问:
Question About Differentiability Of A Piecewise Function
I came across the following question on math stack exchange: Differentiability at a point a implies continuity in a neighborhood of a , and I'm wondering how to prove the function, defined as $x^2$  when x is rational, and 0 when x is irrational, is differentiable at 0. I have had some exposure to epsilon delta proofs, but I have only used them on functions such as polynomial, exponential and trigonometric functions.  I'm not show how to use one to show the function defined above, is differentiable at 0.

Thanks in advance.

答:
Hint.

Consider the Newton quotient
$$
g(x):=\frac{f(x)-f(0)}{x-0}
$$

Note that $g(x)=x$ when $x\ne 0$ is rational and $g(x)=0$ when $x$ is irrational.

You want to show that $\lim_{x\to 0}g(x)=0$. Basically you want an inequality like
$
|g(x)|<\epsilon
$.


问:
Basic Proof of Properties of Kravchuk Polynomials
Fix $n$ and consider the family of polynomials
$$
K_s(\ell) = \sum_{k=0}^s(-1)^k\binom{\ell}{k}\binom{n-\ell}{s-k} \quad \text{defined for} \quad 0 \leq s \leq n
$$

They are called Kravchuk (or Krawtchouk) polynomials.

They are used in a paper I am trying to understand.

The key properties of the polynomials that are used are that if you define the inner product
$$
\langle f, g \rangle = \mathbf{E}_{k \leftarrow Bin(n,\frac{1}{2})}\big[f(k)g(k)\big]
$$
where $Bin(n,\frac{1}{2})$ is the binomial distribution,
on the vector space of functions $\mathbb R^{\{0,1,\ldots,n\}}$ then
$$
\langle K_s, K_r\rangle = \begin{cases} \binom{n}{s} & r=s\\ 0 & r \neq s\end{cases}
$$
so that they form an orthogonal basis, and
$$
\binom{n}{r}K_s(r) = \binom{n}{s}K_r(s)
$$
which allows some computations to be simplified.

However, there seem to be no simple proofs of these facts anywhere; any proof I have found proves this as a special case of some much more general theory and it is too complicated for me to understand.
Can someone give a basic proof of these identities?

答:
First, we give a probabilistic interpretation of $K_s(\ell)$.

Let $A$ be a fixed subset of $\{1,\dots,n\}$ with size $\ell$.

Let $B$ be a randomly chosen subset of $\{1,\dots,n\}$, uniformly chosen from subsets of size $s$.

Then
$$
K_s(\ell)=\binom{n}s \mathbb E\left[(-1)^{|A\cap B|}\right]\tag{1}
$$
This follows by computing $E\left[(-1)^{|A\cap B|}\right]$. A similar computation proves that
$$
\langle K_s,K_r\rangle
=\binom{n}s\binom{n}r\mathbb E\left[(-1)^{|A\cap B|}(-1)^{|A\cap C|}\right],\tag{2}
$$
where

$A$ is randomly chosen from all subsets of $\{1,\dots,n\}$,

$B$ is randomly chosen from all subsets of $\{1,\dots,n\}$ of size $s$, and

$C$ is randomly chosen from all subsets of $\{1,\dots,n\}$ of size $r$,

all independently of each other. Now, imagine you first choosen $B$ and $C$ randomly, and then choose $A$. There are two things that can happen:

$B$ and $C$ are different sets. In this case, it is not hard to show the conditional expectation of $E\left[(-1)^{|A\cap B|}(-1)^{|A\cap C|}\right]$ given $B$ and $C$ is zero. Indeed, choosing any element $x$ of the symmetric difference $B\Delta C$, then flipping whether or not $x\in A$ will flip the sign of $E\left[(-1)^{|A\cap B|}(-1)^{|A\cap C|}\right]$.

$B$ and $C$ are the same set, which is only possible when $r=s$. This happens with probability $1/\binom{n}s$, and in these cases, $E\left[(-1)^{|A\cap B|}(-1)^{|A\cap C|}\right]=1$ surely.

The preceding two bullets, combined with $(2)$, imply that $\langle K_s,K_r\rangle$ is zero when $r\neq s$, and $\binom{n}s$ when they are equal.

The same interpretation proves $$\frac{K_s(r)}{\binom{n}{s}}=\frac{K_r(s)}{\binom{n}r}$$

Both sides are equal to $\mathbb E\left[(-1)^{|A\cap B|}\right]$, where $|A|=r$ and $|B|=s$. The only difference is whether $A$ is fixed and $B$ is random, or the other way  around, but this makes no difference to the expectation.

ContentLicense: CC BY-SA 4.0


问:
$\{x \in H^{\infty} : |x_k-y_k| < \epsilon, k = 1,...,N\}$ is open in $H^{\infty}$ for $y\in H^\infty, \epsilon > 0,N\in\mathbb N$
I know this question has been asked before, but unfortunately, I didn't find the previous answers helpful enough and I am sharing my work here to get help in solving the problem.

Let $H^{\infty}$ be the Hilbert cube (the collection of all real sequences $x := (x_n)_{n \geq1}$ with $|x_n| \leq 1$ for $n = 1,2,...$). For $x,y \in H^{\infty}$ let $d(x,y) = \sum_{n=1}^{\infty}{2^{-n}}|x_n-y_n|$ be the metric on $H^\infty$. You are given $y = (y_n)_{n\ge 1} \in H^\infty,N\in\mathbb N, \epsilon > 0$. Prove that $S = \{x = (x_n)_{n \geq1} \in H^{\infty} : |x_k-y_k| < \epsilon, k = 1,...,N\}$ is open in $H^{\infty}$.

My work:

Take arbitrary $x\in S$, and consider $B(x,r) = \{z\in H^\infty: d(x,z)<r\}$. To show that $S$ is open, we need to find suitable $r$ such that $B(x,r)\subset S$.

$$d(x,z) < r \implies \sum_{n=1}^\infty 2^{-n}|x_n-z_n| < r \implies |x_k - z_k| < 2^k r$$

For $z\in B(x,r)$ to get inside $S$, we require $|z_k-y_k| < \epsilon$ for $k = 1,2,\ldots,N$. We know that for $k=1,2,\ldots,N$,
$$|z_k-y_k| \le |z_k - x_k| + |x_k - y_k| < |z_k - x_k| + \epsilon < 2^kr + \epsilon$$

What do I do now? Thanks!

答:
Fix $x \in S$.

Let $\delta = \epsilon - \max\{|x_1 - y_1|, \ldots, |x_N - y_N|\}.$ Note that $\delta > 0$.

Now, choose $r = \delta/2^{N} > 0$.

We show that $B(x, r) \subset S$. Indeed, let $z \in B(x, r)$.

As you noted, $d(x, z) < r$ implies that $|x_k - z_k| < 2^k r$ for all $k \ge 1$.

Thus, we get $$|x_k - z_k| < \delta$$
for $k = 1, \ldots, N$. For these $k$, we have
$$|z_k - y_k| \le |z_k - x_k| + |x_k - y_k| < \delta + |x_k - y_k| \le \epsilon,$$
where the last inequality follows by choice of $\delta$.

Thus, we have $z \in S$, as desired.


答: Hint: If $x\ne0$, then\begin{align}\left|\frac{f(x)-f(0)}x\right|&=\left|\frac{f(x)}x\right|\\&=\begin{cases}|x|&\text{ if }x\in\Bbb Q\\0&\text{ otherwise}\end{cases}\\&\leqslant|x|.\end{align}

Title: Question About Differentiability Of A Piecewise Function

Body:
I came across the following question on math stack exchange: Differentiability at a point a implies continuity in a neighborhood of a , and I'm wondering how to prove the function, defined as $x^2$  when x is rational, and 0 when x is irrational, is differentiable at 0. I have had some exposure to epsilon delta proofs, but I have only used them on functions such as polynomial, exponential and trigonometric functions.  I'm not show how to use one to show the function defined above, is differentiable at 0.

Thanks in advance.


问:
If $A$ is a matrix with rank($A$)=1,then the minimal polynomial of $A$ is of degree $2$.

If  $n \ge 2$ and if  $A$ is a matrix with rank($A$)=1,then the minimal polynomial of $A$ is of degree $2$.

So, we see that $rank (A) =1$.So, nullity of $A \ge 1$.Now we try to find the range space of $A$.If $y \in R(A)$ then $A(x)=y$.

We see that null space  of $A$ is being spanned by $n-1$ linearly independent vectors. We can extend this to a basis of $\mathbb{R}^n$.Let $x_1,...,x_n$ be the basis where $x_1,..,x_{n-1} \in ker(A)$.We see that $x_n$ must not be in ker($A$) otherwise $A$ is the zero matrix so that $A=0$.

Let $x_n$ be the vector which spans the range space Of $A$ such that $A(\sum(c_ix_i))=x_n$ where $1 \le i \le n$.So from here we can conclude that $0,(c_n)^{-1}$ are the only eigenvalues values. Hence the minimal polynomial becomes $x(x-(c_n)^{-1})$ .

答:
I think it can be easily shown(without assuming the concept of diagonalizability) by using the result that Algebraic multiplicity of an eigen value is greater than or equal to its geometric multiplicity and by case wise determining the Ch eqn and min polynomial.

Proof: Suppose, $T:V\to V$ corresponding to the $n\times n$ matrix $A$, where $V$ is an 'n' dimensional vector space. $\beta$ be the standard basis i.e,  $[T]_{\beta}=A$. Now from rank-nullity theorem,

$dim(Ker T)=(n-1)$(say), Assumed $\beta$ be the extended basis of this.  [since, $dim(V)=n$ and $dim(Im T)=rank(A)=1$(given)]

Now, $A$ is an $n\times n$ matrix with rank $1$ where $n\geq 2$, so, $det(A)=0$. Since, $A$ is matrix of order $n$ , the characteristic eqn, $det(xI-A)=0$ is of degree $n$. Here, $det(A)=0$ so "$0$" is an eigen value because $x=0$ satisfies the ch. eqn. Now trying to find the geometric multiplicity of the eigen value $0$, i.e the dimension of the solution space of $(A-0.I)x=0$ i.e, $\to Ax=0$, which is same with $dim(Ker T)=(n-1)$. So, the eigen value $0$ has geometric multiplicity $(n-1)$. So, the algebraic multiplicity of $0$ is $\geq (n-1)$. Since Characteristic eqn of degree $n$ of the matrix $A$ will look like

i) $x^n=0$  or,

ii) $[x^{(n-1)}\times (x-c)]=0$  , where, $c\neq 0$, the other eigen value.

Now $x^n=0$ cannot be the Ch eqn for all such matrix $A$ , here is the contradictory example ,

take $A=$ $\begin{bmatrix} 1 & 2\\ 2 & 4 \end{bmatrix}$, a $2\times 2$ matrix of rank $1$ , here $x^2=0$ is not the Ch eqn. Check the Ch eqn is $x(x-5)=0$, which is of second type. As we have only two possibilities of Ch eqn then we can conclude that,            The Ch eqn of such matrix $A$ (as given) is $[x^{(n-1)}\times (x-c)]=0$  , $c\neq 0$.

Now $(x-c)=0$ is not the minimal polynomial, because for all such $A$, $(A-cI)\neq 0$, we can assume that same contradictory example again to get contradiction s.t for $A=$ $\begin{bmatrix} 1 & 2\\ 2 & 4 \end{bmatrix}$ (rank=$1$ & order=$2$), $(A-5I)\neq 0$.And also since $rank(A)=1$ $\to$ $A\neq 0$ so $x=0$ is not the minimal polynomial.

Now we will check if the matrix $A$ can satisfy $x(x-c)=0$.  Now $c$ is an non-zero eigen value having algebraic and geometric multiplicity $=1$. So,  the dimension of the solution space of $(A-c.I)x=0$ is $1$, $x\epsilon V$.So, $dim(Ker(A-c.I))=1$ $\to$ $dim(Im(A-c.I))=(n-1)$

[Since, $(A-c.I)$ is $n\times n$ matrix $\simeq$ $T':V\to V]$. So, $rank(A-c.I)=(n-1)$ and $rank(A)=1$ , so $rank(A(A-c.I))=1$ or, $0$ [as, $r(AB)\leq min (r(A),r(B))$].

If $rank(A(A-c.I))=0$ then $A(A-c.I)=0$.

If $rank(A(A-c.I))=1$ we have $(A(A-c.I))$ is $n\times n$ matrix $\simeq$ $T'':V\to V$, so $dim(Ker(T''))=(n-1)$, we can easily check $x\epsilon Ker(T)$ $\to$ $x\epsilon Ker(T'')$ and dimension same so, $Ker(T)=Ker(T'')$. Now, there exist non zero $x$ $\epsilon$ $Ker(T')$ [As $dim(Ker(T'))=dim(Ker(A-c.I))=1$], then $(A-c.I)x=0$ $\to$ $Ax=cx\neq 0$ so, $x$ does not belong to $Ker(T)$, but $A(A-c.I)x=A^2x-cAx=A(cx)-c^2x=c^2x-c^2x=0$, so $x$ $\epsilon$ $Ker(T'')$ , So, $Ker(T)\neq Ker(T'')$ , hence a contradiction. So, $A(A-c.I)=0$ and $x(x-c)=0$ be the minimal polynomial of degree $2$ for all such $A$ (as given). [Hence proved]

I have tried to do in a new way with the basic concepts only.


问:
Analyse the asymptotic behavior of $\sum_{p\leq x} \frac{\log^2(p)}{p}$
This question showed up in a number theory test from previous semesters so I might not know all the material needed to solve this question but here is my try:

The function $f(x)=\frac{\log^2(x)}{x}$ is bounded in $(1,\infty)$ so for some upped bound $C$ on $f(x)$ we have $$\sum_{p\leq x, p\, \text{prime}} \frac{\log^2(p)}{p}\leq C\cdot \pi(x)$$
on the other hand for large enough $x$ we have $\min\left\{\frac{\log^2(p)}{p}|p\leq x, p\, \text{prime}\right\}\geq \frac{\log^2(x)}{x}$ and so:
$$\sum_{p\leq x, p\, \text{prime}} \frac{\log^2(p)}{p}\geq \frac{\log^2(x)}{x} \pi(x)\stackrel{PNT}{=}O(\log(x))$$
Both bounds dont give a tight bound and here I am pretty much stuck.

答:
This isn’t number theory. It really is only analysis.

Note that if $(p_n)_n$ is the increasing sequence of the prime numbers, it’s technically easier to find an equivalent for $S_n=\sum_{k=1}^n{\frac{\ln^2{p_k}}{p_k}}$.

That’s a series, and the general term can be easily (using the PNT) proved equivalent to $\frac{\ln^2{k}}{k\ln{k}}=\frac{\ln{k}}{k}$. In particular, the series is divergent, and thus $S_n \sim \sum_{k=4}^n{\frac{\ln{k}}{k}}$.

Using a series-integral comparison, the sum above can be shown equivalent to $\frac{\ln^2{n}}{2}$.

So the original sum is equivalent to $\frac{\ln^2{\pi(x)}}{2} \sim \frac{\ln^2{x}}{2}$.


问:
Intuition of Addition Formula for Sine and Cosine
The proof of two angles for sine function is derived using $$\sin(A+B)=\sin A\cos B+\sin B\cos A$$ and $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ for cosine function. I know how to derive both of the proofs using acute angles which can be seen here http://en.wikibooks.org/wiki/Trigonometry/Addition_Formula_for_Cosines but pretty sure those who have taken trig know what I'm talking about. So I know how to derive and prove both of the two-angle functions using the acute angles, but what I am completely confused about is where those triangles came from. So for proving the two-angle cosine function, we look at two acute angles, $A$ and $B$, where $A+B<90$ and keep on expanding. So my question is, where did those two triangles come from and what is the intuition behind having two acute triangles on top of each other? </p>

答:
This is a great question, because I'm convinced that all the cool trig identities have simple explanatory diagrams.

In a "unit-hypotenuse" right triangle $\triangle ABC$ (with right angle at $C$), the leg opposite $A$ has length $\sin A$ and the side adjacent to $A$ has length $\cos A$. By proportionality, if the hypotenuse has length $c$, then the leg opposite $A$ has length $c\;\sin A$ and the leg adjacent to $A$ has length $c\;\cos A$. But you know this.

So, for diagramming trig identities,

Wherever you see anything like "this $\cdot \sin(\text{that})$", you look to represent that term by the leg of a right triangle whose hypotenuse is "this" and whose angle opposite the leg is "that". Likewise, represent "this $\cdot \cos(\text{that})$" with a segment adjacent to angle "that" in a triangle with hypotenuse "this".

... and exploit theorems about parallel lines and congruent angles wherever possible.

I'll talk through an example that's slightly simpler than (but strongly related to) the one in your question.

Consider the identity

$$p \sin\theta + q \cos\theta = r \sin\left(\theta+\phi\right)$$
where $p^2+q^2=r^2$ and $\tan\phi = q/p$.

For the left-hand side, we'll need a right triangle with hypotenuse $p$ and one with hypotenuse $q$, each with an acute angle $\theta$. We want to arrange those triangles so that the leg opposite $\theta$ in the $p$-triangle and the leg adjacent to $\theta$ in the $q$-triangle make a straight segment that represents the sum of their lengths. Like so (with $\theta$ represented by a black dot, since I'm recycling an old image):

Looking to the right hand side's $r$, the Pythagorean relation $p^2+q^2=r^2$ suggests we need to show $r$ as the hypotenuse of a right triangle with legs $p$ and $q$. By interesting coincidence, the angle between the hypotenuses of our $p$-triangle and $q$-triangle is itself a right angle! (Why?) So, joining the other ends of these segments gives us $r$.

In fact, we now also have $\phi$ (the white dot), since $q/p$ is clearly the "opposite-over-adjacent" ratio for the marked angle.

To complete the diagram, we need a right triangle with hypotenuse $r$ and acute angle $\theta+\phi$ (strategically placed so that the left-hand and right-hand sides of the identity are clearly equal). Well, we have a right triangle with hypotenuse $r$, but that's not good enough; the angles are wrong. Where can we see $\theta+\phi$ (the sum of the black dot and white dot)? Hmmm ... Look at the top-left corner of the diagram: we have an angle $\theta$ next to and angle $\phi$ ... together they make $\theta+\phi$; and, hey! They're right next to that segment $r$! If only we could make $r$ the hypotenuse of a right triangle with angle $\theta+\phi$.

That's easy: draw a perpendicular!

Then, then new (purple) segment opposite $\theta+\phi$ in a triangle with hypotenuse $r$; it must have length $r\;\sin(\theta+\phi)$.

Since the new segment is clearly congruent to the bottom segment (they're opposite sides of a rectangle), we've demonstrated that the parts of the trig identity are equal. Mission accomplished!

In point of fact, we can do a little bit better. It's not cool that the purple segment is obscuring other parts of the diagram. But how can we move it? We drew it because we needed a triangle with a $\theta+\phi$ in it; to move it, we'd need another angle like that ... oh, wait a minute ... Notice how the sides of the diagram are parallel (being perpendicular to a common segment)? and how that $r$ segment is their transversal? Hmmmm ...

Now we can draw another convenient ---and non-obscuring--- perpendicular to represent $r\;\sin(\theta+\phi)$.

And there you have a diagram of the identity $p\;\sin\theta + q\;\cos\theta = r\;\sin(\theta+\phi)$.

For the angle-sum formulas,

$$\begin{align}
\sin(\alpha + \beta) &= \sin \alpha\;\cos\beta + \cos\alpha\;\sin\beta \\
\cos(\alpha + \beta) &= \cos \alpha\;\cos\beta - \sin\alpha\;\sin\beta
\end{align}$$

it takes a slight leap to know how to represent the component products. Take for instance, $\sin\alpha\;\cos\beta$. Is it a "$\text{this}\cdot\sin(\text{that})$" situation (with "this"$=\cos\beta$ and "that"$=\alpha$), or a "$\text{this}\cdot\cos(\text{that})$" situation? It could be either, but let's assume the former; that means we'd want our diagram to feature a right triangle with hypotenuse $\cos\beta$ and acute angle $\alpha$. But, to create our hypotenuse of length $\cos\beta$, we'll need it to be a leg adjacent to $\beta$ in a triangle with hypotenuse $1$: we must stack one triangle on top of another!

The good news is that the "other" leg in the (blue) triangle with hypotenuse $\cos\beta$ will have length $\cos\alpha\;\cos\beta$, which is convenient, because we'll need one of those. Oh, and hey ... The other leg in the (pink) triangle of hypotenuse $1$ has length $\sin\beta$, which could be the hypotenuse of a convenient new triangle with legs $\cos\alpha\;\sin\beta$ and $\sin\alpha\;\sin\beta$; we need those, too!

Conveniently, we find that we've created (at the bottom-left) the angle $\alpha+\beta$. As before, we can drop a simple perpendicular to bound that angle inside a triangle with hypotenuse $1$; but, as before, we can do a little better:

This gives us a (white) triangle with hypotenuse $1$ and legs $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$, positioned in such a way that the sine segment is clearly the sum of the two vertical segments we constructed, and that the cosine segment is clearly difference of the horizontal segments.

With a very slight adjustment, we turn the diagram into an illustration of the angle-difference identities

$$\begin{align}
\sin(\alpha-\beta) &= \sin\alpha\;\cos\beta - \cos\alpha\;\sin\beta \\
\cos(\alpha-\beta) &= \cos\alpha\;\cos\beta + \sin\alpha\;\sin\beta
\end{align}$$

I encourage you to seek-out more diagrams of identities, both as a simple exercise in understanding your trig, but also as a quest to determine what the identities are really trying to tell you. (Search the web for "trig proof without words", and you'll see how helpful these things can be.)

For instance, here's a simple picture-proof of the Law of Cosines, created using the diagramming strategies described here:

Once you start thinking about Calculus, you might ponder diagrams for the "power series" of sine and cosine and secant and tangent. (If you come up with the diagrams for cosecant and cotangent, let me know!)

ContentLicense: CC BY-SA 3.0


问:
Determine the least number of comparisons, or best-case performance of the following algorithms
Determine the least number of comparisons, or best-case performance,

required to find the maximum of a sequence of n integers.

used to locate an element in a list of n terms with a linear
search.

used to locate an element in a list of n terms using a binary search.

The answer I found says that:

$n-1$ comparisons.

$1$ comparison

$\left\lceil \log(n) \right\rceil$ comparisons.

Question: But in best case, I guess all of them take $\mathcal{O}(1)$. I guess either the question is wrong or my interpretation is incorrect?

答:
For the first problem, suppose the first element is in fact the maximum. Sure, you see it, but how can you be sure it is the maximum? You need to compare this element with the other $n-1$ elements to prove it is the maximum.

However, the best case for the third problem really should be one comparison under the implied assumption that the list is sorted; you determine a "middle" element and it happens to be equal to the search key, you're done.


问:
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

10/inf < 1 ----> lim an=0

答:
Using sandwich theorem is enough for this problem:
\begin{align}
0\le \lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}&\le\lim_{n\to\infty}\dfrac{10^n}{\sqrt{n!}+\sqrt{n!}}\\
&=\dfrac12\times\lim_{n\to\infty}\sqrt{\dfrac{100^n}{n!}}\\
&=\dfrac12\times\sqrt{\lim_{n\to\infty}\left(\dfrac{100^n}{n!}\right)}\\
&=0
\end{align}


问:
If $e^{(k)} = (0,0,\ldots,1,0,0,\ldots)$ then show that $S = \{e^{(k)}: k\ge 1\}$ is a closed subset of $\ell^1$
I saw this problem on MathSE before, and while the proofs given as answers look good - I'm unable to carry forward my own attempts.

If $e^{(k)} = (0,0,\ldots,1,0,0,\ldots)$ with $1$ in the $k$th position, then show that $S = \{e^{(k)}: k\ge 1\}$ is a closed subset of $\ell^1$.

My work:

To show that $S\subset \ell^1$ is closed, we can show that $\ell^1\setminus S$ is open. In other words, for an arbitrary sequence $x\in\ell^1$ we must find $r$ so that $B(x,r)\subset \ell^1\setminus S$. Here, $B(x,r) = \{y\in\ell^1: \|x-y\|_1 = \sum_{i\ge 1}|x_i-y_i| < r\}$. What do I do next?

Another way I thought was to use the following proposition:
$F$ is closed iff if $B(x,\epsilon)\cap F\neq\varnothing$ for every $\epsilon>0$, then $x\in F$. Suppose now for some $x\in\ell^1$, $\forall\epsilon > 0\ (B(x,\epsilon)\cap S\neq\varnothing)$. In other words, $\exists e^{(k)}\in S$ s.t. $$\sum_{i\ge 1} |x_i-e^{(k)}_i| = |x_k-1| +\sum_{i\ge 1,i\ne k}|x_i| < \epsilon, \text{ for all }\epsilon > 0$$ Now how does one conclude that $x\in S$?

Thanks for the help!

答:
Note that

$$\|e^{(n)} - e^{(m)}\|_1 = 2 \text{ whenever } n \neq m$$

This implies that there is no convergent sequence from $S$ converging to some $x \in \ell^1 \setminus S$ because then we would have points from $S$ getting arbitarily close to each other (triangle inequality via limit...) and we just saw that this does not happen. So $S$ is closed and discrete.


问:
Lipschitz constant for $F(t,y,z)=(z,f(y,z)\sin t)$

Let $f\in C^1(\mathbb R^2,\mathbb R)$.
  Prove that all solutions for $x''=f(x,x')\sin t$ such that $x(0)=x(2\pi)$ and $x'(0)=x'(2\pi)$ have period $2\pi$.

I'm in the process of solving the above differential equation.

To do this I define $F\colon \mathbb R\times \mathbb R^2\to \mathbb R^2,F(t,y,z)=(z,f(y,z)\sin t)$.

Then the system is equivalent to $$F(x'(t), f(x(t),x'(t)))=(x'(t),x''(t))$$
Now in order to later be sure there is only one solution, I need to prove that
$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq K\Vert(y_1,z_1)-(y_2,z_2)\Vert$ for some $K>0$.

$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert=\Vert(z_1,f(y_1,z_1)\sin t)-(z_2,f(y_2,z_2)\sin t)\Vert=\Vert (z_1-z_2,(f(y_1,z_1)-f(y_2,z_2))\sin t)\Vert$$ and now I don't know what to do. I don't see how could I possibly related the last expression with $\Vert (y_1,z_1)-(y_2,z_2)\Vert$.

If you're feeling particularly charityful I wouldn't mind having some hints on the rest of the problem.

答:
To obtain uniqueness, it suffices to check that the Lipschitz condition
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq K\Vert (y_1,z_1)-(y_2,z_2)\Vert \tag1 $$
hold locally: that is, every point $(t,y_0,z_0)$ has a neighborhood in which (1) holds with some $K$ (which can depend on the point $(t,y_0,z_0)$ and on the size of neighborhood).

Every continuously diffentiable function is locally Lipschitz. Indeed, because the derivatives in $y$ and $z$ are continuous, they are bounded in some rectangular neighborhood of $(t,y_0,z_0)$. Let $M$ be an upper bound for $|F_y|$ and $|F_z|$. The mean value theorem yields
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_1)\Vert\leq M |y_1-y_2| \tag2 $$
$$\Vert F(t,y_2,z_1)-F(t,y_2,z_2)\Vert\leq M |z_1-z_2| \tag3 $$
From (2) and (3),
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq M (|y_1-y_2|+|z_1-z_2|) \le 2M\|(y_1,z_1)-(y_2,z_2)\| \tag4 $$
which is the desired Lipschitz estimate.

The particular form of $F$ does not matter here; all you have to check is that it has continuous  partial derivatives   in $y$ and in $z$.


问:
What interpretations of matrices are there and how do they connect?
I have been studying linear algebra for quite a while now and I still don't really get what a matrix really is. I have found that there are multiple interpretations for matrices - some of them listed below. What other interpretations are there and how do all these interpretations connect? why do matrices coincidentally (or probably for a reason) represent all of these things?

The way I am thinking of matrices at the moment is that matrices are like a table, much like a spreadsheet in Excel and for some weird reason they also happen to represent a system of linear equations, which comes in handy to solve such systems.

Here are the interpretations I know of:

Geometrically: Matrices transform space

Algebraically: Matrices represent systems of linear equations

As a table: Matrices can store information, for example the adjacency matrix for a graph

答:
I can really recommend the videos by 3Blue1Brown on youtube. I think his playlist "the essence of linear algebra" sums up everything you would like to know:

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

(Simply watch the 3rd video in the series, if you only care about the matrices)


答:
Let us try to complete your attempt 1:

Idea: We would want to pick something like $r = \displaystyle\min_{k \in \Bbb N}d(x^{(k)}, x)$ but the problem is that $\min$ does not make sense and $\inf$ might turn out to be $0$. Thus, we will first show that after a stage, $d(x^{(k)}, x)$ can be explicitly bounded below, by say, $1/2$. In that case, we only have to take $\min$ over a finite range. (This is similar to how we show that a convergent sequence is bounded.)

Let us pick $x \in \ell^1 \setminus S.$ Note that $x \in \ell^1$ implies that $|x_k|$ is eventually strictly less than $1/2$. (Since $x_n \to 0.$)

Suppose that this happens after $k = N$. This means that $$d(e^{(m)}, x) \ge |1 - x_m| \ge |1-|x_m|| > \frac{1}{2}$$
for $m > N$.

Intuitively, this means that $e^{(k)}$ after $k = N$ are at least distance $1/2$ away. Thus, choosing $$r = \min\left\{d(x, e^{(1)}), \ldots, d(x, e^{(N)}), \frac12\right\} > 0$$
does the job.

Title: If $e^{(k)} = (0,0,\ldots,1,0,0,\ldots)$ then show that $S = \{e^{(k)}: k\ge 1\}$ is a closed subset of $\ell^1$

Body:
I saw this problem on MathSE before, and while the proofs given as answers look good - I'm unable to carry forward my own attempts.

If $e^{(k)} = (0,0,\ldots,1,0,0,\ldots)$ with $1$ in the $k$th position, then show that $S = \{e^{(k)}: k\ge 1\}$ is a closed subset of $\ell^1$.

My work:

To show that $S\subset \ell^1$ is closed, we can show that $\ell^1\setminus S$ is open. In other words, for an arbitrary sequence $x\in\ell^1$ we must find $r$ so that $B(x,r)\subset \ell^1\setminus S$. Here, $B(x,r) = \{y\in\ell^1: \|x-y\|_1 = \sum_{i\ge 1}|x_i-y_i| < r\}$. What do I do next?

Another way I thought was to use the following proposition:
$F$ is closed iff if $B(x,\epsilon)\cap F\neq\varnothing$ for every $\epsilon>0$, then $x\in F$. Suppose now for some $x\in\ell^1$, $\forall\epsilon > 0\ (B(x,\epsilon)\cap S\neq\varnothing)$. In other words, $\exists e^{(k)}\in S$ s.t. $$\sum_{i\ge 1} |x_i-e^{(k)}_i| = |x_k-1| +\sum_{i\ge 1,i\ne k}|x_i| < \epsilon, \text{ for all }\epsilon > 0$$ Now how does one conclude that $x\in S$?

Thanks for the help!


答:
Using
$$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$$
we come to limit $\frac{a^n}{n!}$ for $a \gt 1$. Of course exists natural $k=\lfloor a\rfloor \leqslant a \leqslant \lfloor a\rfloor +1$. So we have
$$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \frac{a}{k+1}\cdots \frac{a}{n} \leqslant \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \left(\frac{a}{k+1}\right)^{n-k} \to 0$$

Body:
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

10/inf < 1 ----> lim an=0


问:
If $A$ is open or closed in $(M,d)$, then $(\partial A)^\circ = \varnothing$

$(M,d)$ is a metric space. If $A\subseteq M$ is open or closed in $(M,d)$, then prove that $(\partial A)^\circ = \varnothing$.

We must take two cases: (i) $A$ is open (ii) $A$ is closed.
In either case, we must show that the interior of the boundary of $A$ is empty.

My work:

Let $A$ be open. We want to argue that $(\partial A)^\circ = \emptyset$. Suppose there exists $y\in (\partial A)^\circ$. Then $y\in \partial A$, and there exists some $\epsilon_y > 0$ s.t. $B(y,\epsilon_y)\subset \partial A$. Since $y\in\partial A$, we have $B(y,\epsilon_y)\cap A\neq \emptyset$ and $B(y,\epsilon_y)\cap A^c\neq\emptyset$. I don't know if this helps but we can also say that $\partial A\cap A\neq \emptyset$.

Let $A$ be closed. Then $A^c$ is open. Hoping that the above part would be complete, we get $(\partial A^c)^\circ = \emptyset$. Does this help?

How do I complete my proof attempts - and are there other possibly nicer ways of approaching this? Thanks a lot!

P.S. I came across this related post but I haven't seen many of the results stated by the OP there.

答:
On both cases you state that there exists some $\epsilon_y\in\mathbb{R}^+$ such that $B(y,\epsilon_y)\subset A$. This is never true, as long as $y\in\partial A\implies y\not\in Aº$. So you are having the same proof in both cases because you introduce that fact that is a contradiction always.

You can argue it using that $B(y,\epsilon_y)\subset\partial A$.

If $A$ is closed, then $\partial A\subset A$. This means that $y\in Aº$, because $B(y,\epsilon_y)\subset \partial A\subset A$. This is a contradiction because $B(y,\epsilon_y)\cap (M\setminus A)\neq\emptyset$
.

If $A$ is open, then $B(y,\epsilon_y)\cap A\neq\emptyset$. Using that $B(y,\epsilon_y)\subset\partial A$ and $A=Aº$, we get $\partial A\cap Aº\neq\emptyset$, which is also a contradiction.


问:
Related rates, calculus
Suppose that $k^{2} + h^{3} = 9$. Find $\frac{dh}{dt}$ when $k=1$ and $\frac{dk}{dt} = 3$ ans $= \frac{1}{2}$. I'm differentiating with respect to $t$ but I cannot get the answer if you could show me the steps, or how to approach this question, would be very helpful :)

答:
Treating $k$ and $h$ as functions of $t$, we take the derivative of each side of the given equation with respect to $t$ by using implicit differentiation and chain rule to obtain:
$$\begin{align*}
2k\frac{dk}{dt}+3h^2\frac{dh}{dt} &= 0 \\
3h^2\frac{dh}{dt} &= -2k\frac{dk}{dt} \\
\frac{dh}{dt} &= \frac{-2k\frac{dk}{dt}}{3h^2} \\
\end{align*}$$
Note that when $k=1$, we have $1^2+h^3=9 \iff h^3=8 \iff h=2$. Thus, substitution yields:
$$\frac{dh}{dt} = \frac{-2(1)(3)}{3(2)^2} = \frac{-2}{2^2} = \boxed{\dfrac{-1}{2}}$$</p>

Title: Related rates, calculus


问:
Ratio Problems using equations
Our soccer team won $\frac{2}{9}$ of its game this season. If we lost $15$ more games than we won, how many games did we play this season?  Please give me hints DO NOT GIVE ME THE ANSWERS  I tried to come up with an equation like $\frac{15+l+w}{w+l}=\frac{7}{9}$ This did not work out. Please give me tiny hints.

答:
Hint

Total matches played $=9x$ (let)

Thus, matches won $=2x$ and matches lost $=7x$.

Title: Ratio Problems using equations


问:
Joint filtration of two right-continuous filtrations is right-continuous?
Assume we are given two right-continuous filtrations $\mathbb{F}^X=(\mathcal{F}_t^X)$ and $\mathbb{F}^Y=(\mathcal{F}_t^Y)$ both generated by some stochastic processes $X, Y$. Do we have that the joint filtration $\mathbb{F}^{X,Y}$ is right-continuous?

答:
Added detail: If $G$ is bounded and $\mathcal F^X_\infty\vee\mathcal F^Y_\infty$ measurable, then $E[G\mid\mathcal F^{X,Y}_{t+}]=\lim_nE[G\mid\mathcal F^{X,Y}_{t+1/n}]$, almost surely,  by martingale convergence. If also $G=G_1(X)G_2(Y)$ then by independence this limit is a.s. equal to $\lim_nE[G_1(X)\mid\mathcal F^X_{t+1/n}]\cdot E[G_2(Y)\mid\mathcal F^Y_{t+1/n}]
=E[G_1(X)\mid\mathcal F^X_{t}]\cdot E[G_2(Y)\mid\mathcal F^Y_{t}]$ by the rt-continuity of $\Bbb F^X$ and $\Bbb F^Y$. That is, $E[G\mid \mathcal F^{X,Y}_{t+}]=E[G\mid \mathcal F^{X,Y}_{t}]$, almost surely,  for $G$ of the special form $G_1(X)G_2(Y)$. This equality persists for general $G$ by the monotone class theorem.


问:
Subgroup of $GL(n,\mathbb{C})$ that is not a Matrix Lie group
Given the definition of a Matrix Lie Group as a subgroup of $GL(n,\mathbb{C})$ such that for every sequence $A_n$ its limit $A$ is either in the subgroup or it is not invertible, what are the known examples for which the limit $A$ is invertible but is not contained in the subgroup, hence subgroups of $GL(n,\mathbb{C})$ that are not matrix Lie groups?

答: Well it dawned on me. the group of invertible matrices with rational entries. The limit can have values that are not rational but still be invertible.</p>


问:
What kind of object is the second derivative of a function $f:\mathbb R^n\to\mathbb R^m$?
I wonder what is the meaning of the second derivative or what kind of object it is when we have a function $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$.

The first derivative is the Jacobian matrix, but then, what is the second derivative? How can I treat them when I write $f''$ or $D^2 f$?

Thanks a lot for your help!

答:
The first thing that comes to mind is the Hessian matrix which is the special case for $m=1$:

$$f: \mathbb{R}^n \rightarrow \mathbb{R}$$

The case you asked about occurs in differential geometry, where it is involved in defining the covariant derivative (see this) and related to the connection coefficients.

In a rough sense it is a curvature. Much like the second derivative of a single variabled function tells you how quickly the curve is changing.

As for notation, I usually write this tensor as

$$\left[\frac{\partial^2 f^k(\xi)}{\partial x^i \partial x^j}\right] \;\;\;\ \text{or more simply} \;\;\; \left[\partial_i \partial_j f^k(x)\right]$$

If writing the entries individually, would probably write a Hessian for each $k$.


问:
Counterexample of Borsuk-Ulam theorem (Discontinuous Functions)
I'm looking to build a function $f:S^2 \to \mathbb R^2$ such that $f(x)\neq f(−x)$ for all $x\in S^2$.

By Borsuk-Ulam Theorem, this function must be discontinuous. I was trying to build a not too complicated function, but I always encountered a problem.

I appreciate any help.

答:
Fix two distinct elements $a$ and $b$ in $\Bbb R^2$. Consider an arbitrary $(x, y, z) \in S^2.$

If $x > 0$, map it to $a$ and if $x < 0$, map it to $b$.

If $x = 0$, then do the same as above with $y$ instead of $x$.

If $x = y = 0$, then $z \neq 0$ and thus, we can do the above with $z$ instead of $x$.

Note if $\mathbf x = (x, y, z) \in S^2,$ then both $\bf x$ and $-\bf x$ will follow the rule in the same bullet above and hence, one is mapped to $a$ and the other to $b \neq a$.


答:
Here's an overkill using the Axiom of Choice:

Let $\mathcal A = \{\{\mathbf x, -\mathbf x\} \mid \mathbf x \in S^2\}$ be the collection of all antipodal pairs. From each set in $\mathcal A$, pick exactly one element. Map that to $(1, 0)$ and its antipodal counterpart to $(0, 1)$.

Title: Counterexample of Borsuk-Ulam theorem (Discontinuous Functions)

Body:
I'm looking to build a function $f:S^2 \to \mathbb R^2$ such that $f(x)\neq f(−x)$ for all $x\in S^2$.

By Borsuk-Ulam Theorem, this function must be discontinuous. I was trying to build a not too complicated function, but I always encountered a problem.

I appreciate any help.


答:
As you have pointed out,
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{10+\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}$$ So,
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} (\frac{10}{\sqrt{n+2}+1} + \frac{\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}) = \lim_{n\to\infty} \frac{10}{\sqrt{n+2}+1} + \lim_{n\to\infty} \frac{10}{(\sqrt{n+1})(\sqrt{n+2}+1)} = 0$$ since both limits exist. So, since $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0 < 1$$ you can say that $\lim_{n\to\infty} a_n = 0$.

Title: $\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$


问:
Does this weighted sum of Gaussians converge to $1/x^2$?
By coincidence I discovered that a weighted sum of several Gaussians seems to converge to $1/x^2$. Now I wonder whether that's a well-known property, just a coincidence or if this could be proven.

This is a normal Gaussian with an additional weight factor $1/\sigma$:

$$
f(x, \sigma) = \frac{1}{\sigma} \cdot \frac{1}{\sigma \sqrt{2\pi} } e^{-\frac{1}{2}\left(\frac{x}{\sigma}\right)^2}
$$

If I now accumulate several of these weighted Gaussians with exponentially increasing $\sigma$, I get a curve which closely resembles $1/x^2$. The additional factor $\sqrt{3}$ was determined by trial and error and may not be the correct value.

$$
g(x, n) = \sqrt{3} \cdot \sum_{k=0}^n f\left(x, \frac{2^k}{n}\right)
$$

$$
\lim\limits_{n \to \infty} g(x, n) \stackrel{?}{=} \frac{1}{x^2}
$$

Does anybody have an idea how this could be proven?

答:
Actually, $\lim_{n\to\infty}g(x,n)$ doesn't exist, but $\lim_{n\to\infty}g(x,2^n m)$ exists for any integer $m>0$, and it depends on $m$; however, as $m$ changes, it oscillates around $1/x^2$ with very small amplitude. And, to make its "average value" exactly $1/x^2$, your $\sqrt{3}$ has to be replaced with $\sqrt{2\pi}\log 2$.

For $x\neq 0$ we have $$x^2 g(x,2^n m)=\sqrt\frac{3}{2\pi}\sum_{k=0}^{2^n m}4^{n-k}(mx)^2 e^{-4^{n-k}(mx)^2/2}\underset{n\to\infty}{\longrightarrow}\sqrt\frac6\pi G\left(4,\frac{(mx)^2}{2}\right),$$ where $G(a,b)=\sum\limits_{k=-\infty}^\infty(a^k b)e^{-a^k b}$ for $a>1$ and $b>0$.

Following this answer of mine literally, we obtain (here, $i$ is the imaginary unit) $$G(a,b)=\frac{1}{\log a}\sum_{n=-\infty}^{\infty}\Gamma(1+s_n)b^{-s_n},\qquad s_n=\frac{2n\pi i}{\log a}.\tag{*}\label{gammaseries}$$

As $b$ (i.e. $m$) changes, this oscillates around $1/\log a$, and the amplitudes for our $a=4$ are $$|\Gamma(1+s_{\pm 1})|\approx4.3\cdot10^{-3},\quad|\Gamma(1+s_{\pm 2})|\approx4.9\cdot10^{-6},\quad|\Gamma(1+s_{\pm 3})|\approx4.9\cdot10^{-9},\quad\ldots$$

Another way to get the formula \eqref{gammaseries}. Consider $G(a,a^x)$ as a function of $x\in\mathbb{R}$. It is smooth and has a period $1$, hence a convergent Fourier expansion $G(a,a^x)=\sum_{n\in\mathbb{Z}}c_n e^{-2n\pi ix}$, where
\begin{align*}
c_n&=\int_0^1 G(a,a^x)e^{2n\pi ix}\,dx\\
&=\sum_{k\in\mathbb{Z}}\int_0^1 a^{k+x}e^{-a^{k+x}+2n\pi ix}\,dx\\
&=\sum_{k\in\mathbb{Z}}\int_k^{k+1}a^t e^{-a^t+2n\pi it}\,dt\\
&=\int_{-\infty}^\infty a^t e^{-a^t+2n\pi it}\,dt\qquad\color{gray}{[a^t=z]}\\
&=\frac{1}{\log a}\int_0^\infty z^{s_n}e^{-z}\,dz=\frac{\Gamma(1+s_n)}{\log a}.
\end{align*}


问:
Why is the set $S = \{ (x,y,z) \in \mathbb{N}^3 : x^2 + 4yz = p, p \text{ prime} \}$, finite
I am looking at this proof: http://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2323918/fulltext.pdf

We are given that this set is finite. But it is not immediately obvious to me why.  The rest of the proof makes sense, so I want to understand this part.  Can someone help?

答:
Note that $x \in \{0,1,2,\ldots,\sqrt{p}\}$, $y \in \left\{1,2,\ldots,\dfrac{p-x^2}4 \right\}$ and $z \in \left\{1,2,\ldots,\dfrac{p-x^2}{4y}\right\}$. Hence, clearly we have
$$\# S \leq \sqrt{p} \times \left(\dfrac{p}4 \right)^2 = \dfrac{p^{5/2}}{16}$$</p>


问:
Pointwise + Uniform Convergence:
Let $(f_n)^\infty_{n=1}$ be the seq. of functions $f_{n} : \mathbb{R} \rightarrow \mathbb{R}$ defined by $$f_n(x) = \left(\sin \left(\frac{x}{\pi}\right)\right)^{2n},~~~~\forall x \in \mathbb{R}$$

Find the pointwise limit of the sequence $(f_n)^\infty_{n=1}$ and does it converge uniformly.

So my thoughts were to do as how I would normally would do so, but with a certain limit of it all i.e sub in $x=0$ etc.

We have no limit here so I go to the second method -
If i take the absolute value of $|f_n|$ and realise that $|\sin(x)| \leq 1 $ and then use it I just get

$$-1 \leq f_n(x) \leq 1$$

Which doesn't get me anywhere so I was hoping someone could help with this question thank you

答:
It holds $\sin(\frac x \pi) = \pm 1$ if and only if $x= \pi(\frac \pi 2+ k\pi)$, for $k\in \mathbb Z$.
So for $x_0= \pi(\frac \pi 2+ k\pi)$ you got
$$\lim_{n\to \infty} \left(\sin \left(\dfrac {x_0} \pi \right)\right)^{2n}=
\lim_{n\to \infty} (\pm 1)^{2n}=1.$$
For the other values of $x$ it holds $-1<\sin \left(\dfrac {x} \pi \right)<1$, hence $u:=\sin \left(\dfrac {x} \pi \right)^2\in[0,1)$. Therefore
$$\lim_{n\to \infty} \left(\sin \left(\dfrac {x} \pi \right)\right)^{2n}=
\lim_{n\to \infty} (u)^{n}=0.$$

The sequence converges pointwise to the discontinuous function
$$f:\mathbb R \to \mathbb R\,,
f(x)= \left\{
\begin{array}{ll}
1& \text{ if } x\in \frac{\pi^2} 2+ \pi^2\mathbb Z\\
0& \text{ else.}
\end{array}\right.$$
Since all the function $f_n$'s are continuous, but the limit $f$ is not, the convergence is not uniform.

Title: Pointwise + Uniform Convergence:


问:
The equation of a sphere tangent to a plane at a point and also tangent to another plane

Determine the equation of the sphere that is tangent to the plane
$$(P):6x-3y-2z-35=0$$ at $M(5,-1,-1)$ and also tangent to the plane
$$(Q):6x-3y-2z+63=0$$

My solution:
Let $d$ be the line passing through $M$, $d$ orthogonal to $(P)$. The parametric equations of the line $d$ are:
$$x = 5+6t$$
$$y=-1-3t$$
$$z=-1-2t$$
So the center of the sphere is of the form $C(5+6t, -1-3t,-1-2t)$. The center of the sphere must belong to the line $d$ and the plane $(Q)$, so it is the intersection point between the line and the plane.
$$6(5+6t)+3(1+3t)+2(1+2t)+63=0$$
From this relation we get $t=2$, so the center has the coordinates $C(17,-7,-5)$. Now since it's stated that the plane $(Q)$ is tangent to the sphere, the distance from the center to the plane is equal with the radius. The radius $r=14$, so now we have the equation of the sphere:
$$(x-17)^2+(y+7)^2+(z+5)^2=14^2$$
Is this correct? I'm not sure if my initial condition for the center to belong to the line $d$ and the plane $(Q)$ is right, but I based my assumption off of the fact that they are both tangent to the sphere and so the distance from the center to each plane is equal with the radius.
EDIT: Okay, so my assumption was definitely wrong, I somehow didn't notice the normal vectors were identical

答:
Hint

Your sentence

The center of the sphere must belong to the line $d$ and the plane $(Q)$...

is not correct. As the sphere is supposed to be tangent to the plane $(Q)$ its center does not lie in $(Q)$.

However, you can use the fact that the center of the sphere lies on the line $d$ and that its distance to the plane $(P)$ should be equal to its distance to the plane $(Q)$.


答:
(New user here, so I don't have enough reputation to comment)

Why should $Q$ contain the center of the sphere? In fact, if it does, is $Q$ still tangential?

Here's some motivation for a possible solution: Start with the line $d$. Since the two given planes are parallel, can you see why $d$ must contain the center of the sphere? Now try to use this fact, along with the distance between the planes (which is standard computation) to figure out the center/radius of the sphere.

Title: The equation of a sphere tangent to a plane at a point and also tangent to another plane

Body:
Determine the equation of the sphere that is tangent to the plane
$$(P):6x-3y-2z-35=0$$ at $M(5,-1,-1)$ and also tangent to the plane
$$(Q):6x-3y-2z+63=0$$

My solution:
Let $d$ be the line passing through $M$, $d$ orthogonal to $(P)$. The parametric equations of the line $d$ are:
$$x = 5+6t$$
$$y=-1-3t$$
$$z=-1-2t$$
So the center of the sphere is of the form $C(5+6t, -1-3t,-1-2t)$. The center of the sphere must belong to the line $d$ and the plane $(Q)$, so it is the intersection point between the line and the plane.
$$6(5+6t)+3(1+3t)+2(1+2t)+63=0$$
From this relation we get $t=2$, so the center has the coordinates $C(17,-7,-5)$. Now since it's stated that the plane $(Q)$ is tangent to the sphere, the distance from the center to the plane is equal with the radius. The radius $r=14$, so now we have the equation of the sphere:
$$(x-17)^2+(y+7)^2+(z+5)^2=14^2$$
Is this correct? I'm not sure if my initial condition for the center to belong to the line $d$ and the plane $(Q)$ is right, but I based my assumption off of the fact that they are both tangent to the sphere and so the distance from the center to each plane is equal with the radius.
EDIT: Okay, so my assumption was definitely wrong, I somehow didn't notice the normal vectors were identical


问:
Ratio of corresponding sides of similar triangles, given the areas.
The area of two similar triangles are 72 and 162. what is the ratio of their corresponding sides?</p>

答: When linear dimensions a scaled by the factor $\lambda$, area is scaled by the factor $\lambda^2$. Here, we have $\lambda^2=\dfrac{162}{72}=\dfrac{9}{4}$. So $\lambda=\dfrac{3}{2}$. </p>


问:
Prove that $S^3\setminus S^1$ is connected
As the titles says, I have to prove that $S^3\setminus S^1$ is connected.

I'm having a hard time solving this problem. The best idea I've come up with is to show somehow that, if $S^3\setminus S^1$ wasn't connected, then $S^1$ would be a non-trivial clopen in $S^3$, and that's a contradiction; to do this I've tried to use the fact that in these hypotheses $S^3\setminus S^1=A\cup B$ disjoint open sets and $S^3\setminus S^1=C \cup D$ disjoint closed sets, but I haven't been able to conclude anything (and I'm not sure this is the right way to proceed). Maybe I'm missing something trivial.

Any hints?

答:
Let us consider $\Bbb R^2 \hookrightarrow \Bbb R^4$ by $(x, y) \mapsto (x, y, 0, 0)$.

Under the above, we also get $S^1 \hookrightarrow S^3$.

Consider the map $p : \Bbb R^4\setminus\{\mathbf 0\} \to S^3$ given by $x \mapsto \frac{x}{\|x\|}$.

Under this map, we have $p(\Bbb R^4 \setminus \Bbb R^2) = S^3 \setminus S^1$.

However, $\Bbb R^4 \setminus \Bbb R^2$ is path-connected. (Consider the broken line path.)

Thus, $S^3 \setminus S^1$ is also path-connected, being the continuous image of a path-connected space. In particular, it is connected.


问:
How to Find the Stationary Distribution of this Markov Chain

Given the Markov Chain with state probability matrix $$P = \begin{bmatrix} 0.8 & 0.2 \\ 0.3 & 0.7 \end{bmatrix}$$ Prove that this has the stationary distribution $$\pi = [0.6,0.4]$$

When attempting to solve for $\pi P = \pi$ I obtain the following overdetermined system: $$0.8 \pi_1 + 0.3 \pi_2 = \pi_1 \\ \\ \\\ \\ 0.2 \pi_1 + 0.7 \pi_2 = \pi_2$$ and this has infinitely many solutions so I don't know how to proceed.

答: Solving the system, we have $\pi _1=1.5\pi _2$. Since the two probabilities must sum to 1, $$\pi _1+\pi _2=1\longrightarrow 1.5\pi _2+\pi _2=1\longrightarrow \pi _2=0.4,\:\pi _1=0.6$$


答: No, that is not correct. You already know that the center of the sphere is of the form $(5+6t,-1-3t,-1-2t)$, for some real $t$. Since the planes $P$ and $Q$ and parallel, $C$ has to be the midpoint between $M$ and the point $M'$ which belongs to the line $(5+6t,-1-3t,-1-2t)$ and to $Q$. But $M'=(-7,5,3)$, and therefore $C=(-1,2,1)$ (and the radius is $7$).


问:
Does the following sequence defined recursively converge?
I have a sequence of real numbers defined as follows. Let $n > c > 1$ be integers. We define

$$
a_{i+1} = n - \frac{c}{2 a_i - c}
$$
with $a_0 = n$. If the limit does not exist I am interested in finding minimal, or something close, value of $a_i$. Any help is appreciated! Thank you!

答:
The sequence can be taken for example as:

$$a_1 = \frac{n^2 - c}{n-c} > n $$

since:

$$n^2 - c > n(n-c) = n^2 - nc$$

and for $k = a_i$, we can simply have:

$$ a_{i+1} = \left( \frac{(n-c)k-c}{k-c}\right)$$

as $i \to \infty$ .


问:
Determine whether or not the cube root of x is Lipschitz.
Determine whether or not the cube root of x is Lipschitz.   x^(1/3)

I understand that this is NOT Lipschitz. I am having trouble properly proving this. I know for other problems I have shown functions are not uniformly continuous to prove that they are not Lipschitz, but I can't seem to figure out the right way to go about this problem. Reviewing for a test, Thanks!

答:
Too see it without using derivatives, observe that

$$ x^{1/3}-y^{1/3} = (x-y)/(x^{2/3} + (xy)^{1/3} + y^{2/3} ) $$

Now as x or y goes to zero, you can make the denominator as large as you wish (as cube root is a continuous function at zero). Thus, no single L will suffice.


问:
Is an $L^1$ limit of convex functions convex?
Suppose $f\in L_{loc}^1(\mathbb{R}^n)$ is such that $f_n\in L_{loc}^1(\mathbb{R}^n)$ are convex and converge to $f$ in $L_{loc}^1$. Is $f$ necessarily convex?

答:
Yes, it is. Since the functions $f_n$ are convex in $\mathbb{R}^n$, the sequence $(f_n)$ converges uniformly on compact subsets of $\mathbb{R}^n$ (see e.g. Rockafellar, Convex Analysis, Thms. 10.6 and 10.8).

The convexity of $f$ is now an easy consequence of the inequality
$$
f_n(\lambda x + (1-\lambda)y) \leq \lambda f_n(x) + (1-\lambda) f_n(y),
\qquad \forall n\in\mathbb{N},
$$
that holds for every $x,y\in\mathbb{R}^n$, $\lambda\in [0,1]$.


问:
Dense subset of $L^2(\mathbb{R})$
For $n \in \mathbb{N}$ let $f_n(x) := x^n \exp\left(-\frac{x^2}{2}\right)$. I want to prove that $A := \mathrm{span} \lbrace f_n: n \in \mathbb{N} \rbrace$ is dense in $L^2(\mathbb{R}^n)$.

Clearly, it would be enough to approximate charcteristic functions, but I am not quite sure how to go about this. Maybe $\displaystyle \sum_{n = 1}^\infty \frac{(-1)^n(x)^{2n}}{2 ^nn!} \rightarrow \exp\left(-\frac{x^2}{2}\right)$ pointwise helps...

答:
Take $f \in L^2(\mathbb{R})$ such that :
$$\forall n \geq 0, \quad \displaystyle\int_{\mathbb{R}} f(x) x^n e^{-\frac{x^2}2} dx = 0$$

We want to prove that $f = 0$ everywhere. Define :

$$\forall z \in \mathbb{C},\quad  F(z) = \displaystyle\int_{\mathbb{R}} f(x) e^{izx} e^{-\frac{x^2}2} dx $$

We can prove that $F$ is an entire function and that $F^{(n)}(0) = i^n\displaystyle\int_{\mathbb{R}} f(x) x^n e^{-\frac{x^2}2} dx = 0$ for all $n \geq 0$. Thus, $F = 0$ on all $\mathbb{C}$ (by analyticity). But, $F$ restrited to $\mathbb{R}$ is the fourier transform of  $x \mapsto f(x) e^{-x^2/2}$. By injectivity of the Fourier transform, $f(x) e^{-x^2/2} = 0$ for almost all $x \in \mathbb{R}$. Finally $f = 0$, and the subset is dense in $L^2(\mathbb{R})$.

Title: Dense subset of $L^2(\mathbb{R})$


问:
I can't wrap my head around Peano's 5'th axiom
I'm currently reading about Peano's axiom and the construction of the natural numbers. The fifth Axiom of Peano does not make sense to me:

If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ is also in $T$, then $T$ is the set of Natural numbers $\mathbb{N}$.

I know that this axiom exists in order to exclude elements which could have the following property:
$$s(a) = b$$ $$s(b) = a$$

These elements would not violate axioms $1$ to $4$ however they don't have the desired "natural number" properties hence axiom $5$ has been created to allegedly clean these elements up. I also know that axiom $5$ is supposed to work as follows:

If $0$ is in our set $T$ then $s(0)$ is also in $T$ but if $s(0)$ is in $T$ then $s(s(0))$ is also in $T$ and so on. Thus we declare this set to be the natural numbers.

Now to my actual question: suppose we throw in $a$ and $b$ and $0$ into $T$. Since we do not yet know what the natural numbers are, we can't exclude $a,b$ either (if my thinking is correct). Then for all $n$ in $T$ the succesor $s(n)$ is also in $T$. As previously stated $s(0), s(s(0)), ...$ are all in in $T$ (the actual natural numbers) but if such elements $a$ and $b$ are also in T then their successor $s(a)$ and $s(b)$ are also in $T$ which is  a true statement since $s(a) = b$ and $s(b) =a$.

Axiom $5$ should exclude such elements but the way I am stating it, it seems like these elements do not violate axiom $5$.

答:
(Posted after previous posting was accepted.)

I know that this axiom exists in order to exclude elements which could
have the following property:

$~~~~s(a) = b$

$~~~~s(b) = a$

$\cdots$

Axiom $5$ should exclude such elements but the way I am stating it, it
seems like these elements do not violate axiom $5$.

In this case, $a, b\neq 0$ and $\{a, b\}$ is an isolated, closed loop under the successor function $s$, i.e. neither of the elements of  $\{a, b\}$ are accessible from any other natural numbers. This contradicts the fact that induction holds on $N$.

It can be shown that induction holds on any set $X$ with $x_0$ being the "first" element under the successor function $f$ if and only if there exists no such isolated (unconnected?) subsets $Y\subset X$ under $f$, with $x_0 \notin Y$. (See my formal proof using a form of natural deduction here, 228 lines) .


问:
Find vectors $u$ and $v$ such that the parametrization
Find vectors $u$ and $v$ such that the parametrization $w(t) = u + tv$ describes the line containing the points $(5,2)$ and $(-1,3)$.

答:
(For clarity reasons I'll be using column vector notation) Consider the vectors as
$$u = \begin{pmatrix} u_1 \\ u_2\end{pmatrix}, v = \begin{pmatrix} v_1 \\ v_2\end{pmatrix}.$$
Now, since the line that defines the union of the points $\begin{pmatrix} 5 \\ 2\end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3\end{pmatrix}$ is
$$\begin{pmatrix} -1 \\ 3\end{pmatrix} - \begin{pmatrix} 5 \\ 2\end{pmatrix} = \begin{pmatrix} -1-5 \\ 3-2\end{pmatrix} = \begin{pmatrix} -6 \\ 1\end{pmatrix},$$
we can conclude that your vector $u$ will determine the "position" of your line, and $v$ will determine the "direction" (this is because the vector with the parameter in $w$'s formula is $v$), so we can conclude that $u_1=5,u_2=2,v_1=-6,v_2=1$, to be said:
$$w(t)= u+tv=\begin{pmatrix} 5 \\ 2\end{pmatrix}+t\begin{pmatrix} -6 \\ 1\end{pmatrix}=\begin{pmatrix} 5-6t \\ 2+t\end{pmatrix}$$


问:
Error in Gilbert Strang Differential Equations 2.1 Fundamental Solution to 2nd Order Differential Equation?
In Strang's Differential Equations and Linear Algebra book, Section 2.1, he introduces the fundamental solution / impulse response $g(t)$ to the delta forcing function:

$$m g'' + kg = \delta(t) $$

with given initial conditions $g(0) = 0$ and $g'(0) = 0$. He then says this implies that $g'(0) = 1/m$, since

$$ m g''(0) + k g(0) = \delta(t) \rightarrow m g''(0) = \delta(t) \rightarrow m g'(0) = 1 \rightarrow g'(0) = \frac{1}{m}$$

I'm confused how this doesn't contradict the initial given condition $g'(0) = 0$. Could someone please clarify?

Edit 1: Maybe I'm misunderstanding the text. I don't find Strang easy to follow. Photo below:

答:
I agree that Strang didn't do a great job here... I'm particularly disappointed by his lack of clarification on the domains of his solutions, since this one in particular is implied to be the solution for positive time only. Anyways, on to answering what is happening here (and sorry I'm so late to the game):

For functions with the step function $H(t)$ or the impact function $δ(t)$, there is a strange change in $y$ at $t = 0$ (for $H(t)$, there's a jump from $0$ to $1$, and $δ(t)$ has an instantaneous impact), but besides that point the equation is pretty run-of-the-mill.

For a differential equation $Ay'' + By' + Cy = δ(t)$, it is not inaccurate to say that for $t ≠ 0$, the equation is just the homogeneous version of itself: $Ay'' + By' + Cy = 0$. However, the instantaneous impact at $t = 0$ of course has some effect on the equation, and it can't just be said that we can replace it with zero. But we are certain with this approach that the solution $y$ of $Ay'' + By' + Cy = δ(t)$ for $t < 0$ is a solution of the homogeneous equation above, and that the solution for $t > 0$ is another solution of the homogeneous equation, but is different than that for the solution of $t < 0$ because of the impact. In other words, the effect of the delta function here is to change the initial conditions of the solution.

So, Strang is saying that given initial rest conditions $g(0)=g'(0)=0$ for $mg'' + kg = δ(t)$, the solution specifically for the domain $t > 0$ is identical to the solution of $mg'' + kg = 0$ with $g(0)=0$ and $g'(0)=\frac{1}{m}$.

Since I'm so late to answering this, I can only hope to help someone else who stumbled across this.


问:
Question regarding $\varepsilon$-$\delta$
Consider the following, I try to prove a limit, and get $\frac \varepsilon 3$=$\delta$. So, the length $\delta$ $<$ $\varepsilon$. My question is how will every $\varepsilon$ correspond with a single and unique $\delta$?

答:
Daàvid has already answered the question at hand, but I'll do my best to explain the motivation behind why we define limits in this way. Recall that we say $\lim_{x \to a}f(x) = L$ if and only if

For every $\varepsilon > 0$ there exists $\delta > 0$ such that, for all
$x$, if $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$.

This definition looks pretty complicated, but the idea that it is trying to capture is actually not too tricky. When we say, for instance, that the limit as $x$ approaches $0$ of $\sin x/x$ equals $1$, what we mean is that as $x$ gets closer and closer to $0$, the value of $\sin x/x$ gets arbitrarily close to $1$. In other words, the difference between $\sin x / x$ and $1$ eventually becomes smaller than any tiny number (which we often denote as $\varepsilon$). More precisely, we can make $|\sin x/x - 1|$ smaller than any positive $\varepsilon$ by requiring that $0<|x|<\delta$. If we want $\sin x / x$ to get really close to $1$, then we might have to require that $x$ is really close to $0$ (but not equal to $0$). This shows that depending on how small the $\varepsilon$ is, the value of $\delta$ might have to be small, very small, or extremely small. The point is that no matter how close we want $\sin x / x$ to get to $1$, it is possible to get there by requiring that $\delta$ is sufficiently small. How small that $\delta$ needs to be is dependent on the particular $\varepsilon$.


问:
Are there nonlinear operators that have the group property?
To be clear: What I am actually talking about is a nonlinear operator on a finitely generated vector space V  with dimension $d(V)\;\in \mathbb{N}>1$. I can think of several nonlinear operators on such a vector space but none of them have the requisite properties of a group. In particular, but not exclusively, are there any such nonlinear operator groups that meet the definition of a Lie Group.

答: As some guy once said, dividing operators into linear and nonlinear is like dividing the world into bananas and nonbananas. If you're ignoring the vector space structure on $V$ then it's just a set, and you're just asking for a group action. It is easy to write down group actions, even Lie group actions (see homogeneous space), on a vector space that just don't happen to be linear. Probably the simplest example is $\mathbb{R}$ acting on itself by translation. </p>


问:
Conditional probability proof involving inequality
We have a discrete random variable $X$.
I have a problem showing algebraically (if that is even possible) that $P(X\gt a\mid X \gt a+b) = 1$. Given that $a$ and $b$ are both positive integers.

When I rewrite it as $P(X\gt a\mid X-b\gt a)$ I can see that it makes sense that if we know that $X$ minus anything is more than $a$ then $X$ is definitely more than $a$. But is there a way to show this algebraically somehow?

Body: $$\mathbb{P}(X>a\mid X>a+b)=\frac{\mathbb{P}(X>a;X>a+b)}{\mathbb{P}(X>a+b)}=\frac{\mathbb{P}(X>a+b)}{\mathbb{P}(X>a+b)}=1$$


问:
Given metric spaces $X$ and $Y$, does there exist a metric space containing isometric copies of both $X$ and $Y$?
Let $X$ and $Y$ be metric spaces. Does there exist a metric space $Z$ such that both $X$ and $Y$ can be isometrically embedded into $Z$?

It is easy to see that the answer is positive if $X$ and $Y$ are bounded (consider the disjoint union of the two spaces). But what happens when $X$ or $Y$ is unbounded?

Any help will be appreciated.

答:
Assume $X$ and $Y$ are disjoint. Choose a point $x_0\in X$ and a point $y_0\in Y$. For $x\in X$ and $y\in Y$ define
$$d(x,y)=d_X(x,x_0)+1+d_Y(y_0,y).$$


问:
If $f$ and $g$ are differentiable, then show this limit is $0$.
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable. Show the following limit is true:

\begin{align*}
\lim_{h \rightarrow 0} \frac{|(f(x+h) - f(x))(g(x+h) - g(x))|}{\|h\|} = 0.
\end{align*}

I believe the approach is to use the fact that $f$ and $g$ are continuous to show that $\lim_{h \rightarrow 0} f(x+h) - f(x) = 0$, and analogously for $g$, but I'm confused by the denominator. Is it possible to invoke continuity without doing more manipulations of the quotient?

Note that this limit is part of the proof for the multivariable product rule.

答:
You know that $g$ is differentiable, so you know that
$$
\frac{|g(x+h)-g(x)|}{\|h\|}=\frac{|\sum_{i=1}^n\frac{\partial g}{\partial x_i}(x)h_i+o(\|h\|)|}{\|h\|}\le \left|\sum_{i=1}^n\frac{\partial g}{\partial x_i}(x) \right| + \varepsilon(h)
$$
where $\varepsilon(h)\to 0$ as $h\to 0$. So the second factor in your limit is bounded while the first, as you indicated, approaches zero. Therefore the limit is zero.


问:
Why in an inconsistent axiom system every statement is true? (For Dummies)
I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?

I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.

This previous question is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.

Thanks a lot in advance.

答:
It is somewhat misleading to say "every statement is true" about an inconsistent theory. This might be a point of confusion in this question, and in general the difference between "truth" and "provability" causes many other confusions, so we have to be careful to distinguish them.

"Truth" is a property that a statement has in a particular model. In other words a particular statement is either true or false in a particular model, assuming the statement is written in a formal language compatible with the model.

An inconsistent theory has no models at all. If you have no models, there's no model in which any particular statement can be true. It is correct to say that every statement in the language of the theory is provable from an inconsistent theory, and that every statement in the language of the theory is semantically entailed by the theory. But it's an abuse of language to say that every statement is "true" in the theory: true in what model?

Sometimes, when people are writing an informal proof, they say things like "assume $A$ is true" or "assume $B$ is false". But these are just figures of speech; the actual proof system usually has other ways of dealing with hypotheses than to mark them as "true" and "false". Alternatively, you can view those sayings as abbreviated forms of "assume $A$ is true in some fixed, unspecified model", "assume $B$ is false in our fixed, unspecified model". That interpretation of the informal proof is fine for any consistent theory. But that interpretation is more difficult for inconsistent theories. Because there are no models, it will be a counterfactual statement.

ContentLicense: CC BY-SA 2.5


问:
Proving quadratic formula by considering a function of coefficents
Inspired by the kind of argument discussed in this answer, I begin by considering the following quadratic:

$$ax^2  + bx +c=0$$

And I want a functions $ \zeta(a,b,c)$ which maps from the coefficients to the root, there must be two such functions via the fundamental theorem of algebra. Now, I make a few observations.

$\zeta(\lambda a , \lambda b , \lambda c) = \zeta(a,b,c)$

$\zeta(0,b,c) = - \frac{b}{c}$

I'm not very sure how to construct a polynomial after this but after considering the first property and taking derivative with $\lambda$ we get:

$$ \frac{\partial \zeta}{\partial (\lambda a)} a+ \frac{\partial \zeta}{\partial (\lambda b)} b+ \frac{\partial \zeta}{\partial (\lambda c)} c=0$$

Not sure what to do after this..

Attempt 2.0

Consider a quadratic:

$$ x^2 + Bx + C=0$$

This is equivalent to the original one as we can divide through by $a$, now we have the following properties for the root function $ \zeta(B,C)$:

$\zeta(0,0) = 0$

And I can spot no other significant properties, the problem with this way is that we no longer have the $\lambda$ trick in the previous one.

Any help is appreciated!

答:
The following is probably the best way to formalize the map $\zeta$:

Let $S=\mathbb C^2/\sim$ where $(x,y)\sim(z,w)$ iff $(x,y)=(z,w)$ or $(x,y)=(w,z)$. Also let $\mathbb P_\mathbb C^2$ be the projective space, the quotient $(\mathbb C^3\setminus\{(0,0,0)\})/{\sim}$ where $(a,b,c)\sim(a',b',c')$ iff there exists a $\lambda\in\mathbb C^*$ with $(a',b',c')=(\lambda a,\lambda b,\lambda c)$.

Then we have a well-defined map $\zeta\colon\mathbb P^2_\mathbb C\setminus\{(0,0,1)\}\to S$ sending $(a,b,c)\in\mathbb P^2_\mathbb C\setminus\{(0,0,1)\}$ to the solutions of $ax^2+bx+c=0$. Then the map $\zeta$ becomes continuous (giving $\mathbb P^2_\mathbb C$ and $S$ the quotient topology).


问:
$\mu$-measurable function $f$ and function $\bar{f}$

$(X,A,\mu)$ complete measure space. $f$ $\mu$-measurable on $X$.
Then every function $\bar{f} = f$ $\mu$-almost everywhere $\mu$-measurable.

This is written in my script. In my understanding:
$D, \bar{D} \in A$ and $f: D\to \mathbb{\bar{R}}$ and $\bar{f}: \bar{D} \to \mathbb{\bar{R}}$. $f$ is $\mu$-measurable.

If $f=\bar{f}$ then $D=\bar{D}$ in my understanding. This is trivial. So does that mean: $f=\bar{f}$ on $D\cap \bar{D}$ $\mu$-measurable? And therefore it is $\bar{f}$ almost everywhere $\mu$-measurable?

If needed I can write the definition for $\mu$-measurable down.
Definition of $\mu$-measurable:

$(X,A,\mu)$ measure space, $D\in A$ and $f$ defined on D as $f:D \to$
$\mathbb{\bar{R}}$ is $\mu$-measurable (on X) if $\mu (X\setminus D) =$
$0$ and $f$ $A|_{D}$-measurable. (With $A|_{D} := \{B \cap D | B \in$
$A\}$)

答:
No. you have $f,\bar f:X\to\overline{\mathbb R}$, and $$\mu(\{x:\ f(x)\ne\bar f(x)\})=0.$$
And you need to show that $f$ is measurable, so is $\bar f$,


问:
let $X =\lbrace (a_0, \cdots , a_n) \in S^n \mid \exists (x_1, \cdots , x_n) \in K, a_0 = a_1x_1 +\cdots+a_n x_n \rbrace $ is a closed
Let $K\subset \mathbb{R}^n$ be a compact subset. Prove that
$X =\lbrace (a_0, \cdots , a_n) \in S^n  \mid \exists (x_1, \cdots , x_n) \in K, a_0 = a_1x_1 +\cdots+a_n x_n \rbrace $
is a closed subset of the unit sphere.

I try understand what  is the set $X$ for it i try the case $n=1$ and then it follows in $S^1$ in the plane.

And $X=\lbrace (a_0, a_1) \in S^1  \mid \exists x_1 \in K, a_0 = a_1x \rbrace$ and chose
$K=[a,b]$ with $a<b$.
I note that $(0,1)$ and $(0,-1)\in X$.Now I try characterize the points that lie in $X$ and in fact I suppose that $(x,y)\in X\subset S^1$ since $x^2+y^2=1$ then $x^2=1-y^2$ and $x=\sqrt{1-y^2}$ form here
$1-y^2 \geq 0$ then $1 \geq y^2$ it is $1 \geq y$.

We need that $x=yk$ for $k\in [a,b]$ WLOG suppose $0<a<b$

Now $$yk=\sqrt{1-y^2}$$ for $y\leq 1$
from here $$y^2k^2=1-y^2\Rightarrow (k^2+1)y^2=1 \Rightarrow y^2=\frac{1}{k^2+1}$$
$$y=\sqrt{\frac{1}{k^2+1}} >0, \therefore 0< \frac{1}{k^2+1}\leq 1$$

Then any $k\in [a,b]$ satisfaces the condition and the points that satisface are
$$X=\lbrace \left(\sqrt{1-\frac{1}{k^2+1}},\sqrt{\frac{1}{k^2+1}} \right) \mid k\in [a,b] \rbrace $$

Frome here I want show thaht $X$ is closed, and the more easy way is show that $X$ is compact.
But I can´t see why  is closed since . I think two ways of prove it

Show that $X$ is compact

Show that $S^{n}\setminus X$ is open

*Show that $X=\overline{X}*$

I think that the most easy way is *Show that $X=\overline{X}$ But in case of see that $X$ contains their limit points(adherents points) I don´t see how continue with the general case.

Thanks in advice

答:
Hint

$$\begin{aligned} X &= \lbrace (a_0, \cdots , a_n) \in S^n  \mid \exists (x_1, \cdots , x_n) \in K, a_0 = a_1x_1 +\cdots+a_n x_n \rbrace\\
&=\lbrace (a_0, \cdots , a_n) \in S^n  \mid \exists (x_1, \cdots , x_n) \in K, d((1,x_1, \dots, x_n), P(-a_0, a_1, \dots,  a_n)) = 0\rbrace\\
&= \lbrace (a_0, \cdots , a_n) \in S^n  \mid d(K_1, P(-a_0, a_1, \dots,  a_n)) = 0\rbrace
\end{aligned}$$

where $P(a_0, a_1, \dots,  a_n)$ stands for the plane of equation

$$a_0 \cdot 1 +a_1 x_1 + \cdots + a_n x_n = 0$$ and $K_1$ for the compact $K_1=\lbrace (1, x_1, \dots, x_n) \mid (x_1, \dots, x_n) \in K \rbrace \subset \mathbb R^{n+1}$.

The desired result will be proven, providing that we prove that the map defined on $S^n$ by $(a_0, \cdots , a_n) \mapsto d(K_1, P(-a_0, a_1, \dots,  a_n))$ is continuous.


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
The signs "+" and "-" do not exactly correspond to positive and negative!
The negative of a number (or any element) in a group is just the inverse of that element with respect to the group operation.
We created $i$ as a solution to $x^{2} + 1 =0$ and extended the class of real numbers. However, this extended class had to be made into a group. So, $-i$ is just the additive inverse of $i$ in $\mathbb{C}$ with $0$ as the identity element.</p>

Title: What is $-i$ exactly?


问:
Showing that if $f$ and $g$ are continuous at $a$ then $c_1f + c_2g$ is also continuous at $a$
Let $(X, d)$ be a metric space and $(Y, \left|\left|\cdot\right|\right|)$ be a normed space and $f, g: X \to Y$ be continuous at $a \in X$. I want to show with the $\epsilon - \delta$ definition that if $f$ and $g$ are continuous at $a$ then $c_1\cdot f + c_2\cdot g$ is also continuous at $a$ for arbitrary scalar $c_1, c_2$. I have a hunch that using the preimages and open balls might be the easiest way to go. By continuity of $f$ and $g$, we know that there exists such a $\delta_f$ and $\delta_g$ that $\mathcal{B}_{\delta_f}(a) \subset f^{-1}\left[\mathcal{B}_{\epsilon}\left(f(a)\right)\right]$, $\mathcal{B}_{\delta_g}(a) \subset g^{-1}\left[\mathcal{B}_{\epsilon}\left(g(a)\right)\right]$ for arbitrary $\epsilon$.

But after this I can only wave my hands and state that similarly for $c_1f$ and $c_2g$ there exists $c_1\delta_f$ and $c_2\delta_g$ such that $\mathcal{B}_{c_1\delta_f}(a) \subset \left(c_1f\right)^{-1}\left[\mathcal{B}_{\epsilon}\left((c_1f)(a)\right)\right]$, $\mathcal{B}_{c_2\delta_g}(a) \subset \left(c_2g\right)^{-1}\left[\mathcal{B}_{\epsilon}\left((c_2g)(a)\right)\right]$. I do not know how to progress beyond this point, but I suspect that there is some simple trick I just do not know at the moment.

Edit: I just realized that since the metric in the codomain is $d_{Y}(f(x), f(a)) = \left|\left|f(x) - f(x)\right|\right| \leq \left|\left|f(x)\right|\right| + \left|\left|f(a)\right|\right|$, the easiest way to finish the proof is to first choose such deltas for $f$ and $g$ that $d_Y(f(x), f(a)) \leq \frac{\epsilon}{2c_1}$ and $d_Y(g(x), g(a)) \leq \frac{\epsilon}{2c_2}$, and then apply the triangle inequality for $h = c_1f + c_2g$ to derive the desired $< \epsilon$.

In any case, if you happen to know some other way to finish this proof I am more than happy to read it!

答:
In the Edit: you were progressing nicely. I'll go that way.
First choose an $\epsilon >0$, then by the continuity of $f$ and $g$ at $a$, there exist $\delta_1, \delta_2 >0$ such that $\|f(x)-f(a)\| < \frac{\epsilon}{2|c_1|}$ whenever $d_X(x,a)< \delta_1$ and $\|g(x)-g(a)\| < \frac{\epsilon}{2|c_2|} $ whenever $d_X(x,a)< \delta_2$. Let $\delta=\frac{1}{2}\min \{\delta_1, \delta_2\}$,
then for $d_X(x,a)<\delta$,
$\|(c_1f+c_2g)x-(c_1f+c_2g)a\|=\|c_1f(x)+c_2g(x)-c_1f(a)-c_2g(a)\|\leq |c_1| \cdot \|f(x)-f(a)\|+|c_2| \cdot \|g(x)-g(a)\|<\epsilon$.

If one or both of $c_1, c_2$ is/are zero then it's obvious.


问:
find $f \in L^2(0,1)$ such that $f = g + K(f)$
I'm preparing for my qualifying exam and the question bank had the following question under functional analysis section and I'm stuck with this problem.

Let $L^2(0,1)$ be the Hilbert space of square integrable functions on the interval $[0,1]$ with Lebesgue measure. Define $K : L^ 2 (0,1) \rightarrow L^ 2 (0,1)$ by $K(f)(t) =\int_0^t (t − s)f(s)\,ds$. Show that
$||K|| < 1$. Given $g \in L^ 2 (0,1)$ find $f \in L^ 2 (0,1)$ such that $f = g + K(f)$.

My idea for the second part of the problem: I feel like it is the kernel of some map and that the question is asking to find an element in the coset of $g$ (the coset that you get after you quotient out by the kernel).

I really have no idea how to proceed. Any help would be appreciated.

答:
There is an easier way prove the first part of this statement using the Cauchy-Swartz inequality. Let $f\in L^2[0,1]$ be arbitrary, and define $g\in L^2[0,1]$ by $g(x)=1-x.$ Notice

\begin{eqnarray*}
|K(f)(t)|^2 & = & \Bigg(\int_0^t(t-x)f(x)dx\Bigg)^2
\\ & \leq & \Bigg(\int_0^1(1-x)f(x)dx\Bigg)^2 \\
 & = & \Big|\big<f,g\big>\Big|^2 \\ & \leq & ||f||^2||g||^2 \\ & = & \frac{1}{3}||f||^2
\end{eqnarray*} Therefore,

$$||K(f)||^2= \int_0^1|K(f)(t)|^2dt \leq \frac{1}{3}||f||^2$$ Hence $||K||\leq \frac{\sqrt{3}}{3}<1$.


问:
How can an average decrease in numbers show up as an average percentage increase?
In the list below:

the first column shows the population

the second column shows the number by which the population has increased (so the total new population in the first row would be 27312).

the third column shows the percentage increase i.e. column 2 / column 1

I then took the average of the second column and got an average number.
But then I took the average of the third column and got an average percentage increase.

That doesn't make sense to me. My brain is a little fried. I'm trying to get an intuition for why this is.

Surely an average fall in numbers should mean an average fall in
percentage change?

答:
Look at a two row example:

```<code>  1    9   1000%
100  -50    -50%
</code>```

The numbers in second row are much larger than those in the first, so influence the average numerical change a lot. The percentages in the last column ignore the relative sizes of the rows.

The moral of the story is that averaging percent changes is almost always a bad idea.

Body:
Look at a two row example:

  1    9   1000%
100  -50    -50%

The numbers in second row are much larger than those in the first, so influence the average numerical change a lot. The percentages in the last column ignore the relative sizes of the rows.

The moral of the story is that averaging percent changes is almost always a bad idea.


问:
Finding minimum of value $x^2+y^2+z^2+t^2$

Suppose $x,y,z,t$ are real numbers which holds for these equations:
numbers:

${ \begin{cases}{x^2+6y+2t=2} \\ {y^2+2t+y=-14} \\
 {z^2+4x+2y=-27} \\  {t^2+2z+y=5}\end{cases} }$

What is the minimum
value of $x^2+y^2+z^2+t^2$ ?
$$1)12\quad\quad\quad\quad\quad\quad2)18\quad\quad\quad\quad\quad\quad3)34\quad\quad\quad\quad\quad\quad4)38\quad\quad\quad\quad\quad\quad5)\text{none}$$
$$$$

I added all the equations together:

$$x^2+y^2+z^2+t^2+4x+10y+4t+2z=-34$$
$$(x^2+4x+4)+(y^2+10y+25)+(z^2+2z+1)+(t^2+4t+4)=0$$
$$(x+2)^2+(y+5)^2+(z+1)^2+(t+2)^2=0$$

Since each squares are non negative, each of them should be zero, therefore $x=-2,\quad y=-5,\quad z=-1,\quad t=-2.$

But when I put these numbers in the original equations neither of them holds for the equations. so can we  conclude there is not exist such real numbers $x,y,z,t$ that holds for the equations and therefore the answer is $5)\text{none}$. am I right?

答:
I have both checked your solution, and confirmed it via WolframAlpha. No real solutions exist for this system of equations, and hence, the answer should be none.

Here is the link to the WolframAlpha computation.


答: Consider $f: X \rightarrow \overline{\mathbb{R}}$ a $\mu-$measurable function on the measure space $(X, \mathcal{A}, \mu).$ Also consider $g: X \rightarrow \overline{\mathbb{R}}$ such that $g = f$ on $E \in \mathcal{A}$ with $\mu(E^C) = 0$ (this is what your statement actually says). We need to show that $g$ is also $\mu-$measurable. Indeed, we may write $g = f \chi_E + h \chi_{E^C},$ where $\chi_E$ denotes the indicator function of the set $E.$ Consider $U \subseteq \overline{\mathbb{R}}$ open. Then $g^{-1}(U) = (f^{-1}(U) \cap E) \cup (h^{-1}(U) \cap E^C).$ Since $f$ is measurable and $E \in \mathcal{A},$ we deduce that $f^{-1}(U) \cap E \in \mathcal{A}.$ Moreover, since the measure space we are working with is complete and $h^{-1}(U) \cap E^C \subseteq E^C,$ $\mu(E^C) = 0,$ we also get $h^{-1}(U) \cap E^C \in \mathcal{A},$ so that finally $g^{-1}(U) \in \mathcal{A}.$ Thus, we have showed that $g$ is also measurable.

Title: $\mu$-measurable function $f$ and function $\bar{f}$

Body:
$(X,A,\mu)$ complete measure space. $f$ $\mu$-measurable on $X$.
Then every function $\bar{f} = f$ $\mu$-almost everywhere $\mu$-measurable.

This is written in my script. In my understanding:
$D, \bar{D} \in A$ and $f: D\to \mathbb{\bar{R}}$ and $\bar{f}: \bar{D} \to \mathbb{\bar{R}}$. $f$ is $\mu$-measurable.

If $f=\bar{f}$ then $D=\bar{D}$ in my understanding. This is trivial. So does that mean: $f=\bar{f}$ on $D\cap \bar{D}$ $\mu$-measurable? And therefore it is $\bar{f}$ almost everywhere $\mu$-measurable?

If needed I can write the definition for $\mu$-measurable down.
Definition of $\mu$-measurable:

$(X,A,\mu)$ measure space, $D\in A$ and $f$ defined on D as $f:D \to$
$\mathbb{\bar{R}}$ is $\mu$-measurable (on X) if $\mu (X\setminus D) =$
$0$ and $f$ $A|_{D}$-measurable. (With $A|_{D} := \{B \cap D | B \in$
$A\}$)


问:
How to show $SL_{n}(\mathbb{R})=\bigsqcup_{w\in W}LwU$ where L (or U) are lower(or upper) triangular matrix?
I want to ask you the following problem.

Let W denote the subgroup of permutation matrices in $SL_{n}(\mathbb{R})$. Show the following decomposition.

$SL_{n}(\mathbb{R})=\bigsqcup_{w\in W}LwU$

where L denotes all lower triangular matrices, and U denotes all upper triangular matrices.

I've found some similar statements in Decompose invertible matrix $A$ as $A = LPU$. (Artin, Chapter 2, Exercise M.11) and the http://en.wikipedia.org/wiki/Bruhat_decomposition. Especially, the first one is exactly the same with the problem above except the group the problem deal with : in the link, they deal with $GL_{n}(\mathbb{R})$.

I thought the problem has something wrong by the argument below :

When considering n=2 case, there exists only one permutation matrix - the identity matrix $I$ (if we regard a "permutation matrix" as the one earned by changing the rows of the identity matrix). Then, the decomposition says all the matrices with determinant 1(pick one of those, say A) can be decomposed as

$A = LU$ where  $L = \begin{pmatrix} a_{1} & 0 \\ a_{2} & a_{3} \end{pmatrix}
$ and $U = \begin{pmatrix} b_{1} & b_{2} \\ 0 & b_{3} \end{pmatrix}$ and $a_{1}, a_{3}, b_{1}, b_{3}$ are not zero.

Then,  $A = \begin{pmatrix} a_{1}b_{1} & a_{1}b_{2} \\ a_{2}b_{1} & a_{2}b_{2}+a_{3}b_{3} \end{pmatrix}$, where $a_{1}b_{1}$ should not be zero. But there exists elements in $SL_{n}(\mathbb{R})$ whose first row, frist column entry is zero. After that I leanred that $S_{n}$ can be embedded into $SL_{n}$ natrually. Using this, I examined the case n=2 again. By an elementary eigenvalue argument says that $S_{2}$ can be embedded into $SL_{2}(\mathbb{R})$ in the only one way i.e. to {I, -I}. The only different thing is -I, but by the same argument above, there are elements that cannot be represented by above decomposition even we consider -I.(just the problem of sign, huh?)

After that I rather considered $PSL_{2}(\mathbb{R})$, not $SL_{2}(\mathbb{R})$ when n=2, and when n is odd, but I still don't get it, and at this moment I gave up.

答:
The other $2 \times 2$ permutation matrix is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Here's one way to come up with a proof.  Given any $n \times n$ matrix $A$, starting in the first column and moving rightward, find the highest non-zero entry that is in a row that does not contain a pivot in an earlier column, and this will be your pivot for this column.  Using the two elementary row operations that are not swapping two rows, you can make the pivot in that column equal to $1$ and then clear the entries below this pivot.  The product of these elementary matrices $L$ is lower triangular, and $LA$ has $1$'s in pivot columns, say the pivot in column $j$ occurs in row $w(j)$, and zeroes below all the pivots.  Let $P$ be the permutation matrix with its $1$'s in the entries $(w(j), j)$.  Then $LA = PU$ with $U$ upper triangular.  So $A = L^{-1} P U$.

You also need to check that the union is disjoint, that is if $A = L P U = L' P' U'$, then $P = P'$.  The way to check this is that if you consider the upper left $p \times q$ submatrix $A_{p,q}$ of $A$, then $\text{rank}(A_{p,q}) = \text{rank}(P_{p,q})$.  Moreover, a permutation matrix $P$ is uniquely determined by the values $P_{p,q}$ for all $1 \leq p, q \leq n$.

OwnerDisplayName: anon

LastEditorDisplayName: anon


问:
Is it possible to split the natural numbers into a finite number of sets so that no pair of numbers within a set adds up to a square?
My attempt:

$\lbrace6,19,30\rbrace$ is sufficient to show that two sets are impossible.

Using a computer program with a brute force method I found that separating the numbers $1$ through $85$ into three sets is possible as shown below:

$\lbrace1,4,6,9,13,14,17,18,20,26,28,33,34,37,41,42,49,54,56,57,62,69,70,73,76,78,81,85\rbrace$

$\lbrace2,5,8,10,12,21,22,25,29,30,32,38,40,45,46,48,50,53,58,61,64,65,66,72,74,77,82,84\rbrace$

$\lbrace3,7,11,15,16,19,23,24,27,31,35,36,39,43,44,47,51,52,55,59,60,63,67,68,71,75,79,80,83\rbrace$

but $1$ through $86$ is impossible.

Edit:

WhatsUp in the answer below provides
the set of four numbers: $\lbrace1058, 6338, 10823, 13826\rbrace$ with an explanation of how he got them.  This is a alternative non-brute force way of showing that separating the natural numbers into three sets is impossible.
In a comment of this question a set of five numbers $\lbrace 7442, 28658,148583,177458,763442\rbrace$
is provided by the user Bob Kadylo. This shows that four sets are impossible.

Edit 2:

In a previous version of my post I made a proof showing that the number of sets needed for Natural numbers $1$ to $N$ so that no pair of numbers in the same set sums to a square is no more than $\lfloor\sqrt{2N-1}\rfloor$. I realized that I can do significantly better than this. In order to explain the method that has a smaller upper bound I have to transform the problem into graph theory. An equivalent formulation is to have $N$ vertices labeled from $1$ to $N$. A pair of points are connected iff the two points add up to a square. Then our goal is to color the vertices using the least number of colors so that no two vertices with the same color are connected. The first step in the greedy algorithm for coloring vertices is to make a list of colors with numbers. (ex. RED-1, BLUE-2, GREEN-3, YELLOW-4, etc.) If during the process more colors are required than are on the coloring list, add more colors to the list. The next step is to pick an uncolored vertex and use the lowest color number that isn't connected that the chosen vertex. Repeat the last step until all vertices are colored. The worst case scenario is to use one more color than the degree value of the vertex with the greatest degree (or tied with the greatest degree). If each vertex that is connected to the greatest degree vertex is a different color then the greatest degree vertex has to be a different color from all of those. The vertex with the greatest degree (or tied with the greatest) is $3$. It has degree $\lfloor\sqrt{N+3}\rfloor-1$. Therefore the Number of sets (or colors) required is no more than $\lfloor\sqrt{N+3}\rfloor$. We can do slightly better by using brooke's theorem which states that if a graph is simple, connected, not complete, and not an odd cycle, then the upper bound of the number of colors is equal to the degree of the greatest degree vertex. This means that the new upper bound is $\lfloor\sqrt{N+3}\rfloor-1$ sets. This is the significant improvment from $\lfloor\sqrt{2N-1}\rfloor$ I mentioned at the beginning.

End edits

For each natural number $X$ there are $\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$ numbers that are less than $X$ that when summed to $X$ results in a square. The expression: $\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$ increases as $X$ gets larger, because of this fact my guess is that separating the natural numbers into a finite number of sets so that no pair of numbers in a set doesn't sum to a square is impossible.

答:
Not an answer.

If you want no solutions to $a+b=c^2$ with $a,b,c$ all in the same set, then $16$ sets suffice.

The sets are $$A_i := \{n \in \mathbb{N} : n \equiv i \pmod{5}\}$$ for $i=1,3,4$, $$B_i := \{n \in \mathbb{N} : n = m\cdot 5^{2^{2u}(2v+1)} \text{ with } m \equiv i \pmod{5}, u,v \ge 0\}$$ for $i=1,2,3,4$, $$C_i := \{n \in \mathbb{N} : n = m\cdot 5^{2^{2u+1}(2v+1)} \text{ with } m \equiv i \pmod{5}, u,v \ge 0\}$$ for $i=1,2,3,4$, $$D_i := \{n \in \mathbb{N} : n = m\cdot 5^u+2 \text{ with } m \equiv i \pmod{5}, u \ge 1\}$$ for $i=1,2,3,4$, and $$E := \{2\}.$$

Where on Earth did this come from? Page $10$ of this.


答:
The event $\Big[ X>a\ \&\ X>a+b\Big]$ is the same as the event $\Big[X>a+b\Big],$ i.e. either occurs if and only if they other occurs.

Therefore they both have the same probability.

Title: Conditional probability proof involving inequality

Body:
We have a discrete random variable $X$.
I have a problem showing algebraically (if that is even possible) that $P(X\gt a\mid X \gt a+b) = 1$. Given that $a$ and $b$ are both positive integers.

When I rewrite it as $P(X\gt a\mid X-b\gt a)$ I can see that it makes sense that if we know that $X$ minus anything is more than $a$ then $X$ is definitely more than $a$. But is there a way to show this algebraically somehow?


问:
Let $x_n$ be a bounded sequence. Show that $y_n = n(x_{n+1} - x_n)$ can't have limit $+ \infty$
I have tried to solve this problem by supposing by contradiction that the limit was  $ + \infty$. Then, you'd have $\forall M > 0, \exists N \in \mathbb{N}: \forall n > N, n(x_{n+1} - x_n) > M$. This would mean that $(x_n)$ is increasing after a certain order $N$, which means that it must be convergent. I don't know how to continue past this point.

答:
Suppose $n(x_{n+1}-x_n)>M>0$ for all $n>N$.
Then
$$ x_{n+1}>x_{N+1}+M\sum_{k=N+1}^n \frac 1k$$
for all $n>N$.


答:
If $$\lim_{n \rightarrow +\infty} n(x_{n+1}-x_n) = +\infty$$then there exists $N$ such that for every $n \geq N$,
$$n(x_{n+1}-x_n) \geq 1, \quad \text{i.e.} \quad x_{n+1}-x_n \geq \frac{1}{n}$$

If you sum that from $N$ to an integer $N'$ you get
$$\sum_{n=N}^{N'} x_{n+1}-x_n \geq \sum_{n=N}^{N'} \frac{1}{n}$$

i.e

$$x_{N'+1} \geq x_N +\sum_{n=N}^{N'} \frac{1}{n}$$

Now, let $N'$ tend to $+\infty$ : you get that
$$\lim_{n' \rightarrow +\infty} x_{N'} = +\infty$$

which is absurd.

Title: Let $x_n$ be a bounded sequence. Show that $y_n = n(x_{n+1} - x_n)$ can't have limit $+ \infty$

Body: I have tried to solve this problem by supposing by contradiction that the limit was  $ + \infty$. Then, you'd have $\forall M > 0, \exists N \in \mathbb{N}: \forall n > N, n(x_{n+1} - x_n) > M$. This would mean that $(x_n)$ is increasing after a certain order $N$, which means that it must be convergent. I don't know how to continue past this point.


问:
Determine the expression of the Riemannnian metric of $S^2$ induced by $\mathbb{R}^3$

Consider usual local coordinates  $(\theta, \varphi)$ em $S^2 \subset \mathbb{R}^3$ defined by the parametrization $\phi:(0, \pi) \times (0, 2 \pi) \rightarrow \mathbb{R}^3$ given by
$$\phi(\theta, \varphi) = (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)$$
Using those coordinates determine the expression of the riemannian metric induced over $S^2$ by the euclidian metric of $\mathbb{R}^3$.

Conceptually the riemannian metric will associate each point $p \in S^2$ to a product $\langle \cdot,\cdot \rangle_p$ in its tangent space $T_pM$. I think that what the question means by metric induced by $\mathbb{R}^3$ is taking $\langle \cdot,\cdot \rangle_p$ to be the usual internal product in $\mathbb{R}^3$. I'm having trouble with getting an explicit expression for the metric.

答:
Let $(\theta,\varphi)$ be the coordinates defined by $\phi^{-1}$, and $(x,y,z)$ be the standard coordinates in $\mathbb{R}^{3}$, so that the euclidean metric is

$$g_{0}=dx\otimes dx+dy\otimes dy+dz\otimes dz$$

We identify, for $p\in \mathbb{S}^{2}$, $T_{p}\mathbb{S}^{2}$ as a subspace of $T_{p}\mathbb{R}^{3}$ (via the differential $dj_{p}$ of the inclusion). Now if $U$ is the domain of $(\theta,\varphi)$, and $p\in U$, we have

$$ \dfrac{\partial}{\partial \theta}=\dfrac{\partial x}{\partial \theta}\dfrac{\partial}{\partial x}+\dfrac{\partial y}{\partial \theta}\dfrac{\partial}{\partial y}+\dfrac{\partial z}{\partial \theta}\dfrac{\partial}{\partial z} \\
\dfrac{\partial}{\partial \varphi}=\dfrac{\partial x}{\partial \varphi}\dfrac{\partial}{\partial x}+\dfrac{\partial y}{\partial \varphi}\dfrac{\partial}{\partial y}+\dfrac{\partial z}{\partial \varphi}\dfrac{\partial}{\partial z}
 $$

Since $x=\sin \theta \cos \varphi$, $y=\sin \theta \sin \varphi$, $z=\cos \theta$, we get:

$$ \dfrac{\partial}{\partial \theta}=\cos \theta \cos \varphi\dfrac{\partial}{\partial x}+\cos \theta \sin \varphi\dfrac{\partial}{\partial y}-\sin \theta\dfrac{\partial}{\partial z} \\
\dfrac{\partial}{\partial \varphi}=-\sin \theta \sin \varphi\dfrac{\partial}{\partial x}+\sin \theta \cos \varphi\dfrac{\partial}{\partial y}
 $$

Finally, $g(v,w)=g_{0}(v,w)$ for $v,w\in T_{p}\mathbb{S}^{2}$. Since the basis vectors of $T_{p}\mathbb{R}^{3}$ are orthonormal:

$$ g_{\theta \theta}=\cos^{2}\theta \cos^{2}\varphi+\cos^{2}\theta \sin^{2}\varphi+\sin^{2}\theta=1 \\
g_{\theta \varphi}=g_{\varphi \theta}=-\cos \theta \cos \varphi \sin \theta \sin \varphi+\cos \theta \cos \varphi \sin \theta \sin \varphi=0 \\
g_{\varphi \varphi}=\sin^{2}\theta\sin^{2}\varphi+\sin^{2}\theta\cos^{2}\varphi=\sin^{2}\theta. $$

Hope this helps!


问:
Why is the inverse of this function negative?
$$f(x) = 2x^2 - 12x + 7 \,\,\,,x\le3$$
$$f^{-1}(x) = - \sqrt{(x+11)/2} + 3$$

why is the square root negative? how to know when it's supposed to be negative or positive? Could you please explain the reasoning and logic?

答: In order to know the inverse of $f$, you solve the equation $f(x)=y$. But\begin{align}f(x)=y&\iff2x^2-12x+7=y\\&\iff2x^2-12x+7-y=0\\&\iff x=3\pm\sqrt{\frac{y+11}2}.\end{align}But $x\in(-\infty,3]$. Therefore, if $y\ne-11$, you must take the solution $3-\sqrt{\frac{y+11}2}$; otherwise $x>3$.


答: Essentially, the metric will have the form $${\rm d}s^2 = E(\theta,\varphi){\rm d}\theta^2 +2F(\theta,\varphi){\rm d}\theta\,{\rm d}\varphi + G(\theta,\varphi){\rm d}\varphi^2,$$where $$E(\theta,\phi)=\left\langle\frac{\partial\phi}{\partial \theta}(\theta,\varphi),\frac{\partial\phi}{\partial \theta}(\theta,\varphi)\right\rangle,\quad F(\theta,\varphi)=\left\langle\frac{\partial\phi}{\partial \theta}(\theta,\varphi),\frac{\partial\phi}{\partial \varphi}(\theta,\varphi)\right\rangle,\quad\mbox{and}\quad G(\theta,\varphi)=\left\langle\frac{\partial\phi}{\partial \varphi}(\theta,\varphi),\frac{\partial\phi}{\partial \varphi}(\theta,\varphi)\right\rangle,$$where these inner products are computed with the ambient metric. This is a general mechanism: if $(M^n,g)$ is a Riemannian manifold, $\iota:S \to M$ is a submanifold, and $(u^1,...,u^k)$ are coordinates for $S$, then $\iota^*g$ is described in these coordinates as $$\sum_{i,j=1}^k a_{ij}\,{\rm d}u^i\,{\rm d}u^j,$$where $$a_{ij} = g\left({\rm d}\iota\left(\frac{\partial}{\partial u^i}\right),{\rm d}\iota\left(\frac{\partial}{\partial u^j}\right)\right).$$

Title: Determine the expression of the Riemannnian metric of $S^2$ induced by $\mathbb{R}^3$

Body:
Consider usual local coordinates  $(\theta, \varphi)$ em $S^2 \subset \mathbb{R}^3$ defined by the parametrization $\phi:(0, \pi) \times (0, 2 \pi) \rightarrow \mathbb{R}^3$ given by
$$\phi(\theta, \varphi) = (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)$$
Using those coordinates determine the expression of the riemannian metric induced over $S^2$ by the euclidian metric of $\mathbb{R}^3$.

Conceptually the riemannian metric will associate each point $p \in S^2$ to a product $\langle \cdot,\cdot \rangle_p$ in its tangent space $T_pM$. I think that what the question means by metric induced by $\mathbb{R}^3$ is taking $\langle \cdot,\cdot \rangle_p$ to be the usual internal product in $\mathbb{R}^3$. I'm having trouble with getting an explicit expression for the metric.


问:
How Can I Evaluate $\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$
Question:- Evaluate $$\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Where $a,b,c \in Z$

In order to find Definite Integrals of type

$$\int \frac{ln(x) ln(1+x) ln(1-x)}{(ax+b)(cx+d)}$$

Let $$I_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Differentiate both sides w.r.t $a$

$$I^{'}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x) dx$$

Now Differentiate w.r.t $b$

$$I^{''}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x) dx$$

Again differentiating w.r.t $c$

$$I^{'''}_{a,b,c}=-\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x)ln(1-x) dx$$

Letting different integral values of $a$, $b$ and $c$.we can evaluate numerous integral of above type. But I find that It is very hard to find $I_{a,b,c}$.

答:
lets look at a similar problem:
$$x^a(1+x)^b(1-x)^c=x^a(1+x)^b(1-x)^{b+c-b}=x^a(1-x^2)^b(1-x)^{c-b}$$
then do the binomial expansion of $(1-x)^{c-b}$ and try the substitution $u=x^2$, This should give you a series of incomplete beta functions


问:
explanation of a step in an intuitive proof of Leibniz rule
I was watching this video on an intuitive proof of the Leibniz rule: https://www.youtube.com/watch?v=22JXq09-cqM . I can't understand the step where dt is moved outside of the integral sign. How and why does this happen?

答: From what I understand from the video, it's only a heuristic technique of computing the derivative of the integral $I(t) := \int_{x_1}^{x_2} f(x, t) dx.$ Thus, it is meant to be taken with a grain (or rather a full tablespoon) of salt. It is not rigorous. In his video, $dt$ signifies a finite quantity, however small, thus a constant, it is not to be taken as the $1-$form $dx$ (or measure, what have you, choose your preferred way of looking at integrals), thus can be safely moved outside the integral.


问:
Operation signs as exponents
I sometimes see numbers or constants or variables having exponents the operation signs, such as these:

$a^{+},~ a^{-},~ (1/4)^{+},~ k^{\pm},~ 0^{+},~ 0^{-}$

All I know is the zero to the power of plus and minus, which happens for me to know it from learning limits in calculus.

Can someone explain me what do they mean?

Thanks for the answers.

答:
Something superscripted as $+$ or $-$ means the following:

$a^+$ means that we are approaching $a$ from the positive side of $\text{x axis}$ or from the right side.

$a^-$ means that we are approaching it from the $-\text{x axis}$

Thus, $0^-$ means approaching $0$ from left and $0^+$ means approaching it from right side....

This is the same case with all the numbers.

Title: Operation signs as exponents


问:
Good calculus exercises/problems?
I can't enroll in a university this year, so I'm studying calculus at home, but the only exercises about calculus that I find are the easy ones. Do you know a great page where I can find not only calculus exercises, but problems as well?

I want to find about:

hard limits,

derivatives and integrals,

and some problems

答:
Links-

Spivak: http://www.amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918/ref=sr_1_2?s=books&ie=UTF8&qid=1369543382&sr=1-2&keywords=spivak

Apostal: http://www.amazon.com/Calculus-Vol-One-Variable-Introduction-Algebra/dp/0471000051

Apostal was used at MIT for this open courseware course: http://ocw.mit.edu/courses/mathematics/18-014-calculus-with-theory-fall-2010/assignments/

http://ocw.mit.edu/courses/mathematics/18-024-multivariable-calculus-with-theory-spring-2011/assignments/

Contains pretty good course notes.

The single/multiple variable calculus classes (sans theory) on here often have good free exercizes:

http://ocw.mit.edu/courses/mathematics/

Title: Good calculus exercises/problems?


问:
Meaning of tilde over a vector in Fourier analysis?
A comment on this question says "I've seen ... overtildes ... used for Fourier transforms before."  That question is about a different context, as is this question.

Discrete Fourier transforms do play a big role in the piece I'm reading, which includes expressions such as $\tilde{x}$, where $x$ is a vector.   (Unfortunately, I'm not at liberty to cite or quote the work in question, so I don't want to provide detailed context.)

Is $\tilde{x}$ a common notation for a Fourier transform, as the quotation above implies?

答: Well, not really, no, the most common notation for the Fourier Transform (be it discrete or not) of a function $f$ (let's say $f \in \mathcal{S}(\mathbb{R}^d)$) is $\hat{f} = \mathcal{F}(f).$ Also the dummy variable for the argument of a Fourier transform is $\xi,$ whereas $x$ is kept as a dummy variable for the original function- for example we would write $\hat{f}(\xi) = \mathcal{F}(f)(\xi)$ and $f(x).$ I guess these are only notations that have been widely used over the years, so they became more or less standard nowadays. The tilde notation I have never encountered, but every author has their preferred way of denoting it, I guess. :)