问:
Real world applications of Pythagoras' Theorem
I have a school assignment, and it requires me to list a few of the real world applications of Pythagoras Theorem. However, most of the ones I found are rather generic, and not special at all.

What are some of the real world applications of Pythagoras' Theorem?

答:
Say you're playing frisbee with a few of your friends, and the frisbee gets stuck in a tree. You want to get a ladder to reach it, but you don't know how long the ladder needs to be. You can make a point where you want the ladder to touch the ground, and then measure from there, whereas $ds^2=dx^2+dy^2$ is true, so we can say, as an example, that $dx=3 \text{m}$ long and $dy=4\text{m}$ long. Now, we must solve for $ds$. Here's my calculations:

$ds^2=3^2+4^2=9+16=25$

$\sqrt{ds^2}=ds$

$\sqrt{25}=5$

$ds=5\text{m}$

You'd need a ladder that is $5\text{m}$ long to reach the frisbee.


问:
Show any arbitrary interval has a zero in it
Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$ and whenever $f(a)=f(b)=0$ for some $0\le a,b\le 1$, there exists a $c\in(a,b)$ such that $f(c)=0.$ Show that for all arbitrary open intervals $(y,z)\subseteq(0,1),$  we have $(y,z)$ contains a zero of $f$. Does anyone have any hints

答: Start with the two given zeroes of $f$, $0$ and $1$. Then you get $c \in (0,1)$, a zero of $f$. Now, if $c \notin (x,y)$, $(x,y)$ must be contained in either $(0,c)$ or $(c,1)$ (why?). WLOG, let it be inside $(0,c)$, now we find $d \in (0,c)$, another zero of $f$ and continue like this. Why can this not contninue forever? (Hint: Look at the length of $(x,y)$ and the lengths of the constructed intervals in each stage).

ContentLicense: CC BY-SA 4.0


问:
Definite integration by substitution problem with the substitution $x=e^u−e^{−u}$

I'm stuck at the following integration problem:

a) Given that $x = e^u - e^{-u}$, find an expression for $e^u$ in terms of $x$.

b) Use the substitution $x = e^u - e^{-u}$ to find the exact value of $\int_0^1 \frac {1} {\sqrt {4+x^2}}\, dx$.

My textbook says that the solutions are:

a) $e^u=\frac {x+\sqrt{x^2+4}}{2}$

b) $\ln\frac {1+\sqrt5}{2}$

I have tried some algebraic manipulation in task a) but this did not bring me far. The textbook solutions looks pretty much like the result of the quadratic formula but I do not know if this is a coincidence or not.

I did not understand b) either since I lack the understanding of task a). I might be able to solve it because you could use the substitution $u =\frac x 2 $ because this nicely leads to $\int_0^1 \frac 1 {\sqrt{u^2+1}}\,du $ and then I one could trig-substitution but I cannot solve it with the given substitution.

Could somebody tell me I can solve or even approach the problem?

Body: For b) note $\frac{dx}{\sqrt{x^2+4}}=\frac{(e^u+e^{-u})du}{\sqrt{e^{2u}+2+e^{-2u}}}=du$, so the integral is $\int_0^{\ln\tfrac{1+\sqrt{5}}{2}}du=\ln\tfrac{1+\sqrt{5}}{2}$.


问:
Solve differential equation $\frac{dT}{dz} = \beta(T(z) - T_{0})$
The supposed solution is $T(z) - T_{0} = C_{1}exp(-\beta z) + C_{2}$ but I don't know how to get to it. Any help would be appreciated.

答:
$$\frac{dT}{dz} = \beta(T(z) - T_{0})$$

$$\int \frac{dT}{T(z)-T_0} = \int \beta \, dz$$

$$\log (T - T_0) = \beta z + \text{const.}$$

$$ T - T_0 = \exp(\beta z + \text{const.}) = C_1 \exp(\beta z).$$


问:
I'm teaching a college geometry course. What should I cover?
I've been asked to teach "Foundations of Geometry" at the University of South Carolina. Apparently, professors in the past have all done very different things, and I have a lot of choice in the matter. What course would you like to see?

Some constraints:

I want to say at least a little something about (1) the axiomatic approach (e.g., Euclid's axioms), (2) the modern approach (cool theorems such as Ceva's theorem, the nine point circle, etc., etc.), (3) the constructive approach, with straightedge and compass.

My students differ in background and motivation. Most of them are prospective high school math teachers. Typically they have seen some proofs, but not a lot; for example, they may have proved that the sum of two odd numbers is even. Most of the students won't yet have taken analysis, or algebra, or any other course obliging them to work really hard.

It has been a long time since I have seriously dealt with the subject, so I will definitely want a good book (or books) or other materials to follow.

I won't say anything about projective or other non-Euclidean geometry, because there's another course for that.

I asked a similar question on MathOverflow, but I was a little bit spooked by the answers. I'm sure that the books of Hartshorne and Hadamard are excellent, but I suspect these might be better for stronger students. (And the Hartshorne book discusses "geometry over fields", etc., and most of my students won't have taken abstract algebra.) Some respondents rolled their own solutions -- but are there really not good source materials out there?

My colleague taught a course out of Isaacs' book -- but he really went the extra mile (more like the extra ten miles). He said that the exercises were a bit too difficult, and that he was constantly having to give the students hints, and he slaved over writing up complete solutions. I believe this would not work as well for me as it would for him: he has a very approachable demeanor that I haven't (yet) been able to duplicate, and I'm afraid the students would be unlikely to come to my office hours no matter how much I encouraged them.

I've also looked at other books -- Coxeter and Greitzer is very cool, but perhaps more naturally suited to hotshot high school kids. Posamentier looks promising, a review copy is on its way. Clark's book seems to be somewhat off the beaten path (where I'm not quite sure I want to follow), but it looks very interesting and I ordered a review copy of that also. Other suggestions, big or small, would be very welcome. Thank you so much! --Frank

答:
I personally thought knowing the difference between Euclidean and Non-Euclidean geometry was a big plus in really understanding the axiomatic approach and the idea of mathematics. Especially learning hyperbolic geometry with the Pioncare-model was so intuitive and easy to follow.

I was particularly not a smart student and I have been teaching for 7 years and learning some more math for myself, so I think I know a good book or two. I seriously recommend "Geometry: Euclid and Beyond" from Hartshorne.

ContentLicense: CC BY-SA 3.0


问:
How to solve $x = ab^ x$?
I'm not enough familiar with matrix manipulations, I hope this question is admissible here

Given vectors:
$$a=\begin{bmatrix} a_1 \\\vdots \\a_n\end{bmatrix},\qquad
  b=\begin{bmatrix} b_1 \\\vdots \\b_n\end{bmatrix}$$

where $a^T b = 1$, I want to find the vector $x$ such that:

$$
x = a b^T x
$$

What are the symbolic steps to solve this equation? Thank you!

答:
You are looking for a solution of the system $(I-ab^T)x=0$.

From the following two observations and the rank–nullity theorem, the null space of $I-ab^T$ is one dimensional:

Since $\text{rank}(I) \leq \text{rank}(I-ab^T) + \text{rank}(ab^T)$, $\text{rank}(I-ab^T) \geq n-1$.

Since $a^Tb=1$, $I-ab^T$ is not invertible, so $\text{rank}(I-ab^T) \leq n-1$.

So it suffices to find one nontrivial solution. The matrix $ab^T$ has eigenvector $a$ with eigenvalue $a^Tb=1$. Therefore $a$ is also an eigenvector of $ab^T-I$ (and $I-ab^T$) with eigenvalue $0$.

Title: How to solve $x = ab^ x$?


问:
Fish tank population modelling using differential equations
I have been given the basic differential equation for population modelling, which is
$$p'(t)=rp(t)\left(1-\frac{p(t)}k\right)$$. The coefficient $r$ is $10$ and constant $k$ is $500$, so  the carrying capacity is $500$ fish.
Each fish causes $5$ dollars of expenses per year, and the fish tank costs $1200$ dollars per year to operate. Each sold fish brings $15$ dollars of revenue.

I have been given the task of finding out the best way to harvest the fish without destroying the population and also with highest possible revenue. The assignment says that you can either harvest the fish certain fixed number evey year or harvest the fish based on the population with a certain multiplier.

Assuming that we use the constant amount $700$ and the multiplier $5$, we can get two different equations that model the fish population absed on two different strategies.

$$p'(t)= rp(t)\left(1-\frac{p(t)}k\right)-C$$ ($C$ being constant)

$$p'(t)=rp(t)\left(1-\frac{p(t)}k\right)-Bp(t)$$ ($B$ being a multiplier)

When we include the numerical values, we get:

$$p'(t)= 10p(t)\left(1-\frac{p(t)}{500}\right)-700\\
p'(t)=10p(t)\left(1-\frac{p(t)}{500}\right)-5p(t)$$

So far so good. But my problem is that how am I supposed to make a differential function that shows the rate of revenue?
My attempt was the following:

$$v'(t)=15 \cdot 700-5p(t)-1200\\
v'(t)= 15 \cdot p(t)-5p(t)-1200$$

My other attempt:

$$v'(t)= 15 \cdot 700t-5p(t)-1200t\\
v'(t)= 15 \cdot 5p(t)-5p(t)-1200t$$

I am making this with matlab, but the basic calculations are the thing that bugs me. Have I even done the first parts correctly? How can I form the revenue functions so that they are consistent? Any help would be appreciated.

答:
$v'(t) = \frac {dv}{dt}$ is how much $v(t)$ changes each year. I.e., it is how much profit you make in year $t$.

Consider the constant-sell strategy. In a given year, you sell $C$ fish. That brings in $15C$ in income. You spend $5p(t)$ on upkeep of the fish and $1200$ on upkeep of the tank.
So your profit is income less outgo:
$$v'(t) = (15C) - (5p(t) + 1200)$$

In your second attempt, you get the same result in year one, but in year $t = 2$, you are claiming to bring in $15C\cdot 2 = 30C$ in income, and $15C\cdot 3 = 45C$ in year $3$. Seems a pretty sweet deal, though the equally rising tank maintenance fees are taking a chunk out of it. At least the cost of feeding the fish isn't also rising each year.


问:
Procedure to show martingale property
When one's trying to prove the martingale property (assuming it's adapted and integrable), why does it suffice for a stochastic process $(X_t)_{t\geq0}$ satisfy
$$ \mathbb{E}[(X_t-X_s)\prod_{i=1}^n f_i(X_{s_i})]=0,$$
where $f_i$ are bounded continuous functions and $s_1\leq s_2\leq\dots s_n\leq s<t$.

Is it a simple application of the Monotone Class theorem or is the more/less to it?

答:
From the functional form of the montone class theorem, you can conclude immediately from the identity you have displayed that
$$
\Bbb E[(X_t-X_s)\cdot F]=0
$$
for all bounded $\mathcal F_s:=\sigma(X_u: 0\le u\le s)$-measurable $F$. (See Theorem 1 here: https://almostsuremath.com/2019/10/27/the-functional-monotone-class-theorem/.)

Title: Procedure to show martingale property


问:
How to prove ${2n \choose n} = 2\cdot (2n-1) \cdot \frac{1}{n} {2(n-1) \choose n-1}$?
I've been staring at this identity which appears in my textbook for a while. Plugging in numbers I can verify that this is true, however I have no idea how this was determined or proved? Is there a nice result about binomial coefficients that gives this identity?

The only thing that comes to mind is a creative application of Pascal's identity, but even then I'm getting the result.

答:
Recall that $\displaystyle{a \choose b} = \frac{a \cdot (a-1) \cdots (a-b+1)}{b!}$

The denominator of $\displaystyle{2n \choose n}$ is $n!$.

The denominator of $\displaystyle{2(n-1) \choose n-1}$ is $(n-1)!$.

So $\displaystyle\frac{1}{n} {2(n-1) \choose n-1}$ has the same denominator as ${2n \choose n}$.

The numerator of $\displaystyle{2n \choose n}$ is $2n\cdots (n+1)$.

The numerator of $\displaystyle{2(n-1) \choose n-1}$ is $(2n-2)\cdots n$.

So $\displaystyle\frac{(2n)(2n-1)}{n}{2(n-1) \choose n-1}$ has the same numerator as ${2n \choose n}$.


问:
A discrete subset $Y$ of a topological space $X$ which is Haussdorff is closed?
Question is in the title. Let $X$ be a topological space which is Hausdorff, and let $Y \subset X$ be a subset, so that the induced topology on $Y$ is discrete. Is it then true that $Y$ is closed in $X$?

I have the impression that this is only true if $X$ is locally compact, but I feel like I could be missing something.

答: Any separable metric space can be embedded into $[0,1]^\omega$, the Hilbert cube. $\Bbb Z$ too. Can the image be closed?


Title: How to prove ${2n \choose n} = 2\cdot (2n-1) \cdot \frac{1}{n} {2(n-1) \choose n-1}$?

Body:
I've been staring at this identity which appears in my textbook for a while. Plugging in numbers I can verify that this is true, however I have no idea how this was determined or proved? Is there a nice result about binomial coefficients that gives this identity?

The only thing that comes to mind is a creative application of Pascal's identity, but even then I'm getting the result.

Body:
$$2(2n-1)\frac1n\binom{2n-2}{n-1}=\frac{(2n)(2n-1)}{n^2}\frac{(2n-2)!}{(n-1)!(n-1)!}
=\frac{(2n)!}{n!n!}=\binom{2n}n.$$


答: Here's another approach. Set $A=ab^T$. Then $A$ is idempotent: $$A^2=a(b^Ta)b^T=(b^Ta)ab^T=ab^T=A$$ The column space of any idempotent matrix is precisely equal to the set $$W=\{x:Ax=x\}$$ Here's a quick proof. If $x\in W$ then $$x=Ax\in \text{Col}(A)$$ which shows $W\subseteq \text{Col}(A)$. Meanwhile, if $x\in \text{Col}(A),$ find $v$ so that $Av=x$. Then $$Ax=A(Av)=A^2v=Av=x\in W$$ Hence $W=\text{Col}(A)$. The column space of $A=ab^T$ is clearly $\text{span}\{a\}$ as every column of $A$ is a multiple of $a$ and $b$ is non$-$zero. This means the solution set to $x=ab^Tx$ is precisely equal to the span of $\{a\}$.

Title: How to solve $x = ab^ x$?

Body:
I'm not enough familiar with matrix manipulations, I hope this question is admissible here

Given vectors:
$$a=\begin{bmatrix} a_1 \\\vdots \\a_n\end{bmatrix},\qquad
  b=\begin{bmatrix} b_1 \\\vdots \\b_n\end{bmatrix}$$

where $a^T b = 1$, I want to find the vector $x$ such that:

$$
x = a b^T x
$$

What are the symbolic steps to solve this equation? Thank you!


问:
Show that if $M $ is orthogonal matrix, then $\det M =\pm 1$.
I write like following my prove, and I need to know if it's correct:

If P is an orthogonal matrix, then $$PP^t = I = P^tP$$ Therefore, $$\det(P^tP) = \det (I) \implies \det(P) \det (P^t) \implies \det (P) \det (P) = [\det (P)]^2$$   So $$ [\det(P)]^2 = 1 \implies \det (P) = \pm \sqrt 1 $$  So  $$ \det (P) = \pm 1$$

I prove it writing this, I would like to know if this proof is correct.

答: Since $M$ is orthogonal, we have $$MM^T = I$$ It gives $$\det(MM^T) = \det(I) = 1$$ Therefore, $$\det(M) \det(M^T) = 1$$ now since $\det(M^T) = \det(M)$, we get $$\det(M)^2 = 1 \implies \det(M) = \pm 1$$


问:
Computing logarithm $\log{2}+\log{\left(\frac{p\left(x\right)}{p\left(x\right)+q\left(x\right)}\right)}$
I have the following logarithm to compute,

$$\log{2}+\log{ \left( \frac{p \left( x \right)}{p \left( x \right)+q \left( x \right)} \right)}$$

Then, in the solution,

$$\log{ \left( \frac{p \left( x \right)}{\frac{p \left( x \right)+q \left( x \right)}{2}} \right)}$$

How 2 got inside the log?

Body:
Recall that $\log\left(a\right)+\log\left(b\right)=\log\left(ab\right)$:
$$\log\left(2\right)+\log\left(\frac{p\left(x\right)}{p\left(x\right)+q\left(x\right)}\right)=\log\left(\frac{2p\left(x\right)}{p\left(x\right)+q\left(x\right)}\right)\text{,}$$ which is equivalent to the expression given in the solution:
$$\log{\left(\frac{p\left(x\right)}{\frac{p\left(x\right)+q\left(x\right)}{2}}\right)}=\log{\left(\frac{p\left(x\right)}{1}\div\frac{p\left(x\right)+q\left(x\right)}{2}\right)}=\log{\left(\frac{p\left(x\right)}{1}\times\frac{2}{p\left(x\right)+q\left(x\right)}\right)}$$ $$=\log\left(\frac{2p\left(x\right)}{p\left(x\right)+q\left(x\right)}\right)=\log\left(2\right)+\log\left(\frac{p\left(x\right)}{p\left(x\right)+q\left(x\right)}\right)$$


问:
What is formula for repeating sequence $0, 0, 1, 1, 0, 0, 1, 1, ...$
can somebody help me with this? I was searching through the web and I did not find anything useful. What is the formula for the sequence:

$0, 0, 1, 1, 0, 0, 1, 1, ...$

or

$0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, ...$

I did find solution for $0,1,0,1$ or $1,0,1,0$ or $0,0,0,1$ etc.. But nothing close to this.

Thank you in advance!

EDIT:
Sorry everyone, I am quite new to this. It is a repeating infinite series. Also, I am looking for a more general solution so I can choose the number of zeros and ones - if $n = 3$ then I am looking for a repeating sequence:

$0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, ...$

EDIT2:
Example: I input let say "$k$" into a formula. I'll get $0$ or $1$ as an output. I use "$n$" as a variable for a function that will determine the repairing pattern e.g. If $n=2$ then the pattern is $0,0,1,1,0,0,1,1,0,0,1,1, ...$ and if $k=3$, then the output will be the third position, which is $1$ in this case.

答:
The simplest way:
$$
a_k=\frac{1\pm(-1)^{\left\lfloor\frac {k-1}n\right\rfloor}}2,
$$
where $n$ is the half-period of the sequence and with $\pm$ sign you can choose to start with $0$ or $1$.


问:
Probability of number formed from dice rolls being multiple of 8

A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.

I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$.

The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"

What would be a faster way to solve this using number theory?

答:
Here's another approach. As noted in the problem statement, it all comes down to the last $3$ rolls concatenating to a multiple of $8$.

The last two rolls have to concatenate to a multiple of $4$.

We break these into two groups: The last two are a multiple of $8$, or not:

Group 1: $16, 24, 32, 56, 64$.

Group 2: $12, 36, 44, 52$.

To complete an entry in group 1 to a multiple of $8$, the leftmost digit must be even ($3$ choices). To complete an entry in group 2 to a multiple of $8$, the leftmost digit must be odd ($3$ choices).

So altogether there will be $3\cdot 5 + 3\cdot 4 = 27$ multiples of $8$.

This gives a probability of $\frac{27}{216}=\frac18$.


问:
Prove that $H$ is not a subspace of the $C(-\infty, \infty)$

Show that $$H = \{f(x)\, | \, f \in C(-\infty, \infty) \land f(x) \geq 0\, \forall x \in \mathbb{R}\}$$ is not a subspace of $C(-\infty,\infty)$.

I know that to be a subspace $H$ needs three things:

$H$ needs to include the zero vector

$H$ needs to be closed under vector addition

$H$ needs to be closed under multiplication by scalars

However I am lost when reading the question. I don't even understand the notation. Any help is welcome.

答: Since $f\neq 0\in H$ implies $-f\notin H$, it is not a subspace.


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
We can show a more general result: a (growing) exponential grows faster than any power. Eventually,

$$a^x>x^b.$$

Indeed, taking the logarithm, this is equivalent to

$$x\log a>b\log x$$

or

$$\frac x{\log x}>\frac b{\log a}.$$

The LHS is an unbounded function and the RHS a constant. (For $x>e$,

$$\frac{ex}{\log ex}=e\frac{x}{\log x+1}=e\frac{\log x}{\log x+1}\frac{x}{\log x}\ge\frac e2\frac{x}{\log x}.)$$


问:
Total convergence of a function series
Let us consider the following function series:

$$ \sum_{n=1}^{\infty} \frac{1}{n^x + n^{-x}}.$$

It seems that there is total convergence in $(- \infty, -1-\delta] \cup [1+ \delta, + \infty)$ for $\delta > 0$. Thank-you in advance for any help!

答:
Mathematica 12.2 does it in such a way:

```<code>SumConvergence[1/(n^(1 + \[Delta]) + n^(-1 - \[Delta])), n,
Assumptions -> n >= 1 && \[Delta] > 0, Method -> "IntegralTest"]
</code>```

True

The absolute convergence is obvious since all the terms are positive for each real x.
It is its weakness that the commands

```<code>SumConvergence[1/(n^x + n^-x), n, Assumptions -> x > 0 && n >= 1, Method -> "IntegralTest"]
</code>```

and

```<code>SumConvergence[1/(n^(1 - \[Delta]) + n^(-1 + \[Delta])), n,Assumptions -> n >= 1 && \[Delta] > 0 && \[Delta] < 1, Method -> "IntegralTest"]
</code>```

return the inputs.

Title: Total convergence of a function series

Body:
Mathematica 12.2 does it in such a way:

SumConvergence[1/(n^(1 + \[Delta]) + n^(-1 - \[Delta])), n,
Assumptions -> n >= 1 && \[Delta] > 0, Method -> "IntegralTest"]

True

The absolute convergence is obvious since all the terms are positive for each real x.
It is its weakness that the commands

SumConvergence[1/(n^x + n^-x), n, Assumptions -> x > 0 && n >= 1, Method -> "IntegralTest"]

and

SumConvergence[1/(n^(1 - \[Delta]) + n^(-1 + \[Delta])), n,Assumptions -> n >= 1 && \[Delta] > 0 && \[Delta] < 1, Method -> "IntegralTest"]

return the inputs.


问:
Numerical Approximation: Integral of Exponential to an Exponential power
I am trying to solve a self-consistent equation for the variable $S$.

$$
\mathbf{S} = \int_0^t C_1 \mathbf{S} e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da \\
%
1 = \int_0^t C_1 e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da
$$

I would input values of $C_1, C_2, C_3$ and $C_4$ and I am interested in finding the $\mathbf{S}\neq 0$ that satisfies the above equation.

Things I have tried:

I am thinking numerical approximation but I am unfamiliar with the methods used and not sure what to read up on. Perhaps Runge-Kutta or Eulers method? Would appreciate any tips.

I attempted Bisection Method using a Matlab code, trying to solve for the zero of $\int_0^t C_1 e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da-1$ but it was giving me an error (I think because of the integral.)
Here is the bisection code:

```<code>% Clearing Screen
clc

% Setting x as symbolic variable
syms x;

% Input Section
y = input('Enter non-linear equations: ');
a = input('Enter first guess: ');
b = input('Enter second guess: ');
e = input('Tolerable error: ');

% Finding Functional Value
fa = eval(subs(y,x,a));
fb = eval(subs(y,x,b));

% Implementing Bisection Method
if fa*fb > 0
    disp('Given initial values do not bracket the root.');
else
    c = (a+b)/2;
    fc = eval(subs(y,x,c));
    fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');
    while abs(fc)>e
        fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);
        if fa*fc< 0
            b =c;
        else
            a =c;
        end
        c = (a+b)/2;
        fc = eval(subs(y,x,c));
    end
    fprintf('\nRoot is: %f\n', c);
end
</code>```

In case it is of interest:
I am vaguely familiar with python, Matlab, C++. More familiar with Mathematica.

答:
Just an observation, too much for a comment.

It seems the behavior at $t=0+$ will take a delicate bit of analysis: The left hand side of the equation, the constant $1$, will have to be "balanced" on the right hand side by an integrand that is, or becomes, singular in order for $\lim_{t\rightarrow 0+} \int_0^t$ to keep up, so to speak. A back of the envelope estimate seems to say that for small $t$, $S(t) = C_3^{-1}\log(C_1t) + O(t\log t)$. If nothing else this means any quadrature you end up doing ought to account for this sort of singularity in $S$.


问:
Finding upper and lower limits
What is the upper and lower limit of the given sequence $x_{n}=2n-7\left \lfloor\frac{2n}{7} \right \rfloor, n\in \mathbb{N}$? Again, I'm rather new to real analysis and I'd appreciate it if you explained the steps to find them.

Title: Finding upper and lower limits

Body:
hint

$$x_{n+7}=2(n+7)-7\lfloor \frac{2n+14}{7}\rfloor$$

$$=2n+14-7(\lfloor \frac{2n}{7}\rfloor +2)=x_n$$

So
$$\{x_n, n\in\Bbb N\}=\{x_1,x_2,...,x_7\}$$


问:
Holomorphic vs differentiable (in the real sense).
Why a holomorphic function is infinitely differentiable just because of satisfying the Cauchy Riemann equations, but on the other side, a two variable real function that is twice differentiable is not infinitely differentiable?

I'm asking this for two reasons:

1) $\mathbb{R}^2$ is somehow an analog to the complex plane (they are isomorphic).

2) Holomorphic means that is differentiable in any direction of the complex plane, so why if a function is differentiable in any direction of the real plane then it is not infinitely differentiable?

答: Note that an analytic, complex function $w=f(z)$ implies a function $F:\mathbb R^2\rightarrow\mathbb R^2$ by $r+is=f(x+iy)$.  The reverse is not true, however.  It is not the case that just any differentiable $F:\mathbb R^2\rightarrow\mathbb R^2$ implies a complex, analytic function.  On the contrary, the Cauchy-Riemann equations must be satisfied.  Thus, the real an imaginary parts are tightly tied together.


答: $2n-7\left\lfloor\dfrac{2n}{7}\right\rfloor$ is the remainder you get when you divide $2n$ by $7$. As $n$ increases by $1$, $x_n$ will increase by $2 \mod 7$.

Title: Finding upper and lower limits

Body: What is the upper and lower limit of the given sequence $x_{n}=2n-7\left \lfloor\frac{2n}{7} \right \rfloor, n\in \mathbb{N}$? Again, I'm rather new to real analysis and I'd appreciate it if you explained the steps to find them.


问:
Why we have the formula for covariance the way it is?
I was going through the definition and meaning of variance and covariance. The resources which I have, have only definition and formula without any insight.

For variance, I wrote the formula and asked myself what this formula tells me. I figured out that variance has do to something with the mean of spread. The term in the formula of variance, $(x_i-\bar x)^2$, takes the square of the distance between the $i^{th}$ observation and mean.

Now moving further if we get data on two attributes, for this we can plot a scatter diagram. I compared this situation with the notion of centre of mass in 2D. The mean of the scatter plot will be $(\bar x,\bar y)$. Covariance is expanding the idea of variance to higher dimensions(this was given in book.).

$$\mathbb{Cov}(X,Y)=\mathbb{E}[(x_i-\bar x)(y_i-\bar y)]$$

Above is the formula of covariance. I could not understand why this particular formula?

Second, if we have a point $(x_i,y_i)$ in scatter plot and we have our mean as $(\bar x,\bar y)$ then the square of distance between them will be $(x_i-\bar x)^2+(y_i-\bar y)^2$. So I thought that if we are generalising the concept of variance in two dimensions then we should have the formula of variance as:

$$\mathbb{Cov(X,Y)}=\mathbb{E}[(x_i-\bar x)^2+(y_i-\bar y)^2]=\mathbb{Var(X)}+\mathbb{Var(Y)}$$

Summary of my problem

In the formula of variance we have "square of distance between ith observation and mean" and covariance is same as variance but for two or more dimension then why we are not using "square of distance between the ith observation and mean"? Why we are using something else and in particular why that formula?

Please help I am struggling to digest it. Please.

答:
Covariance of two random variables $X$ and $Y$ is supposed to measure how they covary.

For instance, take two independent dice rolls, and let the result of the first roll be $X$ and the sum of both rolls be $Y$. If $X$ is higher than expected, then $Y$ is probably also higher than expected. If $X$ is lower than expected, then $Y$ is probably also lower than expected. We want this to lead to a positive covariance.

For another example, we again consider two independent dice rolls, $X$ is still the first roll, but $Y$ is now the second roll minus the first roll. This time, if $X$ rolls higher than expected, $Y$ is probably lower than expected, and if $X$ is lower than expected, then $Y$ is probably higher than expected. $X$ and $Y$ vary in opposite directions, so to speak. We want this to lead to a negative covariance. Positive covariance means that the two random variables vary in the same direction, while negative covariance means that they vary in different directions. For this, covariance must allow for negative values, which your formula already doesn't. Basically, yours just measures how much they vary independently, not how they covary.

Now for why this particular formula was chosen, there are multiple reasons. For one, it'd be nice if the covariance of the random variable with itself would just be its variance. It covaries with itself just the way it varies in general. The usual definition accomplished this: $\operatorname{Cov}(X,X)=\operatorname{Var}(X)$. Another reason is that the usual definition has some pretty cool algebraic properties. It's linear in both variables. It's also symmetric and positive definite in the sense that only almost surely constant random variables have nonzero covariance with themselves. Taken together, this means that covariance is an inner product on suitable spaces of random variables, with all the cool results that brings, like the Cauchy-Schwarz inequality.


问:
Counting Questions about Position
Lets say we have a student organization of 8000 students. 5000 are undergrads and 3000 are master students. There is an organization that has a president position, and  three VP positions.

a) How many ways are there to choose people for this organization from all students?

b) How many was are there to choose so that at least one master student and one undergrad get a position?

c) How many was are there to choose so that the VP position has to be an undergrad and president position can be undergrad or masters.

d) How many was are there to choose so that at least one master student is VP?

I am really new to counting and I was wondering if someone could tell me if my logic is right on these.

a) $8000 \cdot 7999 \cdot 7998\cdot7997$ because you choose someone for the first position and then loose that person from the group and continue until you fill all positions.

b) $4! (\binom{5000}{1}\cdot \binom{3000}{1}\cdot \binom{7998}{2})$ because we choose undergrad for one position, master for the second, and then for the rest we can choose from the rest of the students. There is $4!$ ways to arrange this.

c) $3!\bigg( \binom{5000}{3} \bigg) \cdot \binom{7997}{1}$ We choose three undergrads for the VP positions and then there is $3!$ ways to arrange this and then we choose from the left over group to fill the president position.

d) $\binom{8000}{4} - \binom{5000}{4}$ so we are taking all the positions that have undergrads and master students and then subtracting the total positions without master students.

答:
For (b), $\binom{5000}{1}\binom{3000}{1}\binom{7998}{2}$ is indeed
correct for the number of ways to select committee members. But given
four members, there are not $4!$ possible committees. There are only
$4$ since each committe is completely determined by who the president
is, and there are $4$ choices for president.

My apologies, this is incorrect (thank you to G Aker for pointing this out). Since we are choosing from the master and undergrad pool twice, we are double-counting some committees. Instead, you should take the total number of committees and subtract the bad cases, i.e. the cases in which the entire committee is made up of master students or undergraduates. Then once you have the total number of possible committees, think of the possible position assignments.

Similar thing for (c). $\binom{5000}{3}\binom{7997}{1}$ already gives
a committee with $3 $ VP's and a president, so there is no more
re-arranging to do.

Same issue here.

(I am assuming that all of the VP positions are the same, i.e. switching two VP's results in the same committee. It's not quite clear from the problem, but if the three VP positions are distinct, then your answers for (b) and (c) are correct.)

For (d), you forgot to subtract the case when only the president is a master student.

Title: Counting Questions about Position


问:
What is $\mathbb{R}^{n+1}-\mathbb{R}^n$?
In C.H. Edwards's Advanced Calculus of Several Variables he defines the ordinate set $\mathcal{O}_f$ of a function $f:\mathbb{R}^n\to\mathbb{R}$ as the set of points between $\mathbb{R}^n$ and the graph of $f,$ including the points of evaluation, $\mathbf{x}\in\mathbb{R}^n$ and the points in the graph $\left\{\mathbf{x},f\left(\mathbf{x}\right)\right\}\in\mathbb{R}^{n+1}$.  Later on he defines a set $\hat{\mathcal{G}}=\partial\mathcal{O}_f-\mathbb{R}^n,$ where $\partial\mathcal{O}_f$ is the boundary of $\mathcal{O}_f.$  The intent seems clear.  First

$$\mathbb{R}^{n+1}-\mathbb{R}^n=\mathbb{R}^n\times\left(\mathbb{R}-\left\{0\right\}\right)$$

where $\times$ means Cartesian product.  Then

$$\hat{\mathcal{G}}=\left(\mathbb{R}^{n+1}-\mathbb{R}^n\right)\cap\partial\mathcal{O}_f.$$

But long ago I learned that $\mathbb{R}^n$ is the set of all real number n-tuples, and $\mathbb{R}^{n+1}$ is the set of all (n+1)-tuples, so elements of $\mathbb{R}^{n}$ are not elements of $\mathbb{R}^{n+1}$ and $\mathbb{R}^{n}$ is not a subset of $\mathbb{R}^{n+1}.$

So am I correct in concluding that $\mathbb{R}^{n+1}-\mathbb{R}^n$ is not really the relative complement of the two sets?

答: Technically you are correct, but it s customary to identify $\mathbb{R}^n$ with its natural embedding into $\mathbb{R}^{n+1}$, just as the real line is identified with the $x$-axis, (or the $y$-axis) in $\mathbb{R}^2$.


答:
a) If we assume all the VP positions are different then the first is correct. Otherwise you divide by 3! for the duplicate positions.

b) This is incorrect becuase you will be counting the same solutions multiple times. To see this consider there are 5 masters called a, b, c, d, and e and 3 undergraduates f, g, h. When you select your first master and first undergraduat say a and f then fill the rest of the group with say, b and g and permute all the combinations, well the same combinations could have been acheived if you initially picked b and g and filled the rest of the group with a and f.

I would approach this problem by taking your answer to part a. Then working out how many of those combinations have no masters student and subtracting that. Then subtract the number of ways you can have no undergraduate students to get your answer.

c) This is correct again assuming that the vp positions are all diferent.

d)This is wrong that would be the number of ways of having at least 1 master in some position if all the positions are the same. You haven't dealt with the various permutations. Your also not interested in at least 1 master on the committee becuase the master must hold a vp position.

Title: Counting Questions about Position

Body:
Lets say we have a student organization of 8000 students. 5000 are undergrads and 3000 are master students. There is an organization that has a president position, and  three VP positions.

a) How many ways are there to choose people for this organization from all students?

b) How many was are there to choose so that at least one master student and one undergrad get a position?

c) How many was are there to choose so that the VP position has to be an undergrad and president position can be undergrad or masters.

d) How many was are there to choose so that at least one master student is VP?

I am really new to counting and I was wondering if someone could tell me if my logic is right on these.

a) $8000 \cdot 7999 \cdot 7998\cdot7997$ because you choose someone for the first position and then loose that person from the group and continue until you fill all positions.

b) $4! (\binom{5000}{1}\cdot \binom{3000}{1}\cdot \binom{7998}{2})$ because we choose undergrad for one position, master for the second, and then for the rest we can choose from the rest of the students. There is $4!$ ways to arrange this.

c) $3!\bigg( \binom{5000}{3} \bigg) \cdot \binom{7997}{1}$ We choose three undergrads for the VP positions and then there is $3!$ ways to arrange this and then we choose from the left over group to fill the president position.

d) $\binom{8000}{4} - \binom{5000}{4}$ so we are taking all the positions that have undergrads and master students and then subtracting the total positions without master students.


问:
Coordinate geometry: calculating the height of an equilateral triangle
If I have equilateral $\Delta ABC$ with A being $(-x,0)$ and B being $(x,0)$, how can I solve for the coordinates of C in terms of $x$?

I tried the following:

$2x^2 = x^2 + b^2 $ -- pythagorean thm, since we know that one side of the triangle is two times the length of half of the base.

$3x^2 = b^2$ -- simplification

$ x\sqrt3 = b $ -- so this says that the height of point C must be x\sqrt3 units.

This gives me an end result with the coordinates $(0, 2)$. However, frankly, this doesn't seem logical to me...what I'm basically saying with that result is that all equilateral triangles are two units tall...what?! This sounds completely incorrect.

Any help would be great; thanks.

edited: I actually had the right idea, I just made a stupid mistake with my simplification. ._.

答:
Draw a diagram and you will how the triangle works out.

In an equilateral triangle all sides are equal. Thus, the hypotenuse of your right angle triangle is $2x$. Let the third coordinate be represented by $(0,c)$. You should reason out as to why the $X$ coordinate of point $C$ is $0$.

Therefore, it follows that:

$c^2 + x^2 = (2x)^2$

Therefore, $c$ = ----

Hope that is helpful.


问:
Common vectors of a matrix and its orthogonal compliment
Say that we have $W$, which is a subspace of $\mathbb{R}^n$. Would $W$ and $W^⊥$ have any vectors in common?

答: $W$ and $W^{\perp}$ only have the zero in common, because if $y\in W\cap W^{\perp}$, then $y$ is orthogonal to itself.


问:
Volume of a solid using the shell method
The region bounded by $y= 4x^{2}$, $x=0$, $y=16$, about $y=16$

I used the shell and disk methods, but here I will present my work using shells:

$$2\pi\int_a^b {(16-y)(y^\frac{1}{2})} dy =\frac{8192\pi}{15}$$

with $a=0$ and $b=16$.

I used shells here and solved for my function in terms of $y$ which is $x= \left(\frac{y^\frac{1}{2}}{2}\right)$, but this only shows one half of the graph so I doubled it leaving me with just ${y^\frac{1}{2}}$. Maybe I made my mistake here because the answer should be exactly half of what I got.

If anyone spots my mistake please let me know

答: The height of the shell at $y$ is $\sqrt{\frac{y}4}=\frac12y^{1/2}$. The region being revolved about the line $y=16$ is bounded by the line $x=0$, so it lies entirely to the right of the $y$-axis, so no doubling is involved: the mirror image region in the second quadrant is not included.


问:
Composition of functions in inequalities
i'm doing a demonstration, and i am stuck in this part.
If i have an inequality such as:

$f(x)\leq g(x)$

Can i compose both sides? Like this, for example:

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

What would happen if $h(x)$ is monotonically decreasing? I think it would change the inequality, but I'm not sure.

答: This is true of $h$ is increasing. If $h$ is decreasing, the inequality would flip. If $h$ is not monotonic, you cannot compare the compositions.


答:
You have the base vertices of your equilateral triangle at ( -x, 0 ) and ( x, 0 ) , so the altitude of your triangle lies on the y-axis at what you are calling ( 0, b ) .  That makes the base of either of the right triangles you are using $ \ x \ $ , but then the hypotenuse of your triangle (a side of the equilateral triangle) has length $ \ 2x \ $ .  Therefore, the set-up using the Pythagorean Theorem must be

$$ (2x)^2  =  x^2 + b^2 . $$

You need to square that 2x ...

Title: Coordinate geometry: calculating the height of an equilateral triangle

Body:
If I have equilateral $\Delta ABC$ with A being $(-x,0)$ and B being $(x,0)$, how can I solve for the coordinates of C in terms of $x$?

I tried the following:

$2x^2 = x^2 + b^2 $ -- pythagorean thm, since we know that one side of the triangle is two times the length of half of the base.

$3x^2 = b^2$ -- simplification

$ x\sqrt3 = b $ -- so this says that the height of point C must be x\sqrt3 units.

This gives me an end result with the coordinates $(0, 2)$. However, frankly, this doesn't seem logical to me...what I'm basically saying with that result is that all equilateral triangles are two units tall...what?! This sounds completely incorrect.

Any help would be great; thanks.

edited: I actually had the right idea, I just made a stupid mistake with my simplification. ._.


答:
In fact, I would say that this doesn't really have to do with composition of functions at all. Suppose that $v$ and $w$ are real numbers with $v\le w$.

If $h$ is increasing, then we can conclude that $h(v) \le h(w)$.

If $h$ is decreasing, then we can conclude that $h(v) \ge h(w)$.

If neither, then we can't immediately draw any conclusion.

All of this holds with $v=f(x)$ and $w=g(x)$ just the same,  but it doesn't really have to do with the fact that $v$ and $w$ are values of two functions at the same point.

Title: Composition of functions in inequalities

Body:
i'm doing a demonstration, and i am stuck in this part.
If i have an inequality such as:

$f(x)\leq g(x)$

Can i compose both sides? Like this, for example:

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

What would happen if $h(x)$ is monotonically decreasing? I think it would change the inequality, but I'm not sure.


问:
Graph such that each path is induced
I would like to know the name, if there is one, for a graph with the property that every path in it is an induced graph (hence a snake). I'm also interested in any characterization of such a graph.</p>

答:
Given a simple, undirected graph $G$. Define $\mathcal{P}$ as the property that every path $P$ in $G$ is an induced subgraph of $G$. Suppose $G$ satisfies property $\mathcal{P}$. Further assume that $G$ contains a cycle $C_k$ on $k$ vertices with $V(C_k)=\{v_1,v_2,\cdots,v_k\}$ as an induced subgraph, then $P_{k}$ with $V(P_{k})=\{v_1,v_2,\cdots, v_{k}\}$ is a path in $G$. But $(v_{k},v_1) \notin E(P_{k})$ by definition of a path, hence $P_{k}$ is not induced. A contradiction. We can conclude that $G$ is cycle-free.

Therefore, the family $\mathcal{T}$ of graphs satisfying $\mathcal{P}$ are cycle-free (forests).

Title: Graph such that each path is induced


问:
Why does every sequence of real numbers have an upper limit?
Let $\{a_n\}:\mathbb{N}\to\mathbb{R}$, and we define $E$ to be the set of all sub-sequential limits of $\{a_n\}$ as well as possibly the symbols $\infty$ and $-\infty$ if there are some sub-sequences of $\{a_n\}$ that go to $\infty$ / $-\infty$.

Can $E$ be empty? In particular, since $\limsup_{n \to \infty}a_n := \sup E$, does the upper limit of a sequence of real numbers is always defined? From what I found on the Wikipedia page on limit superior and limit inferior, the answer is yes and this have something to do with the fact that $\mathbb{R}$ is complete, but I couldn't find the complete proof, and I'm a bit surprised the issue isn't discussed on Rudin's principles of mathematical analysis.

答: Let $\sigma=\langle a_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$. If $\sigma$ is unbounded, then it has a subsequence converging to $\infty$ or to $-\infty$, so $E\ne\varnothing$. If $\sigma$ is bounded, then by the Bolzano-Weierstrass theorem it has a convergent subsequence, so again $E\ne\varnothing$.


答:
Every sequence of real numbers has a monotone subsequence. This monotone subsequence converges to something in $\mathbb{R} \cup \{-\infty,\infty\}$, i.e. $E \neq \emptyset$.

If you would like to see a proof of this fact, look up the Monotone Subsequence Theorem.

Title: Why does every sequence of real numbers have an upper limit?

Body:
Let $\{a_n\}:\mathbb{N}\to\mathbb{R}$, and we define $E$ to be the set of all sub-sequential limits of $\{a_n\}$ as well as possibly the symbols $\infty$ and $-\infty$ if there are some sub-sequences of $\{a_n\}$ that go to $\infty$ / $-\infty$.

Can $E$ be empty? In particular, since $\limsup_{n \to \infty}a_n := \sup E$, does the upper limit of a sequence of real numbers is always defined? From what I found on the Wikipedia page on limit superior and limit inferior, the answer is yes and this have something to do with the fact that $\mathbb{R}$ is complete, but I couldn't find the complete proof, and I'm a bit surprised the issue isn't discussed on Rudin's principles of mathematical analysis.


问:
Solve $u_t+cu_x=0$ with $c>0$ when $x,t >0$ and contions $u(x,0)=h(x)$ with $x>0$ and $u(0,t)=g(t)$ with $t>0$.
$u_t+cu_x=0$ show  that $u(x,t)=f(ct-x)$.

With conditions, it shows that $$u(x,0)=f(-x)=h(x),$$ $$u(0,t)=f(ct)=g(t)$$
I dont know how to find what $f(ct-x)$ is and what the purpose of condition: $c>0$  when  $x,t >0$ .

答:
We compose the system of characteristics $\frac{dt}1=\frac{dx}c$.
So we have $c dt - dx = 0$. And $c t - x$ is the first integral of system of characteristics. $f(c t - x)$ is the general solution of equation, where $f$ is function from the class $C^1$.

$u(x, t) = f(c t - x)$ is general solution.
From boundary conditions we get:

$u(x, 0) = f(-x) = h(x)$, $x > 0$.

$u(0, t) = f(c t) = g(t)$, $t > 0$.

From (1) $f(z) = h(-z)$, $z< 0$. From $f(z) =g(\frac{z}c)$, $z > 0$.

So, we have the solution $u(x, t) = f(c t - x) = \left\{\begin{matrix}
g(\frac{c t - x}c), & c t - x > 0,\\
h(x - c t), & c t - x < 0.
\end{matrix}\right.$

If $c < 0$ your the first mixed problem will not have a unique solution, or it will have no solution.


问:
Are these simple statements about polygons true?
Let $P$ be an $n$-gon with $n \gt 3$. I'm looking for proofs or counterexamples for the following statements:

there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$

there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$ and the interior angle of $P$ at $B$ is convex

there exist different vertices $X,Y$ of $P$ such that $\overline{XY} \cap \partial P = \{X,Y\}$

For 1. and 2., if there is an offending vertex (must be if any other point is) in the triangle, maybe it could be put in some relation with the considered vertex, forming a finite dag which must have a maximal element, suitable as $B$.

For 3. it's actually easy (see my answer which I'm going to add in a moment) without an additional assumption, so let me also require $\overline{XY}\subset P$. Then, maybe decomposition into polygons (one a subset, one with disjoint interior) with fewer vertices by the external diagonal and applying induction would help?

答: For 3. without the additional requirement, if $P$ is convex, any 2 different non-adjacent vertices will do, otherwise suitable $X$ and $Y$ can be chosen as 2 adjacent vertices of the convex hull of $P$ (with degenerate vertices included, lying on the segment between its neighbours) which aren't adjacent in $P$.


问:
Integer partitioning under constraints on number of parts and maximum elements
I have an use case which requires storing and then looking up the integer partitions of numbers. So before I go and actually store all the integer partitions, I am actually trying to check if they are feasible.I have 3 questions associated with it.

Question 1

I am not actually storing all the partitions of the number rather I am storing only those which have 16 parts and maximum element used is 128 . Hence 2048 has 1 partition and everything greater than that has 0 partitions satisfying the constraints. So is there an upper bound or an approximation to find number of such partitions i.e p(n,k,m) which is number of partitions of n into k partitions with maximum partition element used being m?

Question 2

I found out that

p(16,16,128) = 1 which is 1+1+...+1(16 times)

p(2048,16,128) = 1 which is 128+128+...+128(16 times)

p(17,16,128) = 1 which is 1+1+...+2

p(2047,16,128) = 1 which is 128+128+...+127

And I checked for few more and they were similar. As in p(n,k,m) = p(2048-(n-16),k,m) and we could derive the other partition with the one I had. Is this actually true?

Question 3

What is the best algorithm that can be used to generate such integer partitions under constraints? I had seen some code on the internet which does fairly well but it doesn't eliminate the possibilities fast enough i.e for p(2048.16,128) it takes a lot of time even though there is only one such partition.

答:
I think you are actually defining $p(n,k,m)$ to be the number of partitions of $n$ into $k$ parts of size at most $m$.  One way to compute this is recursively by conditioning on the value $j$ of the largest part:
$$p(n,k,m)=
\begin{cases}
[n = 0] &\text{if $k=0$} \\
[1 \le n \le m] &\text{if $k=1$}\\
\sum_{j=1}^{\min(n,m)} p(n-j,k-1,j) &\text{otherwise}
\end{cases}
$$


问:
$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$.
Definiton.

$f$ is continuous at $a$ provided that for each open set $V$ in $Y$ containing $f(a)$ there is an open set $U$ in $X$ containing $a$ such that $f(U) \subset V$.

Problem.

$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \overline{A}$, $f(a)\in \overline{ f(A)}$.

答:
$\to)$ Hint: Let $A\subseteq X$ and $x\in \overline{A}$ and $V$ be a neighborhood of $f(x)$. It's enough to show $f(A)\cap U\ne \emptyset$.  $f^{-1}(U)$ is a neighborhood of $x$. There's some
$$z\in f^{-1}(U)\cap A$$
$$\Rightarrow f(z)\in U\cap f(A)$$

$\leftarrow)$ Hint: Let $x\in X$ and $V$ be an open neighborhood of $f(x)$. It's enough to show $f^{-1}(V)$ is a neighborhood of $x$. Suppose it is not.
$x$ is in the boundary of $f^{-1}(V)$. So
$$x\in \overline{\big(f^{-1}(V)  \big)^c}$$
$$\Rightarrow f(x)\in f\big(\overline{\big(f^{-1}(V)  \big)^c}\big)\subseteq \overline{f\big(\big(f^{-1}(V)  \big)^c\big)}=\overline{f\big(f^{-1}(V^c)  \big)}\subseteq \overline{V^c}$$
So $f(x)$ is in the boundary of $V$. A contradiction.


问:
Do falling factorial generating functions have the multiplication property?
Both ordinary generating functions and exponential generating functions have the following multiplication property.  If $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n x^n)(\Sigma_n b_n x^n) = \Sigma_n c_n x^n$, then $(c_n)$ is always a sequence of nonnegative integers.  Similarly if $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n \frac{x^n}{n!})(\Sigma_n b_n \frac{x^n}{n!}) = \Sigma_n c_n \frac{x^n}{n!}$, then $(c_n)$ is always a sequence of nonnegative integers.

My question is, do falling factorial generating functions have the same multiplication property?  That is, if $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n (x)_n)(\Sigma_n b_n (x)_n) = \Sigma_n c_n (x)_n$, then is $(c_n)$ always a sequence of nonnegative integers?  If not, does anyone know of a counterexample?

答: Like monomials, falling factorials form a basis spanning the polynomials. Write $x^n=\sum_ku_{nk}(x)_k$ so$$[(x)_n]\sum_ra_r(x)_r\sum_sb_s(x)_s=\sum_{mrs}a_rb_su_{mn}\underbrace{[x^m](x)_r(x)_s}_{\in\Bbb Z},$$where $[f_j]g$ denotes the coefficient of a basis element $f_j$ in $g$. Now we just need to check the $u_{nk}$ are integers. Indeed $u_{nn}=1,\,x^n-(x)_n=\sum_ku_{(n-1)k}(x)_k$ obtains the $u_{nk}$ with $k<n$ as linear combinations over $\Bbb Z$ of the $u_{(n-1)k}$.


问:
Understanding Higher-order Differentiability Conceptually
I understand conceptually that a function $f\colon A\to\mathbf R$ is differentiable at a point $a\in A$ if it can be well approximated by a line there; more precisely, if we can find a constant $f'(a)$ such that $$f(a+h) = f(a) + f'(a)h+o(h).$$

My goal: I want to understand (intuitively) what it means when a function is twice differentiable, or thrice, etc, and likewise what it means when it isn't.

In a very loose sense, I have the intuition that a function is somehow smoother around a point if it has higher order differentiability there, and I suppose this somehow corresponds to the fact that it can be approximated well not only by a line but even better by a polynomial. (e.g. twice differentiability is  $f(a+h) = f(a) + f'(a)h + \tfrac12f''(a)h^2 + o(h^2)$, etc.). Does this mean polynomials canonically define the notion of "smoothness"?

Perhaps I can best get across what I'm asking for with an example. When I look at the graphs of
$$y=x|x| \qquad \text{and} \qquad y=x^3$$
for instance, I see that the latter grows quicker than the former, but around zero, they both basically look "smooth" to me. Yes I know that one is made of the absolute value function which is pointy, but if you just showed me these two pictures:

I wouldn't really feel that one is somehow "smoother" around zero than the other.

Is there a better way I can think about this?

答: I think that there is a sense in which polynomials can indeed be viewed as "defining the notion of smoothness" because a function whose $n$th derivative is identically $0$ is a polynomial of degree $n-1$.


答:
Yes, $(c_n)_{n\ge 0}$ will always be a sequence of nonnegative integers. To prove this, you just need to show that for each $n,m\in \mathbb N$, that $(x)_n (x)_m$ can be written as an $\mathbb N$-linear combination of the falling factorials $(x)_k$. Indeed, it is true that
$$
(x)_n(x)_m=\sum_{k\ge 0}\binom{n}k\binom{m}kk!\,(x)_{n+m-k}\tag1
$$
To prove this polynomial identity, it suffices to prove it is true whenever $x$ is a positive integer. In that case, the identity is equivalent to
$$
\binom{x}{n}\binom{x}{m}=\sum_k \frac{(n+m-k)!}{(n-k)!\,(m-k)!\,k!}\binom{x}{n+m-k}
$$
and there is an easy combinatorial proof. Namely, both sides count the number of ways to take a set of size $x$, paint $n$ of the elements red, and $m$ of the elements blue, allowing an element to be painted both colors (the number of such "purple" elements corresponds to $k$).

Using $(1)$, we get that
$$
\left(\sum_{i\ge 0}a_i(x)_i\right)\left(\sum_{j\ge 0}b_j(x)_j\right)=\sum_{i,j}a_ib_j(x)_i(x)_j=\sum_{i,j,k}a_ib_j\binom{i}{k}\binom{j}kk!(x)_{i+j-k}
$$
Substituting $i+j-n$ for $k$, this becomes $\sum_{i,j,n}a_ib_j\binom{i}{i+j-n}\binom{j}{i+j-n}(i+j-n)!(x)_n$, which implies the following formula for $c_n$:
$$
\boxed{c_n=\sum_{i,j}a_ib_j\binom{i}{n-j}\binom{j}{n-i}(i+j-n)!}
$$
The sum ranges over all $i,j\ge 0$ for which $i+j\ge n$, but the only terms which contribute positively are those for which $\max(i,j)\le n$. Of course, this is always a positive integer if $a_i$ and $b_j$ are.

Title: Do falling factorial generating functions have the multiplication property?

Body:
Both ordinary generating functions and exponential generating functions have the following multiplication property.  If $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n x^n)(\Sigma_n b_n x^n) = \Sigma_n c_n x^n$, then $(c_n)$ is always a sequence of nonnegative integers.  Similarly if $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n \frac{x^n}{n!})(\Sigma_n b_n \frac{x^n}{n!}) = \Sigma_n c_n \frac{x^n}{n!}$, then $(c_n)$ is always a sequence of nonnegative integers.

My question is, do falling factorial generating functions have the same multiplication property?  That is, if $(a_n)$ and $(b_n)$ are sequences of nonnegative integers and $(\Sigma_n a_n (x)_n)(\Sigma_n b_n (x)_n) = \Sigma_n c_n (x)_n$, then is $(c_n)$ always a sequence of nonnegative integers?  If not, does anyone know of a counterexample?


问:
Does the work required to lift an object depend on the acceleration of the object?
I'm confused by an example in Stewart's (8th edition, early transcendentals) Calculus textbook.  Example 1, Section 6.4: "How much work is done in lifting a 1.2kg book off the floor to put it on a desk that is 0.7m high?"

I can understand the computation: (mass) times (acceleration due to gravity) times (distance).  What I don't understand is why the acceleration of the object as it gets lifted does not factor into the equation.  I thought maybe we should just assume it's constant velocity, but that really doesn't make sense if the object is starting on the floor (where presumably the velocity is zero).

Is there a more precise way to phrase this question, like, "How much work is done against the force of gravity in lifting a 1.2kg book off the floor to put it on a desk that is 0.7m high?"

Is my confusion understandable?  I'm not intentionally being pedantic.  (I have a lot of respect for Stewart's Calculus book.)

Overall I was trying to understand (vector) line integrals better, and kept backtracking until I came to this example. I'm posting this in the math forum instead of the physics forum because I'm used to thinking about math definitions.  (I last took a physics class in 11th grade.)

Thank you!

答: This is more like a physics question. But the answer is "It does not matter". If the book starts from rest and it ends at rest it is the same work if you accelerate constantly and decelerate constantly as if you accelerate just a little, stay constant speed, then slow down, or any other combination. This is because the gravitation is a conservative force, and the work done depends only on the initial and final positions (assuming that the velocities are the same - zero in this case)</p>


问:
what is the definition of restricted direct product ?
in an exercise , i'm given a definition of  restricted direct product , but it doesn't make sense for me !

they say , the  restricted direct product of groups $G_i$ is the set of elements of direct product which are the identity in all but finitely many compontents

i really don't understand what it means to be all but finitely !
i googled and found the same defintion with No clearification !

so , any help by example or explanation ?

答:
Just thought I'd mention a more general notion of restricted direct product that is used in some contexts (in particular, I'm thinking of adeles). Given any collection of (perhaps topological) groups or rings or algebras or whatever objects $G_i$, and subobjects $H_i\le G_i$, indexed by $i\in I$, then the restricted product of the $\{G_i\}$ with respect to the $\{H_i\}$ is

$$\prod_{i\in I}^{\large\sf restr}G_i=\big\{(g_i)_{i\in I}:g_i\in H_i ~{\rm for~all~but~finitely~many~}i\in I\big\}\le \prod_{i\in I}G_i.$$

We say that "all but finitely many" elements of a collection satisfy a property $P$ if the number of elements that do not satisfy $P$ is finite. That is, all of the elements in the collection satisfy the property outside of only a finite number of exceptions.

In the case of rings or groups, if all of the $H_i$ are just $\{0\}$, the trivial subobject, then we recover the usual understanding of restricted direct product, which is also often (in fact more often) called the "direct sum" and denoted by the big \oplus symbol as $\bigoplus_{i\in I}G_i$.

In number theory for example we take the restricted direct product of the $p$-adic fields ${\bf Q}_p$ over all primes $p$ (with "${\bf Q}_\infty={\bf R}$" the convention for the "infinite prime $p=\infty$") with respect to the subrings ${\bf Z}_p$ for integer primes $p$ to obtain the ring of adeles ${\bf A}_{\bf Q}$.


问:
2 numbers are chosen from $1,3,5..151$ and multiplied in all possible ways. Find number of ways such that product is a multiple of 5
Number of available elements = 76

Elements that are a multiple of 5 = 15

Now one perfectly acceptable answer for this is
$$\binom {30}{2} +\binom{15}{1}\times \binom {46}{1}$$

I understood how it came, that isn’t the problem.

But why can’t $\binom {15}{1} \times \binom{75}{1}$ be used? It basically means choosing 1 element from the 15 special elements and choosing 1 elements from every other number left.

答:
To summarize the discussion in the comments:

Given that we are choosing from the list $\{1,3,5,\cdots, 151\}$, that is, the odd integers from $1$ to $151$, inclusive, there are $15$ multiples of $5$ available, and $61$ non-multiples of $5$.

We wish to count the ways to choose two distinct terms from this list such that at least one of them is a multiple of $5$.  The spirit of the first calculation is: first count those pairs in which both terms are multiples of $5$ then add the pairs in which exactly one is a multiple of $5$.  The arithmetic is off, however, and the formula ought to read $$\binom {15}2+\binom {15}1\times \binom {61}1=1020$$

The second method is flawed because it double counts those pairs in which both terms are multiples of $5$.  We can fix that by subtracting the number of such pairs, namely:

$$\binom {15}1\times \binom {75}1-\binom {15}2=1020$$

As a third method, in the spirit of the proposed solution from @ChristianBlatter, we can count all possible pairs and subtract off those in which neither term is a multiple of $5$.  Thus:  $$\binom {76}2-\binom {61}2=1020$$


问:
Is $A$ is connected in $\mathbb{R}^2?$ Yes/no
Is  $A=(\{0\} \times \mathbb{R} ) \cup (\{1\}\times  \mathbb{R} )$ is connected in $\mathbb{R}^2?$

My attempt :I think $A$ is connected

$\mathbb{R}$ is homeomorphic to $\{0\}\times \mathbb{R}$ because there is the homeomorphism$$\begin{array}{ccc}\mathbb{R}&\longrightarrow&\{0\}\times \mathbb{R}\\y&\mapsto&(0,y).\end{array}$$

$\mathbb{R}$ is homeomorphic to $\{1\}\times \mathbb{R}$ because there is the homeomorphism$$\begin{array}{ccc}\mathbb{R}&\longrightarrow&\{1\}\times \mathbb{R}\\y&\mapsto&(1,y).\end{array}$$

$$(\{0\} \times \mathbb{R} ) \cup (\{1\}\times  \mathbb{R} \cong \mathbb{R} \cup \mathbb{R}$$

$\implies A$ is connected

答: Observe that $\{0\}\times\mathbb{R}$ and $\{1\}\times\mathbb{R}$ are both relatively open in $A$, non-empty and disjoint. Therefore, $A$ is disconnected.


答:
Your set $A$ is not connected; in fact, as you've noticed, it consists in two (disjoint) copies of $\Bbb R$ "inside" $\Bbb R^2$.
This does not mean that $A\cong\Bbb R$ (it is not, in fact, the case).

Each of the two lines $\{0\}\times\Bbb R$ and $\{1\}\times\Bbb R$ is a closed subspace of $\Bbb R^2$, and hence they are closed subspaces of $A$ (why?); $A$ is the union of the two, so you can safely conclude that $A$ is disconnected (and the two lines are its connected components, since they're both connected spaces).

Title: Is $A$ is connected in $\mathbb{R}^2?$ Yes/no

Body:
Is  $A=(\{0\} \times \mathbb{R} ) \cup (\{1\}\times  \mathbb{R} )$ is connected in $\mathbb{R}^2?$

My attempt :I think $A$ is connected

$\mathbb{R}$ is homeomorphic to $\{0\}\times \mathbb{R}$ because there is the homeomorphism$$\begin{array}{ccc}\mathbb{R}&\longrightarrow&\{0\}\times \mathbb{R}\\y&\mapsto&(0,y).\end{array}$$

$\mathbb{R}$ is homeomorphic to $\{1\}\times \mathbb{R}$ because there is the homeomorphism$$\begin{array}{ccc}\mathbb{R}&\longrightarrow&\{1\}\times \mathbb{R}\\y&\mapsto&(1,y).\end{array}$$

$$(\{0\} \times \mathbb{R} ) \cup (\{1\}\times  \mathbb{R} \cong \mathbb{R} \cup \mathbb{R}$$

$\implies A$ is connected


答:
Since the coordinates of the equilateral triangle is in terms of $x$, the answer will be as well, because it depends on the length of each sides, right ?

It's actually not too hard to find it, but let me show you a trick and an important theorem that we use in order to solve this problem.

1), The altitude of any isosceles triangle is the median, the angle bisector of top angle and the perpendicular bisector.

2), An equilateral triangle is a special case of an isosceles.

This should tell you that the altitude, which lies on the $y$-axis in your case, will not only be perpendicular to the base, but it will also bisect the top angle. This forms a 30-60-90 triangle which you must have learned in geometry.

So, if you want to find the height, just use the $1$-$2$-$\sqrt{3}$ ratio.

Does this help ?


答:
$A$ is indeed the union of two subsets each one homeomorphic to $\mathbb R$... but this is not a reason for $A$ to be connected.

In fact $A$ is the disjoint union of two open subsets for the induced topology from $\mathbb R^2$ and is therefore not connected.

$$A= (\{0\} \times \mathbb R \cap (-\infty,1/2)\times \mathbb R) \cup (\{1\} \times \mathbb R \cap (1/2, \infty)\times \mathbb R)$$


问:
Time reversal of odd signals
This question is about the “reflect about the $y$ axis” method of finding $x(-t)$ for a given $x(t)$. How does this method work if the signal is odd to begin with? Let’s say it is a signal whose equation is not obvious ,as the one shown in figure:

Let’s say that we have were only shown the shape in the positive $x$ axis, how do I know that I have to take $-x(t)$ for the reflected version?

Thanks in advance.

答:
reflection of an odd function about y axis

So I added an image which shows the reflected version($x(-t)$), of $x(t)$ about the $y$ axis.

Suppose you have a domain of definition $\Delta = (a,b)$ for the function $x(t)$. To find $x(-t)$ you must consider $-t \in (a,b)$. So it follows that $t \in (-b,-a)$. If you choose a random value for $t,$ let it be $u_0 = -t_0 \in (-b,-a)$ then  $$x(-u_o) = x(t_0)$$ and $$t_0 \in \Delta  = (a,b)$$ Thus the point $(-t_o, x(t_0))$ of the function $z(t) = x(-t)$ is the symmetric about the $y$-axis of the point $(t_0, x(t_0))$, which belongs to the original signal x(t).

Following this method for every point of $x(t)$, what you manage to accomplish is to reflect the function about the y-axis.

You understand of course all the above I suppose. Until now I have not mentioned anything about the oddness of the signal. It does not make a difference for the method of reflection about the $y$-axis.

However, as you can see, the result of finding $x(-t)$ leads a to the same signal as if we were to take  $$x(-t) = - x(t)$$ i.e. the symmetric of the signal about the $x$-axis.

Note: in the photo I have written $t_0$ as $x_0$ and $x(t_0)$ as $f(x_0)$. Sorry for the inconsistency(from habit).

Title: Time reversal of odd signals


问:
Maximum modulus principle for an unbounded function in the half right plane
Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)>0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$.

I thought about taking the bounded set $H\cap D(0,R), R>0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$.

答:
Using the maximum modulus principle on the intersection of the halfplane with large disks is the correct approach.

Since $\lim_{|z|\to\infty} f(z) =0$ there is an $R > 0$ such that $|f(z)| < 1$ for $|z| \ge R$. For all $r > R$ you can apply the maximum modulus principle to $f$ on $H\cap D(0,r)$ and conclude that $|f(z)| \le 1 $ for $z  \in  H$ with $|z| < r$. It follows that $|f(z)| \le 1$ in $H$.

(For a more general result, see Maximum modulus theorem for holomorphic function on an unbounded region .)


问:
What is a determinant in matrix ? NOT how to get it but WHAT is it in real life
We know to get determinant in a matrix we use sarrus, laplace, x methods and so on

But question : WHAT is it, according to some pages is about permutations the products of element of the matrix , but WHAT IS IT ?

答:
\begin{align}
& \left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} ax+by \\ cx+dy \end{array} \right] \\[8pt]
& \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} a \\ c \end{array} \right] \\[8pt]
& \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] = \left[ \begin{array}{c} b \\ d \end{array} \right]
\end{align}

Draw the arrows representing $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right].$ From the head of the arrow $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right],$ draw another arrow parallel to $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ and from the head of the arrow $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ draw another parallel to $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right].$ There you have a square whose area is $1.$

Then do the same with $\left[ \begin{array}{c} a \\ c \end{array} \right]$ and $\left[ \begin{array}{c} b \\ d \end{array} \right]$ and get a parallelogram. The area of that parallelogram is $\left|\det\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\right|.$ The determinant is positive if turning from $\left[ \begin{array}{c} a \\ c \end{array} \right]$ to $\left[ \begin{array}{c} b \\ d \end{array} \right]$ amounts to turning counterclockwise, and negative if clockwise. In other words, positive if the parallelogram has the same orientation as the square, and negative if the orientation is reversed.


问:
Matrix derivative of expression, mistake in proof
I cannot find the problem with the following proof can someone help?

Prove that for matries X, W and T: $$\frac\partial{\partial W}\frac12 \text{Tr}\left\{\left(XW-T)^T(XW-T)\right)\right\} = X^TXW-X^TT $$

Proof:

$$\frac\partial{\partial W}\frac12 \text{Tr}\left\{\left(XW-T)^T(XW-T)\right)\right\} =\frac\partial{\partial W}\frac12 \sum_{i}\left(\left(XW-T\right)^T(XW-T)\right)_{i,i} $$ $$= \frac\partial{\partial W}\frac12 \sum_{i,j}\left(\left(XW-T\right)^T_{i,j}(XW-T)_{j,i}\right)$$ $$=\frac\partial{\partial W}\frac12 \sum_{i,j} \left((XW-T)^2_{j,i}\right) $$ $$=\frac\partial{\partial W}\frac12 \sum_{i,j,k} \left((X_{j,k}W_{k,i}-T_{j,i})^2\right) $$

Now if we consider just the m,n entry of this matrix

$$ \left[\frac\partial{\partial W}\frac12 \sum_{i,j,k} \left((X_{j,k}W_{k,i}-T_{j,i})^2\right)\right]_{m,n} = \frac\partial{\partial W_{m,n}}\frac12 \sum_{i,j,k} \left((X_{j,k}W_{k,i}-T_{j,i})^2\right) \quad\quad\quad\quad \text{(3)}$$

Now we can put $k = m$ and $ i = n$ as all other terms will have no dependence on $w_{m,n}$ (*). Therefore,

$$ \left[\frac\partial{\partial W}\frac12 \text{Tr}\left\{\left(XW-T)^T(XW-T)\right)\right\}\right]_{m,n} = \frac\partial{\partial W_{m,n}}\frac12 \sum_{j} \left((X_{j,m}W_{m,n}-T_{j,n})^2\right) $$ $$= \sum_{j}(X_{j,m}W_{m,n}-T_{j,n})X_{j,m} $$ $$= \sum_{j} \left( X_{j,m}W_{m,n}X_{j,m} - T_{j,n}X_{j,m}\right) $$ $$= \sum_{j} \left( X_{j,m}W_{m,n}X_{j,m} \right) - \left(X^TT\right)_{m,n} $$

But the first term on the LHS above does not equal $(X^TXW)_{m,n}$

Edit: the problem is with the application of * everything afer * is incorrect becuase it neglects the cross terms from RHS of (3).

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$To
reduce the clutter, define the matrix $Y$ with components
$$\eqalign{
Y_{ij} &= \sum_k X_{ik}W_{kj}-T_{ij} \\
}$$
Calculate the component-wise derivative of $Y$ wrt $W$
$$\eqalign{
\p{Y_{ij}}{W_{mn}}
 &= \sum_k X_{ik}\,\delta_{km}\delta_{jn}
\;=\; X_{im}\,\delta_{jn} \\
}$$
Use this result to calculate the component-wise derivative of the objective function
$$\eqalign{
\phi &= \sum_i\sum_j \tfrac 12Y_{ij}\,Y_{ij} \\
\p{\phi}{W_{mn}}
 &= \sum_i\sum_j Y_{ij}\left(\p{Y_{ij}}{W_{mn}}\right) \\
 &= \sum_i\sum_j Y_{ij}\left(X_{im}\,\delta_{jn}\right) \\
 &= \sum_i Y_{in}X_{im} \\
 &= \left(X^TY\right)_{mn} \\
}$$
The error that you made was to naively $\color{blue}{\rm replace}$ the indices $(k,m=i,n),\,$ instead you must use Kronecker deltas as shown above.


问:
On $C^1$ convex domain
Let $D$ be a $C^1$ domain of $\mathbb{R}^d$. Then we know that there exists a $C^1$ function $\rho:\mathbb R^d\rightarrow \mathbb R$ such that
$$
D=\{x\in \mathbb R^d, \rho(x)<0\}, \quad \partial D=\{x\in \mathbb R^d, \rho(x)=0\},
$$
and
$
x\in \partial D\Longrightarrow d\rho(x)\not=0.
$

Assume now that $D$ is convex. Can we choose $\rho$ to be convex with the same properties?

I know that the first claim follows by the local definition and use of partition of unity, but I can't manage to prove existence of such convex function.

答:
(Original answer deleted)

$\newcommand{\R}{\mathbb{R}}$
Let $D \subset \R^d$ be a compact $C^1$ convex domain that contains the origin in its interior. Its gauge function is defined to be  $g: \R^d \rightarrow [0,\infty)$, where
$$
  g(x) = \inf \{ t > 0 : x/t \in D\}.
$$
Observe that $g$ is homogeneous of degree $1$. Therefore, to prove that $g$ is convex, it suffices to show that it satisfies the triangle inequality:
$$
g(x+y) \le g(x) + g(y).
$$
A proof can be found here: Show that a closed convex ball implies the triangle inequality

Now observe that the boundary of $D$ is parameterized by the map
\begin{align*}
  G: S^{d-1} &\rightarrow \partial D\\
  u &\mapsto \frac{u}{g(u)}
\end{align*}
The boundary $\partial D$ is $C^1$ if and only if $G$ is a $C^1$ map. And $G$ is $C^1$ if and only if $g$ is $C^1$ away from the origin. Moreover, since $\nabla g$ is homogeneous of degree $0$ and continuous, it is uniformly bounded.

Let $\rho = g^2-1$. $D = \{ \rho \le 0\}$ and $\partial D = \{ \rho = 0\}$. Since $g$ is $C^1$ on $\R^d\backslash\{0\}$, so is $\rho$. On other hand, the derivative of $\rho$ is equal to
$$
  \nabla\rho = 2g\nabla g,
$$
which converges to $0$ at the origin and therefore is continuous on all of $\R^d$.

Title: On $C^1$ convex domain


问:
A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$
Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.</p>

答: Hint: Just expand it using the Binomial Theorem, and use the obvious fact that for $i < n$, $i< n$.</p>


问:
Can I take the gradient of a function like this?
Let us say I have a vector function

$$r(u,v) = \langle u,v,1000+u^2+v^2 \rangle.$$

I also have,

$$p(x,y,z) = 5e^{(x + y + z^2)}.$$

Let, $Q(u,v) = (P \circ r)(u,v).$ Then,

$$Q(u,v) = 5e^{(u+v+(1000+u^2+v^2)^2)} $$

Would
$$\frac{\partial Q}{\partial u} = \frac{\partial}{\partial u} (5e^{(u+v+(1000+u^2+v^2)^2)}) ? $$

答:
According to the chain rule, the Jacobian matrix (matrix of derivatives) of $F=f\,o\,g\,:\mathbb{R}^n\rightarrow\mathbb{R}^m$, where $f:\mathbb{R}^p\rightarrow\mathbb{R}^m$ and $g:\mathbb{R}^n\rightarrow\mathbb{R}^p$  is defined as:

$$(Df)(\textbf{x})=(D(f\,o\,g))(\textbf{x})=(Df)(g(\textbf{x}))(Dg)(\textbf{x})$$

Therefore, in this case:

$$(DQ)(u,v)=(D(P\,o\,r))(u,v)=(DP)(r(u,v))(Dr)(u,v)$$

We have:

$$DP(x,y,z)=\begin{pmatrix}\frac{\partial P}{\partial x} & \frac{\partial P}{\partial y} & \frac{\partial P}{\partial z}\end{pmatrix}\Rightarrow (DP)(r(u,v))=DP(r_1,r_2,r_3)=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix}$$

$$Dr(u,v)=\begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

Where $r(u,v)=(r_1(u,v),r_2(u,v), r_3(u,v))$

But also by definition:

$$DQ(u,v)=\begin{pmatrix}\frac{\partial Q}{\partial u} & \frac{\partial Q}{\partial v} \end{pmatrix}$$

Applying the chain rule:

$$DQ(u,v)=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix} \cdot \begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

Then, we know:

$$\begin{pmatrix}\frac{\partial Q}{\partial u} & \frac{\partial Q}{\partial v} \end{pmatrix}=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix} \cdot \begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

And by multiplying for the first column on the right of the expression:

$$\frac{\partial Q}{\partial u}=\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial u}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial u}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial u}$$

And for the second column:

$$\frac{\partial Q}{\partial v}=\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial v}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial v}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial v}$$

Then, the gradient of the function is:

$$\vec{\nabla} Q(u,v) = (DQ(u,v))^T=\begin{pmatrix}\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial u}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial u}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial u} \\ \frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial v}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial v}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial v} \end{pmatrix}$$

This takes simpler computations for very complex functions or compositions of a large number of functions. Of course, in the case posed by the OP the composition may not be so complex as to require this method. You can simply find the explicit representation of the function $Q$ and then find the partial derivatives from there.


问:
Ratio of area of triangle and hexagon
I am looking for the proof of the following claim:

Claim. If the sides of the triangle are  partitioned into $n$ equal segments for $n$ an even integer and each division point adjacent to the corresponding central division point  is connected to the opposite vertex, a central hexagon which has an area of $\frac{32}{(3n-2)(3n+2)}$ of the area of the original triangle  is still formed.

GeoGebra applet that demonstrates this claim can be found here.

I tried to adapt the proof of the Marion's theorem given on this page. The problem I encountered is the following: How to find a trilinear coordinates of the division points : $D,E,F,G,H,I$ . The only formula I am aware of is the formula for the first n-multisection.

答:
We can use the Routh-Steiner theorem, which gives a method to compute the area of a sub-triangle bounded by three cevians.

The three medians of $ABC$ (meeting at the centroid $X$) divide the central hexagon into six triangles.  The triangle $LMX$ is bounded by cevians defined by side ratios $x=y=1,z=(n+2)/(n-2)$.
By Routh-Steiner

$$
\begin{align}
\cfrac{|LMX|}{|ABC|} &=
 \cfrac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)} \\
&= \cfrac{\cfrac{16}{(n-2)^2}}{(3)\left(\cfrac{3n+2}{n-2}\right)\left(\cfrac{3n-2}{n-2}\right)} \\
&= \cfrac{16}{3(3n+2)(3n-2)}
\end{align}
$$   The other five triangles are built using the same ratios, so have the same area as $LMX$.  Thus the area of the hexagon as a fraction of the large triangle is

$$
\cfrac{|JKLMNO|}{|ABC|}=6 \left(\cfrac{16}{3(3n+2)(3n-2)}\right)=\cfrac{32}{(3n+2)(3n-2)}.
$$

Note that the construction and formula also work for non-integer values of $n.$  Here, $n=AC/HR.$

Title: Ratio of area of triangle and hexagon


问:
Taylor's Series problem
Hello I have a problem that I want to solve because it is bugging me a lot. I have this problem in my calculus class, and this is a problem that will be presented by the teacher tomorrow.

$$a e^x + \sqrt{1-b^2 x^2} + c = 1 + 10x + 3x^2 + o(x^2)$$

What is $a, b, c$ ?

I'm really confused about how you solve $a, b, c$, and is it even possible? Is there something that I don't know? Please enlighten me!

答:
Assuming that you are searching for the Taylor's expansion at $x=0$, you can use that
$$
ae^x=a+ax+\frac{ax^2}{2}+o\left(x^2\right)
$$
$$
\sqrt{1-b^2x^2}=1-\frac{b^2x^2}{2}+o\left(x^2\right)
$$

You can deduce from this a taylor expansion of order 2 that you can compare to the one that's given.

Title: Taylor's Series problem


问:
Convolution of uniform distributions
If $X$ is a random variable with a continuous uniform distribution on (0,10), and $Y$ is also from a continuous uniform distribution on (0,10) how do I calculate the CDF of $X+Y$?

Title: Convolution of uniform distributions

Body:
You can compute this directly as follows. First, for any $0<t<20$ write

$$
F(t):=\mathbb{P}(X+Y\leq t) = \int_{0}^{10} \mathbb{P}(X+y\leq t) \frac{1}{10}\, dy
$$

Case I: $20>t\geq 10$, then

$$
F(t)= \int_{0}^{10}  \frac{\min{\{t-y,10}\}}{100}\, dy = 1-\frac{(20-t)^2}{100}.
$$

Case II: $0<t<10$, then
$$
F(t)= \int_{0}^{10}  \frac{\max{\{t-y,0}\}}{100}\, dy = \frac{t^2}{200}.
$$


问:
Holomorphic vs differentiable (in the real sense).
Why a holomorphic function is infinitely differentiable just because of satisfying the Cauchy Riemann equations, but on the other side, a two variable real function that is twice differentiable is not infinitely differentiable?

I'm asking this for two reasons:

1) $\mathbb{R}^2$ is somehow an analog to the complex plane (they are isomorphic).

2) Holomorphic means that is differentiable in any direction of the complex plane, so why if a function is differentiable in any direction of the real plane then it is not infinitely differentiable?

答:
Even though $\mathbb R^2$ and $\mathbb C$ are isomorphic as real vector spaces, they are very different in some algebraic respects, which crucially influence the notion of differentiability.  Recall the key idea that a function is differentiable at a point if it has a best linear approximation (more precisely, a constant plus a linear transformation) near that point.

In the context of functions $\mathbb R^2\to\mathbb R^2$, "linear transformation" means a transformation that respects addition of vectors and multiplication by real scalars.  In other words, the transformation respects the real vector space structure of $\mathbb R^2$.  It is well known that such linear transformations are given by $2\times2$ real matrices (once one has chosen a basis for $\mathbb R^2$).

In the context of functions $\mathbb C\to\mathbb C$, on the other hand, "linear transformation" means a transformation that respects addition of vectors and multiplication by complex scalars.  In other words, the transformation respects the complex vector space structure of $\mathbb C$.  It is well known that such linear transformations are just multiplication by a single complex number.  That's much more restrictive than multiplying by an arbitrary $2\times 2$ real matrix.

Specifically, if we use $\{1,i\}$ as our basis for $\mathbb R^2$, then the $2\times 2$ real matrices that correspond to complex linear transformations are just those of the form $\pmatrix{a & b\\-b & a}$.  Because complex linear transformations are a very special sort of real linear transformations, complex differentiable functions are a very special sort of real differentiable functions.  That "specialness" ultimately accounts for all the miraculous consequences of complex differentiability.


问:
Group theory and Hydrogen-like wave functions
I'm reading Pauling and Wilson's QM, in one place it is written that the $n^2$ wave functions $\psi_{n,l,m}(r,\theta,\phi)$ (for an H-like atom) with the same value of $n$ form a completed group. How? What is the law of composition for which they even form a group? I couldn't get any references online or offline. Any suggestions/references are welcome.

答:
They are a complete set, or grouping (not group), of functions solving the Schroedinger equation for such an atom with energy characterized by that posited n. That is, they are a complete basis of all solutions for that energy.

When it comes to symmetry groups commuting with the hamiltonian, there is first SO(3), with quadratic Casimir ${\mathbf L}\cdot {\mathbf L}$ eigenvalues $\ell (\ell+1)$, $\ell= 0,1,2,...,n-1$, that is, n different irreducible representations of SO(3), each with dimension $2\ell+1$. One of the three generators of SO(3), namely $L_z$ has eigenvalues m, $|m|\leq \ell$, and  connects states in a representation labelled by $\ell$; but cannot connect states for different $\ell$s—to do that you need generators of the larger symmetry of these atoms, SO(4)~ SO(3)×SO(3).

You might easily show the totality of states then is $\sum_{\ell=0}^{n-1} (2\ell +1)=n^2$. They comprise a complete expansion basis for all solutions with that energy.


问:
What's an intuitive way to think about the determinant?
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?

答:
```<code>The determinant of a matrix gives the signed volume of the parallelepiped
that is generated by the vectors given by the matrix columns.
</code>```

You can find a very pedagogical discussion at page 16 of

A Visual Introduction to Differential Forms and Calculus on Manifolds
Fortney, J.P.

google book link, click on "1 Background Material"

Given a parallelepiped whose edges are given by $ v_1 , v_2 , \dots, v_n \in \mathbb{R}^n $. Then if you accept these 3 properties:

$D(I)=1$, where $I=[e_1,e_2,\dots,e_n]$ (identity matrix)

$D(v_1,v_2,\dots,v_n)=0$ if $v_i=v_j$ for any $i\neq j$

$D$ is linear, $$\forall j,\ D(v_1,\dots,v_{j-1},v+cw,v_{j+1},\dots,v_n)=D(v_1,\dots,v_{j-1},v,v_{j+1},\dots,v_n)+cD(v_1,\dots,v_{j-1},w,v_{j+1},\dots,v_n)$$

you can show that $D$ is the parallelepiped signed volume and that $D$ is the determinant.

Body:
The determinant of a matrix gives the signed volume of the parallelepiped
that is generated by the vectors given by the matrix columns.

You can find a very pedagogical discussion at page 16 of

A Visual Introduction to Differential Forms and Calculus on Manifolds
Fortney, J.P.

google book link, click on "1 Background Material"

Given a parallelepiped whose edges are given by $ v_1 , v_2 , \dots, v_n \in \mathbb{R}^n $. Then if you accept these 3 properties:

$D(I)=1$, where $I=[e_1,e_2,\dots,e_n]$ (identity matrix)

$D(v_1,v_2,\dots,v_n)=0$ if $v_i=v_j$ for any $i\neq j$

$D$ is linear, $$\forall j,\ D(v_1,\dots,v_{j-1},v+cw,v_{j+1},\dots,v_n)=D(v_1,\dots,v_{j-1},v,v_{j+1},\dots,v_n)+cD(v_1,\dots,v_{j-1},w,v_{j+1},\dots,v_n)$$

you can show that $D$ is the parallelepiped signed volume and that $D$ is the determinant.


问:
Proof explanation for a sum of sines
Looking through proofs and there's one part I'm confused about, that if solved will get me straight to the answer.

I need to show that $\sum_{m=0}^{N-1} \sin((m+\frac{1}{2})x) = \frac{\sin(Nx/2)^2}{\sin(x/2)}$

How should I go about this?

答:
One way:

Recall that$$\sin p = \frac{1}{2i}[e^{ip}-e^{-ip}]$$

Also, $$\sum_{m=0}^{N-1} a^m = \frac{1-a^N}{1-a}. $$
Thus
$$\sum_{m=0}^{N-1} \sin [(m+\frac{1}{2})x]= \frac{1}{2i} \sum_{m=0}^{N-1}\left\{  \exp [(m+\frac{1}{2})ix]- \exp [-(m+\frac{1}{2})ix] \right\}.$$

Now you have two geometric series.  Compute the sum each of them and combine $\cdots.$

Title: Proof explanation for a sum of sines


问:
Polynomial integral calculation, maybe residue theorem
When evaluating a loop correction for a certain field theory, I arrived at the following integral

$$\int\limits_{\mathbb{R}^{1,3}}\frac{d^4 k}{(2\pi)^4}\frac{1}{\left((k^0)^2-\dfrac{|k|^4}{M^2}-m^2+i\epsilon\right)^2}$$ where $d^4 k = dk^0\,d^3k$ and the $i\epsilon$ prescription is for the definition of the integration path for the propagator. I'm a bit rusty when it comes to evaluating integrals like this and my first thought was to use the residue theorem but I don't really know where to start since one pole is on the upper half plane and the other on the lower half plane!

If there's even some better way of doing it beside using the residue theorem, I'll gladly have a look at them!

Edit

I intended to use the residue theorem first on $k^0$ and then, if necessary, on $k$. I thought even about Wick rotating to simplify the problem but still I don't really know where to start.

答:
For $\varepsilon > 0$ define $\phi_\varepsilon \colon (0,\infty) \to \mathbb{C}$ by
$$ \phi_\varepsilon (\mu) = \int \limits_\mathbb{R} \frac{\mathrm{d} \kappa}{2 \pi} \, \frac{1}{(\kappa^2 - \mu^2 + \mathrm{i} \varepsilon)^2} = \int \limits_\mathbb{R} \frac{\mathrm{d} \kappa}{2 \pi} \, \frac{1}{(\kappa - \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2 (\kappa + \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2}$$
using the principal branch of the square root (and all other power functions). Indeed, the integrand has one second-order pole in the upper and one in the lower half-plane. In order to apply the residue theorem we need to close the integration contour in either half-plane using a semi-circle of infinite radius. In this case both choices work, since the integrand decays fast enough in any direction to make the additional integrals vanish. We choose the upper path and use the formula for the residue at second order poles to obtain
$$\phi_\varepsilon (\mu) = 2 \pi \mathrm{i} \lim_{\kappa \to - \sqrt{\mu^2 - \mathrm{i} \varepsilon}} \frac{\mathrm{d}}{\mathrm{d} \kappa} \frac{1}{2 \pi (\kappa - \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2} = \frac{- 2 \mathrm{i}}{(-2\sqrt{\mu^2 - \mathrm{i} \varepsilon})^3} = \frac{\mathrm{i}}{4 (\mu^2 - \mathrm{i} \varepsilon)^{3/2}} \stackrel{\varepsilon \to 0^+}{\longrightarrow} \frac{\mathrm{i}}{4 \mu^3}$$
for $\mu > 0$. Note that the integral can also be computed using partial fractions.

Your integral is
$$ I(m,M) \equiv \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \phi_\varepsilon \left(\sqrt{m^2 + \frac{\lvert k \rvert^4}{M^2}}\right) \, .$$
The dominated convergence theorem allows us to move the limit inside the integral and introducing spherical coordinates we find
\begin{align}
I(m, M) &= \frac{\mathrm{i}}{8\pi^2} \int \limits_0^\infty \mathrm{d} k \, \frac{k^2}{\left(m^2 + \frac{k^4}{M^2}\right)^{3/2}} \stackrel{k = \sqrt{m M} u}{=} \frac{\mathrm{i}}{8 \pi^2} \left(\frac{M}{m}\right)^{3/2} \int \limits_0^\infty \mathrm{d} u \, \frac{u^2}{(1+u^4)^{3/2}} \\
&\!\!\!\!\!\!\stackrel{1 + u^4 = \frac{1}{t}}{=} \frac{\mathrm{i}}{32 \pi^2} \left(\frac{M}{m}\right)^{3/2} \int \limits_0^1 \mathrm{d} t \, [t(1-t)]^{-1/4} = \frac{\mathrm{i}}{32 \pi^2} \left(\frac{M}{m}\right)^{3/2} \operatorname{B} \left(\frac{3}{4},\frac{3}{4}\right) \\
&= \frac{\mathrm{i} \operatorname{\Gamma}^2 \left(\frac{3}{4}\right)}{16 \pi^{5/2}} \left(\frac{M}{m}\right)^{3/2} = \frac{\mathrm{i}}{8 \sqrt{\pi} \operatorname{\Gamma}^2 \left(\frac{1}{4}\right)} \left(\frac{M}{m}\right)^{3/2} ,
\end{align}
where we have used an integral representation of the beta function, its relation to the gamma function and the gamma reflection formula.


问:
How prove this $\frac{\sqrt{2}}{2}-\frac{1}{15}<F\left(\frac{\pi}{2}\right)<\frac{\sqrt{2}}{2}$

Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$
  and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.

Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$

Find the value $F\left(\dfrac{\pi}{2}\right)$.

答:
I was delayed in posting this, so it is little more than a comment to julien's answer.

Since the series $\sum\limits_{n=1}^\infty\frac1{\sqrt{n^3+n}}$ is convergent, we can integrate term by term
$$
F(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n\sqrt{n^3+n}}
$$
$\sin((2k)\pi/2)=0$ and $\sin((2k+1)\pi/2)=(-1)^k$; therefore,
$$\begin{align}
F(\pi/2)
&=\frac1{\sqrt2}\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt{(2k+1)^3(2k^2+2k+1)}}\\
&=\frac1{\sqrt2}-\frac1{\sqrt{270}}+\dots
\end{align}
$$
By the alternating series test, the final sum is between $\frac1{\sqrt2}$ and $\frac1{\sqrt2}-\frac1{\sqrt{270}}$. That is,
$$
\frac1{\sqrt2}-\frac1{15}\lt\frac1{\sqrt2}-\frac1{\sqrt{270}}\lt F(\pi/2)\lt\frac1{\sqrt2}
$$

This evaluates to $0.6587329279592957$, but the ISC does not find a closed form.


问:
Name of non-complete graph with certain properties
Is there a name for a graph with the following criteria?

not a complete graph

has at least n vertices (let's say n > 3 to exclude trivial cases)

where every vertex's degree is at least n.

Or, is there a class of graphs that tend to have this kind of pattern?
I've run into graphs like this in my research, and I'm hoping to tap into a body of existing knowledge on them.

For example:

答:
The second condition is redundant given the third: if every vertex has degree $n$, there must be at least $n+1$ vertices.

I would call graphs with the third condition "graphs with minimum degree at least $n$" or "graphs $G$ with $\delta(G) \ge n$". This is concise enough that no further terminology has developed.

There isn't a nice way to exclude the complete graph. You could say "other than complete graphs", but first double-check that whatever you're saying isn't also true for complete graphs, just in case. I guess you could also say "graphs $G$ with $n \le \delta(G) \le |V(G)|-2$", since complete graphs are distinguished by having $\delta(G)=|V(G)|-1$.

If you're looking for properties of such graphs, you should look for properties of graphs with minimum degree at least $n$; excluding complete graphs probably won't get you any more properties beyond that.


问:
What is formula for repeating sequence $0, 0, 1, 1, 0, 0, 1, 1, ...$
can somebody help me with this? I was searching through the web and I did not find anything useful. What is the formula for the sequence:

$0, 0, 1, 1, 0, 0, 1, 1, ...$

or

$0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, ...$

I did find solution for $0,1,0,1$ or $1,0,1,0$ or $0,0,0,1$ etc.. But nothing close to this.

Thank you in advance!

EDIT:
Sorry everyone, I am quite new to this. It is a repeating infinite series. Also, I am looking for a more general solution so I can choose the number of zeros and ones - if $n = 3$ then I am looking for a repeating sequence:

$0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, ...$

EDIT2:
Example: I input let say "$k$" into a formula. I'll get $0$ or $1$ as an output. I use "$n$" as a variable for a function that will determine the repairing pattern e.g. If $n=2$ then the pattern is $0,0,1,1,0,0,1,1,0,0,1,1, ...$ and if $k=3$, then the output will be the third position, which is $1$ in this case.

答:
Here is a formula based on another point of view.

Notice that

$\operatorname{base_2}(0,\overline{01})=\frac 13$

$\operatorname{base_2}(0,\overline{0011})=\frac 15$

$\operatorname{base_2}(0,\overline{000111})=\frac 19$

$\operatorname{base_2}(0,\overline{00001111})=\frac 1{17}$

It is not too complicated to verify that with $n$ zeroes and ones ($n$ being the half-period as in user's answer) then (where $x^n$ denotes the digits $\underbrace{xx\cdots x}_\text{n times}$)

$$\operatorname{base_2}(0,\overline{0^n1^n})=\frac 1{2^{n}+1}$$

So if we multiply by $2^k$ then the unit digit is either the $0$ or the $1$ at place $k$ in the binary development, giving the formula below:

$$a_k=\left\lfloor\frac{2^k}{2^n+1}\right\rfloor \bmod 2$$

It is more interesting theoretically than as a practical way of calculating these numbers though...

The formula based on powers of $(-1)^\alpha$ can be written with the floor function directly.

I think one efficient way of computing this sequence could be the following

$$a_k=\left\lfloor\frac kn\right\rfloor-2\left\lfloor\frac k{2n}\right\rfloor$$

The concept is simple, $\lfloor\frac x{2n}\rfloor$ has a period double than $\lfloor\frac x{n}\rfloor$, we multiply by $2$ so that both functions overlap on period $n$, their difference is then $0$, and on the remaining of the period it is $1$, giving birth to a $0^n1^n$ sequence of period $2n$.


问:
Cayley graph with opposite action: is the group abelian?
Let $G$ be a group, let $S$ be a set of generators and let $\Gamma=\Gamma(G,S)$ be the Cayley graph, where there is an edge between $g$ and $h$ if and only if $h=gs$ for some $s\in S$.

We know that the action of $G$ on $\Gamma$ via left multiplication acts on $\Gamma$ via graph automorphisms: if $h=gs$, then for each $a\in G$,
$$
ah = a(gs) = (ag)s ,
$$
and so $ag$ ahd $ah$ are connected by an edge whenever $g$ and $h$ are.

Conversely, the action of $G$ on $\Gamma$ via right multiplication does not in general induce graph automorphisms:
$$
ha = gsa \ne ags ,
$$
unless $G$ is abelian.

Now, is the converse true? If the action of $G$ via right multiplication induces graph automorphisms, can we conclude that $G$ is abelian?

答:
To add a bit to kabenyuk's answer, it's actually easy to show that this is equivalent to the connection set being a union of conjugacy classes.
You want $g\sim h\Longleftrightarrow ga\sim ha$ for all $g,h,a\in G$. But this is equivalent to $g^{-1}h\in S\Longleftrightarrow a^{-1}g^{-1}ha\in S$, which is equivalent to $S$ being closed under conjugation.

(Such Cayley graphs are sometimes called normal Cayley graphs, although that terminology can also mean something else.)