问:
What are some of the Websites/web-tools which must be a part of every mathematician's arsenal?
What are some of the web-tools which must be a part of every mathematician's arsenal? Tools which help mathematicians research, solve problems or even collaborate together!

Here is a list I gathered. Many among you may already know these, But still, in case.

MathQuill--
This is a great website for easy typing in LaTeX.

Integral calculator and the Derivative calculator-- Especially useful when one wants to crack a tough indefinite question. Steps included for those who need them!

OEIS--A list of all mathematically interesting sequences that contains over 3 million of such. Many things are discovered thanks to OEIS.

Desmos--One of the best online graphing tool with loads of computing power.

WolframAlpha and Mathematica-- Who doesn't know them! A unique engine for computing answers and providing knowledge.

MATLAB--Math. Graphics. Programming.
Whether you’re analyzing data, developing algorithms, or creating models, MATLAB is designed for way you think and the work you do.

Software such as Maple, MATLAB, Mathcad, Pari/GP, R, and the WolframAlpha website, etc are famous, and this question is not asking for similar software alternatives.

I haven't added math stack exchange to this list  for obvious reasons :~)

Thanks in advance to all for the over-enthusiastic help :-)

You might want to check out
Online tools for doing symbolic mathematics

PS-No intention of advertisement or promotion of any website or software.

答: You probably want most answers to just give one suggestion, but I'll mention a pair that serve complementary aims. To convert LaTeX to an image, use codecogs; to find out the LaTeX command for something you draw by hand, use detexify (but you have to draw quickly).</p>


问:
Find updated X, Y, Width, and Height
This is based off a previous question that wasn't fully answered. I do not have enough points to reply to the question, and the thread is pretty old.

I have an image with $X, Y$, Width, and Height. The previous answer said to "Multiplying the small $x$-value ($10$) with the ratio between the two heights".

Did he really mean to multiply by the width ratio? If not, does that mean to get $y$-value you multiply by the ratio between the two widths? Does the same apply for width and height? Also does it matter if the desired dimensions are larger or smaller?

答:
In the example both scaling factors in $x$ and $y$ direction (width and height) are equal to $2$, so it doesn't matter.

In general, when width is scaled by $\alpha$ and height by $\beta$, you multiply all $x$-coordinates by $\alpha$ and all $y$-coordinates by $\beta$. (Assuming the $x$-axis is horizontal and the $y$-axis is vertical.)

ContentLicense: CC BY-SA 4.0

Title: Find updated X, Y, Width, and Height


问:
Limit of trigonometric function (cotangent)
I need to solve the following problem:
$\lim_{x\to 3}(x-3) \cot{\pi x}$.
Can anyone give me a hint? I have no idea.

答:
$$\lim_{x\to 3}(x-3) \cot{\pi x}=\lim_{x\to 3} \frac{(x-3)\cos{\pi x}}{\sin{\pi x}}=\frac{0}{0}.$$.

Now, you can use  L'Hopital rule, just one and then you will get your answer.
I hope that helps


问:
Question from Sasane, How and Why of One Variable Calculus
I've been reading through Amol Sasane "The How and Why of One Variable Calculus" and I've got a question from one of the proofs about the sum property of integrals. (Page 199)

I know this doesn't have all the details. We've chosen an arbitrary positive epsilon and found partitions to make this statement of the upper sums of two functions work.

So we have :
$$\overline{S}(f + g) - \epsilon < \overline{S}(f) + \overline{S}(g)$$
Which I can follow when doing the proof myself. Though Sasane's next statement confuses me a little, at least with the technical details. When looking at this proof on my own a thought I had was "Can I take the limit as $\epsilon \to 0$?". I thought about it and it seemed like it might a way of doing this. Though I can't see how it would work rigorously or with change to a non-strict inequality.

Would anyone be able to shed some like on the details or find me material explaining this property?

Edit: If anyone would be so inclined, is this something really obvious and I just didn't see it? Reading through this book I feel like many students starting off would need more explanation than what was given.

答:
Assume the contrary that $\bar{S}(f+g)>\bar{S}(f)+\bar{S}(g)$. Then define $\epsilon = \bar{S}(f+g)-\bar{S}(f)-\bar{S}(g)>0$. It has been shown that $$\bar{S}(f+g)-\epsilon<\bar{S}(f)+\bar{S}(g)$$ since $\epsilon>0$ is positive. However this gives us the contradiction
$$\bar{S}(f+g)-\epsilon<\bar{S}(f)+\bar{S}(g)=\bar{S}(f+g)-\epsilon$$
since no number is strictly less than itself.


答: Starting with the statement $$\overline{S}(f + g) - \varepsilon < \overline{S}(f) + \overline{S}(g)\tag{1}$$ where $\varepsilon > 0$ is arbitrary.  Suppose now that $$\overline{S}(f+g) > \overline{S}(f) + \overline{S}(g)\tag{2}.$$  In this case, set $$\overline{S}(f+g) - \overline{S}(f) - \overline{S}(g) = m > 0$$ and let $\varepsilon' = \frac{m}{2} >0$.  Then we have $$\overline{S}(f+g) -\varepsilon' -\overline{S}(f) - \overline{S}(g) = m - \varepsilon' = m - \frac{m}{2} = \frac{m}{2} > 0.$$  This, however, means that $$\overline{S}(f+g) - \varepsilon' > \overline{S}(f) + \overline{S}(g),$$ which contradicts $(1)$.  Thus, our assumption $(2)$ was incorrect, and we have by contradiction that $$\overline{S}(f+g) \leq \overline{S}(f) + \overline{S}(g).$$

Title: Question from Sasane, How and Why of One Variable Calculus

Body:
I've been reading through Amol Sasane "The How and Why of One Variable Calculus" and I've got a question from one of the proofs about the sum property of integrals. (Page 199)

I know this doesn't have all the details. We've chosen an arbitrary positive epsilon and found partitions to make this statement of the upper sums of two functions work.

So we have :
$$\overline{S}(f + g) - \epsilon < \overline{S}(f) + \overline{S}(g)$$
Which I can follow when doing the proof myself. Though Sasane's next statement confuses me a little, at least with the technical details. When looking at this proof on my own a thought I had was "Can I take the limit as $\epsilon \to 0$?". I thought about it and it seemed like it might a way of doing this. Though I can't see how it would work rigorously or with change to a non-strict inequality.

Would anyone be able to shed some like on the details or find me material explaining this property?

Edit: If anyone would be so inclined, is this something really obvious and I just didn't see it? Reading through this book I feel like many students starting off would need more explanation than what was given.


答:
Quiver lets you draw commutative diagrams easily and export to $\LaTeX$, and MathPix Snip lets you convert screenshots of equations to $\LaTeX$. Both very useful.

Also, libgen and scihub are indispensable with the modern publishing system...

Title: What are some of the Websites/web-tools which must be a part of every mathematician's arsenal?

Body:
What are some of the web-tools which must be a part of every mathematician's arsenal? Tools which help mathematicians research, solve problems or even collaborate together!

Here is a list I gathered. Many among you may already know these, But still, in case.

MathQuill--
This is a great website for easy typing in LaTeX.

Integral calculator and the Derivative calculator-- Especially useful when one wants to crack a tough indefinite question. Steps included for those who need them!

OEIS--A list of all mathematically interesting sequences that contains over 3 million of such. Many things are discovered thanks to OEIS.

Desmos--One of the best online graphing tool with loads of computing power.

WolframAlpha and Mathematica-- Who doesn't know them! A unique engine for computing answers and providing knowledge.

MATLAB--Math. Graphics. Programming.
Whether you’re analyzing data, developing algorithms, or creating models, MATLAB is designed for way you think and the work you do.

Software such as Maple, MATLAB, Mathcad, Pari/GP, R, and the WolframAlpha website, etc are famous, and this question is not asking for similar software alternatives.

I haven't added math stack exchange to this list  for obvious reasons :~)

Thanks in advance to all for the over-enthusiastic help :-)

You might want to check out
Online tools for doing symbolic mathematics

PS-No intention of advertisement or promotion of any website or software.


答:
Overleaf is a good online LaTeX editor, and it particularly nice for collaboration.

CalcPlot3D is a nice tool for 3D graphing, as is Geogebra, though Geogebra can be a bit more finicky.

TeXRendr is a simple real-time online TeX renderer.  It's not meant for anything too advanced, and can't be saved, but it's nice if you just want to check that something will display correctly before compiling.


问:
Vector Line Integral Question
I need to compute the line integral for the vector $\vec{F} = \langle x^2,xy\rangle$, for the curve specified: part of circle $x^2+y^2=9$ with $x \le0,y \ge 0$,oriented clockwise.

Once again, I'm stuck at the setup (this happens a lot with me).  I know that I need to parameterize F, but how would I go about doing this exactly?

答:
Another method is to use Green's theorem:

$$\oint_C \vec{F}\cdot d\vec{r}=\iint_D|\nabla\times \vec{F}|dA$$

Where D is the area within the loop.

$$|\nabla\times \vec{F}|=y - 2x$$

Convert to cylindrical coordinates:

$$y-2x = r\sin{\theta}-2r\cos{\theta}$$

And evaluate the following double integral:

$$\int_\pi^{\frac{\pi}{2}}\int_0^3r(\sin{\theta}-2\cos{\theta})r\space dr \space d\theta$$

ContentLicense: CC BY-SA 3.0

Title: Vector Line Integral Question


问:
Is there a special name given if a number is on the edge of an interval?
For instance, if I have the interval $[0,1]$, is it proper to say that $0$ is on the lower bound of the interval?

I'm looking for the most elegant way to express this.

答:
One relevant term here is "boundary point." Roughly speaking, $x$ is a boundary point of $Y$ (I'm working in $\mathbb{R}$ for simplicity) iff there are points outside of $Y$ "arbitrarily close" to $x$: for all $\epsilon>0$, there is some $z$ such that $z\not\in Y$ but $\vert x-z\vert<\epsilon$. For example, $x=0$ is a boundary point of $Y=[0,1]$ since given $\epsilon>0$ the point $-{\epsilon\over 2}$ is within $\epsilon$ of $x$ but is not in $Y$.

(There is a more general notion of "boundary point," but it's more technical.)

However, this is a somewhat slippery notion. For example, every real number is a boundary point of $\mathbb{Q}$, since there are irrationals arbitrarily close to every real number. Similarly, $0$ is also a boundary point of $(0,1)$ even though $0\not\in (0,1)$. For this reason, terms like "endpoint" may be preferred.

(Why not "minimum" or "maximum?" Well, those terms are great for individual intervals, but consider something a bit more complicated like $A=[0,1]\cup [2,3]$. The point $1$ is an endpoint of $A$, but is neither the minimum nor maximum of $A$.)


问:
Dimension of $(l^\infty)^*$ mod $l^1$.
We can consider $l^1$ as a subspace of $(l^\infty)^*$ because for every $x \in l^1$, there exists a bounded linear functional $T_x: l^\infty \to \mathbb{R}$ defined by $T_x(y) = \sum_n x_n y_n$. Furthermore $l^1$ is closed in $(l^\infty)^*$. I am interested in the dimension of the quotient space $(l^\infty)^*$ mod $l^1$ but I don't know where to even start. I solved a few problems about dimensions of quotient spaces for sequence spaces by considering linearly independent vectors, but here the notion of dual confuses me.

答:
Suppose that $\ell_\infty^*/\ell_1$ is finite dimensional, and let $\{x_1,...,x_n\}$ be a finite subset of $\ell_\infty^*$ such that the set
$$\big\{x_1+\ell_1,...,x_n+\ell_n\bigr\}$$
is basis for the quotient $\ell_\infty^*/\ell_1$. Now, let $(e_n)$ be the standard Schauder basis of $\ell_1$. I.e. $e_n(k)=1$ when $k=n$ and $e_n(k)=0$ when $k\neq n$. Being a Schauder basis means that for every $x=(x(k))_{k=1}^{\infty}\in \ell_1$ we have that
$$\biggl|\biggl|x-\sum_{n=1}^{k}x(n)e_n\biggr|\biggr|_1\to 0$$
as $k\to \infty$. In other words $\overline{Y}=\ell_1$ where $Y=span(e_m:\,m\in \mathbb{N})$ (This shows that $\ell_1$ is separable). Now, its easy to check that
$$\tag{*}\ell_\infty^* = \overline{Y}\oplus Z$$
with $Z=span(x_k:\,1\leq k\leq n)$. Now, since both $\overline{Y}$ and $Z$ are separable by $(*)$ it follows that $\ell_\infty^*$ is separable, hence $\ell_\infty$ must be separable which cant be true. So, the quotient $\ell_\infty^*/\ell_1$ has infinite dimension.

Title: Dimension of $(l^\infty)^*$ mod $l^1$.


问:
For what values of $\alpha$ and $\beta$ the derivative of the function is bounded?
For what values of $\alpha$ and $\beta$ the derivative of the function is bounded?

Let $f_{\alpha,\beta}(t)= \left\{ \begin{array}{lcc}
    t^\alpha \sin(t^\beta) &   si  &  0 < t \leq 1 \\
    \\ 0 &  si & t = 0
   \end{array}
   \right.$

Then $$|f'_{\alpha,\beta}(t)|=|\alpha t^{\alpha-1}\sin(t^\beta)+\beta t^{\alpha + \beta-1 }\cos(t^\beta)|\leq|\alpha t^{\alpha-1}\sin(t^\beta)| +|\beta t^{\alpha + \beta-1 }\cos(t^\beta)| $$

I know that $|\sin(t^\beta)| \leq 1 $ and $|\cos(t^\beta)| \leq 1$ but here i'm stuck. can someone help me?

答: To start you off: if $\alpha > 1$ and $\beta\ge 0$, both terms in $f'_{\alpha,\beta}(t)$ go to $0$ as $t \to 0$.  Now consider the other cases...


问:
What is "progressive" about progressively measurable processes?
So I have just started to learn about stochastic processes, and I got to learn about progressively measurable processes.

Definition
Let $(\Omega, \mathcal{F},P)$ be a probability space and $\{\mathcal{F_t}\}_{t\in[0,\infty)}$ be a filtration. A stochastic process $X:[0,\infty)\times \Omega \to \mathbb{R}$ is said to be $\mathcal{F}$-progressively measurable if, for any $T$, $X|_{[0,T]}$ is $\mathcal{B}([0,T])\otimes\mathcal{F}_T$-measurable.

So that is the definition, and I am failing to see how it is progressive. I would like to know why it was named in such a way.

答: The word "progressive" refers to the progression (in time) of measurability conditions that a progressively measurable process satisfies: for each $t\ge 0$, the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal B([0,t])\otimes\mathcal F_t$-measurable. A progressive process is a little better than just adapted. One consequence is that if $X$ is progressive then $t\mapsto \int_0^t X_s  ds$ is well defined and progressive (even predictable) provided $\int_0^t|X_s| ds<\infty$ for $t>0$, by Fubini's theorem.


问:
Sum of Quasiconvex Functions on Different Domains
Say $f(x)$ and $g(y)$ are both quasiconvex functions. Is the function $h(x,y) = f(x) + g(y)$ also quasiconvex? I know sums of quasiconvex functions on the same domain are not necessarily quasiconvex, but what is the case on different domains?

答: No, consider $f(x) = \sqrt(x)$ and $g(y) = \sqrt{y}$. The set $\{(x,y) : \sqrt{x} + \sqrt{y} \leq 1\}$ is not convex: $(1,0)$ and $(0,1)$ are in the set, but $(0.5, 0.5)$ is not.


答:
If the word is about a single/individual interval (on the real line),

I would just use the terms lower end and upper end to keep things simple.

I am pretty sure I've seen these being used.

So for $[0,1]$ the number zero is the lower end,

and the number one is the upper end of the interval.

Title: Is there a special name given if a number is on the edge of an interval?

Body:
For instance, if I have the interval $[0,1]$, is it proper to say that $0$ is on the lower bound of the interval?

I'm looking for the most elegant way to express this.


问:
Challenging Integrals for High School Students
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.

答:
Not quite an integral, but a differential equation. This one definitely requires some outside of the box thinking!

Consider the solution $y(x)$ of the differential equation
$$
\frac{dy}{dx}=\sqrt{1+y^{2020}}
$$
determined by the initial data $y(0)=0$.

Prove that for all $n \in \mathbb{N}$, $y^{(n)}(0)$ is an integer. Then find the first non-zero value of $y^{(n)}(0)$ with $n>1$.


问:
Graph theory connected graph
I am studying graph thoery and come up with a thought that,

if $G$ is acyclic graph with valency (order of a graph) of every vertex is at least two then $G$ is connected!

The graph is said to be disconnected if at least two vertices of the graph are not connected by a path, otherwise it's said to be connected.

Is this true! if it is then how can i prove that?

All i can imagine is infinite graph which is tree.

(Valency of vertex of graph is number of vertices adjacent to vertex and acyclic means graph without cycle).

答: A finite graph in which every vertex has degree $\geq2$ has a cycle, so the statement is vacuously true.  In the infinite, the statement is false, since a forest of two infinite trees is a counterexample.

Title: Graph theory connected graph


问:
Does inequality follow from Jensen's Inequality
As of now, I am covering some basic material on convex functions from the text Real Analysis by Royden. Further, I have come across Jensen's Inequality, which states,

Let $\varphi$ be a convex function on $(-\infty, \infty)$, $f$ an integrable function over $[0,1]$, and $\varphi \circ f$ also integrable over $[0,1]$. Then

$$\varphi\left(\int_0^1f(x)\,dx\right)\leq\int_0^1\left(\varphi \circ f\right)(x)\,dx.$$

My question is, using this, can it be shown that for an integrable function $f$ over $[0,1]$,

$$\sin\left(\int_0^1f(x)\,dx\right)\leq\int_0^1\sin(f(x))\,dx\ ?\tag{*}$$

I believe this could be argued, or proved rather, by defining $\varphi(x):=\sin(x)$. Moreover, since
$\varphi''(x) = -\sin(x)$, it follows that $\varphi$ is convex on $[\pi,2\pi]$.

Using properties of the sine function,

$$\int_0^1|\left(\varphi \circ f\right)(x)|\,dx = \int_0^1|\sin(f(x))|\,dx\leq\int_0^11\,dx<\infty.$$

That is, $\varphi \circ f$ is integrable over $[0,1]$. Therefore, the inequality $(*)$ follows directly from Jensen's Inequality.

答:
$\sin(x)$ is only convex for some portions of $(-\infty, \infty)$, not for all of it, so Jensen's inequality doesn't hold. For a simple counter-example, try $f(x) = x$. Then $(*)$ is $$\sin\left(\frac{1}{2}\right)\leq1-\cos(1)$$

which is not true.


问:
Prove that if $A$ is symmetric and has a LU-decomposition then $A=LDU' \Rightarrow U'=L^T$
Suppose the matriz $A$ has a LU-decomposition, in other words, suppose there exists matrices $L$ and $U$ such that $A=LU$ where $L$ is lower triangular and $U$ is upper triangular.

We can to prove that $A$ has a LDU'-decomposition, where $D$ is a diagonal (to do so, put $d_{ii}=u_{ii}$ and $u'_{kj}=u_{kj}/u_{kk}$).

Prove that if $A$ is symmetric then $A=LDU' \Rightarrow  U'=L^T$, where $L^T$ is the transpose of matrix $L$.

Yes, $A$ is invertible. Therfore, $LDU'=U'^TDL^T\Rightarrow U'=L^T$. Here is the solution (question 9).

答:
First we need to show that the LU-decomposition of $A$ is unique.

Assume there are two LU-decompositions of $A$.
$LU = \tilde{L}\tilde{U} \Leftrightarrow L\tilde{L}^{-1} = \tilde{U}U^{-1}$. $L$ and $\tilde{L}$ are lower left triangular matrices (with $l_{i,i} = 1$) and $U$ and $\tilde{U}$ are upper right triangular matrices. The inverse matrices of triangular matrices are again triangular matrices and the product of two (lower left/upper right) triangular matrices is again a (lower left/upper right) triangular matrix (You can proof this by calculating the inverse/product). But the only matrix which is a lower left and an upper right triangular matrix is a diagonal matrix. Now $l_{i,i} = 1 \Rightarrow L\tilde{L}^{-1} = I$. Hence $L = \tilde{L}$ and $U = \tilde{U}$. This means the decomposition is unique.

Let $A$ be symmetric and $A = LDU'$ be a LU-decomposition of $A$ where $D$ is a diagonal matrix and $u'_{i,i} = 1$.

$A = LDU' = (LDU')^T = U'^TDL^T$. $U^T$ is a lower left triangular  matrix and $L^T$ is a upper right triangular matrix. This means $(LDU')^T$ is another LU-decomposition of $A$. But the decomposition is unique which implies $L = U'^T$ and $L^T = U'$. Hence $A = LDL^T$.


答:
By contradiction use,
$$f(x)=x^2$$
Then you would get that
$$\sin\left(\int_0^1f(x)dx\right)\approx0.33\nleq0.31\approx\int_0^1\sin(f(x))dx$$
In addition, $\sin$ is not convex on $[0,1]$, it is concave, hence the opposite equality would hold.

Title: Does inequality follow from Jensen's Inequality

Body:
As of now, I am covering some basic material on convex functions from the text Real Analysis by Royden. Further, I have come across Jensen's Inequality, which states,

Let $\varphi$ be a convex function on $(-\infty, \infty)$, $f$ an integrable function over $[0,1]$, and $\varphi \circ f$ also integrable over $[0,1]$. Then

$$\varphi\left(\int_0^1f(x)\,dx\right)\leq\int_0^1\left(\varphi \circ f\right)(x)\,dx.$$

My question is, using this, can it be shown that for an integrable function $f$ over $[0,1]$,

$$\sin\left(\int_0^1f(x)\,dx\right)\leq\int_0^1\sin(f(x))\,dx\ ?\tag{*}$$

I believe this could be argued, or proved rather, by defining $\varphi(x):=\sin(x)$. Moreover, since
$\varphi''(x) = -\sin(x)$, it follows that $\varphi$ is convex on $[\pi,2\pi]$.

Using properties of the sine function,

$$\int_0^1|\left(\varphi \circ f\right)(x)|\,dx = \int_0^1|\sin(f(x))|\,dx\leq\int_0^11\,dx<\infty.$$

That is, $\varphi \circ f$ is integrable over $[0,1]$. Therefore, the inequality $(*)$ follows directly from Jensen's Inequality.


问:
Find the probability that I get exactly $n$ $1$s if I throw an $m$ sided die $k$ times.
Let's say I have an $m$-sided fair die (the sides will have values 1,2,3,.. m) and throw it $k$ times, I want to find the probability to get exactly $n$ $1$'s.

I view the problem as a tree, I throw the die once and the edges lead to every possible outcome with $\frac{1}{m}$ probability and for each outcome I throw the die again and so on.

Define $A$ as the number of ways to get a $1$ the first $n$  times and after that I get no $1$s, which by the counting principle is $$1 \times 1 \times ... (n \ times) \times (m-1) \times (m-1) \times ...$$

Multiply $A$ by $k!$ and divide by $n!$ to get all the possible ways I could get exactly $n$ $1$s,
so the probability of getting exactly $n$ $1$s is $$P=\frac{Ak!}{m^k n!}$$

My question is, is this correct?
If yes is there a better way to write this formula, perhaps by using the notation $\begin{pmatrix} x \\ y \end{pmatrix}$, factorials etc., if no what is the correct answer?

Also what if the die wasn't fair and I had a different probability for every outcome, what would the formula look like then?

答:
You need to  divide by $(k-n)!$ and define $A$ as $$A=\frac{(m-1)^{k-n}}{(k-n)!}$$
So your formula becomes $$P=\frac{(m-1)^{k-n}k!}{(k-n)!m^kn!}$$
Which can be written as $$P=\begin{pmatrix} k \\ n \end{pmatrix} \left( \frac{1}{m} \right)^n \left(1 -\frac{1}{m} \right)^{k-n}$$
Which is a special case of the binomial distribution $$P=\begin{pmatrix} k \\ n \end{pmatrix} p^n \left(1 -p \right)^{k-n}$$ for $p=\frac{1}{m}$.


问:
Why can you transition from a Riemann Sum to an Intergal?
I understand that the integral is defined through a Riemann Sum, as in
$$\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}f\left(a+i\cdot\frac{b-a}{n}\right)\frac{b-a}{n}$$
But I don't understand how this definition is made. I understand the structure of the Riemann Sum which is summing up infinetely many infinietely thin rectangles to approximate the area under a curve, but I don't understand why this turns into an integral and then can be evaluated the way an integral is. Any help is appreciated.

答:
The expression $\int_a^b f(x)\,dx$ itself has no meaning before it is defined. The limit on the right gives a meaning to the expression on the left.

"... then can be evaluated the way an integral is". I assume that you are talking about the fundamental theorem of calculus:
$$
\int_a^b f(x)dx=F(b)-F(a)
$$
where $F$ is an antiderivative of $f$. That is proved (ultimately) using the definition of the expression "$\int_a^b f(x)\,dx$".


问:
Does it make sense to talk about the concatenation of infinite series?
Does a series of numbers defined as the concatenation of two or more infinite series, for example all the positive integers followed by all the negative integers, make mathematical sense?

I came up with this problem while writing an implementation of infinite series in Erlang. I'm not sure how to represent this in code in a way that isn't horrific, and wondered if it even made any sense. I'm a programmer, not a mathematician, so won't understand any answers too steeped in technical terms!

EDIT: Ok, After reading and being confused by a few answers I have discovered that the mathematical definition of concatenation appears to be a bit different to my understanding of it from programming.

When I say "the concatenation of two infinite series", what I actually mean is a new series that consists of all the elements of the first series followed by all the elements of the second series.

So for the example I gave above, the positive integers followed by the negative integers, you could start at the beginning of the series and get only positive numbers, never in fact reaching the second infinite series.

However, if you then applied some sort of 'filter' to this concatenated (in my definition) infinite series, that would somehow make the first part of it finite, then it would be possible to count up to elements of the second series.

I hope this makes more sense now, and if there is a more appropriate mathematical term I can use instead of concatenation, please let me know. Ta!

答:
The answer is, it depends ;-).

It certainly does make sense to talk about the concatenation of two infinite series. If you view an infinite series as a function $f \,:\, \mathbb{N} \to A$ which assigns to each number one element of $A$, then you can view the concatenation of two such series $f_1$, $f_2$ as function $g \,:\, \{1,2\}\times\mathbb{N} \to A$, where $g(1,n) = f_1(n)$ and $g(2,n) = f_2(n)$.

The resulting $g$, however, isn't an infinite series itself, since it doesn't map integers to elements of $A$, but rather pairs of integers.

If you want the result of such a concatenation to be an infinite series itself, you'll have to use a more general definition of what constitutes an infinite series. A way to do that is to use ordinal numbers (http://en.wikipedia.org/wiki/Ordinal_number). Using that, you could define an infinite series of length $\alpha$ to be a mapping $$
  s \,:\, \{\beta \,:\, \text{$\beta$ is an ordinal and }\beta < \alpha\} \to A \text{.}
$$

The original definition of an infinite series would then conincide with this definition if you pick the length $\omega$, i.e. the smallest ordinal larger than all finite ordinals. For $g$, i.e. two concatenation of two series with length $\omega$, you'd have to pick $\alpha = \omega + \omega$. In other words, those would have length $\omega + \omega$.

Ordinals are often constructed in a way that defines an ordinal $\alpha$ to simply be the (set-theoretic) union of all smaller ordinals. That allows you to simplify the definition of a series of length $\alpha$ to just $$
  s \,:\, \alpha \to A \text{.}
$$

You'll have to decide how far (in terms of ordinal sizes) you want to be able to take this. If you only allow finitely many concatenation steps of series of length $\omega$, then all the ordinals you'll encounter will be smaller than $\omega\cdot\omega$. If you allow $\omega$-many concatenations of series of length $\omega$, one such step reaches $\omega\cdot\omega = \omega^2$, the next one $\omega^3$ and so on. After $\omega$-many such steps you reach $\omega^\omega$. A natural limit may thus be the ordinal $\epsilon_0$ (http://en.wikipedia.org/wiki/%CE%95%E2%82%80), which cannot be expressed as a cantor normal form (http://en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form) of smaller ordinals. By limiting yourself to ordinals smaller than that, you're assures that any ordinals you encounter than be written as the sum of finitely many powers of $\omega$, where the exponents obey the same restriction. It should be possible to represent such ordinals quite efficiently in a program, though I've never tried. But something tells me there's a ton on literature on that...


问:
Equivalence and Presupposition
This is a very basic question about the relationship between presupposition and implication. The question is as follows: Let $A$, $B$ and $C$ be three propositions and let $A\Leftrightarrow B$, do we have ($\star$)? $$\textrm{$A$ presupposes $C$ iff $B$ presupposes $C$}\quad\quad\quad(\star)$$ In other words, if two propositions are logically equivalent, must they carry the same presuppositions?

答:
Yes. The definition of presupposition is

$$A \text{ presupposes } C \text{ iff } A \vDash C \text{ and } \neg A \vDash C \text{ and } \nvDash C$$

That is, $A$ presupposes $C$ iff $C$ follows both from $A$ and its negation but $C$ is not tautological.

If $A \Leftrightarrow B$, then $\neg A \Leftrightarrow \neg B$. So

$\begin{align*}
& A \text{ presupposes } C \\
\text{ iff } & A \vDash C \text{ and } \neg A \vDash C \\
\text{ iff } & B \vDash C \text{ and } \neg B \vDash C \\
\text{ iff } & B \text{ presupposes } C
\end{align*}$

So $\textrm{$A$ presupposes $C$ iff $B$ presupposes $C$}$.

Title: Equivalence and Presupposition


答:
Let $(x,y)$ denote the set $\{\alpha\in\Bbb{R}\mid x<\alpha<y\}$ (open interval)

Let $(x,y]$ denote the set $\{\alpha\in\Bbb{R}\mid x<\alpha\le y\}$ (half open)

Let $[x,y)$ denote the set $\{\alpha\in\Bbb{R}\mid x\le\alpha< y\}$ (half open)

Let $[x,y]$ denote the set $\{\alpha\in\Bbb{R}\mid x\le \alpha\le y\}$ (closed)

There are a few different terms used to refer to different extreme points along subsets of the real line:

A lower bound of an interval $I$ is any element $u$ such that for any $e\in\Bbb{R}$, if $e<u$ then $e\notin I$. For example, $-14$ is a lower bound of $(0,1)$. Note that a lower bound does not need to be close to the "lower end" of the interval.

An upper bound of an interval $I$ is any element $u$ such that for any $e\in\Bbb{R}$, if $e>u$ then $e\notin I$. For example, $96$ is an upper bound of $(0,1)$. Note that an upper bound does not need to be close to the "upper end" of the interval.

The minimum of an interval is its least element - i.e. $x$ is the minimum of $[x,y)$ and $[x,y]$. Note that sets which are open on the left do not have a minimum.

The infimum of an interval (open or closed) is its greatest lower bound - i.e. the infimum of all of the above intervals is $x$. Note that every minimum is an infimum, but the converse need not hold.

The maximum of an interval its greatest element - i.e. $y$ is the maximum of $(x,y]$ and $[x,y]$. Not that intervals which are open on the right do not have a maximum.

The supremum of an interval (open or closed) is its least upper bound - i.e. the supremum of all of the above intervals is $y$. Note that every maximum is an supremum, but the converse need not hold.

In the Euclidean topology on $\Bbb{R}$, every limit point of an interval is either an infimum or supremum. It should be noted, however, that limit points in general (even when working in the reals) do not need to behave this way. For a full discussion of limit points, I would suggest looking through any suitably thorough undergraduate textbook on topology.


问:
How does this simplification of this derivative work?
So there is a simplification I don't understand it goes like this:

\begin{alignat}{2}
&\text{Original equation:}\qquad 2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}&=0\\
&\text{Simplification:}\qquad\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right)&=0
\end{alignat}

The reason why I don't understand it:
if I try to differentiate the simplification I won't get the original equation.
When I differentiate it I get:
$$\frac{1}{2r}\left(2r\left(\frac{d\theta}{dt}\right)+r^2\left(\frac{d^2\theta}{dt}\right)\right)$$
I'm just using the product rule. Could someone explain to me what I'm doing wrong. It would be very much appreciated.

Thanks in advance.

答:
You applied the product rule incorrectly. You should have
$$
(uv)'=u'v+uv'
$$
where $u=r^2$ and $v=d\theta/dt$.

Note that
$$
u'=2r r'\quad v'=\frac{d^2\theta}{dt^2}
$$
where $\displaystyle(\cdot)'=\frac{d(\cdot)}{dt}$.

Notes.

\begin{align}
\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right)
&=\frac{1}{2r}(2r\frac{dr}{dt}\frac{d\theta}{dt}+r^2\frac{d^2\theta}{dt^2})\\
&=\frac{dr}{dt}\frac{d\theta}{dt}+\frac{r}{2}\frac{d^2\theta}{dt^2}
\end{align}


问:
Jacobian of a transformation on a Manifold
Let $a,b\geq{0}$. I am trying to find the Jacobian of the following transformation

\begin{equation}
\begin{aligned}
\mathbb{R}^3\times\mathbb{S}^2&\longmapsto\mathbb{R}^3\times\mathbb{R}^3\\
(v,w)&\longmapsto (p,q),
\end{aligned}
\end{equation}
where
\begin{equation}
\begin{aligned}
p&=\frac{v}{2}+\sqrt{av^2+b}\,\,w,\,\,\, and\\
q&=\frac{v}{2}-\sqrt{av^2+b}\,\,w\,.
\end{aligned}
\end{equation}
Well, the fact that $w\in\mathbb{S}^2$ yields to a constraint satisfied by $p$ and $q$, as
$$(p-q)^2-a(p+q)^2=b,$$
and thus the final and initial spaces have the same dimension.

I computed the Jacobian of the same transformation, but starting from $\mathbb{R}^3\times\mathbb{R}^3$, and it was $(av^2+b)^{\frac{3}{2}}$.

Could anyone please give me an idea on how to proceed? I'm very grateful.

答:
You cannot take the Jacobian without considering the constraint. Otherwise the Jacobian at a point of a manifold would be the Jacobian of the surrounding flat space.

I'm not sure what can be assumed from the question, but here is an idea of how to proceed:

Parametrize the sphere using your favorite mapping, say $w=(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$. Then $f$ becomes a function $\Omega\subset\mathbb{R}^5\to\mathbb{R}^6$. The five tangent vectors $f_v$, $f_\theta$, $f_\phi$ live in $\mathbb{R}^6$.

Form the first fundamental form $5\times5$ matrix $g_{ij}=f_i\cdot f_j$, where $i,j$ range over the five vectors. Finally, the 'Jacobian' is $\sqrt{\det[g]}$.

Using the given function, I calculate (without double checking) $\sqrt{\det[g]}=\sqrt{\frac{1}{2}(av^2+b)(4a^2v^2+av^2+b)}\cos\theta$.


答:
Careful, $r$ is also a function of $t$, so you should have:
$$\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right) = \frac{1}{2r}\left(2r\frac{dr}{dt}\frac{d\theta}{dt} + r^{2}\frac{d^{2}\theta}{dt^2}\right) = \frac{dr}{dt}\frac{d\theta}{dt} + \frac{r}{2}\frac{d^{2}\theta}{dt^2}.$$  This gives us $$\frac{dr}{dt}\frac{d\theta}{dt} + \frac{r}{2}\frac{d^{2}\theta}{dt^2}=0,$$ which upon multiplying by $2$ gives us $$2\frac{dr}{dt} \frac{d\theta}{dt} + r\frac{d^{2}\theta}{dt^{2}} = 0.$$

Title: How does this simplification of this derivative work?

Body:
So there is a simplification I don't understand it goes like this:

\begin{alignat}{2}
&\text{Original equation:}\qquad 2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}&=0\\
&\text{Simplification:}\qquad\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right)&=0
\end{alignat}

The reason why I don't understand it:
if I try to differentiate the simplification I won't get the original equation.
When I differentiate it I get:
$$\frac{1}{2r}\left(2r\left(\frac{d\theta}{dt}\right)+r^2\left(\frac{d^2\theta}{dt}\right)\right)$$
I'm just using the product rule. Could someone explain to me what I'm doing wrong. It would be very much appreciated.

Thanks in advance.


问:
Defining a function by a series
We let the function $f(x)$ be defined by the series  $\sum_{n=1}^\infty \frac{n4^n}{n^2+1}x^{2n}$ and we need to show that $$\lim_{x\to 0} \frac{f(x)-x\sin(2x)}{x^4} = \frac{32}{5} + \frac{4}{3} = \frac{116}{15}$$

I'm really confused as to how we are supposed to define the function by a serie, I mean I can go the other way around and try to represent a function as a serie, but I'm not quite sure about representing a series as a function. I would really appreciate some help in figuring this one out, thanks in advance!

答:
When you have a series like that and prove (or are told) that it converges for some values of $x$ then the value that it converges to is the value of the function the series defines. You may or may not be able to write that function in terms of other functions you know.

In this case you can use the series for $\sin$ in the expression inside the limit to write that limit as the limit of a simple power series. Then the limit will be the constant term.

Final grammatical point. In English, "series" is both singular and plural, so you write "define a function by a series", not "define a function by a serie".

Title: Defining a function by a series


问:
Fin the greatest and least value of $(4\sin θ-6\cos θ)^2-3$ where $0°<θ<90°$
This question I found the tan by doing 4^2/6^2 and put it in the form of 2tan of 13 sin(θ-56.3) and I know the maximum is sin inverse of 1 for maximum and sin inverse -1 for minimum but I know the answers are 49,-3 but I don’t know how to get to it any tips ang help much appreciate.</p>

答:
HINT for the maximum: $(4 \sin \theta - 6 \cos \theta)^2-3$ takes its maximum on the interval $(0,90^{\circ})$ at the value of $\theta$ where either:

the maximum of $4 \sin \theta - 6 \cos \theta$ is attained in $(0,90^{\circ})$.

the minimum of $4 \sin \theta - 6 \cos \theta$ is attained $(0,90^{\circ})$.

Do you see why this is?

In fact eye-balling this, $4 \sin \theta - 6 \cos \theta$  achieves its maximum and its minimum it will achieve at the interval $(0,90^{\circ})$ at the endpoints: $\theta \rightarrow 0$ or $\theta \rightarrow$ $90^{\circ}$. See what you can do with this.

HINT for the minimum: This function takes its global minimum value [at least on the real numbers that is] at $\theta$ for which $4 \sin \theta - 6 \cos \theta =0$. Why is this? Does $4 \sin \theta - 6 \cos \theta$ attain a value of 0 for some $\theta \in (0, 90^{\circ})$?


问:
Is there any Simplification to the sum
Can we simplify the following sum $$\sum_{k=0}^{10}{20\choose k}(1/4)^k (3/4)^{20-k}$$
MY TRY: I tried to reduce this expression using the fact that $20\choose k$ = $20\choose20-k$.

Going by the expansion of $(1/4+3/4)^{20}$, we get

$$(1/4+3/4)^{20}=1= \sum_{k=0}^{20}{{20\choose k}(1/4)^k (3/4)^{20-k}}$$  $$1=
 \Bigr(\sum_{k=0}^{9}{{20\choose k}\bigr((1/4)^k (3/4)^{20-k}\bigl)+\bigr((1/4)^{20-k} (3/4)^{k}\bigl)}\Bigr)+ {20\choose 10}(1/4)^{10} (3/4)^{10}$$

I am stuck over here. How to simplify this from here?

Any hint would be a great help!

答:
Consider that the terms of the binomial expansion
$$
\Pr (m\;\left| {\;n} \right.,p) = \left( \matrix{  n \cr  m \cr}  \right)p^{\,m} q^{\,n - m}
$$
reperesent the Binomial distribution
which is the probability of having $m$ successes in $n$ Bernoulli trials, with success probability $p$ and failure probability $q= 1-p$..

Therefore the partial sum represents the corresponding Cumulative distibution function
$$
Q(m\;\left| {\;n} \right.,p)
 = \sum\limits_{k = 0}^m {\left( \matrix{  n \cr   k \cr}  \right)p^{\,k} q^{\,n - k} }
$$
which is known to be expressible through the Regularized Incomplete Beta function
$$
\eqalign{
  & Q(m\;\left| {\;n} \right.,p)
 = \sum\limits_{k = 0}^m {\left( \matrix{  n \cr  k \cr}  \right)p^{\,k} q^{\,n - k} }
  = I_{\,q} (n - m,m + 1) =   \cr
  &  = \left( {n - m} \right)\left( \matrix{  n \cr   m \cr}  \right)
\int_0^q {t^{\,n - m - 1} \left( {1 - t} \right)^{\,m} dt}  \cr}
$$

So in your case we have
$$
\sum\limits_{k = 0}^{10} {\left( \matrix{  20 \cr  k \cr}  \right)p^{\,k} q^{\,20 - k} }
  = 10\left( \matrix{  20 \cr   10 \cr}  \right)\int_0^{3/4} {t^{\,9} \left( {1 - t} \right)^{\,10} dt}
$$
and the value can be derived from computing by a CAS the incomplete Beta to obtain
$$
\sum\limits_{k = 0}^{10} {\left( \matrix{  20 \cr   k \cr}  \right)p^{\,k} q^{\,20 - k} }
  \approx \left( \matrix{  20 \cr  10 \cr}  \right) \cdot 5.39 \cdot 10^{\, - 6}  \approx 0.996
$$

Title: Is there any Simplification to the sum


问:
Arc Length Clarification Question
So I was given the following prompt:

What is the length of the curve $y=1-\cos(x)$ from $x=0$ to $x=4\pi$?

I was able to set up an integral and find a numerical answer for this problem, but I just wanted some clarification as to whether or not this is correct. I set up the integral of: $$L=\int_0^{4\pi}\sqrt{1+[\sin(x)]^2}\,dx$$ and I found the answer of $15.281$ from this integral, any clarification about whether or not this is correct would be appreciated!

答:
Yes, you have applied the arc length formula correctly:
$$
\int_a^b\sqrt{1+(f'(x))^2}\,dx
$$

You then got an elliptic integral and you have the correct approximate value.

Title: Arc Length Clarification Question


问:
Find min of $IA + IB + IC +ID$ in tetrahedron $ABCD$
Let the point $I$ in tetrahedron $ABCD$. Find $\min\{IA + IB + IC + ID\}$.

I can't solve this problem, even in the case ABCD regular. Please help

答:
Summary

For regular tetrahedron, the answer isn't that hard to figure out because under very general assumptions, the point $I$ that minimize the sum of distances is unique.

A regular tetrahedron has many axis of rotation symmetry, all of them pass through the centroid. By the uniqueness of $I$, $I$ lies on the intersection of all these axis of rotation symmetry and hence must coincides with the centroid.

Similar arguments work for any tetrahedron which has more than one axis of rotation symmetry.

Existence and Uniqueness of I

Let $S = \{ \vec{x}_1, \vec{x}_2, \ldots, \vec{x}_m \}$ be any collection of $m \ge 3$ distinct points in $\mathbb{R}^n$ such that no three of them are collinear. Let $d_S: \mathbb{R}^n \to \mathbb{R}$ be the sum of distances to points in $S$:

$$d_S(\vec{p}) = \sum_{i=1}^{m} |\vec{x}_i - \vec{p}|$$

It is clear $d_S(\vec{p})$ is a continuous function in $\vec{p}$. In fact, it is $C^{\infty}$ over $\mathbb{R}^{n} \setminus S$ with gradient:

$$ \vec{\nabla} d_S(\vec{p}) =\; -\sum_{i=1}^{m} \hat{n}_i(\vec{p}) \;\stackrel{def}{=}\; -\sum_{i=1}^{m}\frac{\vec{x}_i-\vec{p}}{|\vec{x}_i-\vec{p}|}\tag{*1}$$

Notice $d_S(\vec{p}) \to \infty$ as $|\vec{p}| \to \infty$ and bounded below by $0$ over $\mathbb{R}^n$. $d_{S}$ achieves
its absolute minimum at some finite $\vec{p}_{min}$. What we want to show is this $\vec{p}_{min}$ is unique.

For any two $\vec{p}_1, \vec{p}_2 \in \mathbb{R}^n$ and $\vec{x}_i \in S$, it is easy to check:

$$\frac{|\vec{x}_i - \vec{p}_1| + |\vec{x}_i - \vec{p}_2|}{2} \ge \left| \vec{x}_i - \frac{\vec{p}_1+\vec{p}_2}{2}\right|$$
and the equality holds if and only if $\vec{p}_1$ and $\vec{p}_2$ lies on a ray start at $\vec{x}_i$. Since no three of $\vec{x}_i$ are collinear and $m \ge 2$, there is at least one $\vec{x}_i \in S$ which makes above inequality strict. This implies:

$$\frac{d_S(\vec{p}_1) + d_S(\vec{p}_2)}{2} > d_S(\frac{\vec{p}_1+\vec{p}_2}{2})\tag{*2}$$

i.e. $d_S$ is a strictly convex function. If $d_S(\vec{p}) = d_S(\vec{p}_{min})$ for any other $\vec{p}$, then $(*2)$ will leads to the contradiction that $d_S(\frac{\vec{p} + \vec{p}_{min}}{2}) < d_S(\vec{p}_{min})$.
As a result, the point $\vec{p}_{min}$ that minimize $d_{S}(\cdot)$ is unique.

Apply this to a non-degenerate tetrehedron $T = \left<ABCD\right>$, we get the uniqueness of $I$.

Where can $I$ be?

If $\vec{p}_{min} \notin S$, then being an absolute minimum, $\vec{p}_{min}$ satisfies:

$$\vec{\nabla} d_{S}(\vec{p}) = \vec{0}\tag{*3}$$

Conversely, if any $\vec{p} \notin S$ satisfies $(*3)$, then $\vec{p}$ cannot be a local maximum nor saddle point because $d_S$ is convex. Using the strict convexity of $d_S$, it is not hard to show $\vec{p}$ is actually the unique absolute minimum $\vec{p}_{min}$.

In short, $\vec{p}_{min}$ either belongs to $S$ or a point $\notin S$ which satsifies $(*3)$.

Apply this to non-degenerate tetrahedron $T = \left<ABCD\right>$ again. It is easy to
see

$$\vec{\nabla} d_{S}(\vec{p}) \neq \vec{0}\quad\text{ for }\quad \vec{p} \in
\mathbb{R}^3 \setminus ( \operatorname{int}(T) \cup S )$$
because the unit vectors $\vec{n}_i(\vec{p})$ in $(*1)$ falls into some half space and can never add up to $\vec{0}$.

This means for the non-degenerate tetrahedron $T$, $I$ is either:

one of the vertex $A, B, C, D$

or inside the interior of $T$ and satisfies $\vec{n}_A(I) + \vec{n}_B(I) + \vec{n}_C(I) + \vec{n}_D(I) = \vec{0}$.

For the second case, I'm not aware of any analytic expression which will tell you where
$I$ is. However, there is a simple algebraic/geometric criterion to tell us whether $I$ coincides with one of the vertices.

Consider the vertex $A$, if it satisfies one (and hence both) of following equivalent
conditions:

algebraic: $\left|\vec{n}_B(A) + \vec{n}_C(A) + \vec{n}_D(A)\right| \le 1$

geometric: the solid angle span by the tetrahedron at $A$ is $\ge \pi$.

One can show that $A$ is a local minimum of $d_T$ and hence $I = A$. If $A$ doesn't
satisfy these conditions, then one can show that $A$ cannot be a local minimum of $d_T$
and we need to look for I elsewhere.

In the case where none of the vertices are $I$, then $I$ falls inside the interior of $T$
and satisfy:

$$-\vec{\nabla} d_T(I) = \vec{n}_A(I) + \vec{n}_B(I) + \vec{n}_C(I) + \vec{n}_D(I) = \vec{0}$$

Similar to the geometric criterion above, this has the geometric interpretation that at
$I$, the four solid angles span by any 3 of the unit vectors $\vec{n}_A(I), \vec{n}_B(I), \vec{n}_C(I), \vec{n}_D(I)$ all equal to $\pi$.

Both of these are direct analogues of the 2 dimension results about Fermat Torricelli point. Namely:

Given a triangle $ABC$ and a point $I$ that minimize $IA + IB + IC$.

If one of the angle, say $\angle{A} \ge 120^\circ$, then $I = A$.

If no angles $\ge 120^\circ$, then $I$ falls inside interior of $\triangle ABC$
and $\angle{AIB} = \angle{BIC} = \angle{CIA} = 120^{\circ}$.


问:
Finding the roots of two polynomials from their sum
Problem: given that $a, b, c, d > 0$ and the sum of all real roots of the polynomials
$dx^{865}+cx^{65}+bx^5+a$ and
$ax^{865}+bx^{860}+cx^{800}+d$ is $-4.25$, what are these roots?

So apparently the powers of $x$ under the same coefficient sum up to 865 here. I can also see that polynomial $1$  has only one solution. Can't really think of anything else.

答:
Hint: If $\alpha$ is a root of the first, then $ \frac{ 1}{ \alpha}$ is a root of the second.

Hint: How many real roots are there in the first equation? Use the fact that $ a, b, c, d > 0$.

Hence, conclude that

 $\alpha_1 + \frac{1}{ \alpha_1 } = -4.25$


问:
What is the set of fixed point under the action of T in this example
Let $M= \lbrace g\lambda g^{-1} , $ $ g \in SU(3)\rbrace $ such that $\lambda = diag(i\lambda_1, i\lambda_2, i\lambda_3) $, $\lambda_i \in \mathbb{R}$ and $\lambda_1 > \lambda_2 > \lambda_3 .$

Let  $T= \lbrace diag (t,t,t') , t,t' \in U(1)\rbrace$ be the subgroup of SU(3) of diagonal matrices such that the first and the second entries are equal. Consider the action of T on M by conjugation.

What is the set $M^T=\lbrace m \in M , t.m=m \forall t \in T\rbrace $ of fixed point in $ M $ under the action of T on M ?

I greatly appreciate any help in this.

答:
Hint: We have $A \cdot B = B$ if and only if $AB = BA$, where $AB$ denotes the matrix product of the matrices $A$ and $B$. Now, consider the specific matrix
$$
A_0 = \pmatrix{-1 \\ & -1\\ && 1} \in T.
$$
If $B$ is any matrix such that $A_0B = BA_0$, what can we say about the matrix $B$? Show that if $B$ is of this required form, then we have $AB = BA$ for every matrix $A \in T$.

Then, note that the set $M$ can be nicely characterized with the help of the spectral theorem.


问:
Show that convex hull can be represented as an intersection of half-spaces
Yesterday I've stuck with the following problem:

Suppose that $a_0, a_1, ..., a_n$ are affine independent points of $n$-dimensional affine space.

$H_i$ is the hyperplane that passes through all these points except for $i$ ($i=0, 1, \dots, n$).

$H_i^+$ is the half-space bounded by $H_i$ that contains $a_i$. Show that
$$
\text{conv} \{a_0, a_1, \dots, a_n\} = \bigcap_{i=0}^n H_i^+
$$

My attempt(s):

Firstly let's note that $H^+:=\bigcap_{i=0}^n H_i^+$ is the convex set contains all points $a_0, a_1, ..., a_n$ (as an intersection of such sets).
Then I see two opportunities that can bring us to solution:

Let S be an arbitrary convex set that contains $a_0, a_1, ..., a_n$. We need to show that $H^+ \subseteq S$

Show that $H^+=\{ \sum_{i=0}^n \mu_i a_i: \sum_{i=0}^n \mu_i=1 \space \text{and} \space \mu_i \ge0, \space i=0, 1, \dots, n \}$

If it helps, I'll describe my thoughts on both of these ways. But actually I didn't solve the problem.

Any ideas?

HINT from the TextBook: consider the barycentric coordinate system associated with $a_0, a_1, \dots, a_n$

答:
Let $\;g_i=0\;$ be an equation of the hyperplane $H_i\;$ $(i=0,1,\ldots,n)$, that is

$$ g_i : E \to \Bbb R \quad \text{an affine function with}\quad H_i=g_i^{-1}(0), $$

where $\;E\;$ is the affine (real) space in which we are working. The half-space $\;H_i^+\;$ is described by the inequality $\;g_i\geq0\;$ or $\;g_i\leq0$, and we can assume, for simplicity, that it is of the form $\;g_i\geq0\;$, changing the sign of $\;g_i\;$ if necessary. We have, for each $\;i=0,1,\ldots,n$:

$$ g_i(a_i)>0 \qquad \text{ and }\qquad g_i(a_j)=0 \text{$\;\;$ for$\;\;$}j\neq i. \tag{1}$$

Note that it cannot be $\;g_i(a_i)=0\;$ because otherwise it would be $\;a_i\in H_i$.

Now let $\;x\in H^+:=\displaystyle \bigcap_{i=0}^nH_i^+$. This implies that $\;g_i(x)\geq0, \forall i$. With respect to the barycentric coordinate system associated with $(a_0,a_1,\ldots,a_n)$, we can write

$$ x=\sum_{j=0}^n \lambda_ja_j\qquad\text{with}\quad \sum_{j=0}^n\lambda_j=1 $$

and therefore we only need to prove that all the $\lambda's\;$ are $\;\geq0\;$. This follows from $(1)$:
$$ g_i(x) = g_i\big(\sum_{j=0}^{n}\lambda_ja_j\big) = \sum_{j=0}^{n}\lambda_jg_i(a_j) = \lambda_ig_i(a_i) $$

and the conclusion follows from $\;g_i(x)\geq0\;$ and $\;g_i(a_i)>0$.


答:
From MacLaurin expansion we get
$$x \sin 2x=2 x^2-\frac{4 x^4}{3}+\frac{4 x^6}{15}+\ldots[\text{ terms containing higher power of }x]+\ldots$$
furthermore we are given that
$$f(x)=2 x^2+\frac{32 x^4}{5}+\frac{96 x^6}{5}+\ldots[\text{ terms containing higher power of }x]+\ldots$$
so we can substitute in the limit
$$\lim_{x\to 0} \frac{f(x)-x\sin(2x)}{x^4} = \lim_{x\to 0}\frac{2 x^2+\frac{32 x^4}{5}+\frac{4 x^6}{15}+\ldots+\ldots-\left(2 x^2-\frac{4 x^4}{3}+\frac{96 x^6}{5}+\ldots\right)}{x^4}=\lim_{x\to 0}\frac{x^4\left[\left(\frac{32 }{5}+\frac{4 }{3}\right)-(284/15)x^2+ \text{ higher powers of }x\right]}{x^4}=$$
$$=\lim_{x\to 0} \left[\left(\frac{32 }{5}+\frac{4 }{3}\right)-(284/15)x^2+ \text{ higher powers of }x\right]$$
the terms containing $x$ tend to zero so we are left with
$$=\frac{32}{5} + \frac{4}{3} = \frac{116}{15}$$

Title: Defining a function by a series

Body:
We let the function $f(x)$ be defined by the series  $\sum_{n=1}^\infty \frac{n4^n}{n^2+1}x^{2n}$ and we need to show that $$\lim_{x\to 0} \frac{f(x)-x\sin(2x)}{x^4} = \frac{32}{5} + \frac{4}{3} = \frac{116}{15}$$

I'm really confused as to how we are supposed to define the function by a serie, I mean I can go the other way around and try to represent a function as a serie, but I'm not quite sure about representing a series as a function. I would really appreciate some help in figuring this one out, thanks in advance!


问:
Is there a way to simplify this expression with complex numbers
Let $z\in\mathbb{C}$ and $a,b\in\mathbb{R}$. Is there a way to simplify the expression
$$ \frac{a+bz+z^2}{a-bz+z^2}
 $$
without having to calculate everything by setting $z=x+iy$ and then making the denominator real by multiplying with the complex conjugate? In particular: is there a way to tell if the expression above lies inside the complex unit ball immediately?

EDIT:
Now suppose further that the roots of the bottom polynomial $z_1,z_2$ have a positive real part. Is this expression in the complex unit ball if we limit $z$ to the complex numbers with negative real part? Specifically I'm interested in the case $a=12, b=6$.

答:
Let $z_1,z_2$ be the roots of of the bottom polynomial. Then the expression factors to

$$\frac{z+z_1}{z-z_1}\cdot\frac{z+z_2}{z-z_2}$$

which in general is unbounded as $z$ approaches either root.


问:
The existence of a graph homomorphism
Is there a graph homomorphism between the 6-hypercube (https://en.wikipedia.org/wiki/File:6-cube_graph.svg) and the cube (https://en.wikipedia.org/wiki/File:3-cube_graph.svg)?</p>

答:
A graph $G$ is bipartite if and only if there is a homomorphism from $G$ to $K_2$. So trivially there is a homomorphism from any cube to another. The interesting question for bipartite graphs is whether there is a surjective homomorphism from
one to another.

Edit: View the vertices of $(d+1)$-cube as binary vectors with length $d+1$.
Each vector can be written as $0u$ or $1u$ and the vectors of the form $0u$
form a subgraph isomorphic to $Q_d$. Consider the map that sends $0u$ to $u$
and $1u$ to $u+e_1$, where $e_1$ is the vector of length $d$ with first entry 1 and all other entries 0. Then this map is a surjective homomorphism from
$Q_{d+1}$ to $Q_d$.

Title: The existence of a graph homomorphism


问:
Definition of a Boundary Action
I am reading An Intrinsic Characterization of $C^*$-Simplicity by Matthew Kennedy and have a question about the definition of a boundary action. On page 4, he says that the definition of a boundary action is given in section 2.3. I've read through section 2.3 several times but I don't see a definition of boundary action. Am I going crazy? If someone can spot it, would you please point it out? If section 2.3 does not in fact have the definition, would someone please kindly provide the definition of boundary action as used by the author (I can't suss out the meaning).

答:
My interpretation of this text is that a $G$-space $X$ is called a $G$-boundary if (...definition that follows...). When $X$ is a $G$-boundary, the action of $G$ on $X$ should be called a boundary action. I don't see any other plausible interpretation of this text.

Hope that helps,

Title: Definition of a Boundary Action


问:
Using set notation, define the set of even natural numbers between 100 and 500.

Using set notation, define the set of even natural numbers between 100 and 500.

This is what I have so far:

$P$ is even numbers so the set of natural numbers between 100 and 500 would be

$$P = \{x:x \in\mathbb  N, 100 < x < 500\}$$

Would this be correct?

答:
This is a more longwinded approach to help clarify what's going on. But be explicit in how you define 'between'. In this sense, between could be interpreted as $100 \leq x \leq 500$ or $100 < x < 500$. As long as you are explicit about your assumption, you're fine.

The sentence "Define the set of even natural numbers between 100 and 500" can be broken into several components:

P = {even natural numbers between 100 and 500}

P = {elements of x in natural numbers such that (x is even) AND (x is greater than 100) AND (x is less than 500) }

P = $ \{ x \in \mathbb{N} : (\exists k \in \mathbb{N} \quad x = 2k) AND (x > 100) AND (x < 500) \} $

P = $ \{ x \in \mathbb{N} : (\exists k \in \mathbb{N} \quad x = 2k) \land [(x > 100) \land (x < 500)] \} $

This way makes the inequality $100 < x < 500$ explicit using the logical and operator.


问:
Explanation for geometric proof of arc length cosh
I find this video, showing a way to find curve length of $cosh(x)$.
https://www.youtube.com/watch?v=0i1vecdN_pw

However, I fail to understand how exactly this works. Why is the length of the origin to intersection of circle with x-axis the curve length?

Isn't it via pythagoras $\sqrt{y(N)^2-1}$ ?

答:
For reference, take a look at this demonstration of $\cosh(x)$ and the definitions of $\cosh(x)$ and $\sinh(x)$ below:

$$
\begin{array}{l}
\sinh x:=\frac{e^{x}-e^{-x}}{2} \\
\cosh x:=\frac{e^{x}+e^{-x}}{2}
\end{array}
$$

From the demonstration, we see that the arclength $MN$ originates at $(0,1)$ and terminates at $(x,r)$ as per the definitions above. Notice that $r=\frac{e^{x}+e^{-x}}{2}=\cosh(x)$ and is the radius of the circle. To answer your question, it must be shown that the arclength of $\cosh(x)$ is the same as the bottom leg of the right triangle shown in the demonstration. Notice that this triangle has a vertical (smaller) leg of $1$ and a hypotenuse of $r$. From simple trigonometry, the long leg, or the distance along the $x$-axis, is $\sqrt{r^2-1}$.

Mathematically, this can be summarized as (using the arclength formula):

$$
\int_0^x \sqrt{1+(\frac{d}{du}\cosh(u))^2}\;du=\sqrt{r^2-1}
$$

Taking the arclength expression first, note that $\frac{d}{dx}\cosh(x)=\sinh(x)$. This makes the argument of the square root $1+\sinh^2(x)$. To continue integrating this, this argument can be simplified to $\cosh^2(x)=1+\sinh^2(x)$ from the identity $\cosh^2(x)-\sinh^2(x)=1$. Our integral then becomes:

$$
\int_0^x \sqrt{\cosh^2(u)}\;du=\int_0^x \cosh(u)\;du
$$

From inspection, also note that $\int \cosh(x)=\sinh(x)+C$. Thus, our result is:

$$
\sinh(x)-\sinh(0)=\sinh(x)
$$

Now, we must show that $\sinh(x)=\sqrt{r^2-1}$:

$$
\begin{align*}
r^2-1&=\frac{e^{2x}+2e^xe^{-x}+e^{-2x}}{4}-\frac{4}{4}\\
&=\frac{e^{2x}+e^{-2x}-2}{4}
\end{align*}
$$

$$
\begin{align*}
\sinh^2(x)&=\frac{e^{2x}-2e^xe^{-x}+e^{-2x}}{4}\\
&=\frac{e^{2x}+e^{-2x}-2}{4}
\end{align*}
$$

$$\therefore r^2-1=\sinh^2(x)\implies \boxed{\sqrt{r^2-1}=\sinh(x)}$$

This shows that the length of the segment along the $x$-axis intersecting the circle ($\sqrt{r^2-1}$) is the same as the arclength of $\cosh(x)$.


问:
Does the Weierstrass M-test show analyticity?
I'm trying to show (textbook exercise) that the riemann-zeta function is analytic. The solution is here:

Why does the proof say that the zeta series converges to an analytic function? Doesn't the M-test merely show uniform convergence? The zeta series (whose term is inside the first modulus in the above solution) isn't a power series, so I can't argue that convergence implies analyticity either.

答:
The limit of a sequence of analytic functions that converges uniformly is an analytic function.

This can be proved by combining uniform convergence with Morera's theorem because  uniform convergence allows you to switch limit and integral.

http://en.wikipedia.org/wiki/Morera%27s_theorem#Uniform_limits


问:
Suppose that $x_1 = \frac{1}{2}$, and $x_{n+1} = (x_n)^2$. Show that $\left\lbrace x_n\right\rbrace$ converges and find lim $x_n $
Suppose that $x_1 = \frac{1}{2}$, and $x_{n+1} = (x_n)^2$. Show that $\left\lbrace x_n\right\rbrace$ converges and find lim $x_n $.

The hint that the problem gives is "You cannot divide by zero". I want to base the proof on this hint but don't know how to proceed. Thank you!

答:
Notice that $$x_1=\frac{1}{2},\\x_2=x_1^2=\frac{1}{4},\\x_3=x_2^2=\frac{1}{16}$$ and so on. This is a monotonically decreasing sequence, and also it converges. Why?

Think if the sequence $\{x_n\}$ can attain a value less than $0$ or not as the sequence is decreasing. Then apply the Monotone Convergence Theorem to know that the sequence converges.

Now, as the sequence converges, thus limit exists (say $L$). Thus, $$\lim_{n\rightarrow\infty}x_n=L$$

Can you now apply the inductive method and find the limit now?

$$\begin{align}x_{n+1}&=x_n^2\\\lim_{n\rightarrow\infty}x_{n+1}&=\lim_{n\rightarrow\infty}x_n^2\\L&=L^2\\ L&=1,0\end{align}$$Now, $L$ cannot be $1$. Thus, $L=0$.


问:
How to compute the class group?
I am not sure how to find the class group with a given discriminant d. First, we need to find the set of all primitive positive definite forms. Then we reduce all of them to split them into equivalence classes which is the class group.

I am not sure how to find the set of all primitive positive definite forms. Say $(a,b,c)$ is a form, then we know $b^2-4ac=d$ and $a,b,c$ don't have a common divisor other than 1. In theory, there could be infinitely many forms since there are no other restrictions...

Any help will be very appreciated!

答: $(a,b,c)$ is equivalent to a unique reduced form, say $(a,b,c)$. In that form, $|b|\le a$, so that $b^2\le a^2$, and $a\le c$, so that the discriminant $D<0$ satisfies $-D = 4ac-b^2 \ge 4a^2-a^2 = 3a^2$. It follows that $a\le \sqrt{\frac{-D}{3}}$. This puts a bound on $a$, and therefore on $b$. Further, you know that $b\equiv D\mod{2}$. So there are only a finite number, and you can find them by trying the various cases.

Title: How to compute the class group?


问:
How to show that this formula is a tautology using resolution?

I know how to show a CNF formula is a tautology using resolution, I've come across this DNF now and have to show its a tautology, is it the same technique or do I need to change it to a CNF? if so how?

```<code>             F := (A ∧ B) ∨ (A ∧ ¬B) ∨ (¬A ∧ B) ∨ (¬A ∧ ¬B)
</code>```

答:
You can consider a truth table for the statement:

$$\begin{bmatrix}
 & | & A & \neg A
\\ \hline B & | & T \lor F \lor F \lor F = T &  F \lor F \lor T \lor F = T
\\ \neg B & | & F \lor T \lor F \lor F = T & F \lor F \lor F \lor T = T
\end{bmatrix}$$


问:
$U_{p^2}$ a group under multiplication.
Let $p\in\mathbb{N}$ be prime and define $U_{p^2} = \{[a]_{p^2} \in \mathbb{Z}_{p^2} : p \nmid a\}$. Note $U_{p^2}$ is a group under multiplication.

(i) What is the identity element of $U_{p^2}$? ( I think this could be $[0]_{p^2}$)

(ii) Use the statement in (a)(i) to give a proof that axiom (G3) holds for $U_{p^2}$.
((G3) Given $h\in H$, we need to check that $h^{-1}\in H$.)

(iii) Show that $|U_{p^2}|= p(p−1)$.

I really don't understand this topic and the text book gives the example above for us to try out but i just don't know where to start. Would really appreciate any help or understanding of the above question or topic. Thanks

答:
Equivalently, $U_{p^2}=\{a\in\Bbb Z_{p^2}\mid\operatorname{gcd}(a,p^2)=1\}$ is the multiplicative group or the integers modulo $p^2$. Therefore, $U_{p^2}$ is made of the integers (equivalence classes) from $1$ to $p^2-1$ which are not multiple of $p$ (as for these latter $\operatorname{gcd}(a,p^2)=p\ne1$). Now, $np\le p^2-1\iff n\le p-\frac{1}{p}$, so for $n=1,\dots,p-1$. From this, it follows that $|U_{p^2}|=p^2-1-(p-1)=p(p-1)$. This answers iii).

See here for further reference on this group.

Title: $U_{p^2}$ a group under multiplication.


问:
A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$
I saw this twitter post that reads:

Find all the pairs of positive integers $(x,y)$ satisfying $$  x^2 + y^2 - 5xy + 5 = 0 . $$

I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.

For example, the first general solution presented reads:

$$ \begin{array} { r c l }
 x &=&
\dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n
- \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 21 \left(55 + 12\sqrt{21} \right)^n
+ \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\
 y &=&
\dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n
- 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 63 \left(55 + 12\sqrt{21} \right)^n
+ 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.

Plugging $n=0,1,2,3$ gives the first few solutions

$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449)
$$

I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.

Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?

答:
It's $(5y-2x)^2-21y^2=-20$, which is a Pell type equation.

I got that by completing the square:

$x^2-5xy+y^2=-5\implies \left(x-\frac52y\right)^2-\frac{21}4y^2=-5\implies (5y-2x)^2-21y^2=-20$.

So it's $X^2-21Y^2=-20$, with $X=5y-2x$ and $Y=y$.


问:
Limit of trigonometric function (cotangent)
I need to solve the following problem:
$\lim_{x\to 3}(x-3) \cot{\pi x}$.
Can anyone give me a hint? I have no idea.

答: $$\lim_{x\to 3}(x-3) \cot{\pi x}=-\lim_{x\to 0}x \cot{\pi x}=-\frac1\pi\lim_{x\to 0}\frac{\pi x}{\sin \pi x} \cos{\pi x}.$$


问:
Integration of a function should equal the area
It seems a silly question but I can't see what's wrong

\begin{align*}
\sqrt{\dfrac{2}{T} \int_0^{T/2} \Big(\dfrac{-2V_mt}{T}+V_m \Big)^2dt}
=&\sqrt{\dfrac{2V^2_m}{T} \int_0^{T/2} \Big(\dfrac{4t^2}{T^2}-4t/T+1 \Big)dt}\\
&=\sqrt{\dfrac{2V_m^2}{T}{\int}_0^{\tfrac{T}{2}}\Bigg(\dfrac{4t^2}{T^2}-\dfrac{4t}{T}+1 \Bigg) dt}\\
  & =\sqrt{\dfrac{2V^2_m}{T}  \Bigg( \dfrac{4\Big(\dfrac{T}{2} \Big)^3}{3T^2} - \dfrac{4\Big(\dfrac{T}{2} \Big)^2}{2T} +  \dfrac{T}{2}\Bigg)} = \dfrac{V_m}{\sqrt{3}}
\end{align*}

Now, the answer should be related to the area of the triangle, and hence it should be simply $\sqrt{\frac{1}{T}(\dfrac{1}{2}V^2_mT)}=V_m/\sqrt{2}$, but the answers don't match.

The goal is to find the RMS value

答: The inner integral is $[-V_mt^2/T+V_mt]_0^{T/2}=-V_mT/4+V_mT/2=V_mT/4$, which as expected is the area of a triangle of base $T/2$, height $V_m$. So for $T\ne0$ the surd is $\sqrt{V_m/2}$, which equals the right-hand side for $V_m\in\{0,\,3/2\}$.


问:
Existence of unit speed geodesics
I am reading Lee's book "Riemannian manifolds: An Introduction To Curvature". Let $(M,g)$ be a Riemannian manifold. Fix $p\in M.$ It is well known that the exponential map $\exp_p$ map is a local diffeomorphism at $0.$ We can use the identification $T_pM$ to be $\mathbb R^n$ by fixing an orthonormal frame of $T_pM.$ Then we can use the exponential map to define a smooth coordinate chart on a neighbourhood of $p\in M.$ A coordinate chart obtained in this fashion is called a normal coordinate chart. Let $(\mathcal U,x^i)$ be a normal coordinate system. Exercise 5.6 of Lee asks to prove that
``At any point $q\in\mathcal U -p,$ $\partial/\partial r$ is he velocity vector of he unit speed geodesic from $p$ to $q$, and therefore has unit length with respect to $g.$"

Lee defines $r(x):=(\sum_i{x^i}^2)^{\frac{1}{2}}$ and $\partial/\partial r:=\frac{x^i}{r}\partial/\partial x^i$. I can show the following. Let $V:=\exp_p^{-1}(q).$ Then consider the maximal geodesic $\gamma_V$ passing through $p$ and initial velocity $V.$ We know that in normal coordinates $\gamma_V(t)=(tV^1,\dots,tV^n)$. Therefore, after easy calculation $\partial/\partial r|_{\gamma_V(t)}=|V|^{-1}\gamma_V(t)$ and if $|V|=1$ probably we have what Lee asks as $\gamma_V$ is also constant speed. But how can we guarantee that there exists $V$ with $|V|=1.$ This can be shown if there exists a $|V|>1$. Then we can scale it by using star-convexity. But what if $|V|<1?$ Also what about uniqueness Lee is mentioning?

答:
As you've written, $\gamma_V(t) = \exp_p(tV)$. This doesn't have unit speed unless $|V| = 1$ so we rescale, defining $\gamma(t) = \exp_p(t|V|^{-1}V)$. By the chain rule this now was constant unit speed (first evaluate $\dot\gamma(0)$ then use that $\lambda \mapsto \exp_p(\lambda v)$ has constant speed). Now, in coordinates, this is given by the curve
$$
\tilde{\gamma}(t) = |V|^{-1}(tV^1, \ldots, tV^n)
$$
and therefore its tangent vector is $\dot{\tilde{\gamma}}(t) = |V|^{-1}(V^1, \ldots, V^n)$. But, as $\partial/\partial x^i$ is just the $i$th standard basis vector in coordinates, we have
$$
\frac{\partial}{\partial r}  = \frac{x^i}{r}\frac{\partial}{\partial x^i} = \frac{1}{r}x,
$$
where $x = (x^1, \ldots, x^n)$ is the coordinates of a point on the manifold. Thus for instance when $x = \tilde{\gamma}(t) = t|V|^{-1}V$, we have $\frac{\partial}{\partial r} = r^{-1}x = |t|^{-1}t|V|^{-1}V = |V|^{-1}V = \dot{\tilde{\gamma}}(t)$. In this last line I have abused notation, writing $V$ for the coordinate representation $(V^1, \ldots, V^n)$.

Title: Existence of unit speed geodesics


问:
Need help understanding what is a free factor of a free group.
Let $F_n$ be the free group with $n$ generators $\{ x_1 , \dots , x_n \}$. From my understanding, we say that a subgroup $K \leq F_n$ is a free factor of $F_n$ if there exists a subgroup $H \leq F_n$ such that $F_n = H * K$. Furthermore, if I understand correctly, if we write that $K = \langle f_1 , \dots , f_s \rangle$ and $H = \langle g_1 , \dots , g_m \rangle$, then we have that $H * K = \langle f_1 , \dots , f_s , g_1 , \dots , g_m \rangle$.

Now my confusion is as follows - with this definition, isn't EVERY subgroup of a free group a free factor? For a silly example, let $n=2$, $F_2 = \langle a, b \rangle$ and let $K = \langle aba , b \rangle$. Then can't we just take $C = \langle a \rangle$, which gives $K * C = \langle aba, b , a \rangle = \langle a, b \rangle = F_2$ ? Isn't it always possible to do this for any $K$, by just letting $C$ to be the subgroup of $F_n$ generated by the $x_i$'s that don't occur in $K$ (so possibly $C = F_2$)?

答:
There's an (understandable) confusion between the elements $aba$, $b$, and $a$ generating $F_2$, versus generating $F_2$ freely (that is, with no relations among them). The notation $H*K$ is a free product, so no relations are allowed between their elements; however, there's an obvious relation among $aba$, $b$, and $a$.

For an even simpler example, consider $F_1 \cong \Bbb Z$, and let $K=\langle 2\rangle$. Then we can similarly ask, can't we just take $C=\langle 5\rangle$, since $\langle 2,5\rangle = \Bbb Z$? No, because $\Bbb Z$ is not the free product of $\langle 2\rangle$ and $\langle 5\rangle$.


答:
You can simplify the equations by $V_m$ and WLOG, $T=2$. Now,

$$\sqrt{\int_0^1(1-t)\,dt}=\frac1{\sqrt2}$$ which matches the other formula.

The square in the first integral makes a different question.

Title: Integration of a function should equal the area

Body:
It seems a silly question but I can't see what's wrong

\begin{align*}
\sqrt{\dfrac{2}{T} \int_0^{T/2} \Big(\dfrac{-2V_mt}{T}+V_m \Big)^2dt}
=&\sqrt{\dfrac{2V^2_m}{T} \int_0^{T/2} \Big(\dfrac{4t^2}{T^2}-4t/T+1 \Big)dt}\\
&=\sqrt{\dfrac{2V_m^2}{T}{\int}_0^{\tfrac{T}{2}}\Bigg(\dfrac{4t^2}{T^2}-\dfrac{4t}{T}+1 \Bigg) dt}\\
  & =\sqrt{\dfrac{2V^2_m}{T}  \Bigg( \dfrac{4\Big(\dfrac{T}{2} \Big)^3}{3T^2} - \dfrac{4\Big(\dfrac{T}{2} \Big)^2}{2T} +  \dfrac{T}{2}\Bigg)} = \dfrac{V_m}{\sqrt{3}}
\end{align*}

Now, the answer should be related to the area of the triangle, and hence it should be simply $\sqrt{\frac{1}{T}(\dfrac{1}{2}V^2_mT)}=V_m/\sqrt{2}$, but the answers don't match.

The goal is to find the RMS value


问:
Fin the greatest and least value of $(4\sin θ-6\cos θ)^2-3$ where $0°<θ<90°$
This question I found the tan by doing 4^2/6^2 and put it in the form of 2tan of 13 sin(θ-56.3) and I know the maximum is sin inverse of 1 for maximum and sin inverse -1 for minimum but I know the answers are 49,-3 but I don’t know how to get to it any tips ang help much appreciate.</p>

答:
$$(4 \sin t-6 \cos t)^2-3;\;t\in[0,\pi/2]$$
is maximum at $t=0$ where the value is $33$

to find the minimum, set the derivative $=0$.

$$(4 \sin t-6 \cos t) (6 \sin t+4 \cos t)=0$$

$$4\sin t -6 \cos t=0$$

so the minimum value is
$(4 \sin t-6 \cos t)^2-3=0^2-3=-3$

while $6 \sin t+4 \cos t=0$ is not true in $[0,\pi/2]$


问:
Verify that a set is $\sigma$-weakly dense in a von Neumann algebra

The above lemma is from Takesaki's book (Vol 2, Chapter VIII).
I have two questions.

1.How to conclude that $e_nm_{\varphi}e_n$ is $\sigma$-weakly dense in $e_nMe_n$?

From Lemma 2.4, we only know that $m_{\varphi}$ is a $M_a^{\varphi}$-bimodule.

2.How to prove that $\cup e_nMe_n$ is $\sigma$-weakly dense in $M$?

答:
The question is why $m_\varphi$ is dense in $M$.

The assumption is that $\varphi$ is fns. Being semifinite, there exists projections $\{p_j\}$ with $\varphi(p_j)<\infty$ for all $j$ and $p_j\to 1$ weakly. Then $p_j\in m_\varphi$ for all $j$. As $m_\varphi$ is an $M_a^\varphi$-module, given $y\in M_a^\varphi$ we have $a=\lim p_ja\in \overline{m_\varphi}$. As $M_a^\varphi$ is dense in $M$ (Lemma 2.3), we get that
$$
\overline{m_\varphi}=\overline{M_a^\varphi}=M.
$$

$e_n\nearrow 1$


问:
Why can you transition from a Riemann Sum to an Intergal?
I understand that the integral is defined through a Riemann Sum, as in
$$\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}f\left(a+i\cdot\frac{b-a}{n}\right)\frac{b-a}{n}$$
But I don't understand how this definition is made. I understand the structure of the Riemann Sum which is summing up infinetely many infinietely thin rectangles to approximate the area under a curve, but I don't understand why this turns into an integral and then can be evaluated the way an integral is. Any help is appreciated.

答:
Perhaps a good way to think about that formula is the following:

There exist some functions f defined over the interval [a,b] which have the property that the limit on the right hand side exists and the sequence of sums converges to that limit. In the case where that happens we will name that limit, or call that limit by the symbology on the left hand side.

So then in the Analysis class or Advanced Calculus class, (depending on how the classes are named, in the old days and still in some places it goes by the name Calculus), you will prove that these limits exist for various families of functions, and you will also prove what is known as the fundamental theorem of calculus.

When you say evaluation you are applying the fundamental theorem of calculus and some of the other theorems which you will prove in the Analysis or the Advanced Calculus class.

So when you say:

"but I don't understand why this turns into an integral and then can be evaluated the way an integral is."

I interpret that to mean that you are not confused as to what the evaluation rules are and how to evaluate per se, but rather why you are allowed to use those rules to evaluate.

The answer to that query is the subject matter of what could be a first Analysis course and exceeds the length and detail allowed in this format. I encourage you to look into it further as time allows.

If time is an issue and you are trying to get through your current course, you can be satisfied applying the evaluation rules to the best of your ability to all of the functions you will encounter in your course and later, should you perchance progress further in your study, the mystery of why you are allowed to use those evaluation rules will be cleared up.


问:
Proving that if $a^n \equiv 1 \pmod m$ then $d\mid n$
I'm having some trouble proving the following:

Let $d$ be the smallest positive integer such that $a^d \equiv 1 \pmod m$, for $a \in \mathbb Z$ and $m \in \mathbb N$ and with $\gcd(a,m) = 1$. Prove that, if $a^n \equiv 1 \pmod m$ then $d\mid n$.

The first thing that came to my mind was Euler's theorem but I couldn't conclude anything because I'm not very skilled when it comes to using Euler's totient function. Can someone give me any tips or show me how to solve this?

答:
Hint:

Prove the contrapositive assertion: if $d$ does not divide $n$, $d$ is not the smallest positive integer such the $a^d\equiv 1 \bmod n$.

Note: actually, it is a particular case of a more general assertion: for an element $g$ of finite order  in a group , the set $\bigl\{n\in\mathbf Z\::\: a^n=1\bigr\}$ is an ideal of $\mathbf Z$, and as $\mathbf Z$ is a P.I.D.…


答:
Let $n = dq + r$ with $0 \le r < d$, then

$$
a^r \equiv a^{dq+r} \equiv a^n \equiv 1 \pmod m
$$

which means that $r = 0$ since otherwise we have a contradiction with the fact that $d$ is the $smallest$ positive integer such that $a^d \equiv 1 \pmod m$.

Title: Proving that if $a^n \equiv 1 \pmod m$ then $d\mid n$

Body:
I'm having some trouble proving the following:

Let $d$ be the smallest positive integer such that $a^d \equiv 1 \pmod m$, for $a \in \mathbb Z$ and $m \in \mathbb N$ and with $\gcd(a,m) = 1$. Prove that, if $a^n \equiv 1 \pmod m$ then $d\mid n$.

The first thing that came to my mind was Euler's theorem but I couldn't conclude anything because I'm not very skilled when it comes to using Euler's totient function. Can someone give me any tips or show me how to solve this?


答: It is hard to read the mind of the book author or to see an error in the question, but for the same reason that zeta converges absolutely for $\Re s > 1$, its derivatives converge absolutely.  By uniformity of convergence, whatever the limit of zeta is, it is complex differentiable in $\Re s > 1$ (because summation and differentiation operations commute under uniform convergence), and therefore analytic.</p>

Title: Does the Weierstrass M-test show analyticity?

Body:
I'm trying to show (textbook exercise) that the riemann-zeta function is analytic. The solution is here:

Why does the proof say that the zeta series converges to an analytic function? Doesn't the M-test merely show uniform convergence? The zeta series (whose term is inside the first modulus in the above solution) isn't a power series, so I can't argue that convergence implies analyticity either.


答:
Given a group $G$ with subgroups $K$ and $H$, we denote by $G=K * H$ that $G$ is the internal free product of $G$ and $H$. This equivalent to the following two properties being satisfied:

Any element of $G$ can be written as finite product with factors taken from $K$ and $H$.

Any finite product of non-identity factors alternating between $K$ and $H$ is not the identity in $G$.

Alternatively, you can define this as the homomorphism $K*H\to G$ being an isomorphism, where $K*H$ denotes the external free product and the map sends alternating words to the corresponding product in $G$. Surjectivity of this homomorphism is equivalent to (1), while injectivity is equivalent to (2).

When $G=\langle a,b\rangle$ is the free group on two generators, $H=\langle aba,b\rangle$ and $K=\langle a\rangle$, the first property is satisfied, but the second is not:
$$
(aba)(a^{-1})(b^{-1})(a^{-1}) = 1
$$
and $aba\in H$, $a^{-1}\in K$, $b^{-1}\in H$ and $a^{-1}\in K$ are all non-identity elements.

In terms of the map on the external free product, the homomorphism $K*H\to G$ would be surjective but not injective in this example, since the alternating word $(aba)(a^{-1})(b^{-1})(a^{-1})$ is not the identity in the external free product $K*H$ but its image in $G$ is. Hence, the kernel is non-trivial.

Title: Need help understanding what is a free factor of a free group.

Body:
Let $F_n$ be the free group with $n$ generators $\{ x_1 , \dots , x_n \}$. From my understanding, we say that a subgroup $K \leq F_n$ is a free factor of $F_n$ if there exists a subgroup $H \leq F_n$ such that $F_n = H * K$. Furthermore, if I understand correctly, if we write that $K = \langle f_1 , \dots , f_s \rangle$ and $H = \langle g_1 , \dots , g_m \rangle$, then we have that $H * K = \langle f_1 , \dots , f_s , g_1 , \dots , g_m \rangle$.

Now my confusion is as follows - with this definition, isn't EVERY subgroup of a free group a free factor? For a silly example, let $n=2$, $F_2 = \langle a, b \rangle$ and let $K = \langle aba , b \rangle$. Then can't we just take $C = \langle a \rangle$, which gives $K * C = \langle aba, b , a \rangle = \langle a, b \rangle = F_2$ ? Isn't it always possible to do this for any $K$, by just letting $C$ to be the subgroup of $F_n$ generated by the $x_i$'s that don't occur in $K$ (so possibly $C = F_2$)?


问:
Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$
How to get a closed form for the following sum:

$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$

Where $F_k$ is the $k$th Fibonacci number.

I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$

Body: Write $\varphi:=\tfrac{1+\sqrt{5}}{2}$ so $F_n=\tfrac{\varphi^n-(-\varphi^{-1})^n}{\sqrt{5}}$ in the convention $F_0=0,\,F_1=1$. Since $1+\varphi^2=\sqrt{5}\varphi$, the sum is$$\begin{align}\tfrac15\sum_{k=0}^n((-1)^k(\varphi^{2k}+\varphi^{-2k})-2)&=\frac{\tfrac{1-(-\varphi^2)^{n+1}}{1+\varphi^2}+\tfrac{1-(-\varphi^{-2})^{n+1}}{1+\varphi^{-2}}-2(n+1)}{5}\\&=\frac{-(-\varphi^2)^{n+1}-\varphi^2(-\varphi^{-2})^{n+1}-\sqrt{5}\varphi(2n+1)}{5\sqrt{5}\varphi}\\&=\frac{(-1)^n\varphi^{2n+1}+(-1)^n\varphi^{-2n-1}-\sqrt{5}(2n+1)}{5\sqrt{5}}\\&=\frac{(-1)^nF_{2n+1}-(2n+1)}{5}.\end{align}$$


问:
Is $O(4)\cong SO(4)\times C_2$?
My claim is $O(4)\ncong SO(4)\times C_2$, since there is no "$-I$" role as in the proof that $O(3)\cong SO(3)\times C_2$, as $\det(-I)=1$ in $\mathbb{R}^4$. However, I cannot proceed like $O(2)$ case that both $SO(2)$ and $C_2$ are abelian while $O(2)$ is not. Are there any properties I missed?

答: Hint. Show $Z(O(4))=Z(SO(4))$ in order to prove $Z(O(4))\not\cong Z(SO(4)\times C_2)$.

Title: Is $O(4)\cong SO(4)\times C_2$?


问:
Equivalence of Definitions of Principal $G$-bundle
I've finally gotten around to learning about principal $G$-bundles.

In the literature, I've encountered (more than) four different definitions.  Since I'm still a beginner, it's unclear to me whether these definitions are equivalent or not.  I would appreciate any clarification.

All maps and group actions are assumed continuous.

Definition 1: A principal $G$-bundle is a fiber bundle $F \to P \xrightarrow{\pi} X$ together with a right action of $G$ on $P$ such that:

(1) $G$ acts freely and transitively on fibers.

(2A) $G$ preserves fibers.

Definition 2: A principal $G$-bundle is a fiber bundle $F \to P \xrightarrow{\pi} X$ together with a left action of $G$ on $F$ (note $F$ here) such that:

(1) $G$ acts freely and transitively on $F$.

(2B) There exists a trivializing cover with $G$-valued transition maps.

Definition 3: A principal $G$-bundle is a fiber bundle $F \to P \xrightarrow{\pi} X$ together with a right action of $G$ on $P$ such that:

(1') $G$ acts freely on $P$ and $X = P/G$ and $\pi\colon P \to X$ is $p \mapsto [p]$.

(2C) There exists a trivializing cover that is $G$-equivariant.

Definition 4: A principal $G$-bundle is a fiber bundle $F \to P \xrightarrow{\pi} X$ together with a right action of $G$ on $P$ such that:

(2A) $G$ preserves fibers.

(2C) There exists a trivializing cover that is $G$-equivariant.

Thoughts: It seems to me that Definition 4 is not equivalent to the other three.  More than anything else, I am unclear as to why the existence of a trivializing cover that is $G$-equivariant is equivalent (is it?) to the existence of one that has $G$-valued transition functions.

I've also seen a fifth definition which assumes only condition (1).

Thanks in advance.

答:
For the equivalence of these definitions, I would look here: Local triviality of principal bundles.

The existence of a $G$-equivariant cover is equivalent to the existence of $G$-valued transition functions:

Suppose $(U_\alpha,\Phi_\alpha)$, $\Phi_\alpha : P\vert_{U_\alpha} \to U_\alpha\times F$, is a trivializing cover.  This defines a collection of maps $\phi_\alpha : P\to F$ by
$$
\Phi_\alpha(p) = (\pi(p), \phi_\alpha(p)).
$$
For a right principal $G$-bundle, this covering is $G$-equivariant if $\phi_\alpha(pg) = \phi_\alpha(p)g$.  Now we have
$$
\Phi_\alpha \circ \Phi_\beta^{-1} : U_\alpha \cap U_\beta \times F \to U_\alpha \cap U_\beta \times F
$$
is an isomorphism of trivial $G$-bundles and so takes the form
$$
(x, f) \mapsto (x, h_{\alpha\beta}(x,f)).
$$
If the covering is $G$-equivariant then so is this map, which means that $h_{\alpha\beta}(x,fg) = h_{\alpha\beta}(x,f)g$.  Since $G$ is acting freely and transitively, fixing a point of $F$ identities $F$ with $G$ and $h_{\alpha\beta}$ is entirely determined by the function $g_{\alpha\beta}: U_\alpha\cap U_\beta \to G, x \mapsto h_{\alpha\beta}(x,e)$.  Thus the transition functions are given by left-multiplication by $g_{\alpha\beta}$.  This is what is meant by the transition functions being $G$-valued.

Conversely, if the transition functions are $G$-valued then the trivializations will be $G$-equivariant.  This is because
$$
P = \sqcup_\alpha U_\alpha \times F/\sim, ~~ (x, f) \sim (x, g_{\alpha\beta}(x)f) \text{ for } x \in U_\alpha\cap U_\beta.
$$
The equivariance then comes from the fact that the transition functions are operating by left-multiplication, while the $G$-action is right multiplication.


问:
Regular Cardinal of a Power set and its Chains

My attempt(almost sure that it is not quite working):

It is known that the greatest cardinal is P (ω) which is not a well-orderable set. Another fact is that P (ω) is a regular cardinal which by the definition give us :every unbounded subset A<P (ω) has the same cardinality as P (ω). But by the property of cardinals, namely that if two sets P (ω) and A have the same cardinal numbers, then they are equinumerous,the problem is solved.However, I have just used some well known propertis. Is there any way to complete the proof using that:

A chain in <P (ω), < > will be such a subset A< P(ω)that for every two elements in it they will be (<=) comparable.  With other words, A<P (ω) is a chain, if  is linear ordered.

答:
None of what you've written works.

First of all, none of the three facts you claim at the outset (that $\mathcal{P}(\omega)$ has the greatest cardinality, that it is not well-orderable, and that its cardinality is regular) is justified. The first is provably false: for the first consider e.g. $\mathcal{P}(\mathcal{P}(\omega))$ and more generally note that there is no largest cardinality. The second is independent of $\mathsf{ZF}$, and in particular it is consistent with $\mathsf{ZF}$ that every set can be well-ordered (this is exactly the axiom of choice). Finally, the third is even undecidable from $\mathsf{ZFC}$ axioms: it is consistent, for example, that $\vert\mathcal{P}(\omega)\vert=\aleph_{\omega_1}$, which is singular.

Moreover, even if we were to grant those assumptions your subsequent claim that every unbounded set in $(\mathcal{P}(\omega), \subseteq)$ has size continuum would also unjustified. Indeed, that claim is false: consider e.g. $\{\omega\}$, which is a one-element unbounded subset of $\mathcal{P}(\omega)$.

The actual solution to the problem, incidentally, is constructive: we can whip up an explicit example of a size-continuum chain in $\mathcal{P}(\omega)$. This problem has appeared repeatedly on this site. One hint I have is to replace $\omega$ by $\mathbb{Q}$ (since they have the same cardinality, the two powerset-posets will be isomorphic) and think about a particular kind of set of rational numbers which you've already seen in real analysis ...

 Dedekind cuts. Think about how the left cuts corresponding to reals $r$ and $s$ compare if $r<s$.


问:
Co-finite and Co-countable topologies are not normal
I’ve been investigating these two topologies on $\mathbb R$ and believe that they are both not normal as not every 2  disjoint closed sets have disjoint open neighbourhoods but I’m struggling to find an example that shows this (a counter example to the definition of normal).

I believe the same example will work for both cofinite and cocountable topology.

Can anyone help please?

Thanks

答: Take for example the closed subsets $A=\{0\}$ and $B=\{1\}$. Suppose you have neighborhoods $U$ of $0$ and $V$ of $1$ such that $U\cap V=\varnothing$. That would mean that $U^c\cup V^c=\mathbb{R}$. But $U^c$ and $V^c$ are finite (countable) in the co-finite (co-countable) topology, leading to a contradiction. So that would be a counterexample.


答: Every two non-empty open sets have non-empty intersection, therefore any two disjoint non-empty closed sets are instances that disprove normality.</p>

Title: Co-finite and Co-countable topologies are not normal

Body:
I’ve been investigating these two topologies on $\mathbb R$ and believe that they are both not normal as not every 2  disjoint closed sets have disjoint open neighbourhoods but I’m struggling to find an example that shows this (a counter example to the definition of normal).

I believe the same example will work for both cofinite and cocountable topology.

Can anyone help please?

Thanks


答:
I think you misunderstand something. The RMS is the square root of the integral of the square of the function.

So your first calculus $$\begin{align*}
\sqrt{\dfrac{2}{T} \int_0^{T/2} \Big(\dfrac{-2V_mt}{T}+V_m \Big)^2dt}
= \dfrac{V_m}{\sqrt{3}}
\end{align*}$$

is correct. But it should not be the area below the curve ! It is the area below the curve of the squared function.


问:
Limit to infinity with absolute value
I am learning math all by myself, no school, no teacher, and I come across a problem that I do not understand:

$\lim _{x \to -\infty } \frac{  | x^{2} - x  | +2 x^{2} }{1 - 3x^{2}}$

From my understanding, $|x²-x|$ should be translated as $-(x² - x)$, as we are approaching negative infinity. That will result in $-\frac{1}{3}$.
The book solutions is  $-1$, and I can confirm the solution by graphing the function. If I translate as $(x²-x)$ I get the $-1$.

So, why is it not $-(x²-x)$?

Thanks.

答:
As $x\to-\infty$, you have that $x^2\to\infty$.

Even easier, $x^2-x>0$ for all $x<0$.

Title: Limit to infinity with absolute value


问:
Complex numbers order property
As a complex number set/field isn't an ordered set/field.

Now $1 \in \mathbb{C}  $ & $2 \in \mathbb{C}  $ .

How is $2>1$ ?

答: Any two numbers that are elements of an ordered set can be compared to one another. $2$ and $1$ are elements of the set of integers, so they are comparable. The fact that they are also members of the set of complex numbers is immmaterial.</p>

Title: Complex numbers order property


问:
$\{r_n: n = 1,2,...\}$ enumeration of the rational numbers in $[0, 1]$. Show $\sum_{n = 1}^\infty \frac{1}{2^n}f(|x - r_n|)$
$f$ is a non-negative Lebesgue integrable function on $[0, 1]$, and $\{r_n: n = 1,2,...\}$ is an enumeration of the rational numbers in $[0, 1]$. I want to show:
\begin{equation}
       \sum_{n = 1}^\infty \frac{1}{2^n}f(|x - r_n|)
   \end{equation}

converges for a.e. $x \in [0, 1]$.

I know that since $f$ is Lebesgue integrable, it follows that $\int_0^1 |f|dx < \infty$. However, I'm not sure how I would use this to solve the problem.

答:
For each $n$ is
$$
\int_0^1 f(|x - r_n|)\, dx = \int_0^{r_n} f(r_n-x) \,dx + \int_{r_n}^1 f(x-r_n) \, dx \le 2 \int_0^1f(x) \, dx \, .
$$
The function
$$
 F(x) = \sum_{n = 1}^\infty \frac{1}{2^n}f(|x - r_n|)
$$
is nonnegative, measurable, and
$$
 \int_0^1 F(x) \, dx = \sum_{n = 1}^\infty \frac{1}{2^n}\int_0^1  f(|x - r_n|) \, dx
\le 2 \int_0^1f(x) \, dx < \infty
$$
which implies that $F(x) < \infty$ a.e. on $[0, 1]$ (see for example If f is integrable, is it finite almost everywhere?).


问:
Set $M_t = 4B_t^2 + e^{4B_t -8t} -4t$. Find $\mathbb{E}(M_\sigma)$ for $\sigma = \inf(t \geq 0 : |B_t| = 1)$.
Let $B = (B_t)_{t \geq 0}$ be a standard Brownian motion and set $M_t =  4B_t^2 + e^{4B_t -8t} -4t$. Find $\mathbb{E}(M_\sigma)$ for $\sigma = \inf\{t \geq 0 : |B_t| = 1\}$.

I have shown that $M_t$ is a martingale and am looking to use the optional stopping theorem. I can see that

$|M_{t \wedge \sigma}| = |4B_{t \wedge \sigma}^2 + e^{4B_{t \wedge \sigma} -8({t \wedge \sigma})} -4({t \wedge \sigma})|$

$\hspace{1.25cm}$ $\leq 4|B_{t \wedge \sigma}^2| + |e^{4B_{t \wedge \sigma}}| +4|{t \wedge \sigma}|$

$\hspace{1.25cm}$ $\leq 4 + e^4 +4|{t \wedge \sigma}|$.

I'm not entirely sure how to deal with the $|{t \wedge \sigma}|$ part. If I can find a bound for this part then I can use the optional stopping theorem to conclude that

$\mathbb{E}(M_\sigma) = \mathbb{E}(M_0) = 1$.

I feel there might be something obvious that I'm missing but I'm not too sure, thanks in advance.

答: You can show that $B_t^2-t$ is a martingale, so $0 = \mathbb{E}[B_{t \wedge \sigma}^2 - t\wedge \sigma]$ and hence $\mathbb{E}[t \wedge \sigma] = \mathbb{E}[B_{t \wedge \sigma}^2] \le 1$.  By the monotone convergence theorem, we have $\mathbb{E}[\sigma] = \lim_{t \rightarrow \infty} \mathbb{E}[t \wedge \sigma] \le 1$ and therefore $$|M_{t \wedge \sigma}| \le 4 + e^4 + 4(t \wedge \sigma) \le 4 + e^4 + 4 \sigma.$$  Now we can use the dominated convergence theorem to conclude $$\mathbb{E}[M_\sigma] = \lim_{t \rightarrow \infty} \mathbb{E}[M_{t \wedge \sigma}] = \lim_{t \rightarrow \infty} \mathbb{E}[M_0] = 1.$$


问:
Find the constant $c$ if you have a given limit

If $\displaystyle\lim_{x\to11}\frac{|x^2-6x|-|x^2+cx|}{x-11}.=22$, find the value of the constant $c$.

I have a problem getting the correct result which is $-16$. I tried absolutely everything and the closest I got was $-6$. My idea was to just check limits from both sides (although one side would be enough because I was sure an answer is a final number, thus I chose $11$ from the positive side so the absolute values cancel out easier). When I simplify as much as I can, I really don't get anywhere. I don't know what to do with the $x-11$ in the denominator.

答:
Note that near $x=11$ one has $x^2-6x>0$ so $|x^2-6x|=x^2-6x$. If we also had $x^2+cx>0$ near $x=11$, then the numerator would simplify to $-(6+c)x$ and the limit of the entire fraction would not exist. Therefore $c$ must be such that $x^2+cx<0$ near $x=11$. Then the original expression is equal to
$$
\frac{2x^2-x(6-c)}{x-11}=\frac{x(2x-(6-c))}{x-11}.
$$
The singularity will cancel out only if $2x-(6-c)=0$ for $x=11$.
Thus
$$
22-(6-c)=0
$$
and solving for $c$ we get $c=-16$.


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
I'm not sure, but to me your specific question doesn't seem to have been given the simple answer to why assuming Pythagoras as an axiom is wrong in that situation.

The reason is: because you're actually being asked "Given the set of axioms you've been taught, derive Pythagoras." The question implicitly assumes some particular axiom system.

In general a proof could be considered formally a set of symbols obeying some rules (of logic) which begins with a set of axioms and assumptions and ends with the statement you want to prove.

Title: What is a proof?


问:
Simplifying $\sqrt{34+15\sqrt2}$
$$\sqrt{34+15\sqrt2}$$
If we want $34+15\sqrt2$ to be a nice square $(a+b)^2=a^2+2ab+b^2$, most likely it is the case that $15\sqrt2$ corresponds to $2ab$. I don't know what to do from here. Is there any general approach that I can use to determine if it can be further simplified?

答:
Should be $\left(a+\sqrt{2} b\right)^2=34+15 \sqrt{2}$

$a^2+2b^2=34;\;2\sqrt{2}ab=15\sqrt 2\to 2ab=15\to b=\frac{15}{2a} $

$a^2+2\left(\frac{15}{2a}\right)^2=34$

$2 a^4-68 a^2+225=0$

which has no nice solutions, thus $\sqrt{34+15 \sqrt{2}}$ cannot be simplified further.

Title: Simplifying $\sqrt{34+15\sqrt2}$


答:
As an exploratory tactic,

the left hand side would be $O(x)$ if $c \le -11$

and the right hand side looks like $2 \times 11$

so I might start by considering whether there was a $c$ which gave $x^2-6x +x^2+cx = 2x(x-11)$

and there is one, namely $c=-16$.  Checking this, it works.

Looking for other solutions

If $c>-11$ then the left hand side is the limit of $\frac{x^2-6x-x^2-cx}{x-11} = \frac{-x(6+c)}{x-11}$

which cannot have a finite limit as $x \to 11$ when $c\not=-6$

and has a zero limit when $c=-6$.

So that fails and there is no other solution

Title: Find the constant $c$ if you have a given limit

Body:
If $\displaystyle\lim_{x\to11}\frac{|x^2-6x|-|x^2+cx|}{x-11}.=22$, find the value of the constant $c$.

I have a problem getting the correct result which is $-16$. I tried absolutely everything and the closest I got was $-6$. My idea was to just check limits from both sides (although one side would be enough because I was sure an answer is a final number, thus I chose $11$ from the positive side so the absolute values cancel out easier). When I simplify as much as I can, I really don't get anywhere. I don't know what to do with the $x-11$ in the denominator.