问:
How to find Chord BD in diagram below?
A Circle is drawn with center as A, radius as 12 units. A smaller incircle is drawn as shown with center E and which passes through A. BD is the tangent to the smaller circle. Find BD?

EDIT: Line DC need not be tangent to smaller circle.

I tried following thigs:

1] Constructed CD , making angle BDC as 90 = > Angle subtended by semicircle.

2] Using Pythagoras I see $BD^2 + CD^2 = 24^2$

3] I tried many things, but cannot find CD ?

e.g. I joined points A and D in hope of getting angle DAC which might help in finding Chord CD , which in turn could help in getting BD, but I could not progress in that direction?

How could I find length of BD ? IS there any other approach to solve this problem than what I have tried so far?

答:
If you call $E$ te tangent point of $BD$ and the small circle, AND if you call $O$ the center of the small circle, you have that the triangles $BOE$ and $BCD$ are similar (recall that $BD\perp OE$ since $OE$ is the radius at the tangent point and $BD$ is the tangent line).

So $\frac{BE}{BO}=\frac{BD}{BC}$.

BUT, $BO=12+6=18.$ $BC=24$ and $BE^2+OE^2=BO^2$, whence $BE=\sqrt{BO^2-OE^2}=\sqrt{18^2-6^2}=\sqrt{288}=12\sqrt2$.

Now, it is easy to determine the length of $BD$.

Title: How to find Chord BD in diagram below?


问:
Does the converse of the statement hold?
I've proved the following statement: "Given $(Z, T_Z)$ the product topological space of $(X, T_X)$ and $(Y, T_Y)$. If $A \in T_X$ and $B \in T_Y$ then $A \times B \in T_Z$".

Converse statement: "Given $(Z, T_Z)$ the product topological space of $(X, T_X)$ and $(Y, T_Y)$. If  $A \times B \in T_Z$ then $A \in T_X$ and $B \in T_Y$".

Now, I'm trying to prove the converse of the statement is not always true but I cannot take find an example to show it's false.

Is the converse of the statement true? If not, please find an example to show it's false.

P/S: I'm thinking the statement is not true because of the union operation of cartesian.

答: If $A$ is empty and $B$ is not open it follows that $A\times B$ is open in the product space. So the converse is false. But if $A$ and $B$ are nonempty and $A \times B$ is open then $A$ and $B$ are open. Proof: Define $f: X \to Z$ by $f(x)=(x,b)$ where $b$ is  fixed element of $B$. Show that this map is continuous. The inverse image under this map of $A\times B$ is exactly $A$ so $A$ is open. Similarly, $B$ is also open.

ContentLicense: CC BY-SA 4.0


问:
A condition for convergence of a sequence
Consider the sequence $s_n=\sqrt{n}$. Although $(s_{n+1}-s_n)\rightarrow 0$ as $n\rightarrow \infty$, the sequence is not convergent (since it is unbounded).

(This is a very common example in study of  Cauchy sequences where, by this example, it is remarked that making consecutive difference smaller is not sufficient for Cauchy condition.)

Suppose we put such a condition on a sequence; i.e. suppose $(x_n)$ is a sequence such that $(x_{n+1}-x_n)\rightarrow 0$ as $n\rightarrow\infty$ and $(x_n)$ is bounded. Can we say that $(x_n)$ must be convergent? If not, what condition, instead of boundedness, will guarantee the convergence of the sequence $(x_n)$?

Body:
No. Think about
$$0, 1, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 1, \frac{3}{4}, \frac{1}{2}, \frac{1}{4}, 0, \frac{1}{5}, ...$$


问:
Calculate the area on a sphere of the intersection of two spherical caps
Given a sphere of radius $r$ with two spherical caps on it defined by the radii ($a_1$ and $a_2$) of the bases of the spherical caps, given a separation of the two spherical caps by angle $\theta$, how do you calculate the surface area of that intersection?

To clarify, the area is that of the curved surface on the sphere defined by the intersection. At the extreme where both $a_1,a_2 = r$, we would be describing a spherical lune.

Alternatively define the spherical caps by the angles
$\Phi_1 = \arcsin(a_1/r)$ and $\Phi_2 = \arcsin(a_2/r)$.

答:
The formula for the area is complicated (see other answers), but it has a simple Legendre expansion
$$
A(\theta) = \sum_{l=0}^{\infty} a_l P_l(\cos\theta)
$$
with coefficients given by the associated Legendre functions,
$$
a_l = \pi \, (2l+1) \sin(\alpha_1) \sin(\alpha_2) \, P_l^{-1}(\cos\alpha_1) \, P_l^{-1}(\cos\alpha_2)
$$
which may be helpful.

ContentLicense: CC BY-SA 3.0


问:
$(1,2,...,n-1)$ and $(1,2,...,n)$ are generators of the symmetric group $S_n$
How can you prove that the permutations $(1,2,\ldots,n-1)$ and $(1,2,\ldots,n)$ generate the symmetric group $S_n$? I have found proofs for other generators like $(1,2)$ and $(1,2,\ldots,n)$, but not this one.  A good paper about this topic is https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf, but does not contain what I am looking for. Thank you in advance!

答: Let $\sigma=(1,2,\dots,n)$ and $\tau=(1,2,\dots,n-1)$. Then $\tau^{-1}\sigma$ (where permutations are composed right-to-left) is $(n-1,n)$. If you can show that $(1,2)$ and $(1,2,\dots,n)$ are a generating set, you now have a proof that $\sigma$ and $\tau$ are also a generating set.


问:
Are my examples of False $\implies$ False being true correct?
I know that $p \implies q$ is true if $p$ is false. I am trying to wrap my brain around this by thinking about examples. I have two examples that I would like to understand.

Example 1: Consider the set

$$
X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}.
$$

This set is equal to the interval $(-\infty, 0]$ because, for example, $-1 \geq 0 \implies -1\leq 0$ is a true statement. Is that correct?

<s>Example 2: I know that the definition of limit is
$$
\forall \epsilon>0 \exists \delta > 0: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon.
$$

Would this make this a true statement for all functions $f$? Say $\epsilon$ is given, then pick $\delta = -1$. Then $0<\lvert x- a\rvert < \delta$ is false, so it is true that it does imply that $\lvert f(x) - L\rvert < \epsilon$. Have I just misunderstood the definition?</s>

EDIT: I just realized that in my second example the requirement is that $\delta$ be positive. I guess I am down to the first example.

答:
Yes, the statement in your first example is indeed a true statement.

Note that when $p$ is false, the constraint is essentially removed from $q$, therefore we can't really say much about $p\rightarrow q$. However, $p$ being false doesn't necessarily contradict the material implication either. This is why, by convention, $p\rightarrow q$ is true when $p$ is false. I believe this gets to the heart of the matter.


问:
When fitting a polynomial to data points, how to determine the reasonable degree to use?
I have wondered the following: Suppose that there is a set of data points $(x_i,y_i)$. Then I would like to know if it is more reasonable to assume if there is a polynomial relation of degree $m$ between them or of degree $n$. Is there way to measure it? I know that Lagrange's polynomial gives the exact relation but for example physics formula $F=ma$ says that sometimes it is correct to choose linear polynomial to model the phenomenon.</p>

答:
If the criteria of fitting is not specifed on unambiguous mathematical form one cannot definitively answer to your question. For example specifying "as close as possible" is not mathematical.

First, one have to specify the criteria of fitting : Least Mean Square Errors, or Least Mean Absolute Errors, or Least Mean Square Relative Errors or another criteria.

Second, one have to define the maximum acceptable value of error (in above sens) according to the criteria chosen.

On a practical viewpoint in case of polynomials, one can proceed to successive regressions with increassing $n$ until the error (in above sens) becomes equal or lower that the specified acceptable value.


问:
What is Lie Theory/ a Lie Group, simply?
I'm studying physics, and I continually come across mentions of "Lie Theory" and "Lie Groups" as they relate to such topics as particle physics and String Theory, as well as vague mentions of "symmetry".

I've attempted to read some texts on the topic and, while I feel I could probably sift through it in due time, it is very terse and esoteric. I've had group theory, calculus up to university calc II, and a teensy bit of analysis. Until I'm able to study this proper, what is Lie Theory/ a Lie Group, put simply? What is with these vague mentions of "symmetry"? What can be understood in terms of what I know now?

答:
There is already an answer, so let me try to give you a few simple, naive and hand-wavy ideas.

About symmetry:

"Symmetry" refers to the fact that some things can transform into themselves without changing too much. For example, if someone shows you a cube on a table, ask you to close your eyes, and then to open them, you will not be able to tell if this person has rotated the cube or not. This is the fact that a cube possesses a lot of symmetries. Now, let us assume that this cube has coloured faces (each face has a different colour). Then, you will be able to detect any rotations: the colours will change. Thus, colouring breaks the symmetry; otherwise said, a coloured cube is less symmetric than a cube.

Now let us look at a physical theory as a mathematical object. Then one often makes a lot of symmetry assumptions: you cannot tell where in the universe or when in time you are just making experiments and using newtonian mechanics, since it is a space and time invariant theory! I mean that in an empty universe, two balls will attract themselves in the same way, no matter where they are, nor what time it is.

Now, history has shown that one can understand objects by understanding its symmetries. And group theory is a way to formulate symmetry. This is quite a big topic!

About the Lie things:

Differential calculus is powerful, so Sophus Lie thought that instead of studying symmetry in its full generality, it would be easier to study symmetry with the help of differential calculus. Hence Lie theory.


问:
To find the position of the ant after 2020 moves is $(p, q)$,
An ant is moving on the coordinate plane. Initially it was at (6, 0). Each move
of the ant consists of a counter-clockwise rotation of 60◦ about the origin fol-
lowed by a translation of 7 units in the positive x-direction. If the position of
the ant after 2020 moves is $(p, q)$, find the value of $p^2$ + $q^2$

For this problem,I just drew some and tried to figure out how this is going to go... But I have absolutely no idea what is the trajectory of the ant after 2020 moves.
I need help.

答:
In the complex plane, a rotation of $60^\circ$ is achieved by multiplying by $e^{i\frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$ For ease of notation, let $\alpha = e^{i\frac{\pi}{3}}$.

Let $z_0 = 6$ be the initial position of the ant and $z_k$ be the location of the ant after $k$ moves. Then

$$\begin{align}
z_1 &= z_0 \alpha + 7 \\
z_2 &= z_1 \alpha + 7 &&= z_0 \alpha^2 + 7 \alpha + 7 \\
z_3 &= z_2 \alpha + 7 &&= z_0 \alpha^3 + 7 \alpha^2 + 7 \alpha + 7 \\
& \;\; \vdots \\
z_n &= z_{n-1} \alpha + 7 &&= z_0 \alpha^n + 7 \alpha^{n-1} + 7 \alpha^{n-2} + 7 \alpha^{n-3} + \ldots + 7 \alpha + 7 \\
\\
& &&= z_0 \alpha + 7 \left[ \alpha^{n-1} + \alpha^{n-2} + \alpha^{n-3} + \ldots + \alpha + 1 \right] \\
\\
& &&= z_0 \alpha + 7 \left[ \frac{\alpha^n - 1}{\alpha - 1} \right]
\end{align}$$

In particular, noting that $\alpha^6 = 1$, so $\alpha^7 = \alpha$, we have that

$$\begin{align}
z_7 &= z_0 \alpha^7 + 7 \left[ \frac{\alpha^7 - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 \left[ \frac{\alpha - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 = z_1
\end{align}$$

This means that every six moves returns the ant to its starting position. Thus, since $2020 \equiv 4 \bmod{6}$, we have that $z_{2020} = z_4$.

We now use the fact that $\alpha^n = \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}$, and in particular, $\alpha^3 = -1$. Then

$$\begin{align}
z_{2020} = z_4 &= z_0 \alpha^4 + 7 ( \alpha^3 + \alpha^2 + \alpha+ 1 ) \\
\\
&= 6 \alpha^4 + 7 ( \alpha^2 + \alpha ) \\
\\
&= 6 ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} ) + 7 \left[ ( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} ) + ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} ) \right] \\
\\
&= 6 ( \frac{-1}{2} - i \frac{\sqrt{3}}{2}) + 7 \left[ ( \frac{-1}{2} + i \frac{\sqrt{3}}{2} ) + ( \frac{1}{2} + i \frac{\sqrt{3}}{2} ) \right] \\
\\
&= -3 -3 \sqrt{3}i + 7 \sqrt{3}i \\
\\
&= -3 + 4 \sqrt{3}i
\end{align}$$

Thus $p = -3$ and $q = 4 \sqrt{3}$, so the answer is

$$ p^2 + q^2 = 9 + 48 = \boxed{57} $$


问:
Finding SD by only knowing the mean and one other piece of information
"The weights of bags of vegetables are approximately normally distributed with mean of 1kg. One quarter have weight greater than
1.1kg. Find the standard deviation."

I know this question might be simple to most of you, but probability/statistics really isn't my strong suit. I usually understand the standard deviation to be $\sigma=\sqrt{\sum(x_i-\mu)^2\cdot N^{-1}}$. In this case we aren't concerned with specific values of $x$ I don't think, and we don't know the sample size. All I can extrapolate from the question is that since it's 25% that have weight greater than 1.1kg then that 25% starts less than one standard deviation away from the mean.

If I were to guess, I would say that since this is normally distributed then I know that 34% of the bags fall within one standard deviation above the mean, and if 25% have a weight greater than 1.1kg, then 25% have a weight between 1kg and 1.1kg. So the deviation for 25% is 0.1kg which is less than 1 standard deviation, so I think I should multiply 0.1 by $\frac{34}{25}$ to get $0.136$ which, if my assumptions are correct, is equal to one standard deviation?

This doesn't feel concrete, I'm sure there's a more rigorous way to find the answer, any help is appreciated.

答:
One quarter have weight greater than 1.1kg

I translate it for you: $75\%$ of bags has a weight that is less or equal to $1.1$ kg. Assuming normality with mean 1 this can be formalized in the following way

$$\Phi\Bigg(\frac{1.1-1}{\sigma}\Bigg)=0.75$$

$$\frac{1.1-1}{\sigma}\approx 0.6745$$

$$\sigma\approx 0.1483$$


问:
in a group G where all elements are of order 3 show that $(ba)^2=a^2b^2$
This is a problem from my algebra book stated as:

In a group G all elements is of order 3 except the neutral element. Show that $(ba)^2=a^2b^2$ for all $a,b \in G$

So I have $a$, $a^2$, $a^3=1$, $a^4=a$, same for $b$ and $a^3=1=b^3$.

At first I thought "except the neutral element" just meant that $a^3=1$ but now I'm not even sure about that so I guess I need some clarification about that.

Then I tried to solve it as follows:

$$(ba)^2=(ba)(ba)=b(ab)a=b(a^4b^4)a=..... $$

where I tried to manipulate the expression with various combinations of:

$a=a^4=a  a^3 $

Using inverses for example $a^2=a^{-1} $

But I'm not really getting anywhere except going around in a circle showing that $(ba)(ba)=(ba)(ba)$

答:
Since every element, except the identity, has order $3$, $\forall a\in G\ a^3=1$ (for $a\neq 1_G$ this follows from the assumption on the order, for $1_G$ this is trivial)
$$(ba)^2\cdot a^2b^2=baba\cdot a^2b^2=baba^3b^2=bab^3=ba=(ba)(ba)^3=(ba)^4\\
(ba)^2a^2b^2=(ba)^4\\
a^2b^2=(ba)^2
$$


问:
Evaluate $\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}dx=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}} $

(1) Prove that$$\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}\mathrm{d}x=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}$$
(2) Prove that$$\int_0^\infty \frac{\mathrm{e}^{-ax^2}-\mathrm{e}^{-bx^2}}{x^2}\mathrm{d}x=\frac{1}{2}\ln\frac{b}{a} $$

I just learned improper integral with parameters and worked out a basic example below
$$
\int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{d}x=\ln\frac{b}{a}
$$
But I do not know how to compute the two integrals above. I think the basic example may be of some help and tried integration by parts, inspired by $\int_0^\infty\frac{\sin x}{x}\mathrm{d}x$ and $\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm{d}x$.

Any hint is appreciated.

答: Write $\frac{1}{x^2}=\int_0^\infty ye^{-xy}dy$ so the LHS of (1) is$$\begin{align}&\int_0^\infty dy\:y\int_0^\infty dx(e^{-(2a+y)x}-2e^{-(a+b+y)x}+e^{-(2b+y)x})\\&=\int_0^\infty dy\:y\left(\frac{1}{2a+y}-\frac{2}{a+b+y}+\frac{1}{2b+y}\right).\end{align}$$You can handle (2) the same way, thought it helps to first replace $\int_0^\infty$ with $\tfrac12\int_{-\infty}^\infty$, so you just have to evaluate Gaussians on $\Bbb R$.


答:
All I can extrapolate from the question is that since it's 25% that
have weight greater than 1.1kg

That´s true. So the equation $P(X\geq 1.1)=0.25$, where $X\sim \mathcal N(1, \sigma^2)$ Next we can standardize the random variable: $Z=\frac{X-\mu}{\sigma}$

So the equation is $P\left(Z\geq \frac{X-\mu}{\sigma}= \frac{1.1-1}{\sigma}\right)=0.25$. Thus

$$1-\Phi\left(\frac{0.1}{\sigma} \right)=0.25$$
Can you proceed?

Title: Finding SD by only knowing the mean and one other piece of information

Body:
"The weights of bags of vegetables are approximately normally distributed with mean of 1kg. One quarter have weight greater than
1.1kg. Find the standard deviation."

I know this question might be simple to most of you, but probability/statistics really isn't my strong suit. I usually understand the standard deviation to be $\sigma=\sqrt{\sum(x_i-\mu)^2\cdot N^{-1}}$. In this case we aren't concerned with specific values of $x$ I don't think, and we don't know the sample size. All I can extrapolate from the question is that since it's 25% that have weight greater than 1.1kg then that 25% starts less than one standard deviation away from the mean.

If I were to guess, I would say that since this is normally distributed then I know that 34% of the bags fall within one standard deviation above the mean, and if 25% have a weight greater than 1.1kg, then 25% have a weight between 1kg and 1.1kg. So the deviation for 25% is 0.1kg which is less than 1 standard deviation, so I think I should multiply 0.1 by $\frac{34}{25}$ to get $0.136$ which, if my assumptions are correct, is equal to one standard deviation?

This doesn't feel concrete, I'm sure there's a more rigorous way to find the answer, any help is appreciated.


问:
if $p\implies q$ is the same as $\lnot p \lor q$, then...
If $p\implies q$ is the same as $\lnot p \lor q$, then what is $p\implies \lnot q$?

I'm not sure if this is $\lnot p \lor \lnot q$, or $\lnot p \lor q$.

I'm trying to figure this out, because i have a problem:

~(q v p) --> ~r). I use demorgans law on this to make it ~p ^ ~q --> ~r. Then I need to make it simpler, my understanding would be that it MIGHT be ~(~p ^ ~q) v ~r) which might be (p ^ q) v r. However, I'm almost 100% sure i'm wrong.

I can't wrap my head around logic, It's so difficult for me to comprehend all of these rules; could someone please explain the answer to my original question, and provide the correct answer for the ~(q v p) --> ~r) question; and where I went wrong? I would be very grateful.

p.s. if you know of any easy to comprehend resources for logic that would be great.

Thanks.

答:
Well, $p \to r$ is the same as $\neg p \vee r$. Thus if $r=\neg q$, this means that $p \to \neg q$ is the same as $\neg p \vee (\neg q)$, which by order of logical operations can be written $\neg p \vee \neg q$.

No need to worry about De Morgan's laws!


Title: in a group G where all elements are of order 3 show that $(ba)^2=a^2b^2$

Body:
This is a problem from my algebra book stated as:

In a group G all elements is of order 3 except the neutral element. Show that $(ba)^2=a^2b^2$ for all $a,b \in G$

So I have $a$, $a^2$, $a^3=1$, $a^4=a$, same for $b$ and $a^3=1=b^3$.

At first I thought "except the neutral element" just meant that $a^3=1$ but now I'm not even sure about that so I guess I need some clarification about that.

Then I tried to solve it as follows:

$$(ba)^2=(ba)(ba)=b(ab)a=b(a^4b^4)a=..... $$

where I tried to manipulate the expression with various combinations of:

$a=a^4=a  a^3 $

Using inverses for example $a^2=a^{-1} $

But I'm not really getting anywhere except going around in a circle showing that $(ba)(ba)=(ba)(ba)$

Body:
We have
$$
a^2b^2=a^{-1}b^{-1}=(ba)^{-1}=(ba)^2.
$$


问:
Sturm bound - order at $\infty$ for $f \in M_k(\Gamma_0({N}))$

Let $f \in M_k(\Gamma_0({N}))$ and $\text{ord}_{\infty}(f)>\vert \Gamma: \Gamma_0 (N) \vert \frac{k}{12}$, then $f=0$

My idea is the following. Let $L$ be the representatives of the  left cosets, than

$$
g:=\prod_{\alpha \in L}{f\vert_{k\alpha^{-1}}} \; \in M_{nk}(SL2(\mathbb{Z}))
$$
where $n=\vert \Gamma: \Gamma_0 (N) \vert$.

By the valence formula for modular forms in $M_k(SL2(\mathbb{Z})))$, $f$ vanishes if $\text{ord}_{\infty}(g)> \vert \Gamma: \Gamma_0 (N) \vert \frac{k}{12}$.

How do I go from here ?

Would appreciate any help.

答:
For $f\in M_k(G)$, $$g = \prod_{\gamma\in G\backslash SL_2(\Bbb{Z})} f|_k \gamma$$

$m= k |SL_2(\Bbb{Z})/G|$

$g^{12}$ is a weight $12 m$ modular form for the full modular group, if it has a zero of order $>m$ at $i\infty$ then $\Delta^{-m} g^{12}$ is a weight $0$ modular form, thus it is constant, and since it vanishes at $i\infty$ it is zero, ie. $g = 0$ and $f=0$.


问:
Riesz Representation Theorem and Continuous Map
The following is similar to a problem I found on some analysis quals:

$X, Y$ are compact metric spaces and $F: X \to Y$ is surjective and continuous, if there is a finite measure $\nu$ on the Borel subsets of $Y$, then there is a finite measure $\mu$ on the Borel subsets of $X$ such that
$$
\int_{Y} f d \nu = \int_{X} f \circ F d \mu
$$
for all continuous functions on $Y$.

I'm guessing that the Riesz Representation Theorem needs to be used here, but for that I'd need a functional defined on the continuous functions on $X$ although the statement is about continuous functions of $Y$. What is a good approach here?

Edit: Is the following true: If $g: X \to \mathbb{R}$ is continuous, then there is a continuous $f: Y \to \mathbb{R}$ such that $g = f \circ F$? We could simply define $f(F(x)) = g(x)$, but then do we know that $f$ is continuous on $Y$? If we can do this then the claim will follow:
Let $H: X \to \mathbb{R}$ be the functional
$$
H(g) = \int  f d \nu
$$
where $f$ the corresponding continuous function on $Y$ as desrcibed above. Then Riesz Representation implies the result.

答:
Consider $\{g \in C(X): g=f\circ F \,\, \text {for some} \, \, f \in C(Y)\}$. Since $F$ is surjective there is only one $f$ for each $g$ in this space.If we show that this is a closed subspace of $C(X)$ then we can use Hahn-Banach Theorem to show that the map $g=f\circ F \to \int f d\nu$ can be extended to an element of the dual of $C(X)$  and  the proof  is finished by Riesz Theorem.

Suppose $g_n=f_n\circ F \to g$ uniformly. Note that $\sup_{y\in Y} |f_n(y)-f_m(y)| \leq \sup _{x \in X} |g_n(x)-g_m(x)| \to 0$. This implies that $f_n$ converges uniformly to some $f \in C(Y)$ and it is clear that $g=f\circ F$.


答:
Given that the normal distribution is symmetric and that 1/4 of the vegetables have weight larger than 1.1 you can say:

$\int_{-\infty}^{0.9} f(x)· \,dx = 1/4 $

$\int_{0.9}^{1} f(x)· \,dx = 1/4 $

$\int_{1}^{1.1} f(x)· \,dx = 1/4 $

$\int_{1.1}^{+\infty} f(x)· \,dx = 1/4 $

The inter-quantile range (IQR) is defined as:

$IQR = Q_{75} - Q_{25} = 1.1 - 0.9 = 0.2$

It can be shown that for a normal distribution:

$IQR = 2 ·\phi^{-1}(0.75)·\sigma$

Where $\phi^{-1}$ is the inverse cumulative density function or quantile function of the standard normal distribution. Hence:

$\sigma = \frac{IQR}{2 ·\phi^{-1}(0.75)} \approx \frac{0.2}{2 · 0.6745} = 0.1483 $

To check the answer, you can ran in R:

```<code>x = rnorm(10000000, 1, 0.2/(2*qnorm(0.75)))
plot(density(x))
quantile(x)
# 0%       25%       50%       75%      100%
# 0.2135559 0.9000991 1.0001097 1.0999769 1.7831438
</code>```

Body:
Given that the normal distribution is symmetric and that 1/4 of the vegetables have weight larger than 1.1 you can say:

$\int_{-\infty}^{0.9} f(x)· \,dx = 1/4 $

$\int_{0.9}^{1} f(x)· \,dx = 1/4 $

$\int_{1}^{1.1} f(x)· \,dx = 1/4 $

$\int_{1.1}^{+\infty} f(x)· \,dx = 1/4 $

The inter-quantile range (IQR) is defined as:

$IQR = Q_{75} - Q_{25} = 1.1 - 0.9 = 0.2$

It can be shown that for a normal distribution:

$IQR = 2 ·\phi^{-1}(0.75)·\sigma$

Where $\phi^{-1}$ is the inverse cumulative density function or quantile function of the standard normal distribution. Hence:

$\sigma = \frac{IQR}{2 ·\phi^{-1}(0.75)} \approx \frac{0.2}{2 · 0.6745} = 0.1483 $

To check the answer, you can ran in R:

x = rnorm(10000000, 1, 0.2/(2*qnorm(0.75)))
plot(density(x))
quantile(x)
# 0%       25%       50%       75%      100%
# 0.2135559 0.9000991 1.0001097 1.0999769 1.7831438


问:
Limits of an increasing function
Let $a<b$ and let $f: [a,b] \to R$ be an increasing function.

I'm asked to provide an example where $\lim_{x\to a^{+}}f(x)$ does not equal

$$\inf\{\,f(x) \; | \;x \in [a,b]\,\},$$

and I'm having trouble understanding what exactly is going on in this situation and was wondering if anyone can enlighten me.

Background

Prior to this question I was asked to prove that there exists an $S$ in $R$ such that

$$ S = \inf\{\,f(x)\; |\; x \in (a,b]\,\}.$$

I used the fact that $F$ is an increasing function and showed $F$ is bounded from below and therefore must have an infimum.

Then I was asked to prove that the $\lim_{x\to a^{+}}f(x)$ exists, and is equal to S.

I showed this by definition of limit, using the fact that if $F$ is bounded from below, then there exists $L$ such that $L$ is less than or equal to $F(x)$.

I have a feeling my professor wants me to use the things I showed above to find the example to the initial question, but I just can't connect the dots…

答:
This is a picture of what can happen at the endpoint (what needs to happen, in fact).

Remember to keep your counterexamples as simple as possible!

For the limit part ($L$ is the unfilled circle), you can perhaps use the sequential definition of limit, with the version of the completeness axiom that applies to monotonic sequences.

Title: Limits of an increasing function


答:
The big difference between the two infimums is that one includes $a$ while the other doesn't. To take advantage of it we use cases. For example,
$$
f(x) = \begin{cases}g(x), &x \neq a\\
-1 &x=  a\end{cases}
$$
What will happen if $g(x) > -1\ \forall x \in (a,b]$?

Title: Limits of an increasing function

Body:
Let $a<b$ and let $f: [a,b] \to R$ be an increasing function.

I'm asked to provide an example where $\lim_{x\to a^{+}}f(x)$ does not equal

$$\inf\{\,f(x) \; | \;x \in [a,b]\,\},$$

and I'm having trouble understanding what exactly is going on in this situation and was wondering if anyone can enlighten me.

Background

Prior to this question I was asked to prove that there exists an $S$ in $R$ such that

$$ S = \inf\{\,f(x)\; |\; x \in (a,b]\,\}.$$

I used the fact that $F$ is an increasing function and showed $F$ is bounded from below and therefore must have an infimum.

Then I was asked to prove that the $\lim_{x\to a^{+}}f(x)$ exists, and is equal to S.

I showed this by definition of limit, using the fact that if $F$ is bounded from below, then there exists $L$ such that $L$ is less than or equal to $F(x)$.

I have a feeling my professor wants me to use the things I showed above to find the example to the initial question, but I just can't connect the dots…


问:
Application of Green's theorem for $f(x,y) = (-y, x)$

Let $\partial D$ be the boundary for a $C^1$ curve. Show that the area of $D$ is $$\frac12 \int_{\partial D} (x \ dy-y \ dx) = \int_{\partial D}x \ dy = -\int_{\partial D}y \ dx$$

I'm trying to apply Green's theorem to $f(x,y)=(-y, x)$ here and getting $$\frac12 \int_{\partial D} (x \ dy - y\ dx) = \frac12 \int_{\partial D} \frac{\partial f_2}{\partial x} -\frac{\partial f_1}{\partial y} \ dx \ dy = \frac{1}{2}\int_{\partial D} 2 \ dx \ dy$$

but not sure how to continue from here. How do I get the limits for the integral?

答:
The question to me seems like it's asking you to verify, using Green's theorem, the result you have in your question.

Greens theorem states an alternative way to calculate a line integral $\int_C F \cdot ds$.

If, for example, we are in two dimension, $C$ is a simple closed curve, and $F(x,y)$ is defined everywhere inside $C$, we can use Green's theorem to convert the line integral into to double integral via
\begin{align*}
  \iint_D \left(\frac{\partial F_2}{\partial x} -
    \frac{\partial F_1}{\partial y}\right) dA
\end{align*}

Pick a vector field  ${\bf F(x,y)}=(−y/2,x/2)$ and verify that
$$\frac{\partial F_2}{\partial x} -
    \frac{\partial F_1}{\partial y} = 1$$
Now, if  $C$ is a counterclockwise oriented simple closed curve that bounds a region where you can apply Green's theorem, the area of the region $D$ bounded by $C=\partial D$ is

\begin{align*}
  \text{Area of $D$} = \int_C F \cdot ds = \frac{1}{2} \int_C x\,dy - y\,dx,
\end{align*}


问:
How to find the integral for $\int^1_{\frac{1}{1+e^\sigma}}(\frac{y}{1-y})^{1/r}\frac{1}{y-1} dy$
How can I find the solution for following integral.

$$\int^1_{\frac{1}{1+e^\sigma}}(\frac{y}{1-y})^{1/r}\frac{1}{y-1} dy$$

where $r,\sigma$ are positive constants.

答:
To make the story short since @Z Ahmed already adressed the problem in terms of the beta function, you could write
$$\int_a^1  y^{\frac{1}{r}}\,(1-y)^{1-\frac{1}{r}}\,dy=\frac{\pi  (r-1) }{2
   r^2}\csc \left(\frac{\pi }{r}\right)-B_a\left(1+\frac{1}{r},2-\frac{1}{r}\right)$$ Now, if $a$ is small, you could use
$$B_a\left(1+\frac{1}{r},2-\frac{1}{r}\right)=a^{\frac{1}{r}}\left(\frac{ r}{r+1}a+\frac{ (1-r)}{2 r+1}a^2+\frac{ (1-r)}{2r(3 r+1)}a^3+O\left(a^4\right)\right) $$

Using $\sigma=\pi$ and $r=e$, the "exact" value should be
$$I=0.38992993433\cdots$$while the above truncated expansion gives
$0.38992993411$


问:
Use probability generating function to calculate probability

Let $N$ be a discrete random variable having a probability generating
function

$G _ { N } ( z ) = e ^ { - 3 + 2 z + z ^ { 2 } }$

Find $\Pr(N ≤ 2)$

If the PGF is in the form of $\sum z ^ { x } p ( x )$, I can work out the PMF of $N$. But how can I deal with this case?

答:
Break up $G_N(z)=e^{-3}e^{2z}e^{z^2}$ and work out the Cauchy product of the last two factors:
$$G_N(z)=e^{-3}(1+2z+2z^2+\cdots)(1+z^2+\cdots)=e^{-3}(1+2z+3z^2+\cdots)$$
where $\cdots$ indicates $z^3$ or higher terms which we don't need. Therefore $P(N\le2)=e^{-3}(1+2+3)=6e^{-3}$.


问:
$\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q)=(\Bbb Z/2\Bbb Z)^n$

Let $p_1,...,p_n\in\Bbb N$ be distinct primes. Compute $\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q)$

Intuitively, $G:=\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q) = (\Bbb Z/2\Bbb Z)^n$. My first idea is that if $\sigma\in G$, then $\sigma(\sqrt{p_i})=\{\sqrt{p_i},-\sqrt{p_i}\}$ for each $i$. Hence, from this, pick $\sigma_i\in G$ such that $\sigma_i(\sqrt{p_i})= -\sqrt{p_i}$ and $\sigma_i(\sqrt{p_j}) = \sqrt{p_j}$ for all $j\neq i$. Then, $\sigma_i$'s generates $G$ and each $\sigma_i$ does not generates $\sigma_j$ if $i\neq j$, $G = (\Bbb Z/2\Bbb Z)^n$. Is this valid?

Another way to do this is using the fact that 'if $K/F,L/F$ are both Galois and $K\cap L =F$ then $\text{Gal}(KL/F)\simeq\text{Gal}(K/F)\times\text{Gal}(L/F)$', we inductively apply this. The problem is whether $\sqrt{p_i}\in\Bbb Q(\sqrt{p_1},...,\sqrt{p_{i-1}})$. Any hint or comment will be appreciated.

答:
Every element of that Galois group is $\pm\sqrt{p_i}$ for every $p_i$. There are $n$ $p_i$'s and we have choice of $2$ values on every of them.
Hence we have $2^n$ automorphisms in that Galois group. because every of them is of order $2$, this gorup is ismorphic to product of $n$ copies of $\mathbb{Z}/2\mathbb{Z}$
(You should earlier prove that for every $j+1$, $\sqrt{p_{j+1}}$ is not contained in $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_j})$.)


问:
Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ $\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$
Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ the following does hold:

$$\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$$

I tried to simplify the inequality to $$na^n(b-a)<b(b^n-a^n)$$
$$na^n(b-a)<b(b-a)(\sum_{k=0}^{n-1}b^{k}a^{n-1-k})$$

How to continue? (Notice that I don't know if $b-a$ is necessarily positive).

Body:
It is equivalent to
$$a^{n+1}-b^{n+1}< (n+1)(a-b)a^{n}.$$
Assume $a>b$, then
$$\begin{aligned}
a^{n+1}-b^{n+1}&=(a-b)\cdot \sum_{i=0}^{n} a^i b^{n-i}\\
&< (a-b) (n+1)a^{n}.
\end{aligned}$$
And also
$$\begin{aligned}
a^{n+1}-b^{n+1}&=(a-b)\cdot \sum_{i=0}^{n} a^i b^{n-i}\\
&> (a-b) (n+1)b^{n},
\end{aligned}$$
that is
$$b^{n+1}-a^{b+1}< (b-a)(n+1)b^{n}.$$


问:
If a family of sets $C$ is a subset of a $σ$-algebra $F$, then are the subsets of $C$ in the $σ$-algebra as well?
Recently I asked this question
If a set is in a σ-algebra, then are the subsets in the σ-algebra as well?

involving a set in F rather than a family of sets.

Now the question is

Let $A \subseteq F$ where $F$ is a σ-algebra. If $B \subseteq A$, is it true that $B \subseteq F$ ?

My answer would be that if $E \in B$ then $E \in A$ and therefore$E \in F$ $\forall E$. This means $B \subseteq F$.

So am I correct to assume that the claim is correct?

答: Yes, you are correct. You don't need $F$ to be a $\sigma$-algebra. It is always true that if $A \subseteq  F$ and $B \subseteq A$ then $B \subseteq F$. The proof is as you state it (and you have not used that $F$ is a $\sigma$-algebra).


问:
$|f|$ has a global minimum where $f:\mathbb{C} \to \mathbb{C}$ is any polynomial function
The following Lemma is used in a proof for the Fundamental Theorem of Algebra, but I cannot understand some steps.

Lemma: Let  $f:\mathbb{C} \to \mathbb{C}$  be any polynomial function.
Then there exists a point  $z_0 \in \mathbb{C}$  where the function
$|f|$  attains its minimum value in  $\mathbb{R}$.

The proof starts like this:

If  $f$  is a constant polynomial function, then the statement of the
Lemma is trivially true since  $|f|$  attains its minimum value at
every point in  $\mathbb{C}$ . So choose, e.g.,  $z_0=0$.

If  $f$  is not constant, then the degree of the polynomial defining
$f$  is at least one. In this case, we can denote  $f$  explicitly by

$f(z)=a_n z^n + ... + a_1 z + a_0$

with $a_n \neq 0$. Now, assume  $z \neq 0$ , and set
$A=max\{|a_0|,…,|a_{n−1}|\}$ . We can obtain a lower bound for
$|f(z)|$  as follows:

$f(z)=|a_n| |z|^n |1+\frac{a_{n-1}}{a_n} \frac{1}{z}+...+\frac{a_0}{a_n} \frac{1}{z^n}| \leq |a_n| |z|^n (1 - \frac{A}{|a_n|} \sum \limits_{k=1}^{\infty} \frac{1}{|z|^k})=|a_n| |z|^n (1 - \frac{A}{|a_n|} \frac{1}{|z|-1})$.

I don't see how these two steps follow. We can distribute the absolute value over all the factors, but how does one turn the second factor from $1+...$ into a $1-...$ using an infinite sum? After that it seems like an application of a geometric series, but how do we know that $|z|>1$?

I've been stuck for quite some while now with no notable progress. Am I missing anything obvious here?

Edit:

The next step of the proof is a follows (so it justifies the use of the formula for the geometric series):

For all $z \in \mathbb{C}$ such that $|z| \geq 2$, we can further
simplify this expression and obtain

$ (*) |f(z)| \geq |a_n| |z|^n (1 - \frac{2A}{|a_n| |z|})$.

It follows from this inequality that there is an $R > 0$ such that $|f(z)| > |f(0)|$, for all $z \in \mathbb{C}$ satisfying $|z| > R$.

Now this confuses me once again. We have $|f(0)|=a_0$, so how does this follow from the inequality given above? I think it ought to have something to do with the definiton of $A$, but I can't quite figure it out.

Could anyone please provide a hint or an explanation?

答:
$$\left|1+\frac{a_{n-1}}{a_n} \frac{1}{z}+...+\frac{a_0}{a_n} \frac{1}{z^n}\right|\ge1-\left|\frac{a_{n-1}}{a_n}\right|\frac1{|z|}-\ldots-\left|\frac{a_0}{a_n}\right|\frac1{|z|^n}\ge$$

$$\ge1-\frac A{|a_n|}\sum_{k=1}^n\frac1{|z|^k}\ge1-\frac A{|a_n|}\sum_{k=1}^\infty\frac1{|z|^k}$$

where $\;A=\max\left\{\,|a_0|,\,|a_1|,\ldots,|a_{n-1}|\,\right\}\;$

About the $\;|z|>1\iff\frac1{|z|}<1\;$ : you're right, there's no data about this in the part you copied. Where did you get this proof from?


问:
Proving a sequence involved in Apéry's proof of the irrationality of $\zeta(3)$, converges
I am trying to understand Apery's 1978 proof that $\zeta(3) = \displaystyle \sum_{n=1}^\infty \frac{1}{n^3}$ is irrational. The idea behind the proof is to find an 'accelerated' series for $\zeta(3)$ which converges too fast to $\zeta(3)$, thus proving that $\zeta(3)$ cannot be rational. In particular, a particular quantity is defined:

$$e_{n,k} = \displaystyle \sum_{m=1}^k \frac{(-1)^{m-1} (m!)^2 (n-m)!}{2m^3 (n+m)!},\quad k \leq n.$$

The key is to show that $\displaystyle \lim_{n \rightarrow \infty} e_{n,k} = 0$ uniformly in $k$, and I have no idea why this sum converges to 0. Any ideas?

答:
By the triangular inequality, $|e_{n,k}|\leqslant\frac12 a_n$ with
$$
a_n=\sum_{m=1}^nb_{n,m},\qquad b_{n,m}=\frac{(m!)^2(n-m)!}{m^3(n+m)!},
$$
hence the desired uniform convergence holds as soon as $a_n\to0$.

To prove that $a_n\to0$, note that
$$
b_{n,m}=\frac{(m-1)!(m-1)!(n-m)!}{m(n+m)!}\leqslant\frac{(m-1)!(m-1)!(n-m)!}{(n+m)!},
$$
and use twice the fact that $i!j!\leqslant (i+j)!$ for every nonnegative integers $i$ and $j$. This yields
$$
b_{n,m}\leqslant\frac{(n+m-2)!}{(n+m)!}=\frac1{(n+m-1)(n+m)}=\frac1{n+m-1}-\frac1{n+m}.
$$
Thus $a_n$ is bounded by a telescoping sum, namely
$$
a_n\leqslant\sum_{m=1}^n\left(\frac1{n+m-1}-\frac1{n+m}\right)=\frac1{n}-\frac1{2n}=\frac1{2n},
$$
This shows that $|e_{n,k}|\leqslant\frac1{4n}$ for every $n\geqslant1$ and uniformly over $k\leqslant n$, hence the proof is complete.

Edit One can refine the estimates above, taking into account the variations of the sequence  $(b_{n,m})_{1\leqslant m\leqslant n}$ for some fixed $n$, which is nonincreasing on $1\leqslant m\leqslant m_n$ for some $m_n\approx n/\sqrt2$ and nondecreasing on $m_n\leqslant m\leqslant n$. Thus, $|e_{n,k}|\leqslant\frac12  b_{n,1}+\frac12 b_{n,n}$. Hence, $n^2e_{n,k}\to\frac12 $ uniformly over $k$.


答: By assumption (first equality), $e=(ba)^3=(ba)^2(ba)$, and then $(ba)^2=(ba)^{-1}=a^{-1}b^{-1}$. Now, $a^2b^2=a^{-1}b^{-1}$ if and only if $a^3b^3=e$, and this latter holds by assumption. Then indeed $(ba)^2=a^2b^2$.


问:
Central Limit (distribution of scaled random walk)
I have been reading Steven E Shreve's Stochastic Calculus for Finance Volume II Continuous-time models. I'm trying to understand a theorem from Chapter 3 that says that distribution of scaled random walk $W^{n}(t)$ converges to normal distribution, basically a version of central limit. I can't seem to figure out how he did the following calculation:

.

答:
$X_j=1$ with probability $1/2$ and $X_j=-1$ with probability $1/2$. So
$E[e^{\frac{u}{\sqrt{n}}X_j}] = \frac{1}{2}e^{\frac{u}{\sqrt{n}}1}+ \frac{1}{2}e^{\frac{u}{\sqrt{n}}(-1)}.$


问:
Show that for $0<a<1$ and $n\in \mathbb N^+$: $\left(1-a\right)^{n}<\frac{1}{1+an}$
Show that for $0<a<1$ and $n\in \mathbb N^+$ the following does hold:

$$\left(1-a\right)^{n}<\frac{1}{1+an}$$

I know that for $a \ge -1$ and $n \in \mathbb N^+$ we have that $(1+a)^n \ge 1+an$,on the other hand since $0<a<1$ we have that $0<(1-a^2)^n=(1-a)^n(1+a)^n<1$ which follows that $$(1-a)^n<\frac{1}{(1+a)^n}\le \frac{1}{1+an}$$
$$\implies (1-a)^n< \frac{1}{1+an}$$
Is my solution right?

答:
Your proof is nicely done,we could do it like this also :

For $n=1$ its obvious.Suppose its true for $n=k$ then $${(1-a)}^k(1+ak)<1$$ if $n=k+1$ then $$\begin{align*}{(1-a)}^{k+1}(1+ak+a)={(1-a)}^k(1+ak)(1-a)\frac{(1+ak+a)}{1+ak}&\\ <\frac{(1-a)(1+ak+a)}{1+ak}&\\= 1-\frac{a^2k+a^2}{1+ak}<1 \end{align*}$$

Hence its true for $n=k+1$ our proof by induction is over.


问:
prove the following compact result about $GL(n,\mathbb{R})$
Let $K$ be a compact subset of $\mathbb{R}^n$:

prove for any $A\in GL(n,\mathbb{R})$ (endowed with Euclidean topology),there exist some neighborhood $U(A)\subset GL(n,\mathbb{R})$ near $A$ and some compact set $F\subset \mathbb{R}^n$(which only depends on choice of $A$) . such that $U(A)K$ always contains in $F$.

My attemp since $\{A\}$  is compact in $GL(n,\mathbb{R})$,there always exist some $U(A)$ as neiborhood of $A$ that is relative compact in $GL(n,\mathbb{R})$.We only need to show for set of relative compact matrix if acts on compact set $K$ it is still contains in compact set.

答:
Let $U(A)=\{B\in GL(n,\mathbb R)\mid \frac{||A||}{2}<||B||<2||A||\}$, where $||\cdot||$ is the operator norm.

($U(A)$ is open follows from the fact that all norm on $\mathbb R^{n\times n}$ are equivalent.)

Let $R=\max\{||x||\mid x\in K\}<\infty$.

Then $||Bx||\leq ||B||\cdot ||x||\leq 2||A||\cdot  R$ for all $B\in U(A)$ and $x\in K$.

Take $F=\{x\in \mathbb R^n\mid ||x||\leq 2||A||\cdot R\}.$


问:
Prove that any dihedral group can be generated by symmetry and reflections.
One definition of a dihedral group $D_n$ is

$$\langle s,r : s^2=r^2=(sr)^n=1 \rangle.$$

where $s$ is a symmetry along an axis and $r$ is a rotation of $2\pi/n$. I was just thinking this is how dihedral groups are defined but is it possible to show that the dihedral groups can be generated by two generators of order 2?

答:
Rotation by 2*pi/n generates all rotations, so it is left to prove that we can obtain also all symmetries.
Check what is $rsr^{-1},r^{2}sr^{-2},r^{3}sr^{-3}$,... up to $r^{n-1}sr^{-(n-1)}$


问:
Domain of $\sqrt{x}$ function when solving $x^2 - 2=0$
I've been going through this solution and got stuck on explaining myself why  $x^2 = 2 \Rightarrow \sqrt{x^2} =  \sqrt{2}\Rightarrow x = \pm \sqrt2$.  Domain of $f(x)=\sqrt{x}$ is $[0,\infty)$. How taking square root both sides give $\pm \sqrt{2}$ ?

答:
What exactly is not clear to you?

In general $\sqrt{x^2} = |x|$

So your equation is equivalent to $|x| = \sqrt{2}$

And from that last equation you find the two roots

$x=-\sqrt{2}$ and $x=\sqrt{2}$


答:
I think whoever wrote the text was referring to the real-valued identity $\sqrt{x^2}=\lvert x\rvert$, hence $\sqrt{x^2}=\sqrt2\Leftrightarrow \lvert x\rvert=\sqrt2\Leftrightarrow x=\sqrt2\lor x=-\sqrt2.$

Personally, I'm something of a fan of $$x^2-2=0\Leftrightarrow (x+\sqrt2)(x-\sqrt2)=0\Leftrightarrow x-\sqrt2=0\lor x+\sqrt2=0.$$

Title: Domain of $\sqrt{x}$ function when solving $x^2 - 2=0$

Body: I've been going through this solution and got stuck on explaining myself why  $x^2 = 2 \Rightarrow \sqrt{x^2} =  \sqrt{2}\Rightarrow x = \pm \sqrt2$.  Domain of $f(x)=\sqrt{x}$ is $[0,\infty)$. How taking square root both sides give $\pm \sqrt{2}$ ?


问:
Positive realness conditions for solutions of a quadratic equation
The following system of algebraic equations arises from simple eigenvalue problem in mechanics:
\begin{align*}
\frac{{{c}_{1}}^2+{x _{1}}^2}{{{c}_{1}}^2+{x _{2}}^2}-\frac{1}{a_1^2}&=0,\\
\frac{{{c}_{2}}^2+{x _{1}}^2}{{{c}_{2}}^2+{x _{2}}^2}-\frac{1}{a_2^2}&=0.
\end{align*}
We are looking for real (finite) solutions $x_1,x_2$ of the system such that
\begin{equation}
x_1>0,\quad x_2>0,\quad x_1^2\neq x_2^2.\tag{1}\label{eq01}
\end{equation}
For the parameters $a_1,a_2,c_1,c_2$ we assume that they are all real and positive.

Now, the solutions $x_1,x_2$ are directly found to be:
\begin{align*}
x_1&=\sqrt{\dfrac{\left(1-a_1^2\right)c_1^2-\left(1-a_2^2\right)c_2^2}{a_1^2-a_2^2}},\\[1em]
x_2&=\sqrt{\dfrac{a_2^2\left(1-a_1^2\right)c_1^2-a_1^2\left(1-a_2^2\right)c_2^2}{a_1^2-a_2^2}},
\end{align*}
The questions then are: (i) What conditions must be placed on $a_1,a_2,c_1,c_2$ so that the conditions in \eqref{eq01} hold? Is there a single condition on the parameters that must be true in both the solutions such that the conditions in \eqref{eq01} hold?

答:
I think I cracked the problem via some elemntary calculations. All in all we need to consider four cases: (i) $a_2<a_1<1$; (ii) $a_1<a_2<1$; (iii) $1<a_1<a_2$; and (iv) $1<a_2<a_1$. The statement then reads as follows:

In order that $x_1,x_2>0$ and $x_1^2\neq x_2^2$ the following conditions must hold:
\begin{equation*}
\begin{array}{ll}
&\dfrac{c_1^2}{c_2^2}>\dfrac{a_1^2\left(1-a_2^2\right)}{a_2^2\left(1-a_1^2\right)}&\text{under condition (i)}\\[1em]
&\dfrac{c_1^2}{c_2^2}<\dfrac{a_1^2\left(1-a_2^2\right)}{a_2^2\left(1-a_1^2\right)}&\text{under condition (ii)}\\[1em]
&\dfrac{c_1^2}{c_2^2}>\dfrac{1-a_2^2}{1-a_1^2}&\text{under condition (iii)}\\[1em]
&\dfrac{c_1^2}{c_2^2}<\dfrac{1-a_2^2}{1-a_1^2}&\text{under condition (iv)}\\[1em]
\end{array}
\end{equation*}
I hope the statement is correct.


问:
Prove that $U$ and $V$ are Isomorphic
Let there be $V$ vector space of all Converging sequences with real numbers:

$V$ = {($a_1,a_2,a_3,...) \in\ V|$ $\forall i \in \mathbb {N}$: $a_i \in\ \mathbb{R}$ & $\exists$ $lim_{ n\to\infty}a_n$ }

$U$ = {($a_1,a_2,a_3,...) \in\ V|$ $\lim_{n\to\infty}a_n=0$ },

In other words - $U$ is the subset in $V$ that contains all of real sequences converging to $0$

Prove that $V$ and $U$ are Isomorphic

MY METHOD:
 I tried to come up with this linearic transformation:

($a_1,a_2,a_3,...)$ $\rightarrow $ ($a_1 - \lim_{n\to\infty}a_n$
,$a_2$ - $\lim_{n\to\infty}a_n$,$a_3$ -
$\lim_{n\to\infty}a_n$,...)
 But this is NOT an INJECTIVE linearic
transformation

 Why?:
 For Example:
 Lets us take
two different Converging sequences: $(1,$ $\frac12$, $\frac13$,
$\frac14$,$...)$ and $(2,$ $\frac32$, $\frac43$, $\frac54$,$...)$ are
being sent to the same output.

答:
Combining the proposals in the comments with your attempt, we can actually come up with an isomorphism:

For a sequence $(a_n)\in V$ let $\alpha:=\lim_na_n$, and define
$$\varphi((a_1,a_2,\dots)):=(\alpha,\,a_1-\alpha,\,a_2-\alpha,\,\dots)\,.$$
You can easily write up its inverse.

Title: Prove that $U$ and $V$ are Isomorphic


问:
Limit of difference of pseudo-arcsinh and arcsinh
$$\DeclareMathOperator{\erf}{erf}$$
$$\DeclareMathOperator{\arcsinh}{arcsinh}$$
Let $f$ a function defined by : $$f(x) = \int_0^x \frac{\erf(u)}{u}du$$Where $\erf$ is the error-function. This function look like $\arcsinh(x)$ :

In black there is $f(x)$ and in red this is $\arcsinh(x)$. Now, we can see (or at least conjecture) that $$\kappa := \lim_\limits{x \rightarrow +\infty} \left[ \int_0^x \frac{\erf(u)}{u}du - \arcsinh(x)\right]$$exists and is finite. That would imply that $f(x) \sim \kappa + \arcsinh(x)$ and more accuratly $f(x) \approx \kappa + \arcsinh(x)$. By using $\arcsinh(x) \sim \ln(2x)$ we would have $$f(x) \sim \kappa + \ln(2x)$$Now, heres the graph of $f(x)-\arcsinh(x)$ :

In black this is $f(x)-\arcsinh(x)$. We can easily see that it converges toward $\kappa$ (in red). Here, we have $\kappa \approx 0.288607829951$.

My question is, what is the exact value of $\kappa$ ?

答:
$$f(x) = \int_0^x \frac{\text{erf}(u)}{u}du=\frac{2 x }{\sqrt
   {\pi }}\,
   _2F_2\left(\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};-x^2\right)$$Using series for large values of $x$
$$f(x)-\sinh ^{-1}(x)=\frac{\gamma }{2}-\frac{1}{4 x^2}+O\left(\frac{1}{x^3}\right)$$ which is not much.

In fact, $\gamma$ appears because the first terms of the expansion of $f(x)$ are
$$\log \left({x}\right)-\frac{1}{2}\psi \left(\frac{1}{2}\right)=\log (2 x)+\frac{\gamma }{2}$$

If you want much closer, instead of $\sinh ^{-1}(x)$ use
$\sinh ^{-1}\left(e^{\gamma /2} x\right)$ which would give
$$f(x)-\sinh ^{-1}\left(e^{\gamma /2} x\right)\sim -\frac{e^{-\gamma }}{4 x^2}$$

This would be a very idea for an approximation of the solution of $f(x)=k$.

The initial estimate would then be
$$x_0=e^{-\gamma /2} \sinh (k)$$

Trying for $k=3.456$, Newton iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 11.86091721 \\
 1 & 11.87273377 \\
 2 & 11.87273966
\end{array}
\right)$$

Edit

We can improve the estimate of the solution of equation $f(x)=k$ at the price of a single evaluation of the hypergeometric function. Using the original Householder method, we should have
$$x_1=x_0+\frac{3 f(x_0) \left(f(x_0) f''(x_0)-2 f'(x_0)^2\right)}{f(x_0)^2 f'''(x_0)+6 f'(x_0)^3-6 f(x_0) f'(x_0) f''(x_0)}$$ with
$$f'(x_0)=\frac{\text{erf}(x_0)}{x_0} \qquad f''(x_0)=\frac{2 e^{-x_0^2}}{\sqrt{\pi } x_0}-\frac{f'(x_0)}{x_0}$$
$$f'''(x_0)=-\frac{4 e^{-x_0^2} \left(x_0^2+1\right)}{\sqrt{\pi } x_0^2}+2\frac{f'(x_0)}{x_0^2}$$

Applied to the working case, this would give
$$x_0=\color{red}{11.8}609172053431$$
$$x_1=\color{red}{ 11.87273965533}87$$ while the "exact" solution is
$$x =\color{red} { 11.8727396553392}$$ and we could continue for better and better using the first iteration of an higher order method.


答:
We want to prove that, for any $n\in\mathbb{N}^+$ and $a,b>0;\;a\ne b$ the following inequality holds
$$a^n (a-(n+1) (a-b))-b^{n+1}<0$$
Proof by induction. It is true for $n=1$
$$a^1 (a-(1+1) (a-b))-b^{1+1}=-(a-b)^2<0$$
Now suppose it is true for $n$ that is
$$P(n)=-n a^{n+1}+b a^n+b n a^n-b^{n+1}<0\tag{1}$$
and let's prove it for $(n+1)$
$$P(n+1)=2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}$$
write the quotient and perform the long division wrt $b$
$$\frac{P(n+1)}{P(n)}=\frac{2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}=\\=\frac{b(-n a^{n+1}+b a^n+b n a^n-b^{n+1})+(-2 b a^{n+1}-2 b n a^{n+1}+n a^{n+2}+a^{n+2}+b^2 a^n+b^2 n a^n)}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}$$

$$b+\frac{a^n \left(a^2 n+a^2-2 a b n-2 a b+b^2 n+b^2\right)}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}=b+\frac{a^n (n+1) (a-b)^2}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}$$
as the result is positive and the denominator is negative, the numerator must be negative, that is $P(n+1)<0$

Title: Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ $\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$

Body:
Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ the following does hold:

$$\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$$

I tried to simplify the inequality to $$na^n(b-a)<b(b^n-a^n)$$
$$na^n(b-a)<b(b-a)(\sum_{k=0}^{n-1}b^{k}a^{n-1-k})$$

How to continue? (Notice that I don't know if $b-a$ is necessarily positive).


问:
solution to second order ode (not in standard form)
So I used separation of variables to solve a linear PDE.

Thus, I arrived at the following ODES:

$$
u_{t} = - \lambda^{2} u = 0
$$

which has the solution:

$$
u = A \exp{(- \lambda^{2} t)}
$$
the 2nd ODE is given by:
$$ k u_{xx} + \frac{2k}{x} u_{x} + \lambda^{2} u = 0 $$
Here, $k, \lambda > 0$.
the boundary conditions to be applied to the 2nd ODE are:
$$u_{x} (0,t) = u(a,t) = 0$$

Just asking how do you solve the 2nd ODE and apply its B.Cs as the ODE is not a 2nd order ODE with constant coefficients.

After this, everything else seems quite easy.

答:
With the substitution $u(x) = v(x)/x$, the second order ODE becomes:
$$
k v_{xx} + \lambda^2 v = 0
$$

So the original ODE has the fundamental solution (for $k \lambda \neq 0$):
$$
u(x) = \frac{A \cos(\omega x) + B \sin(\omega x) }{x}, k \omega^2 = \lambda^2
$$

The function $u$ has the following expansion close to $x = 0$:
$$
u(x) = \frac{A}{x} + B\omega - \frac{A \omega^2}{2} x + \mathcal{O}(x^2)
$$

So, enforcing $u_x(0) = 0$ implies that $A = 0$ (otherwise we don't even have a derivative there). The second boundary condition requires:
$$
0 = B \frac{\sin(\omega a)}{a}
$$

So non-trivial solutions exist, when $\omega \in \frac{\pi}{a} \mathbb{N}$.

The case $\lambda = 0$ gives:
$$
u(x) = \frac{A + B x}{x}
$$

But this is impossible for the boundary conditions $u_x(0) = 0$ implies $A=0$ and then $u(a) = B$.

Summarizing, the general solution to your full PDE is then ($t \geq 0$):
$$
u(x,t) = \sum_{n=1}^\infty B_n \frac{\sin\left(\frac{n \pi}{a} x \right)}{x} \exp\left( - k t \frac{n^2 \pi^2}{a^2} \right)
$$


问:
If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is abelian.

Theorem 9.3 *$G/Z$ Theorem.

Let $G$ be a group and let $Z(G)$ be the center of $G$. If $G/Z(G)$ is cyclic, then $G$ is Abelian.

$\mathsf{PROOF}$ Since $G$ is Abelian is equivalent to $Z(G)=G$, it suffices to show that the only element of $G/Z(G)$ is the identity coset $Z(G)$. To this end let $G/Z(G)=\langle gZ(G)\rangle$ and let $a\in G$. Then there exists an integer $i$ such that $aZ(G)=(gZ(G))^i = g^iZ(G)$. Thus, $a=g^iz$ for some $z\in Z(G)$. Since both $g^i$ and $z$ belong to $C(g)$, so does $a$. Becaus $a$ is an arbitrary element of $G$ this means that every element of $G$ commutes with $g$ so $g\in Z(G)$. Thus, $gZ(G)=Z(G)$ is the only element of $G/Z(G)$. $\Box$

A few remarks about Theorem 9.3 are in order. First, our proof shows that a better result is possible. If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is Abelian. Second, in practice, it is the contrapositive of the theorem that is most often used - that is, if $G$ is non-Abelian, then $G/Z(G)$ is not cyclic. For example, it follows immediately from the statement and Lagrange’s Theorem that a non-Abelian group of order $pq$, where $p$ and $q$ are prime, must have trivial center. Third, if $G/Z(G)$ is cyclic, it must be trivial.

In the book Contemporary abstract algebra, page 194, there's a result corresponding to theorem 9.3 which is

If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is abelian.

My question is - is there any way to prove this result using the same method used to prove the Theorem 9.3 ?
I've attached an image of the proof of Theorem 9.3.

My proof: [In short]
$$\begin{align*}
 G/H &= \ \langle gH\rangle \\
   &\implies \text{for any} \ a,b \in G,a\in g^iH \ \text{and} \ \ b\in g^jH.
\\ \therefore a&=g^ih_1 \ \text{and} \ b=g^jh_2.
\\ \therefore  ab &= (g^ih_1)(g^jh_2)
\\ &= g^{i+j}h_1h_2
\\ &= (g^jh_2)(g^ih_1)
\\ &= ba.
\end{align*}$$

答:
The more general result can indeed be proved using the same method. May I present a few additional, general results that would help in organising the proof:

Lemma 1. Let $G$ be an arbitrary group, $H \trianglelefteq G$ a normal subgroup and $S, T \subseteq G$ subsets such that $H=\langle S \rangle$ ($S$ is a generating system for $H$) and $G/H=\langle \sigma(T) \rangle$ (the canonical image of $T$ generates the quotient), where $\sigma \colon G \to G/H$ denotes the canonical surjection. Then $G=\langle S \cup T \rangle$, i.e. $G$ is generated by the union $S \cup T$.

Lemma 2. Let $G$ be an arbitrary group and $S \subseteq G$ a generating system consisting of mutually commuting elements (in formal expression, $G=\langle S \rangle$ and $S \subseteq \mathrm{C}_G(S)$). Then $G$ is abelian.

These two lemata suffice to prove the following:

Proposition. Let $G$ be an arbitrary group and $H \leqslant \mathrm{Z}(G)$ a central subgroup such that $G/H$ is cyclic. Then $G$ is abelian.

Proof. As the quotient $G/H$ is cyclic, there exists $a \in G$ such that $\sigma(a)$ generates $G/H$ (where $\sigma$ denotes the canonical surjection, as above). It then follows by virtue of lemma 1 that $G$ is generated by $H \cup \{a\}$ ($H$ generates itself, of course) and since $H$ is central -- consisting of elements which commute with anything in $G$ -- $H \cup \{a\}$ is easily seen to consist of pairwise commuting elements and hence lemma 2 applies. $\Box$

Let me know if you would like to see proofs for the two lemata (they are very simple).


问:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \}$
I've tried to solve this limit:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} $

$x-x^2 \ln (1+ \sin \frac{1}{x}) \sim x-x^2 \sin \frac{1}{x} \sim x-x = 0 $

but the final result should be $\frac{1}{2}$

答:
$$\lim x-x^2 \ln (1+ \sin \frac{1}{x})\\=\lim x-x^2\left(\sin \frac{1}{x}-\frac 1 2 \sin^2 \frac{1}{x}+\frac1 3 \sin^3 \frac{1}{x}+\cdots\right)\\= \lim x-x^2 \sin \frac{1}{x} +\frac1 2\\=\lim x-x^2\left(\frac 1 x-\frac 1{6x^3}+\frac 1 {120x^5}-\cdots\right)+\frac 1 2\\=\frac 1 2$$.

I just used expansion of $\ln(1+x),\sin (x)$ and the fact $\lim_{x\to 0}\frac {\sin x}{x}=1$


问:
Is there a smooth compact set between any two compact sets?
today I saw the following statement and of course believe that this is true but I don't know how to prove it rigorously (and neither do my colleagues).

Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $K\subset \Omega$ compact. Then there is a set $A$ such that $K\subset A\subset \Omega$ with $\partial A$ smooth (smooth means at least $C^1$).

However, my first ansatz was: Mollify the characteristic function of $K$ such that its support is in $\Omega$ and denote this functions by $g$. Now my intention was to use the regular value theorem on $g$. Like $g^{-1}(1/2)$ should be a smooth manifold. But I can't show whether this point is a regular point nor do I know that $\{x :g(x)\geq 1/2\} \supset K$. Does anyone have a different or additional idea on this?

答:
For each $p\in K$, let $B(\delta_p,p)\subset\Omega$ where $B(\delta_p,p)$ is an ball of ratio $\delta_p$ and ceter $p$. Because $K$ is compact, you can find a finite number of balls $B(\delta_{p_i},p)$ such that $$K\subset\bigcup_{i=1}^mB(\delta_{p_i},p)\subset\Omega$$

To finish in we have to consider the points where any two balls intersect. Denote by $A=\cup_{i=1}^mB(\delta_{p_i},p)$.

Without loss of generality consider that $n=2$. The procedure is the same for $n>2$. Take two balls and consider their boundarys that are the circles $\partial B_1$ and $\partial B_2$. We can suppose that the points in the intersections of the circle lie whitin the axis $y$ in $\mathbb{R}^2$. Take some $\epsilon>0$ small and erase the part of the circles contained in $(-\epsilon,\epsilon)\times\mathbb{R}$. We now have two arcs of circles. We join these two arcs by two lines (if we were in dimension $n>2$, we could use a cylinder).

Now we have something like this. By using a argument of partition of unity, you can regularize all the part that is the intersection between the lines and the circles. If you take $\epsilon>0$ small you still have that $A\subset \Omega$.


答:
Set $\dfrac1x=h$ to find

$$\lim_{h\to0^+}\dfrac{h-\ln(1+\sin h)}{h^2}$$

$$=\lim_{h\to0^+}\dfrac{h-\sin h+\dfrac{\sin^2h}2+O(h^3)}{h^2}$$

$$=\lim_{h\to0^+}\dfrac{h-\sin h}{h^2}+\dfrac12\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2+\lim_{h\to0^+}\dfrac{O(h^3)}{h^2}$$

For the first part, use the second problem of Are all limits solvable without L'Hôpital Rule or Series Expansion

Alternatively,

$$\lim_{h\to0^+}\dfrac{h-\ln(1+\sin h)}{h^2}=\lim_{h\to0^+}\dfrac{h-\sin h}{h^2}+\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2\cdot\lim_{h\to0^+}\dfrac{\sin h-\ln(1+\sin h)}{\sin^2h}$$

For the last part, use $(3)$ problem of  Are all limits solvable without L'Hôpital Rule or Series Expansion

Title: $\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \}$

Body:
I've tried to solve this limit:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} $

$x-x^2 \ln (1+ \sin \frac{1}{x}) \sim x-x^2 \sin \frac{1}{x} \sim x-x = 0 $

but the final result should be $\frac{1}{2}$


问:
prove that if $u \equiv w \bmod \Phi(m)$ and $\operatorname{gcd}(a, m)=1$ then $a^{u} \equiv a^{w} \bmod m$
Let $u \equiv w \bmod \Phi(m)$.
if $\operatorname{gcd}(a, m)=1$ then how to elementary prove that $a^{u} \equiv a^{w} \bmod m$ ?

答:
Note that if $\gcd(a, m) = 1$,  by Euler's theorem $a^{\varphi(m)} \equiv 1 \pmod{m}$.

Then as $\varphi(m) \mid (u-w)$, we have $a^{u-w} \equiv 1 \pmod{m}$ and $a^u \equiv a^w \pmod{m}$.


问:
Determine whether these structures are is isomorphic
Determine whether these structures are is isomorphic ? ,

if they are isomorphic Show the isomorphism function and proved

if they are no isomorphic proved by an appropriate verse

A. $M_2= \langle(1,\infty),+\rangle$ ,  $M_1 =  \langle(0,\infty),+\rangle$

attempt:

if i want to show isomphism in A. or B.  i need to show .

for all sign of function $f$

1.$h(F^{M}(a_1,a_2,...a_n))=F^N(h(a_1),h(a_2),...(h(a_n))$

2.Injective function

3.Surjective function

if i try to prove isomorphic in B. between $M_2$ , $M_1$
i dont know how to prove them with in open segment

答:
Question A

The sentence $\phi \equiv \forall x \exists y(y+y=x)$ is satisfied in $M_1$ but not $M_2$. Therefore the two structures can't be isomorphic.


问:
Cube roots of the complex numbers 1+i?
I cant get any good reference in my books regarding cube of complex numbers. Please help me find cube roots of the Complex number i+1??</p>

答:
Since $1+i = \sqrt{2} e^{i\pi/4}$, we have

$$(1+i)^{1/3} = 2^{1/6} e^{i\pi/12}.$$

Alternatively, $1+i$ could be written $\sqrt{2} e^{9i\pi/4}$ or $\sqrt{2} e^{17i\pi/4}$, providing two more cube roots.  WolframAlpha provides a nice illustration of this:

Title: Cube roots of the complex numbers 1+i?


问:
How can $\sin(90) = 1$?
I know this question has been asked before, but the answers don't really make intuitive sense.

As angle $\theta$ approaches $90$ degrees, how can $\sin(90) = 1$ if there IS NO triangle?

This breaks all other ratios as well, because the other ratios can be written in the form of $\sin(\theta)$ as well.

A simple answer (if it exists) s'il vous plait.

Thanks in advance.

答:
To spell out @Vercingetorix's point:

The radius-$1$ circle centred on the origin in $2$ dimensions has equation $x^2+y^2=1$ by the Pythagorean theorem. (If it's not obvious this holds when $x$ and/or $y$ is negative, reflect in an axis so they're not, and note $(-u)^2=u^2$.)

For acute $\theta$, a right-angled triangle exists containing $\theta$ with vertices $(0,\,0),(\cos\theta,\,0),\,(\cos\theta,\,\sin\theta)$. The first and last of these are the ends of the hypotenuse, which is a radius, and the side adjacent to $\theta$ is the radius ending in $(1,\,0)$. Rotating $\theta$ anticlockwise from that radius gives the hypotenuse.

Any radius can be used the same way, even if the resulting angle isn't acute, and it meets the circle at $(\cos\theta,\,\sin\theta)$. This defines the trigonometric functions for any $\theta\in\Bbb R$. The choice $\theta=90^\circ$ gets you to the radius ending at $(0,\,1)$, so $\cos90^\circ=0,\,\sin90^\circ=1$.

Title: How can $\sin(90) = 1$?


问:
Prove $\int_2^\infty{\frac{\ln(t)}{t^{3/2}}},\mathrm{d}t$ converges
Show, using a comparison test, that $\displaystyle \int_2^\infty{\frac{\log{t}}{t^{\frac32}}}\mathrm{d}t$ converges.

All the answers I've tried shows it diverges, taking $\log{t} \le t^{1/2}$ and $\log{t} \le t$.

Cheers

答:
Solving $\log t = t^{1/4}$ we get $t_1=4.177, t_2=5503.66$ from W|A.

Differentiating $(\log t)' = \frac{1}{x}$ and $(t^{1/4})' = \frac 1 4 t^{-3/4}$. And  $t>t^{3/4}\implies 1/t<t^{-3/4}$.

for $t \ge t_2$, $\log t \le t^{1/4}$.

$$\int_2^\infty \frac{\log t}{t^{3/2}}dt = \int_2^{t_2} \frac{\log t}{t^{3/2}}dt + \int_{t_2}^\infty \frac{\log t}{t^{3/2}}dt \le \int_2^{t_2} \frac{\log t}{t^{3/2}}dt + \int_{t_2}^\infty \frac{t^{1/4}}{t^{3/2}}dt$$.

We know that $\displaystyle \int_{t_2}^\infty \frac{t^{1/4}}{t^{3/2}}dt$ converges. So the integral converges by comparison.

Integrating by parts::
$$\int_2^\infty \frac{\log x}{x^{3/2}}dx = -\frac{2 \log x}{\sqrt x} \big |_2^\infty - \int_2^{\infty} \frac 1 x \frac{-2}{\sqrt x}dx = \sqrt 2 \log 2  + 2 \sqrt 2 $$


答:
First let $\Delta\Omega$ be the boundary of $\Omega$ as $\Omega$ is open and bounded $\Delta\Omega$ is compact and disjoint from $\Omega$ (hence does not intersect $K$).

Now as $K$ and $\Delta\Omega$ are non intersecting compact sets there exists some $\varepsilon>0$ such that no pair of points $x\in K, y\in\Delta\Omega$ have $\|x-y\|<2\varepsilon$ Furthermore we may choose a finite number of points $x_1,\dots, x_n\in K$ and $y_1,\dots y_m\in\Delta\Omega$  such that the open $\varepsilon$ balls $\mathcal B(\varepsilon, x_i)$ and $\mathcal B(\varepsilon, y_i)$ cover $K$ and $\Delta\Omega$ respectively.

Now let $A'\subset\Omega$ be the set of points $p$ such that there exists some $i$ for which $\|p-x_i\|<\|p-y_j\|$ for every $j\leq m$.

Then for every point $p\in\Delta A'$ there must exist some pair $i,j$ such that $\|p=x_i\| = \|p=y_j\|>\varepsilon$.

Therefore $\Delta A'$ is a simplex and for every $p\in\delta A$ the ball $\mathcal B(p,\frac \varepsilon 2)$ is contained in $\Omega\setminus K$.

I can't remember a theorem off the top of my head, but I'm pretty sure there's a smooth modification of $\Delta A'$ that stays within $\frac\varepsilon 2$ of $A$.

Title: Is there a smooth compact set between any two compact sets?

Body:
today I saw the following statement and of course believe that this is true but I don't know how to prove it rigorously (and neither do my colleagues).

Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $K\subset \Omega$ compact. Then there is a set $A$ such that $K\subset A\subset \Omega$ with $\partial A$ smooth (smooth means at least $C^1$).

However, my first ansatz was: Mollify the characteristic function of $K$ such that its support is in $\Omega$ and denote this functions by $g$. Now my intention was to use the regular value theorem on $g$. Like $g^{-1}(1/2)$ should be a smooth manifold. But I can't show whether this point is a regular point nor do I know that $\{x :g(x)\geq 1/2\} \supset K$. Does anyone have a different or additional idea on this?


答: $\sin x$ is the $y$ coordinate on the unit circle and $\cos x$ is the $x$ coordinate. If you use your intuition a $90$ degrees $\cos x$ must be $0$ because its a straight line up and $\sin x$ must be $1$.

Title: How can $\sin(90) = 1$?

Body:
I know this question has been asked before, but the answers don't really make intuitive sense.

As angle $\theta$ approaches $90$ degrees, how can $\sin(90) = 1$ if there IS NO triangle?

This breaks all other ratios as well, because the other ratios can be written in the form of $\sin(\theta)$ as well.

A simple answer (if it exists) s'il vous plait.

Thanks in advance.


答: If you a want to know how is $\sin(90^{\circ})=1$ from a triangle's point of view, well you can take any right angled triangle and fix the base length. Now increase the perpendicular side. You will see that as you increase the side, length of hypotenuse also increases. If the angle opposite to the perpendicular side is $\theta$, so now as you increased that side, notice that $\theta \to 90^{\circ}$, and the length of hypotenuse tends to the length of the perpendicular side. So, $\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \to 1 $ as $\theta \to 90^{\circ}$.

Title: How can $\sin(90) = 1$?


问:
Completeness of two metrics in the same space
Let $d, d'$ two metrics on the same space $X$, and suppose that $(X,d)$ is complete. I've read that if every ball
$B'_r(x_0)=\left\{x\in X | d'(x_0,x)<r\right\}$
is contained in a ball
$B_R(x_0)=\left\{x\in X | d(x_0,x)<R\right\}$,
then $(X,d')$ is also complete.
How does the completeness follow from the above property?

答: Suppose that $d$ is the discrete metric. Then it is complete of course. Moreover, for any metric $d'$, we have the property you described above (letting $R = 1.5$ for instance). But clearly there are non complete metrics. So what you say is not true.


问:
Circle hard problem
In a recent test, I was given this exercise.
A point P is 25cm away from the center of a circle. Form this point two tangents were drawn to a circle AP and PB. The distance between AB is 24cm. Find the radius. Can someone explain plz?</p>

答: Hint: The blue triangle is a right triangle

Title: Circle hard problem


问:
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$
I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$

Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$

But the final result should be $-\frac{4}{3}$.

Any help would be appreciated.

答:
I think the mistake is from the first step.Emm...in fact,$$\lim \frac{{\ln (1 + x\arctan x)}}{{\sqrt {1 + 2{x^4}}  - 1}} = \frac{{\ln (1 + x\arctan x)(\sqrt {1 + 2{x^4}}  + 1)}}{{2{x^4}}} = \infty , \\ \lim \frac{{{e^{{x^2}}} - 1}}{{\sqrt {1 + 2{x^4}}  - 1}} = \frac{{({e^{{x^2}}} - 1)(\sqrt {1 + 2{x^4}}  + 1)}}{{2{x^4}}} =  \infty $$
However,when $${f_1} \sim \alpha ,{f_2} \sim \beta $$ but $$\frac{\alpha }{g} = \infty or\frac{\beta }{g} = \infty $$
you can't apply
$$\frac{{{f_1} - {f_2}}}{g} \sim \frac{{\alpha  - \beta }}{g}$$

I don't know if I made it clear to you...


问:
Inequality: $|1+4(x^4-x^2)|\leq 1$ implies $|x|\leq1$
My question comes from here: Lax-Wendroff method for linear advection - Stability analysis, but nobody needs to know anything about Lax-Wendroff. My question is very easy.

Basically, in the last line of the answer, it is claimed that

$$|1+4(x^4-x^2)|\leq 1$$
is satisfied if $|x|\leq1$.

And I drew a plot to confirm this:

My question is, how do I obtain the desired inequality $|x|\leq1$ algebraically?

Body: Note $|(2x^2-1)^2|\le1$ iff $-1\le2x^2-1\le1$, i.e. $0\le x^2\le1$.


问:
What is the inverse function of $f(x)=x-\log x (0<x<1)$
What is the inverse function of $f(x)=x-\log(x)$ where $0<x<1$?

答: By solving the equation $y=x-\log x$ in $(0,1)$, we get $x= -W\left(-e^{-y}\right)$, where $W$ is the Lambert W function. You can verify this result by definition.


答: Assume that $|x| > 1$, now we have $x^4 > x^2$, so $|1+\text{something positive}| > 1$ always. Hence, the opposite.

Title: Inequality: $|1+4(x^4-x^2)|\leq 1$ implies $|x|\leq1$

Body:
My question comes from here: Lax-Wendroff method for linear advection - Stability analysis, but nobody needs to know anything about Lax-Wendroff. My question is very easy.

Basically, in the last line of the answer, it is claimed that

$$|1+4(x^4-x^2)|\leq 1$$
is satisfied if $|x|\leq1$.

And I drew a plot to confirm this:

My question is, how do I obtain the desired inequality $|x|\leq1$ algebraically?


答:
Another hint:

In a right triangle, the altitude is the geometric mean of the segments it delimits on the hypotenuse. If you denote these lengths $a$ and $b$, you obtain the standard high school problem
$$ab= 12^2,\qquad a+b=25.$$

Title: Circle hard problem

Body:
In a recent test, I was given this exercise.
A point P is 25cm away from the center of a circle. Form this point two tangents were drawn to a circle AP and PB. The distance between AB is 24cm. Find the radius. Can someone explain plz?</p>


Body:
Convert this limit by substituting $$x=\frac{1}{t}$$
as $x \rightarrow +\infty, t \rightarrow 0$ therefore,
$$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} = \lim_{t \to 0} \left \{ \frac{1}{t}-\frac{1}{t^2} \ln (1+ \sin t) \right \}  $$
$$\lim_{t \to 0} \left \{ \frac{t - \ln (1+ \sin t)}{{t^2}} \right \} =\lim_{t \to 0} \left \{ \frac{t - (\sin t - \frac{\sin^2t}{2} + \frac{\sin^3t}{3} -\frac{sin^4t}{4}...)}{{t^2}} \right \}  $$
$$\lim_{t \to 0} \left \{ \frac{t - \sin t + \frac{\sin^2t}{2} - \frac{\sin^3t}{3} +\frac{sin^4t}{4}...}{{t^2}} \right \}  =\lim_{t \to 0} \left \{ \frac{t - \sin t}{t^2} +  \frac{\sin^2t}{2t^2}+\frac{ - \frac{\sin^3t}{3} +\frac{sin^4t}{4}...}{{t^2}} \right \} $$
$$\lim_{t \to 0}{\frac{t-\sin t}{t^2}}= \lim_{t \to 0}\frac{t-(t-\frac{t^3}{3!}-\frac{t^5}{5!}...)}{t^2}=0$$
$$\lim_{t \to 0}{\frac{\sin^2t}{2t^2}=\lim_{t \to 0}\frac{1}{2}(\frac{\sin t}{t})^2}=\frac{1}{2}$$
$$\lim_{t \to 0}{\frac{-\frac{\sin^3t}{3!}+\frac{\sin^5t}{5!}...}{t^2}=0}$$
So you get:
$$\lim_{t \to 0} \left \{ \frac{t - \ln (1+ \sin t)}{{t^2}} \right \}=0+\frac{1}{2}+0 = \frac{1}{2}$$
Hope this helps...


Body: You can do it as follows:\begin{align}\bigl|1+4(x^4-x^2)\bigr|\leqslant1&\iff\bigl(1+4(x^4-x^2)\bigr)^2\leqslant1\\&\iff8(x^4-x^2)+16(x^4-x^2)^2\leqslant0\\&\iff(x^4-x^2)\bigl(1+2(x^2-x^2)\bigr)\leqslant0\\&\iff(x^4-x^2)\bigl(x^4+(x^2-1)^2\bigr)\leqslant0\\&\iff x^2(x^2-1)\leqslant0\\&\iff x^2\leqslant1\\&\iff x\in[-1,1].\end{align}


问:
Two properly homotopic maps induce the same map on end-cohomology
I am trying to solve the following problem.

Two properly homotopic maps induce the same map on end-cohomology.

Let $X$ be a locally compact Hausdorff space and $R$ be a commutative ring. Note that end-cohomology is defined as follows: Consider the cochain complex whose $q$-th cochain is $\frac{C^q(X;R)}{C^q_{\textbf{c}}(X;R)}$. Here, $\textbf{c}$ stands for compactly supported. The Cohomology groups of this complex will be denoted by $H_\textbf e^q(X;R)$. Here, $\textbf e$ stands for the end.

Now, let Let $f,g:X\to Y$ be two proper maps between two locally compact Hausdorff space, and $\mathscr H:f\simeq g$ be a proper homotopy. I want to show, $f_*=g_*:H^q_\textbf e(X;R)\to H^q_\textbf e(Y;R)$.

My Thoughts: I am trying to using the long exact sequence $$\cdots \to H_\textbf c^q(X;R)\xrightarrow{i_X}H^q(X;R)\to H_\textbf e^q(X;R)\to H_\textbf c^{q+1}(X;R)\to \cdots$$ obtained from the short exact sequnce $$0\to C_\textbf c^q(X;R)\to C^q(X;R)\to \frac{C^q(X;R)}{C^q_\textbf c(X;R)}\to 0.$$ Since, $\mathscr H$ is a proper homotopy we have $f_*=g_*:H^q_\textbf c(X;R)\to H^q_\textbf c(Y;R)$. Also, $\mathscr H$ is a homotopy, then $f_*=g_*:H^q(X;R)\to H^q(Y;R)$. Now, I am trying to use some algebra stuff like, Snake Lemma or $5$-lemma to prove the problem. But cannot proceed further.

Any help will be appreciated. Thanks in advance.

答:
This strategy (using maps you know and the five lemma) will NEVER work to show that some pair of maps are the same. Indeed consider the pair of maps of short exact sequences from $V \to V^2 \to V$ to itself, which is zero on the outer factors and either zero in the middle or $\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}.$ As you can see, only the maps in the middle differ. Another way to say this is "maps which between filtered vector spaces whose associated graded map is zero need not themselves be zero maps." Think about this long enough and you will see that in fact this is geometrically obvious.

Now for your question show that a proper homotopy induces a chain homotopy on $C^*$ which preserves the subcomplex of compactly supported things (this is where you use properness), so that it induces a chain homotopy on the quotient.


问:
How to show that $f(x)=g(x)$ for all $x\in[a,b]$?
Let $f$ and $g$ be continuous functions on $[a,b]\subset\mathbb{R}$. Suppose also that $f(x)=g(x)$ almost everywhere on $[a,b]$. Show that infact $f(x)=g(x)$ for all $x\in[a,b]$.

There every some results I had proved in doing metric space that $E=\{x:f(x)=g(x)\}$ is a closed set(I don't know if I can use this fact in this problem or not). I think I have to use the compactness of $[a,b]$ somehow which I couldn't figure out. Any hint will be enough. Thank you.

答:
$f - g$ is continuous since $f, g$ are. By continuity then $(f-g)^{-1}([a, b]\setminus \{0\})  = \{x : f(x) \not = g(x)\}$ is open.

But the latter is of measure $0$ (as $f \equiv g$ a.e); every non empty open set has measure $>0$. So it must be empty, i.e, $f = g$.


答: Hint: if you let $\theta\in[0,2\pi)$, then every complex number can be uniquely represented by $re^{i\theta}$. Now the cube roots are just this number raised to the $1/3$ power.</p>

Title: Cube roots of the complex numbers 1+i?

Body: I cant get any good reference in my books regarding cube of complex numbers. Please help me find cube roots of the Complex number i+1??</p>


问:
Proving two paths are not path homotopic
If I wanted to prove that two paths $f:[0,1] \rightarrow X$ and $g:[0,1] \rightarrow X$ are not path homotopic, I can prove that $f * \bar{g}$ is not path homotopic with the constant path that stays at $f(0)$. Here $\bar{g}$ denotes the reverse of $g$.

I'm not sure why showing $f * \bar{g}$ is not nulhomotopic tells us that $f$ and $g$ is not path homotopic. Can someone explain this?

答:
Commenter: That is what the term "path-homotopic" means in every reference.

OP: If f and g are path-homotopic, then f * (g-bar) is based homotopic to g * (g-bar) which is based homotopic to the identity, as you saw when you introduced fundamental groups. Contradiction.


问:
Show that for $0<a<1$ and $n\in \mathbb N^+$: $\left(1-a\right)^{n}<\frac{1}{1+an}$
Show that for $0<a<1$ and $n\in \mathbb N^+$ the following does hold:

$$\left(1-a\right)^{n}<\frac{1}{1+an}$$

I know that for $a \ge -1$ and $n \in \mathbb N^+$ we have that $(1+a)^n \ge 1+an$,on the other hand since $0<a<1$ we have that $0<(1-a^2)^n=(1-a)^n(1+a)^n<1$ which follows that $$(1-a)^n<\frac{1}{(1+a)^n}\le \frac{1}{1+an}$$
$$\implies (1-a)^n< \frac{1}{1+an}$$
Is my solution right?

Body:
Let
$$f(a)=(1-a)^n(1+an),a\in[0,a].$$
Compute its derivative:
$$f'(a)=-n(n+1)a(1-a)^{n-1}<0,\forall a\in(0,1).$$
So
$$0=f(1)<f(a)=(1-a)^n(1+an)<f(0)=1,\forall a\in(0,1),$$
which implies
$$\left(1-a\right)^{n}<\frac{1}{1+an}.$$


问:
Index of submodule generated by multiplication by one element of free $\mathbb{Z}$-module
This question appears in an algebraic number theory notes with $M=\mathcal{O}_K$ for a number field $K$.

Let $M$ be a free $\mathbb{Z}$-module of rank $n$ which is a ring itself. Let $x\in M$, then multiplication by $x$, $\phi_x:y\mapsto xy$ is a $M$-endomorphism, hence can be represented by a $n\times n$ $\mathbb{Z}$-matrix C. Let $\{v_1\,\dots,v_n\}$ be a basis of $M$, and let $N$ be the submodule generated by $\{xv_1,\dots,xv_n\}$ over $\mathbb{Z}$. I want to show that $[M:N]=|\det(C)|$, where $[M:N]$ is the index of $N$ as a subgroup of $M$. Could anyone please help?

答: From the isomorphism $M\to \Bbb{Z}^n$ given by the $v_j$ then $M/N\cong \Bbb{Z}^n/ C \Bbb{Z}^n$ has $Vol( C[0,1]^n) = |\det(C)|$ elements.


答: Let $A=\{x\in [a,b],f(x)\neq g(x)\}$, then $A$ is an open subset of $[a,b]$ because $f-g$ is continuous. If $A\neq\emptyset$, let $x\in A$, since $A$ is open there exists $\delta>0$ such that $]x-\delta,x+\delta[\subset A$, thus $\lambda(A)\geqslant 2\delta>0$ which is not because $\lambda(A)=0$ ($f=g$ a.e.). Therefore $A=\emptyset$ and $f(x)=g(x)$ for all $x\in[a,b]$.

Title: How to show that $f(x)=g(x)$ for all $x\in[a,b]$?

Body:
Let $f$ and $g$ be continuous functions on $[a,b]\subset\mathbb{R}$. Suppose also that $f(x)=g(x)$ almost everywhere on $[a,b]$. Show that infact $f(x)=g(x)$ for all $x\in[a,b]$.

There every some results I had proved in doing metric space that $E=\{x:f(x)=g(x)\}$ is a closed set(I don't know if I can use this fact in this problem or not). I think I have to use the compactness of $[a,b]$ somehow which I couldn't figure out. Any hint will be enough. Thank you.


问:
Why $\log_b(x)$ = $-\log_{1/b}(x)$?
IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.

I tried to prove this to myself on paper.

I can write both equations in exponential form. Pretend $x=4$:

$$\log_{1/2}(4)=\frac{1}{2}^y=4$$

Then write the same for the other version:

$$-\log_2(4)=-2^y=4$$

If I know these are equivalent then the two exponent versions should be the same no?

$$4=\frac{1}{2}^y=-2^y$$

It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?

答: Let $x>0$, then $\log_b(x)$ is the unique real number such that $b^{\log_b(x)}=x$, but $b^{-\log_{1/b}(x)}=\frac{1}{b^{\log_{1/b}(x)}}=\left(\frac{1}{b}\right)^{\log_{1/b}(x)}=x$, therefore $-\log_{1/b}(x)=\log_b(x)$.

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?


问:
Permutation Group Problem
Hi I have an exercise in book, i.e.

Let $P \in S_n$ be a subgroup of prime order and suppose $x \in S_n$ normalizes but does not centralize P. Show that $x$ fixes at most one point in each orbit of $P$.

I understand the first sentence. But i don't know where to start to answer the question. Any hint ?

答:
First notice that any $x\in S_n$ will have a fixed point in its action on $P$ by conjugation, namely $e$ (the identity element).

So we have to look at the orbits of $x$ on $P-\{e\}$. Now, since $P$ is of prime order it is cyclic, and if $x$ has a fixed point $\alpha\in P-\{e\}$ then $x\alpha = \alpha x$ and we might assume without loss of generality that $P=\langle \alpha\rangle$. Can you complete the proof?

Title: Permutation Group Problem


问:
How to state precisely this requirement relative to the application of natural deduction rules?

Contrary to replacement  rules, in natural deduction, rules of infrence need to be applied " to an entire line" .

For example, I cannot use Modus Ponens in the following way

Premise (1)  : $( A \rightarrow (B\rightarrow C)) $

Premise (2) :  $B$

Conclusion : $C$  ( wrongly applying MP to the consequent in premise (1) ) .

What is the ( standard) formal statement of this requirement?

答:
First, and as an aside, let me just say that the example:

Premise (1)  : $( A \rightarrow (B\rightarrow C)) $

Premise (2) :  $B$

Conclusion : $C$  ( wrongly applying MP to the consequent in premise (1) ) .

is really weird!  If this is supposed to be an application of an inference rule as applied to part of a larger statement, one would expect that one applies the $B$ to the $B \to C$ as it occurs within the larger statement, and thus have the result $C$ to be still part of that same larger statement, thus resulting in $A \to C$, rather than just $C$.

I mean, this is how it works for rules of replacement. That is, when applying double negation on $A \to \neg \neg B$, the result is $A \to B$, rather than merely $B$.

So, the real analogue should be:

Premise (1)  : $( A \rightarrow (B\rightarrow C)) $

Premise (2) :  $B$

Conclusion : $A \to C$  ( wrongly applying MP to the consequent in premise (1) ) .

Unfortunately, this argument is in fact valid!

So ... this is a really bad demonstration as to why one should not apply inference rules to partial statements.

Here is a better example to demonstrate why you should not apply inference rules to parts of statements the way you can with rules of replacement.

Let's take the simple inference rule of Simplification (also known as $\land$ Elim):

$P \land Q$

$\therefore P$

Now suppose we try to apply this as follows:

$Premise (1)  : (A \land B) \to C$

$Conclusion: B \to C $ ( wrongly applying Simplification to the antecedent of premise (1) )

OK, now we really have an invalid inference on our hands: If $B$ is True but $A$ and $C$ are false, then the premise is true, but the conclusion is false.

And so it is with almost all inference rules: sometimes when you try to apply them to part of a statement you actually end up with a valid argument ... but other times you end up with an invalid argument. So ... better to just not allow the application of inference rules to parts of a statement at all, and that is how inference rules are typically defined.

OK, so that gets us to your actual question: How do inference rules formalize the fact that they can only be applied to whole lines only?  Well, actually that is really easy.

Modus Ponens, for example, says that if you have any statement of the form $\varphi \to \psi$, and another of the form $\varphi$, then you can infer $\psi$.

With that, can we apply Modus Ponens to go from $A \to (B \to C)$ and $B$ to $A \to C$? No, for the very simple reason that $A \to (B \to C)$ is not of the form $B \to ...$

That is, when the inference rule says "If you have a statement of the form ....", then it is already implied that that statement is a line in your proof, i.e. a whole statement. So with that, inference rules can by definition only be applied to whole statements.  And there is no further formalization needed for how the rule should not be applied.


问:
Limits of the digits of Squared Numbers
Hey I don’t know what to call the post I think the name doesn’t reflect the question very well (considering limits are a separate thing in maths), but the question, I think, is very intriguing.

When a 1-digit number is squared the number of digits that could be in the square is 1 or 2. In other words the first odd number and first even number. (Yes, I am aware 0 is an even number as well but I am going to ignore that fact because a number can’t have 0 digits. If you want to you can say “The number of digits in the square of a 1-digit number is the first odd number or the second even number”)

And this is true in general, the number of digits in square of a number of n digits can have either the nth odd number of digits or nth even number of digits.

Now here comes the interesting part in the domain of 1 digit number, 3 is the last number which gets me an odd number of digits when squared. 31 is the last number which gives an odd number of digits, when considering only two digit numbers and 316 when squared gives me an odd number of digits and it is the last number in the three digit realm that does this. And I found that this pattern continues for at least seven digits. The pattern here is - 3, 31, 316, 3162, 31622, etc...
The numbers that come onwards are just the previous number with a digit attached to the back of it.

Why is this happening?

答:
This is happening

because the largest number which gives an odd number ($2n+1$) of digits when squared

is the largest number $m$ such that $m^2<10^{2n+1}$,

which is the largest number $m$ such that $m<10^n\sqrt{10}$.

So $m$ has the first $n$ digits of $\sqrt{10}=3.1622...$

Title: Limits of the digits of Squared Numbers


问:
Using small angle approximation to turn x(t) and y(t) to the usual projectile kinematic equation
Given the horizontal and vertical position-time functions:
\begin{align*}
        x(t) &= \frac{v_0\cos\ \theta\sin\ \omega t}{\omega}\\
        y(t) &= \dfrac{v_0\sin\ \theta\sin(\omega t)}{\omega} + \frac{mg}{k} \cos(\omega t) - \frac{mg}{k}
\end{align*}

Show that both functions reduce to usual projectile kinematic equations for very small $\omega$.

答:
Use $$\sin x\approx x\\\cos x\approx 1-\frac12x^2$$
So $$\frac{\sin(\omega t)}{\omega}\approx t$$and $$\cos(\omega t)-1\approx-\frac12(\omega t)^2$$


问:
Solving the domain and range of a region satisfying two inequalities?
The question I was provided was:

"Find the domain and range of the region satisfied by the following inequalities:
i) $y \ge (x-1)^2$
ii)$y \le2x+1$

Any help would be greatly appreciated. Would you recommend graphing or solving algebraically?

答:
I explain more detail:

first ,let $y=2x+1,y=(x-1)^2$, you will have two points$(0,1)$ and $(4,9)$

$ y \le 2x+1$ means all points below the red line, for example, $(0,0),(1,0),(1,1)$

$y \ge (x-1)^2$ means all points within the Parabola，for example $(1,1)(2,2)$

so two curves will have a closed area. the domain and range are: $ [0,4]$ and $[0,9]$


答:
You've written your formulas in a way that is confusing you.  You wrote  $$\log_{1/2}(4)=\frac{1}{2}^y=4$$ but what you actually meant was $$\log_{1/2}(4)=y \quad\text{ because }\quad \frac{1}{2}^y=4.$$

Then for the other version you have:

$$-\log_2(4)=-2^y=4$$

but it should be:

$$-\log_2(4)=y\quad\text{ because }\quad ???$$

Let's move that minus sign:

$$\log_2(4)=-y\quad\text{ because }\quad???$$

Okay, now it looks like the first one, and we get:

$$\log_2(4)=-y\quad\text{ because }\quad2^{-y} = 4.$$

And $2^{-y} = 2^{-1\cdot y} = \left(2^{-1}\right)^y = \left(\frac12\right)^y$.

I'm not sure that gives any intuition, but at least it shows where your mistake was.

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?

Body:
IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.

I tried to prove this to myself on paper.

I can write both equations in exponential form. Pretend $x=4$:

$$\log_{1/2}(4)=\frac{1}{2}^y=4$$

Then write the same for the other version:

$$-\log_2(4)=-2^y=4$$

If I know these are equivalent then the two exponent versions should be the same no?

$$4=\frac{1}{2}^y=-2^y$$

It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?


问:
Proof help - Calculus - infimum of $f(x)$
I need some help proving the following question:

Let $a < b$ and let $f : [a, b] \to \mathbb R$ be an increasing function.

Prove that there exists an $s\in \mathbb R$ such that $s = \inf \{f (x) ~| ~a < x ≤ b\}$.

So, I would like to say that due to the definition of increasing function, for every $x_{1} < x_{2}$, we get $f(x_{1}) < f(x_{2})$.

From here I am a little confused, because it seems so obvious that there exists an infimum for this set.

Every hint will be so appreciated!

Thanks a lot!

答:
The number $f(a)$ is a lower bound of the set $\{f(x)\,|\,a<x\leq b\}$ because $f$ is increasing.

Therefore the set has a real infimum as a consequence of the completeness of the real numbers.

Note that $f$ is not assumed to be continuous, so the infimum is not necessarily $f(a)$.


答:
This comes from the following logarithmic identities:

$$\log(\frac{x}{y}) = \log(x)-\log(y)$$
$$\log_b(a)=\frac{\log_x(a)}{\log_x(b)}$$

where $x$ is any positive number (except 1).

Then you can calculate

$$\log_{1/b}(x) = \frac{\log_b(x)}{\log_b(1/b)} = \frac{\log_b(x)}{\log_b(1) - \log_b(b)}$$

Now remember that $\log_b(1)=0$ for any allowed base and $\log_b(b)=1$. So this simplifies to

$$\frac{\log_b(x)}{0 - 1} = -\log_b(x)$$

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?


问:
What are the values of $\alpha \in \mathbb{R}$ for which the series $\sum_1^{\infty} \frac{(-1)^n}{n^{\alpha}} $ converges?
What are the values of $\alpha \in \mathbb{R} $ for which the series $$ \sum_1^{\infty} \frac{(-1)^n}{n^{\alpha}} $$ converges?

I've used Liebnitz Test and proved that for all $\alpha \gt 0$ the series converges. Now for $\alpha \lt 0 \ $ I can sense that the sequence
$$\frac{(-1)^n}{n^{\alpha}}$$ becomes oscillatory but how do I formally prove that in this case the series diverges?

答: If $\alpha\leqslant0$, then$$\left|\frac{(-1)^n}{n^\alpha}\right|=n^{-\alpha}\geqslant1,$$since $-\alpha\geqslant0$. Therefore, you don't have $\lim_{n\to\infty}\frac{(-1)^n}{n^\alpha}=0$, and so your series diverges.


问:
Calculating integral of $x^4 \cos(x^{10})$ from $0$ to infinity by residue theorem
Calculating integral of $x^4 \cos(x^{10})$ from $0$ to infinity by residue theorem.

$$\int_{0}^{\infty }x^{4}\cos x^{10} \; dx$$

答:
Let $ R>1 $, define a function $ f:z\mapsto\exp{\left(-z^{2}\right)} $, and consider a contour $ \mathscr{C}_{R} $ defined as follows :

Applying the residue theorem to $ f $ over $ \mathscr{C}_{R} $, we get : $$ \oint_{\mathscr{C}_{R}}{f\left(z\right)\mathrm{d}z}=0 $$

Since : \begin{aligned}\oint_{\mathscr{C}_{R}}{f\left(z\right)\mathrm{d}z}&=\int_{\Gamma_{R}}{f\left(z\right)\mathrm{d}z}+\int_{\Delta_{R}}{f\left(z\right)\mathrm{d}z}+\int_{0}^{R}{f\left(x\right)\mathrm{d}x} \\ &=\mathrm{i} R\int_{0}^{\frac{\pi}{4}}{f\left(R\,\mathrm{e}^{\mathrm{i}\theta}\right)\mathrm{e}^{\mathrm{i}\theta}\mathrm{d}\theta}-\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\int_{0}^{R}{f\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}x\right)\mathrm{d}x}+\int_{0}^{R}{f\left(x\right)\mathrm{d}x}\\ \oint_{\mathscr{C}_{R}}{f\left(z\right)\mathrm{d}z}&=\mathrm{i} R\int_{0}^{\frac{\pi}{4}}{f\left(R\,\mathrm{e}^{\mathrm{i}\theta}\right)\mathrm{e}^{\mathrm{i}\theta}\mathrm{d}\theta}-\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\int_{0}^{R}{\mathrm{e}^{-\mathrm{i}x^{2}}\mathrm{d}x}+\int_{0}^{R}{\mathrm{e}^{-x^{2}}\mathrm{d}x}\end{aligned}

And : \begin{aligned} \left|\mathrm{i} R\int_{0}^{\frac{\pi}{4}}{f\left(R\,\mathrm{e}^{\mathrm{i}\theta}\right)\mathrm{e}^{\mathrm{i}\theta}\mathrm{d}\theta}\right|\leq R\int_{0}^{\frac{\pi}{4}}{\left|f\left(R\,\mathrm{e}^{\mathrm{i}\theta}\right)\right|\mathrm{d}\theta}&=R\int_{0}^{\frac{\pi}{4}}{\mathrm{e}^{-R^{2}\cos{\left(2\theta\right)}}\,\mathrm{d}\theta} \\ &\leq R\int_{0}^{\frac{\pi}{4}}{\frac{\mathrm{d}\theta}{1+R^{2}\cos{\left(2\theta\right)}}}=\frac{R}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}\theta}{1+R^{2}\cos{\theta}}}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{R\arctan{\left(\frac{R^{2}-1}{\sqrt{R^{4}-1}}\right)}}{\sqrt{R^{4}-1}}\underset{R\to +\infty}{\longrightarrow}0\end{aligned}

Taking $ R $ to $ +\infty $, we get : $$ \int_{0}^{+\infty}{\mathrm{e}^{-\mathrm{i}x^{2}}\,\mathrm{d}x}=\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}\int_{0}^{+\infty}{\mathrm{e}^{-x^{2}}\,\mathrm{d}x} $$

Which means : $$ \int_{0}^{+\infty}{\cos{\left(x^{2}\right)}\,\mathrm{d}x}=\frac{\sqrt{\pi}}{2\sqrt{2}} $$

Substituting $ \left\lbrace\begin{matrix}x=y^{5}\ \ \ \ \\ \mathrm{d}x=5y^{4}\,\mathrm{d}y\end{matrix}\right.$, we get : $$ \int_{0}^{+\infty}{y^{4}\cos{\left(y^{10}\right)}\,\mathrm{d}y}=\frac{\sqrt{\pi}}{10\sqrt{2}} $$


问:
Prove that a given block matrix is symmetric indefinite

Let matrix $A\in\mathbb{R}^{n\times n}$ be symmetric positive definite and let fat matrix $B\in\mathbb{R}^{m\times n}$, where $m<n$, have full row rank. Let \begin{equation}\mathcal{C}=\begin{bmatrix}A&B^T\\B&0\end{bmatrix}.\end{equation} Prove that block matrix $\mathcal{C}$ is symmetric indefinite.

I have been able to show the symmetry of the given matrix as follows:

When taking a tranpose of a block matrix we must also tranpose each individual block, but note that $A$ is SPD, hence $A^T=A$;

\begin{equation}\mathcal{C}^T=\begin{bmatrix}A^T&B^T\\(B^T)^T&0\end{bmatrix}=\begin{bmatrix}A&B^T\\B&0\end{bmatrix}\end{equation}

and hence $\mathcal{C}=\mathcal{C}^T$ and $\mathcal{A}$ is symmetric.

However, I do not know how to show the indefiniteness of this matrix.

答:
$$ \begin{bmatrix} x^T & y^T \end{bmatrix}\begin{bmatrix}A&B^T\\B&0\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = x^TAx + x^TB^Ty +y^TBx = x^TAx + 2y^TBx.
$$

We know that when $y = 0$ and $x \neq 0$, we have $x^TAx > 0$. So all we need to do is argue that this expression can also be negative.

Note that $B$ has full row rank means that $y^TB = 0$ if and only if $y = 0$.

So pick an $x$ not in the kernel of $B$ and then look at the function
$$f(y) = y^TBx.$$
Argue that you can make this sufficiently negative to make the whole expression at the top negative.


问:
Projection of linear combination of real numbers
Suppose a real number $\alpha$ is a rational linear combination of two linearly independent real numbers, such as $\pi$ and $\sqrt 3$. For instance, say $$\alpha = \tfrac12\pi + 5\sqrt 3.$$

By a simple linear algebra argument ($\{\pi,\sqrt3\}$ is a $\mathbb Q$-basis), it's clear that the coefficients of $\alpha$ are uniquely determined.

Question: Given $\alpha\in\operatorname{span}_\mathbb Q(\{\pi,\sqrt 3\})$, how do we determine the coefficients of $\pi$ and $\sqrt 3$?

It feels like there is enough information, but at the same time, I imagine that we can find different coefficients which give us numbers arbitrarily close to $\alpha$, so it is important that we have an exact expression for $\alpha$.

(In other words, if we only know that $\alpha\approx 10.23105$, say, I imagine it would be hopeless to expect that we can determine the coefficients, even if we restrict our view to the module $\operatorname{span}_{\mathbb Z}(\{\pi,\sqrt 3\})$.)

The reason I ask this question is because for each $n$, the integral $$I(n)=\int_0^{\pi/3}\sin^{2n}x\,dx$$ is a number of the form $a\pi+b\sqrt 3$. It's easy to obtain the recurrence relation
$$I(n) = \begin{cases} \hfil\frac\pi3\hfil & \text{if $n=0$}\\[3pt]
(1-\tfrac1{2n})\,I(n-1)-\frac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1} & \text{otherwise}
\end{cases}$$
for $I(n)$, but I want to be able to compute the coefficients of $\pi$ and $\sqrt 3$ separately (in exact form) using a computer program.

I appreciate any help with solving either of the two problems, I'm assuming a solution to one necessarily sheds light on the other, that's why I haven't posted these as two separate questions.

Body:
I will answer your specific question about $I_n$.

Let $J_n=\dfrac{2^{2n}(n!)^2}{(2n)!}I_n$. We have  $J_0=I_0=\dfrac{\pi}{3}$ and

$J_n=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n-1}{2n}I_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n-1}{2n}\dfrac{(2n-2)!}{2^{2n-2}((n-1)!)^2}J_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n}{2n}\dfrac{2n-1}{2n}\dfrac{(2n-2)!}{2^{2n-2}((n-1)!)^2}J_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=J_{n-1}-\dfrac{2^{2n}(n!)^2}{(2n)!}\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})$,

thus

$J_n=J_{n-1}-\dfrac{1}{2n}\dfrac{3^{n}(n!)^2}{(2n)!}\dfrac{\sqrt{3}}{3}=J_{n-1}-\dfrac{3^{n-1}}{2n}\dfrac{(n!)^2}{(2n)!}\sqrt{3}$.

Summing relations yields $J_n=\dfrac{\pi}{3}-(\displaystyle\sum_{k=1}^n\dfrac{3^{k-1}}{2k}\dfrac{(k!)^2}{(2k)!})\sqrt{3}$.

Consequently, $I_n=\dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{\pi}{3}-\dfrac{(2n)!}{2^{2n}(n!)^2}(\displaystyle\sum_{k=1}^n\dfrac{3^{k-1}}{2k}\dfrac{(k!)^2}{(2k)!})\sqrt{3}$.


问:
Can someone help me with this limit?
$ \lim_{x \to 0}x^{|\tan x |} \overset{??}{=} 1 ??$

If you consult the graph it is intuitive to think that when $ x \to 0 $ the function tends to $1$. I have tried to lower the exponent with logarithms and apply L'Hospital's rule, but nothing. The only solution I see is to apply infinitesimal equivalents, which I have not tried yet.

Thank you all!

答:
Hint:

Determine the limit of the logarithm, using that $\tan x\sim_0 x$.

Title: Can someone help me with this limit?


问:
$p^{3}+m^{2}$ is square of a number.
Well i thought it is a nice problem so i will post it here.

1) Prove that for every natural numbers $m$, There is at most two primes $p$ where $p^{3}+m^{2}$ is a perfect square.

2) Find all natural numbers $m$ so that $p^{3}+m^{2}$ is the square of a number for exactly $2$ prime numbers $p$.

答:
I have my own answer so tell me if you had a better answer :)

Suppose that $p^{3}+m^{2}=b^{2}$ for some natural numbers $b$. So, $(b-m)(b+m)=p^{3}$ We know that the number $p^{3}$ can only be written as $(1)(p^{3}),(p)(p^{2}),(p^{2})(p),(p^{3})(1)$. Knowing $b+m>b-m$ makes the only answers of $(b-m,b+m)$ be $(1,p^{3}),(p,p^{2})$.

Case 1: $b-m=1$ and $b+m=p^{3}$

This results in $p^{3}-1=2m$ and so $p=\sqrt[3]{2m+1}$

Case 2: $b-m=p$ and  $b+m=p^2$
This one results in $p^2+p=2m$ and so $p=\frac{1+\sqrt{8m+1}}{2}$ (since the negative one is unacceptable.)

This itself solves the first part since for every $m$, $p$ can only be first one, second one, both, or non.
Now we have to find all numbers $m$ so that both $\frac{1+\sqrt{8m+1}}{2}$ and $\sqrt[3]{2m+1}$ are prime. We first find all values of $m$ which $8m+1$ is a square. Suppose $8m+1=n^2$ for some numbers $n$. Then, $(4m+1)^2=n^2+(4m)^2$. Using pythagorean theorem, we find that all possible values of $m$ are in the form of $\frac{x(x+1)}{2}$. Putting $m$ in the first case results in $2*\frac{x(x+1)}{2} +1=p^{3}$ and so $x^2+x+1=p^3$. Now using the solutions which are given in  here, the only possible values of $(x,p)$ becomes $(18,7)$, and so $m=\frac{x(x+1)}{2}=\frac{18*19}{2}=171$.Checking this answer in both cases, we obtain $p=19$ and $p=17$. So the only possible $m$ becomes $171$.

Title: $p^{3}+m^{2}$ is square of a number.