问:
Behavior near the regular point of a smooth vector field
I'm reading Warner's Foundations of Differentiable Manifolds and Lie groups and come up with some trouble in understanding the proof of the theorem about the regular point of a smooth vector field. Here are the statement and the proof:

I barely understand the last part of the proof. I don't understand why we can know $d\sigma$ is non-singular by just computing the values of the differential at each tangent vectors $\frac{\partial}{\partial r_i}|_0$ in $T_0\mathbb{R}^n$. I mean we don't really know what $\frac{\partial}{\partial r_i}|_0$ is for $i\geq 2$, do we? Also, I don't understand why, by choosing the new coordinate system, we can obtain $d\sigma \left( \frac{\partial}{\partial r_1}|_\left( t,a_2,\cdots,a_n\right)\right)=X_{\sigma \left( t,a_2,\cdots,a_n\right)}$ and thus we obtain the conclusion.

Any help or comment are hugely appreciated!!

答:
For your first question:

I don't understand why we can know $d \sigma$ is non-singular by just computing the values of the differential at each tangent vectors $\frac{\partial}{\partial r_i}$ in $T_0\Bbb{R}^n$

The differential $d \sigma$ is a linear map. It is just a basic fact from linear algebra that if the images of basis vectors under a linear map are linearly independent, then the map is non-singular.

Second:

I mean we don't really know what $\left. \frac{\partial}{\partial r_i} \right|_0$ is for $i \geq 2$, do we?

I assume $r_i$ are the standard coordinates on $\Bbb{R}^n$, so that the $\frac{\partial}{\partial r_i}$ are just the standard coordinate vector fields on $\Bbb{R}^n$.

Lastly:

Also, I don't understand why, by choosing the new coordinate system, we can obtain $d \sigma \left( \frac{\partial}{\partial r_1} \right) = X_{σ(t,a_2,⋯,a_n)}$ and thus we obtain the conclusion.

Maybe you will have to look in your book at what Corollary (a) of 1.30 says, but the proof says that this implies that $\sigma$ is a coordinate map (in other words, $t,a_2,\dots,a_n$ are local coordinates near $m$). It is noted earlier in the proof that $d \sigma \left( \left. \frac{\partial}{\partial r_1} \right|_0 \right) = X_m$. To see why this is true not just at zero, think back to what the parameter $t$ represents:

You take the surface defined by the equation $y_1=0$, and what $\sigma$ does is "flow" the surface for time $t$ along the vector field $X$. Since $t$ is the first coordinate (corresponding to $r_1$), $d \sigma \left( \frac{\partial}{\partial r_1} \right)$ is the vector field pointing in the $t$-direction. In other words, it points along the flow lines of $X$.


问:
Find the norm of the bounded linear functional $F(x)=\int_{-1}^1 x(t)dt-x(0)$.
Let $C([-1,1])$ be the set of continuous functions $f:[-1,1]\to \mathbb{R}$. Consider the functional, $F:C([-1,1])\to \mathbb{R}$ given by, $$F(x)=\int_{-1}^1 x(t)dt-x(0).$$ Show that $F$ defines a bounded linear functional and find $\lVert F\rVert$.

My attempt:

I've already shown that $F$ is a bounded linear functional such that $\lVert F\rVert\leq 3$. My idea for proving that $\lVert F\rVert\geq 3$ is to define a sequence of functions $x_n:[-1,1]\to \mathbb{R}$ satisfying $\lVert x_n\rVert=1$ and $|f(x_n)|>3-\frac{1}{n}$, for all $n\in \mathbb{N}$, but so far I couldn't figure out how to define such sequence. Any suggestion?

答: Take $x_n(t)=1$ for $\frac 1 n \leq |t| \leq 1$ and $x_n(t)=1+2n(|t|-\frac 1 n)$ for $|t| \leq \frac 1 n$. Then $\|x_n\| = 1$ and $F(x_n) \to 3$.

ContentLicense: CC BY-SA 4.0


问:
Alternating Direction Method of Multipliers (ADMM) special case
I was trying to prove some special cases of ADMM, but had some trouble getting this one so please help. Thank you in advance

Considering x-update when A=I

$x^+ = argmin_x(f(x)+({\rho}/2){\mid\mid}x-v{\mid\mid}_2^2) = prox_{f,\rho}(v)$.

The special case to be proved is

$f(x)={\lambda}{\mid\mid}.{\mid\mid}_1$,   $x_i^+ = S_{\lambda/\rho}(v_i)(S_a(v))= (v-a)_+-(-v-a)_+) $

答: This is soft-thresholding operator as prox of the $1$-norm. Try to compute it yourself: Start with the case $n=1$, i.e., $f(x) =|x|$. Then observe that the prox of $\|\cdot\|_1$ can be computed component-wise.


问:
Isomorphism $f:\mathbb{Z}/17\mathbb{Z}\times \mathbb{Z}/561\mathbb{Z} \to \mathbb{Z}/51\mathbb{Z}\times \mathbb{Z}/187\mathbb{Z}$
I would like to find an group isomorphism $f:\mathbb{Z}/17\mathbb{Z}\times \mathbb{Z}/561\mathbb{Z}  \to \mathbb{Z}/51\mathbb{Z}\times \mathbb{Z}/187\mathbb{Z} $. By the fundamental theorem of a finite abelian group and the Chinese remainder theorem, we know that those groups are isomorphic, but I want to show it by constructing an isomorphism.

However, I don't know what the first step is. The only thing I know is that $f(0,0)=(0,0)$ since an isomorphism maps an identity element to an identity element.

Then I saw How to construct an isomorphism? and tried to mimic the way, like $f(x,y)=(x\mod{51},y\mod{187})$, but it is obviously not a surjection.

Now I am stuck here. Any help?

答:
We introduce an intermediate group $\mathbb Z_{17}×\mathbb Z_3×\mathbb Z_{187}$. Represent an arbitrary element of this group as $(a,b,c)$ where the indices are residues modulo $17,3,187$ respectively.

There is an isomorphism from this group to the domain of $f$: $(a,b,c)\mapsto(a,187b+c)$. There is also an isomorphism to the codomain of $f$: $(a,b,c)\mapsto(3a+b,c)$. Put these two isomorphisms together and you have the required $f$.


问:
Regression coefficient when eliminating a row
I have a problem understanding the proof of the following proposition that gives the expression of the regression coefficients when removing a row from the design matrix.
The demo says the following:
It's known that,
$X^tX=\sum_{k=1}^n \overrightarrow{x_k}\overrightarrow{x_k^t}=X^t_{-i}X_{-i}+\overrightarrow{x_i}\overrightarrow{x_i}$
applying the Sherman-Morrison-Woodbury theorem,
$(X_{-1}^tX_{-i})^{-1}=(XX^t)^{-1}+\frac{(X^tX)^{-1}\overrightarrow{x_i}\overrightarrow{x_i^t}(X^tX)^{-1}}{1-h_{ii}}$
my problem comes now, since from here it concludes directly,
$\widehat{\beta_{-i}}=(X_{-1}^tX_{-i})^{-1}X^t_{-i}\overrightarrow{y_{-i}}=\widehat{\beta}+\frac{1}{1-h_{ii}}(X^tX)^{-1}\overrightarrow{x_i}e_i$

and I am not able to understand that last step.
Please if someone can explain to me why that product gives that expression at the end, I try to introduce the inverse expression, but I don't see where the beta is coming from, and I'm very involved with the issue of X and y Without the i-th row, it seems to me that something needs to be added or done to solve it, which I am not able to see.
Thank you very much in advance.

答:
Here is a careful derivation. Let $X$ be the $n\times p$ design matrix, and let the $p\times 1$ column vector $x_i$ hold the $i$th row of $X$. Let $X_{-i}$ be the design matrix with row $i$ removed. Check that
$$X^\top X=\sum_{k=1}^n x_k x_k^\top\tag1$$ and $$
X_{-i}^\top X_{-i}=\sum_{k=1}^n x_k x_k^\top-x_ix_i^\top=X^\top X-x_ix_i^\top\tag2
$$
so by the Sherman-Morrison formula
$$
(X_{-i}^\top X_{-i})^{-1}=(X^\top X)^{-1} + \frac{(X^\top X)^{-1}x_ix_i^\top (X^\top X)^{-1}}{1-x_i^\top (X^\top X)^{-1} x_i},\tag3
$$
For brevity write $A:=(X^\top X)^{-1}$. Introduce the hat matrix
$ H:=XAX^\top$
and observe that the $(i,i)$ element of $H$ is
$$h_{ii}=x_i^\top Ax_i\tag4
$$ so we can rewrite (3) as
$$ (X_{-i}^\top X_{-i})^{-1}=A + \frac{Ax_ix_i^\top A}{1-h_{ii}}.\tag5
$$ Let $y$ be the $n\times 1$ vector of responses $(y_1,y_2,\ldots,y_n)^\top$. If $y_{-i}$ denotes the vector of responses with element $i$ removed, then
$$
X_{-i}^\top y_{-i}=\sum_{k=1}^n x_ky_k-x_iy_i=X^\top y-x_iy_i.\tag6
$$
Combining (5) and (6) we have
$$
\widehat{\beta_{-i}}=(X_{-i}^\top X_{-i})^{-1}X_{-i}^\top y_{-i}=\left( A + \frac{Ax_ix_i^\top A}{1-h_{ii}}\right)\left(X^\top y-x_i y_i\right).\tag7$$
Expand the RHS of (7) as a sum of the following four terms:
$$
\begin{aligned}
AX^\top y &= \hat\beta\\
\frac{Ax_ix_i^\top A}{1-h_{ii}}X^\top y &=\frac{Ax_ix_i^\top\hat\beta}{1-h_{ii}}\\
-Ax_iy_i&=-\frac{(1-h_{ii})A x_iy_i}{1-h_{ii}}\\
-\frac{Ax_ix_i^\top A}{1-h_{ii}}x_iy_i&\stackrel{(4)}=-\frac{Ax_ih_{ii}y_i}{1-h_{ii}}
\end{aligned}\tag8
$$
Collecting these and cancelling yields
$$
\widehat{\beta_{-i}}=\hat\beta + \frac{Ax_ix_i^\top\hat\beta}{1-h_{ii}}-\frac{A x_iy_i}{1-h_{ii}}
=
\hat\beta + \frac{Ax_i(x_i^\top\hat\beta-y_i)}{1-h_{ii}}.\tag9
$$
In (9) we recognize the $i$th residual $e_i:=y_i-x_i^\top\hat\beta $. This demonstrates
$$
\widehat{\beta_{-i}}=\hat\beta -\frac{(X^\top X)^{-1}x_ie_i}{1-h_{ii}}.\tag{10}
$$


问:
How to show that $\det(A+I)\ne 0$
How to show that for any skew symmetric real matrix $A$, we have $\det(A+I)\ne 0?$

Where to begin? I'm looking for some clue only.

答: let $\lambda$ be an eigen value of a skew symmetric matrix $A$ and $X\ne 0$ be an eigen vector correspomding to this $\lambda$ so $AX=\lambda X\Rightarrow X^*A^*=\bar{\lambda}X^*\Rightarrow X^*(-A)X=\bar{\lambda}X^*X\Rightarrow  -X^*\lambda X=\bar{\lambda}X^*X\Rightarrow (\lambda+\bar{\lambda})X^*X=0\Rightarrow \lambda=\bar{\lambda}$ as $X$ was non null, so Eigen values of real skew symmetric matrices arepurely imaginary  or $0$, The problem you have stated says that $-1$ is an eigen value of $A$ so contradiction. </p>

ContentLicense: CC BY-SA 3.0

Title: How to show that $\det(A+I)\ne 0$


问:
If $B=5I_2-3A+A^2$ can I show this matrix is invertible?
Let $D=\begin{pmatrix} 2 & 0\\ 0 & 7 \end{pmatrix}$, then $A=PDP^{-1}$ and $B=5I-3A+A^2$. Show that B is diagonalizable.

I have tried algebraically, trying to get $B^{-1}$, but get nothing.

答: Let me give you another simple proof. From $D$, we can find 2 and 7 are eigenvalues of $A$. By using the equation, we can find $B$ have eigenvalues 3 and 33. Since they are distinct, $B$ is also diagonalizable.


问:
How are (pre)sheaves even well-defined on sites?
I am trying to learn a bit about étale cohomology from Milne’s book and I have run into some early problems. I learned ordinary abelian sheaf cohomology from Hartshorne. There, sheaves are defined on the category of open subsets of a scheme $X$ in which the morphisms are just the open inclusions. A presheaf on $X$ is then just a contravariant functor on this category. To motivate the étale site and étale cohomology, many sources first introduce sites via the Zariski open immersions, but I am already confused by this.

The Zariski site on $X$ is defined to have objects $f \colon U \rightarrow X$ which are open immersions and morphisms given by maps $V \rightarrow U$ which commute with the structure morphisms of $U$ and $V$. Then a presheaf $\mathscr{I}$ is a contravariant functor on this category. With this definition, many sources, including Milne, call the group $\mathscr{I}(U)$ the sections over $U$. But already I don’t know what this means. In the case of Hartshorne this is fine. But in the case of the Zariski site, there could be many distinct open immersions from $U$ to $X$. So what then does $\mathscr{I}(U)$ actually refer to? Is there any reason to believe that $\mathscr{I}(f \colon U \rightarrow X)$ is equal or even isomorphic to $\mathscr{I}(g \colon U \rightarrow X)$ if $f$ and $g$ are distinct open immersions? I suspect in the sheaf case one can use the sheaf axioms to argue that they are isomorphic, but what about in the presheaf case?

The problem seems to be even worse when it comes to the étale site. In the case of the Zariski site, at least open immersions factor into an isomorphism and an open inclusion. But two étale morphisms $f \colon U \rightarrow X$ and $g \colon U \rightarrow X$ need not be related in any way as far as I can see. So what does $\mathscr{I}(U)$ mean, and is there any reason to think it is independent of the structure morphism?

I realise this may be a completely silly question, but I can’t see an obvious answer after a while thinking about it.

答: This is an abuse of notation. $F(U)$ is not just a function of $U$ as an object, it's a function of the morphism $f : U \to X$, which is being dropped from the notation. That is, it's shorthand for $F(U \xrightarrow{f} X)$.


问:
What could be better than base 10?
Most people use base 10; it's obviously the common notation in the modern world.

However, if we could change what became the common notation, would there be a better choice?

I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?

I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.

Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:

First and foremost, 12 is a highly composite number — the smallest
  number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
  and 12). As noted, 10 has only two. Consequently, 12 is much more
  practical when using fractions — it's easier to divide units of
  weights and measures into 12 parts, namely halves, thirds, and
  quarters.

And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.

So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?

答:
In order to answer this question, it is first necessary to ask: What makes one base “better” than other?  Some reasonable things to consider are:

Size

There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/\log{b})$).

If the base is too small, then numbers explode into cumbersome long strings of digits.  For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100.  Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.

On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese.  And forget about learning the mulitplication table, whose size is $O(b^2)$.  The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.

So, what we want is a happy medium.

Fraction-friendliness

This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).

If a base has a lot of factors, it makes fractions easier to work with.  For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems.  But in base-12, 1/3 is a nice simple 0.4.

Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”.  And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions.  But 1/3 is more common than 1/5.

Title: What could be better than base 10?


问:
Finding the rank and dimker of a matrix.
Hi so im working on a question about finding the $rank(A)$ and the $dim(Ker(A))$ of a 7x5 Matrix. Without being given an actual matrix to work from.

I have been told that that the homogeneous equation $A\vec x=\vec0$ has general solution $\vec x=\lambda \vec v$ for some non zero $\vec v$ in $R^{5}$.

So my thinking so far is that I know for an $m*n$ matrix we know that:

$rk(A)+dimker(A)=n$
which must mean that $rk(A)+dimker(A)=5$

but this is where I get stuck and dont know how to proceed.

Any help is greately appreciated.

.

This is the exact question for the person who asked.

答: Is $v$ fixed? If yes, then kernel is generated by only this vector, so 1 dimensional. By the formula you wrote, rank is 4.


问:
From a pack of 52 cards, two cards are drawn randomly. What is the probability of both the cards being kings?
How do you do this? So, I was doing a mini maths test and the question popped up. Because there were 4 kings, so I did 4/52(there were 52 cards) and then simplified it. I got it wrong. So confused! (No replacement, but it will be great if you guys show me with replacement as well).

答:
Hint: this can be solved using a “probability tree”

start by drawing the first card. What are the odds of it being a king? Simply $4/52$.

Now that you were lucky enough to draw the first king there are only ... kings left in the deck of only ... cards.

Conclude.


答:
I will give another way to think about this problem, which might help the OP.

The sample space in this problem is ${52}\choose{2}$ since there are ${52}\choose{2}$ ways to pick a pair of cards. There are 4 kings, therefore there are ${4}\choose{2}$ ways of picking a pair of kings. From here we see that the probability of getting 2 kings is

$$P(\textrm{getting 2 kings }) = \frac{\textrm{number of ways the desired event can occur}}{\textrm{number of possible outcomes}}=\frac{{4}\choose{2}}{{52}\choose{2}}$$

Title: From a pack of 52 cards, two cards are drawn randomly. What is the probability of both the cards being kings?

Body: How do you do this? So, I was doing a mini maths test and the question popped up. Because there were 4 kings, so I did 4/52(there were 52 cards) and then simplified it. I got it wrong. So confused! (No replacement, but it will be great if you guys show me with replacement as well).


问:
Evaluating a series with the Möbius function and greatest common divisor.

Problem:  Let $\gcd(a,b,c,d)$ refer to the largest integer $r$ such that $r$ divides each of $a,b,c,d$.  Evaluate the series $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty}\sum_{d=1}^{\infty}\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^{2}b^{2}c^{2}d^{2}}\gcd(a,b,c,d)^{4},$$ where $\mu(n)$ is the Möbius function.

I tried several tricks, but I eventually got stuck.  I think it should be possible to rewrite the entire thing as an Euler Product. It looks very similar to
the double series $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{\mu(a)\mu(b)}{a^{2}b^{2}}\gcd(a,b)^{2}=\frac{6}{\pi^2}.$$

Any help is appreciated.

Body:
Your sum can be re-written in terms of an Euler product:

$$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})\approx .16544\cdots$$

A proof can be given as follows:

First note that,

$$d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4\iff d\mid \gcd(x_1,x_2,x_3,x_4)$$

So we get:$$\sum_{d=1}^\infty f(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid \gcd(x_1,x_2,x_3,x_4)}$$
$$=\sum_{d\mid \gcd(x_1,x_2,x_3,x_4)}f(d)$$

Where $1_{A}=[A]$ in Iverson bracket notation

Then define:

$$\phi_s(n)=n^s\prod_{p\mid n}(1-\frac{1}{p^s})$$

So that we have: $(\phi_s*1)(n)=n^s$

Now set $f=\phi_4$ in the aforementioned equality and we get that:

$$\sum_{d=1}^\infty \phi_4(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\gcd(x_1,x_2,x_3,x_4)^4$$

Now we note that:

$$\sum_{x_i=1}^\infty\frac{\mu(x_i)}{x_i^2}1_{d\mid x_i}=\sum_{n=1}^\infty\frac{\mu(dn)}{(dn)^2}=\frac{6}{\pi^2}\frac{\mu(d)}{\phi_2(d)}$$

Then multiplying both sides of the previous series by $\frac{\mu(x_1)}{x_1^2}\frac{\mu(x_2)}{x_2^2}\frac{\mu(x_3)}{x_3^2}\frac{\mu(x_4)}{x_4^2}$ and rearranging gives:

$$\sum_{d=1}^\infty \phi_4(d)(\frac{\mu(x_1)}{x_1^2}1_{d\mid x_1})(\frac{\mu(x_2)}{x_2^2}1_{d\mid x_2})(\frac{\mu(x_3)}{x_3^2}1_{d\mid x_3})(\frac{\mu(x_4)}{x_4^2}1_{d\mid x_4})$$
$$=\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$$

Thus:

$$\sum_{x_1=1}^\infty\sum_{x_2=1}^\infty\sum_{x_3=1}^\infty\sum_{x_4=1}^\infty\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$$

$$=(\frac{6}{\pi^2})^4\sum_{d=1}^\infty \phi_4(d)\frac{\mu(d)^4}{\phi_2(d)^4}=\frac{1296}{\pi^8}\sum_{n=1}^\infty\frac{\phi_4(n)}{\phi_2(n)^4}|\mu(n)|=\frac{1296}{\pi^8}\prod_{p}(1+\frac{\phi_4(p)}{\phi_2(p)^4})$$

So we have:

$$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})$$

A similar argument will give your second sum as $\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{a^2b^2}\gcd(a,b)^2=\frac{6}{\pi^2}$.

In addition to formula for other such similar generalizations.


问:
Variance of conditional expectation and constant term using LIE
I'm trying to solve an exercise using Law of iterated expectations, but I'm stuck.
I'm not sure if I'm on the right track as expanding the RHS more will not result in answer.
I believe I'm looking for an expression that looks like,
$$E[Y^2]-E[Y]^2+E[\epsilon^2]-E[\epsilon]^2 = V[Y]+V[\epsilon]$$
but not completly sure.

(e) If $Y = E(Y|X) + \epsilon$ and $E(\epsilon|X)$ is constant then $V(Y)=V(E(Y|X))+V(\epsilon))$.
\begin{align*}
    V[Y] &= E[Y^2]-(E[Y])^2 \\
    &= E[E[Y^2|X]+\epsilon]-(E[E[Y|X]+\epsilon])^2 \\
    &= E[Y^2]+E[\epsilon]-(E[Y+\epsilon])^2 \\
    &= E[Y^2]+E[\epsilon]-(E[Y]^2+E[Y\epsilon]+E[\epsilon]^2) \\
\end{align*}

Any guidance or help would be highly appreciated.

答:
Using the identity
$$
\operatorname{Var}(U+V)=\operatorname{Var}(U) + \operatorname{Var}(V) + 2\operatorname{Cov}(U,V)\tag1$$

you just need to show that the covariance between $U:=E(Y\mid X)$ and $V:=\epsilon$ is zero. To do so, apply LIE to write
$$
E (UV) = E[ E(UV\mid X) ]= E [ U E(V\mid X) ],\tag2
$$
the last equality following from the fact that $U$ is $X$-measurable. But we are told $E(V\mid X)$ is a constant, say $c$. Upon taking expectations we deduce $c=E(V)$. Continuing (2) we get
$$
E (UV) = E[ U E(V) ] = E(U) E(V)
$$
and we've proved $U$ and $V$ have zero covariance.


问:
Finding an example the fundamental theorem of calculus fails when $f$ is not continuous.
Is the following statement true or not?

If $f$ is Riemann integrable on $[a,b]$, then the function $F(x)=\int_a^xf(t)dt$ is differentiable on $(a,b)$ and $F'(x)=f(x)$ for $x\in (a,b)$

I think it is not. Since $F$ is differentiable at the point where $f$ is continuous, but I can not think a counter example, is anyone has a counterexample? Thanks a lot.

答:
False: Take $f(x)=sgn(x)$ (sign function) on the interval $[-1,1]$ (that is, $a=-1$). Then

$$F(x)=|x|-1$$

and is not differentiable at $x=0$. The theorem as stated is true if $f$ is a continuous function however. Indeed, that is the full statement of the fundamental theorem of calculus (at least the first part).


问:
Finding partial derivatives for equations expressed in terms of $z$ where $z=f(x,y)$ to find tangent plane
I am having troubles finding partial derivatives.

If $f(x,y)=2x^2+y^2$ then,
$$f_x=4x$$
$$f_y=2y$$

That's simple enough. But when I see a $z$ in the equation, I get stumped. I know $z=f(x,y)$. I don't really see the process.

For example, if $z=2x^2+y^2$ then do we differentiate both sides with respect to $x$ like this?

$$f_x:\frac{dz}{dx}=4x$$
$$f_y:\frac{dz}{dy}=2y$$

Even worse, what am I supposed to do for something like this? Same thing?

$$x+y^2+z^3=3$$
$$z^3=-x-y^2+3$$

$$f_x:3z^2\frac{dz}{dx}=-1$$
$$f_x:\frac{dz}{dx}=\frac{-1}{3z^2}$$

Likewise,

$$f_y:\frac{dz}{dy}=\frac{-2y}{3z^2}$$

I'm pretty sure that's wrong but I don't know why. Can someone please help me understand? Thanks.

Edit
Some context for the last example (it's from a homework problem):

Find the equation of the tangent plane to the surface with equation $x+y^2+z^3=3$ at the point (2,1,0).

I know that the equation for the tangent plane is
$$z=f(x_0,y_0)+[f_x(x_0,y_0)](x-x_0)+[f_y(x_0,y_0)](y-y_0)$$

Since the $f_x$ and $f_y$ found above contain z, do I plug in $z_0$?

So I'm trying to find $f_x$ and $f_y$.

Edit 2

I get it now

答:
What you are looking for is called the chain rule. If you have:
$$g(z(x,y))=0$$
Then:
$$\frac{\partial g}{\partial x}=\frac{dg}{dz}\cdot\frac{\partial z}{\partial x}$$

So if $x + y^2 + z^3 = 3$, you take the derivative according to $x$:
$$1+0+3z^2 \cdot \frac{dz}{dx}=0 \ \ \to \ \ \frac{dz}{dx}=\frac{-1}{3z^2}$$
What does mean? this is a differential equation that may or may nor be easy to solve for $z$, but if you do know what $z$ is, you could just plug n' play:
$$\frac{dz}{dx}=\frac{-1}{3z^2}=\frac{-1}{3(3-x-y^2)^{2/3}}$$


问:
Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0$
Find the Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0 $ tanx$ \neq -1$

I've tried to find the derivative:
$$ f'(x)= \frac{-(\cos x+\sin x)(\sin x+\cos x)-(\cos x-\sin x)^2}{(\cos x+\sin x)^2}$$
$$f'(x)=\frac{-2}{  (\cos x+\sin x)^2  }$$

$f'(x)<0$ as tan x $\neq$ -1

$x\geq 0 \Rightarrow f(x)<f(0)$

$f(x)<1$.

So maximum value is 1. Is this correct? What is the minimum value?

答:
I would try it another way:

$$
\cos(x)-\sin(x) = \sqrt{2}\cos(x+\frac{\pi}{4}) \\
\cos(x)+\sin(x) = \sqrt{2}\cos(x-\frac{\pi}{4})
$$

then

$$
f(x)=\frac{\cos(x+\frac{\pi}{4})}{\cos(x-\frac{\pi}{4})}=-\frac{\sin(x-\frac \pi 4)}{\cos(x-\frac \pi 4)}=-\tan(x-\pi/4) \\
$$

As you might know, the function $\tan(x)$ is not bounded. Hence your $f(x)$ has no global maxima or minima. E.g. it takes all values from $-\infty$ to $\infty$.


问:
How to determine the 90% confidence interval for the population proportion of students who found the math test easy?
Out of $120$ students, $85%$ found the math test easy. How to determine the $90%$ confidence interval for the population proportion of students who found the math test easy?

I am not sure about my computation but I tried to solve it in the following way.

The $85%$ of 120 students is 102 students.

Since the confidence level is 90%, there will be 92 students

My answer is not an interval but a fixed percentage which is $76.5%$.

Any comments and suggestions will be highly appreciated.

答:
No, that is not correct.

You want to find the real percentage of students who find this test easy.

Your empirical percentage you got is $85/120 \approx 0.708$. So you could say, the percentage of students who find this test easy is 70.8%.

However, this could be just because you the first students who left the test, who might find the test easier as the other ones who are still writing. Or you could ask only the last people, who might find the test harder than others.

In other words, you have no way to check, if your empirical percentage is equal to the real one.

Thus, you need a model for the probability, e.g. each of the student has the same percentage of finding the test easy and they dont interact with each other.
Then, you can now look at the empirical mean value and standard deviation and see what real percentage would agree with a probability of 90% with your empirical finding of 85 out of 120.

I dont want to spoil the whole computation here, but hopefully this helped you understand what you are trying to achieve.


问:
Difference Between Definitions of an Affine Variety and an Algebraic Affine Variety
I'm used to the definition of Affine Variety as
$$V(f_1, ..., f_s)=\{a=(a_1, ..., a_n) \in k^n \hspace{2mm}| \hspace{2mm} f_i(a)=0 \hspace{3mm} \text{for every} \hspace{3mm} 1 \leq i \leq s\}$$
for some field $k$. However, I've read recently (on a French book) the definition of an Affine Variety $V$ as a subset of some vector space such that, for every $x, y \in V$, the line that intersects $x$ and $y$ are entirely contained in $V$. Formally speaking, an affine variety $V$ is a set such that
$$ x, y \in V, t \in \mathbb{R} \implies (1-t)x+ty \in V.$$
These two definitions, of course, are different, but could someone explain me why there is this conflict in definitions? It seems to me that in French Affine Varieties are called Algebraic Affine Varieties, and what they call by Affine Variety is what I just wrote here. What would be the equivalent definition in english?

Thanks.

答:
Let me upgrade my comment in to an answer. Your first definition is one of many for affine varieties. The second definition is for an "affine space", which is a vector space where we forget where $0$ is - that means we're allowed to add any two elements of our space as long as the coefficients sum to one, but we don't have scalar multiplication, a choice of a $0$ vector, etc.

One thing that may be contributing to your confusion is that the French word "variété" has multiple meanings - this has been discussed here on MSE in the past.


问:
Function which the limit is the same for every point, but the function is not the same for every point
Let $L \in \mathbb R$.

Let $f:\mathbb R\to\mathbb R$.

$\lim_{x\to x0}f(x)=L$ for every $x0\in \mathbb R$, so there exists $x\in \mathbb R$ such that $f(x) \neq L$.

I tried finding such function that for every point the limit is the same, but I couldn't find any function besides constant functions and then the second part of the claim is false.

Maybe this claim is false?

答: You can take$$f(x)=\begin{cases}0&\text{ if }x\ne0\\1&\text{ otherwise.}\end{cases}$$


问:
Finding $d(\sin(3x))/dx$ with the definition of derivative.
In this post a user claims
\begin{align*}
f'(x)
&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
&=\lim_{h\to 0}\frac{\sin 3(x+h)-\sin(3x)}{h}\\
&=\lim_{h\to 0}\frac{2\cos (3x+\frac32h)\sin \frac32h}{h}\\
&=\lim_{h\to 0}\frac{2\cos (3x+\frac32h)\frac32h}{h}\\
&=\lim_{h\to 0}3\cos (3x+\frac32h)\\
&=3\cos(3x)
\end{align*}

How did he get $\frac 32h$ from $\sin(\frac 32h)$?

I don't understand.

答:
Small angle approximation, probably. $\sin(x) \approx x$ for $x$ small.

More rigorously, asymptotic equivalence. Note the well-known limit

$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$

This gives the asymptotic equivalence $\sin(x) \sim x$ as $x \to 0$. This means, within some circumstances, you can interchange $x$ and $\sin(x)$ in limits where $x \to 0$. (In this case, $x=3h/2$.)

An alternative angle of approach as pointed out by player3236 in the comments is that they could have multiplied to get a cancellation as so:

$$\require{cancel}\frac{2\cos (3x+\frac32h) \cancel{\sin \frac32h}}{h} \cdot \frac{\frac 3 2 h}{\cancel{\sin \frac 3 2 h}} = \frac{2\cos (3x+\frac32h)\frac32h}{h}$$


问:
Nested Interval Theorem: Another part
Let [a1, b1] ⊇ [a2, b2] ⊇ · · · be a sequence of intervals. If lim(bi − ai) → 0, then ∩[an, bn] is a singleton set.

I have proved the first part which says that the infinite intersection of such subsets is not empty and is a closed interval. But how do I prove this part? I tried solving it by assuming that the intersection contains two unequal numbers x and y. And then by generating real numbers between them I was trying to reach some kind of contradiction but was not able to proceed.
Can someone help me with this proof?

答:
Conclude the sequences $a_n$ and $b_n$ are increasing and decreasing, respectively.

Since $[a_n, b_n]$ is a closed interval, $a_n \leq b_n$. So conclude that $a_n$ and $b_n$ are bounded above and below, respectively.

By the Monotone Convergence Theorem, they converge. That is, $a_n \rightarrow a$ and $b_n \rightarrow b$.

Then $\bigcap_n [a_n, b_n]=[a,b]$

But since $b_n - a_n \rightarrow 0$, by the algebra of limits, we have that $a=b$.

Therefore, $[a,b]=\{ x\}$ for some $x \in \mathbb{R}$. We can also conclude that $x$ is unique.

Title: Nested Interval Theorem: Another part


问:
Enriched category has to be locally small
I am reading material on enriched categories. The requirements for the category over which one enriches (be it $\textbf{Grp}$, $\textbf{Vect}_k$ are very clear, that they need to be monoidal and so). But what are the conditions on the category $\mathscr{C}$ that is being enriched? Some references say that one replace the hom-sets by hom-objects, which may imply that $\mathscr{C}$ is at least locally small. Is this correct, and why?

答:
The comments from EBP and Berci are correct - I just want to give some more detail and an example. I quote EBP's comment here, since it's important:

When considering an enriched category, this is not necessarily done by enriching an ordinary category which already exists. We are simply defining an enriched category to be a class of objects together with hom objects for each pair of objects.

A locally small category is one in which the hom-sets are indeed sets: there is a set of morphisms between any pair of objects.

In an enriched category, one replaces these sets with an object in some other category, $\mathscr{V}$. If $\mathscr{V}$ has the form of "a set with some extra structure", such as the examples you mentioned, then it could make sense to think of the enriched category as "locally small", in the sense that you can still think of its hom-sets as being sets of morphisms. As Berci noted in the comments this can be made formal by specifying a faithful functor from $\mathscr{V}$ to $\mathsf{Set}$.

However, that need not be the case at all. A good example is a Lawvere metric space, which is a category enriched in the monoidal poset $([0,\infty],\ge,+)$. Then the thing that's assigned to each pair of objects isn't a set of morphisms but just a number, which can be interpreted as a distance. (It turns out that these "distances" are guaranteed to obey the triangle inequality, although they relax some of the other axioms of a metric space.) A good gentle introduction to Lawvere metric spaces and similar poset-enriched categories is found in chapter 2 of Spivak and Fong's Seven Sketches in Compositionality.

In the case of a Lawvere metric space it doesn't make much sense to ask whether the category is locally small, because its hom-objects are neither sets nor proper classes but something else entirely. So I would say that counts as a counterexample - enriched categories don't necessarily have to be locally small.


问:
How to evaluate the limit of this following expression $\color{black}{\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}}$?
I am not able to find the limit of the following expression, after trying a bit hard:$$\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}$$ Can some one help with some approach?

答:
Taking $n^2$ common from denominator we have $\frac{6n}{9n^2-r^2}=\frac{6}{9-\frac{r^2}{n^2}}\frac{1}{n}$ . Now observing the fact that it's a Riemann sum we have $\lim_{n\to \infty} \sum_1^{n} \frac{6n}{9n^2-r^2}=\int _0^1 \frac{6}{9-x^2}dx$ .
I hope you can continue from here.


问:
Reasons why division by zero is not infinity or it is infinity.
So, here is a fundamental question, division by zero is undefined, there are some ways to prove this. Lets take the following limit

$$\lim_{x \to 0^+} \frac{1}{x} = +\infty $$

but

$$\lim_{x \to 0^-} \frac{1}{x} = -\infty $$

So the limit does not exist, thus division by zero is undefined.

But let's go way more basic. What is division? Division can be though of as a number of succesive substractions until you reach zero or before you hit a negative number.

Say you want to divide 8/3, you start by

$$8-3 = 5$$

Do it once more

$$5-3 = 2$$

And then you stop, because if you substract 2-3 = -1 and we said we were going to stop at zero or before it got negative.

So how many substractions did we had to perform before we reached zero or a negative number? 2 of them, what number was left after the last substraction? also 2, so 8/3 is 2 with a remainder of 2. Simple as that.

Let's do the same but now lets divide 8/0

So we start

$$8-0=8$$

One more time

$$8-0 = 8$$

and again...

$$8 - 0 = 8$$

Do you see were I am going with this? So we can say, ok, substracting zero from eight will never reach zero so division by zero is not defined, or we can go crazy and count the number of times we can keep doing this and the result is infinity, so could we say that 8/0 is infinity with a remainder of 8?

答:
Here are four different counter-examples/arguments to disprove what you said:

$1:$

Division by $0$ is literally impossible and you cannot use this operation. Why? Suppose $\frac{n}{0}=k$. This would result in $n=0$, which is a contradiction (I don't even want to talk about why $\frac{0}{0}$ doesn't exist)

$2:$

AssuminG you can indeed work with $\frac{k}{0}$, by the same algorithm, you could say that for any $n$, $8$ divided by $0$ is $n$ with a remainder of $8$, thus leading in a contradiction.

Why is this a contradiction? Well (in our case), division is a bijection, a function, from $\mathbb{N}$ to $\mathbb{Q}$ (actually from $\mathbb{N^*}$ but we assume dividing by $0$ is possible) and if $\frac{k}{0}$ could take multiple values, this would defeat the purpose of a function.

Remember, division was first and then people found out we cannot divide by $0$

$3:$

The algorithm you defined is not at all division. It is finding the integer part of $\frac{m}{n}$ where $m,n>0$. Just to put in evidence another flaw of your so called division, look at this:

What is $\frac{8}{-1}$? (it is $-8$)

Well $8-(-1)=9$. We need to do it again... $9-(-1)=10$ uh we need to do it again... $10-(-1)=11$ oh wait we never reach a negative number!

So does this imply that $-8=\frac{8}{-1}=+\infty$?

Just as another observation, people often belive that division is some sort of subtraction and multiplication is some sort of adittion. This is wrong, exactly because of $0$ (which led to people struggling with this in the ancient time), because of negative numbers, because of irrational numbers, because of complex numbers etc.

$4:$

Assuming you still belive in division by $0$, you don't care that division is a bijection and you still belive in your algorithm, then take a look at this:

$$\frac{1}{0}=\infty=\frac{2}{0}\Rightarrow\frac{1}{0}=\frac{2}{0}\Rightarrow\frac{1}{0}\cdot 0=\frac{2}{0}\cdot 0\Rightarrow 1=2$$

I am sorry if at this points your eyes are bleeding, reader


答:
Hint: Divide by $\cos x$.

$$\dfrac{\cos x-\sin x}{\cos x+\sin x}$$
$$=\dfrac{1-\tan x}{1+\tan x}$$
$$=\dfrac{\tan\left(\dfrac{\pi}{4}\right)-\tan x}{1+\tan\left(\dfrac{\pi}{4}\right)\cdot\tan x}$$
$$=\tan\left(\frac{\pi}{4}-x\right)$$

Therefore the range of this function extends from $-\infty$ to $+\infty$.

Title: Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0$

Body:
Find the Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0 $ tanx$ \neq -1$

I've tried to find the derivative:
$$ f'(x)= \frac{-(\cos x+\sin x)(\sin x+\cos x)-(\cos x-\sin x)^2}{(\cos x+\sin x)^2}$$
$$f'(x)=\frac{-2}{  (\cos x+\sin x)^2  }$$

$f'(x)<0$ as tan x $\neq$ -1

$x\geq 0 \Rightarrow f(x)<f(0)$

$f(x)<1$.

So maximum value is 1. Is this correct? What is the minimum value?


问:
Is a part of stationary process is stationary process?
Problem

Assume the we are given a stationary process $x_t$. That is $$E(X) = const$$ $$ Cov(x_t, x_{t+k}) = f(k)$$   It is known that $E(x_t) = 4$, $Var(x_t) = 16$ and $Cov(x_t, x_{t-1}) = 4 $.

We go through each $t$ and with probability $0.3$ delete $x_t$. Let's call new sequence $y_t$. (In other words, $(y_t)$ will contain some of elements of $(x_t)$)

Is $y_t$ a stationary process?

Find $ \Bbb Var(y_t), \Bbb Cov(y_t,y_{t-1})$.

My thoughts

Of course, the expected value will stay the same because it doesn't depent on other $y_t$. However, I think, the covariance won't be the same. My counterexample, assume there are $$(x_t) = x_1, x_2, x_3, x_4, x_5,\dots$$ $$(y_t) = x_1, x_2, x_3, x_4, x_6, \dots  $$

that is we droped only $x_5$ among first five variables.
We know that as $(x_t)$ is stationary process

$$ Cov(x_1,x_3) = Cov(x_3,x_5) = f(2)$$
however as if $(y_t)$ is stationary process:
$$ Cov(x_1,x_3) = Cov(x_3,x_6) = f(3) \implies f(2) = f(3)$$
It may happend the equality $f(2) = f(3)$ is not held. Hence, $y_t$ is not stationary process.
Is my logic right? I don't proceed to the variance utill I find out whether $y_t$ is stationary or not.

答:
Hint

We don't know which $x_i$ will be in $(y_t)$ and which won't.
Let us use indicator $a_i = a_i(i = j)$:

Then some $y_t$ is $y_t = x_j = \sum_i a_i x_i = 0 \cdot x_1 + \dots + 1\cdot x_j + 0\cdot \dots + \dots$

Then $$\Bbb E(y_t) =  \Bbb E(\sum_i a_i x_i) = \sum_i \Bbb E( a_i x_i) = \sum_i \Bbb E( a_i)  \Bbb E(x_i)$$
Remember that $\Bbb E(a_j) = (1 - 0.3)\cdot1 + 0.3\cdot0 = 0.7$

This should help to find out what the process is stationary.


答:
Elaborating on @Archis Welankar's answer:

Once you've solved the Limit of Sum definition of the series - limit, you then simply solve for the integral.

$ \int _0^1 \frac{6}{9-x^2}dx = \mathbf{\ln(2)}$

Title: How to evaluate the limit of this following expression $\color{black}{\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}}$?

Body: I am not able to find the limit of the following expression, after trying a bit hard:$$\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}$$ Can some one help with some approach?


问:
Complete lattices and sublattices – which requirement is more stringent?
I'm studying from Michael Carter's "Foundations", and on page 29 he makes the comment, "Note that the requirement of being a sublattice is more stringent than being a complete lattice in its own right".

This seems counterintuitive. Some lattice $L$ is complete if every nonempty subset $S\subseteq L$ has a least upper bound and greatest lower bound in $L$. This is a more stringent requirement than merely needing bounds for every combination of two points in $L$. By the same token, requiring that all subsets $T \subseteq S$ have meets and joins in $S$ (thereby making $S$ complete) ought to be more stringent than merely requiring $S$ to be a lattice, right? In what sense are the requirements for being a sublattice more stringent on $S$?

One more related question – suppose $X= \{ (1,1),(2,1),(3,1),(1,2),(1,3),(3,3) \}$. The book says $X$ is a complete lattice, but not a sublattice of $\{1,2,3\}\times \{1,2,3\}$. The point $(3,3)$ is an upper bound for all pairs of points in $X$, so shouldn't that be enough to qualify $X$ as a sublattice of the set above? Why is $X$ not considered a sublattice?

答: The meet and join operations in a sublattice must agree with the meet and join operations in the lattice. Consider your example. The least upper bound, in $X$, of $(1,2)$ and $(2,1)$ is $(3,3)$. In the space $\{ 1,2,3 \} \times \{ 1, 2, 3 \} $ the least upper bound of $(1,2)$ and $(2,1)$ is $(2,2)$.</p>


问:
coin flipping question. Question about Bayes vs conditional probability formula.
A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).

I tried doing this problem by hand again and I think I understand most important parts except this one very easy one:

What is the difference here between Bayes and Conditional prob?

Rules and Formulas:

Conditional probability formula is :

P(A given B) = P(A and B) /P(B)

Bayes:

P(A given B) = P(B given A) * A / P(B) (this can be expanded out I know)

Also, regardless of dependency:

P(A and B) = P(A) * P(B given A) = P(B) * P(A given B)

So Bayes is just a more fleshed out version of conditional probability... used when P(B) isn't as simply stated as P(B).

My question... do we use Bayes when we can't use conditional probability because P(A and B) isn't simple. In the coin problem, we use Bayes because why exactly? Is it because A and B are dependent variables in the coin problem?

So this was what I was writing when this confused occurred to me:
Let’s do this coin problem using Bayes:
So first things’s first, this is a two step problem. We need to first figure out how likely the coin is an unfair coin given information of flipping 5 heads. Given that it’s heads 5 times, how likely is the coin unfair? Then you need to figure out, given the likelihoods of fair vs biased coin, the next coin flip's likelihood of heads.

P(A) = coin is biased

P(B) = coin is heads 5 times.

So let’s try to find P(A given B)

Remember P(A and B) = P(A) * P(B given A) = P(B) * P(A given B)

That’s true regardless of dependency. But we’re trying to find P(A given B) here. And P(B given A) is much easier to find... its just 1. So use P(B given A) when trying to calc P(A and B). Remember, we’re working with dependent variables here so you can’t just use P(A and B) or the conditional formula.

Bayes:
P(A given B) = to P(B given A) * P(A) / P(B)
= 1 * 1/5 / P(B)

P(B)= (1*1/5) + ((1/2)^5 (4/5))
=(1/5 + (1/32 * 4/5)) =
(1/5 + (4/160)) = 9/40

So P(A given B) = (1/5) / (9/40) = 8/9

If conditional probabiltiy and Bayes are the same formula, why can't I do this problem using conditional formula?

P(A given B) = P(A and B) / P(B)

= P(A)*P(B) / P(B)

= 1/2 *9/40 / 9/40

= wrong answer

What error in thinking am I making here? It seems like the conditional prob formula isn't taking account dependency... which is odd since the formula seems to be made to do that.

oh right... it's that P(A and B) = P(A) * P(B) only when the variables are independent. So it seems like that should always be the first question I ask. Or I can just always use P(A and B) = P(A) * P(B given A) or P (B) * P(A given B) because that's the equation to use regardless of dependence... right...stupid...

Perhaps my question is about independence. P(A and B) = P(A)*P(B) if and only if A and B are independent right? If the question involves dependent variables, you have to use P(A and B) = P(A) * P(B given A) or = P(B) * P(A given B) right? And usually the choice of which formula to use is apparent P(A) * P(B given A) or P(B) * P(A given B). Like... if you're trying to find P(A given B)... you can't use P(B) * P(A given B) to represent P(A and B).

Is everything I've written made sense?

答:
You've done a great job.  $P(A\cap B)=\frac{1}{5}$, because that is the probability that you have both selected the unfair coin and flipped 5 heads.  $P(A)P(B)=\frac{1}{5}\frac{9}{40}=\frac{9}{200}$.  Hence, you have proved conclusively that $A,B$ are not independent.

In fact, if $P(A|B)=\frac{P(A)P(B)}{P(B)}=P(A)$, that is exactly the intuitive understanding of independence; $A$ is equally likely to occur regardless of whether or not $B$ has occurred.


答:
Let $[z]$ denote integer GIF, then $f(x)=x+x[x], x\in [0,2)$ is  discontinuous at $x=1$, If $g(x)=\int_{0}^{x} t(1+[t]) dt$ then $g(x)=\frac{x^2}{2}, x \in [0,1]; g(x)=x^2-\frac{1}{2} \in x \in [1,2)\implies g'(x)=x, x\in [0,2) \implies g'(0.3)=0.3, g'(1.3)=1.3$
But by fundamental theorem of Integration, we get
$$g'(x)=x(1+[x]) \implies g'(0.3)=0.3, g'(1.3)=2.6$$
Hence, there is a contradiction for $x=1.3$

Title: Finding an example the fundamental theorem of calculus fails when $f$ is not continuous.

Body:
Is the following statement true or not?

If $f$ is Riemann integrable on $[a,b]$, then the function $F(x)=\int_a^xf(t)dt$ is differentiable on $(a,b)$ and $F'(x)=f(x)$ for $x\in (a,b)$

I think it is not. Since $F$ is differentiable at the point where $f$ is continuous, but I can not think a counter example, is anyone has a counterexample? Thanks a lot.


问:
Question in Proof of Every Tychonoff space X can be embedded as a subspace of a cube
I am learning Topology from Dolciaini Expository text in Topology by S.G Krantz.

I have question in proof of following theorem:

Here $I^{C}$ denote the collection of all functions from C to I.

I am not able to prove $f$ to be embedding.

For proving 1. $f$ is 1-1 .

$f$ is continuous.

$f^{-1}$ is continuous on image of $f$.

$f(x)$ is clearly continuous. let $f(x)=f(y)$ , then if $f^{-1}$ exists we can prove it 1-1 . But I also can't deduce on why $f^{-1} $ exists. I don't have any other thoughts except these.

Can you please tell how to prove 1 and 3 in defination of embedding.

答:
It is a good idea to set up proper notations for the map. For $x \in X$ write $F_x$ for the map  defined on $C$ by $F_x(f)=f(x)$. The the map under consideration is $x \to F_x$.

If the image of $x$ is same as image of $y$ then $F_x=F_y$ so $F_x(f)=F_y(f)$ of $f(x)=f(y)$ for all $f \in C$. By complete regularity $x \neq y$ implies there exists $f\in C$ such that $f(x)=1$ and $f(y)=0$. This contradiction shows that $x=y$. Hence our map is one-to-one.

Hint for continuity of the inverse: Let $U$ be an open set in $X$ . It is enough to show that the image of $U$ is an open subset of the range of our map. Let $x \in U$ There exists $f \in C$ such that $f(x)=1$ and $f(y)=0$ for all $y \in X\setminus U$. Consider $\{F_t: F_t(f) >\frac1 2\}$. Verify that this is open, contains $F_x$ and it is contained in the image of $U$.


问:
How many vertices can a hypercube-hyperplane intersection have?
Consider the n-hypercube $[-1, 1]^n$, and intersect it with an n-hyperplane. The intersection is in general an $(n-1)$-dimensional polytope. How many vertices can it have?

For example, when $n=2$, it must be a line segment, so it can have at most 2 vertices. When $n=3$, it can have at most 6 vertices. I'm having difficulty with $n\ge 4$.

When $n=4$, we can intersect the cube by $\{x | \sum_{i=1}^4 x_i = 0\}$ to get a rhombic dodecahedron with 14 vertices, so that's a lower bound at least.

答: Consider the hyperplane $x_1 = 1$. The intersection has $2^{n-1}$ vertexes, isn't it ? So the maximum number in the general case is at least that!


问:
How to solve trigonometric equations with an absolute value?
To solve$$|\cos 2x|\geq|\sin x|,(x\neq \frac{n\pi}{2})$$

Taking cases doesn't seem efficient to me here and it is the only method I know to approach these problems.

Is there a better way to solve this? Perhaps squaring it?

答:
Hint:

Squaring and setting $t:=(\sin x)^2\ge0$, you need to solve$$(1-2t)^2\ge t$$

or
$$\bigg(t-1\bigg)\left(t-\frac14\right)\ge0.$$

The first factor cancels at $x=k\pi$ or is negative. The rest of the discussion reduces to

$$\sin^2x\le\frac14$$ or $$-\sin\frac\pi6\le\sin x\le\sin\frac\pi6.$$


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:
To prove convergence of steepest descent, $\alpha\in(0,1)$ is enough.
Additional restrictions on $\alpha$ are required in order to prove additional results as, e.g.,

existence of step-sizes for Wolfe-Powell step-size strategy

convergence of nonlinear CG

acceptance of full step-size $t=1$ in globalized Newton


问:
Prove $\frac{k[x,y]}{I}$ and $k[x]$ are not isomorphic where $I=<xy-1>$.
This question is related to my previous question Existence of surjective but not injective map between rings imply being non isomorphic?

Here my question is, how can i show $\frac{k[x,y]}{I}$ and $k[x]$ are not isomorphic. I showed there exist a ring hom. which is surjective but not injective. However this is not enough to prove the being non isomorphic.

答:
If you're considering your rings as $k$-algebras, the argument is easy.

Suppose $\phi \colon k[x,y]/(xy-1) \to k[x]$ is a $k$-algebra isomorphism. Because $\bar x$ is invertible, its image $\phi(\bar x)$ is invertible as well. But $k[x]^* = k^*$, so $\phi(\bar x) \in k^*$. This contradicts the fact the $\phi$ is an isomorphism of $k$-algebras.

If you're considering your rings as general rings, it takes a bit more work to derive a contradication from $\phi(\bar x) \in k^*$.

After you conclude that $\phi(\bar x) \in k^*$, look at $\phi(1 + \bar x)$. Note that $\phi(1 + \bar x) = 1 +\phi(\bar x) \in k$. Now there are two cases: either $1 + \phi(\bar x) \in k^*$ or $1 + \phi(\bar x) = 0$. The first case gives a contradiction because $1 + \bar x$ is not invertible in $k[x,y]/(xy-1)$, so its image under $\phi$ also cannot be invertible. The second case gives a contradiction because it means that $\phi(\bar x) = -1$ and therefore $\bar x = -1$, which is not the case.

As an aside, to see that $1 + \bar x$ is not invertible, consider the $k$-algebra homomorphism from $k[x,y]/(xy - 1)$ to $k$ induced by $x \mapsto -1$ and $y \mapsto -1$. (Note that this is a homomorphism because $xy - 1 \mapsto 0$). Under this homomorphism $1 + \bar x$ maps to $0$, which is not invertible, so $1 + \bar x$ is not invertible either.


问:
The Sum of Geometric-Factorial Series
I was pondering over my Sequences & Series homework, when the following series struck my mind:

$0!+1!x+2!x^{2}+3!x^{3}+.......+n!x^{n}$

And believe me, I am serious that I couldn't think of any approach, but it looked beautiful.I tried to look up all possible summation methods, but it seemed like they didn't offer a break to me.  So guys, any ideas or approach?

答:
It is not so bad if you are aware of the incomplete gamma function since
$$f_n(x)=\sum_{k=0}^n k! \,x^k$$
$$f_n(x)=-\frac{e^{-1/x}}{x}\Bigg[(-1)^n (n+1)!\, \Gamma \left(-(n+1),-\frac{1}{x}\right)+\Gamma
   \left(0,-\frac{1}{x}\right) \Bigg]$$ For $x=1$ appears the subfactorial functions and, for $x>1$ the exponential integral function.

Title: The Sum of Geometric-Factorial Series


问:
Integration by parts on manifolds with Hessian
I'm working on the following exercise in my Riemannian manifolds book: prove an "integration by parts" formula for the expression

$\int_{\Omega} f_{1} \text{Hessian}(f_{2})(X,Y) d \mu_{g} $

where $\Omega$ is a subset of a Riemannian manifold $(M,g)$, $X,Y$ are two tangent vectors and $f_{1},f_{2}$ are smooth functions on $(M,g)$. Here, $d \mu_{g}$ is the standard measure on $(M,g)$.

Also: $\text{Hessian}(f)(X,Y)=g(\nabla_{X}\text{grad}_{g}(f),Y)$.

Thanks in advance!

答:
For those you encounter the same problem, here is the solution:

Proposition 6.2 (Integration by parts formula)
page 27
https://arxiv.org/abs/1505.04817v1


问:
Show that $\angle PAB = \angle CAQ$
Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.

Prove that $\angle PAB = \angle CAQ $

I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$... wait this looks like an apolonian circle.

Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.

Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$

答:
Let $M, N$ be two intersection points of the perpendicular bisector of $BC$ with the circumcircle $(O)$ of three points $A,B,C$

WLOG, we assume further that the point $M$ is in the interior of the angle $\angle BAC$, i.e $M$ is the midpoint of the arc $BC$ which doesn't pass through the point $A$

Also, WLOG, we assume $P$ near $A$ than $Q$.

By some argular arguments, we can see that:
$$\angle MBP= \angle MBQ  $$
Indeed, by symmetry, we have

$\angle QBC= \angle QCB = \angle ACQ- \angle ACB= 180^o-\angle ABP -\angle ACB $

$\angle MBQ= \angle QBC- \angle MBC =\angle QBC- \frac{1}{2} \angle BAC$
$= \frac{1}{2}( \angle QBC +\angle QBC-\angle BAC )=\frac{1}{2}( \angle QBC +180^o-\angle ABP -\angle ACB-\angle BAC )$
$= \frac{1}{2}(\angle QBC+\angle ABC-\angle ABP)=\frac{1}{2}(\angle QBA-\angle ABP)$
$=\frac{1}{2}\angle QBP$

Besides, $BM \perp BN$,
Thus $(O)$ is the Apollonian circle for two points $P,Q$ and the ratio $\frac{MP}{MQ}$

AS $(O)$ passes through $A$, which makes $AM$ be the bisector of $\angle PAQ$

Hence, the conclusion
$\square$

P/s: Oh my rusty brain cells of Olympiad toolbox.

Title: Show that $\angle PAB = \angle CAQ$


问:
Show that cancellation law does not hold for $<\mathbb{R},•>$.
Please help me in showing that cancellation law does not hold for Reals under $•$ operation, i.e. $<\mathbb{R},•>$.

Let, $R_1=a, R_2=b, R_3=c$

$R_1•R_3=R_2•R_3 \nRightarrow R_1 = R_2$

$a.c= b.c \nRightarrow a = b$

I can not find a such example, nor can prove as shown above.

答: The reals, being a field, honour something close to a cancellation law: if $(a-b)\cdot c=0$ then $a-b=0$ or $c=0$, i.e. if $ac=bc$ then $a=b$ or $c=0$.


答:
Seeing as this question is tagged as precalculus and contest math, here's a solution that avoids Riemann sums:

Write
$$
\frac{6n}{9n^2-r^2}=\frac{6n}{(3n+r)(3n-r)}=\frac1{3n+r}+\frac1{3n-r}
$$
and sum over $r=1,\ldots,n$ to find that
$$
\frac1{3n} + \sum_{r=1}^n \frac{6n}{9n^2-r^2}=\frac 1{2n} + \frac1{2n+1} +\cdots + \frac1{4n-1}+\frac1{4n}.
$$
If you're familiar with sums of the form $S_n:=\frac1{an}+\cdots+\frac1{bn}$ (for $b>a$) you know that the RHS converges to $\log (b/a)=\log 2$.

For completeness, here's an elementary derivation.

Claim: Assume $b>a$. Then $$\lim_{n\to\infty} \left(\frac 1{an}+\cdots+\frac1{bn}\right) = \log\left(\frac ba\right).$$

Proof: Take the inequality
$$
\frac{x-1}x\le \log x \le x-1
$$
and substitute $x=1+\frac1k$. This yields for every $k$
$$
\frac1{k+1}\le \log(k+1)-\log k\le \frac 1k.
$$
Sum from $k=an$ to $bn$ and use telescoping to get
$$S_n-\frac1{an}+\frac1{bn+1}\le \log(bn+1)-\log(an)\le S_n.
$$
Rearrange this to the form
$$
\log\left(\frac{bn+1}{an}\right)\le S_n\le \log\left(\frac{bn+1}{an}\right)+\frac1{an}-\frac1{bn+1}.
$$
Take the limit as $n\to\infty$ and apply the squeeze theorem to obtain the result.


问:
Decorating And Cutting A Cake Fairly
Imagine an N x N square cake you are tasked with decorating with N types of toppings (colors). You then with to share the cake equally amongst N friends. All pieces must be rectangular and have same the area. It's also only fair that no matter what kind of piece your friends have, they each have the exact same type of toppings. How should you decorate the cake so that no matter how you cut it, it's always a fair division?

For example below, the 4 x 4 case of 4 different colors for 4 friends:

Take any equal area rectangular cut and each will contain all four different colors:

It's not hard to see that any cut that's equal is a fair division by the definition above. Call any N x N coloring of this kind a fair division coloring.

Below are fair division colorings for 2 x 2, 3 x 3, and 5 x 5, 7 x7, and 9 x 9:

For N=p, p a prime, the coloring is trivial. However, I found it is impossible for N=6 and N=8. I used brute force methods and found that any coloring of 6 different colors for the 6 x 6 grid, there will exist a cut that doesn't create a "fair" division. The same is true for N=8.

My question: is there a general rule to determine for what N a fair division coloring of the square exists?

EDIT: I am absolutely confident this kind of problem has been studied before, but I'm not sure where to find a paper on this. I appreciate any and all thoughts on where I can look.

答:
Just an observation

Suppose such a coloring works for some $n$. Let $p$ and $q$ pe integers such that $q>p$ and $p\cdot q=n$ Consider the following $2$ rectangles ($ABCD$ and $AB'C'D'$, in the corner):

(image didnt work)

(Note that $AB'=AD=p$ and $B'B=D'D=q-p$ and $A,D,D'$ are on a line and $A,B,B'$ are on a line and $O$ is the intersection of $B'C'$ and $CD$ )

This means that rectangles $B'BCO$ and $DD'C'O$ have the same $p(q-p)$ colors inside them.

Now consider the rectangle $D'XYZ$ such that $X$ is on the line $AD$ and $Y,Z$ are on the line $BC$ and $D'X=p$. If $2p>q$, this means that $D'XYZ$ intersects $B'BCO$, thus leading to a contradiction, because $DD'C'O$ is inside $D'XYZ$ so $DD'C'O$ already has the colors in $B'BCO$.

Thus, the coloring is NOT possible for any $n$ such that there exist $p,q$ $p\cdot q=n$ and $2p>q>p$ (or $2n>q^2>n$)

This idea can probably lead to some better results. For example, using this observation, $6$ is clearly not valid, while $10$ might be valid. Also, prime powers might be valid too.

Title: Decorating And Cutting A Cake Fairly


问:
Prove that the $n$th positive nonsquare integer is $n+\{ \sqrt{n} \}$ where $\{x\}$ is denotes the integer closest to $x$.
For any positive integer $n$, let $f(n)$ denote the $n$th positive nonsquare integer, i.e.,

$f(1)=2$

$f(2)=3$

$f(3)=5$

$f(4)=6$

etc.

Prove that $f(n)=n+\{ \sqrt{n} \}$ where $\{x\}$ denotes the integer closest to $x$ (for example, $\{\sqrt{1}\}=1$, $\{\sqrt{2}\}=1$, $\{\sqrt{3} \}=2$, $\{\sqrt 4 \}=2$,...)

答:
This was fun! Thanks for asking it ^_^

The key observation is this:

$$f(n^2) = n^2 + n$$

After all, there are $n$ squares $\leq n^2$ that we had to skip along the way.

I'll write this somewhat more suggestively as

$$f(n^2) = n^2 + \left \{ \sqrt{n^2} \right \}.$$

Now let's see what happens as we increment $n^2$. We want to understand what happens everywhere between $f(n^2)$ and $f((n+1)^2) = f(n^2 + 2n + 1)$. So we want to look at $f(n^2 + k)$ for $1 \leq k < 2n + 1$.

The next square happens at $(n+1)^2 = n^2 + 2n + 1$, which is $n+1$ steps away from $f(n^2)$. So we'll split this region into two pieces:

First we'll look at $1 \leq k < n+1$

Then we'll look at $n+1 \leq k < 2n+1$

For $k < n+1$, we have no fears! We don't see any more square numbers, and this makes counting the nonsquares particularly easy:

$$f(n^2 + k) = f(n^2) + k = n^2 + k + n$$

So we're led to a natural question: For $k < n+1$

$$n \overset{?}{=} \left \{ \sqrt{n^2+k} \right \}$$

And the answer is easily seen to be yes. Intuitively, this is because $n^2+n$ is closer to $n^2$ than it is to $(n+1)^2$. I'll leave it to you to formalize the argument. (As a hint: $\sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}}$, and you might imagine $\sqrt{1 + \frac{1}{n}}$ doesn't change the value when we round to the nearest integer).

For $n+1 \leq k < 2n+1$, we've seen exactly one new square. So

$$f(n^2 + k) = f(n^2) + k + 1 = n^2 + k + n + 1$$

Again, you see the question coming: For $n+1 \leq k < 2n+1$

$$n + 1 \overset{?}{=} \left \{ \sqrt{n^2 + k} \right \}$$

And again, it's easy to see the answer is yes. This is (intuitively) because $n^2 + n + 1$ is now closer to $(n+1)^2$ than to $n^2$. You can do a similar manipulation to my previous hint to formalize this intuition.

I hope this helps ^_^


Body:
We can even compute the partial sums
$$S_n=\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}=\sum_{r=1}^n\frac{1}{3 n+r}+\sum_{r=1}^n\frac{1}{3 n-r}$$
$$S_n=\big[\psi (4 n+1)-\psi (3 n+1) \big]+\big[\psi (1-3 n)-\psi (1-2 n)\big]$$ Using asymptotics
$$S_n=\log (2)+\frac{1}{24 n}+\frac{1}{64 n^2}+O\left(\frac{1}{n^4}\right)$$


问:
Finding the number of solutions for $e^{\tan x}=\sin x+ \cos x$ graphically
The number of solutions of the equation
$$e^{\tan x}=\sin x+ \cos x$$ In the interval $[-\pi,\pi]$

I tried to approach this question graphically but hit a dead end when trying to graph $e^{tanx}$

For graphing sinx+cosx I graphed $\sqrt 2(sin(x+\frac{\pi}{4}))$

I still feel this question can be solved graphically,Is there a method to graph $e^{tanx}$ to some accuracy ?

答:
There is the trivial solution $x=0$.

For tho other root start plotting from $$x=\frac \pi 2 +\text{some small number}$$ to $x=\pi$. Now, zoom more and more.


问:
How to prove that $1/n!$ is less than $1/n^2$?
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

答:
Hint:

$n^2 <n! \implies n(n-(n-1)!) <0$, and note that $(n-1)! >n$ for all $n>3$


问:
What will be the Ito derivative?
Suppose we have the SDE $~~$ $dV=A_t ~dt+ B_t~  dW_t $. $~~$ Let $Z=(1+V)^\theta$. Then what will be $L(e^{\alpha t} Z).$

My opinion: We know that by Ito formula $LZ= \theta (1+V)^{\theta-1} dV +\frac{\theta(\theta-1)}{2} (1+V)^{\theta-2} B_t^2 dt.$ Then, I suppose $L(e^{\alpha t} Z)=\alpha e^{\alpha t} Z +e^{\alpha t}LZ$ will be the answer. But I am not not sure how can I get that? Is there any particular condition on Z only when this will hold. Please help.

答:
First we have using Ito's lemma with $Z_t$ we get

$$dZ_t = [A_t\theta(1 + V_t)^{\theta - 1} + \frac{B_t^2}{2}\theta(\theta - 1)(1 + V_t)^{\theta - 2}]dt + B_t \theta(1 + V_t)^{\theta - 1}dW_t.$$

We can then use the formula for the product of a deterministic function, $f(t)$, and a stochastic process, $X_t$, which is
$$d(f(t)X_t) = f(t)dX_t + X_tdf(t) + df(t)dX_t.$$

Letting $f(t) = \exp(\alpha t)$ and $X_t = Z_t$ gives the solution.
If $X_t$ is an Ito diffusion, $$dX_t = a(t, W_t)dt + b(t, W_t)dW_t,$$ then in the above formula we have $df(t)dX_t = f'(t)dtdX_t = 0$.

Title: What will be the Ito derivative?


问:
How to compute limit of $(1/a+2/a^2+...+n/a^n)$ when $n\to\infty$
I'm doing some calculus, and I found this specific problem

lim(1/a+2/a^2+...+n/a^n) when n->inf, and I have not a single idea how to compute it.

答:
This is arithmetico–geometric series.

The answer is $\dfrac a{(a-1)^2}$ when $a>1$.

See that link for a proof.


问:
Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$
I need to find a closed-form for the following integral. Please give me some ideas how to approach it:
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$</p>

答:
My calculation shows that

\begin{align*}
\int_{0}^{\infty} \frac{x^3}{(x^4 + 1)(e^x - 1)} \, dx
&= \frac{\gamma}{2} - \log\sqrt{2\pi} + \frac{\pi}{4} \frac{\sin\frac{1}{\sqrt{2}}}{\cosh\frac{1}{\sqrt{2}} - \cos\frac{1}{\sqrt{2}}} \\
&\quad - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n\left(1 + (2\pi n)^4\right)} \\
&\approx 0.389075976914101580976629 \cdots.
\end{align*}

My solution is divided into several steps:

Step 1. Let us introduce the function

$$ I(s) = \int_{0}^{\infty} \frac{x^{s}}{(x^4+1)(e^{x}-1)} \, dx $$

It is easy to see that

$$ I(s) + I(s+4) = \Gamma(s+1)\zeta(s+1). \tag{1} $$

This allows us to extend $I(s)$ as a meromorphic function on $\Bbb{C}$.

Step 2. Consider a contour $C$ starting from $\infty + \epsilon i$, making a counter-clockwise turn around $z = 0$ and going back to $\infty - \epsilon i$ as follows:

If we use a logarithm function with the branch cut $[0, \infty)$, we see that

$$ (e^{2\pi i s} - 1)I(s) = \int_{C} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz. \tag{2} $$

Also, for $ 1 < s < 2$ we can confirm that $(2)$ is rewritten as

$$ (e^{2\pi i s} - 1)I(s) = \lim_{n\to\infty} \int_{C_{n}} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz, $$

where $C_n$ is the contour given by

Here, the condition $1 < s < 2$ is introduced in order to create an appropriate decay speed for the integral along the contour $C - C_{n}$. By applying the Cauchy integration formula, we have

$$ (e^{2\pi i s} - 1)I(s) = -2\pi i \sum_{\omega^4 = -1} \operatorname{Res}_{z=\omega} \frac{z^{s}}{(z^4+1)(e^z - 1)} - 2\pi i \sum_{n\neq 0} \operatorname{Res}_{z=2\pi i n} \frac{z^{s}}{(z^4+1)(e^z - 1)}.  $$

Simplifying,

$$ I(s) = \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s+1}}{e^{\omega} - 1} - \frac{2^{s}\pi^{s+1}}{\sin \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{1+(2\pi n)^4} \tag{3} $$

Since both sides define a meromorphic function for $\Re s < 3$, they coincide on this range.

Step 3. Combining $(1)$ and $(3)$, we have

\begin{align*}
I(s+3)
&= \Gamma(s)\zeta(s) - I(s-1) \\
&= \Gamma(s)\zeta(s) - \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s}}{e^{\omega} - 1} - \frac{(2\pi)^{s}}{2\cos \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{n \left( 1+(2\pi n)^4 \right)}
\end{align*}

Taking $s \to 0$, we obtain the desired result.


问:
How to find the sum of the areas of all odd-sided squares in a $9*8$ Grid.

Find the sum of the areas of all odd-sided squares in a $9*8$ Grid.

What I Tried: As a general formula, I know that the number of squares in a $9*8$ Grid would be :-
$$\rightarrow \frac{M(M + 1)(3N - M + 1)}{6}$$
Where $N \geq M$. So putting $N = 9$ and $M = 8$ gives :-
$$\rightarrow \frac{8(9)(20)}{6}$$
Which is a total of $240$ squares.

Now, I am wondering whether there is a formula for finding the number of odd-sided squares, or even-sided squares, or both. For example in a $2*2$ grid you will have $4$ odd-sided and $1$ even sided square, assuming each small square's length to be $1$ unit.

I also have no idea on finding the area, because in that case I have no choice but to count the different odd-sided and sized squares independently, and then find out their area and add then up. The answer is given to be :-
$$\rightarrow \sum\limits_{r=1}^4 (11-2r)(10-2r)(2r-1)^2 $$

How do we do this? How are we even getting the answer?

Can anyone help me? Thank You.

答:
Hint: Take cases: I.e. when the side of square is $1,3,5,7$ units

More Bigger Hint:-

 To generalise let the square be of side $2r-1$ . As the grid is $9*8$  the number of such squares possible is the ways in which we can select $2r$ lines which are consecutive (visualising helps) from $9+1$ lines times the number of ways in which we can select $2r$ lines consecutively from $8+1$ lines which is $(11-2r)(10-2r)$.


答:
I'd make a few observations about $f(x) =e^{\tan x}$.

First, it is always positive.

Second, it is periodic, so if you can sketch it for $0 < x < \pi$ then you can repeat that for $-\pi < x < 0$

Third, let's think about what happens when $x$ is $-\pi$, $- \pi/2$, $0$, $\pi / 2$, $\pi$:

$f(-\pi) = f(0) = f(\pi) = e^0 = 1$

For $x <\pi/2$ as $x\to \pi/2, f(x)\to \infty$, while for $x>\pi/2$ as $x\to \pi/2, f(x)\to 0$.  Same behaviour around $-\pi/2$.

Fourth, $f(\pi/4) = e$, $f(3\pi/4) = e^{-1}$.

So, to sketch, start at $(0,1)$ and draw a curve through $(\pi/4,e)$ that shoots off to infinity at $\pi/2$.

Then sketch a curve from $(\pi/2,0)$ through $(3\pi/4,e^{-1})$ to $(\pi,1)$.

Title: Finding the number of solutions for $e^{\tan x}=\sin x+ \cos x$ graphically

Body:
The number of solutions of the equation
$$e^{\tan x}=\sin x+ \cos x$$ In the interval $[-\pi,\pi]$

I tried to approach this question graphically but hit a dead end when trying to graph $e^{tanx}$

For graphing sinx+cosx I graphed $\sqrt 2(sin(x+\frac{\pi}{4}))$

I still feel this question can be solved graphically,Is there a method to graph $e^{tanx}$ to some accuracy ?


问:
find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$
I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$

Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$

Since $e^{x^4-2x^3+x^2-1}>0$ for all $x\in \Bbb R$ and $2x-1<0\iff x<\frac{1}{2}$. $f$ is decreasing on $(-\infty,\frac{1}{2})$.

Furthermore, $f$ is increasing on $(\frac{1}{2},\infty)$, $f$ is differentiable at $x=\frac{1}{2}$, and $f'(\frac{1}{2})=0$, $f$ attains its minimum value at $x=\frac{1}{2}$.

Am I right?

答:
Everything is fine ! A little bit more can be said:

$f$ is strictly decreasing on $(-\infty,\frac{1}{2}]$

and

$f$ is strictly increasing on $[\frac{1}{2}, \infty).$


答:
$f(x)=\int_{0}^{x^2-x} e^{t^2-1} dt \implies f'(x)=  (2x-1) e^{(x^2-x)^2-1} >0 ~if~ x>1/2$. Hence $f(x)$ in increasing for $x>1/2$ and decreasing for $x<1/2$. Yes you are right. there is a min at $x=1/2$. This one point does not matter, you may also say that $f(x)$ is increasing in $[1/2,\infty)]$ and decreasing on 4(-\infty, 1/2]$.

Note: whether a function increasing or decreasing is decided by two points (not one). For instance, $x_1>x_2 \leftrightarrows f(x_1) > f(x_2).$ If $f(x)$ is decreasing.

Title: find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$

Body:
I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$

Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$

Since $e^{x^4-2x^3+x^2-1}>0$ for all $x\in \Bbb R$ and $2x-1<0\iff x<\frac{1}{2}$. $f$ is decreasing on $(-\infty,\frac{1}{2})$.

Furthermore, $f$ is increasing on $(\frac{1}{2},\infty)$, $f$ is differentiable at $x=\frac{1}{2}$, and $f'(\frac{1}{2})=0$, $f$ attains its minimum value at $x=\frac{1}{2}$.

Am I right?


问:
Is there a simple interpretation I' in which these are all false?
The axioms of Q are extra-logical assertions true in the intended interpretation I^A of elementary arithmetic. Consider the first three axioms:
Q1: ∀x∀y(x′ = y′ → x = y),     Q2: ∀x o ≠ x′,     Q3: ∀x(x ≠ o → ∃y x = y′)
Is there a simple alternative interpretation I' in which they are all false?</p>

答: Sure - consider the natural numbers with the symbol "$0$" interpreted as usual and the function $'$ interpreted as $$x'=0$$ for all $x$. This violates all three rules at once.


问:
Proving that $\inf\{U(f,P)\}=\inf\{U(f,P\cup\{c\})\}$
I'm not sure if the following statement is true or not, but I intuitively I think it is, I was wondering if I could get some pointers on how to prove it if it is in fact true:

Suppose that $f$ is integrable on $[a,b]$. Then
$$ \inf\{U(f,P)\} = \inf\{U(f,Q)\}, $$
where $P$ is an arbitrary partition of $[a,b]$, $Q = P\cup\{c\}$, and $c$ is an arbitrary point between $a$ and $b$.

答:
Yes, equality holds.

Let $S_1:=\text{All partitions on}\ [a,b]$ and $S_2:= \text{All partitions on}\ [a,b] \ \text{with atleast 3 elements in it.}$ Then $S_1=\{(a,b)\} \cup S_2\ $  i.e $S_2 \subset S_1$
$$\implies \operatorname{inf}\{U(f,P):P \in S_1\} \leq \operatorname{inf}\{U(f,P):P \in S_2\}$$

If $P \in S_1$ then $P=(a,b) \ \text{or} \ P \in S_2$. Clearly $\{U(f,P):P \in S_1\}$ contains elements less than $U(f,(a,b))$. So definitely inf of this set is lower than $U(f,(a,b))$. Using $S_1=\{(a,b)\} \cup S_2\ $, every element of $\{U(f,P):P \in S_1\}$ can be bounded from below by an element of $\{U(f,P):P \in S_2\}$ which in turn is bounded by $\operatorname{inf}\{U(f,P):P \in S_2\}$. So we obtain
$$\implies \operatorname{inf}\{U(f,P):P \in S_1\} \geq \operatorname{inf}\{U(f,P):P \in S_2\}$$
Hence equality.

Also in general $\operatorname{Sup}\{U(f,P):P \in S_1\} \neq \operatorname{Sup}\{U(f,Q):Q \in S_2\}$.

Consider the function $f:[0, 2] \rightarrow \mathbb{R}$ defined as
$$f(x) := \begin{cases}
                   x & 0 \leq x \leq 1 \\
                   0 &  1<x \leq 2 \\
                 \end{cases}$$


问:
What is the (parametric) intersection of a plane and a sphere?
Can someone please show me how to prove that the intersection of the plane

$$x+y+z=0$$  and the sphere

$$x^2+y^2+z^2=1$$

can be expressed as

$$x(t)=\frac{\cos t-\sqrt3 \cdot\sin t}{\sqrt6}$$

$$y(t)=\frac{\cos t+\sqrt3 \cdot\sin t}{\sqrt6}$$

$$z(t)=\frac{-2\cos t}{\sqrt6}$$

Ps: Also, why am I not getting the correct notation? I am using a macbook pro (safari) if that is a concern?

答:
You can solve for $z=-x-y$ and plug in $x^2+y^2+(-x-y)^2=1$, which rearranges to $2x^2+2xy+2y^2=1$.  Hence for any fixed $z$ you have a rotated ellipse.

Let $u=(x+y/2)$.  Then $u^2=x^2+xy+y^2/4$, so the equation becomes $2u^2+\frac{3}{2}y^2=1$.  In standard form this is:  $$\frac{u^2}{(\frac{1}{\sqrt{2}})^2}+\frac{y^2}{(\frac{\sqrt{2}}{\sqrt{3}})^2}=1$$

We can solve this parametrically as $u=\frac{1}{\sqrt{2}}\sin t$, $$y=\sqrt{\frac{2}{3}}\cos t$$  We can now find $$x=u-y/2=\frac{1}{\sqrt{2}}\sin t-\frac{1}{\sqrt{6}}\cos t$$ and $$z=-x-y=\frac{-1}{\sqrt{2}}\sin t +(-\frac{1}{\sqrt{6}}-\sqrt{\frac{2}{3}})\cos t$$

Alas, you posted your revisions after I'd made my choices so my solution will not agree with yours.  There are three choices at the first step (solve for $z,y,x$), and two choices at the second step (replace $x$ or $y$).  One of those six choices might give the result you have.  :-)

Followup: A cross term ($xy$) is a rotation and can always be eliminated by a change of variables, pointing in the directions of the major and minor axes.  There is a systematic way to do this; see here or here, or you can do it by the seat of your pants, like I did.  I didn't directly rotate, instead I sheared, which made the computations a bit simpler.

Edit: fix sign error, thanks @Mhenni.

Edit 2: add general method


问:
How to determine the 90% confidence interval for the population proportion of students who found the math test easy?
Out of $120$ students, $85%$ found the math test easy. How to determine the $90%$ confidence interval for the population proportion of students who found the math test easy?

I am not sure about my computation but I tried to solve it in the following way.

The $85%$ of 120 students is 102 students.

Since the confidence level is 90%, there will be 92 students

My answer is not an interval but a fixed percentage which is $76.5%$.

Any comments and suggestions will be highly appreciated.

答:
The issue with your question is that there are several ways to construct a confidence interval for a binomial proportion.

Let $n$ be the total number of observations (in your case, $n = 120$ students observed).  Let $x$ be the number of observations for which the event of interest occurred (here, $x = 85$ represents the number of students that found the test easy).

Now let $\hat p = x/n$ be the observed proportion of students that found the test easy.

Let $\alpha$ be chosen such that the confidence level of the interval is $100(1-\alpha)\%$.  In your case, $\alpha = 0.10$ for a $90\%$ confidence interval.

Let $z_{\alpha/2}$ represent the upper $\alpha/2$ percentile of the standard normal distribution; that is to say, it is a $z$-score for which $$\Pr[Z > z_{\alpha/2}] = \alpha/2$$ where $Z$ is a standard normal random variable.  In your case, $z_{0.05} \approx 1.645$ because for $Z \sim \operatorname{Normal}(0,1)$, $\Pr[Z > 1.645] \approx 0.05$.

With all this, there is the Wald interval:

$$\hat p \pm z^*_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

There is the Wilson score interval:

$$\frac{1}{1+z_{\alpha/2}^2/n} \left(\hat p + \frac{z_{\alpha/2}^2}{2n}\right) \pm \frac{z_{\alpha/2}}{1 + z_{\alpha/2}^2/n} \sqrt{\frac{\hat p (1 - \hat p)}{n} + \frac{z_{\alpha/2}^2}{4n^2}}.$$

Then there is also the exact Clopper-Pearson interval, whose formula involves the evaluation of the $F$-ratio distribution or incomplete beta function, or a binomial sum.

The Wikipedia article also includes the Agresti-Coull interval and arcsine interval.  The first two are the most popular in use.  Which one do you want to use?


问:
Ask for some help to understand the notations
The following is taken from a text which tries to proves a relation between dimensions of subspaces.

My first question comes from

the canonical injections $in_1:M\to M\oplus N$

in line 3 and 4. My understanding of direct sum $M\oplus N$ is that, when we write down this notation, $M$ and $N$ must be independent, i.e., $M\cap N=0$. But since $M$ and $N$ are two arbitrary subspaces as indicated in line 1 an 2, they are not necessarily independent. What does the author mean here? Does $\oplus$ mean something other than direct sum? Furthermore, If we assume $M\oplus N$ denotes direct sum, problems comes out immediately. In line 4, the author wrote

injections $f:M\cap N\to M\oplus N$

If $M\cap N=0$, then the function $f$ is just the trivial $0\to 0$. So is the following $g$. I believe this is not what the author intended to define.

The second question arises from the third line if Lemma 2.14. I don't really know what that line of symbols and "a short exact sequence" in the subsequent line mean. Does that line of arrows mean something arcane in the circle of math? Of course, the question may due to I don't understand what $f$ means in the first place. I wish I could understand what the author wanted to say, after I have tried hard, so could you please help me understand all these stuff? Thanks a lot.

答:
Leaving this as an answer because it might be too long for a comment.

You might be confused by what it means to be a direct sum. A direct sum $M \oplus N$ is as in Wikipedia. What you might be referring to is that $M + N = M \oplus N = V$ if and only if $M \cap N = \mathbf{0}$ the zero subspace. See for example this older MSE question. There is often a bit of trouble in the terminology if you're unfamiliar with it, since the direct sum is 'essentially' the direct product (for a finite collection of vector spaces - for an infinite collection, a direct sum is usually imposed by a 'local finiteness' condition, that in any sum there are only finitely many nonzero coefficients).
This definition should answer your first two questions.

As for the second question, a short exact sequence is defined here in Wikipedia.
It is a sequence of five terms, two of which are zero, and it is exact at each component (it's probably helpful to understand what the definition of 'exactness' means - again, look at Wikipedia).

In the setting of your given lemma, you have injections $f, g: M \cap N \to M \oplus N$ coming from the canonical inclusions $M, N \hookrightarrow M + N$ (can you see why this is the case? This is very straightforward, spelled out in the text, and also follows from our definition above).
Once you've understood this, hopefully the statement should make sense.


问:
Drawing numbers from a set, quantile
We have a set containing $20$ numbers from $1$ to $20$. Each time we draw only one number, and repeat it $15$ times (without replacement). Let's denote $X-$ the largest drawn number. Find the smallest ${{16} \choose {15}}/{20 \choose 15}$-fractile  of a random variable $X$.

So, we look for such $x_p$ that $F(x_p)\geq {{16} \choose {15}}/{20 \choose 15}.$ If $X$ is maximum of all drawn numbers, then $F_X(t)=\Bbb P(X\leq t)=\Bbb P(X_1\leq t, ...,X_m\leq t)$ and, if all of $X_1,...X_m$ are iid then we have $\Bbb P(X_1\leq t)\cdot...\cdot\Bbb P(X_m\leq t)$. But... how to proceed?

答:
Verifying by simulation of a million iterations in R:

```<code>set.seed(2020)
q = 16/choose(20,15); q
[1] 0.001031992
x = replicate(10^6, max(sample(1:20, 15)))
mean(x <= 16)
[1] 0.001032
</code>```

Agrees with the elegant Answer of @peterwhy (+1), which IMHO @itsme should Accept and upvote.

Note: Vector x contains a million observed
maximums and  x <= 16 is a logical vector of
a million TRUEs and FALSEs, the mean of which is its proportion of TRUEs.

Title: Drawing numbers from a set, quantile

Body:
Verifying by simulation of a million iterations in R:

set.seed(2020)
q = 16/choose(20,15); q
[1] 0.001031992
x = replicate(10^6, max(sample(1:20, 15)))
mean(x <= 16)
[1] 0.001032

Agrees with the elegant Answer of @peterwhy (+1), which IMHO @itsme should Accept and upvote.

Note: Vector x contains a million observed
maximums and  x <= 16 is a logical vector of
a million TRUEs and FALSEs, the mean of which is its proportion of TRUEs.


问:
Checking where a series of functions is uniformly convergent, an application of Weierstrass M-Test?
Let $\sum_{n=1}^{\infty}$fn(x) = $\sum_{n=1}^{\infty} \frac{(1-x^2)^n}{(1+x^2)^n}sin(nx)$

I am attempting to figure out where this series is uniformly and absolutely convergent.

I am trying to use M-Test,

if 0 < x < 1  then   $0<x^2<1$

so |$\frac{(1-x^2)^n}{(1+x^2)^n}sin(nx)|<\frac{1}{(1+x^2)^n} $

the RHS is convergent geometric series, so I applied M-test.

Is this the correct approach to use?

And how do I check that this series is not uniformly convergent on x>1.

答:
Partial answer. The series is not uniformly convergent on $(1,\infty)$ and $(-\infty, -1)$: If it is uniformly convergent there then the general term must tend to $0$ uniformly. But then $\sin (nx) \to 0$ uniformly. Take $x=\frac {2m\pi} n+\frac {\pi} {2n}$ with $m$ large to get a contradiction.

By M-test the series converges uniformly on $\{x: \epsilon \leq |x| \leq 1\}$ for any $\epsilon \in (0,1)$.


答:
We can do the comparison with $\dfrac{2}{n^2}$. It is certainly true that $\dfrac{1}{1!}\le \dfrac{2}{1^2}$. </p>

<p>And for $n \gt 1$, we have $n!\ge (n)(n-1)$. But $n-1\ge \dfrac{n}{2}$, and therefore $n!\ge \dfrac{n^2}{2}$. It follows that $\dfrac{1}{n!}\le \dfrac{2}{n^2}$ for all positive $n$.  </p>

Title: How to prove that $1/n!$ is less than $1/n^2$?

Body:
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

Body:
We can do the comparison with $\dfrac{2}{n^2}$. It is certainly true that $\dfrac{1}{1!}\le \dfrac{2}{1^2}$.

And for $n \gt 1$, we have $n!\ge (n)(n-1)$. But $n-1\ge \dfrac{n}{2}$, and therefore $n!\ge \dfrac{n^2}{2}$. It follows that $\dfrac{1}{n!}\le \dfrac{2}{n^2}$ for all positive $n$.


问:
Show normality of a process.
I am given $(X_t)_{t \geq 0} = e^{-t}B_{e^{2t}}$ where $(B_{t})_{t \geq 0}$ is a standard Brownian Motion. It is also defined $M_t = X_t - X_0 + \int_0^tX_sds$.

I want to show that $M_t$ is a Gaussian random variable. I started computing the char. function:
$$E[e^{iX_tu - iX_0u + iu\int_0^tX_sds}]$$
I know that the process $(X_t)_{t \geq 0}$ is stationary, would this be useful for computing the above expectation? How should I proceed? Any hint on how to compute that integral?

答: All linear combinations of $B_t$'s are normally distributed, so the same is true of linear combinations of $X_t$'s. Use Riemann sums to see that $M_t$ is almost sure limit of  sequence of normal random variables which implies that $M_t$ is also normal.

Title: Show normality of a process.


答:
We have
\begin{align}
\lim_{n\to\infty}\left(\frac{1}{a}+\frac{2}{a^2}+...+\frac{n}{a^n}\right)&=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{a^k}\\
&=\sum_{k=1}^{\infty}\frac{k}{a^k}
\end{align}
Let $1/a=x,$ so we have
\begin{align}\sum_{k=1}^{\infty}\frac{k}{a^k}&=\sum_{k=1}^{\infty}kx^k\\
&=x\sum_{k=1}^{\infty}kx^{k-1}
\end{align}
Now, let
$$f(x)=\sum_{k=1}^{\infty}kx^{k-1}$$
So
$$\int f(x)dx=\sum_{k=1}^{\infty}\int kx^{k-1}dx$$
The RHS equals
$$\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}\text{  for $|x|$<1, i.e., $|a|>1$}$$
So
$$f(x)=1/(x-1)^2$$
i.e.,
$$\lim_{n\to\infty}\left(\frac{1}{a}+\frac{2}{a^2}+...+\frac{n}{a^n}\right)=\frac{x}{(x-1)^2}=\frac{a}{(a-1)^2}\text{  for $|a|>1$}$$
$\frac{x}{(x-1)^2}=\frac{a}{(a-1)^2}$ can be easily found using simplification, so don't be confused.

Title: How to compute limit of $(1/a+2/a^2+...+n/a^n)$ when $n\to\infty$

Body:
I'm doing some calculus, and I found this specific problem

lim(1/a+2/a^2+...+n/a^n) when n->inf, and I have not a single idea how to compute it.


答:
There are two types of direct sums.

Given a subspaces $U_1,U_2$ of a vector space $V$ we can define their inner sum $U_1+U_2$ to be the smallest subspace of $V$ containing both $U_1$ and $U_2$. We say this inner sum is direct, if $U_1\cap U_2=0$.

Meanwhile there is the (external) direct sum of arbitrary vector spaces $U$ and $V$. It is given by their product $U\times V$ with componentwise vector space structure. The notation $U \oplus V$ is there to remind us that the direct sum of vector spaces has some very special properties, which other types of products like the product of sets don’t share.

It may be confusing to have the same symbol $\oplus$ for both the inner and external variant, but this has a reason: Up to isomorphism they are the same. Given two subspaces $U_1,U_2$ of $V$ their sum $U_1+U_2$ is internally direct if and only if it is isomorphic to the external direct sum $U_1\oplus U_2$.

Now an exact sequence is a sequence of vector spaces and homomorphisms
$$...\rightarrow V_1 \overset{\phi_1}{\rightarrow} V_2 \overset{\phi_2}{\rightarrow} V_3 \rightarrow...$$
with the property $\operatorname{Im} \phi_n = \operatorname{ker} \phi_{n+1}$ for all $n$. I cannot go into details here but they are very useful. Anyway it is a good exercise in the definitions to check that for subspaces $U_1, U_2$ of $V$ the sequence
$$ 0 \rightarrow U_1 \cap U_2 \rightarrow U_1 \oplus U_2 \rightarrow U_1+U_2 \rightarrow 0$$
proposed by your author is exact.

Title: Ask for some help to understand the notations

Body:
The following is taken from a text which tries to proves a relation between dimensions of subspaces.

My first question comes from

the canonical injections $in_1:M\to M\oplus N$

in line 3 and 4. My understanding of direct sum $M\oplus N$ is that, when we write down this notation, $M$ and $N$ must be independent, i.e., $M\cap N=0$. But since $M$ and $N$ are two arbitrary subspaces as indicated in line 1 an 2, they are not necessarily independent. What does the author mean here? Does $\oplus$ mean something other than direct sum? Furthermore, If we assume $M\oplus N$ denotes direct sum, problems comes out immediately. In line 4, the author wrote

injections $f:M\cap N\to M\oplus N$

If $M\cap N=0$, then the function $f$ is just the trivial $0\to 0$. So is the following $g$. I believe this is not what the author intended to define.

The second question arises from the third line if Lemma 2.14. I don't really know what that line of symbols and "a short exact sequence" in the subsequent line mean. Does that line of arrows mean something arcane in the circle of math? Of course, the question may due to I don't understand what $f$ means in the first place. I wish I could understand what the author wanted to say, after I have tried hard, so could you please help me understand all these stuff? Thanks a lot.


问:
Solve integrals $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$ and $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$?
Can the solution to the integrals

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$$

and

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$$

be computed analytically - or approximated? I have only figured out the trivial solution for $\beta=1$ where I can compute the second integral by

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-v}}~\mathrm{d}v = \left[e^{-e^{-v}}\right]_{\alpha}^{\infty} = 1 - e^{-e^{-a}}.$$

答:
With $s:=e^{-\beta v}$

$$\int e^{-v}e^{-e^{-\beta v}}dv=-\frac1\beta\int s^{1/\beta-1}e^{-s}ds$$

is an incomplete Gamma integral.

The other integral is more difficult.


问:
How to solve $\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}$?
Part of an exercise, I got the following sum:
$$
\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}
$$
I'm trying to evaluate it and get the answer. What is the easiest way to do it?

答:
$$ ( \binom{100}{0} + \binom{100}{1} ) + ( \binom{100}{2} + \binom{100}{3} ) ... + ( \binom{100}{58} + \binom{100}{59} )$$

$$ =\binom{101}{1} + \binom{101}{3} + \binom{101}{5}...\binom{101}{59}$$

Not much better you can do than this I Think, there is no closed form for the above acc. to wiki


问:
Morphism between quotient polynomial rings
Le $\Bbb{K}$ be a field.
I want to show wether $\frac{\Bbb{K}[X]}{\langle X^2 \rangle}$ and $\frac{\Bbb{K}[X]}{\langle X^2  + 1\rangle}$ are isomorphic.

First thing to note is that there is an nilpotent element $[X]$ in the first quotient. But in the second one I think it will depend on the roots of $X^2+1$ because for example in $\Bbb{F}_2$, $x^2+1=(X-1)^2$ so $[X-1]$ is nilpotent. So in the case where it can't be factorized there is no nilpotent element (how to prove that ?). Otherwise I don't see any morphism we can exhibit.

Any hint ?

答:
I will change your $\Bbb K$ to $K$, for simplicity. All isomorphism below are isomorphisms of rings (or $K$-algebras, actually).

The structure of $K[X]/(X^2 + 1)$ depends on the number of different roots of the polynomial $X^2 + 1$ in $K$. There can be three situations:

There are two different roots $r_1, r_2$. We then have $$K[X]/(X^2 + 1) \simeq K[X]/(X - r_1) \times K[X]/(X - r_2)$$ by chinese remainder theorem, and each factor $K[X]/(X - r_i)$ is canonically isomorphic to $K$. Hence $K[X]/(X^2 + 1) \simeq K^2$ in this case.

There is one double root $r$. It is easy to see that this happens iff the characteristic of $K$ is equal to $2$, and in that case we have $X^2 + 1 = (X + 1)^2$, hence $$K[X]/(X^2 + 1) \simeq K[X]/(X + 1)^2 \simeq K[X]/X^2$$ via evident isomorphisms.

There is no root, i.e. $-1$ is not a square in $K$. In this case, $K[X]/(X^2 + 1)$ is a field, which is the quadratic extension $K[\sqrt{-1}]$ of $K$.

We see that in the second case, we do have $K[X]/(X^2 + 1) \simeq K[X]/X^2$.

In the remaining two cases, $K[X]/(X^2 + 1)$ is a product of fields, hence does not have any nilpotent element, while $X$ is a nilpotent element in $K[X]/X^2$. Thus they are not isomorphic in these two cases.

To summerize, we have $K[X]/(X^2 + 1) \simeq K[X]/X^2$ iff the characteristic of $K$ is equal to $2$.