问:
Finding the limit when denominator = 0
I'm having trouble solving this limit:

$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$

How do I (algebraically) determine if it is positive or negative?

答: Given any $M > 0$, we have that $\forall x \in \left( -2 - \frac1{\sqrt{M}},-2 \right)$, we have $\frac1{(x+2)^2} > M$.</p>

Title: Finding the limit when denominator = 0


问:
How can I fold paper into 3 x 4 grid? Or prove that it can't be done?
I am trying to fold paper so that it looks like 3 x 4 grid of 12 rectangles of equal size.

Like this https://www.wolframalpha.com/input/?i=3+x+4

Its easy to get 4 rectangles. Just fold twice. But how to get 3?

答:
Let the paper be $[0,a]\times[0,b]$
By halving you find $(\frac a2,0)$, make a crease to find the line through $(\frac a2,0)$ and $(0,b)$. This intersects the diagonal at $(\frac a3,\frac b3)$.</p>

ContentLicense: CC BY-SA 3.0


问:
I need to develop $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ using the binomial coefficients formula and show a few properties
In the previous questions I've proved that $(1+i)^{2n+1}=a_n+ib_n$ where $a_n$ and $b_n$ are $\pm 2^n$ and that $(1-i)^{2n+1}=a_n - ib_n$ and that $|P_n(1)|=2^n$ using the previous statements. But now I need to show that the polynomial $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even (the coefficient in front of $X^n$ is non-zero only when $n$ is even, with real coefficients, and with $2n+1$ as a dominant coefficient, using the binomial coefficient formula.

I also need to show that the coefficient $X^{2n-2}$ in $P_n$  is $\frac{n(1-2n)(1+2n)}{3}$.

I developed $(X+i)^{2n+1}$ and I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$

When I develop $(X-i)^{2n+1}$ I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$, so the same thing.

So obviously when I subtract the two parts I get $0$ which won't let me prove the properties. Where is my mistake ? Do I use the binomial coefficient the wrong way ?

EDIT: I think I have made a mistake in the second development, the second term should has a + sign when the subtraction is made so I get $\frac{2(2n+1)X^{2n}i}{2i} = (2n+1)X^{2n}$

How do I show it is even though ? Is the end result  enough to show this property?

答:
In the expansion
$$(X+i)^{2n+1} - (X - i)^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}X^ki^{2n+1-k}- \sum_{k=0}^{2n+1}\binom{2n+1}{k}X^k(-i)^{2n+1-k}$$
the terms  such that $2n+1-k$ is even, i.e. k is odd, cancel pairwise, so there remains the terms with $k=2\ell+1\enspace (0\le \ell\le n)$, and the final expansion is
\begin{align}(X+i)^{2n+1} &- (X - i)^{2n+1}=\sum_{\ell=0}^{n}\binom{2n+1}{2\ell} X^{2\ell}2\,i^{2(n-\ell)}\\
&=2\sum_{\ell=0}^{n}\binom{2n+1}{2\ell} X^{2\ell}(-1)^{n-\ell}.
\end{align}

ContentLicense: CC BY-SA 4.0


问:
Evaluate $\iint xy $ over a parallelogram given with $2x-y=1$, $2x-y=3$, $x+y=-2$, $x+y=0$
I have to compute an integral: $$\iint_R xy \; dxdy$$ over a parallelogram given with $2x-y=1$, $2x-y=3$, $x+y=-2$, $x+y=0$, in a way that I use variable substitution:$u=2x-y,v=x+y$.
If I substitute x and y and multiply by Jacobian determinant I get the following integral:
$$\int_{-2}^0\int_1^3\frac{(1-uv)(2v-u)(v+u)}{9}\;dudv$$ which is not correct. Did I do something wrong?

答:
$\displaystyle x = \frac{u+v}{3}, y = \frac{2v-u}{3}$

$\displaystyle x_u = \frac{1}{3}, y_v = \frac{2}{3}, x_v = \frac{1}{3}, y_u = -\frac{1}{3}$

So your Jacobian for the transformation should be $\, J = \frac{1}{3} \times \frac{2}{3} - \frac{1}{3} \times (-\frac{1}{3}) = \frac{1}{3}$.

So your integral becomes

$I = \frac{1}{3} \displaystyle \int_{-2}^{0}\int_1^3\frac{(2v-u)(u+v)}{9} \, du \, dv$


问:
triangle combination edge colored
Stacked triangles

hi, im stucked at this problem and i dont know, how to move on.

The problem sounds like:

we have N stacked triangles (picture) We color edges of these triangles,that way where at least one edge in each triangle is colored.How many ways we have to  color these edges.

In the picture you can see, there is help. That sounds:

Finding recursion divide into two cases- a)edge 1,2 is not colored.

b) atleast one of 1,2 edge is colored

I appreciate every help.

答:
Consider that the stack of triangles is growing from left to right, so that the next triangle added will include edge $2$.  We call the edge that will be shared with then next triangle (edge $2$ in the picture) the "last edge".

Let $c_n$ be the number of admissible stacks of $n$ triangles in which the last edge is colored, and let $u_n$ be the number of admissible stacks of $n$ triangles in which the last edge is uncolored.

Now if the last edge is colored, all $4$ ways to color the two remaining edges of the next triangle give an admissible stack of $n+1$ triangles.  In two cases the last edge of the augmented stack is colored, and in two it is not.  On the other hand, if the last edge is uncolored, then there are only $3$ ways to color the two remaining edges, as we must color at least one of them.  Two ways give a stack whose last edge is colored, and one gives a stack with the last edge uncolored.

We can summarize the last paragraph as $$\begin{align}
c_{n+1}&=2c_n+2u_n\tag1\\
u_{n+1}&=2c_n+u_n\tag2
\end{align}$$

The initial conditions are $$c_1=4\\u_1=3$$ as may easily be verified.

The number of admissible $n$-triangle stacks is $a_n:=c_n+u_n$

As Calvin Lin suggests in a comment, rewrite $(2)$ as $$2c_n=u_{n+1}-u_n$$ and substitute this in $(1)$: $$\frac{u_{n+2}-u_{n+1}}2
= u_{n+1}-u_n+2u_n\implies u_{n+2}=3u_{n+1}+2u_n\tag3$$

Subtracting $(2)$ from $(1)$ gives $$\begin{align}
c_{n+1}
&=u_{n+1}+u_n\\
&=3u_n+2u_{n-1}+3u_{n-1}+2u_{n-2}\\
&=3u_n+5u_{n-1}+2u_{n-2}\\
&=3(u_n+u_{n-1})+2(u_{n-1}+u_{n-2})\\
&=3c_n+2c_{n-1}\tag4
\end{align}$$

Now the recurrences $(3)$ and $(4)$ clearly give

$$a_{n+2}=3a_{n+1}+2a_n$$

Given the initial conditions, this can be solved in the to give an explicit formula for $a_n$.  It's a second-order linear difference equation (recurrence relation) with constant coefficients.

Computation give the initial sequence $$7, 25, 89, 317, 1129, 4021, 14321, 51005, 181657, 646981, 2304257$$

I put this into OEIS and found A007484. The page gives the explicit formula, so you can check.

Title: triangle combination edge colored


问:
Is a random variable constant a.e. iff it is trivial sigma-algebra-measurable?
In the following question

Is a random variable constant iff it is trivial sigma-algebra-measurable?,

whose answer I copy here it is proven that

If $F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is a constant

If $f==c$ is constant it is ALWAYS measurable (for any sigma-algebra).
This holds as $f^{-1}[C]$ is $X$ if $c \in C$ and empty if $c \notin
> C$. And both sets are in any sigma-algebra.

On the other hand, if $f$ is $F$-measurable and non-constant, then it
assumes at least two values $c_1$ and $c_2$. The set $f^{-1}[{c_1}]$
must be in $F$ (by being $F$-measurable, as ${c_1}$ is a closed set)
but this set is non-empty (as $c_1$ IS a value of $f$) and not $X$ (as
the points $x$ where $f$ assumes the value $c_2$ are not in it). So
this set cannot be in $F$, and so $f$ must be constant.

However in my probability lecture this property was given like this

$F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is an a.e (almost everywhere) constant,
f being a random variable

Is this a generalization? How do I prove it? I ended up proving it without the a.e, just as in the reference question, and I don't know how to deal with the a.e to make it more general.
For instance in the converse: If $f=c$ a.e  then there should be sets $N$ with null measure for wich the function  is not equal to c, So when I take a borel set B and do the preimage $f^{-1}(B)$ what happens if N \subset B?  this leads to the additional question of what is the preimage of a negligible set?

答:
The almost-everywhere version is false: it's indeed the case that a function $f:X\to \Bbb R$ is constant if and only if it's $\{\emptyset,X\}$-measurable.

If you have a measure space $(X,\mathcal E,\mu)$ and a map $f:X\to \Bbb R$, it is true that $f$ is equal almost everywhere to a $\{\emptyset,X\}$-measurable function if and only if $f$ is almost everywhere constant, but it's hard to say if this was the intention of the claim, or if the statement you cite is just an error.


问:
Simplify binomial sum
It is well-known that for any integer $j$
$$
\sum _{i=0}^k (-1)^{i} \binom{k}{i} (j-i)^k = k!
$$
But I wonder if there's a closed form for the following sum
$$
\sum _{i=j}^k (-1)^{i} \binom{k}{i} (j-i)^k
$$

The following table gives the value for this $k = 1,\cdots, 6$ and $j = 0, \cdots k$.

```<code>1   0
2   1   0
6   5   1   0
24  23  12  1   0
120 119 93  27  1   0
720 719 662 360 58  1   0
</code>```

答:
This answer is a supplement to @MarkoRiedel's interesting derivation. We provide a somewhat different approach to derive the identity
\begin{align*}
\color{blue}{\sum_{p=0}^{q-1} \left\langle k \atop p \right\rangle = \sum_{p=0}^{q-1} (-1)^p {k\choose p} (q-p)^k}\tag{1}
\end{align*}
which is the crucial one in @MarkoRiedels answer. I also use his  notation to ease comparison.

The numbers $ \left\langle k \atop p \right\rangle$ are the Eulerian numbers as they are defined in Concrete Mathematics
, Section 6.2 by D.E. Knuth, R.L. Graham and O. Patashnik.

Double generating function: In order to show (1) we start with a double generating function $A(t,u)$ of the Eulerian numbers stated as formula [14v] in Advanced Combinatorics by L. Comtet.

We obtain according to formula [5g] in Comtet:
\begin{align*}
\color{blue}{A(t,u)}&=1+\sum_{k=1}^\infty\sum_{q=1}^k \left\langle k \atop q-1 \right\rangle \frac{t^k}{k!}u^q\tag{2}\\
&\,\,\color{blue}{=\frac{1-u}{1-ue^{t(1-u)}}}\\
&=(1-u)\sum_{p=0}^\infty u^pe^{pt(1-u)}\\
&=\sum_{p=0}^\infty u^p\sum_{l=0}^\infty \frac{p^l}{l!}t^l(1-u)^{l+1}\\
&\,\,\color{blue}{=\sum_{p=0}^\infty u^p\sum_{l=0}^\infty\frac{p^l}{l!}t^l\sum_{j=0}^{l+1}\binom{l+1}{j}(-1)^ju^j}\tag{3}
\end{align*}

Note that Comtet uses a slightly different definition of Eulerian numbers, which results in the lower index $q-1$ of $\left\langle k \atop q-1 \right\rangle$ in (2).

Coefficient comparison:

Coefficient comparison of (2) and (3) gives
\begin{align*}
\color{blue}{\left\langle k \atop q-1 \right\rangle}
&=k![t^ku^q]\sum_{p=0}^\infty u^p\sum_{l=0}^\infty \frac{p^l}{l!}t^l\sum_{j=0}^{l+1}\binom{l+1}{j}(-1)^ju^j\\
&=[u^q]\sum_{p=0}^\infty u^p p^k\sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^ju^j\tag{4}\\
&=\sum_{p=0}^q[u^{q-p}]p^k\sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^ju^j\tag{5}\\
&=\sum_{p=0}^q[u^{p}](q-p)^k\sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^ju^j\tag{6}\\
&\,\,\color{blue}{=\sum_{p=0}^{q-1}(q-p)^k\binom{k+1}{p}(-1)^p}\tag{7}
\end{align*}

Comment:

In (4) we select the coefficient of $t^k$.

In (5) we use the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$. We also set the upper index of the outer sum to $q$, since other terms do not contribute.

In (6) we change the order of summation: $p\to q-p$.

In (7) we select the coefficient of $u^p$.

The identity (7) is stated in Concrete Mathematics as formula (6.38). The sum in (7) is pretty close to the wanted sum (1) besides the upper index $k+1$ which should be $k$. In order to derive (1) we do some

Telescoping:

We obtain from (7)
\begin{align*}
\left\langle k \atop q-1 \right\rangle&=\sum_{p=0}^{q-1}(q-p)^k\binom{k+1}{p}(-1)^p\\
&=\sum_{p=0}^{q-1}(q-p)^k\binom{k}{p}(-1)^p+\sum_{p=1}^{q-1}(q-p)^k\binom{k}{p-1}(-1)^p\\
&=\underbrace{\sum_{p=0}^{q-1}(q-p)^k\binom{k}{p}(-1)^p}_{a_{k,q}}-\sum_{p=0}^{q-2}(q-1-p)^k\binom{k}{p}(-1)^p\\
&=a_{k,q}-a_{k,q-1}\\
\end{align*}
It follows via telescoping
\begin{align*}
\sum_{p=1}^{q-1}\left\langle k \atop p \right\rangle
&=\sum_{p=1}^{q-1}\left(a_{k,p+1}-a_{k,p}\right)\\
&=\sum_{p=2}^{q}a_{k,p}-\sum_{p=1}^{q-1}a_{k,p}\\
&=a_{k,q}-a_{k,1}\\
\end{align*}
and since
\begin{align*}
a_{k,1}=\sum_{p=0}^0(1-p)^k\binom{k}{p}(-1)^p=1=\left\langle k \atop 0 \right\rangle
\end{align*}
we derive
\begin{align*}
\color{blue}{\sum_{p=0}^{q-1}\left\langle k \atop p \right\rangle} = a_{k,q}
\color{blue}{= \sum_{p=0}^{q-1}(q-p)^k\binom{k}{p}(-1)^p}
\end{align*}
and the claim (1) follows.

Title: Simplify binomial sum


问:
What the probability of rolling a number?
A die has been "loaded" so that the probability of rolling any even number is $\frac{3}{15}$ and the probability of rolling any odd number is $\frac{2}{15}$. (Assume the die is six sided with each side numbered one through six. Enter your probabilities as fractions.)

What is the probability of rolling a $2$, given that an even number is rolled?

I tried to solve it by assuming that:

$A$ is the event of rolling an even number

$B$ Is the event of rolling a $2$
Then I used the conditional probability form:
$$P(A|B) = \frac{P(A \cap B )}{P(B)},$$
where
$P(A) = \frac{3}{15}$ is given, $P(B) = \frac13 \cdot\frac{3}{15}$.

I am not pretty sure of $P(B)$ and I do not know how to find $P(A \cap B )$.
Any help please?

答:
I do not know how to find $P(A \cap B )$

That part doesn't have to be hard. If $A$ is the event of rolling an even number and $B$ is the event of rolling a 2, then $B$ is a subset of $A$. So $A\cap B=B$ and in particular, $P(A\cap B)=P(B)$.

That said, before plugging in those numbers, you should still double-check your computation of $P(B)$!


问:
Why this line integral over this straight line isn't possible without parametrization.
There is a line integral over a straight line as follows (the problem here: https://tutorial.math.lamar.edu/Solutions/CalcIII/LineIntegralsPtI/Prob1.aspx):

Line Integral: $$\int 3x^2 - 2y\ ds$$
Equation of the line: $$2y=7x-9$$

In the solution, The author suggests that due to the reverse direction of line, parametrization is necessary but I don't know why.
In fact, I can't figure out why my solution (below) leads to a wrong result (negative the correct one), despite the fact that I'm defining the integration interval($x$) from 3 to 1 so it is supposed to be consistent with the direction.
$$\int 3x^2 - 2y\  ds \rightarrow  \int_{3}^{1}  3x^2 - 2y \sqrt{1+\left (\frac{7}{2}  \right )^2} dx $$

Edit: once I simplified the main function, but now it is as in the problem.

答:
You are traversing the line in the opposite direction than the "natural one" (usually $x$ increases), this is why the result should be negative.

If you want your answer to be correct, you need to do $\int_3^1$, not $\int_1^3$, since you should start from the point where $x$ is 3, and finish where $x$ is 1.

Edit: The line segment is the set of points $(x,\frac{7x-9}2)$ from $x=3$ to $x=1$.  Thus $ds=-\sqrt{1+\big(\frac72\big)^2}\,dx$, and we have
$$\int_\mathcal C 3x^2-2y\,ds = \int_3^1\Big(3x^2-2\Big(\frac{7x-9}{2}\Big)\Big)\Big(-\sqrt{1+\Big(\frac72\Big)^2}\,dx\Big) = 8\sqrt{53}.$$
Notice that $ds$ has a minus sign because the orientation is reversed. If this is confusing, you can just pick a parametrisation where the directions are aligned with the integral. The strangeness here occurs because we are going backwards. (Intuitively, it should make sense that $\Delta s \approx - c\,\Delta x$.)


问:
Maximising the variance in PCA - Trace Proof
Quite urgent request for a proof here as my professor has not run through it.

Scenario is the PCA problem where we are trying to maximise the variance in the latent space. I want to prove the following $\sum_k^d W_k^TS_tW_k = Trace[W^TS_tW]$ where $S_t$ is the covariance matrix of dimension $F \times F$. $W_k$ is the $k^{th}$ eigenvector of $S_t$ $\in F \times 1$ and $W = [W_1, W_2, .., W_k]$.

I sort of know the intuition behind it, the eigenvectors are orthogonal and thus $W_k^TW_j$ where $k \neq j = 0$. I am having issue with the formal proof.

I have tried the following

$\sum_k^d \big(\sum_i^F W_{k,i}St_{i,j} W_{j,k}\big)$ and stated that if $i \neq j, W_{j,k}W_{k,i} = 0$

Does anyone have a better proof for this please? or provide a hint as to where I am going wrong

答: The trace of $W^\top S_t W$ is the sum of is diagonal entries. Show that the $k$th diagonal entry is $W_k^\top S_t W_k$. (Think about how to compute the matrix multiplication $W^\top S_t W$.)


问:
Simplifcation of $e^{(-3 \pi i)/4}$
I am trying to figure out how my textbook simplifies this:

$$\frac{1}{4}e^{(-3 \pi i)/4} = -\frac{1}{4}e^{(\pi i)/4}$$

I know that $e^{\pi i} = -1$ but what is going on here?

答: Multiply the left-hand side by $-e^{4\pi i/4} = -e^{\pi i} = 1$.

Title: Simplifcation of $e^{(-3 \pi i)/4}$


问:
Finding solutions of a system of linear equations
I was asked the following:

$$\begin{align}2x_1 - 4x_2 + 8x_4 &= 10\\
3x_1 + 2x_2 &= 3\\
3x_1 + 2x_3 + 6x_4 &= 11\\
8x_1 + 6x_2 + 2x_3 + 14x_4 &= 24\end{align}$$
  Find the solution set $W$. Is $W$ a vector space?

I know I could express one variable from one of the equations and substitute into the remaining ones. For example, from the second equation I get
$x_1=1-\frac23x_2$.

After substituting this into the remaining equations I have
$$
\begin{align*}
-\frac{16}3x_2+8x_4&=8\\
-2x_2+2x_3+6x_4&=7\\
\frac23x_2+2x_3+14x_4&=16\\
\end{align*}
$$
I could continue by expressing another variable and plug into other equations. But this approach seems to be rather cumbersome and if I have more than three equations I am prone to make mistakes. Is there a more elegant way to do this?

I also do not know what to say about the part of the assignment asking about vector space.

答:
If you set up the augmented matrix for the system and perform RREF (Gaussian Elimination), you end up with:

$$
\left[\begin{array}{cccc|c}
1 & 0 & 0 & -2 & -1\\
0 & 1 & 0 & 3 & 3\\
0 & 0 & 1 & 6 & 7\\
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$

There are other methods (substitution, inverses, ....), but you did not indicate what you are currently familiar with.

For example, if using substitution, here are hints:

Use the second equation to solve for $x_1$ and substitute that in to the first.

Solve the first equation for $x_4$ and substitute into the third equation.

Solve the third equation for $x_3$ and substitute into the fourth equation.

Solve the fourth equation for $x_2$ and then solve the system.

You should get the same answers I show in my augmented matrix.

I'll let you figure out the details and answer for the solution $W$ and is $W$ a vector space.


问:
$\lim\limits_{n\to \infty} \int_0^1 |f(x)-a_nx-b_n| dx=0$ implies $(a_n)_n,(b_n)_n$ are convergent

Let $f:[0,1] \to \mathbb R$ be a continuous function and the sequences $(a_n)_n,(b_n)_n$ s.t. $$\lim_{n\to \infty}  \int_0^1 |f(x)-a_nx-b_n| dx=0.$$ Prove that $(a_n)_n,(b_n)_n$ are convergent.

I know that $$\left|\int_0^1f(x)dx\right| \le \int_0^1|f(x)|dx.$$ Can somebody help me, please?

答:
Here is a elementary proof without Fubini, or any measure theory or TVS theory at all. Set

$\tag1 F(t)=\int_0^tf(x)dx\ \text{and}\ F_n(t)=\int_0^t(a_nx+b_n)dx.$

Then, $|F_n(t)-F(t)|\le \int_0^t|f(x)-(a_nx+b_n)|dx\le \int_0^1|f(x)-(a_nx+b_n)|dx\to 0$ so $F_n$ converges uniformly to $F$. That is,

$\tag2 \frac{a_n}{2}t^2+b_nt\to \int_0^tf(x)dx$

so it follows that

$\tag3 \frac{a_n}{2}+b_n\to \int_0^1f(x)dx$

$\tag4 \frac{a_n}{8}+\frac{b_n}{2}\to \int_0^{1/2}f(x)dx$

from which it follows that $a_n\to a$ and $b_n\to b$ for the real numbers $a,b$ which can be calculated in terms of the integrals on the RHS of these expressions. Now

$\int_0^1|f(x)-(ax+b)|dx=\int_0^1|f(x)-(ax+b)+(a_nx+b_n)-(a_nx+b_n)|dx\le$

$\int_0^1|f(x)-(a_nx+b_n)|dx+\int_0^1|(ax+b)-(a_nx+b_n)|dx\to 0$ as $n\to \infty.$ It follows that

$\tag5 \int_0^1|f(x)-(ax+b)|dx=0$

and since the integrand is continuous, that

$\tag6 f(x)-(ax+b)=0.$


问:
A tensor over $\mathbb R$ is sum of two decomposable tensors
Let $W$ be a vector space over $\mathbb R$ with basis $\{a,b\}$. Consider the tensor
$$a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b.$$

a) Show that this tensor cannot be presented as sum of two decomposable tensors.

b) Show that such presentation is possible in complexification of $W$.

My only attempt was to write an equation
$$(\alpha a + \beta b)\otimes (\gamma a + \delta b) \otimes (\epsilon a + \zeta b) + (\alpha' a + \beta' b)\otimes (\gamma' a + \delta' b) \otimes (\epsilon' a + \zeta' b) =$$
$$= a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b,$$
and from equality of coefficients deduce a system of 8 equalities and 12 variables. I think I can brute force it some way, but I was wondering if there are any other approaches.

答:
I'll start by stating some common theorems with tensor products.

1. Preliminaries

Theorem 1 (Zero-sum)

Let $U$,$V$ be two vector spaces on $F$, if for $n$ linearly independent vectors $x_1,x_2,...,x_n \in U$, there are $n$ vectors $v_1,...,v_n \in V$ such that:
$$x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n= 0 $$
then $v_1=v_2=...=v_n=0$
$\square$

Theorem 2 (Span)

Let $U$,$V$ be two vector spaces on $F$. $W$ is a subspace of $V$.

Let $ t $ be a tensor in $U \otimes W$ and
if for $n$ linearly independent vectors $x_1,x_2,...,x_n \in U$, there are $n$ vectors $v_1,...,v_n \in V$ such that:
$$t=x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n $$
then $\text{span}(v_1,...,v_n) \subset W$

$\square$

Remark 3: We will use mainly theorem 2, the proof of theorem 2 is based on theorem 1. Furthermore, we only need this theorem in the special case where $\dim W=2$

2. Main answer

Assume there are vectors $u_1 , u_2, u_3,v_1,v_2,v_3 \in W$ such that:
$$ a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b= u_1\otimes u_2 \otimes u_3+u_1\otimes u_2 \otimes u_3 =:t$$
On one hand, we see that:
$$ t \in W\otimes( \underbrace{\text{span}( u_2\otimes u_3, v_2 \otimes v_3)}_{=:U})$$
On the other hand,
$$ t= a \otimes ( a \otimes a +b \otimes b)+ b\otimes ( -b \otimes a + a \otimes b) $$
Then by theorem 2, we imply that $a \otimes a +b \otimes b$ and $ -b \otimes a + a \otimes b  $ lie in $U$.

By theorem 1, none of two above vector is null (and clearly they are independent) , whereas $\dim(U)\le 2$.

Therefore, $\dim(U)=2$ and $ U = \text{span}( a \otimes a +b \otimes b , -b \otimes a + a \otimes b ) $

So there are two elements $p,q \in F$ ( $F$ is the field on which we define $W$, in this case it is either $\mathbb{R}$ or $\mathbb{C}$) such that:
$$p (a \otimes a +b \otimes b ) +q( -b \otimes a + a \otimes b   ) =u_2 \otimes u_3$$

Or ,
$$ a \otimes( pa +qb) +b \otimes ( bp-qa) = u_2 \otimes u_3$$
Again, by theorem $2$ ( or by any other straightforward reasoning), as $a,b$ are independent, $pa+qb$ and $bp-qa$ cannot be independent ( as $\dim( \text{span}(u_3)) = 1$)

As $a,b$ are independent, the two above vector are linearly dependent iff $p^2+q^2=0$
This cannot occur in $\mathbb{R}$ unless $p=q=0$ which also leads to a contradiction.

Hence the conclusion for $\mathbb{R}$.

The conclusion for $\mathbb{C}$ (if true) is just a result from some simple calculations.

Remark Indeed, many arguments I presented above can be omitted or shortened given that we are familiar with the tensor product. I just tried to show this point as clear as possible.


答:
Here is a solution that reduces the problem to linear algebra. (There are probably other solutions.)

Reduction to linear algebra

Let $\{\varphi,\chi\}$ be a basis of $W^*$ with $\varphi(a) = \chi(b) = 1$ and $\varphi(b) = \chi(a) = 0$.
Suppose that
$$ a \mathbin{\otimes} a \mathbin{\otimes} a - b \mathbin{\otimes} b \mathbin{\otimes} a + a \mathbin{\otimes} b \mathbin{\otimes} b + b \mathbin{\otimes} a \mathbin{\otimes} b = u \mathbin{\otimes} v \mathbin{\otimes} w + x \mathbin{\otimes} y \mathbin{\otimes} z. $$
Applying $\varphi$ and $\chi$ in the first coordinate, we find
\begin{align*}
  a \mathbin{\otimes} a + b \mathbin{\otimes} b &= \varphi(u) v \mathbin{\otimes} w + \varphi(x) y \mathbin{\otimes} z;\\[1ex]
  -b \mathbin{\otimes} a + a \mathbin{\otimes} b &= \chi(u) v \mathbin{\otimes} w + \chi(x) y \mathbin{\otimes} z.\tag*{$(1)$}
\end{align*}
Therefore $C := v \mathbin{\otimes} w$ and $D := y \mathbin{\otimes} z$ are rank one matrices in $W \mathbin{\otimes} W$ with the property that both $a \mathbin{\otimes} a + b \mathbin{\otimes} b$ and $-b \mathbin{\otimes} a + a \mathbin{\otimes} b$ are linear combinations of $C$ and $D$. We will show that this is impossible over $\mathbb{R}$ and possible over $\mathbb{C}$. In the latter case, the obtained matrices $C$ and $D$ can easily be used to solve the original problem.

Solution

The preceding remarks show that the problem is equivalent to the following: find $2 \times 2$ matrices $C = (c_{ij})$ and $D = (d_{ij})$, each of rank one, and scalars $\lambda_1,\lambda_2,\mu_1,\mu_2$ such that
$$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \lambda_1 C + \lambda_2 D \qquad \text{and} \qquad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \mu_1 C + \mu_2 D. \tag*{$(2)$} $$
We prove that this is impossible over $\mathbb{R}$. Since $C$ and $D$ have rank one, we must have $\lambda_1,\lambda_2,\mu_1,\mu_2 \neq 0$, and we may choose some non-zero $p \in \ker(C)$. But then it follows from $(2)$ that
\begin{align*}
 p &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}p = (\lambda_1 C + \lambda_2 D)p = \lambda_2 Dp;\\[1ex]
 p' &:= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}p = (\mu_1 C + \mu_2 D)p = \mu_2 Dp;
\end{align*}
where $p'$ is $p$ rotated by $-90^\circ$. But then $Dp$ is simultaneously parallel ($Dp = \frac{1}{\lambda_2} p \neq 0$) and perpendicular ($Dp = \frac{1}{\mu_2} p' \neq 0$) to $p$, with $Dp \neq 0$. This is impossible.

Finally, over $\mathbb{C}$ one may take
$$ C = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \qquad \text{and} \qquad D = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}. $$
In order to turn this into a solution of the original problem, choose $u,x \in W$ such that $\varphi(u),\varphi(x),\chi(u),\chi(x)$ are the desired scalars in equation $(1)$.

Closing remarks

It can be shown that the approach taken here always works to determine the rank of a $3$-tensor:

Proposition ([Lan12, Thm 3.1.1.1]). Let $T \in U \mathbin{\otimes} V \mathbin{\otimes} W$. Then $\text{rank}(T)$ equals the number of rank one matrices needed to span $T(U^*) \subseteq V \mathbin{\otimes} W$.

Here we interpret $T$ as a linear map $U^* \to V \mathbin{\otimes} W$ in the natural way: if $T = \sum_{i=1}^r u_i \mathbin{\otimes} v_i \mathbin{\otimes} w_i$ and $\psi \in U^*$, then $T(\psi) = \sum_{i=1}^r \psi(u_i) v_i \mathbin{\otimes} w_i$. The proposition is not hard to prove, and generalizes the "reduction to linear algebra" paragraph above to arbitrary $3$-tensors.

References.

[Lan12] J. M. Landsberg, Tensors: Geometry and Applications, Graduate Studies in Mathematics 128, American Mathematical Society, 2012.

Title: A tensor over $\mathbb R$ is sum of two decomposable tensors

Body:
Let $W$ be a vector space over $\mathbb R$ with basis $\{a,b\}$. Consider the tensor
$$a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b.$$

a) Show that this tensor cannot be presented as sum of two decomposable tensors.

b) Show that such presentation is possible in complexification of $W$.

My only attempt was to write an equation
$$(\alpha a + \beta b)\otimes (\gamma a + \delta b) \otimes (\epsilon a + \zeta b) + (\alpha' a + \beta' b)\otimes (\gamma' a + \delta' b) \otimes (\epsilon' a + \zeta' b) =$$
$$= a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b,$$
and from equality of coefficients deduce a system of 8 equalities and 12 variables. I think I can brute force it some way, but I was wondering if there are any other approaches.


问:
Height of Pyramid from Two Angles and Side

Full Disclosure: I am a high school student.

Suppose that I have a right triangular pyramid, with triangle BCD being the base and A being the top point. D is a right angle. Angle ABD = 37deg, angle ACD = 32deg. What is the height of AD (the height of the pyramid)? Thanks!

I managed to get three equations and combine them, but each time I do that, it proves that 1=1 or that a side is = to a side, which is useless.

答:
Let $a=BD$, $b=CD$.

$$a^2+b^2=108^2 \tag{1}$$

$$h=b \tan 32° \tag{2}$$

$$h=a \tan 37° \tag{3}$$

Therefore $a \tan 37°=b \tan32°$ ($h$ has been eliminated, getting

$$a=b \dfrac{\tan 32°}{\tan 37°}$$

You have now a second equation with $a$ and $b$ only, that you can combine with (1) to obtain $b$ for example as a solution  a quadratic equation.

Then in a second step, you will obtain $h$ from (2) for example.


问:
Proof of validity of Maclaurin Series Expansions
In the derivation of the formula given by the Maclaurin series expansion of a function, given below,
$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(x)}{3!}x^3+\cdots$$
we begin by assuming that the function $f(x)$ can be written in the form
$$a_0+a_1x+a_2x^2+a_3x^3+\cdots$$

and then go on to differentaite repeatedly etc.

My question is, how do we prove that this assumption is correct and reasonable in the first place? As far as I can see, all we have shown is if our intial assumption is correct the $f(x)$ can be written as the power series stated above.

Thank you for your help.

答:
The assertion that there is a power series $\sum_{n=0}^\infty a_nx^n$ such that, near $0$, you have$$f(x)=\sum_{n=0}^\infty a_nx^n\tag1$$is actually equivalent to the assertion that (again, near $0$),$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots\tag2$$In that sense, yes, the assumption is correct and reasonable.

Here's a proof of the fact that they are equivalent. First of all, if a power series $\sum_{n=0}^\infty$ converges on some open interval centered at $0$, then its sum is a $C^\infty$ function, and its derivative is $\sum_{n=0}^\infty na_{n+1}x^n$. So, since we have $(1)$, then $f(0)=a_0$. Then $f'(x)=a_1+2a_2x+3a_3x^2+\cdots$, and therefore $a_1=f'(0)$. And so on… So, $(2)$ holds. And if $(2)$ holds, just take $a_n=\frac{f^{(n)}(0)}{n!}$ and you have $(1)$.

Note that if $f$ is a $C^\infty$ function, then the series $(2)$ still makes sense. And then an interesting problem is whether or not its sum is or not $f(x)$ near $0$. That doesn't always occur.


答:
Hint

$\ CD=h\cot 32\ $, $\ BD=h\cot 37\ $.  If you plug these identities into Pythagoras's theorem for the triangle $BCD$, you should be able to obtain a quadratic equation for $\ h\ $.

Title: Height of Pyramid from Two Angles and Side

Body:
Full Disclosure: I am a high school student.

Suppose that I have a right triangular pyramid, with triangle BCD being the base and A being the top point. D is a right angle. Angle ABD = 37deg, angle ACD = 32deg. What is the height of AD (the height of the pyramid)? Thanks!

I managed to get three equations and combine them, but each time I do that, it proves that 1=1 or that a side is = to a side, which is useless.


答:
http://upload.wikimedia.org/wikipedia/commons/8/8d/Haga_theorem_1.svg

If your piece of paper is a square ABCD of side length 1, then from the diagram, if AP=1/2, then QC=1/3.

Title: How can I fold paper into 3 x 4 grid? Or prove that it can't be done?

Body:
I am trying to fold paper so that it looks like 3 x 4 grid of 12 rectangles of equal size.

Like this https://www.wolframalpha.com/input/?i=3+x+4

Its easy to get 4 rectangles. Just fold twice. But how to get 3?


答:
Hint:

Welcome!
You have the relations
$$\frac h{BD}=\tan 37^{\circ}, \quad \frac h{CD}=\tan 32^{\circ}\iff BD=h\cot 37^{\circ}\quad CD=h\cot 32^{\circ}$$
Can you proceed?


答:
Functions that can be written as a power series are called analytic functions.

Given a function $f:\mathbb{R}\to\mathbb{R}$, one can ask, when can we write
$$
f(x)=\sum_{n=0}^{\infty} a_nx^n,\quad |x|<R
$$
for some $R>0$.

In a mathematical statement, we do not "prove" assumptions, which are given. Though it is natural to ask when such assumptions can be indeed satisfied. (A statement that has assumptions that can never be true is boring. For example, "if $1>2$, then $5>9$" is a true statement. But one does not "prove" that the assumption "$1>2$" is true.)

A necessary condition for having the power series above is that $f$ must be infinitely differentiable on $(-R,R)$. But this is not sufficient, as the classical example of bump functions show: https://en.wikipedia.org/wiki/Bump_function.

You seem to confuse yourself with what the statement is.

Suppose $f:(-R,R)\to\mathbb{R}$ can be written as
$$
f(x)=\sum_{n=0}^\infty a_nx^n, |x|<R
$$
Then, it is true that
$$
f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n, |x|<R
$$

In order to prove this assertion, one does NOT prove that $$
f(x)=\sum_{n=0}^\infty a_nx^n, |x|<R
$$
is true. One uses it as a given assumption.

Proof. By the assumption, $f$ is infinitely differentiable on $(-R,R)$. On the other hand, it is a property of the power series that one can differentiate term by term, namely,
$$
f'(x)=\sum_{n=1}^\infty a_n nx^{n-1},\quad |x|<R
$$
which implies that $f'(0)=a_1$. Similarly, one can show that
$$
f^{(n)}(0)=a_n n!
$$

Newly added.

As for when a given $C^\infty$ function can be written as a power series, one has the following theorem (see Stewart's Calculus).

Suppose $f:\mathbb{R}\to\mathbb{R}$ is an infinitely differentiable function. Let $a$ be a real number. Define $a_n=\frac{f^{(n)}(a)}{n!}$ for each $n$ with the convention that $f^{(0)}(a):=f(a)$ and $0!=1$. If for each $x$ with $|x-a|<R$,
$$
\lim_{n\to\infty} f(x)-T_n(x)=0
$$
where $T_n(x):=\sum_{k=0}^n a_n(x-a)^n$,
then
$$
f(x)=\sum_{n=0}^\infty a_n(x-a)^n,\quad |x-a|<R.
$$

This theorem is not so interesting in that it is simply unwrapping the definition of an infinite series.

You may also want to look at this post: Theorems that give sufficient condition for a $C^{\infty}$ function to be analytic

Title: Proof of validity of Maclaurin Series Expansions

Body:
In the derivation of the formula given by the Maclaurin series expansion of a function, given below,
$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(x)}{3!}x^3+\cdots$$
we begin by assuming that the function $f(x)$ can be written in the form
$$a_0+a_1x+a_2x^2+a_3x^3+\cdots$$

and then go on to differentaite repeatedly etc.

My question is, how do we prove that this assumption is correct and reasonable in the first place? As far as I can see, all we have shown is if our intial assumption is correct the $f(x)$ can be written as the power series stated above.

Thank you for your help.


问:
Find the sum of $\sum \frac{1}{k^2 - a^2}$ when $0<a<1$
So I have been trying for a few days to figure out the sum of

$$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical
analysis and CAS that this sum equals

$$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$

But I have not been able to prove this yet. Anyone know how? My guess is that the
sum of this series is related to fourier-series but nothing particalr comes to mind.

For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
S & \equiv \bbox[5px,#ffd]{\sum_{k = 1}^{\infty}
{1 \over k^{2} - a^{2}}} =
\sum_{k = 0}^{\infty}
{1 \over \pars{k + 1 - a}\pars{k + 1 + a}}
\\[5mm] & =
{\Psi\pars{1 - a} - \Psi\pars{1 + a} \over -2a} =
{\Psi\pars{1 - a} - \bracks{\Psi\pars{a} + 1/a} \over -2a}
\\[5mm] & =
{1 \over -2a}\bracks{\pi\cot\pars{\pi a} - {1 \over a}} =
\bbx{{1 \over 2a}\bracks{{1 \over a} - \pi\cot\pars{\pi a}}}
\\ &
\end{align}


Title: I need to develop $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ using the binomial coefficients formula and show a few properties

Body:
In the previous questions I've proved that $(1+i)^{2n+1}=a_n+ib_n$ where $a_n$ and $b_n$ are $\pm 2^n$ and that $(1-i)^{2n+1}=a_n - ib_n$ and that $|P_n(1)|=2^n$ using the previous statements. But now I need to show that the polynomial $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even (the coefficient in front of $X^n$ is non-zero only when $n$ is even, with real coefficients, and with $2n+1$ as a dominant coefficient, using the binomial coefficient formula.

I also need to show that the coefficient $X^{2n-2}$ in $P_n$  is $\frac{n(1-2n)(1+2n)}{3}$.

I developed $(X+i)^{2n+1}$ and I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$

When I develop $(X-i)^{2n+1}$ I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$, so the same thing.

So obviously when I subtract the two parts I get $0$ which won't let me prove the properties. Where is my mistake ? Do I use the binomial coefficient the wrong way ?

EDIT: I think I have made a mistake in the second development, the second term should has a + sign when the subtraction is made so I get $\frac{2(2n+1)X^{2n}i}{2i} = (2n+1)X^{2n}$

How do I show it is even though ? Is the end result  enough to show this property?

Body:
Since

$\begin{array}\\
(a+b)^n
&=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}\\
(x+i)^{2n+1}
&=\sum_{k=0}^{2n+1} \binom{2n+1}{k}x^ki^{2n+1-k}\\
&=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}i^{2n+1-2k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}i^{2n+1-(2k+1)}\\
&=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}i^{2(n-k)+1}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}i^{2(n-k)}\\
&=i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\\
(x-i)^{2n+1}
&=\sum_{k=0}^{2n+1} \binom{2n+1}{k}x^k(-1)^{2n+1-k}i^{2n+1-k}\\
&=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{2n+1-2k}i^{2n+1-2k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{2n+1-(2k+1)}i^{2n+1-(2k+1)}\\
&=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{2(n-k)+1}i^{2(n-k)+1}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{2(n-k)}i^{2(n-k)}\\
&=-i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\\
\end{array}
$

If we subtract the terms
the second term cancels out and we are left with

$\begin{array}\\
(x+i)^{2n+1}-(x-i)^{2n+1}
&=i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}-(-i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k})\\
&=2i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}\\
\text{so}\\
\dfrac{(x+i)^{2n+1}-(x-i)^{2n+1}}{2i}
&=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}\\
\end{array}
$

The highest order term has
$k = n$
so it is
$(2n+1)x^{2n}
$.

If we add the terms
instead of subtracting,
the first term cancels out and we are left with

$\begin{array}\\
\dfrac{(x+i)^{2n+1}+(x-i)^{2n+1}}{2}
&=\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\
\end{array}
$

The highest order term has
$k = n$
so it is
$x^{2n+1}
$.


问:
Compute $\frac{4}{5}\pi \int_0^{2\pi}\int_0^1\sqrt{v^2-2 v\cos\theta+1}\>v{\rm d}v\, {\rm d}\theta$
The integral

$$I=\frac{4}{5}\pi \int_0^{2\pi}\int_0^1\sqrt{v^2-2 v\cos\theta+1}\>v{\rm d}v\, {\rm d}\theta$$

arises from the calculation of the expected distance between two random points inside a unit circle. Can't figure out the change in variables to transform it to

$$I=\frac{4}{5}\pi \int_{-\pi/2}^{\pi/2}\int_0^{\sqrt{2(1+\cos2\theta')}} v'^2\,{\rm d}v'\,{\rm d}\theta'$$
Change of  variable

${}$

答:
Referring to the graph above, the variable changes used are

$$v = \sqrt{{v'}^2-2{v'}\cos\theta'+1},\>\>\>\>\>\>\>\frac{\sin\theta}{\sin\theta'}=\frac{v'}{v}
$$

Then, according to the Jacobian determinant

$$dv d\theta = \frac{v'}{v}dv' d\theta'$$

with the ranges $\theta'\in[-\frac\pi2, \frac\pi2]$ and $v'\in [0, \sqrt{2(1+\cos2\theta')}]$. Thus,
the integral is transformed to

$$I=\frac{4}{5}\pi \int_{-\pi/2}^{\pi/2}\int_0^{\sqrt{2(1+\cos2\theta')}} v'^2\,{\rm d}v'\,{\rm d}\theta'=\frac{128\pi}{45}$$


问:
How to write a point on an ellipse using r and theta
We can write any point on the circle as $(r\cos\theta,r\sin\theta$), Can we do samething for the ellipse?

答:
The parametrization of ellipse:

$$
\begin{cases}
x=a\cos\theta, \\
y=b\sin\theta.
\end{cases}
$$

For the circle $a=b=r$.


问:
Make an analysis of each of the following derivations in $ \mathbb S_1^= $, that is...
Make an analysis of each of the following derivations in $ \mathbb S_1^= $, that is, mark the axioms, the rules of inference, $ \mathbb S_1^= $ - theorem and, if applicable, the hypotheses auxiliaries used. However, the derivation may not be $ \mathbb S_1^= $ - deduction, that is, it may prove to be a deductive fallacy in $ \mathbb S_1^= $.

Let $ \lnot \exists x \alpha [x] $ and $ \forall x \lnot \alpha [x] \mathbb L_1^= $, such that

$ \lnot \exists x \alpha [x] \vdash_{S_1^=} \forall x \lnot \alpha [x] \mathbb L_1^= $.

I-) $ \lnot \exists x \alpha [x] $

II-) $ \lnot \lnot \alpha [c] $

III -) $ \alpha [c] $

IV -) $ \exists x \alpha [x] $

V -) $ \exists x \alpha [x] \land \lnot \exists x \alpha [x] $

VI -) $ \lnot \alpha [c] $

VII -) $ \forall x \lnot \alpha [x] $

V-) It is wrong, right?

答:
Well, it is valid, however you do not need to assume a double negation, when a position will do.

It also helps to indent or otherwise mark your subproofs.

$\def\fitch#1#2{\begin{array}{|l}#1\\\hline#2\end{array}}
\\\begin{array}{|l}~~1.~\lnot\exists x~\alpha[x]\hspace{16ex}\textsf{premise}\\\hline\begin{array}{|l}~~2.~\alpha[c]\hspace{20ex}\textsf{assume (with a fresh arbitrary variable)}\\\hline~~3.~\exists x~\alpha[x]\hspace{16.5ex}\textsf{existential generalisation}\\~~4.~\exists x~\alpha[x]\land\lnot\exists x~\alpha[x]\hspace{5ex}\textsf{conjunction}\end{array}\\~~5.~\lnot\alpha[c]\hspace{19.5ex}\textsf{indirect proof}\\~~6.~\forall x~\lnot\alpha[x]\hspace{16ex}\textsf{universal generalisation (since c is arbitrary)}\end{array}$


问:
I need to develop $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ using the binomial coefficients formula and show a few properties
In the previous questions I've proved that $(1+i)^{2n+1}=a_n+ib_n$ where $a_n$ and $b_n$ are $\pm 2^n$ and that $(1-i)^{2n+1}=a_n - ib_n$ and that $|P_n(1)|=2^n$ using the previous statements. But now I need to show that the polynomial $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even (the coefficient in front of $X^n$ is non-zero only when $n$ is even, with real coefficients, and with $2n+1$ as a dominant coefficient, using the binomial coefficient formula.

I also need to show that the coefficient $X^{2n-2}$ in $P_n$  is $\frac{n(1-2n)(1+2n)}{3}$.

I developed $(X+i)^{2n+1}$ and I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$

When I develop $(X-i)^{2n+1}$ I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$, so the same thing.

So obviously when I subtract the two parts I get $0$ which won't let me prove the properties. Where is my mistake ? Do I use the binomial coefficient the wrong way ?

EDIT: I think I have made a mistake in the second development, the second term should has a + sign when the subtraction is made so I get $\frac{2(2n+1)X^{2n}i}{2i} = (2n+1)X^{2n}$

How do I show it is even though ? Is the end result  enough to show this property?

Body:
From the binomial theorem,
$$(x+a)^{2n+1}=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}$$
so
$$\begin{align}
(x+a)^{2n+1}-(x-a)^{2n+1}&=\sum_{m=0}^{2n+1}{2n+1\choose m}x^m(a^{2n+1-m}-(-a)^{2n+1-m})\\
&=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}(1-(-1)^{2n+1-m})\\
&=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}(1+(-1)^{m}).
\end{align}$$
Then, when $m$ is even, $(-1)^m=1$, and $1+(-1)^m=1+1=2$. Thus all the terms where $m=2k$ is even are non-zero, and we have
$$\sum_{0\le m\le2n+1\\ m \text{ is even}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})=\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n+1-2k}(1+(-1)^{2k})\\
=2\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n-2k+1}.$$
Then, when $m$ is odd, $(-1)^m=-1$, and $1+(-1)^m=1-1=0$. Thus all the terms where $m=2k+1$ is odd are $0$, and we have
$$\sum_{0\le m\le2n+1\\ m \text{ is odd}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})=\sum_{k=0}^{n}{2n+1\choose 2k+1}x^{2k+1}a^{2n+1-2k-1}(1+(-1)^{2k+1})\\
=0.$$
Then,
$$\begin{align}
(x+a)^{2n+1}-(x-a)^{2n-1}&=\sum_{0\le m\le2n+1\\ m \text{ is even}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})\\
&+\sum_{0\le m\le2n+1\\ m \text{ is odd}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})\\
&=\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n+1-2k}(1+(-1)^{2k})\\
&=2\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n-2k+1}.
\end{align}$$
If you set $a=i$ you see that $a^{2n-2k+1}=(i^2)^{n-k}\cdot i=(-1)^{n-k}\cdot i$. Thus
$$(x+i)^{2n+1}-(x-i)^{2n+1}=2i\sum_{k=0}^{n}(-1)^{n-k}{2n+1\choose 2k}x^{2k},$$
and, finally,
$$p_n(x)=\frac1{2i}((x+i)^{2n+1}-(x-i)^{2n+1})=\sum_{k=0}^{n}(-1)^{n-k}{2n+1\choose 2k}x^{2k}.$$
You should be able to finish from here.


问:
A clarification on the definition of an interval
This an example of a definition of an interval as given by K.G. Binmore: Mathematical Analysis: A Straightforward Approach.

A (real) interval is a subset I of the real numbers such that:

$\forall x, y \in I: \forall z \in \mathbb{R} : ({x \le z \le y
 \implies z \in I})$

I'm wondering how does this definition fit together with open intervals where endpoints $x, y$ are not included in the interval. Does it still hold that $x, y \in I$?

Furthermore, how should I understand $[3,2] = \emptyset$ which is a closed interval? Shouldn't closed intervals by definition contain endpoints and thus the set is not empty?

答:
Open intervals are still fine by this definition.  In less compact language, the definition states that an interval $I$ is composed of all the numbers such that if you pick two numbers $x$ and $y$ in the interval, then every $z$ between them (including endpoints) is also in the interval.  So, if you have an open interval, meaning the endpoints are not included, then you can't take either of the endpoints as your $x$ and $y$.

As for the set $[3,2]$, remember that $$[3,2] = \{x\in \mathbb{R} : 3 \leq x \leq 2\}$$ by definition of the notation $[a,b]$.  However, in this case, as you've pointed out, $[3,2] = \emptyset$.  However, we can still think of it as the empty interval, because it is vacuously true that if you pick any two numbers $x,y$ in the empty set then every number in between those two numbers are also in the empty set (because no two such number exist in the first place).


问:
Find the volume generated by rotating the given region about the specified line
I am really unsure how to solve this problem. I was hoping someone can give me some helpful hints so I can figure it out. I also looked at some other posts but they weren't helpful

Find the volume generated by rotating the given region about the specified line.
$R1$ about $x = 0$

答: $\dfrac  { \pi}3$, because the slope is $x$, then you square it, then you take the antiderivative of $x^2$ from $0$ to $1$ times $\pi$, and get $\dfrac {x^3}3$ from $0$ to $1$ times $\pi$, and when you plug it in, you get $\dfrac {\pi}3$.


答:
A point on the ellipse is typically represented as $(a \cos(\theta), b \sin(\theta))$, where $a$ is the length of the semi-major (typically $X$) axis and $b$ is the length of the semi-minor (typically $Y$) axis. If you want to write the equation in polar form, then we have
$$r = a \sqrt{1-e^2 \sin^2(\theta)}$$
where $e$ is the eccentricity defined as $e = \sqrt{1-\left(\dfrac{b}a \right)^2}$, where $b \leq a$.

Title: How to write a point on an ellipse using r and theta

Body: We can write any point on the circle as $(r\cos\theta,r\sin\theta$), Can we do samething for the ellipse?


问:
Prove $\sin x < x < \tan x$ for $0 < x < \pi/2$ algebraically
I know there are some proofs showing that, for  $0 < x < \pi/2$:

$\sin x < x < \tan x $

with the use of the unit circle, however, can I prove this algebraically in some way as well?

答:
Since
$\sin' = \cos
$ and
$\cos' = -\sin$
and
$0 < \cos < 1$
for $0 < x < \pi/2$,
$\sin(x)
=\int_0^x \cos(t) dt
\lt x
$.

Since
$\tan'(x)
=\sec^2(x)
=\dfrac1{\cos^2(x)}
$,
$\tan(x)
=\int_0^x\dfrac1{\cos^2(t)}dt
\gt\int_0^x 1dt
=x
$.


问:
Use of definition of limit to prove $\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}$
I know that by definition I have to prove that $$\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹|\sqrt[3]{x}-2|<ϵ.$$

I have this:

$ |\sqrt[3]{x}-2|= |\dfrac{(\sqrt[3]{x}-2)((\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)}{(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2}|= \dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|}$.

I know that |x-2|<$\delta$. But I don't know what $\delta$ choose and how narrow down the denominator.

Body: If $x >0$ we have $  \dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|} \leq \frac {|x-2|}  {2^{2/3}}$ so you can take $\delta=\min \{1,\epsilon {2^{2/3}}\}$. [Note that $x>0$ for $|x-2| <\delta$].


答: With $a$ the major and $b$ the minor semiaxis, you get $(a\cos\theta,b\sin\theta)$ for your orientation. A rotated ellipse will be a bit harder, for this I'd compute the point in its unoriented form and then rotate afterwards. Note that the angle $\theta$ in the above is a parameter, but is not actually the angle as it is in the circular case.</p>


问:
Determinant of $I+aa^T+bb^T$
Given vectors $a$ and $b$, is there any good formula to find an equivalent form for

$$\det(I+aa^T+bb^T)$$

similar to the Sylvester’s determinant identity of $\det(I+aa^T)=1+a^Ta$?

答:
The Weinstein–Aronszajn identity states that $\det(I_m+AB) = \det(I_n+BA)$ where $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix.

If we apply this formula for $A = \begin{bmatrix}a & b\end{bmatrix}$ and $B = A^T = \begin{bmatrix}a^T \\ b^T\end{bmatrix}$, then we have $$\det(I+aa^T+bb^T) = \det(I+AB) = \det(I_2+BA) = \det\left(\begin{bmatrix}a^Ta+1 & a^Tb \\ b^Ta & b^Tb+1\end{bmatrix}\right) = (a^Ta+1)(b^Tb+1)-(a^Tb)(b^Ta) = (\|a\|_2^2+1)(\|b\|_2^2+1)-\langle a,b\rangle^2$$

Title: Determinant of $I+aa^T+bb^T$


答:
Let $$ A= \begin{pmatrix} & \\ a & b \\ & \end{pmatrix} $$
Then $I+aa^T+bb^T = I+AA^T$ and, according to the Weinstein–Aronszajn identity,
$$
\det(I+AA^T) = \det(I_2 +A^T A)
=\det\begin{pmatrix} 1+a^Ta & a^T b \\   b^T a & 1+b^T b \end{pmatrix}
\\
= (1+a^Ta)(1+b^T b) - (a^T b)^2
$$

Title: Determinant of $I+aa^T+bb^T$

Body:
Given vectors $a$ and $b$, is there any good formula to find an equivalent form for

$$\det(I+aa^T+bb^T)$$

similar to the Sylvester’s determinant identity of $\det(I+aa^T)=1+a^Ta$?


问:
Number of Zonotope Edges Parallel to Generator
Suppose we have a zonotope $Z$ that is the Minkowski sum of line segments $U_1+\dots +U_n$. All the edges of $Z$ are parallel to some $U_i$. Is it also true that the number of edges parallel to $U_i$ equals the number of edges parallel to $U_j$ for every $j\neq i$? Surely this question has been asked and probably answered before. I looked through a page of zonohedra and verified that they all answer the question true.

https://www.ics.uci.edu/~eppstein/junkyard/ukraine/ukraine.html

It is known that $Z$ is a zonotope iff $Z$ is the projection of a cube from $\mathbb{R}^n$. Of course, a cube in $\mathbb{R}^n$ is the sum of $n$ orthonormal vectors and has the same number of edges parallel to each of its generators. I suspect there may be a good induction proof down this way by adding one generator, $v$, at a time and  but haven't found one. The new generator divides the old zonotope into two pieces. The edges parallel to $v$ will stretch between the two halves.

Any help is appreciated!
Thanks! Aaron

答:
Partial answer.

This is true as long as no three of the $n$ vector summands are coplanar. In that case all the two dimensional faces are parallelograms. For each summand you build a zone of parallelograms by following an edge across to its opposite edge. Each pair of zones meets twice. That means there are $2(n-1)$ edges in each direction.

The truncated octahedron and the other examples in the page you link mean that this condition is not necessary. But something is, since the hexagonal prism fails, as @MWinter points out.


问:
Backtracking line search: Are the bounds on $\alpha$ $(0,1)$ or $(0, 1/2)$?
In my optimization class we are using the Boyd and Vandenberghe Convex Optimization textbook, which gives the following stopping criterion for backtracking line search (p. 464):

$$f(x + t \Delta x) \leq f(x) + \alpha t \nabla f(x)^T \Delta x$$

Here $\alpha$ is a parameter on $(0, 1/2)$. These bounds are used when proving the following fact related to gradient descent: If $0 \leq t \leq 1/M$ and $\alpha \lt 1/2$, then

$$f(x - t \nabla f(x)) \leq f(x) - \alpha t || \nabla f(x) ||^2$$

The proof given is

$$\begin{align}
f(x - t \nabla f(x)) & \leq f(x) - t || \nabla f(x) ||^2 + \frac{Mt^2}{2} || \nabla f(x) ||^2 &\text{(Taylor's theorem)}\\
& \leq f(x) - \frac{t}{2} || \nabla f(x) ||^2 & (\text{because} -t + Mt^2/2 \leq -t/2)\\
& \leq f(x) - \alpha t || \nabla f(x) ||^2
\end{align}$$

But considering the geometry of the method, it seems to me like it should be "safe" to pick any $\alpha \lt 1$. And this more general limit is mentioned on Wikipedia and this wiki.

Beyond the simplicity of the proof above, is there a deeper reason for restricting $\alpha$ to $(0, 1/2)$?

答:
First, for gradient descent (or steepest descent in general), $\alpha\in (0,1)$ is enough. It is easy to adapt the proof presented in the book to this case. I show below such a proof for gradient descent, but for steepest descent it is similar.

We will show that for $0\le t\le \frac{2(1-\alpha)}{M}$ and $\alpha\in (0,1)$ we have
$$f(x - t \nabla f(x)) \leq f(x) - \alpha t \| \nabla f(x) \|_2^2.$$
Indeed,
\begin{align}
f(x - t \nabla f(x)) & \leq f(x) - t \| \nabla f(x) \|_2^2 + \frac{Mt^2}{2} \| \nabla f(x) \|_2^2\\
& = f(x) - \left(1 - \frac{Mt}{2} \right)t \|\nabla f(x) \|^2 \\
& \leq f(x) - \alpha t || \nabla f(x) ||^2.
\end{align}
Now the proof proceeds as presented in the book. Update for clarification: You should replace accordingly the values in the remaining part of the proof. For example, instead of "Therefore the backtracking line search terminates either with $t = 1$ or with a value $t\ge \beta/M$", it should now reads "Therefore the backtracking line search terminates either with $t = 1$ or with a value $t\ge 2(1-\alpha)\beta/M$". And so on.

Second, and more interestingly:

Beyond the simplicity of the proof above, is there a deeper reason for
restricting $\alpha$ to $(0,1/2)$.

The reason why the authors restricted $\alpha \in (0,1/2)$ is, apparently, because they want to use the same bounds of $\alpha$ for all the methods presented in that chapter of the book. The tricky part here is that the analysis of the damped Newton method requires $\alpha < 1/2$ (pages 490-491, and later on page 505 for self-concordant functions). For the other ones, $(0,1)$ is enough (but the authors took advantage of their assumption $\alpha < 1/2$ to simplify expressions, so their proofs need to be adapted to the $(0,1)$ case).

A better presentation would have been using $(0,1)$ for the general case, and restricting to $(0,1/2)$ when necessary. (Or at least making it clear when $(0,1/2)$ is necessary instead of "The reason for requiring $\alpha$ to be smaller than $0.5$ will become clear later.")


问:
If matrix $B$ is skew-symmetric with eigenvalue $r$, then $-r$ is also an eigenvalue of $B$
I've considered the determinants of both $B$ and $-B^T$, and found that

$\det(B) = (-1)^n \det(B)$, where $B$ is a $n \times n$ matrix.

I've also tried the approach where $r$ is an eigenvalue of $B$ iff $\det(B-rI) = 0$, but as determinants are not preserved over addition/subtraction,
I'm not really sure what to do from there.

答: Hint: Note that $r$ is an eigenvalue of $B$ if and only if $\det(B - rI) = 0$. Note also that $\det(A) = \det(A^T)$ for any matrix $A$.


问:
Whether a Conditional Multinomial Distribution is still a Multinomial Distribution?
Consider $(X_1, X_2, X_3, X_4)$ is a multinomial distributed with parameters $n$ and p $= (p_1, p_2, p_3, p_4)$. What is the distribution of $X_1$ given $X_3+X_4 = m$?

From the conditional probability, we know $$
P(X_1 = x_1 | X_3 + X_4 = m) = \frac{P(X_1 = x_1, X_3 + X_4 = m)}{P(X_3 + X_4 = m)},
$$ then what we need to figure out are the denominator and numerator.

For the denominator, we can write it as (not sure)$$
P(X_3 + X_4 = m) = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(1-p_3-p_4)^{n-m} = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m},
$$ since $\sum p_i = 1$.

For the numerator, we can write it as (not sure)$$
P(X_1 = x_1, X_3 + X_4 = m) = P(X_1 = x_1, X_3 = m - x_4, X_4 = x_4, \sum_{k/\{1,3,4\}}X_k = n-m-x_1) = \frac{n!}{x_1!(m-x_4)!x_4!(n-m-x_1)!}p_1^{x_1}p_3^{m-x_4}p_4^{x_4}p_2^{n-m-x_1}.
$$

Plug 1. and 2. into the conditional probability and do some algebric calculation, it eventually yields to $$
P(X_1 = x_1 | X_3 + X_4 = m) = \frac{m!(n-m!)}{x_1!(m-x_4)!x_4!(n-m-x_1)!}(\frac{p_1}{p_2})^{x_1}(\frac{p_3}{p_3+p_4})^m(\frac{p_2}{p_1+p_2})^{n-m}(\frac{p_4}{p_3})^{x_4}.
$$ But this one seems it is not a multinomial distribution, so I am wondering whether the calculation is not correct or the conditional distribution of multinomial is not needed to be multinomial distribution?

Body: For the denominator you have already found $$P(X_3 + X_4 = m) = \frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m}$$ so the numerator would be $$P(X_1=x, X_3 + X_4 = m) = \frac{n!}{m!x!(n-m-x)!}(p_3+p_4)^{m}p_1^x p_2^{n-m-x}$$ making $$P(X_1=x \mid X_3 + X_4 = m) = \dfrac{\frac{n!}{m!x!(n-m-x)!}(p_3+p_4)^{m}p_1^x p_2^{n-m-x}}{\frac{n!}{m!(n-m)!}(p_3+p_4)^{m}(p_1+p_2)^{n-m}} \\= \dfrac{{(n-m)!}p_1^x p_2^{n-m-x}}{{x!(n-m-x)!}(p_1+p_2)^{n-m}} \\= {n-m \choose x}\left(\frac{p_1}{p_1+p_2}\right)^x\left(1-\frac{p_1}{p_1+p_2}\right)^{n-m-x}$$  which is a binomial distribution with parameters $n-m$ and $\frac{p_1}{p_1+p_2}$.


问:
Why is the following Calculus Theorem True?
If $f_x$ and $f_y$ exist and are continuous near $(a,b)$ for a function $f(x,y)$, then the function $f$ is differentiable at $(a,b)$.

Could someone explain a geometric meaning or give an intuitive explanation of why
this theorem holds true? I don't understand why these two conditions are
sufficient on their own for a differentiable function.

答: For $f_x$ and $f_y$ to be exist and be continuous it means that there are no "pointy" corners near f(a,b). Corners like you would get from $f(x,y) = |x| + |y|$ at (0,0).  The hard part to visualize is could there be a corner at some other axis, like a diagonal.  Basically the theorem is saying an fold in the fabric of f(x,y) near (a,b) would cause a non-continuous partial differential in either $f_x$, $f_y$ or both.


问:
Let $H\unlhd G$ and $a\in G$. If the coset $aH$ has order $3$ in $G/H$ and $|H| = 10$, what are the possible $|a|$ in $G$?

Context

I am an undergrad student taking Abstract Algebra I, and this is a problem on a homework assignment.

Problem

Let $H$ be a normal subgroup of $G$, and let $a ∈ G$. If the coset $aH$ has order 3 in
the factor group $G/H$, and $|H| = 10$, what are the possible orders of $a$ in $G$?

What I know

Since $H$ is a normal subgroup, its left coset should equal its right coset.

Since $H$ has 10 elements, the highest order of an element should be 10.

答:
Hint:

The order of $a^3$ is a divisor of the order of $H$ (Lagrange's theorem). On the other hand, if the order of $a$ is $d$, the order of $a^3$ is $\dfrac d{\gcd(d,3)}$. Can you deduce the possibilities for $d$?


问:
Ring of polynomials become a field
The K vector space K[α] is of finite dimension.
Then The ring K[α] is field.
Where α is an algebraic element..
I want an explanation and an example, i thought x doesn't have an inverse thus K[x] can never be a field just a ring, how come its a field here???</p>

答:
Let's take an example, maybe that will make things clearer.

Consider $\Bbb Q[\sqrt2]$. It is, in some sense, the polynomial ring over $\Bbb Q$ in the variable $\sqrt2$. A typical element looks like
$$
5\sqrt2^3-\frac75\sqrt2^2+2\sqrt2-\frac{140}{23}
$$
However, it is also the case that $\sqrt2^2=2$, which means that the above can be simplified to
$$
12\sqrt2-\frac{1022}{115}
$$
This simplification allows us to construct inverses to any non-zero element of $\Bbb Q[\sqrt2]$. For instance, the above element has inverse
$$
\frac{13\,225}{2\,764\,316}\left(12\sqrt2+\frac{1022}{115}\right)
$$

Title: Ring of polynomials become a field


问:
If I flip four coins, what is the probability that the first coin is heads AND there are more heads than tails?
I don't know why this answer is eluding me, but I had to figure it out somehow. I know that the probability of flipping heads first is 1/2. 8/16 outcomes have H first. The probability of having more heads than tails is 5/16: HHHH, HHHT, HHTH, HTHH, THHH. I realize that I can count it and that four of these results have H first, meaning that the answer is 4/16 or 1/4. However, how is this calculated mathematically? You can not simply multiply the two probabilities as they are inclusive. I know that they are independent as well. The first coin flip doesn't matter to having more heads than tails as it is still possible regardless. What probability formula would I need to solve this?</p>

答:
You are correct, you cannot multiply the probabilities. However this is because the events are not independent.

The outcomes for obtaining heads first and more heads are: $\sf\{HHHH, HHHT,HHTH,HTHH\}$, which are four from the sixteen total outcomes. (Thus the answer is $1/4$).

The outcomes for obtaining tails first and more heads are: $\sf\{THHH\}$, which is one from the sixteen total outcomes.

Thus there is a dependency between the event of "the first is a head" and "there are more heads than tails."

You seek the probability that you have obtained a head first, and two or three heads among the remaining three. That is expressing the favored event as an intersection of independent events, so...$$\dfrac 12\cdot\left(\binom 32+1\right)\dfrac 1{2^3}=\dfrac 14$$


问:
Show that the inclusion $\mathbb{RP}^2\to\mathbb{RP}^3$ induced by the inclusion $S^2\to S^3$ is not homotopic to the constant map
I'm dealing with the following question:

Consider the inclusion $i : \mathbb{R}P^{2} \rightarrow \mathbb{R}P^{3}$ induced by the inclusion of $S^{2}$ into $S^{3}$ as its equator.
Show that $i$ is not homotopic to a constant map.

It's a practice problem, but I'm stuck on how to show this. I feel as if it's a very simple step that I'm missing?

答:
It is well-known that $\mathbb RP^n$ has a CW-structure with one cell in each dimension. In  fact, $\mathbb RP^n$ is obtained from $\mathbb RP^{n-1}$ by attaching an $n$-cell. See CW complex structure of the projective space $\mathbb{RP}^n$. Thus the $k$-skeleton of $\mathbb RP^n$ is $\mathbb RP^k$. By cellular approximation we see that the inclusion $i : \mathbb RP^2 \to \mathbb RP^3$ induces an isomorphism
$$i_* : \pi_1(\mathbb RP^2) \to \pi_1(\mathbb RP^3).$$
This is just a special case of the fact the the inclusion $i : X^2 \to X$ of the $2$-skeleton $X^2$ of a CW-complex $X$ induces an isomorphism $i_* : \pi_1(X^2) \to \pi_1(X)$.

But is well-known that $\pi_1(\mathbb RP^2) \approx \mathbb Z_2$, thus $i$ cannot be null-homotopic (in that case we would have $i_* = 0$ which is not an isomorphism).


问:
Maximising the variance in PCA - Trace Proof
Quite urgent request for a proof here as my professor has not run through it.

Scenario is the PCA problem where we are trying to maximise the variance in the latent space. I want to prove the following $\sum_k^d W_k^TS_tW_k = Trace[W^TS_tW]$ where $S_t$ is the covariance matrix of dimension $F \times F$. $W_k$ is the $k^{th}$ eigenvector of $S_t$ $\in F \times 1$ and $W = [W_1, W_2, .., W_k]$.

I sort of know the intuition behind it, the eigenvectors are orthogonal and thus $W_k^TW_j$ where $k \neq j = 0$. I am having issue with the formal proof.

I have tried the following

$\sum_k^d \big(\sum_i^F W_{k,i}St_{i,j} W_{j,k}\big)$ and stated that if $i \neq j, W_{j,k}W_{k,i} = 0$

Does anyone have a better proof for this please? or provide a hint as to where I am going wrong

答:
With help of the trace identities $a^Ta = \text{tr}(aa^T)$ and $\text{tr}(ABC) = \text{tr}(BCA)$ for vector $a$ and suitable martrices $A,B,C$ we get

$$\text{tr}\left(W^TS_tW\right)
  = \text{tr}\left(S_tWW^T\right)
  = \sum_{k=1}^d\text{tr}\left(S_tW_kW_k^T\right)
  = \sum_{k=1}^d\left(S_tW_k\right)^TW_k
 = \sum_{k=1}^dW_k^TS_tW_k.
$$


问:
Show that if $ad-bc \ne 0$ then a system of equations has a unique solution. Know answer. Trouble understanding it.
I am starting through the free Hefferon Linear Algebra book. I am stuck on the problem below. What I don't get is the "We take 3 cases" part right at the beginning. Why those 3 cases? What significance do those 3 cases have? What are they related to? Why are we just plugging in zero and non-zero, and not say every kind of number? This is my first real exposure to proofs and it shows. Any help with this is greatly appreciated!

Problem

Solution

答:
We can actually avoid all this splitting into cases. If we multiply the first equation by $d$ and the second by $b$, the equations become
$$\begin{cases}
  ad\,x+bd\,y=d\,j\\
  \kern3.8mu bc\,x+bd\,y=b\,k.
\end{cases}$$
Subtracting one from the other yields
$$(ad-bc)x = dj-bk\implies x=\frac{dj-bk}{ad-bc},$$
where you'll notice that I could divide by $ad-bc$ precisely since it wasn't zero.

By precisely similar reasoning (instead multiplying the first by $c$ and the second by $a$), we can subtract them and find the explicit unique solution $y=(cj-ak)/(ad-bc)$ for $y$.

On the other hand, if $ad-bc$ is zero, then the equation we got for $x$ above is just $0x=dj-bk$. This either has no solutions (if $dj-bk$ is non-zero, we want to find $x$ such that $0x=\text{a non-zero number}$, which is impossible), or infinitely many (if $dj-bk$ is zero, then the equation is $0x=0$, which is true for any $x$), and the exact same goes for the equation we would get for $y$.

With regards to all the casework, there is no reason why doing it that way is fundamentally important.  The simple fact is that if you want to divide by $a$, you need to assume that $a\neq 0$, and then think about the specific case when $a=0$ separately afterwards. Similarly for $c$. This is why the author decided to split into cases. But as I showed above, it isn't necessary to prove what you want. (Also as saulspatz pointed out, you usually realise you need to split in to cases half way through, not at the beginning.)


问:
Test if population standard deviation is less than 7.2.
With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon.

With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers.

Id like to test the hypothesis that the waiting times for a single line are longer.
For this problem, I get Chi critical value of 36.41502850. Which is much larger than noted in the solution. If Chi squared test statistic value is larger than the critical value, then the null hypothesis ( that the waiting times for singe line has a more variation could be rejected? ) What if one would like to determine if one has to wait longer than 7.2 minutes? Then I guess the appropriate F score would be 1-$/alpha$ quantile?

答:
@Tomik (+1) has already indicated that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = 25-1 =  24).$
Notice that your problem gives standard deviations,
not variances. You should already have enough hints to
do do the test. So I just show output from a statistical
software package that does this test without requiring
the data instead of just the SD.

Output from a recent release of Minitab.

```<code>Test and CI for One Variance

Method

Null hypothesis         σ = 7.2
Alternative hypothesis  σ < 7.2

The chi-square method is only for the normal distribution.

Statistics

 N  StDev  Variance
25   3.50      12.3

95% One-Sided Confidence Intervals

            Upper
            Bound
              for   Upper Bound
Method      StDev  for Variance
Chi-Square   4.61          21.2

Tests

                 Test
Method      Statistic  DF  P-Value
Chi-Square       5.67  24    0.000
</code>```

The value 13.85 cuts probability 0.05 from
the lower tail of $\mathsf{Chisq}(\nu = 24).$
[You may be able to get that value from a printed chi-squared CDF table; or with software.]
The P-value of your test is the area under the density curve of this distribution to the left of 5.67; it's 0.000042.
(about 0). [This computation requires software.]

Body:
@Tomik (+1) has already indicated that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = 25-1 =  24).$
Notice that your problem gives standard deviations,
not variances. You should already have enough hints to
do do the test. So I just show output from a statistical
software package that does this test without requiring
the data instead of just the SD.

Output from a recent release of Minitab.

Test and CI for One Variance

Method

Null hypothesis         σ = 7.2
Alternative hypothesis  σ < 7.2

The chi-square method is only for the normal distribution.

Statistics

 N  StDev  Variance
25   3.50      12.3

95% One-Sided Confidence Intervals

            Upper
            Bound
              for   Upper Bound
Method      StDev  for Variance
Chi-Square   4.61          21.2

Tests

                 Test
Method      Statistic  DF  P-Value
Chi-Square       5.67  24    0.000

The value 13.85 cuts probability 0.05 from
the lower tail of $\mathsf{Chisq}(\nu = 24).$
[You may be able to get that value from a printed chi-squared CDF table; or with software.]
The P-value of your test is the area under the density curve of this distribution to the left of 5.67; it's 0.000042.
(about 0). [This computation requires software.]


问:
triangle combination edge colored
Stacked triangles

hi, im stucked at this problem and i dont know, how to move on.

The problem sounds like:

we have N stacked triangles (picture) We color edges of these triangles,that way where at least one edge in each triangle is colored.How many ways we have to  color these edges.

In the picture you can see, there is help. That sounds:

Finding recursion divide into two cases- a)edge 1,2 is not colored.

b) atleast one of 1,2 edge is colored

I appreciate every help.

答:
Instead of linear recurrence relation, one can analyze this problem using transfer matrix.

Let $T_1,\ldots,T_n$ be the $n$ triangles from left to right.

For $1 \le k \le n$, let $e_{k-1}$ be the left edge, $e_k$ be the right edge and $h_k$ be the horizontal edges of triangle $T_k$.

For $0 \le k \le n$,  let $u_k = 1$ if $e_k$ is colored and $0$ otherwise.

For each triangle, let $N_k(u_{k-1},u_k)$ be the possible choices to coloring of $h_k$ to meet the constraint (at least one edge of  $T_k$ is colored) given $u_{k-1}$ and $u_{k}$.

It is easy to see

$$N_k(u_{k-1},u_k) = \begin{cases}1, & u_{k-1} = u_k = 0\\
2,& \text{otherwise}\end{cases}$$

The total number of triangles we want, let's call it $\mathcal{N}_n$, equals to
following sum over all possible choices of $u_0,\ldots,u_n$.
$$\mathcal{N} = \sum_{u_0,\ldots,u_n \in \{0,1\}} N_1(u_0,u_1)N_2(u_1,u_2)\cdots N_n(u_{n-1},u_n)$$

Let $\Delta$ be the $2 \times 2$ matrix $\begin{bmatrix}1 & 2\\2 & 2\end{bmatrix}$ (with indices running over $0$ and $1$). In terms of $\Delta$, above sum can be rewritten as

$$\mathcal{N}_n = \sum_{u_0,\ldots,u_n \in \{0,1\}} \Delta_{u_0u_1}\Delta_{u_1u_2}\ldots\Delta_{u_{n-1}u_n}$$

Replace inner sums by matrix product and let $U$ be the $2 \times 2$ matrix with all entries $1$, we can rewrite above sum as

$$\mathcal{N}_n = \sum_{u_0,u_n\in\{0,1\}} (\Delta^n)_{u_0u_n} = {\rm Tr}(\Delta^n U)\tag{*1}$$

$\Delta$ has two simple eigenvalues $\lambda_{\pm} = \frac{3 \pm \sqrt{17}}{2}$. By Cayley Hamiltion theorem, we have

$$(\Delta - \lambda_+ I_2)(\Delta - \lambda_- I_2) = 0
\quad\implies\quad \Delta (\Delta - \lambda_{\pm}I_2) = \lambda_{\mp}(\Delta - \lambda_{\pm}I_2)\tag{*2}
$$
Decompose $I_2$ as $\frac{1}{\lambda_{-}-\lambda_{+}}((\Delta - \lambda _{+}I_2) - (\Delta - \lambda_{-}I_2))$ and apply $\Delta^n$ on both sides. $(*2)$ tell us

$$\Delta^n = \frac{1}{\lambda_{-} - \lambda_{+}}\left(\lambda_{-}^n(\Delta - \lambda_{+}I_2) - \lambda_{+}^n(\Delta - \lambda_{-}I_2)\right)$$

Substitute this into $(*1)$ and notice
${\rm Tr}(\Delta U) = 7$ and ${\rm Tr}(U) = 2$, we obtain following closed form of $\mathcal{N}_n$:

$$\begin{align}\mathcal{N}_n
&=
\frac{1}{\lambda_{+} - \lambda_{-}}
\left(\lambda_{+}^n (7 - 2\lambda_{-}) - \lambda_{-}^n(7 - 2\lambda_{+})\right)\\
&= \frac{1}{\sqrt{17}}
\left[(4 + \sqrt{17})\left(\frac{3+\sqrt{17}}{2}\right)^n -
(4-\sqrt{17})\left(\frac{3-\sqrt{17}}{2}\right)^n\right]
\end{align}
$$

For the first few $n$, this redproduces the number on OEIS A007484
$$(\mathcal{N}_1,\mathcal{N}_2,\ldots) = (7,25,89,317,1129,4021,14321,51005,181657,646981,...)$$

Title: triangle combination edge colored


问:
Inner measure of the Irrationals inside $[0,1]$?
Here, I am taking Lebesgue measure. There is a notion that a set $E$ is measruable iff $m_*(E) = m^*(E)$. Let $E' = [0,1] \cap \mathbb I$, where $\mathbb I$ denote the set of irrational numbers.

By definition, $m_*(E) = \sup_{F}\{|F| \mid F \subset E, F \text{ compact}\}$. Then, $m_*(E') = \sup_{F}\{m^*(F) \mid F \subset E', F \text{ compact}\}$. Take some compact $F \subset E'$ such that $m^*(E') < m^*(F) + \epsilon$. Then,

$F$ must not contain a closed interval.

$F$ is nowhere dense in $[0,1]$ since if it is dense in some $O \subset [0,1]$, then compactness $F$ implies that $\mathbb Q \cap F \neq \varnothing$.

$F$ has strictly positive measure.

How is this possible?

答:
You actually know that there exists by other routes, in fact you know that $|[0,1]\cap \mathbb{Q}|=0$ so it is measurable in the Lebesgue complete measure and it implies that $|[0,1]\cap \mathbb{I}|=1$.

So you know it is possible, the only thing is "how?"; well, we know there exist some ugly sets out there, like the Cantor set, actually you may be able to use a generalized Cantor set.

An other idea is to use the fact that the sets $\{ x \}$ with $x\in [0,1]\cap \mathbb{I}$ are closed and bounded, so compact, and also every finite union of these singletons. It is true that their measure is $0$, but we are talking about their $\sup$, so there is hope.


问:
How to show that the definition of differential of a map between two regular surfaces is independent of the choice of curve?
We define for $q \in S$ where $S$ is arbitrary regular surface, the tangent space at $q$ as
$T_q(S) = \{\alpha'(0) \;|\; \alpha :(-\epsilon, \epsilon) \to S \text{ a smooth curve such that } \alpha(0) = q \}$

Let $S_1$ and $S_2$ be two regular surfaces. Let $p \in S_1$ and $f:S_1\to S_2$ be differentiable at $p$. Then we define the differential map as

$Df_p : T_p(S_1) \to T_{f(p)}(S_2)$ defined as following:

Let $w$ in $T_p(S_1) \rightarrow \exists\;$$\alpha : (-\epsilon, \epsilon)\to S_1$ such that $\alpha(0) = p$ and $w = \alpha'(0)$. Now we map $w$ to $(f\circ \alpha)'(0)$, ie, $Df_p(w) = (f\circ \alpha)'(0)$. This is meaningful since, $(f\circ\alpha)(0) = f(p)$ and hence, $(f\circ \alpha)'(0) \in T_{f(p)}(S_2)$

Now what I want to show is that, $Df_p(w) = (f\circ \alpha)'(0) = (f\circ \gamma)'(0)$ for any $\gamma:(-\epsilon, \epsilon)\to S_1$ such that $\gamma(0)= p$ and $\gamma'(0) = w$

How do I show this precisely?

答:
Essentially it is because of the chain rule.

For functions $F \colon M \to N$ and $G \colon N \to K$, the differential of the composite map $G \circ F \colon M \to K$ is
$$ d(G \circ F)_m = dG_{F(m)} \circ dF_m $$

In your case, you have
$$ (f \circ \gamma)'(0) = df_{\gamma(0)} \circ \gamma'(0) $$

So if $\gamma'(0) = \alpha'(0)$, then you get $(f \circ \alpha)'(0) = (f \circ \gamma)'(0)$.


问:
Venn diagram method?
I'm not that bad at Venn diagram but I got confused on this question.

At a college, $90$ percent of students who take physics also take math and $60$ percent of students who take math also take physics. If $30$ percent of students at the college take physics, what percent of students take math?

I don't mind to learn a new way to do this kind of problems except sketching venn diagram. Any suggestion would be appreciated.

答:
Let $n$ be the number of students at the college, let $p$ be the number of students taking physics, and let $m$ be the number of students taking math. Then:

$p = 0.3n$ students take physics.

Out of these $p$ physics students, $0.9p$ of them also take math.

But we also know that $0.6m$ students take both physics and math so that $0.6m = 0.9p$.

Thus, we find that:
$$
\frac{m}{n}
= \frac{0.6m}{0.6n}
= \frac{0.9p}{0.6n}
= \frac{0.9(0.3n)}{0.6n}
= 0.45
$$
So $45\%$ of students take math.

Title: Venn diagram method?


问:
Proving weak convergence for a specific function.
So the function is
$$u_n(x)=\begin{cases} a&\mbox{ if } x\in (\frac{i}{n},\frac{i+1/2}{n}], i=0,\dots,n-1,
\\ b&\mbox{ if } x\in(\frac{i+1/2}{n},\frac{i+1}{n}],i=0,\dots,n-1.\end{cases}$$
in $\textit{L}^1((0,1))$ and i want to show $u_n \rightharpoonup (a+b)/2$.

Is anybody able to give a cleaver, or anyways simple proof I couldn't see?
I think i have successfully proven it, although it took way to much effort on a specific step of my "proof".

For anybody who is interested this is my attempt:

Using density of the simple functions in $\textit{L}^{\infty}((0,1))$, i must prove the definition for weak converges holds, but i can restrict the argument to simple functions.

So given a simple function $\overline{g}$ i know there exist costants $c_1=0<c_2<\cdots<c_k=1$ such that $\overline{g}$ is costant on these, so you can come to this partial conclusion:
$$\lim_{n\to +\infty} \int_{(0,1)}\overline{g}u_ndm=\lim_{n\to +\infty}\sum_{j=1}^{k-1}\overline{g}(c_i)\int_{c_j}^{c_{j+1}}u_n(z)dz=\sum_{j=1}^{k-1}\overline{g}(c_j)\lim_{n\to +\infty}\int_{c_j}^{c_{j+1}}u_n(z)dz=$$
$$= \sum_{j=1}^{k-1}\overline{g}(c_j)\biggl[ \lim_{n\to +\infty}\int_{(0,1)}a\chi_{E_n^{(j)}}(z)dz + \lim_{n\to +\infty}\int _{(0,1)}b\chi_{F_n^{(j)}}(z)dz \biggl] =$$
$$= (\star)\sum_{j=1}^{k-1}\overline{g}(c_j)\biggl[ a \lim_{n\to +\infty}|E_n^{(j)}| + b \lim_{n\to +\infty}|F_n^{(j)}| \biggl](\star)$$
where
$$E_n^{(j)}:=\biggl{\{} \bigcup_{i\le n-1}\biggl(\frac{i}{n},\frac{i+1/2}{n} \biggr] \bigcap (c_j,c_{j+1}) \biggr{\}} \mbox{ e } F_n^{(j)}:=\biggl{\{} \bigcup_{i\le n-1}\biggl(\frac{i+1/2}{n},\frac{i+1}{n} \biggr] \bigcap (c_j,c_{j+1}) \biggr{\}} \mbox.$$

These next step is where i think i put to much effort for a proof like this, but i am simply finding the limit of the measure of these two sets.

Lets take  $I_n^{(j)}:=\{ i\le n-1:\bigl(\frac{i}{n},\frac{i+1}{n}\bigr]\subseteq (c_j,c_{j+1}) \}$, by definition
$$(\star \star)E_n^{(j)} \supseteq \bigcup_{i\in I_n^{(j)}}\biggl(\frac{i}{n},\frac{i+1/2}{n}\biggr] \mbox{ e } F_n^{(j)} \supseteq \bigcup_{i\in I_n^{(j)}}\biggl(\frac{i+1/2}{n},\frac{i+1}{n}\biggr] (\star \star)\mbox,$$

and we can observe that for $n \ge \frac{2}{(c_{j+1}-c_j)}$ we get
$$\bigcup_{l=0}^{\lfloor n(c_{j+1}-c_j)\rfloor-1} \biggl( c_j+\frac{l}{n},c_j+\frac{l+1}{n} \biggr]  \mbox{, with measure } \le |(c_j,c_{j+1})|$$
and $\exists$ $\overline{i}\in I_n^{(j)}$ such that
$$\frac{\overline{i}}{n}\in \biggl[c_j,c_j + \frac{1}{n}\biggr)\mbox,$$
We can observe that we can obtain our sets
$$ \bigcup_{i\in I_n^{(j)}}\biggl(\frac{i}{n},\frac{i+1}{n}\biggr] \mbox,$$
by a translation $\le 1/n$ to the right of the sets $\bigcup_{l=0}^{\lfloor n(c_{j+1}-c_j)\rfloor-1} \biggl( c_j+\frac{l}{n},c_j+\frac{l+1}{n} \biggr]$. With this process i might loose the last of my intervals, but I loose at most one.

So i can deduce this: $|I_n^{(j)}| \ge \lfloor n(c_{j+1} - c_j) \rfloor -1 \ge n(c_{j+1}-c_j)-2$.

Now, using $(\star \star)$,

$$\lim_{n\to +\infty}|E_n^{(j)}| \ge \lim_{n\to +\infty}\sum_{i\in I_n^{(j)}}\biggl| \biggl( \frac{i}{n},\frac{i+1/2}{n}  \biggr] \biggr| = \lim_{n\to +\infty}\sum_{i\in I_n^{(j)}}\frac{1}{2n} = \lim_{n\to +\infty}\frac{|I_n^{(j)}|}{2n} \ge$$
$$\ge \lim_{n\to +\infty}\frac{n(c_{j+1}-c_j)-2}{2n} = \frac{(c_{j+1}-c_j)}{2}$$
and in the same way
$$\lim_{n\to +\infty}|F_n^{(j)}| \ge \frac{(c_{j+1}-c_j)}{2} \mbox,$$
but since we have also that $|E_n^{(j)} \cup F_n^{(n)}|=|E_n^{(j)}|+|F_n^{(j)}|=|(c_j,c_{j+1})|=(c_{j+1}-c_j)$, we can deduce that
$$\lim_{n\to +\infty}|E_n^{(j)}|=\lim_{n\to +\infty}|F_n^{(j)}|=\frac{(c_{j+1}-c_j)}{2} \mbox.$$

This holds $\forall 1\ge j\ge k-1$, so now if we take
$$n\ge \max_{j}\biggl{\{} \frac{2}{(c_{j+1}-c_j)} \biggr{\}}$$
we can conclude our first series of equalities in $(\star)$ with:
$$=\lim_{n\to +\infty}\int_{(0,1)}\overline{g}u_ndm = \sum_{j=1}^{k-1} \overline{g}(c_j)\frac{(a+b)}{2}(c_{j+1}-c_j)=\sum_{j=1}^{k-1}\int_{c_j}^{c_{j+1}}\overline{g}(z)\frac{(a+b)}{2}dz=\int_{(0,1)}\overline{g}(z)\frac{(a+b)}{2}dz \mbox,$$
which is exactly what we wanted.

答:
Okay, a possible solution can be like this.

For simplicity, assume $a,b>0$

Let $\mu$ be the limit measure to which we want to prove the convergence.

Firstly, we can prove that $$ \int f d\mu_n \rightarrow  \int fd\mu$$
for all continuous function $f$ on $[0,1]$.

The passage from continuous functions to measurable ones is in the following.

Because $\mathcal{C}([0,1])$  is a subset dense of $L^1$, for all $f \in L^1$ and $\epsilon>0$, there is a function $g \in \mathcal{C}([0,1])$ such that:
$$\| f-g\|_1 < \epsilon $$
Therefore, for all $n$,
$$ \left|\int f d\mu_n - \int f d\mu\right| \le  \left|\int g d\mu_n- \int g d\mu\right|  +\underbrace{\left| \int (f-g)d\mu_n \right|}_{\le (a+b)\epsilon} + \underbrace{\left| \int (f-g)d\mu \right|}_{\le (a+b)\epsilon} $$
Thus,
$$ \limsup \left|\int f d\mu_n - \int fd\mu\right| \le 2(a+b)\epsilon$$
Hence,
$ \lim  \int f d\mu_n = \int f d\mu$
for all $f \in L^1$
As $L^{\infty}[0,1] \subset L^{1}[0,1]$, we have the conclusion.
$\square$


问:
A Variant of the QR Decomposition
This is exercise P5.2.6 from Matrix Computations 4ed, Golub and Van Loan:

Suppose $A \in \mathbb{R}^{n \times n}$ and $D = \mathrm{diag}(d_1, \cdots, d_n) \in \mathbb{R}^{n \times n}$. Give a procedure to calculate an orthogonal $Q \in \mathbb{R}^{n \times n}$ s.t.
$$
R := Q^TA - DQ^T
$$
is upper triangular.

I tried different ideas:

Use induction on the dimension $n$, where in each step the Householder reflection / Givens rotation is applied;

Split $Q = [q_1 | q_2 | \cdots | q_n]$ into column vectors, substitute into $Q^TA - DQ^T = R$ , then solve for $Q$ and $R$;

Make use of the original QR decomposition;

but none of them seems the right path. Any hint is appreciated !

答:
Rewrite the equation

$$ R = Q^{T}A - DQ^{T}$$
$$ QR = A - QDQ^{T} $$
$$ A = QR + QDQ^{T}$$

Then the basic $QR$ decomposition corresponds to when $D = \mathbb{0}$.

Title: A Variant of the QR Decomposition


问:
A question about  binomial-coefficients
How to compute the following

$\sum_{t=0}^{k}{2m-1-t \choose t}{2m-1-(k-t) \choose k-t}$, where $1\leq k\leq 2m-1$?

Please point me to some references if this has already been studied. Thanks for your time!

答:
To elaborate on Dennis's response, a useful way for dealing with sequences and series is by using generating functions.  Given a sequence $(a_n)$, you can associate the power series $\displaystyle\sum_{n=0}^{\infty} a_n x^n$.  Assuming that the sequence doesn't grow too quickly, you can view the power series as an analytic function defined near $0$, and you can get a correspondence between operations on sequences and operations on functions.  Further, if you know the function associated to a sequence, you can then use calculus and the theory of Taylor series to determine what the sequence is.

A classic use if with the Fibonacci sequence, defined by $a_{n+2}=a_{n+1}+a_n$ and $a_0=0, a_1=1$.  If we defined $f(x)=\sum a_n x^n$, then taking the recurrence relation, multiplying it by $x^n$, and summing over all values of $n$, yields that
$$ (f(x)-x)/x^2 = f(x)/x + f(x). $$
This allows you to solve for $f(x)$, and partial fractions allows you to solve for the coefficients.

Back to the case at hand, we can solve this problem in a few steps.

 Find a nice expression for $f(x)=\sum_k \binom{(2m-1)-k}{k} x^k$

 Note that the generating function for your sequence (take your sum, multiply by $x^k$, and sum over all values of $k$) is $f(x)^2$.

 Determine the coefficients in the Taylor series expansion of $f(x)^2$.

For more details on this method and other applications of generating functions, I recommend that fantastic (free) book by H. Wilf, Generatingfunctionology.  Note that, starting on page 52, there is a list of useful power series which might be a good jumping off point for thinking about solving part (1).  You might also want to check out the "snake oil method."

Title: A question about  binomial-coefficients


问:
Prove that the derivative is unique
Can someone maybe elaborate on what this question wants me to do exactly  and perhaps give me a hint?

Prove that $f:I\to \mathbb {R}$ has a derivative at $c\in I$ if and only if there exist a number
$A\in \mathbb {R}$ and a function $\eta:I\to \mathbb {R}$ such that
$f(x) - f(c) = A (x-c) + \eta(x)$ for every $x\in I$ and $\lim_{x\to c} \left(\frac{\eta(x)}{x-c}\right) =0$.
Show directly that when they exist, $A$ and $\eta$ are uniquely determined, thus $A = f'(c)$.

EDIT:
Assume that $A$ is not uniquely determined then $f^{'}(c)= A_{1}$ and $A_{2}$ $f(x)-f(c)=A_1(x-c)+\eta_1(x)=A_2(x-c)+\eta_2(x)$ then
$ A_1-A_2=\frac{\eta_2(x)-\eta_1(x)}{x-c}$.

Further $  A_1-A_2=\lim_{x \to c}\frac{\eta_2(x)-\eta_1(x)}{x-c}=0$( if it did not then $f^{'}(c)$ would not exist contradicting the question. thus $A_{1}=A_{2}$ Hence A must be uniquely determined. thus $\eta$ is also unique determined.

答:
Hint: If $f(x)-f(c)=A_1(x-c)+\eta_1(x)=A_2(x-c)+\eta_2(x)$ then
$ A_1-A_2=\frac{\eta_2(x)-\eta_1(x)}{x-c}$. Now take the limit as $x\to c$.

Title: Prove that the derivative is unique


问:
Determining that a vector is not a least squares solution.
Can I conclude that, if $Au\cdot(b-Au)≠0$, then $u$ is not a least square solution?

When is it not true that $Ax\cdot(b-Ax)≠0$ but $x$ is a least squares solution?

答: Yes, you can indeed conclude that if $Au \cdot (b - Au) \neq 0$, then $u$ is not a least squares solution. Conversely, if $Au \neq 0$ and $Au \cdot (b - Au) = 0$, then $u$ is necessarily a least squares solution.


问:
Newton's law of cooling ODE with time-varying environment temperature
I am attempting to solve the equation of Newton's law of cooling using an uncertainty in temperature $(22 \pm 3)\,^{\circ} \,\rm{C}$. I used a sinusoid to represent that but nevertheless got to a point where I could further do analytically.

$$\frac{dy}{dt} + ky = k \left (22+ 3 \sin \frac{\pi t}{12} \right), \quad y(0) = 37.5, \quad y(2) = 34$$

How would we solve this equation analytically? Please let me know.

答:
Observe that you are solving a  first order linear ordinary differential equation and you can use an integrating factor here.

Multiplying both sides of the equation by $e^{kt}$, we have

$$\frac{dy}{dt} e^{kt} + yke^{kt} = k \left (22+ 3 \sin \frac{\pi t}{12} \right)e^{kt}$$

Or, $$\frac{d}{dt}\left(ye^{kt} \right) = k \left (22+ 3 \sin \frac{\pi t}{12} \right)e^{kt}$$

Or, $$ye^{kt} = k \int \left (22+ 3 \sin \frac{\pi t}{12} \right)e^{kt} dt $$

Or, $$y = \frac{2 \left(216k^2 \sin \frac{\pi t}{12} + 11(144k^2 + \pi^2) - 18 \pi k \cos \frac{\pi t}{12} \right)}{144k^2 + \pi^2} + kc e^{-kt}$$

Now $$y(0) = kc + \frac{2(11(144k^2 + \pi^2) - 18 \pi k)}{144k^2 + \pi^2} = 37.5$$

$$y(2) = \frac{2 \left(216k^2 \sin \frac{\pi}{6} + 11(144k^2 + \pi^2) - 18 \pi k \cos \frac{\pi}{6} \right)}{144k^2 + \pi^2} + kc e^{-2k} = 34$$

We'll have to solve the above two equations for $k$ and $c$ using numerical techniques.

I got the following values:

$$k = 0.135985, c = 123.008$$

Finally, $$y(t) = 0.6374 \sin \frac{\pi t}{12} - 1.2272 \cos \frac{\pi t}{12} + 16.7272 \exp(-0.135985t) + 22$$


问:
Marginal Density and Conditional Density Problem
Suppose (X, Y) is uniformly distributed over the region defined by $0 \leq y \leq \sqrt{1-x^2}$ and $-1 \leq x \leq 1$.

a. Find the marginal densities of X and Y

b. Find the conditional densities of $f_{X|Y = y}(x) and f_{Y|X = x}(y)$.

c. Are X and Y independent?

So I understand that the region is essentially a half circle and since it's a uniform distribution $f(x, y) = 1$ when $0 \leq y \leq \sqrt{1-x^2}$ and $0 \leq x \leq 1$ and $f(x, y) = 0$ otherwise.

For part A I'm a little confused on how to get the bounds for the integral to get the marginal densities, $f_X(x)$  and $f_Y(y)$. Would $f_X(x) = \int_0^\sqrt{1-x^2}dy$ and $f_Y(y) = \int_0^\sqrt{1-y^2}dx$?

For part B I know that $f_{X|Y=y}(x) = \frac{f(x,y)} {f_X(x)}$ and $f_{Y|X=x}(y) = \frac{f(x,y)} {f_Y(y)}$ If I am correct in part A, I believe $f_{X|Y=y}(x) = \frac{1}{\sqrt{1-y^2}}$ and $f_{Y|X=x}(y) = \frac{1}{\sqrt{1-x^2}}$ but I'm confused what the bounds are for this function to hold.

答:
but I'm confused what the bounds are for this function to hold.

The support, $\{\langle x,y\rangle: -1\leqslant x\leqslant 1~, 0\leqslant y\leqslant \surd(1-x^2)\}$, is the upper unit half-circle.

This may also be represented by: $\{\langle x,y\rangle: 0\leqslant y\leqslant 1~, -\surd(1-y^2)\leqslant x\leqslant \surd(1-y^2)\}$

[Notice: for any $y$, the values for $x$ range from negative to positive . ]

You have that $f_{\small X,Y}(x,y)= c\,\mathbf 1_{-1\leqslant x\leqslant 1, 0\leqslant y\leqslant\surd(1-x^2)}$ for some constant $c$ such that $\int_{-1}^1\int_0^{\surd(1-x^2)} c\,\mathrm d y\,\mathrm d x = 1$.

(Note: $c$ is the inverse of the support's area.)

You now have enough to correct your calculations.


答:
It's easy to construct such an $F$.

Let $(q_k)_{k=1}^{\infty}$ be an enumeration of the rationals in $[0,1]$, let $0 < c < 1$ be a constant, and let $I_k = (q_k - c2^{-(k+1)}, q_k + c2^{-(k+1)})$.

Let $U = \bigcup I_k$ and $F = [0,1] \setminus U$. Note that $U$ is open, so $F$ is compact. Also note that the measure of $U$ is at most $\sum_{k=1}^{\infty} c2^{-k} = c$, so the measure of $F$ is at least $1-c$.

Finally, since $U$ covers $[0,1] \cap \mathbb Q$, it follows that $F$ is contained in $[0,1] \setminus \mathbb Q$.

So, $F$ is compact, has positive measure, and contains only irrationals.

Title: Inner measure of the Irrationals inside $[0,1]$?

Body:
Here, I am taking Lebesgue measure. There is a notion that a set $E$ is measruable iff $m_*(E) = m^*(E)$. Let $E' = [0,1] \cap \mathbb I$, where $\mathbb I$ denote the set of irrational numbers.

By definition, $m_*(E) = \sup_{F}\{|F| \mid F \subset E, F \text{ compact}\}$. Then, $m_*(E') = \sup_{F}\{m^*(F) \mid F \subset E', F \text{ compact}\}$. Take some compact $F \subset E'$ such that $m^*(E') < m^*(F) + \epsilon$. Then,

$F$ must not contain a closed interval.

$F$ is nowhere dense in $[0,1]$ since if it is dense in some $O \subset [0,1]$, then compactness $F$ implies that $\mathbb Q \cap F \neq \varnothing$.

$F$ has strictly positive measure.

How is this possible?


问:
Bounding entries of the inverse of a matrix with bounded entries
Let $A$ be an $n$-by-$n$ matrix with integer entries whose absolute values are bounded by a constant $C$. It is well-known that the entries of the inverse $A^{-1}$ can grow exponentially on $n$. (See the replies to Estimations for the size of the biggest entry in an inverse Matrix .) Can they grow more rapidly than exponentially? Or are they bounded by $(C+O(1))^n$?

答:
The inverse of a nonsingular square $n \times n$ matrix containing only 0s and 1s can be as large as $n^{n/2(1 + o(1))}$, and can be no larger. This is proved in:

N. Alon and V. H. Vu, Anti-Hadamard matrices, coin weighing, threshold gates and indecomposable hypergraphs, J. Combinatorial Theory, Ser. A 79 (1997), 133-160.

Credit here: https://mathoverflow.net/a/182393/92003


问:
Use of definition of limit to prove $\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}$
I know that by definition I have to prove that $$\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹|\sqrt[3]{x}-2|<ϵ.$$

I have this:

$ |\sqrt[3]{x}-2|= |\dfrac{(\sqrt[3]{x}-2)((\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)}{(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2}|= \dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|}$.

I know that |x-2|<$\delta$. But I don't know what $\delta$ choose and how narrow down the denominator.

Body:
(Note that $a^2+ab+b^2=(a+\frac b2)^2+\frac 34 b^2>\frac 34 b^2$.)

$$\dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|} = \frac{|x-2|}{\left(\sqrt[3]{x}+\frac{\sqrt[3]{2}}{2}\right)^2 + \frac 34 (\sqrt[3]{2})^2 } < \frac{|x-2|}{0+\frac 12 \cdot 1}= 2|x-2|$$

So $|x-2| < \delta: = \frac{\varepsilon}{2} \implies |\sqrt[3]{x}-\sqrt[3]{2}| < \varepsilon$.


问:
Integration of a $k$-form over chains
In Spivak's Calculus on Manifolds, he defines the integral of a $k$-form over a $k$-chain, and proves a version of Stokes' theorem for this situation, before moving on to discuss the integral of a differential form over a manifold.  Other books, such as Introduction to Smooth Manifolds by Lee and Analysis on Manifolds by Munkres, seem to skip this step.  They define the integral of a $k$-form over a $k$-manifold without mentioning or proving theorems about $k$-chains.  (Am I correct about that?)

What is the advantage (if any) of Spivak's approach?  To me, $k$-chains appear to be an unnecessary weird definition in a subject that already has too many unfamiliar definitions.

答: In Lee the integral is only defined for an n-form on an n-manifold. If you want to integrate a lower dimensional k-form (k < n), you need to use a submanifold of dimension k and pull the form back with inclusion. I believe one purpose of chains is to avoid talking about submanifolds and orientations of manifold boundaries. Instead We can integrate lower dimensional forms in a higher dimensional space using singular chains.</p>

Title: Integration of a $k$-form over chains


问:
Consequence of induced homomorphism of covering map being trivial?
Suppose $p: A \rightarrow B$ is a covering map and it has been shown that the induced homomorphism $p_*$ between the two fundamental groups of $A$ and $B$ must be trivial. Can someone provide some intuition as to why the fundamental group of $B$ must be trivial?

答:
This is clearly not true. The map $p:\mathbb{R}\to S^1$ defined by $p(t)=e^{2\pi it}$ is a covering map. Since $\mathbb{R}$ is simply connected we have $\pi_1(\mathbb{R}, 0)\cong\{e\}$, and so $p_*$ is a trivial homomorphism. However, $\pi_1(S^1,1)\cong\mathbb{Z}$, it is not trivial.

What is true is that if $p:E\to B$ is a covering map with $p(e_0)=b_0$ then $p_*:\pi_1(E,e_0)\to\pi_1(B,b_0)$ is injective. Indeed, if $[f]_E\in Ker(p_*)$ then $p\circ f$ is homotopic to the constant loop $b_0$ in $B$. Note that $f$ is exactly the (unique) lifting of $p\circ f$ to $E$ which begins at the point $e_0$. Liftings of homotopic paths are homotopic. So if $[p\circ f]_B=[e_{b_0}]_B$ then $[f]_E=[e_{e_0}]_E$. So the kernel of $p_*$ is trivial.

So in particular, if $p_*$ is trivial then $\pi_1(E,e_0)$ is trivial. Anyway, the fundamental group of the covering space is "smaller", it is isomorphic to a subgroup of $\pi_1(B,b_0)$.


问:
How to remove divergence from any vector field?
I am trying to simulate fluids in 2D. I have a vector field which represents the flux; a continuous function that maps 2D positions to 2D velocity. I do some simulation steps and the flux changes. The new flux may have non-zero divergence, which is undesired, because I want to simulate non-compressible fluid. Therefore I want to have zero divergence at all points by changing the flux as little as possible. How can I calculate this most optimally and accurately?

My failed attempts:

I found Helmholtz decomposition which can be useful, but I don't think it tries to stay as close as possible to the original vector field with non-zero divergence.

I could not get this to work. The formula there does not even make sense to me.

答:
The solution that worked for me has been Hodge decomposition.

I have found exactly what I was looking for here. It even has an example implementation.

In hindsight, I should have asked this question in gamedev. What I have been trying to achieve was similar to plasma pong.


问:
$i^{-1}\mathcal F$ presheaf is a sheaf?
Let $X$ be a closed subset of $Y$ and $i: X\to Y$ be the embedding map. Is it true that for a sheaf $\mathcal F$ its $i^{-1}\mathcal F$ presheaf is a sheaf?

答:
Let $U_j$ be an open covering of $U$, an open of $X$. A section of $i^{-1}F$ over $U_j$ is the data of an open $V_j$ of $Y$ such that $V_j \cap X = U_j $ and a section $s_j \in F(V_j) $, identified over restriction to "smaller and smaller $V_j$'s ".

Now take compatible sections over each $U_j$. This means having pairs as above $(V_j, s_j) $ such that for any $i, j$ there exist a $U_i \cap U_j \subset V_{ij} \subset V_i \cap V_j$ with the property that

$$ s_i | V_{ij} = s_j | V_{ij} $$

This is the problem: we have no obvious candidate to take here.
Here:

https://mathoverflow.net/questions/75580/does-one-need-to-sheafify-when-defining-the-inverse-image-of-a-sheaf-with-respec

There is a comment with a counterexample, in which it is actually not a sheaf. However, I think that if you have a closed embedding of manifolds $i:N \to M$, then it should be true. I can think about this if you are interested in.

Title: $i^{-1}\mathcal F$ presheaf is a sheaf?