbb{Q} \to A_0$. Check that this is $\mathbb{Z}$-bilinear, so it yields a homomorphism $A \otimes_{\mathbb{Z}} \mathbb{Q} \to A_0$ determined by $a \otimes p/q \mapsto [(pa,q)]$. Check that this is inverse to the homomorphism we constructed previously.

Corollary $a \otimes 1 = 0$ in $A \otimes_{\mathbb{Z}} \mathbb{Q}$ if and only if $na = 0$ for some $n \in \mathbb{Z} \setminus \{0\}$.

Proof. The element $a \otimes 1 \in A \otimes_{\mathbb{Z}} \mathbb{Q}$ corresponds to $[(a,1)] \in A_0$ via the aforementioned isomorphism, so $a \otimes 1 = 0$ if and only if $[(a,1)] = 0 = [(0,1)]$. By definition, this is equivalent to the existence of some $n \in \mathbb{Z} \setminus \{0\}$ such that $na = n(1 \cdot a - 1 \cdot 0) = 0$, as desired.

Title: $0$ element of $A\otimes \mathbb{Q}$

Body:
I am studying tensor product and I feel that it is not easy to distinguish $a\otimes b=0$ or $\neq 0$.

I'm thinking the following special case and it should be true.

Let $A$ be an abelian group, in other word a $\mathbb{Z}$ module, and  $a\otimes 1 \in A\bigotimes_{\mathbb{Z}} \mathbb{Q}$.
Then, $a\otimes 1 =0$ if and only if there exists some $0\neq n\in \mathbb{Z}$ such that $na=0$.

I think "if part" is trivial, but the other part is difficult. I want a proof of this claim or a counter examaple.


问:
Applying the Inverse Laplace Transform to both sides of an equation
In the Laplace domain, we have this:
$$X_1(s) = \frac{1}{s+4}(U(s)-X_2(s))$$
I want to get this by:
$$\dot x_1(t) = -4x_1(t) + u(t)-x_2(t)$$
This is done by rearranging the first equation into:
$$(s+4)X_1(s) = U(s)-X_2(s)$$
Then take the Inverse Laplace Transform on both sides.
However, I would like know if I could achieve the same thing without rearranging the equation 1 first. Taking the Inverse Laplace Transform directly on both sides of the first equation resulting:
$$x_1(t) = e^{-4t}(u(t)-x_2(t))$$
Then take derivative with respect to $t$ on both sides:
$$\dot x_1(t) = -4e^{-4t}(u(t)-x_2(t)) + e^{-4t}(\dot u(t)-\dot x_2(t))$$
$$\dot x_1(t) = -4x_1 + e^{-4t}(\dot u(t)-\dot x_2(t))$$

I got stuck here. If this cannot be converted to the desired form, what is the reason for this?

答:
The Mistake

Your mistake is here:

$$
X_1(s) = \frac{1}{s+4} \cdot(U(s)-X_2(s)) \xrightarrow{\mathscr{L}^{\,-1}} x_1(t) = e^{-4t}\cdot(u(t)-x_2(t))
$$

In doing so, you have implicitly assumed the following equality as true:

$$
\color{red}{
\mathscr{L}^{-1} \left[ F(s) \cdot G(s) \right] = f(t) \cdot g(t)} \qquad \text{(Wrong!)}
$$

The Right Way

The correct version involves the convolution integral:

$$
\bbox[5px,border:1.1px solid black] {
\mathscr{L}^{-1} \left[ F(s) \cdot G(s) \right] = \left( {f * g} \right)\left( t \right) = \int_{{0}}^{{t}}{{f( {t - \tau } )g\left( \tau  \right)\,d\tau }}
}
$$
(Note that a convolution operation in time domain corresponds to a multiplication operation in Laplace complex frequency domain)

So, for your example:

$$
\begin{alignat}{1}
\mathscr{L}^{-1}[X_1(s)] = \mathscr{L}^{-1}\left[ \frac{1}{s+4} \cdot (U(s)-X_2(s)) \right]
\end{alignat}
$$

Then:

$$
\begin{alignat}{1}
x_1(t) = \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( u(\tau) - x_2(\tau)  \right)\,d\tau }}
\implies
\dot{x}_1(t) &= \dfrac{d}{dt} \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( u(\tau) - x_2(\tau)  \right)\,d\tau }}
\end{alignat}
$$

I don't think there is a way to get to the original differential equation from here without knowing nothing about $u(t)$ and $x_2(t)$. But if you choose two functions and solve this integral, it will lead to the same result given by the Inverse Laplace Transform operation without convolution.

An Example

Let's pick $u(t) = cos(t)$ and $x_2(t) = e^{-t}$. Then, we have $\mathscr{L} \left[u(t)-x_2(t) \right] = \dfrac{s-1}{s^3+s^2+s+1}$

Anti-transforming without using convolution:

$$
X_1(s) = \dfrac{s-1}{(s+4)\cdot(s^3+s^2+s+1)} = \frac{-1/3\,}{s+1} + \frac{5/51}{s+4} + \frac{(4s+1)/17}{s^2+2}
\\[30px]
\begin{alignat}{1}
x_1(t) &= -\frac{1}{3}e^{-t} + \frac{5}{51}e^{-4t}+\frac{1}{17}\left(4\cos(t) + \sin(t) \right)
\\[5px]
\dot{x}_1(t) &= +\frac{1}{3}e^{-t} + \frac{20}{51}e^{-4t}+\frac{1}{17}\left(\cos(t) -4\sin(t) \right)
\end{alignat}
$$

Anti-transforming using convolution:

$$
X_1(s) = \bigg(\frac{1}{s+4}\bigg) \cdot \bigg(\frac{s-1}{s^3+s^2+s+1}\bigg)
\\[30px]
\begin{alignat}{1}
\dot{x}_1(t) = \dfrac{d}{dt} \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( \cos(\tau) - e^{-\tau}  \right)\,d\tau }}
\,=\, \frac{1}{3}e^{-t} + \frac{20}{51}e^{-4t}+\frac{1}{17}\left(\cos(t) -4\sin(t) \right)
\end{alignat}
$$


答: The proof is to explicitly construct the localization $M = (\mathbf{Z} - \{0\})^{-1}A$. This carries a $\mathbf{Q}$-vector space structure. Then there is an obvious $\mathbf{Z}$-bilinear mapping $A \times \mathbf{Q} \to M$ carrying $(a,1)$ to a nonzero element.

Title: $0$ element of $A\otimes \mathbb{Q}$


问:
How calculate the trace norm via convex optimization?
I am taking the convex optimization course by CMU (though I am not a CMU student), and got stuck on this problem.

Formally, show that computing $\left \| X \right \|_{tr}$ can be expressed as the following convex optimization problem:
$$\begin{array}{ll} \underset{{Y \in \mathbb{R}^{m \times n}}}{\text{maximize}} & \mbox{tr} \left( X^T Y \right)\\ \text{subject to} & \begin{pmatrix}  I_{m} & Y \\ Y^{T} & I_{n} \end{pmatrix} \succeq 0\end{array}$$
where $I_{p}$ is the $p \times p$ identity matrix.

Any help would be appropriated.

答:
What are the eigenvalues of $$\begin{pmatrix}  I_{m} & Y \\ Y^{T} & I_{n} \end{pmatrix}?$$ Well, consider an eigenvector $[p, q].$ Then:

$$ p + Y q = \lambda p,$$
and

$$Y^{T}p + q = \lambda q.$$

We see that $$Y q = (\lambda - 1) p,$$ while $$Y^{T} p = (\lambda - 1) q.$$ Applying $Y$ to the second equation we see that $$YY^{T} p = (\lambda -1)^2.$$

So the block matrix is positive semidefinite if and only if $I\succeq YY^{T},$ which is exactly the condition in the question pointed out by Rodrigo.


问:
Non-trivial solutions implies row of zeros?
If there exist non trivial solutions, the row echelon matrix of homogenous augmented matrix A has a row of zeros.

True or False?

I'm not sure where to begin as to see why this would be true or false. I know that if there are a row of zeros it means that there are infinitely many solutions, but not sure how I can tell if that means there are non-trivial solutions though. Any help would be appreciated. Thanks in advance.

答:
The trivial solution is the unique solution where every variable equals zero hence if you have infinitely many solutions, you have non-trivial solutions.

However, one thing you'll need to be careful of is how many equations you have and how many variables you have.


问:
Formulas for percentage gain/loss equivalent to formula for percentage change?
Can you please tell me whether the percentage change formula below can be used in place of both the percentage gain formula and the percentage loss formula below? Thank you in advance.

I have seen that percentage gain can be calculated using this formula:

$$\frac{2\text{nd value}}{\text{original value}} (-1) (\cdot 100)$$

e.g. original value = 10, 2nd value = 12
12/10 -1 x 100 = 20% percentage gain

and percentage loss as follows:

$$1 - \frac{\text {second value}}{\text{original}} \cdot 100$$

e.g. original value = 12, 2nd value = 10
1- 10/12 x 100 = 20% loss

Do their results differ from the result the following percentage change formula gives? Or can this be used to replace both formulas? :
$$\frac{\text{new value} - \text{old value}}{\text{new value}}$$

答:
Percentage change is always calculated with respect to original value.

Hence percentage change=$\frac{(new\ value)-(old\ value)}{old\ value}\times 100$

As for whether this formula can replace both for loss and gain remember that this would be you formula for gain i.e., if there is a gain then there is no change of formula while if there is a loss then to calculate loss percentage you need to take negative of this formula

%Gain=$\frac{(new\ value)-(old\ value)}{old\ value}\times 100$

%Loss=$\frac{(old\ value)-(new\ value)}{old\ value}\times 100$ to ensure that you get a positive output.


问:
Let $(a_n)_{n=1}^\infty$ be an infinite sequence of complex numbers. Prove the following limit.
Given $$\lim_{x\to \infty} \frac 1x \sum_{n\le x} a_n = k,$$
I want to prove that
$$\lim_{x\to \infty} \frac {1}{\log x} \sum_{n\le x} \frac {a_n}{n} = k.$$

I'm much more interested in learning the technique(s) necessary in order to prove this rather than a direct proof. Specifically, we have learned about asymptotic estimates and summation by parts, but I don't see how I can use those techniques to prove the problem statement. Thank you for any insight!

答:
Let $A(x)$ be defined as

$$
A(x)=\sum_{n\le x}a_n=kx+o(x)
$$

then we have

$$
\begin{aligned}
\sum_{n\le x}{a_n\over n}
&=\int_{1^-}^x{\mathrm dA(t)\over t} \\
&=\int_1^x{\mathrm d(kt)\over t}+\int_1^x{\mathrm d[o(t)]\over t} \\
&\sim k\log x
\end{aligned}
$$


问:
Find a function $f$ which is Riemann-integrable on $[0,T]$, and so that $\int_0^T f(t)^2 dt$ is infinite.
I am reading "Linear Algebra, Signal Processing, and Wavelets - A Unified Approach: Python Version" by Øyvind Ryan.

There is the following exercise in this book.

Exercise 1.18: Riemann-Integrable Functions Which Are Not
Square Integrable

Find a function $f$ which is Riemann-integrable on $[0,T]$, and so that
$\int_0^T f(t)^2 dt$ is infinite.

$\int_0^1 \frac{dt}{\sqrt{t}} = 2$, but $\int_0^1 \frac{dt}{t} = \infty$.

But do we say $f(x) = \frac{1}{\sqrt{x}}$ is riemann integrable on $[0,1]$?

I think if $f$ is riemann integrable on $[0,T]$, $f$ must satisfy the condition that $f$ is bounded on $[0,T]$ at least.

Is there really $f$ such that $f$ is bounded on $[0,T]$ and $\int_0^T f(t) dt$ exists and $\int_0^T f(t)^2 dt$ is infinite?

答:
A function $g$ is Riemann integrable on $[0,T]$ iff $g$ is bounded and continuous almost everywhere on $[0,T]$. If your $f$ is Riemann integrable, then it satisfies the latter two properties, then so does $f^2$ and there is no way that $f^2$ cannot be Riemann integrable. It is impossible for the integral of a Riemann integrable function on a compact interval to take the value $\infty$.

So perhaps you are looking for improper Riemann integrability. In such a case, we can take exactly your function $f(x)=\dfrac{1}{\sqrt{x}}$.


问:
How do we solve the integral $\int \frac{1}{x^2-y}\mathrm{d}x$?
The integral to be solved is given by:

$$ I = \int \frac{1}{x^2-y}\mathrm{d}x$$

I was wondering what integral substitution I would need to make. I looked at symbolab and it directed me to use $x = u\sqrt{y}$ as a substitution, but where does it come from?

答: If you follow the substitution advice "y" will go out of the integral as a constant the integration will concern another function of "u" variable that is not easy to integrate as the first one , I will recommend to you to go as x=cos(a) a trigonometric function is very flexible tool you can find your way to a form of integral with a multiplication of two trigonometric function then you can integrate by part ,sorry I don't know your mathematic level.this is as far as I can go</p>


问:
Geometric Expectation
Two coins with heads probabilities $\frac{1}{3}$ and $\frac{1}{4}$ are alternately tossed, starting with the $\frac{1}{3}$ coin, until one of them turns up heads. Let $X$ denote the total number of tosses, including the last. Find:

$$E(X)$$

I have seen a recent post that describes the answer as:

$$\mathbb{E}(X) = \displaystyle 1 + \frac{2}{3} (1 + \frac{3}{4} \mathbb{E}(X))$$

This still confuses me because how can you solve for $E(X)$ with $E(X)$ in the equation?

答: That's not the answer, but it's very close. All you need to do is to solve the equation $u = 1 + (2/3)[1 + (3/4)u]$, which is a linear equation in one variable.

Title: Geometric Expectation


问:
$G = G(p_1) \times G(p_2) \times \cdots \times G(p_k)$ and $|G(p_i)| = p_i ^{n_i}$.
After the proof of this lemma:

Let $G$ be a finite Abelian group of order $p^nm$, where $p$ is a prime that
does not divide $m$. Then $G = H \times K$, where $H = \{x \in G \mid x^{p^n}= e\}$
and $K = \{x \in  G \mid x^m = e\}$. Moreover, $|H| = p^n$.

They tell me the following:

Given an Abelian group $G$ with $|G|= p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k}$, where the
$p’s$ are distinct primes, we let $G(p_i)$ denote the set $\{x \in G \mid x^{{p_i}^{n_i}} = e\}$. It then follows immediately from Lemma 1 and induction that $G = G(p_1) \times G(p_2) \times\cdots \times G(p_k)$ and $|G(p_i)| = p_i ^{n_i}$.

My problem is that I don't really know how to do this induction. Any hint please?

I know this question is not very elaborate. But I have had quite a bit of difficulty solving group problems using induction. I would appreciate any help.

答:
The base case $k=1$ is immediate,  by Lagrange.

If we assume the result for given  $k $, then the result  for  $k+1$ follows immediately from the lemma, since $|G|=p_{k+1}^{n_{k+1}}m $, where $p_{k+1}\not|m $.


答:
By using elementary row operations to change a row (say the bottom row of a matrix $M$) into zeros, you have shown that the bottom row is a nonzero combination of the other rows.

By backtracking and figuring out the coefficients and putting them in a vector $\vec{x}$, it shows that $\vec{x}M=0$, where $\vec{x}\neq \vec{0}$. That $\vec{x}$ is a nontrivial solution to the matrix equation..

Title: Non-trivial solutions implies row of zeros?

Body:
If there exist non trivial solutions, the row echelon matrix of homogenous augmented matrix A has a row of zeros.

True or False?

I'm not sure where to begin as to see why this would be true or false. I know that if there are a row of zeros it means that there are infinitely many solutions, but not sure how I can tell if that means there are non-trivial solutions though. Any help would be appreciated. Thanks in advance.


问:
Calculate the variance of days of stock goes up
the price of a stock increases with the probability $1/3$,  and decreases with probability $3/2$ daily. Let U be the number of days it goes up, after 5 days. What is $var(U)$?

I think the expected value of U is 5 times 1/3 equals to 5/3, but I am not sure if using e.g. $(1 - 5/3) ^ 2 * 1/3 + （2 - 5/3）^ 2 * 1/9 + ...$ is a correct way of finding the variance.

答:
The probability is going up is the same everyday.

Hence the variance is just $5\left( \frac13\right) \left( \frac23\right)$ according to binomial distribution.


问:
Convert a range of integers to another range of integers while keeping an even distribution
I have a random integer x in the range [A, B] (inclusive).

Is it always possible to convert x to another integer y in the range [C,D] (inclusive) while keeping an even distribution?

This only seems possible if the size of [A,B] and [C,D] are divisible without rest.

Is there any trick I can use to make this possible?

Practical example of what I would like to do:

I get a random integer x in the range [0, 1048576].

I convert x to a new integer y in the range [10, 19].

I repeat the experience a large amount of times.

At the end, I get an even distribution of y=10, y=11, y=12... (≈10% of y=10, ≈10% of y=11, ≈10% of y=12 ...)

Thank you for bearing with me and my questionable math skills.

答:
Yes, you're right, if you perform the experiment only once.

If you do an experiment with 1048577 events each with probability $1/1048577$, there's no way you're going to be able to combine those outcomes into larger sets each with probability exactly $1/10$. The best you can get is some approximation.

One possibility would be to repeat the experiment until you get a result that works. For example, if you get a result in the range $[0,1048569]$, take the remainder after division by $10$, and add $10$. Otherwise, repeat the experiment until you get an answer in this range. It is unlikely you'll have to do the experiment many times. In principle there is no guaranteed upper bound to the number of times you'll have to do it, but despite this the average number of experiments you'll need is finite.


答: If you want it to be exactly even, you are right that you need the size of $[C,D]$, which is $D-C+1,$ to divide the size of $[A,B]$ evenly.  If it doesn't divide evenly but the size of $[A,B]$ is much larger than the size of $[C,D]$, as in your example, it will be very close.  Here you start with $1048577$ possibilities and you want to sort them into $10$ bins.  You can start by putting $\lfloor \frac {1048577}{10}\rfloor=104857$ in each bin and you have $7$ left over.  You can put those each in separate bins.  Now some bins have $104858$ and some have $104857$.  Although that is not perfectly even, it is only one part in $104857$ off, which is rather small.  If that is close enough for you, you are done.  If not, you can pull another random number with a range of $10$ to assign each of the left overs to a bin.

Title: Convert a range of integers to another range of integers while keeping an even distribution

Body:
I have a random integer x in the range [A, B] (inclusive).

Is it always possible to convert x to another integer y in the range [C,D] (inclusive) while keeping an even distribution?

This only seems possible if the size of [A,B] and [C,D] are divisible without rest.

Is there any trick I can use to make this possible?

Practical example of what I would like to do:

I get a random integer x in the range [0, 1048576].

I convert x to a new integer y in the range [10, 19].

I repeat the experience a large amount of times.

At the end, I get an even distribution of y=10, y=11, y=12... (≈10% of y=10, ≈10% of y=11, ≈10% of y=12 ...)

Thank you for bearing with me and my questionable math skills.


问:
Solutions to the following equation including a Mobius function
Let $\mu$ be the Mobius function. Now, how would one solve the following equation with respect to the variables $a\in \mathbb{N}$ and $b\in \mathbb{N}$?

$\sum_{d=1}^{\infty}\mu(d)\lfloor \frac{a}{d}\rfloor\lfloor \frac{b}{d}\rfloor = a\cdot b$?

Alternatively, how could we find the local maximums and minimums of the function $\sum_{d=1}^{\infty}\mu(d)\lfloor \frac{a}{d}\rfloor\lfloor \frac{b}{d}\rfloor$ with respect to $a$ or $b$ when we actually know the other variable? Can we do this by differentiation?  Is there any way of doing this?

答:
All the questions you asked can be answered with the exact formula
\begin{align*}
\sum_{d=1}^\infty \mu(d) \lfloor \tfrac ad \rfloor \lfloor \tfrac bd \rfloor
&= \sum_{d=1}^\infty \mu(d) \sum_{\substack{m\le a \\ d\mid m}} 1 \sum_{\substack{n\le b \\ d\mid n}} 1 \\
&= \sum_{m\le a} \sum_{n\le b} \sum_{\substack{d\ge 1 \\ d\mid m \\ d\mid n}} \mu(d) \\
&= \sum_{m\le a} \sum_{n\le b} \sum_{\substack{d\ge 1 \\ d\mid\gcd(m,n)}} \mu(d) = \sum_{m\le a} \sum_{\substack{n\le b \\ \gcd(m,n)=1}} 1.
\end{align*}
(Definitely avoid differentiation with functions that aren't even continuous.)


问:
What is feature selection?
I have been participating in the kaggle digit recognizer competition and after some initial success someone suggested that I look into feature selection.  I've done a bit of research but I am having a hard time wrapping my head around what exactly feature selection is.

Below is a partial description of the data I am working with:

Each image is 28 pixels in height and 28 pixels in width, for a total
  of 784 pixels in total. Each pixel has a single pixel-value associated
  with it, indicating the lightness or darkness of that pixel, with
  higher numbers meaning darker. This pixel-value is an integer between
  0 and 255, inclusive.

So if we arrange the pixels 28x28 we get them ordered like below:

```<code>000 001 002 003 ... 026 027
028 029 030 031 ... 054 055
056 057 058 059 ... 082 083
 |   |   |   |  ...  |   |
728 729 730 731 ... 754 755
756 757 758 759 ... 782 783
</code>```

So based on the above, if a number was represented it is probably safe to say that the pixel at position 000 (aka the pixel at the very top left of the image) will most likely be unimportant when determining what the number is because it will probably be the same for all numbers.  I make this assumption based on the fact that the top left pixel is so far out from the center where most of the number would be.

My question: Would me excluding 000 when doing my calculation be considered a very basic form of feature selection?  If not, could someone try to explain it to me?

答:
Sure, that's an example of feature selection. But generally the term refers to a more formal process of deciding which features are relevant, and which are either irrelevant or redundant to some other features. When you want to pick only a few features, it's a good idea to determine and evaluate your choice based on examining the actual data.

Incidentally, feature selection can tell you which features to throw out, but it can't tell you where to get your features from in the first place. Tossing a bunch of pixels into a black box classifier will virtually always produce crappy results, so you'd probably want to add additional features, like edge detection at each pixel, or the average value of the top third of the image, or the location of the darkest area after blurring, or.... etc. The number of potential features is gigantic (and consists mostly of useless data), which is why feature selection is useful.

Title: What is feature selection?


问:
An estimate for required $n^{\alpha}\ge \ln(n^2)$
I am trying to find an estimate in terms of $\alpha\in(0,1)$ of how large $n$ should be so that the below inequality is satisfied for all large $n$.
$$ n^{\alpha}\ge \ln(n^2)$$
Is there a good way to go about this?

答:
For the equality$$n^{\alpha}= \log(n^2)$$ let $x=n^2$ and solve
$$x^{\frac \alpha 2} =\log(x)\implies x=\Bigg[-\frac{2 }{\alpha }W\left(-\frac{\alpha }{2}\right)\Bigg]^\frac 2 \alpha\implies\color{blue}{n_*=\Bigg[-\frac{2 }{\alpha }W\left(-\frac{\alpha }{2}\right)\Bigg]^\frac 1 \alpha}$$ where $W(.)$ is Lambert function.

$$n^{\alpha}\ge \log(n^2) \qquad \text{if} \qquad n\le n_*$$


问:
Cyclic property of the root of a cubic polynomial
Let $f(x) = x^3 - 3x + 1$. Let $\alpha, \beta, \gamma \in \mathbb{R}$ be the roots of $f$ with $\alpha > \beta > \gamma$. Let $g(x) = x^2 - 2$. We see that $\beta = g(\alpha)$, $\gamma = g(\beta)$ and $\alpha = g(\gamma)$ by calculation using $f$.

Why does this cyclic property hold? What is the relation between $f$ and $g$? Could you tell me its background or generalization?

答:
This answer is factually incorrect: see the comments below.

Let $a$ be any root of $f(x)$. Then $a^3 - 3a + 1 = 0 \Rightarrow a^2 - 3 = -1/a \Rightarrow a^2 - 2 = -1/a +1$. Thus $g(a) = -1/a + 1$.

Let us find $g(g(g(a)))$, which is  $g(g(-1/a + 1)) = g\left(-\frac{1}{-1/a+1} + 1 \right) = g\left(\frac{-a}{-1+a} + \frac{-1+a}{-1+a} \right) $ $ = g\left(\frac{-1}{-1+a}\right) = \frac{-1}{-1/(-1+a)} +1 = -1+a+1=a$. From this process, the roots of $f(x)$ in terms of $a$ are $a$, $-\frac{1}{a} + 1$ and $\frac{-1}{1+a}$. This means that cycling any root three times through $g(x)$ gives the original root, and these are the only three roots, since a cubic polynomial has at most three real roots.

The tricky part is the condition that $\alpha > \beta > \gamma$, where $\alpha$ is the largest root. In other words, how do we know that the cycling order isn't actually $\alpha, \gamma, \beta$? Observe that $(\alpha, \beta, \gamma)$, $(\beta, \gamma, \alpha)$ and $(\gamma, \alpha, \beta)$ represent the same cyclic permutation, and the other permutation $(\alpha, \gamma, \beta)$ also makes up three permutations, and because there are only $3! = 6$ in total, these are the only permutations.

So if the other permutation $(\gamma, \beta, \alpha)$ is true, then $\beta > \alpha$ (any other inequality would also hold for $(\gamma, \alpha, \beta)$), or $-\frac{1}{\alpha} + 1 > \alpha$, which leads to $\alpha < 0$ (*). By Descartes' rule of signs, $f(-x) = -x^3 + 3x + 1$ has one sign change, hence there is only one negative root. However, since there are three real roots from the question, the other two roots are positive ($0$ is not a root). Therefore $\alpha$ is not the largest root: contradiction, and the cycling order is $\alpha, \beta, \gamma$. Hence proved.

(*) Rearrange to get $\frac{1}{a} < 1- a$ as the inequality sign flips. You will have to split it into positive and negative cases to prove this algebraically, but this can be shown by a sketch.

I haven't actually shown that the values that $g(x)$ cycles through are the roots of $f(x)$! Vieta's formulas can be used to do this: prove that $a + g(a) + g(g(a)) = 0$, and do the same for $\alpha \beta + \beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$.


问:
How to understand this proof of the countability of N*N?
These days, I am reading Prof.Liviu Nicolaescu's lecture notes on honors calculus.But the proof of an example in Chapter3 makes me confused.

In this example, he constructs a function, and try to prove that it is a bijection. However, I can't understand his method, though he states that domain of the function is $N*N$, I find that firstly, (1,1),(2,1),(2,2),(3,1),(3,2),(3,3),...which in general can be denoted as (m,n) can not represent all elements of $N*N$, he just considered the situation that m>=n, he didn't consider the situation that m<n.Secondly, he said that $\phi$ (m,n)=#$S_1$+...+#$S_{m-1}$+n, however, when m=1,2, the results are wrong. Thirdly, I don't think that it is clear that $\phi$(m,n) is a bijection. I can't prove that it is injection and I think proving it is surjection is even harder for me.

I am a self learner, so I don't know either it is that I'm wrong or the proof itself has some problems. Hope that someone can help me.

I'm not an English native speaker, so please forgive my poor English grammer.

答:
Yes, that proof (although not the result of course!) is incorrect, the big error being the first one: that not all pairs show up in the claimed enumeration.

The general idea of proceeding by "blocks," though, is perfectly fine. Here's one way to do this which actually works: have the $n$th block consist of those pairs whose entries add up to $n+1$. (If we were including $0$ in $\mathbb{N}$, "$n+1$" would get replaced by "$n$" there.) Then each block itself can be ordered (say) by comparing left coordinates. This yields:

$$(1,1),\quad (1,2), (2,1),\quad (1,3), (2,2), (3,1),\quad (1,4), (2,3), (3,2), (4,1),\quad ...$$

(This is very close to the Cantor pairing function and I think may actually have been what Nicolaescu had in mind.) Now to get the corresponding $\phi$, note that $(m,n)$ shows up in the $(m+n-1)$th block and is the $m$th element of that block. Also, the $k$th block has $k$ elements in it. Putting these observations together you can get an explicit formula for $\phi$:

 The number of blocks appearing before $(m,n)$ shows up is $m+n-2$, and the total number of pairs in all these blocks is therefore the sum of the first $m+n-2$ many natural numbers. So the pair $(m,n)$ shows up in position $${(1+m+n-2)(m+n-2)\over 2}+m.$$

Incidentally, this is actually a great motivating moment. It should be clear at this point that actually finding an exact bijection between two infinite sets can be annoying. (Another exercise at this point is to find a bijection between $\mathbb{R}$ and $[0,1]$ - it's not as simple as it may look at first!) Fortunately, there is an extremely powerful tool which makes things much nicer for us, namely the Cantor-Schroeder-Bernstein (or whatever) theorem. This says that it's enough to just find a pair of injections. This is often vastly easier: for example, to show that there is a bijection $\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}$ we can just consider the injections $$i\mapsto(i,i)\quad\mbox{and}\quad (m,n)\mapsto 2^m3^n.$$ Moreover, CBS doesn't use the axiom of choice - it's a perfectly direct argument.


问:
Change of variables mixing dependent and independent variables
I have a system of two ordinary differential equations with two independent variables: $x,y$. And only dependent variable: $t$

Can I make the following change of variables: $t=ay$, such that the equation in $y$ turns into an algebraic equation and there remains only one differential equation in $x$ now depending on $y$?

Thanks!

答: Sadly no. In general there's no reason to believe that the solution would have $t=a y$ so you can't just assume this to be the case.</p>


问:
Is my substitution correct?
I have to solve $$\int_{0}^{1}x^5\arcsin{x}dx$$.

I substituted $$\arcsin x= t$$
I got,

$$\int_{0}^{\pi/2}t\sin^5t\;\cos t\;dt$$
How do I go further? And is this correct substitution? Thanks in advance.

答:
HINT

What about integration by parts?

\begin{align*}
\int x^{5}\arcsin(x)\mathrm{d}x = \frac{x^{6}\arcsin(x)}{6} - \frac{1}{6}\int\frac{x^{6}}{\sqrt{1-x^{2}}}\mathrm{d}x
\end{align*}

Then your problem is reduced to solve the last integral.

This can be done through the substitution $x = \sin(t)$:
\begin{align*}
\int\frac{x^{6}}{\sqrt{1-x^{2}}}\mathrm{d}x = \int\sin^{6}(t)\mathrm{d}t
\end{align*}

Can you take it from here?

Title: Is my substitution correct?


问:
Represent total variation of continuous function by integration of counting function
$f : [a,b] \to \mathbb R$ is continuous, let $M(y)$ be the number of points $x$ in $[a,b]$ such that $f(x)=y$. prove that $M$ is Borel masurable and $\int M(y)dy$ equals the total variation of $f$ on $[a,b]$

This is an exercise in 'real analysis for graduate students'(Richard.F.Bass)

At first time, I thought  $M(y)=\mu(\{x \mid f(x)=y\})$ where $\mu$ is counting measure. But I cannot find the relation between Borel measurability and $\mu(\{x \mid f(x)=y\})$

Is there anyone would help me?

答:
This is the Banach indicatrix Theorem, which he proved in 1925:

[1] S. Banach, "Sur les lignes rectifiables et les surfaces dont l'aire est finie" Fund. Math. , 7 (1925) pp. 225–236
http://matwbn.icm.edu.pl/ksiazki/fm/fm7/fm7116.pdf

An exposition in English of Banach's proof is given in

Banach Indicatrix Function

Generalizations are in

[2] S.M. Lozinskii, "On the Banach indicatrix" Vestnik Leningrad. Univ. Math. Mekh. Astr. , 7 : 2 pp. 70–87 (In Russian)

[3] https://projecteuclid.org/journals/real-analysis-exchange/volume-27/issue-2/Generalization-of-the-Banach-Indicatrix-Theorem/rae/1212412867.full


问:
$0$ element of $A\otimes \mathbb{Q}$
I am studying tensor product and I feel that it is not easy to distinguish $a\otimes b=0$ or $\neq 0$.

I'm thinking the following special case and it should be true.

Let $A$ be an abelian group, in other word a $\mathbb{Z}$ module, and  $a\otimes 1 \in A\bigotimes_{\mathbb{Z}} \mathbb{Q}$.
Then, $a\otimes 1 =0$ if and only if there exists some $0\neq n\in \mathbb{Z}$ such that $na=0$.

I think "if part" is trivial, but the other part is difficult. I want a proof of this claim or a counter examaple.

答:
Here is another solution: Tensor the exact sequence $0 \rightarrow (a) \rightarrow A \rightarrow A / (a) \rightarrow 0$ with $\mathbb{Q}$,  which is flat over $\mathbb{Z}$. $(a)$ is the subgroup generated by $a$.

If such an $n$ exists, then $a \otimes_{\mathbb{Z}} \mathbb{Q} \sim (\mathbb{Z}/n\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Q} = 0$. Therefore, $a \otimes 1$ (which is in the image of $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow A \otimes_{\mathbb{Z}} \mathbb{Q}$) must be 0.

If no such $n$ exists, then $(a) \sim \mathbb{Z}$ , so $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \sim \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q} \sim \mathbb{Q}$, with the isomorphism function being $na \otimes p/q = np /q$. Under this isomorphism, $1 \in \mathbb{Q}$ corresponds to $a \otimes 1$, so $a \otimes 1$ is not 0 in $(a) \otimes_{\mathbb{Z}} \mathbb{Q}$. Therefore, by injectivity of $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow A \otimes_{\mathbb{Z}} \mathbb{Q}$, its image in $A \otimes_{\mathbb{Z}} \mathbb{Q}$ is not 0 either.

Title: $0$ element of $A\otimes \mathbb{Q}$


问:
The derivative of $\int_0^xxf(t)dt$ and why we cannot treat $xf$ as continuous.
This comes from Calculus by Michael Spivak, 3rd ed., chapter 14, Problem 11.

Find $F'(x)$ if $F(x) = \int_0^x xf(t)dt$. The answer is not $xf(x)$; you should perform an obvious manipulation on the integral before trying to find $F'$.)

Here, Spivak fails to include the assumption that $f$ is continuous.

With this assumption, the answer is $\int_0^xf(t)dt + xf(x)$, and is solved here, here, and here.

I understand the solution. I was able to do it myself. It makes sense to me. I also solved it using the $\lim_{h\to 0}$ approach, and that worked too.

Yet, naively, it seems like the answer should be $xf(x)$. It seems like since $x$ "behaves like a constant" in that integration, and because $f$ is continuous, then $xf$ must also be continuous, and so $F'(c) = cf(c)$.

I'm hoping to build better intuition about why this is wrong.

I was thinking perhaps one way of looking at it is that $xf$ cannot be assumed to be continuous because we are free to change the value of $x$ somewhat arbitrarily. In fact, at any fixed point $c$, $xf(c)$ varies with $x$, so certainly we cannot guarantee anything about the value of $xf(t)$ simply by restricting $t$ to certain values.

Does this make any sense? I'm sorry to use such sloppy reasoning/language/arguments.

There's a good chance I just don't yet have the tools to fully "get it" yet, but I appreciate any insight you might be able to share.

Edit

Ian's answer helps explain something that confused me in Spivak's solution to a later problem.

The next two problems are:

Prove that if $f$ is continuous, then
$$\int_0^xf(u)(x-u)du = \int_0^x\left(\int_0^uf(t)dt\right)du.$$
Hint: Differentiate both sides, making use of Problem 11.

Use Problem 12 to prove that

$$\int_0^xf(u)(x-u)^2du = 2\int_0^x\left(\int_0^{u_2}\left(\int_0^{u_1}f(t)dt\right)du_1\right)du_2.$$

These each require fairly straightforward calculations, the details of which we can skip here. However, in the published Answer Book solution for the latter, problem 13, Spivak does something that initially I found baffling. He begins:

Applying Problem 12 to $g(u) = f(u)(x-u)$, we obtain
$$\int_0^xf(u)(x-u)^2du =\int_0^x[f(u)(x-u)](x-u)du$$
$$=\int_0^x\left(\int_0^uf(t)(x-t)dt\right)du$$

This substitution is misleading. The result he's using from problem 12. applies only to functions that do not depend on $x$, while Spivak's $g$ here clearly does depend on $x$.

I think he gets away with it here only because $\frac{\partial}{\partial x}[f(u)(x-u)] = f(u).$

答: The Fundamental Theorem of Calculus  says that the derivative of $\int_a^{x}  g(t)dt$ is $g(x)$ if $g$ is  continuous. This does not apply to the derivative of $\int_a^{x}  g(t,x)dt$. You cannot apply the theorem when the integrand depends on $x$ even if it continuous in both variables.


问:
$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists
$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists  could any one tell me which of the following statements are correct?

$1. \text{if } \lim_{x\to\infty} f(x) \text{ exists, then it is 0}$

$2.  \lim_{x\to\infty} f(x) \text{ must exists, and it is 0}$

$3. \text{ in case if f is non negative } \lim_{x\to\infty} f(x) \text{ exists, and it is 0}$

$4. \text{ in case if f is differentiable } \lim_{x\to\infty} f'(x) \text{ exists, and it is 0}$

I solved one problem in past which says: if $f$ is uniformly continuos and $\int_{0}^{\infty} f(x)dx$ exists then $\lim_{x\to\infty} f(x)=0$, so the condition in one says $f$ is uniformly continuous? and hence $1$ is true? well I have no idea about the other statements. will be pleased for your help.

答:
The point is that the limit need not exist. You can construct something according to the comment by J.J. For instance, you can consider a function such that at any integer the height is one but the width is $1/n^2$. Then the area of each rectangle is $1/n^2$ and

$$
\int_0^\infty f(x)dx=\sum\frac{1}{n^2}<\infty.
$$

But the limit does not exists. You get

$$
\lim_{n\rightarrow\infty} f(n)=1,
$$

$$
\lim_{n\rightarrow\infty} f(n-\delta)=0,\;\; 0<\delta<<1
$$

Thus, 1 is true. If the limit exists, it should be zero (if it isn't then the function can't be integrable, right?). 2 is not true (see the hint by J.J. and the first part of my answer above). I think that 3, 4 are not true due to the same reason.

Edit: To prove that 4 is not true you can construct a (more decaying) example and then integrate it.


问:
How to find a basis for orthogonal complement, given the linear map?
Given a linear map  $\langle \cdot\,, \cdot \rangle:\, P_3(\mathbb{R}) \times P_3(\mathbb{R}) \to \mathbb{R}$ defined by:
$$
⟨p,q⟩ = \int_{0}^{2} p(x) q(x) \, dx
$$
for all $p,q \in P_3(\mathbb{R})$.

If $⟨\cdot\,,\cdot⟩$ is an inner product on $P_3(\mathbb{R})$, what is the basis for $\{ x, x^2 \}^\bot$?

答:
HINT

We say that two vectors are orthogonal when their inner product is zero.

Moreover, a vector is orthogonal to a vector space iff it is orthogonal to one of its basis.

At your case, take a polynomial $f\in\textbf{P}_{3}(\mathbb{R})$, apply the restrictions $\langle f,x\rangle = \langle f,x^{2}\rangle = 0$ and solve the corresponding system of linear equations.

Can you take it from here?


问:
Upper bound for the exponential integral Ei(x)
I'm trying to find an upper bound for

$f(m, \alpha) := \frac{m\cdot \mathrm{Ei}(m(1-\alpha))}{\exp(m(1-\alpha))}$. Here $m \ge 1$ and $\alpha \in (0, 1)$.

When I compute values of $f$, I see that $f(m, \alpha) \le \frac{2}{1-\alpha}$.

However, I cannot show this rigorously. Could you please help to obtain a tight upper bound? Thank you.

答:
$$f(m, \alpha)=m\frac{ \mathrm{Ei}(m(1-\alpha))}{\exp(m(1-\alpha))}$$ Let $x=m(1-\alpha)$ to make
$$g(x, \alpha)=\frac 1{1-\alpha}\, x\,\mathrm{Ei}(x) \,e^{-x}$$
The maximum value of $x\,\mathrm{Ei}(x) \,e^{-x}$ is a bit lower than $1.5$ ($1.4841$ to be precise). So
$$f(m, \alpha) \lt \frac {1.4841}{(1-\alpha)}$$


问:
What should I substitute?
$$\int_{0}^{1}x^5\sqrt{\frac{1+x^2}{1-x^2}}dx$$
Should I substitute $$x=\sqrt{\cos2u}$$

Thanks in advance!

答:
Hint:

For $$F(n)=\int_0^1 x^{2n+1}\sqrt{\dfrac{1+x^2}{1-x^2}}\ dx$$

set $x^2=\cos2t\implies2x\ dx=-2\sin2t\ dt,0\le2t\le\dfrac\pi2$

$x=1\implies t=0,x=`0\implies t=\dfrac\pi4$

$\sqrt{\dfrac{1+x^2}{1-x^2}}=+\cot t$

$$F(2)=-\int_{\pi/4}^0\cos^22t\cot t\sin 2t\ dt$$

$$=-\int_{\pi/4}^0(1+\cos4t)\cos^2t\ dt$$

$$=\int_0^{\pi/4}\dfrac{(1+\cos4t)(1+\cos2t)}2\ dt\text{ as }\int_a^bf(x)\ dx=-\int_b^af(x)\ dx$$

Use  Werner Formulas to complete the assignment

Title: What should I substitute?


问:
Complex numbers - separate real/imaginary parts
$$K(\omega) = \frac{1}{1 + j\omega RC}$$

Uhm...How do I separate the real part from the imaginary part here? :U And how can I find the argument? I mean, if the document I got this formula from is right, the argument should be negative...

答:
If you have $\frac{z_1}{z_2}$ a ratio of complex numbers, multiplying
$$
\frac{z_1}{z_2}\cdot\frac{\overline{z_2}}{\overline{z_2}} = \frac{z_1\overline{z_2}}{z_2\overline{z_2}}
$$
will give you a real denominator ($\overline{z_2}$ is the complex conjugate of $z_2$). Once you do that, you need only write $z_1\overline{z_2} = x + jy$, $x,y\in\Bbb R$, as then
$$
\frac{z_1}{z_2} = \frac{x + jy}{z_2\overline{z_2}} = \frac{x}{z_2\overline{z_2}} + j\frac{y}{z_2\overline{z_2}}.
$$</p>


问:
Where's the flaw in this construction of a set that's both countable and uncountable?
Say we constructed a set of all finite math expressions, which is countable. Then we take the subset of expressions that evaluate to a single real number. That subset should still be countable. Now we apply the Cantor argument to this subset to construct a new real number. I'm assuming all of this can probably be done using a finite expression, so this number should've been in the subset already.

So the subset is both countable and uncountable at the same time. Where's the flaw in this logic? Are there multiple?

答:
I keep finding near-duplicates but can't find an exact duplicate, so here goes:

Note that you haven't  been clear about what "finite expression" means. This is in fact the problem that turns into the fundamental error:

"I'm assuming all of this can probably be done using a finite expression."

Your whole argument hinges on the claim that the "antidiagonal real" corresponding to whatever system of definitions you're using is definable by a definition within that system itself. This may feel obvious, especially since we're generally not very careful about the precise nature of mathematical language. However, it is in fact false: a general theorem of Tarski, usually stated for first-order logic but much more broadly applicable than that, says that this is in fact almost never the case!

I think the main obstacle to understanding this is the fact that we generally only have one "mathematical language" in mind, namely the informal-but-"rigorous" natural language we actually use to talk about mathematics. To de-mystify things a bit, consider "local" versions of Cantor's theorem: given a "complexity class" (in a very broad sense) $\mathfrak{C}$ we can talk about $(i)$ reals definable by an expression in $\mathfrak{C}$ and $(ii)$ arrays definable by an expression in $\mathfrak{C}$. The process of forming the antidiagonal corresponding to a given array is simple enough that, for "reasonable" $\mathfrak{C}$, it's something $\mathfrak{C}$ lets us do. The "local Cantor's theorem," then, is this:

The $\mathfrak{C}$-definable reals can never be put into a single $\mathfrak{C}$-definable array. Put another way, the set of $\mathfrak{C}$-definable reals is not "$\mathfrak{C}$-countable.

For example:

The polynomial-time-computable reals, while obviously countable, are not polynomial-time-computably countable.

The computable reals, while obviously countable, are not computably countable.

The reals which are first-order definable over $(\mathbb{R};+,\times)$ without parameters, while obviously countable, are not $(\mathbb{R};+,\times)$-definably-without-parameters countable.

And so on. In my opinion this "local" analysis will, once understood, clear up a lot of the issues one may have with both Cantor's argument and general ideas about definability broadly construed. (That's not to say that everything about definability and countability suddenly becomes intuitive - see e.g. here for an example which trips plenty of people up! - but it's a good first step.)


问:
Best Intro to Number Theory books for non mathematicians?
I want to know if there are any books that can teach me or introduce me to number theory requiring no prerequisites. This is coming from someone who wants to broaden their mathematical horizons and investigate number theory and cryptology.

Any suggestions are welcome.

答: You don't make it clear what your background is. Some of the other suggestions would probably be good for someone who has at least some experience with proofs. If your background is high school math and this is your first real exposure to proofs, I'd recommend Invitation to Number Theory by Ore.


问:
selecting 4 letters from word BANANA (i am confused between 15 or 30)
in how many ways 4 letters be selected to form words from the letters BANANA?</p>

答:
Assuming order doesn't matter, you can organize your thoughts by listing the outcomes in dictionary order: AAAB, AAAN, AABN, AANN, ABNN, there are only the five outcomes. If order does matter, for each of these listed prior you can count the number of arrangements of them, giving 4+4+12+6+12=38

-@JMmoravitz

OwnerDisplayName: sugam Pratap


问:
m/m/2 question in queueing theory
Customers arrive at a serving-system according to a Poisson process with rate 1. In the system there are two serving stations, A and B, which only take care of one customer at a time. If a customer come to a full system, both A and B are busy, then he leaves. If a customer come to an empty system he chooses A. When a customer has been served he leaves.
The serving time are independent of the arrivals and mutually independent following an exponential distribution with rate 1. Let Xt take the value (0; 0) if the system is empty, (1; 0) if A is being used but not B, (0; 1) if A is free and B is being used and (1; 1) if both A and B are busy. Then $(X_t)_{t \geq 0}$ is an irreducible Markov-chain with state-space {(0; 0); (1; 0); (0; 1); (1; 1)}. Determine the Q-matrix of the
chain and the equilibrium distribution.</p>

答:
In state

(0,0) arrivals happen at rate 1 and take you to the state (1,0)

(1,0) arrivals happen at rate 1 and take you to the state (1,1) while services happen also at rate 1 and take you to the state (0,0)

(0,1) arrivals happen at rate 1 and take you to the state (1,1) while services happen also at rate 1 and take you to the state (0,0)

(1,1) no arrivals can happen (they are not admitted to the system) and two services are happening in parallel. A service at rate 1 which will take you to (0,1) and another at rate 1 which will take you to (1,0).

So
$$ Q = \begin{pmatrix}
-1 & 1 & 0 & 0 \\
1 & -2 & 0 & 1 \\
1 & 0 & -2 & 1 \\
0 & 1 & 1 & -2
\end{pmatrix}.$$

We compute the stationary distribution of $Q$ by solving $\pi Q=0$ with the requirement that the elements of $\pi$ sum to 1. This gives
$$\pi=\left(\frac{2}{5},\frac{3}{10},\frac{1}{10},\frac{1}{5}\right).$$

Title: m/m/2 question in queueing theory


问:
Calculating CDF of summation of normal variable
I have iid normal random variables $X_i$ from i = 1,..n. And I want to calculate pdf of:

$ \sum_{i=1}^{n} a_i x_i $

So I know that I should be calculating the MGF and then get the product of those variables. But I could not calculate the CDF of this normal distribution. How can I calculate the CDF of this normal distribution by using the properties of Moment Generating Function ?

答:
If your $X_i\sim N(\mu;\sigma^2)$ the pdf of your sum is

$$Y\sim N\left(\mu\Sigma_i a_i;\sigma^2\Sigma_i a_i^2 \right)$$

But the CDF of a Gaussian cannot be expressed by elementary function.

$$F_Y(y)=\int_{-\infty}^y f_Y(t)dt$$

This integral can only be written in terms of the special function erf (error function)


问:
Are all open sets open intervals?
Are all open sets open intervals? If not, provide an example. I think that all open intervals are open sets but not the other way around, but couldn't find an example.</p>

答: This question ultimately depends on the set and topology. Since the tag is real-analysis, I take it your talking about $\mathbb{R}$ with the standard topology. In that case, you're correct that all open intervals are open sets, whereas not every open set is an open interval. Take for example $E:=(1,2)\cup(4,5)$. Then $E$ is open, as the union of open sets, but is not an interval.

Title: Are all open sets open intervals?


答:
You haven't mentioned it, but I understand that you're talking about intervals on $\mathbb R$.

Are all open intervals open sets? Yes.

Open intervals on $\mathbb R$ are of the form $(a,b)$, where we let $a,b$ take on values in $[-\infty,\infty]$.  You can check that these are open sets.

Are all open sets open intervals? No!

The union of two open sets is open (more generally, any arbitrary union of open sets is open). So, take $(0,1)$ and $(2,3)$. These are open intervals, hence open sets. From what we just said, $(0,1) \cup (2,3)$ is also an open set, but not an open interval (let alone an open interval, this isn't even an interval).

P.S. What you should know:
When is $A\subset \Bbb R$ open?

$A\subset \Bbb R$ is open if for every $x\in A$, there exists some $\epsilon > 0$ so that $(x-\epsilon,x+\epsilon) \subset A$. In other words, for every point in $A$, you can find a neighborhood of that point which is contained in $A$.

Title: Are all open sets open intervals?

Body: Are all open sets open intervals? If not, provide an example. I think that all open intervals are open sets but not the other way around, but couldn't find an example.</p>


问:
interesting set of integrals
When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?</p>

答:
I recommend you to prove by induction with all odd integers since $$\int \frac{\sin3x}{\sin x}dx$$
will depend on $$\int \frac{\sin x}{\sin x}dx$$

using integration by parts.

Title: interesting set of integrals


问:
If $\lim_{x \to 0}g(x)=L$. How do I prove $\lim_{h \to 0} g(f(x+h)-f(x)) = \lim_{y \to 0}g(y)$ using $\epsilon-\delta$
If I know $\lim_{x \to 0}g(x)=L$. For my particular problem $L=0$.
So I've been trying to prove: $\lim_{h \to 0} g(f(x+h)-f(x)) = \lim_{y \to 0}g(y)=0$ using $\epsilon-\delta$.

(Also I'm assuming the $g(x)$ is continuous at $0$ and $f(x)$ is continuous at $x$)

So I know that $\forall\epsilon$, there is some $\delta$ such that
$0<|k|<\delta\Rightarrow|g(k)-L|<\epsilon$    (1)

I have to prove that $\forall\epsilon_1$,$\exists\delta_1$ such that
$0<|h|<\delta_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon_1$

If $\exists\epsilon=\epsilon'$, and $\delta=\delta'$ satisfies (1), such that
$0<|k|<\delta'\Rightarrow|g(k)-L|<\epsilon'$    (2)

If I to find value for $\delta_1$, when $\epsilon_1=\epsilon'$ such that
$0<|h|<\delta_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$  (3)

I can prove $\forall\epsilon_1,\exists\delta_1$.

If I take $f(x+h)-f(x)=k$,  then RHS of (2) and (3) are same.
$$\Rightarrow0<|f(x+h)-f(x)|<\delta'\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$$
So I have to find $\delta_1$ such that
$0<|h|<\delta_1\Rightarrow0<|f(x+h)-f(x)|<\delta'$?

This seems similar to another $\epsilon-\delta$ type condition but...I'm not sure.

I know $\lim_{h\to0}f(x+h)=f(x)$, since I assumed $f(x)$ is continuous at $x$.

Since $|h|>0\Rightarrow f(x+h)\neq f(x)\Rightarrow|f(x+h)-f(x)|>0$? If this is true, I think it has been proved.

(Also please check if there are any other mistakes or misassumptions)

答:
Your statement $(1)$ is correct but incomplete as the way it stands it doesn't use continuity of $g$. Note that since $g$ is continuous, it's safe to write $|k|\lt \delta$ rather than restricting to $0\lt |k|\lt \delta$.

Statement $(2)$ is not required as it is same as $(1)$

Then "if I take $f(x+h)-f(x)=k$, then RHS of $(2)$ and $(3)$ are same" Why? That's not true $f(x+h)-f(x)$ may not be a constant. You think that it is always true that $f(x+10^{-Googol})=f(x+10^{-100})?$

You may proceed as below also:

$g$ is continuous at $0$ and $\lim_{x\to 0}g(x)=L$

$\begin{align}
&\implies \forall\epsilon\gt 0, \exists \delta\gt 0: |x|\lt \delta\implies |g(x)-L|\lt\epsilon \end{align}$$\tag 1$

$f$ is continuous at $x$ and therefore, $\lim_{h\to 0} f(x+h)-f(x)=0$.  It follows that $\exists \delta'\gt 0: |h|\lt \delta'\implies |f(x+h)-f(x)|\lt\delta \tag 2 $

From (1) and (2) above, it follows that $\forall\epsilon \gt 0,\exists \delta': |h|\lt \delta'\implies |g(f(x+h)-f(x))-L|\lt \epsilon $. This proves that $\lim_{h\to 0}g(f(x+h)-f(x))=L$.


问:
Typo mistake in Rudin Book
I have some confusion in Rudin book

My confusion is given below marked in red colour

Here  both $f$ and $g$ don't look like characteristics function since charcteristic function

look like $$f_{r}(x)=\begin{cases}1, & x\in V_r \\ 0, & x\notin V_r. \end{cases} $$

$$g_{s}(x)=\begin{cases}1, & x\in \bar{V_s} \\ 0, & x\notin \bar{V_s}. \end{cases} $$

My confusion: why  $f$ is lower semicontinious  and $g$ is upper semicontinious?

My attempt : $$f_{r}(x)=\begin{cases}r, & x\in V_r \\ 0, & x\notin V_r. \end{cases} $$

$$\implies f^{-1}(-\infty , r)=\{ x : f(x) < r\}= \begin{cases}X &\text{if r >1} \\ X \setminus V_r &\text{ if 0 <r} \le 1, \\ \emptyset,& \text{r <0}. \end{cases} $$

Here  $V_r$  is open $\implies X \setminus V_r$ is closed

so  by the theorem  characteristics function  of closed  set are upper semi-continious

we conclude that $f$ is upper- semicontnious

similarly for $g(x)$

$$\implies g^{-1}(-\infty , s)=\{ x : g(x) < s\}= \begin{cases}X &\text{if s >1} \\ X \setminus \bar{V_s} &\text{ if 0 <s} \le 1, \\ \emptyset,& \text{ s<0}. \end{cases} $$

Here  $\bar{V_s}$  is closed   $\implies X \setminus \bar{V_s}$ is open

so  by the theorem  characteristics function  of open  set are lower semi-continious

we conclude that $g$ is lower- semicontnious

Edits : I think  rudin made typo mistake  and my   attempt  is almost correct

答:
I was wrong

$\implies f^{-1}(-\infty , r]$ is closed   mean $f$ is lower semicontinious

$\implies g^{-1}(-\infty , s)$ is  open mean $g$ is  upper semicontnious

Title: Typo mistake in Rudin Book


问:
Invertible endomorphism with purely imaginary eigenvalues
Let V be real vector space and let T be an invertible endomorphism on V. Suppose that T has only purely imaginary eigenvalues, why does this imply that V is of even dimension?</p>

答: If $T$ is an endomorphism of the real vector space $V$, the degree of the characteristic polynomial of $T$ is equal to the dimension of $V$; thus if $\dim V$ is odd, this characteristic polynomial is an element of $\Bbb R[x]$ of odd degree.  Any such polynomial has at least one real root, hence $T$ has a real eigenvalue;  thus, by way of contraposition, the hypothesis that $T$ have purely imaginary eigevalues implies that $\dim V$ is even.


问:
Solving restricted/one-sided limits

Is the function $f(x) = \sqrt{x^{2} - 9}$ continuous at $x = 3$?

I tried to check by solving each requirement for continuity, which are the following:

The limit at the point must exist;

The function at the point must exist;

The limit at the point must be equal to the function at the point.

The second requirement can be easily checked as 'satisfied' as $f(3) = 0$. Thus, I only need to check if the first requirement is satisfied.

I started by solving for the left-hand limit as follows:
\[L^{-} = \lim_{x \to 3^{-}}\left[\sqrt{x^{2} - 9}\right]\]

Then, I eventually got $L^{-}$ to be zero.

My question is this: Is it correct to rewrite the problem by substituting $x = 3 - a$, where $a \geqslant 0$, then solving for the limit as $a$ approaches zero?

Edit: This is a question about solving one-sided limits, and not about the continuity of the square root function.

答:
About substitutions in limits in general:

In general, that's absolutely right (here, in particular, something more important is to be noted. See the Addendum). $x\to 3^-$ can be thought of as $x = 3-h$ where $h\to 0^+$ or $x = 3+h$ where $h\to 0^-$. Guess what, that's exactly the substitution you made.

Subtle point to note: You say $x = 3-a$, where $a\ge 0$. While that is perfectly fine (in general it is fine, but for this particular case please check the addendum), I just want to mention that $a > 0$ is no problem either. When talking about $\lim\limits_{a\to 0^+}f(a)$ for some $f$, $a$ is approaching $0$ from the positive side, but is never equal to $0$. Hence, $a>0$ suffices and the substitution $x = 3-a$ works.

Before making substitutions though, one must be extremely careful about the domain of the function.

Addendum/Warning:

The domain of $f(x) = \sqrt{x^2-9}$ is $x^2 \ge 9$, i.e. $|x| \ge 3$ or $x \in (-\infty,-3]\cup [3,\infty)$. So, it does not make sense to talk about the left hand limit at $3$, since values immediately to the left of $3$ are not in the domain of $f$. You may ask, does $$\lim_{x\to 3} \sqrt{x^2-9}$$ still exist? Yes, it does! This is because the only one-sided limit we can even talk about at $3$ is the right-handed one. We define the limit at $3$ to be equal to this one-sided limit.

Title: Solving restricted/one-sided limits


答:
You do not need to evaluate for left-hand limit simply because it does not belong to the domain of the function. In general for obtaining continuity of boundary point we consider the limiting value of the function as the value of the one-sided limit. Thus if,

$$\lim_{x\rightarrow 3^+}f(x)=f(3)$$

you can say that the function is continuous at $x=3$. as the function is defined only for $|x|\ge3$ assuming you are working over $\mathbb R$.

Title: Solving restricted/one-sided limits

Body:
Is the function $f(x) = \sqrt{x^{2} - 9}$ continuous at $x = 3$?

I tried to check by solving each requirement for continuity, which are the following:

The limit at the point must exist;

The function at the point must exist;

The limit at the point must be equal to the function at the point.

The second requirement can be easily checked as 'satisfied' as $f(3) = 0$. Thus, I only need to check if the first requirement is satisfied.

I started by solving for the left-hand limit as follows:
\[L^{-} = \lim_{x \to 3^{-}}\left[\sqrt{x^{2} - 9}\right]\]

Then, I eventually got $L^{-}$ to be zero.

My question is this: Is it correct to rewrite the problem by substituting $x = 3 - a$, where $a \geqslant 0$, then solving for the limit as $a$ approaches zero?

Edit: This is a question about solving one-sided limits, and not about the continuity of the square root function.


问:
Which of these variables affect y more?
I have two equations:

1: $\displaystyle y=\frac{0.0029}{\sqrt{k}}$

2: $\displaystyle y=\frac{0.0060}{\sqrt{m}}$

Which variable has a greater effect on $y$? i.e., does doubling $m$ have a greater effect on $y$ than doubling $k$? How do I describe/prove that?

Thanks!

(I asked a similar question previously, here. However I feel that the square root nature of this question makes it an entirely different question, and so I've posted a new question.)

答: $k$ is linear with $m$. On the other hand, $k$ is linear with $y^2$. So $y$ affects $k$ more than $m$ does.</p>

Title: Which of these variables affect y more?


问:
If a finite field has characteristic 2 why is every element a square?

If a finite field has characteristic 2 why is every element a square?

I've attempted this problem by calculating the square root as follows: $\mathbb{F}$ has $q = 2^m$ elements, so if $a \in \mathbb{F}$ then $a = a^q = (a^{2^{(n-1)}})^2$. Since, $a \in \mathbb{F}$ is a square elements if there exist $b \in \mathbb{F}$ s.t $b^2 = a$, I believe the next step would be to show that the square root $b = a^{2^{(n-1)}} \in \mathbb{F}$. Is this reasonable or should I take a different approach to the problem (and if so what approach?)?

答:
You can do it like this.

If $a^2=b^2$ then $a=\pm b$, but in characteristic $2$ we have $b=-b$ so that if $a^2=b^2$ then $a=b$.

Now square all the $n$ distinct elements of your finite field. You get $n$ different answers. And since there are only $n$ elements in the field each of them must appear in the list of squares.

You can also do it by noting that the multiplicative group of non-zero elements has odd order, and every element of a group of odd order is a square. Working with $q$ in your case is a simple case of this underlying fact.


问:
A fair die and some ties in a wardrobe - formal argument needed

A wardrobe $W_1$ contains $m$ black and $m$ blue ties. A fair die with $m$ sides is rolled. If the $r$th face appears, then $r$ ties are removed from the wardrobe $W_1$, and placed in another wardrobe $W_2$ (empty to begin with). Now, pick a tie from $W_2$ at random. What is the probability that it is a tie of black color?

Quick guess: $\displaystyle \frac{1}{2}$

Suppose $r$ ties are removed from $W_1$, and the tuple $(x,y)$ denotes how many ties of each color are removed. Here, $x+y = r$, $x$ represents black and $y$ represents blue ties. The possibilities for $(x,y)$ are as follows:
$$(0,r)\\ (1,r-1)\\ (2,r-2)\\ \vdots \\(r-2,2)\\(r-1,1)\\(r,0)\\\text{where } 1\le r\le m$$
There is an obvious inherent symmetry involved, i.e. if $(x_1,y_1)$ is a possibility, then so is $(y_1,x_1)$. This is what led me to think that the probability of picking a tie of black color from the second wardrobe is exactly $\frac{1}{2}$, since I found no reason to bias myself towards either of the colors. Is that some good intuition to count upon?

How would we formally evaluate $$\mathbf P(\text{Black tie is picked from }W_2) = ?$$
The following comes to mind:
$$\mathbf P(\text{Black tie is picked from }W_2) = \sum_{r=1}^m \mathbf P(\text{Black tie is picked from }W_2|r^{\text{th}} \text{ face appears on the die})$$
Edit: Still looking for a formal argument.

答:
Apply the formula for the Expected Value for a binomial random variable r ~ E(r) =
P(r) * r , where P(r) is probability of success, r is number of trials.

When we remove r ties randomly from W1 with m black and m blue ties,  we expect $\frac{m}{m+m}r $=$\frac{1}{2 }r $ blue ties and $\frac{m}{m+m}r$=$\frac{1}{2 }r $ black ties removed.

Similarly, if we pick up 1 tie randomly from W2, we expect $\frac{r/2}{r/2+r/2}x1=\frac{1}{2 }$ black tie picked, or the probability to pick a black tie is $\frac{1}{2}$.


问:
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?

I know that $\overline{A \cap B}$ equals to $\overline{A}\cup \overline{B}$ which can be read as "not A or not B". But it is not what I am looking for.

答: I'd say "not both A and B"</p>

Body: I'd say "not both A and B"


问:
What is the sum of the series with terms 1/(2/(3/(4/(5/(6/...?
I wondered how continually compounding a fraction in the denominator would behave as the successive denominators increment. What I was doing was essentially:

$\dfrac 1 {\dfrac 2 {\dfrac 3 {\dfrac 4 {\dfrac 5 {\dfrac 6 {\dfrac 7 {\dfrac 8 {\dfrac 9 {\vdots}}}}}}}}}$

I noticed that the result oscillated between numbers that were above 1 and below 1, and I realized that the pattern was actually progressing with respect to adding two numbers, i.e. the series is 1/2, 1/(2/(3/4)), etc.

Following this trail it appears that the limit as the number of (pairs of) denominators we add approaches infinity, this number approaches zero, at least as far as I can tell.

Consulting OEIS (A001147, A000165) resulted in my inspiration to find the limit of the infinite sum as we travel down the fraction, and I was able to create what I believe is the correct formula:

$\displaystyle S_\infty=\sum_{i=1}^\infty{\prod_{n=1}^i{\dfrac{2n-1}{2n}}}$

So, what I am wondering is,

what is the solution to this, assuming it converges?

how can I (properly) prove that the expression above is also equal to the expression below (I am confident it is)?

$\displaystyle \sum_{n=1}^\infty{\dfrac{1}{4^n} \binom{2n}n}$

Personally hoping it's convergent, since I believe the terms tend to zero.

答:
Consider
$$ S_p=\sum_{i=1}^p{\prod_{n=1}^i{\dfrac{2n-1}{2n}}}=\sum_{i=1}^p  \frac{\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }\,\Gamma (i+1)}=\frac{2 \,\Gamma \left(p+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma (p+1)}-1$$
Using Stirling approximation
$$S_p=-1+2\sqrt{\frac p \pi}\left(1+\frac{3}{8 p}-\frac{7}{128 p^2}+O\left(\frac{1}{p^3}\right)\right)$$
On the other hand
$$T_p=\displaystyle \sum_{n=1}^p{\dfrac{1}{4^n} \binom{2n}n}=-1+2^{-2 p-1} (p+1) \binom{2p+2}{p+1}=\frac{2 \Gamma \left(p+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (p+1)}-1$$