答: We can do a proof by “ contradiction “. Assume $s_n + t_n \to L < \infty $, then $s_n + t_n$ is bounded, say by $M$. Thus $|s_n| = |s_n + t_n - t_n| \le |s_n + t_n| + |t_n| \le M + T$. So $s_n$ is bounded, but it is unbounded since $s_n \to \infty $. Therefore $s_n + t_n \to \infty $.

Title: Help proving that $\lim(s_n+t_n)=+\infty$

Body:
If $\lim s_n=+\infty$ and if $(t_n)$ is a bounded sequence, then $\lim(s_n+t_n)=+\infty$.

My proof:

Suppose $\lim s_n=+\infty$ and $(t_n)$ is bounded.

Since $\lim s_n=+\infty$, by definition we have that for all $M\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies that $s_n>M$.

Since $(t_n)$ is bounded, we have that for all $n\in\mathbb{N}$, $k\leq t_n\leq k'$ for some $k,k'\in\mathbb{R}$.

This is where I get stuck. I'm not sure how to put these two definitions together coherently. It's obvious to me that $t_n$ being bounded means that $s_n + t_n >M$. However, I'm having trouble arriving at this result from the definitions that I've stated.

Maybe since $\lim s_n=+\infty$, I could say that $s_n>M-t_n$ which implies that $s_n+t_n>M$. But I'm not sure how the definition of $t_n$ being bounded plays into this.

Am I on the right track? Any hints/advice would be helpful. Thank you!


问:
If you roll a fair 6-sided die and then flip a fair coin that number of times, what is the probability that you will get at least two heads?
My idea is to use disjoint events and calculating the probability of getting at least two heads for each number rolled. For example, if I roll a 3, I would calculate the probability with the expression $(\frac{1}{6}) (\frac{1}{2})^3 \binom{3}{2} + (\frac{1}{6}) (\frac{1}{2})^3\binom{3}{3})= \frac{1}{12}$ and then add up the probabilities of getting at least two for each rolls, since the events are disjoint, summing to $\frac{67}{128}$. Is this a valid solution? Is there a better approach to solving this problem?

答:
There is a 1/6 chance you will roll a 1. Then you CAN'T flip two heads.  The probability is 0.

There is a 1/6 chance you will roll a 2. The probability of flipping two head in two flips is (1/2)(1/2)= 1/4.  The probability is (1/6)(1/4)= 1/24.

There is a 1/6 chance you will roll a 3. The probability of flipping two heads,and then a tail,  HHT,  (1/2)(1/2)(1/2)= 1/8.  But the probability of HTH or THH is the same.  The probability of two heads in three flips is 3/8.  The probability is (1/6)(3/8)= 1/16.

There is a 1/6 chance you will roll a 4. The probability of flipping two heads,and then two tails, HHTT, (1/2)(1/2)(1/2)(1/2)= 1/16. There are $\frac{4!}{2!2!}=
6$ permutations, HHTT, HTHT, THHT, THTH, and THHT, all having the same probability, 1/16.  The probability of two heads in four flips is 6/16= 3/8.  The probability is (1/6)(3/8)= 1/16.

There is a 1/6 chance you will roll a 5.  The probability of flipping two heads, and then three tails is $(1/2)^5= 1/32$.  There are $\frac{5!}{2!3!}= 10$ permutations (I won't write them all) so the probability of two heads and three tails in any order is $\frac{10}{32}= \frac{5}{16}$.  The probability is (1/6)(5/16)= 5/96.

There is a 1/6 chance you will roll a 6. The probability of flipping two heads then four tails is $(1/2)^6= 1/64$. There are $\frac{6!}{2!4!}= 15$ permutations so the probability of two heads and four tails in any order is [tex]\frac{15}{64}[/tex].  The probability is (1/6)(15/64)= 5/32.

If you roll a die then flip a coin that number of times the probability you get exactly two heads is 1/24+ 1/16+ 1/16+ 5/96+ 5/32= 4/96+ 6/96+ 6/96+ 5/96+ 15/96= 36/96= 3/8.

ContentLicense: CC BY-SA 4.0


问:
Graph structure on graph homomorphisms
Let $G$ and $H$ be directed graphs. I want to form the following new graph $N$:

The vertices of $N$ are the graph homomorphisms from $G$ to $H$;

Given $f,g:G\to H$, there is an edge from $f$ to $g$ if and only if for all vertices $x$ of $G$, there is an edge of $H$ from $f(x)$ to $g(x)$.

(Note that this does not use the edges of $G$ in the construction.)

My question is: is this construction known? If yes, how is it called, and where can I read more about it?

答: I don't know a name for this construction but it is adjoint to the "Cartesian product" of directed graphs (warning: despite the name, this is not the categorical product in the category of directed graphs).  That is, writing $[G,H]$ for your graph of homomorphisms from $G$ to $H$, there is a natural bijection between homomorphisms $G\to [H,K]$ and homomorphisms $G\mathbin{\square}H\to K$ where $\square$ denotes the Cartesian product.  This makes the category of directed graphs a closed symmetric monoidal category with respect to the Cartesian product.

Title: Graph structure on graph homomorphisms


问:
Investigating a $1/2$-dimensional sphere
The volume of an Euclidian sphere in $n$-dimensions with radius $r$ equals
$$V_n=r^n \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}.$$
This formula is valid for $n\in\mathbb{N_0}$. Now I am asking myself if I can extend the definition to the rational numbers. Let's define $p=1/2$ and let $r=1$. The formula above, regardless of what it says about spheres, can be evaluated to
$$V_{1/2} = 1^{1/2}\frac{\pi^{1/4}}{\Gamma(\frac{1}{4}+1)}=\frac{4\sqrt[4]\pi}{\sqrt{2 \varpi\,\sqrt{2 \pi}}},$$
where $\varpi$ is the lemniscatic constant. (It is equal to $\pi\cdot G$, $G$ is the Gauss constant). So let's assume that the volume of a $1/2$-dimensional unit sphere is equal to the expression above. Keep all of this in mind I will come back to it.

To measure dimensions of "objects" (sorry for the non-rigorosity) we can use the Hausdorff dimension. The Hausdorff dimension of a circle is two, the Hausdorff dimension of a cube is three etc., you get it. The nice thing about the Hausdorff dimension is, that it also can measure non-integer dimensions, for example the Koch snowflake that has the dimension $\frac{\log 4}{\log 3}$. Assume we can define a $1/2$-dimensional sphere with the Hausdorff dimension. With enough axioms we can then find properties of this sphere. So here is my question:

Can we find the volume of a $1/2$-dimensional sphere by using the Hausdorff dimension, and if yes, is it equal to the value $V_{1/2}$ we arrived at earlier?

答:
Since no one has answered this question yet, let me write what I think about this interesting, but ill posed (in my opinion) question. First of all, In mathematics there are multiple meaning of dimension depending on the object we are working with. Here you are trying to incorporate two different "dimensions" (dimensional of a real vector space and  Hausdorff dimension of a metric space) and see if one extend the other while making sense of volume. This may or may not possible, and most likely not. For a simple example, consider the embedding $\mathbb{N}\hookrightarrow\mathbb{Z}.$ Using $\mathbb{N}$ we count thing, like $5$ apples while $-5$ apples makes no sense anymore in the extension $\mathbb{Z}.$ Similarly the meaning of dimension turn out to be different when one looks at more general spaces.

If we merely start with a space with Hausdorff dimension $1/2,$ the volume of a sphere must means its Hausdorff_measure. Consider the generalized Cantor set obtained by  removing at the middle $1/2$ segment of $[0,1],$ and in the $n$th iteration remove the central interval of length $1/2L_{n-1}$ from each remaining segment. This is an example for a this kind of space. Now you can test your conjecture $$V_{1/2} =2^{5/4}\varpi^{-1/2}$$ by an explicit computation. I will leave the rest for someone familiar with Geometric measure theory. But keep in mind that there is no canonical space with Hausdorff dimension $1/2$ that we could declare to be $\mathbb{R}^{1/2}.$ So, this is more likely to be a dead-end.

Another less promising way to approach this problem is understand what is a vector spaces of fractional dimension. Clear this will not be a natural generalization of the existing concept. For example, a linear transformation between two vector spaces is a matrix (after choosing a basis for the domain). Then, what would be a factional dimensional matrix?

Under the assumption that there is a some "fractional" analogue vector spaces: A priori, there's no reason to believe that they'd look like typical fractals as this is an algebraic approach oppose to previous geometric approach. In general, the dimension of a vector space is the cardinality of any basis. Rephrasing this definition in other ways (such as  trace of the identity operator, etc.) allow us to generalize it in some cases. For example, there is already a well established notion of super vector spaces with negative integer dimensions in this modified sense (and generalizes to dualizable object in a symmetric monoidal category; see). But so far there is no sensible distinguished class of "fractional dimensional vector spaces". Even if one manage to identify them, then we need a nice Lebesgue measure to measure volumes of subsets. So, lot more things are still waiting in this road to be discovered, if possible.

Near the end of writing this answer, I found this question with two well written answers along the same line as mine. Hope it will help to clear the scenario more.


问:
Complexity of computing the spectral radius of a non-symmetric square matrix
Unfortunately, I was not able to find appropriate literature that describes the computational complexity (in O notation) for computing the spectral radius of a square matrix (not symmetric!). In particular, my matrix is of the form $A \in \mathbb{R}^{N \times N}$, where $N$ is a finite integer.

While there is some literature on graph theory, and therefore, for symmetric matrices (adjacency matrices), in the case of non-symmetric square matrices my search was unsuccessful.

I would be very grateful for any leads, also regarding spectral radius approximations.

答:
The eigenvalues of an $N$ by $N$ matrix (and thus the spectral radius) can be computed in $O(N^{3})$ time by Francis's algorithm (aka the implicitly shifted QR method.)

The Arnoldi method for computing the largest magnitude eigenvalue might be faster than computing all of the eigenvalues, particularly if the matrix is sparse.  If the matrix is dense, then each iteration takes $O(N^2)$ time, but the number of iterations depends on the spectrum of $A$ and can be bad in some cases.  I'm not aware of a worst-case analysis of this, but practically, this is another $O(N^{3})$ method for dense matrices.

You can play with the two approaches in MATLAB using:

```<code>tic; U=eig(A); max(abs(U)), toc
</code>```

versus

```<code>tic; abs(eigs(A,1)), toc
</code>```

For random dense matrices from 1000 by 1000 up to 10000 by 10000, eigs() is roughly twice as fast as eig() but both are scaling as $O(N^{3})$.

The results returned by eigs() with default tolerance are accurate to about 5 digits.  Furthermore, there's quite a bit of variance in the time taken by eigs() depending on the particular matrix.

Body:
The eigenvalues of an $N$ by $N$ matrix (and thus the spectral radius) can be computed in $O(N^{3})$ time by Francis's algorithm (aka the implicitly shifted QR method.)

The Arnoldi method for computing the largest magnitude eigenvalue might be faster than computing all of the eigenvalues, particularly if the matrix is sparse.  If the matrix is dense, then each iteration takes $O(N^2)$ time, but the number of iterations depends on the spectrum of $A$ and can be bad in some cases.  I'm not aware of a worst-case analysis of this, but practically, this is another $O(N^{3})$ method for dense matrices.

You can play with the two approaches in MATLAB using:

tic; U=eig(A); max(abs(U)), toc

versus

tic; abs(eigs(A,1)), toc

For random dense matrices from 1000 by 1000 up to 10000 by 10000, eigs() is roughly twice as fast as eig() but both are scaling as $O(N^{3})$.

The results returned by eigs() with default tolerance are accurate to about 5 digits.  Furthermore, there's quite a bit of variance in the time taken by eigs() depending on the particular matrix.


问:
If $\lim_{x \to \infty} \frac{f'(x)}{x}=2$ does it follow that $\lim_{x \to \infty} \frac{f(x)}{x^2}=1$?
I need to show that the following statement is true or false. $$\displaystyle\lim_{x \to \infty} \frac{f'(x)}{x}=2 \Rightarrow \displaystyle\lim_{x \to \infty} \frac{f(x)}{x^2}=1$$

I considered $f(x)=x^2$, and it seems that the above statement is true. If so, how do I go about proving this? Or maybe there is a counterexample?

I appreciate your help.

答: $\lim\limits_{x\to\infty}\dfrac{f'(x)}{x}=2$, therefore $\lim\limits_{x\to\infty}{f'(x)}$  equals $\infty$. Now apply De L'Hospital's rule to the second limits and the result follows.</p>

ContentLicense: CC BY-SA 3.0


答: You can compute the spectral radius using Gelfand's formula. Otherwise, obviously computing the spectral radius is at least as hard as finding the largest (or smallest) root of a polynomial, which has been studied, and engendered great confusion.</p>

Title: Complexity of computing the spectral radius of a non-symmetric square matrix

Body:
Unfortunately, I was not able to find appropriate literature that describes the computational complexity (in O notation) for computing the spectral radius of a square matrix (not symmetric!). In particular, my matrix is of the form $A \in \mathbb{R}^{N \times N}$, where $N$ is a finite integer.

While there is some literature on graph theory, and therefore, for symmetric matrices (adjacency matrices), in the case of non-symmetric square matrices my search was unsuccessful.

I would be very grateful for any leads, also regarding spectral radius approximations.


问:
What curves subtend a constant fraction of angle at center.
It is known that if the angle subtended from two foci $ \text{ in the sketch} (\pm 3, -4)  $ at origin and at the arc are $ (\gamma, \gamma/2 ) $ respectively then the curve is a circular arc.

What are curves if the angles subtended on them are $ \gamma/4, \gamma/6 $ etc. instead?

Are these curves known?

Plotted here are  boundary angle fractions found  by dot product of the enclosing vectors.

答:
They're circular arcs as well: referencing your diagram, if you want the angle to be $k\gamma$ continue vertically from your point $O$ until you hit the angle $\widehat{O'}=2k\gamma$ at point $O'$. Then, your locus is an arc of the circle with center $O'$ and passing trhough your two original "foci" (and symmetrically for the part in the lower half-space).

Addendum: Removing the unnecessary dependence on $\gamma$, more generally, given a segment $AB$, the locus of the points that see the segment by an angle $0<\alpha<\pi$ may be devised as such:

call $U_1$ and $U_2$ the two points such that $\triangle AU_iB$ is an isosceles triangle with basis $AB$ and $\angle AU_iB=\frac\pi2-\alpha$.

your locus is the set of points $P$ satisfying the identity $\min\left(\overline{PU_1},\overline{PU_2}\right)=\overline{U_1A}$, which is the union of two circles, minus the arcs lying inside either of the disks.

This assumes the limit-case convention that $A$ and $B$ should see $AB$ by any angle.

Notice that the set $\{U_1,U_2\}$ is completely determined by the fact that $\overline{U_iA}=\overline{AB}\csc \alpha$. Namely, if we set cartesian coordinates such that $A=\left(-\frac12,0\right)$ and $B=\left(\frac12,0\right)$, then we may take $U_i=\left(0,\frac{(-1)^i\cot\alpha}{2}\right)$ and the locus is described by the equation $$\min\left(\sqrt{x^2+\left(y-\frac{\cot \alpha}2\right)^2},\sqrt{x^2+\left(y+\frac{\cot \alpha}2\right)^2}\right)=\csc \alpha$$
or, after some algebra, $$x^2+\left(\lvert y\rvert-\frac12\lvert \cot\alpha\rvert\right)^2=\csc^2\alpha$$


问:
Law of large numbers for nearest neighbors (proof feedback)
Let $X$ be a random variable taking values in $\mathbb R^d$ and $S_n$ an iid sample of size $n$ distributed according to $X$. Let $X_{(i)}(x)$ be the $i$th nearest neighbor to $x$ in the set $S_n$ (in $\ell_2$ norm). Show that $X_{(i)}(X)$ converges to $X$ in probability for $i \leq k$ where $k$ is such that $k/n \rightarrow 0$ as $n\rightarrow \infty$.

What I've tried

Here's my attempt for $k=1$:

$$\mathbb P \left ( \left \|X_{(1)}(X) -X\right \| < \epsilon \right ) = \mathbb P \left ( \min_i \left \|X_i -X\right \| < \epsilon \right )$$

Since the $X_i \in S_n$ have the same support as $X$, there is an $N_\delta$ large enough such that for all $n \geq N$, $S_n$ contains an element within $\epsilon$ of $X$ with probability at least $1- \delta$.

For the general case we can extend the argument to requiring $N_\delta$ to be large enough such that at least $k$ elements are within $\epsilon$ of $X$.

I could use some feedback on my reasoning.

Perhaps it's also possible to show this by contradiction. If the elements of $\{X_{(i)}(X)\}^k$ don't converge to $X$ in probability, then they remain bounded away from $X$ in probability for all $n$.

答:
Below, we assume:

a) $i=i(n)\in {\mathbb N}$ satisfies $i(n)/n \to 0$;

b) Conditional on $X$, the RVs $X_1,X_2,\dots$ are IID with the same support as $X$ (note: they do not need to have the same distribution as $X$). This automatically holds if $X,X_1,X_2,\dots$ are IID.

Fix $\epsilon>0$.

If $|X_{(i)}(X) - X| > \epsilon$, then the number of the samples among the first $n$ which are farther then $\epsilon$ from $X$ is at least $n-i+1$. The latter event is contained in the event $S^{X,\epsilon} _n > n-i+1$, where for $z \in {\mathbb R}^d$, $S^{z,\epsilon}_n=\sum_{j=1}^n {\bf 1}_{\{\|X_j-z\|>\epsilon\}}$.

Let $K$ be the support of $X$. Then:

For every $z\in K$, $P(\|X_1-z\|>\epsilon)<1$ and therefore by LLN,  $P(S^{z,\epsilon}_n > n-i+1) \to 0$

$P(X \not \in K)=0$.

Thus

\begin{align*} P( |X_{(i)}(X) - X|>\epsilon) &\le P(S^{X,\epsilon}_n > n - i+1 ) \\
 & = E [ \underset{(*)}{\underbrace{P(S^{X,\epsilon}_n > n-i+1 | X)}}{\bf 1}_K(X)]
\end{align*}

By 1., $(*)\to 0$ a.s. on the event $\{X \in K\}$. Using this, bounded convergence and 2. give that the RHS tends to $0$ as $n\to\infty$, completing the proof.


问:
How to find the time for a treatment when counting the difference of pills taken?
The problem is as follows:

Louis took three and a half type $A$ pills every twelve hours and a
half type $B$ pills every $6$ hours. He did this until the difference
in the number of pills taken was $25$. If he starts taking both types
of pills together, how long does the treatment last and how many pills
had he taken in total?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{4.75 days and 44 pills}\\
2.&\textrm{4.75 days and 45 pills}\\
3.&\textrm{3.75 days and 43 pills}\\
4.&\textrm{4.5 days and 45 pills}\\
\end{array}$

How exactly should I solve this problem?.

What I've attempted so far was to use this formula which is based on the fact that the number of pills taken can be found by computing the division of the total time with that of the interval between each dose plus $1$ which guarantees to account all together and without falling in the off by one error.

Thus the labels are as follows:

$t_{1}$: total elapsed time

$t_{2}$: interval time between dose

$\textrm{total of A pills: a}$

$\textrm{total of B pills: b}$

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{t_2}+1\right)=a$

$\left(\frac{1}{2}\right)\left(\frac{t_1}{t_2}+1\right)=b$

$a-b=25$

Replacing with the given information:

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{12}+1\right)=a$

$\left(\frac{1}{2}\right)\left(\frac{t_1}{6}+1\right)=b$

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{12}+1\right)-\left(\frac{1}{2}\right)\left(\frac{t_1}{6}+1\right)=25$

Then solving this yields:

$t=\frac{528}{5}$

But this number is not an integer and it doesn't seem to help me to get the requested time. Thus I need help in the right approach for this question. Can someone help me here? It would help me a lot a wordy answer so I can understand what is going on.

答:
Hmm. So every $12$ hours, he takes $3.5A + B$ pills. In $24$ hours, he takes $7A + 2B$ pills, for a difference of $5$ pills. That suggests $5$ days, but now we can look at that last $24$ hours.

After $4$ days, he's taken $28A + 8B$ pills. Oh, but this is one of those tricky ones, isn't it? His first set of pills was at $t = 0$, which means he's taken these $28A + 8B$ pills at $t = 3.5$ (measuring $t$ in days). Then at $t=3.75, P = 28A + 8.5B$. At $t = 4, P = 31.5A + 9B$. At $t=4.5, P = 35A + 10B$.

$4.5$ days and $45$ pills seems to be choice number 4. I think algebra just... isn't actually the best way to solve this, partly because these are stepwise functions. But for an algebraic solution, I'd probably use rates so I don't have an "elapsed time" as well as two "interval times." So measuring $t$ in hours,

$$A = 3.5 + \frac{3.5}{12} t$$
$$B = 0.5 + \frac{0.5}{6} t$$
$$A-B = 25 = 3.5 + \frac{3.5}{12} t - 0.5 - \frac{0.5}{6} t$$

And that also yields an answer of $4.4$ days, which is what you had. So then you have to look at where the stepwise function is at $4.25$ and $4.5$ days, and $4.5$ wins as above.


问:
Reducing DE to Bessel Equation
We're using Kreyszig's Advanced Engineering Mathematics in the second part of my "online" Applied Analysis course. He introduces the theory, with minimal examples, then often assigns exercises that, when I learn the solution, leave me wondering how on earth he expected me to figure out the technique he had in mind.

I'm struggling with the following problem, for instance. We're asked to solve the differential equation $x^2y''+xy'+(4x^4-1)y=0$. My lecturer said to find a real number such that $z=x^{p}$ reduces the DE to a Bessel equation for some function in $z$, but I'm struggling to see how to do it. I was hoping someone could help.

答:
When substituting a variable, use the chain rule. If we define $z=x^p$, then using the chain rule we get that (see the wiki here for more information https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives)
\begin{align}
\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dz}{\mathrm dx}\frac{\mathrm dy}{\mathrm dz}=pz^{1-1/p}\frac{\mathrm dy}{\mathrm dz},
\end{align}
(not sure how much you've done with ODE's, I calculate $z'$, then substitute $x=z^{1/p}$ to get $z'$ in terms of $z$, he same goes for the higher order derivatives) and that
\begin{align}
\frac{\mathrm d^2y}{\mathrm dx^2}&=\left(\frac{\mathrm dz}{\mathrm dx}\right)^2\frac{\mathrm d^2y}{\mathrm dz^2}+\frac{\mathrm d^2z}{\mathrm dx^2}\frac{\mathrm dy}{\mathrm dz}\\
&=p^2z^{2(1-1/p)}\frac{\mathrm d^2y}{\mathrm dz^2}+p(p-1)z^{1-2/p}\frac{\mathrm dy}{\mathrm dz}.
\end{align}
Substituting the two give that
\begin{align}
z^{2/p}\left(p^2z^{2(1-1/p)}\frac{\mathrm d^2y}{\mathrm dz^2}+p(p-1)z^{1-2/p}\frac{\mathrm dy}{\mathrm dz}\right)+z^{1/p}\cdot pz^{1-1/p}\frac{\mathrm dy}{\mathrm dz}+(4z^{4/p}-1)y=0.
\end{align}
Collect your terms,
\begin{align}
p^2z^{2}\frac{\mathrm d^2y}{\mathrm dz^2}+p^2z\frac{\mathrm dy}{\mathrm dz}+(4z^{4/p}-1)y=0.
\end{align}
Note that $p$ could be any number (except 0 or 1), so choose it based on what you need.
Edit: $p$ could be 1 it just wouldn't do anything.

Title: Reducing DE to Bessel Equation


问:
Hensel's Lemma when P cannot be raised to an integer power
I'm trying to solve the problem $x^2 ≡ 10\pmod{135}$. I know that I need to use both Hansel's Lemma and the Chinese Remainder Theorem to solve this.

So far, I believe I need to start with Hansel's Lemma as $135$ does not have $3$ co-prime factors. $135$ can be simplified to $5\cdot 3^3$. How do I account for this leading $5$? Is it legal to divide the entire equation by 5 here?

答:
I’m not sure what you mean. If you want to solve this using the Chinese Remainder Theorem, then you solve $x^2\equiv 10\pmod{5}$, and $x^2\equiv 10\pmod{27}$; then you take the (up to) four combinations of solutions you get to find the solutions modulo $135$ that correspond to the pairs.

To solve $x^2\equiv 10\pmod{5}$, note that this is $x^2\equiv 0\pmod{5}$, so you just get $x\equiv 0\pmod{5}$.

Now, to solve $x^2\equiv 10\pmod{27}$, you can try to use Hensel’s Lemma to lift the solutions. You first solve $x^2\equiv 10\pmod{3}$, and then you lift the each solution to modulo $9$ and then each of those to modulo $27$. In fact, it’s easy to just start at $9$: $x^2\equiv 1\pmod{9}$ has solutions $x=1$ and $x=-1$. Then to lift them to modulo $27$, we note that $f(x)=x^2-10$ has $f’(x)=2x$. This is not zero modulo $9$ at $1$ or at $-1$, so you can lift both solutions uniquely to solutions modulo $27$. With the usual formula you get that the solution $x\equiv1\pmod{9}$ lifts to $x\equiv 19\pmod{27}$.

You can now use the Chinese Remainder Theorem for this solution: you are looking for an $x$ such that
$$\begin{align*}
x\equiv 0&\pmod{5}\\
x\equiv 19&\pmod{27}
\end{align*}$$
which gives you a solution modulo $135$.

You do the same with the solution $x\equiv -1\pmod{9}$. Say you get that this lifts to $x\equiv a\pmod{27}$. Then you use the Chinese Remainder Theorem to look for an $x$ such that
$$\begin{align*}
x&\equiv 0\pmod{5}\\
x&\equiv a\pmod{27}
\end{align*}$$
to get the corresponding solution modulo $135$.


答: In the time interval $[0,12)$ he takes $3\frac 12\ A$ pills and $1\ B$ pill because the half $B$s are taken every six hours.  That means the $A$ pills pull ahead by $2\frac 12$ pills every $12$ hours.  It takes $10$ of those $12$ hour intervals to have $A$ ahead by $25$, but no pills have been taken for the last $6$ hours.  The first time $A$ was exactly $25$ ahead was $6$ hours before, which is after $4.75$ days.  The $A$ pills were taken at $0,0.5,1,1.5\ldots 4.5$ days, which is $10$ times and $35$ pills.  The $B$ pills were taken at $0,0.25,0.5,0.75\ldots 4.75$ days, which is $20$ times and $10$ pills.

Title: How to find the time for a treatment when counting the difference of pills taken?

Body:
The problem is as follows:

Louis took three and a half type $A$ pills every twelve hours and a
half type $B$ pills every $6$ hours. He did this until the difference
in the number of pills taken was $25$. If he starts taking both types
of pills together, how long does the treatment last and how many pills
had he taken in total?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{4.75 days and 44 pills}\\
2.&\textrm{4.75 days and 45 pills}\\
3.&\textrm{3.75 days and 43 pills}\\
4.&\textrm{4.5 days and 45 pills}\\
\end{array}$

How exactly should I solve this problem?.

What I've attempted so far was to use this formula which is based on the fact that the number of pills taken can be found by computing the division of the total time with that of the interval between each dose plus $1$ which guarantees to account all together and without falling in the off by one error.

Thus the labels are as follows:

$t_{1}$: total elapsed time

$t_{2}$: interval time between dose

$\textrm{total of A pills: a}$

$\textrm{total of B pills: b}$

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{t_2}+1\right)=a$

$\left(\frac{1}{2}\right)\left(\frac{t_1}{t_2}+1\right)=b$

$a-b=25$

Replacing with the given information:

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{12}+1\right)=a$

$\left(\frac{1}{2}\right)\left(\frac{t_1}{6}+1\right)=b$

$\left(3+\frac{1}{2}\right)\left(\frac{t_1}{12}+1\right)-\left(\frac{1}{2}\right)\left(\frac{t_1}{6}+1\right)=25$

Then solving this yields:

$t=\frac{528}{5}$

But this number is not an integer and it doesn't seem to help me to get the requested time. Thus I need help in the right approach for this question. Can someone help me here? It would help me a lot a wordy answer so I can understand what is going on.


问:
spectrum of compact forward weighted shift
Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward weighted shift so that $Te_n=t_n e_{n+1}$.

Suppose $T$ is compact. Find $\sigma(T)$.

Things I know:

I know that $\sigma(T)$ is always compact and hence closed. So I'm suspecting that   $\sigma(T)$ is the closure of some set, maybe the eigenvalues?

Other than that, I'm not sure how to continue.

Any help or ideas will be appreciated!

答:
Hints: as mentioned in the comments, $t_n\to 0$. The weights I assume are positive by definition. Now, since $\|Tx\|^²=\sum|t_nx_n|^2$ it is not hard to show that $\|T\|=\max \{t_n\}.$ In fact, inductively, show that

$\tag1 \|T^n\|=\max_k\{t_kt_{k+1}\cdots t_{k+n-1}\}$

Now, choose $n$ large enough so that $t_n<\epsilon$ whenever $n\ge N$ so

$\tag2 t_{k+N}\cdots t_{k+n-1}\le \epsilon^{n-N-1}$

Finally, combine $(1)$ and $(2)$ with the spectral radius formula to show that $\sigma (T)=0.$


问:
Tossing a coin with time evolving bias
Consider the following random process starting at time $t = 1$ with a coin such that $\Pr(\text{Heads}) = p_1$. The bias of the coin, $p_t$, evolves with time as follows for $t \geq 1$:

At time $t$, the coin is tossed (it has $\Pr(\text{Heads}) = p_t$). Let $X_t$ denote the outcome of the toss.
\begin{align}
p_{t+1} = \begin{cases} \max\{p_t - \Delta_1, 0\} & \text{ if } X_t = \text{Heads}, \\
\min\{p_t + \Delta_2, 1\} & \text{ if } X_t = \text{Tails},\end{cases}
\end{align}
where $0 < \Delta_1 < \Delta_2$ are some small known constants. Thus, the bias of the coin evolves with time such that it always attempts to make the coin a fair one based on the observed outcome of a toss.

I am interested in bounding the following expression: $\displaystyle \Pr\left( \bigcup_{t = 1}^T \{p_t < \epsilon\} \right)$ for some $\epsilon > 0$. The value of $T$ is known and $\Delta_1$ and $\Delta_2$ are small constants, roughly $O(1/T)$, so that the $p_t$ does not tend to saturate to extreme values often.

Intuitively, this probability should be small because of the "corrective" nature of the underlying random process. However, I am unable to obtain an elegant upper bound on the same. The union bound is resulting in a very weak bound. I have also tried some Martingale based approaches but haven't been to find a concrete answer.

Any hints or references will be highly appreciated. Thank you.

答:
This is too long to be a comment, and it will necessarily be incomplete, but I'll start writing up some of what I know whenever I get a chance.  (I apologize in advance that I'll edit this post a few times, but I won't have a chance to write it all in one sitting.)

In terms of references, I had a hard time finding anyone else who had studied this problem, because the literature uses different terminology than I would have.  I searched for random walks, biased coin tosses, avalanche processes, and branching processes.  But the main results in this field are from Karlin and McGregor studying "birth-death processes" in the 1950s.

They study a case where the probability of births/deaths is linearly biased, which is similar to our problem, but I don't think it's exactly the same, because the population is not bounded from above.  It's been a long time since I read these papers, so I could have misremembered something.  The best short summary that I remember reading is in the following book, under the topic "random walk polynomials":  Recurrence Relations, Continued Fractions and Orthogonal Polynomials by Richard Askey and Mourad Ismail.

As you've already seen and I will make explicit below, the problem is equivalent to a random walk where the probability of moving left/right is linearly biased with respect to the current position.  (I have jokingly called this "a drunkard in a valley," since the drunkard is more likely to move downhill toward the center of the valley.)

There are a number of famous results that apply to random walks with arbitrary move coefficients.  For example, there is a series representation due to Touchard.  There is also an extremely easy way to represent the time generating function of the first-return time as a continued fraction, and there is a method for handling such continued fractions due to Flajolet from around 1990.  But I wasn't able to use any of these to get something useful out of the problem.  I was successful using the poor physicist's standard trick of applying transforms.  I suspect that I don't have enough to solve your problem but definitely enough to get started.

Most of my existing results are in my office, where I can't access them right now because of the pandemic.  But I have enough at home that I should be able to give the really basic results.

My formulation of the problem is that there is a random walk with position $X(t)$ where the probability of moving left/right when at position $x$ is
$$ \begin{align} \ell_x &= 1 - \frac{x}{N} \\ r_x &= \frac{x}{N} \end{align} $$
where $N$ is a positive integer.

Let $p_{x,t}^s$ be the probability that $X(t)=x$ given that $X(0)=s$ (the "start" position).  Then $p_{x,t}^s$ satisfies
$$ p_{x,t}^s = (1 - \delta_{x,0}) r_{x-1} p_{x-1,t-1}^s + (1 - \delta_{x,N}) \ell_{x+1} p_{x+1,t-1}^s + \delta_{x,s} \delta_{t,0} $$
where $\delta$ is the Kronecker delta.  The first term represents a walk moving right from $x-1$ to $x$.  The second is a walk moving left from $x+1$ to $x$.  The third is the initial condition.

Let $P_x^s(z)$ be the generating function of $P_{x,t}^s$ with respect to $t$, i.e.,
$$ P_x^s(z) = \sum_{t=0}^\infty p_{x,t}^s z^t $$
Then one can show from the time-evolution equation of $p$ that
$$ P_x^s(z) = (1 - \delta_{x,0}) z r_{x-1} P_{x-1}^s(z) + (1 - \delta_{x,N}) z \ell_{x+1} P_{x+1}^s(z) $$
Now let $P^s(u, z)$ be the generating function of $P^s(z)$ with respect to $x$, i.e.
$$ P^s(u, z) = \sum_{x=0}^N P_x^s(z) u^x $$
The time-evolution equation becomes a differential equation in $P^s(u,z)$ in $u$:
$$ \frac{z}{N} (u^2 - 1) \frac{dP^s}{du} + (1 - uz) P^s = u^s $$

This is a linear first-order inhomogeneous differential equation, so the solution is straightforward.  The homogeneous part has solution
$$ \begin{align}
H(u, z) &= \exp\left( \frac{N}{z} \int \frac{1 - uz}{1 - u^2} du \right) \\
        &= C_s(z) (1 - u)^{-\alpha(z)} (1 + u)^{\beta(z)}
\end{align} $$
where $\alpha(z) = \frac{N}{2z}(1-z)$ and $\beta(z) = \frac{N}{2z}(1+z)$.  $C_s(z)$ is an integration constant that can depend on $z$ since the integration is with respect to $u$.  The integral is easy (just decompose into partial fractions).

The inhomogeneous part can be written as an integral in the usual way.  The closed form solution is an Appell $F_1$ function.  (The full expression is in my office notes.)  But in many applications, including mine but possibly not yours, there is one really important thing to observe.  The inhomogeneous solution must be $O(u^{s+1})$.  You can see this by writing the solution $P^s(u, z) = u^{s+1} f(u, z)$ and then show that $f(0, z)$ is a constant.

This has a hugely important corollary: For $x \le s$, the homogeneous part is the entire solution.  I'll use this below.  (Everything below is based on the assumption that $x \le s$.)  Moreover, if you need to consider $x > s$, then you can note that the original problem has a symmetry where the substitutions $x \to N - x$ and $s \to N - s$ do not affect the probabilities.  So you can essentially figure out quite a lot about the probabilities without ever calculating the inhomogeneous solution.

On the other hand, there is still $C_s(z)$.  There isn't an obvious boundary condition to determine $C_s(z)$ without the full inhomogeneous solution.  (For example, we know that $P^s(1,1)=\frac{1}{1-z}$, but this involves terms where $x>s$.)  It turns out that $C_s(z)$ is closely related to the inhomogeneous solution.  I believe that the relationship is something like $C_s(z)=-I_s(1,z)$, where $I_s(u,z)$ is the inhomogeneous solution.  This is based off of my memory, but I think that there is a simple argument to show this based on requiring $I_s(u,z)$ to be well-behaved near $u=1$.  This means that you can derive a lot of information about the ratios of probabilities at different values of $x$, but you can't get the absolute probabilities unless you solve the inhomogeneous problem.

With all this in mind, we can start inverting the generating functions.
The series for $H(u,z)$ in $u$ is
$$ \begin{align}
H(u,z) &= C_s(z) \sum(1 - u)^{-\alpha(z)} (1 + u)^{\beta(z)} \\
       &= C_s(z) \sum_{j,k} (-1)^j \binom{-\alpha}{j} \binom{\beta}{k} u^{j+k} \\
       &= C_s(z) \sum_x u^x \sum_j (-1)^j \binom{-\alpha}{j} \binom{\beta}{x-j} \\
       &= C_s(z) \sum_x u^x \binom{\beta}{x} {_2}F_1\left(\begin{matrix} \alpha, -x \\ \beta - x + 1 \end{matrix}; -1 \right)
\end{align} $$
where ${_2}F_1$ is the Gauss hypergeometric function.  So for $x \le s$ then we have
$$ P^s_x(z) = C_s(z) \binom{\beta}{x} {_2}F_1\left(\begin{matrix} \alpha, -x \\ \beta - x + 1 \end{matrix}; -1 \right) $$

This may be the end of the results that I have that are useful to you.  I'm not really sure.  I'm happy to keep discussing here when you have thought about it.  I am mainly interested in deriving simple numerical approximations for the first-passage time to $x=s-1$, which is related to the ratio $P^s_s(z)/P^s_{s-1}(z)$.  If this is useful I can probably provide some more information.  But you may need absolute probabilities near $x=0$ or $x=N$, which is something I haven't devoted a lot of time to thinking about this.

One more piece of advice.  The hypergeometric representation above is hard to invert, but it is very versatile and you can extract a lot of information from it.  (There are a bunch of equivalent hypergeometric representations using Euler and Pfaff transforms, and of course there are also integral representations.)  And there are other approaches to the same problem, like the orthogonal polynomial representation of Karlin and McGregor mentioned above, that are closely related.

Disclaimer: All of the above is based on very messy notes where I had re-derived some of the main expressions at home, and there may be typos, so some of it may need fixing.  Please let me know if you find errors.

Title: Tossing a coin with time evolving bias


问:
Prove that the map $M:C(Y,Z) \times C(X,Y) \rightarrow C(X,Z)$ is continuous.
Let $X,Y,Z$ locally compact Hausdorff spaces.Prove that the map $M:C(Y,Z) \times  C(X,Y)  \rightarrow C(X,Z)$ given  by $M(f,g)=f\circ g$ is continuous.

Attempt

Let $[K,U]$ open element of the subbase for $C(X,Z)$ where $K$ is compact subsetof $X$ and  $U$ open such that $U \subset Y$ we must prove that $M^{-1}([K,U])$

$$M^{-1}([K,U])=\lbrace (f,g)\in C(Y,Z) \times  C(X,Y)  \mid f\circ g\in [K,U]\rbrace $$

From here $f(g(K))\subset U$ for $K\subset X$ compact and open $U\subset Z$.

Frome here I don´t know how to follow the proof since I don´t know how it condition restringes our $f$ and $g$.Any Idea of how I should start was
very helpful.

答:
It turns out that you only need the fact that $X, Y$ are locally compact and Hausdorff $(*)$. Here are two key properties that I'll be using. For reference, I am using the notation and ideas from Tammo tom Dieck's Algebraic Topology. Specifically, this is chapter 2 section 4.

If $X$ is locally compact and Hausdorff and $Y$ is any space then the evaluation map $e_{X, Y}: C(X, Y) \times X \longrightarrow Y$ via $(f, x) \mapsto f(x)$ is continuous

For any two spaces $Y, Z$ where $e_{Y, Z}$ is continuous, there is a bijection between $C(X \times Y, Z) \longrightarrow C(X, C(Y, Z))$ via "currying." That is, a map $f: X \times Y \longrightarrow Z$ yields a map $f^{\wedge}: X \longrightarrow C(Y, Z)$ via $f^{\wedge}(x)(y) = f(x, y)$. The inverse is given by $g \mapsto g^\vee$ where $g^\vee: X \times Y \longrightarrow Z$ is defined as $g^\vee(x, y) = g(x)(y)$. In other words, $g^\vee = e_{Y, Z} \circ (g \times id_Y)$. These maps are easily checked to be inverses between $Set(X \times Y, Z) \longrightarrow Set(X, Set(Y, Z))$, where $Set(A, B)$ is the set maps between $A$ and $B$. So what this statement really says is that these inverse functions restrict to the continuous case, i.e. that $f$ continuous implies $f^\wedge$ continuous and $g$ continuous implies $g^\vee$ continuous.

I'll defer the proof of these two facts until the end of my post. You can also read the reference yourself here. Taken these as a given, let's proceed with the problem at hand.

Let $X, Y$ be locally compact and Hausdorff and $Z$ arbitrary. Then we want to show continuity of the map $M: C(Y, Z) \times C(X, Y) \longrightarrow C(X, Z)$. By fact 1 above, the evaluation map $e_{X, Z}$ is continuous as $X$ is locally compact and Hausdorff. Then by fact 2, for any space $A$, we have our bijection $C(A, C(X, Z)) \longrightarrow C(A \times X, Z)$ taking $\phi \mapsto \phi^\vee$. In particular, take $A = C(Y, Z) \times C(X, Y)$. Then we have that $M: C(Y, Z) \times C(X, Y) \longrightarrow C(X, Z)$ is continuous if and only if $M^\vee: C(Y, Z) \times C(X, Y) \times X \longrightarrow Z$ is continuous.

Let's compute then what $M^\vee$ is. For $(f, g) \in C(Y, Z) \times C(X, Y)$ and $x \in X$, the definition above tells us that $M^\vee((f, g), x) = (M(f, g))(x) = f(g(x))$. We can decompose $M^\vee$ as the following composition:

\begin{align*}
C(Y, Z) \times C(X, Y) \times X &\longrightarrow C(Y, Z) \times Y &\longrightarrow Z\\
(f, g, x) &\mapsto (f, g(x)) &\mapsto f(g(x))
\end{align*}

Sorry for the poor formatting. The first map $(f, g, x) \mapsto (f, g(x))$ is just $id_{C(Y, Z)} \times e_{X, Y}$, which is continuous by local compactness and Hausdorffness of $X$ and fact 1 above. The second map is $e_{Y, Z}$, which is continuous by local compactness and Hausdorffness of $Y$ and fact 1 above. Hence, $M^\vee = e_{Y, Z} \circ (id_{C(Y, Z)} \times e_{X, Y})$ is continuous, so $M = M^{\vee \wedge}$ is continuous by fact 2.

Now let's prove these two facts, which are largely why the compact open topology is so useful.

Let's consider $e_{X, Y}: C(X, Y) \times X \longrightarrow Y$. Take some $f(x) = e_{X, Y}(f, x)$ and an open neighborhood $f(x) \in U \subseteq Y$. By continuity, $f^{-1}[U] \subseteq X$ is open. By local compactness and Hausdorffness of $X$, there is some compact neighborhood $x \in K \subseteq f^{-1}[U]$. Hence, $f[K] \subseteq U$ so $f \in [K, U]$. Then $e_{X, Y}^{-1}[U]$ contains the neighborhood $[K, U] \times K$ of $(f, x)$. Thus, we have continuity.

As discussed, we need only show that $f$ continuous implies $f^\wedge$ continuous and $g$ continuous implies $g^\vee$ continuous. Let's begin with some $f: X \times Y \longrightarrow Z$ continuous and consider $f^\wedge: X \longrightarrow C(Y, Z)$. Take a subbasic element $[K, U] \subseteq C(Y, Z)$ and suppose that $x \in (f^\wedge)^{-1}[[K, U]]$. Then $f^\wedge(x) \in [K, U]$, so for all $y \in K$ $f^
\wedge(x)(y) \in U$. Of course, this means that $f[\{x\} \times K] \subseteq U$. Thus $\{x\} \times K \subseteq f^{-1}[U]$. As $f^{-1}[U]$ is open in the product topology, for every $(x, y)$ with $y \in K$ there is a basic open neighborhood $(x, y) \in V_x \times V_y \subseteq f^{-1}[U]$. By compactness, we can cover $K$ by finitely many $V_{y_1}, \dots, V_{y_n}$. Let $V = V_{x_1} \cap \dots \cap V_{x_n}$. This is an open neighborhood of $x$. Then $V \times (\bigcup_{i=1}^n V_{y_i})$ is an open neighborhood of $\{x\} \times K$ Furthermore, take some $(a, b) \in V \times (\bigcup_{i=1}^n V_{y_i})$. Then $b \in V_{y_i}$ some $i$ so $(a, b) \in V_{x_i} \times V_{y_i} \subseteq f^{-1}[U]$. Point being, we have found a neighborhood $\{x\} \times K \subseteq V \times (\bigcup_{i=1}^n V_{y_i}) \subseteq f^{-1}[U]$. Thus, $V \times K \subseteq V \times (\bigcup_{i=1}^n V_{y_i}) \subseteq f^{-1}[U]$ so $f^\wedge[V] \subseteq [K, U]$. Thus, $x \in V \subseteq (f^\wedge)^{-1}[[K, U]]$ so $f^\wedge$ is continuous. Note that we made no assumptions on $X, Y, Z$ is this direction.

Thankfully the other direction is much easier. Let $g: X \longrightarrow C(Y, Z)$ be continuous. Then $g^\vee = e_{Y, Z} \circ (g \times id_Y)$. Indeed, evaluating the right hand side at $(x, y)$ yields $e_{Y, Z}(g(x), y) = g(x)(y) = g^\vee(x, y)$. We assumed that $e_{Y, Z}$ was continuous so $g^\vee = e_{Y, Z} \circ (g \times id_Y)$ is continuous as well.

EDIT: $(*)$ By the way, depending on your definition of locally compact, you can drop the Hausdorff assumption. Since you explicitly mention Hausdorffness, I assume that to you a locally compact space is one such that every point has a compact neighborhood. If we in addition have Hausdorffness then every neighborhood of a point contains a compact neighborhood within it. We only need this property, not the stronger fact that $X, Y$ are Hausdorff as well. In the book I reference by Tammo tom Dieck, a locally compact space is one where every neighborhood of a point contains a compact nieghborhood within it. There are non-Hausdorff spaces satisfying this form of local compactness, such as the Sierpinski space.


问:
Upper bounds for a sequence of integers defined recursively
Given $\alpha\geq0$ we consider the sequence
$$
C_k=k^\alpha\sum_{j=0}^{k-1}C_jC_{k-1-j}
$$
with $C_0=1$. I'm interested in upper bounds (in terms of $\alpha$) for such a sequence. I know that when $\alpha=0$ the previous sequence reduces to the Catalan numbers where the bound is

$$
O(4^k/k^{3/2}),
$$
but I wonder whether something is known for positive $\alpha$.

答:
There are three cases.

If $\alpha=0$, then we have the standard Catalan numbers, which you know are $O\left(k^{-\frac{3}{2}}4^k\right)$.

If $\alpha=1$, then ${\log(k)}2^{k-2}k!^{\alpha}\leq C_k\leq O(\sqrt{k})\cdot2^kk^{\alpha!}$.

If $\alpha>1$, then $C_k=\Theta(2^kk!^{\alpha})$ and there are explicit and simple bounds.

To see this, first let $U_k=\frac{C_k}{2^kk!^{\alpha}}$.  This sequence satisfies the recurrence $U_0=1$, \begin{align*}
U_{k+1}&=\frac{(k+1)^{\alpha}\sum_{j=0}^k{C_jC_k}}{2^{k+1}(k+1)!^{\alpha}} \\
&=\frac{1}{2}\sum_{j=0}^k{\frac{C_j}{2^jj!^{\alpha}}\frac{C_{k-j}}{2^{k-j}(k-j)!^{\alpha}}\cdot\frac{j!^{\alpha}(k-j)!^{\alpha}}{k!^{\alpha}}} \\
&=\frac{1}{2}\sum_{j=0}^k{\binom{k}{j}^{-\alpha}U_jU_{k-j}}
\end{align*}  These binomial coefficients grow very rapidly.  For $\alpha>0$, this sum is dominated by its outermost terms.  Our goal will be to expand out the first few terms and use them to control the growth of $U_k$.

First, an elementary property.  Note that $U_0=1$ and $U_1=\frac{1}{2}$ and apply the recursion: $$U_{k+1}\geq\frac{1}{2}\left(\binom{k}{0}^{-1}U_0U_k+\binom{k}{k}^{-1}U_kU_0\right)=U_k$$  Thus $\{U_k\}_k$ is increasing after $U_0>U_1$.  In particular, $U_k\geq U_1=\frac{1}{2}$ for all $k$.

Second, estimating the recursion.  Since $\{U_k\}_k$ is increasing, for $k\geq6$, we have \begin{align*}
U_kU_0+\frac{1}{k^{\alpha}}\cdot U_{k-1}U_1&\leq U_{k+1} \\
&\leq U_kU_0+\left(\frac{1}{k}\right)^{\alpha}\cdot U_{k-1}U_1+\left(\frac{2}{k(k-1)}\right)^{\alpha}U_{k-2}U_2+{} \\
&\phantom{\leq}\quad\quad\frac{1}{2}\sum_{j=3}^{k-3}{\binom{k}{j}^{-\alpha}U_{k-3}^2}
\end{align*}  What is this sum of binomial coefficients?  Well, for $k\geq6$, \begin{align*}
\sum_{j=3}^{k-3}{\binom{k}{j}^{-\alpha}}&\leq\sum_{j=3}^{k-3}{\binom{k}{j}^{-1}} \\
&\leq(k-6)\cdot\frac{6}{k(k-1)(k-2)} \\
&\leq\frac{6}{k(k-1)}
\end{align*}

Also, $U_2=\frac{1}{2}$.  Thus for $k\geq6$, $$
\frac{U_{k-1}}{2k^{\alpha}}\leq U_{k+1}-U_k\leq\frac{U_{k-1}}{2k^{\alpha}}+\frac{2^{\alpha-1}U_{k-2}}{(k(k-1))^{\alpha}}+\frac{3U_{k-3}^2}{k(k-1)}\quad\quad\quad(1)$$

This is where we must split off the $\alpha=1$ case.  Suppose $\alpha>1$.  The right-hand-side of (1) is decreasing in $\alpha$, so we can reduce it to $$\frac{U_{k-1}}{2k^{\alpha}}\leq U_{k+1}-U_k\leq\frac{U_{k-1}}{2k^2}+\frac{U_{k-2}}{2k(k-1)}+\frac{3U_{k-3}^2}{k(k-1)}\quad\quad\quad(2)$$  (Yes, this is not sharp.  I only need $\omega(k^2)$ denominators and higher orders make for more complicated algebra.)

Now, $\{U_k\}_k$ is increasing, so it suffices to estimate the supremum and show it finite.  Suppose there exists $S$ such that $U_k\leq S$ for all $1\leq k\leq K$.  Then, equation (2) simplifies to $$\frac{1}{4k^{\alpha}}\leq U_{k+1}-U_k\leq\frac{S}{2k^2}+\frac{S+6S^2}{2k(k-1)}$$  Iterating, \begin{align*}
\sum_{j=6}^K{\frac{1}{4j^{\alpha}}}\leq U_{K+1}-U_6&\leq\sum_{j=6}^K{\frac{S}{2j^2}+\frac{S+6S^2}{j(j-1)}}\\
&\leq\sum_{j=6}^{\infty}{\frac{S}{2j^2}+\frac{S+6S^2}{j(j-1)}} \\
&=S\left(\frac{\pi^2}{12}-\frac{5269}{3600}\right)+\frac{S+6S^2}{5} \\
&=S\left(\frac{\pi^2}{12}-\frac{4549}{3600}+\frac{6}{5}S\right)
\end{align*}  In particular, if $\alpha=2$, we can verify that, for all $1\leq k\leq6$, we have $$U_k\leq\frac{136613}{230400}\approx0.593$$  The $\{U_k\}_k$ are decreasing in $\alpha$, so this in fact shows that we may take $S=\frac{136613}{230400}$ for any $\alpha$.  But, for that value, $$\frac{\pi^2}{12}-\frac{4549}{3600}+\frac{6}{5}S\leq1$$  So we can substitute and observe a closed recursion: $$\frac{\zeta(\alpha)-\sum_{j=1}^5{j^{-\alpha}}}{4}+O(k^{1-\alpha})\leq U_{K+1}-U_6\leq S\cdot1$$

Thus $U_{K+1}$ has a finite supremum between $U_6+\frac{\zeta(\alpha)-\sum_{j=1}^5{j^{-\alpha}}}{4}$ and $\frac{613271740721}{398131200000}-\frac{136613\pi^2}{3317760}+U_6$.  But we already have estimates on $U_6$.  Putting it all together, there is some $$\lim_{k\to\infty}{U_k}\in\left[\frac{1}{2}+\frac{\zeta(\alpha)-\sum_{j=1}^5{j^{-\alpha}}}{4},\frac{849339004721}{398131200000}-\frac{136613\pi^2}{3317760}\right]$$ where the upper bound is $\approx1.73$  (Numerically, it appears the limit is $\approx0.5$.)

If $\alpha=1$, then our argument is a little trickier, to the point that I haven't tried to work out all the details.  First, suppose $U_k\leq f(k)$, for some increasing but slowly-varying $f$ (we will end up with $f(j)=\Theta(\sqrt{j})$).  Let $H_n=\sum_{j=1}^n{\frac{1}{j}}=\log{(n)}-\gamma+o(1)$.  Then equation (1) shows that $$\frac{H_K-H_5}{4}\leq U_{K+1}-U_6\leq\sum_{j=6}^K{\left(\frac{f(j)}{2j}+\frac{f(j)+3f(j)^2}{2j(j-1)}\right)}$$  The left-hand side shows that we must have at least logarithmic growth: the claimed lower bound.  To close this recursion on the right, we need $$U_6+\sum_{j=6}^K{\frac{f(j)}{2j}\left(1+\frac{1+3f(j)}{j-1}\right)}\leq f(j)$$  If $f$ is sharp, there will be equality.

This looks awful, but since $f$ is slowly-varying, we can apply a quantitative form of the integral test  (see exercise 30(a)): $$\sum_{j=6}^K{\frac{f(j)}{2j}\left(1+\frac{1+3f(j)}{j-1}\right)}\approx\int_{j=6}^K{\frac{f(j)}{2j}\left(1+\frac{1+3f(j)}{j-1}\right)\,dj}$$  Here the error term is proportional to the largest value of the integrand.  We will see that $f(j)=(1+o(1))C\sqrt{j}$, so $\frac{f(j)}{j}=(1+o(1))Cj^{-\frac{1}{2}}\to0$ as $j\to\infty$.

If we apply this integral approximation and differentiate, then we have a differential equation that we can solve for $f$: \begin{align*}
f'(j)&=\frac{f(j)}{2j}\left(1+\frac{1+3f(j)}{j-1}\right) \\
f(6)&=U_6
\end{align*}  Here I have neglected the error term, because this has an exact solution (according to Mathematica).  A naive inclusion of the derivative of the error term does not, and is the finicky details that I have skipped for this part.Fn. 1  In any case, the exact solution is $$f(j)=\frac{\sqrt{j-1}}{\frac{640\sqrt{5}}{531}-3\left(\tan^{-1}{(\sqrt{j-1})}-\tan^{-1}{(\sqrt{5})}\right)}\leq.698\sqrt{j-1}$$  Thus $U_k=O(\sqrt{j})$, the claimed upper bound.

Fn. 1: If you want to make this rigorous, a good place to start is to approximate the derivative of the error term with $\frac{f(j)}{2j^2}$.  Then there is another exact solution, albeit not exactly in closed form….


问:
Probabilities of no ace and exactly not one ace in throwing 10 dies
The probabilities i am looking for are used for a problem in Feller's introduction to probability that asks for the probability $p$ of finding $2$ or more aces in throwing $10$ dies. For this problem i considered that:

$p = 1-x-y$

Where $x$ is the probability of finding no ace at all, and $y$ being the probability of finding exactly one ace. Of course $x = \frac{5^{10}}{6^{10}}$... i think. But i am a little more confused about finding $y$, more because the answer to this exercise is that: $p = 1- 10\times\frac{5^9}{6^{10}-5^{10}} $ but i don't really understand where this result can come from, more considering the result i got for $x$

答:
Hint

If the probability of an event occurring is $p \in (0,1)$ and $q = (1-p)$ then

the probability of having exactly $k$ successes in $n$ trials : $k \in \{0,1,2,\cdots, n\}$ is

$$\binom{n}{k}p^k q^{(n-k)}.$$


问:
Why does $\sin(x)=\sin(y)$ not equal $\sin^{-1}\left(\sin\left(x\right)\right)=y$?
The equation $\sin(x) = \sin(y)$ does not produce the same output as $y=\sin^{-1}(\sin(x))$. Can anyone explain why this is?

答:
Note that the range $y$ in the explicit function is

$$y=\sin^{-1}(\sin(x))\in \left[-\tfrac\pi2, \tfrac\pi2\right]
$$

while the range of $y$ in the implicit function
$$\sin y= \sin x,\>\>\>\>y\in (-\infty, \infty)$$


问:
Losing solutions when dividing by an expression
There's something that's been  aggravating me when solving implicit differentiation. For instance, take the implicitly defined curve

$$\cos x+\cos y=\frac{1}{2}$$

(Differentiate both sides w.r.t x)

$$-\sin x-\sin y \frac{dy}{dx}=0$$

$$-\frac{dy}{dx}\sin y= \sin x$$

$$\frac{dy}{dx}=-\frac{\sin x}{\sin y}$$

The last step is the one which really confuses me, which when we divide by $(-\sin y)$: aren't we saying by doing this step that $(-\sin y)$ won't be equal to zero; aren't we kind of erasing or canceling certain values on the graph / function?

Like when we do in questions such as:

$3\cos x-12\sin x\cos^2x=0$,

$0<x<\pi$

We can make the equation look neater by dividing by $\cos x$ but we will lose solutions, as $\cos x$ won't be equal to zero and $x=\pi / 2$ won't be among our answers.

答:
I always had the same weird feeling when I solve elementary differential equations or differentiate implicit functions. But the following will resolve your concerns thoroughly.

If you have plotted the graph of $\cos x + \cos y = 1/2$, you could see that the derivative at $\sin y = 0$ is actually undefined (vertical). Therefore you can only differentiate assuming that $y$ is differentiable. This differentiability condition gives us the fact that $\sin y\neq 0$.

A useful theorem is the implicit function theorem. We only need the 2-dimensional result in our case. The theorem (2-dimensional) states that if $f$ is a continuously differentiable function (in your case $f(x, y)=\cos x +\cos y - 1/2$) that defines a curve $f(x, y)=0$, then at any point $(a, b)$ on the curve, there exists an open set of $U\ni (a, b)$ and a unique continuously differentiable function $g:U\to \mathbb{R}^2$ such that $y=g(x)$ if the Jacobian of the implicit function, namely

$$\frac{\partial f}{\partial y}(a, b)\neq 0$$

in our 2-dimensional case. The differentiability of $f$ ensures the existence of this partial derivative.

Now, for our $f(x, y)=\cos x +\cos y - 1/2$, we have $y$ is a continuously differentiable function of $x$ if $\frac{\partial f}{\partial y}(a, b)=-\sin y\neq 0$. This leads to the assumption $\sin y\neq 0$ directly.

The problem you had is not with factoring or moving factors. At the moment you apply the operator $\frac{d}{dx}$, you automatically assume $y$ is a differentiable function of $x$ (which requires $\sin y\neq 0$). Thus, you can write $\frac{dy}{dx}=-\frac{\sin x}{\sin y}$ at the points where $y$ is differentiable.


问:
Intuition behind multiplicative unitaries
A multiplicative unitary on a Hilbert space $H$ is a unitary $V: H \otimes H \to H\otimes H$ such that the pentagon identity
$V_{[12]} V_{[13]}V_{[23]}= V_{[23]} V_{[12]}$ holds. Here the leg-numbering notation is employed. For example,
$$V_{[12]}= V \otimes 1 \in B(H \otimes H \otimes H).$$

Is there any intuition behind this pentagon identity? I can't memorise the identity, but I need it often (in the context of quantum groups), and I think if I would have some intuition I would be able to memorise it better.

答:
If $G$ is a locally compact group and $H=L^2(G)$ (for the right invariant Haar measure), then $H\otimes H$ identifies naturally
with $L^2(G\times G)$ and one may  consider the so called Kac-Takesaki
operator
$$
  V:L^2(G\times G)\to L^2(G\times G)
  $$
defined by
$$
  (V\xi )(x, y)=\xi (xy, y),\quad \forall \xi \in  L^2(G\times G) ,\quad \forall x,  y\in G.
  $$
One may easily prove that $V$ is unitary.

Notice that  $V$ codifies the multiplication operation of $G$, so it may be seen as an incarnation of the very
group structure of $G$, so much so that  $G$ itself can  be recovered from $V$.

It is easy to show  that both
$V_{[12]} V_{[13]}V_{[23]}$ and $V_{[23]} V_{[12]}$ send the square integrable function
$\xi $ on $G\times G\times G$ to the function $\eta $ given by
$$
  \eta (x, y, z)=\xi (xyz, yz, z),
  $$
so $V$ satisfies the pentagon identity.

When $H$ is an abstract Hilbert space, and $V$ is an operator on $H\otimes H$ satisfying the pentagon identity, one may therefore
suspect that something resembling a group must be in the background!

[1] Baaj, Saad; Skandalis, Georges, Multiplicative unitaries and duality for crossed products of (C^*)-algebras, Ann. Sci. Éc. Norm. Supér. (4) 26, No. 4, 425-488 (1993). ZBL0804.46078.

[2] Sołtan, Piotr M.; Woronowicz, Stanisław L., From multiplicative unitaries to quantum groups. II, J. Funct. Anal. 252, No. 1, 42-67 (2007). ZBL1134.46044.


问:
Is the answer to this problem really an answer?
Looking through some long-forgotten math texts of mine, I discovered this problem:

Suppose a spherical floating buoy has radius 1 m and density $\frac{1}{4}$ that of sea water. Given that the formula for the volume of a spherical cap is $V_{\text{cap}} = \frac{\pi{h}}{6}(3r^2 + h^2)$, to what depth does the buoy sink in water?

After puzzling over this for a half hour I gave up and looked at the answer. To my surprise, it didn't actually seem directed at the question; it was: "When $x \approx 0.65$, or when the sphere is at a depth of approximately 0.65 metres." Can the problem be solved as stated? I suspect that the authors neglected to some information.

答:
I recommend looking up the formula on Wikipedia, which makes clear that the $r^2$ in your formula is not the squared radius of the sphere. Rearranging gives the last expression in the article of
$$V_{\text{cap}}=\pi \frac{h^2}{3}(3r_\text{radius}-h)$$
Here $h$ is the depth by which the sphere is submerged.

The mass of the whole buoy is $\frac 4 3 \pi r_\text{radius}^3 * \frac 1 4 \rho_{\text{water}}$.

The mass of water displaced is $V_{\text{cap}} \rho_{\text{water}}$.

When are these equal? You should find an equation $h^3-3h^2+1=0$. This equation has three real roots, of which the physical one is the one in $[0,2]$, namely $0.652704\ldots$. (This has an exact expression involving complex numbers but I suspect you were meant to approximate the solution somehow.)


问:
Let $A$ be finite. Suppose $B$ is countably infinite. Is the set $B^A$ uncountably infinite?

Let $A$ be a finite set. Suppose $B$ is countably infinite. Is the set $B^A$ uncountably infinite?

I am new to infinite cardinality arguments, but my initial thinking is yes. However, I am uncertain if the claim requires the Axiom of Choice... I am concerned that I am using the Axiom of Choice without realizing it, and I would like to not invoke the axiom if possible. The following is my pseudo-informal argument.

Because $B$ is countably infinite, there exists an injection (possibly bijection) from  $B$ to $\mathbb N$. (i.e. $B \preceq \mathbb N)$. Let $F$ be one such function so that $F:B \xrightarrow{1-1} \mathbb N$.

Knowing that $\mathbb N$ is well-ordered, I can conclude that all subsets of $\mathbb N$ are well-ordered. Thus every member of $\mathcal P(\mathbb N) \setminus \{\emptyset\}$ has a minimal element. Further, by Cantor's Diagonal argument $\mathcal P (\mathbb N)$ is uncountably infinite, which implies that $\mathcal P(\mathbb N) \setminus \{\emptyset\}$ is uncountably infinite as well. I believe a similar argument can be made for $\mathcal P(B)$.

Let $p_{_B}$ be an arbitrary element of $\mathcal P( B) \setminus \{\emptyset\}$. I know that restricting $F$'s domain to $p_{_B}$ will produce a subset in $\mathbb N$. Call this subset $\mathbb N_{p_B}$. (i.e. $F[p_{_B}]=\mathbb N_{p_B}$)

Let $n_{0}$ be the minimal element of $\mathbb N_{p_B}$. Let $b \in p_{_B}$ be the element that is mapped to through $F^{-1}$...i.e. $F^{-1}(n_0)=b$. Because $F$ is injective, this is unambiguous.

Now, consider the construction of a function $g_{p_B}$. Let $g_{p_B}$ be the function that takes all elements of $A$ and maps them to the $b \in p_{_B}$ that corresponds to minimal element $n_0$ for a given $\mathbb N_{p_B}$. (This is where I am concerned...I think I am using a valid invocation of the Axiom of Replacement, but perhaps it is actually the Axiom of Choice)

Because there are an uncountably infinite number of $p_{_B}$'s in  $\mathcal P( B) \setminus \{\emptyset\}$, there must necessarily be an uncountably infinite number of $g_{p_B}$'s that satisfy the above definition. The collection of all such functions is a subset of $B^A$. Therefore, if a subset of a given set is uncountably infinite, the set itself must be uncountably infinite.

Edit

Just to highlight an issue in my reasoning, here is one of my afterthoughts (confirmed by Arturo Magidin):

"Now that you mention it, it does seem like there is necessarily a chance that many of the $g_{p_B}$'s are redundant...i.e. there may be a $p_{_B}\neq p'_{_B}$ that can be represented by the same $g$ function. Perhaps that is an issue as well."

答: The answer is negative: if $A$ is finite and $B$ is countably infinite then $B^A$ is either finite or countably infinite, and hence is certainly not uncountably infinite. You can prove this by induction on the size $|A|$ of $A$. If $|A| = 0$, then $A$ is empty and $B^A$ comprises the single function $\emptyset$. If $|A|= k + 1$, pick $a_0$ in $A$ and observe that any function $f_0 : A \setminus \{a_0\} \to B$ has only countably many extensions to a function $f : A \to B$ (because we have only countably many choices for $f(a_0)$). By the inductive hypothesis there are only countably many such $f_0$, and then using the fact that $X \times Y$ is countable if both $X$ and $Y$ are, you have that there are at most countably many such $f$.


问:
Show $\frac{1}{\log(x)}$ is in $L^p ((0,0.5))$ for $p \geq 1$
Show $\frac{1}{\log(x)}$ is in $L^p ( (0,0.5))$ for $p \geq 1$ without using the $\operatorname{Li}(x)$ function or using $u = \log(x)$ substitution (that approach gives an infinite series and is not good for evaluating $x=0$).

答: $f\colon x\mapsto\frac1{\lvert\log x\rvert}$ is continuous on $(0,\frac12)$ and approach a finite limit on either end.  So $f$ is uniformly continuous hence $f^p$ is integrable (and the case $p=\infty$ is immediate from uniform continuity of $f$).


问:
Question about Twin Primes.
Let:

```<code>P = (11, 17, 29, 41, 59, 71, 101, ...} and Q = {13, 19, 31, 43, 61, 73, 103, ...}.
</code>```

It is well known that both P and Q are infinite sets (please, view the following two references):

Math Forum

Math Exchange

The map defined by:

```<code>  f : P -> Q , f(K) = (K + 2);

  f^(-1) : Q -> P , f^(-1)(K) = (K - 2);
</code>```

defines a bijection between P and Q.

My question is :
Why can' t  we  conclude (by the above facts ) that the set:

```<code> Twin Primes = {(11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103) ...}
</code>```

is an infinite set?

Thank you!

答:
It is well known that both P and Q are infinite sets

If you are defining the sets P and Q so that for any positive integer $n$, the $n$-th element in P and the $n$-th element in Q form a pair of twin primes, then the above statement represents circular reasoning, where you are assuming that there are an infinite number of pairs of twin primes.

If you are not so defining the sets P and Q, then you can not use the bijection argument to establish that there are an infinite number of pairs of twin primes.

Title: Question about Twin Primes.


问:
Why is this relation $R=\{(a,b), (b,c), (c,a)\}$ transitive?
I have a set of relations, shown below:

$R=\{(a,b), (b,c), (c,a)\}$ for $A= \{a,b,c\}$

According to my professor, this relation is transitive but I don't understand why. I was under the impression that any given three relations of a set of relations have to follow the $aRb, bRc, aRc$ pattern but since $aRc$ isn't in the set then it'd fail the test since $aRc$ and $cRa$ aren't equivalent. I'm sure I'm missing something, but I don't understand what.

答:
As expressed, the relation $R$ fails to be transitive.  Indeed, we need $(a, c) \in R$ and instead of $(c, a) \in R$, as $(c, a)$ poses a number of problems in terms of the relation being transitive:

As written:

$(a, b), (b, c) \in R,$ but $(a, c) \notin R$, as you observe.

Furthermore: $(c, a), (a, b) \in R$ but $(c, b) \notin R$.

And, $(b, c), (c, a) \in R$, but $(b, a)\notin R$.

For each reason above, transitivity fails.

So $(c, a) \in R$ seems to be the ill-placed, perhaps misprinted pair in $R$, which if replaced by $(a, c)$ would alleviate these failures, and the relation would then be transitive.


问:
Golden Mean - Prove, in general, that For all $\in N | τ^{n+1} = τ^n + τ^{n-1} $
Prove in general that For all $\in N | τ^{n+1} = τ^n + τ^{n-1}  $

I've been trying to work on this problem over the past few days and I seem to be missing something. I know that $ τ = (1 + √5)/2$

I also know τ is a root of $p(x) = x^2 - x -1 $

However even with this information I can't seem to figure it out. How would I prove this?

So far this is what I have

$0=t^2 + t^{n-1}  + t^{n+1} \rightarrow 0 = t^n(1 + t^{-1} + t) \rightarrow$

$ 0 = t^n(1 + \frac{1}{t} + t)  \rightarrow 0 = t^n(\frac{t + 1 + t^2}{t})$

答:
Divide each term in your equation by $\tau^n$ and you'll get
$$\tau = 1+\frac{1}{\tau}.$$
Multiply both sides with $\tau$:
$$\tau^2 = \tau+1.$$
If you solve this easy quadratic equation, you will see that $$\tau_1=\varphi:=\frac{1+\sqrt 5}{2},\,\tau_2=\Phi:=\frac{1-\sqrt 5}{2}.$$


问:
Constructing a proper dense subalgebra from a subalgebra inside an ideal
Given a C* algebra $A$ and *-sub-algebra $B$, where $B$ is contained in a two-sided, closed ideal $I$, can we construct a proper, dense sub-algebra that contain $B$?

I am considering the following construction: Given $A$, we can always find a proper dense set $D$($A$ is impossible to equip with discrete topology), then we adjoin $D$ to $B$,so $B[D]:=\{$polynomials of $D$, with coefficients in $B$.$\}$ is a dense sub-algebra of $A$. The only question is I am not sure if it is proper. Is that a possible construction?

答:
If $A/I$ is infinite dimensional, choose a proper dense sualgebra $D\subseteq A/I$ (can you prove that such a thing exists?) and let $E=q^{-1}(D)$.  Then $E$ is a proper
dense sualgebra containing $I$, and hence also $B$.


问:
Existence of norm for C*-Algebra
I was wondering wether one can always find/construct a norm which turns an involutive algebra into a C*-algebra. For sure, if it exists it is unique, but does it always exist. If not can you provide a simple example of an involutive algebra such that there's no norm which turns it into a C*-algebra. Thanks alot!
Kind regards, Alex</p>

答:
(Below by "algebra" I mean "complex algebra.")

You can't even always find a norm that turns an algebra into a Banach algebra. For example, the Weyl algebra $\mathbb{C} \langle x, y \rangle/(xy - yx - 1)$ can't embed into a Banach algebra (see this math.SE question).

Another constraint on being a Banach algebra is that the spectral radius (a purely algebraic concept) of every element of a Banach algebra is finite (in fact bounded by its norm), and this fails in general. For example, the spectral radius of every nonzero element of $\mathbb{C}[x]$ is infinite.

Yet another constraint is that, by the Gelfand-Mazur theorem, the only Banach algebra which is also a division algebra is $\mathbb{C}$. This means that, for example, $\mathbb{C}(x)$ cannot be a Banach algebra, but it means more generally that any commutative algebra $A$ with a maximal ideal $m$ such that $A/m$ is not isomorphic to $\mathbb{C}$ cannot be a Banach algebra (since if $A$ were a Banach algebra then $m$ would be closed, so $A/m$ would also be a Banach algebra).

In the case of C*-algebras there are even heavier restrictions. For example, the finite-dimensional C*-algebras are products of matrix algebras, and these are precisely the semisimple algebras. Most finite-dimensional algebras are very far from semisimple; the smallest example is $\mathbb{C}[x]/x^2$ (which can be made into a Banach algebra).

More generally, the Jacobson radical of a C*-algebra is always zero (so a C*-algebra is always semiprimitive). This is a strong and purely algebraic constraint, and it is easy to write down many examples of algebras which are not semiprimitive, e.g. a commutative algebra with nilpotent elements can't be semiprimitive.

Title: Existence of norm for C*-Algebra


问:
Powers of transcendental numbers that lead to integers
For a given real number $x>0$ define $a_k(x)$ as follows for $k \geq 1$:
$a_1(x)=x$, $a_{k+1}(x)=x^{a_k(x)}$, so that $a_2(x)=x^x$, $a_3(x)=x^{x^x}$....

Question 1: Is there a explicit transcendental number $x$ such that $a_k(x)$ is an integer for a $k>0$?

Question 2: Let $Y:= \{x \in \mathbb{R} \mid x>0$, $x$ is transcendental and there exists $k >0$ such that $a_k(x)$ is an integer $\}$. Is $Y$ measurable and if yes, what is its measure?

The question is motivated by https://www.youtube.com/watch?v=BdHFLfv-ThQ where it is discussed whether $a_4(\pi)$ is an integer.

答: For question 2, the answer is that the set is countable, so it is measurable and has measure $0$.  Given any $n\gt 1$ and $k\gt 1$ we can see $a_k(x)$ is monotonically increasing with $x$, so we can implicitly invert it.  We can find $x$ numerically to whatever precision we want, but generally you won't find a formula for it.


问:
Help finding linear transformation Equation T:V -> V
Let $V$ be a vector space, and $$T:V→V$$ a linear transformation such that $$T(2v_1−3v_2)=−3v_ 1+2v_ 2$$ and  $$T(−3v_ 1+5v_2)=5v_ 1+4v_2$$

Solve $$T(v_1), T(v_2), T(−4v_ 1−2v_2)$$

I tried to put it into an Augmented matrix, take the inverse, and apply the inverse in order to work backwards from there but I still don't have any luck.. Does anyone have tips?

答:
You list some very overpowered things that you don't need for this problem. Remember that linearity is a very strong condition.

Hint: how do you "make" $v_1$ out of copies (linear combinations) of $2v_1-3v_2$ and $-3v_1+5v_2$?

 Multiply the first thing by $5$, and the second thing by $3$

 $5(2, -3) + 3(-3, 5) = (1, 0)$

 $T(v_1) = T(5(2v_1-3v_2)+3(-3v_1+5v_2))= 5(T(2v_1-3v_2))+3(T(-3v_1+5v_2))=5(-3v_1+2v_2)+3(5v_1+4v_2)$ where the last inequality comes from your definition of $T$, and linearity of $T$

Now, can you do the other problems?


问:
Number of polygons in a random collection of dots
I assumed it was $\sum\limits_{i=3}^n \binom{n}{i}$. Am I right?  I'm not sure if I'm using the right terminology to search online.  Does this problem have a name? Could someone point me to a standard proof and/or recommend some good tests for it.

答:
You are assuming that any selection of dots makes exactly one polygon.  You need to clarify what sorts of polygon you allow.  If three dots are in a straight line the triangle is degenerate.  The four corners of a square also allow a second polygon if you allow sides to cross.  If you insist the polygon be convex, there are sets of points that do not make a polygon.  Even if you prohibit self-intersections there are groups of points that determine more than one polygon.  The two below can be translated so the vertices match.

If you allow self-intersecting polygons and prohibit three colinear points it is simple.  Pick a set of points as in your sum, but multiply each term by $(i-1)!$ for the ways to order the points in a circle and divide by $2$ for the number of ways to go around the circle.  This gives http://oeis.org/A002807, which begins $1, 7, 37, 197, 1172, 8018, 62814, 556014, 5488059, 59740609, 710771275$


答: Not an answer (so don't downvote), but a demonstration that @crystalsnowlion's "answer" is incorrect.  After all, once you've selected $i$ points, there are typically more than one polygon determined by them:

Title: Number of polygons in a random collection of dots

Body: I assumed it was $\sum\limits_{i=3}^n \binom{n}{i}$. Am I right?  I'm not sure if I'm using the right terminology to search online.  Does this problem have a name? Could someone point me to a standard proof and/or recommend some good tests for it.


问:
Why cant we apply the square root of a negative number to equate more imaginary numbers?
In mathematics, polynomials like $x^2-1$ would have a clear solution of $x=\pm 1$. However, without complex numbers could you solve $x^2+1$? No, there would be no possible solution without adding a new axis of imaginary numbers. Simply, we added this axis because it provided us with solutions within our field of mathematics, so then we assumed them to exist.

So we know that you can only assume something to exist if it produces a solution (the most important reason for the imaginary axis to exist), then my question is why don't we do the same for an additional axis called $j$?

For example, we created the imaginary numbers by $\sqrt{-1}=i$, which then we unioned this imaginary axis to our real number axis to form the complex plane $\mathbb{C}=a+bi$. So then, why can't we square root a negative complex number to equate $j$. Visually:
\begin{gather*}
\sqrt{-(a+bi)}=j\\
\textrm{We then union this new axis with the previous to form the new set of numbers:}\\
\mathbb{C}_2=a+bi+cj:a,b,c\in\mathbb{R}\\\\
\textrm{We can then further extend this to even greater numbers:}\\
\mathbb{C}_3=a+bi+cj+dk: a,b,c,d\in\mathbb{R}\\
\mathbb{C}_\infty=a+bi+cj+dk+...+z(\infty): a,b,c,d,...,z\in\mathbb{R}
\end{gather*}
And this repeats on and on, where you take the square root of a negative number and from this method you keep on formulating a new type of  axis of a number.

Why don't we already have this in mathematics? We've done it once with solely the imaginary plane, why not extend it to a three axis field? Why do we limit ourselves?

I'm more than sure there are problems out there that can be solved with a $\mathbb{C}_2$ number!

I must say the beauty of this is mathematically speaking we can then equate infinity or at least have a new symbol for it. $\mathbb{C}_\infty=\infty$

答:
+1 to your query.

Generally, you only create complications (i.e. $\mathbb{R} \to \mathbb{C}$) to facilitate solving problems.  When attacking for example $\sqrt{x + iy}$ there is absolutely no need to extend the Complex Numbers.

Any non-zero complex number $z = (x + iy)$ can be converted to polar coordinates
$re^{i\theta} : r \in \mathbb{R^+}$.

Therefore, by DeMoivre's theorem, $\sqrt{z} = \pm\sqrt{r} \times e^{(i\theta/2)}.$


答: To answer your question: the complex numbers are algebraically closed. But this should not stop you from defining new "complex" numbers, such as quaternions, octonions, sedenions etc. This is called the Cayley-Dickson construction.</p>

Title: Why cant we apply the square root of a negative number to equate more imaginary numbers?

Body:
In mathematics, polynomials like $x^2-1$ would have a clear solution of $x=\pm 1$. However, without complex numbers could you solve $x^2+1$? No, there would be no possible solution without adding a new axis of imaginary numbers. Simply, we added this axis because it provided us with solutions within our field of mathematics, so then we assumed them to exist.

So we know that you can only assume something to exist if it produces a solution (the most important reason for the imaginary axis to exist), then my question is why don't we do the same for an additional axis called $j$?

For example, we created the imaginary numbers by $\sqrt{-1}=i$, which then we unioned this imaginary axis to our real number axis to form the complex plane $\mathbb{C}=a+bi$. So then, why can't we square root a negative complex number to equate $j$. Visually:
\begin{gather*}
\sqrt{-(a+bi)}=j\\
\textrm{We then union this new axis with the previous to form the new set of numbers:}\\
\mathbb{C}_2=a+bi+cj:a,b,c\in\mathbb{R}\\\\
\textrm{We can then further extend this to even greater numbers:}\\
\mathbb{C}_3=a+bi+cj+dk: a,b,c,d\in\mathbb{R}\\
\mathbb{C}_\infty=a+bi+cj+dk+...+z(\infty): a,b,c,d,...,z\in\mathbb{R}
\end{gather*}
And this repeats on and on, where you take the square root of a negative number and from this method you keep on formulating a new type of  axis of a number.

Why don't we already have this in mathematics? We've done it once with solely the imaginary plane, why not extend it to a three axis field? Why do we limit ourselves?

I'm more than sure there are problems out there that can be solved with a $\mathbb{C}_2$ number!

I must say the beauty of this is mathematically speaking we can then equate infinity or at least have a new symbol for it. $\mathbb{C}_\infty=\infty$


答:
A very simple answer to question $1$

Question $1$: Is there a explicit transcendental number x such that $a_k(x)$ is an integer for a $k>0$?.

Let $k=1$, thus we have $a_1(x) = x$. Thus, you may have $x = \pi, e,$ or $ln(N)$ for any positive natural number $N$.  There may be more, but those are just some examples because any transcendental number may be substituted in for $x$.

Title: Powers of transcendental numbers that lead to integers

Body:
For a given real number $x>0$ define $a_k(x)$ as follows for $k \geq 1$:
$a_1(x)=x$, $a_{k+1}(x)=x^{a_k(x)}$, so that $a_2(x)=x^x$, $a_3(x)=x^{x^x}$....

Question 1: Is there a explicit transcendental number $x$ such that $a_k(x)$ is an integer for a $k>0$?

Question 2: Let $Y:= \{x \in \mathbb{R} \mid x>0$, $x$ is transcendental and there exists $k >0$ such that $a_k(x)$ is an integer $\}$. Is $Y$ measurable and if yes, what is its measure?

The question is motivated by https://www.youtube.com/watch?v=BdHFLfv-ThQ where it is discussed whether $a_4(\pi)$ is an integer.


问:
If $\lim_{x \to \infty} \frac{f'(x)}{x}=2$ does it follow that $\lim_{x \to \infty} \frac{f(x)}{x^2}=1$?
I need to show that the following statement is true or false. $$\displaystyle\lim_{x \to \infty} \frac{f'(x)}{x}=2 \Rightarrow \displaystyle\lim_{x \to \infty} \frac{f(x)}{x^2}=1$$

I considered $f(x)=x^2$, and it seems that the above statement is true. If so, how do I go about proving this? Or maybe there is a counterexample?

I appreciate your help.

答:
Since

$$\lim_{x \to \infty} \frac{f'(x)}{x} =2,$$

we have that $f'(x) > x$ for sufficiently large $x$.  With this, we may show that $f(x)$ increases without bound (i.e. $\lim_{x \to \infty} f(x) = \infty$).  Thus the limit of $f(x)/x^2$ meets the hypotheses of l'Hopital, so that
$$\lim_{x \to \infty} \frac{f(x)}{x^2}= \lim_{x \to \infty} \frac{f'(x)}{2x} = \frac{1}{2} \lim_{x \to \infty} \frac{f'(x)}{x} = 2 \cdot \frac{1}{2}=1.$$

As Peter Tamaroff mentions in the comments elsewhere on this page, the assumption (in l'Hopital's rule) that our numerator tends to infinity is actually superfluous: it suffices for the denominator to go to infinity.


问:
Given Planes, $\pi_1:x+2y-z=0,\pi_2:-x+2y-3z-4=0$, and $\pi_3:x+6y-5z-4=0$. Determine nature and co-ordinates of intersection of planes.
$\require{enclose}$
Consider the planes below

$\color{chocolate}{\pi_1:x+2y-z=0\\\pi_2:-x+2y-3z-4=0\\\pi_3:x+6y-5z-4=0}$

a. Without solving the system, and by examining the normals, determine the nature of the intersection of the three planes. Include diagrams with your answers.

b. Find the intersection of the planes.

These were given as Assignment. I attached my solution. Can somebody check is if the solution is correct? And is it good?

NOTE: In the image, Coefficient of $x$ is $A$, $y$ is $B$ and $z$ is $C$. The equation of $\pi_1$ is numbered as ${\enclose{circle}{\kern .06em 1\kern .06em}}$, $\pi_2$ as ${\enclose{circle}{\kern .06em 2\kern .06em}}$ and $\pi_3$ as ${\enclose{circle}{\kern .06em 3\kern .06em}}.$

答:
Your solutions are correct. A tip on how to check these answers for yourself: dot your solution line vector with the normals of the planes. If you get zero, it means they are perpendicular, so the line is in the plane (if at least one point on the line is in the plane).

Details: Your solution has vector $(-1,1,1)$, while your plane $\pi_1$ has normal vector $(1,2,-1)$. And note that $(-2,1,0)$ is on your line, but obviously that satisfies $x+2y-1z=0$.


问:
Arithmetic mean of $\left(1-\frac{1}{n^5}\right)$ with an $n-1$ number of $1's$
Okay, here's the question in a nutshell (The entire question is much longer but I've simplified it to the equation below, I'll explain what it is):

$$1-\frac{\left(1-\frac{1}{n^5}\right) + (n-1)}{n}$$

There are $n$ numbers of which $1$ of them is $\left(1-\frac{1}{n^5}\right)$  and all the others are $1's$. How much  is the arithmetic mean of these numbers less than $1$?

First of all, to me, the equation above seems impossible to simplify, but it might be easy for some of you, so could you show me step-by-step how to simplify it?

Second of all, if it IS indeed impossible, where have I gone wrong at interpreting the question? I suspect I did something wrong with the first half of the question, so please point out where I've gone wrong (IF impossible)

Thanks in advance!

Body: $$1 - \frac{(1 - n^{-5}) + (n-1)}{n} = 1 - \frac{n - n^{-5}}{n} = 1 - 1 + n^{-6} = \frac{1}{n^6}.$$  I'm not sure what else you need to do.


问:
"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-...$
"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$

Pose $$S=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$
$$C=\cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)+...$$

$$C+iS = e^{ix}-\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3-\dfrac{1}{4}(e^{ix})^4+...$$

Let $t$ = $e^ix$

Then we have a series of $t-\dfrac{t}{2}+\dfrac{t}{3}-\dfrac{t}{4}+...=\log(1+t)$

Which is $\log(1+e^{ix})=\log(1+\cos(x)+i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$

$\log([1+\cos(x)]^2+i\sin(x))=\dfrac{1}{2}\log(\log([1+\cos(x)]^2+i\sin^2(x))+\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$. Since we are interested only in the imaginary part, we have the sum for $S$ is:

$$\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$$

I don't know what to do next.

In his paper "Subsidium Calculi Sinuum", Euler wrote the series has the "sum"

$$\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...=\dfrac{x}{2}$$

We obtained this by multiplying $dx$ integrating (his words is Illa autem series per $dx$ multiplicata et integrata dat:"

$$\cos(x)-4\cos(2x)+9\cos(3x)-16\cos(4x)...=0$$

I don't know how he obtains $\dfrac{x}{2}$, since the left hand side is $0$, how can it be $x/2$. Returning to my own "sum", how can I obtain the $1/2$

This series is quite important because it appears in Fourier "Analytical Theory of Heat".

答:
You are almost there! We have:

$$\ln(1 + \cos(x) + i\sin(x))= \ln(A + iB) = \frac{1}{2}\ln(A^{2}+B^{2})+i\arctan\bigg(\frac{B}{A}\bigg)$$

$$=\frac{1}{2}\ln\big((1 + \cos(x))^{2} + \sin^{2}(x)\big)+i\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Then, the imaginary part is:

$$\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Using the identity $\displaystyle\tan\bigg(\frac{x}{2}\bigg) = \frac{\sin(x)}{1 + \cos(x)}$:

$$ = \arctan\bigg(\tan\bigg(\frac{x}{2}\bigg)\bigg) = \boxed{\frac{x}{2}}$$

Of course, with this approach you will run into the usual domain problems with inverse trig functions and complex logarithms.


答:
Note

$$S’(x) = \cos x - \cos2x +\cos 3x -\cos 4x+ \cdots\\
= Re(e^{i x} - e^{i 2x} + e^{i 3x} -e^{i 4x} +  \cdots)
= Re \frac 1{1+ e^{-i x}}= \frac 12
$$

Thus

$$S(x) = \int_0^x S’(t) dt = \frac12x$$

Title: "Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-...$

Body:
"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$

Pose $$S=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$
$$C=\cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)+...$$

$$C+iS = e^{ix}-\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3-\dfrac{1}{4}(e^{ix})^4+...$$

Let $t$ = $e^ix$

Then we have a series of $t-\dfrac{t}{2}+\dfrac{t}{3}-\dfrac{t}{4}+...=\log(1+t)$

Which is $\log(1+e^{ix})=\log(1+\cos(x)+i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$

$\log([1+\cos(x)]^2+i\sin(x))=\dfrac{1}{2}\log(\log([1+\cos(x)]^2+i\sin^2(x))+\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$. Since we are interested only in the imaginary part, we have the sum for $S$ is:

$$\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$$

I don't know what to do next.

In his paper "Subsidium Calculi Sinuum", Euler wrote the series has the "sum"

$$\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...=\dfrac{x}{2}$$

We obtained this by multiplying $dx$ integrating (his words is Illa autem series per $dx$ multiplicata et integrata dat:"

$$\cos(x)-4\cos(2x)+9\cos(3x)-16\cos(4x)...=0$$

I don't know how he obtains $\dfrac{x}{2}$, since the left hand side is $0$, how can it be $x/2$. Returning to my own "sum", how can I obtain the $1/2$

This series is quite important because it appears in Fourier "Analytical Theory of Heat".


问:
Function with compact support has uniformly continuous integral over $(-\infty,x]$?
If $f$ is a bounded Lebesgue measurable function that has compact support, then is $\int_{-\infty}^x f(t)dt$ uniformly continuous?

I have just learned of compact support and do not see how compact support can lead to uniform continuity since the integral here is not over a compact set? I believe this is a condition for a distribution function in probability and would like to understand why. Also, what if the function $f$ was just integrable, then would this integral still be uniformly continuous?

答:
Hint:

First, show that the function is continuous. Then, prove the following.

Let $g:\mathbb R\to\mathbb R$ be continuous. Suppose $\displaystyle \lim_{x\to \infty}g(x)$ and $\displaystyle\lim_{x\to -\infty}g(x)$ both exist and are finite. Then $g$ is uniformly continuous.

Apply this to $g(x):=\int_{-\infty}^x f(t)dt. $


问:
Paradox: increasing sequence that goes to $0$?
It is $10$ o'clock, and I have a box.

Inside the box is a ball marked $1$.

At $10$:$30$, I will remove the ball marked $1$, and add two balls, labeled $2$ and $3$.

At $10$:$45$, I will remove the balls labeled $2$ and $3$, and add $4$ balls, marked $4$, $5$, $6$, and $7$.

$7.5$ minutes before $11$, I will remove the balls labeled $4$, $5$, and $6$, and add $8$ balls, labeled $8$, $9$, $10$, $11$, $12$, $13$, $14$, and $15$.

This pattern continues.

Each time I reach the halfway point between my previous action and $11$ o'clock, I add some balls, and remove some other balls. Each time I remove one more ball than I removed last time, but add twice as many balls as I added last time.

The result is that as it gets closer and closer to $11$, the number of balls in the box continues to increase. Yet every ball that I put in was eventually removed.

So just how many balls will be in the box when the clock strikes $11$?

$0$, or infinitely many?

What's going on here?

答:
In my oppinion the answer is pretty simple.

If we have a sequence $A_n$ of sets, the cardinality doesn't always commute with the limit:

$$\lim_n \, {\rm card}(A_n) \neq {\rm card}(\lim_n (A_n)) \,.$$

And somehow this simple result is counterintuitive for us.


问:
Probabilities of no ace and exactly not one ace in throwing 10 dies
The probabilities i am looking for are used for a problem in Feller's introduction to probability that asks for the probability $p$ of finding $2$ or more aces in throwing $10$ dies. For this problem i considered that:

$p = 1-x-y$

Where $x$ is the probability of finding no ace at all, and $y$ being the probability of finding exactly one ace. Of course $x = \frac{5^{10}}{6^{10}}$... i think. But i am a little more confused about finding $y$, more because the answer to this exercise is that: $p = 1- 10\times\frac{5^9}{6^{10}-5^{10}} $ but i don't really understand where this result can come from, more considering the result i got for $x$

答:
Use $\begin{pmatrix}n \\ k\end{pmatrix}p^kq^{n-k}$
with n= 10, k= 1, p= 5/6 and q= 1/6.


答: The compact support is irrelevant: as long as $f$ is measurable, bounded and Lebesgue integrable (actually, you could relax the latter to $1_{(-\infty,0]}f$ being integrable), the function $g$ will be Lipschitz continuous because for $s<t$ $$\lvert g(t)-g(s)\rvert\le\int_s^t\left\lvert f(x)\right\rvert\,dx\le M\lvert t-s\rvert$$

Title: Function with compact support has uniformly continuous integral over $(-\infty,x]$?

Body:
If $f$ is a bounded Lebesgue measurable function that has compact support, then is $\int_{-\infty}^x f(t)dt$ uniformly continuous?

I have just learned of compact support and do not see how compact support can lead to uniform continuity since the integral here is not over a compact set? I believe this is a condition for a distribution function in probability and would like to understand why. Also, what if the function $f$ was just integrable, then would this integral still be uniformly continuous?


答:
You are thinking too hard.

The actual sum is $\underbrace{1+1+...... + 1}_{n-1\text{ ones}} + (1- \frac 1{n^5})=$

$\underbrace{1+1+......+1 (+ 1}_{n \text{ ones}}- \frac 1{n^5})=$

$n -\frac 1 {n^5}$.

So the mean is $\frac {n-\frac 1{n^5}} n = 1 - \frac 1{n^6}$.

And how much less than $1$ is that?

Well, it literally is $\frac 1{n^6}$ less than $1$.

Title: Arithmetic mean of $\left(1-\frac{1}{n^5}\right)$ with an $n-1$ number of $1's$

Body:
Okay, here's the question in a nutshell (The entire question is much longer but I've simplified it to the equation below, I'll explain what it is):

$$1-\frac{\left(1-\frac{1}{n^5}\right) + (n-1)}{n}$$

There are $n$ numbers of which $1$ of them is $\left(1-\frac{1}{n^5}\right)$  and all the others are $1's$. How much  is the arithmetic mean of these numbers less than $1$?

First of all, to me, the equation above seems impossible to simplify, but it might be easy for some of you, so could you show me step-by-step how to simplify it?

Second of all, if it IS indeed impossible, where have I gone wrong at interpreting the question? I suspect I did something wrong with the first half of the question, so please point out where I've gone wrong (IF impossible)

Thanks in advance!


问:
How to apply Trig Substitution with this question?
I'm working on this integral.
I know that we need to apply the rule $\sqrt{x^2-a^2}$ identity, however I'm not sure how to deal with the $-x^2$.

$$ \int \frac{1}{\sqrt{12x-x^2}} \text{d}x $$


问:
The derivative of $\int_0^{\sin x} \sqrt{1-t^2} dt$ is given as $\lvert \cos x\rvert \cos x$. But why the absolute value?
The given problem: Find the derivative of $$ F(x)=\int_0^{\sin x}\sqrt{1-t^2}dt $$ I know the answer is supposed to be $\lvert \cos x\rvert \cos x$, but I don't understand where the absolute value comes in.

Using the fundamental theorem, my working out was as follows:

\begin{align}
F'(x) & =\cos x \sqrt{1-sin^2x}\\
& = \cos x \sqrt{\cos^2x} \\
& = \cos x (\pm \cos x)\\
& = \pm \cos^2x
\end{align}

Which was evidently wrong, but I don't understand why the sqrt of $\cos^2x$ must be an absolute value. It is a square, so what would be the problem if it were negative?

Am I missing something very simple? I'm a long distance student so I can't exactly raise my hand in class and ask what's what.

Afterthought: Perhaps I'm grabbing the wrong end of the stick and it's the other cos term that's been made an absolute value, not the sqrt...even if that were the case though I'd still be confused.

答:
We can use Fundamental Theorem of Calculus and chain rule. We know that
$$ \frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}{f{(t)} \, \mathrm{d}t} = f{(x)}. $$
So,
$$
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{\sin{x}}{\sqrt{1 - t^{2}} \, \mathrm{d}t} &= \cos{x} \cdot \sqrt{1 - \sin^{2}{x}} \\
&= \cos{x} \cdot \sqrt{\cos^{2}{x}}
\end{align*}
$$
Because $ \sqrt{z^{2}} = z $ for every real number $ z $, so
$$ \cos{x} \cdot \sqrt{\cos^{2}{x}} = \left|\cos{x}\right| \cos{x}. $$


问:
Solving the equation $10^{-x} = 5^{2x}$ with logarithms
$$10^{-x} = 5^{2x}$$

I'm having trouble isolating $x$. I get both logs on one side and then I'm stuck because I have nothing to divide with on the other side, and I can't factor it.

Thanks

答: If $10^{-x} = 5^{2x}$, then $-x\log(10) = 2x\log(5)$ thus either $x=0$ or $-\log(10)=2\log(5)$ (or both). Since $-\log(10)\neq2\log(5)$ we must conclude that $x=0$.</p>


问:
Alternative Quadratic Formula
Well the formula for solving a Quadratic equation is :

$$\text{If }\space ax^2+bx+c=0$$

then

$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$

But looking at this : [Wolfram Mathworld] (And also in other places)

They give An Alternate Formula:

$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$

How does one get this?

Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?

Please help, Thanks!

答:
The alternate form can be very useful to improve numeric precision or stability in some cases; for instance, if $a$ is much smaller in magnitude than $b$ and $c$, the formula is very close to being linear. In such cases, one root is much smaller in magnitude than the other, and in practical cases, that's often the one of interest. But, in the conventional formula, that 'near' root is found by taking the difference of two values which are very close in magnitude, which can result in loss of precision.

Consider an example which arises in optics: given a sphere radius $R$ centred at $(0,0,R)$ we want to find where it intersects a line described parametrically as $\mathbf{x} = \mathbf{x}_0 + s\mathbf{d}$; and (relative to $R$) we know the ray is reasonably close to, and roughly parallel to, the $z$ axis, and we are interested in the intersection closer to the $z=0$ plane. It's assumed that $\mathbf{x}_0$ is close to this plane, too, so this will be the solution with a smaller magnitude of $s$.

Solved by plugging the parametric line into ${x_x}^2 + {x_y}^2 + (x_z-R)^2=R^2$, and expanding;
this leads to a quadratic in $s$ :
$$\begin{align}
  as^2 + bs + c & = 0 \\
  a & = \mathbf{d} \cdot \mathbf{d} \\
  b & =2(\mathbf{x}_0 \cdot \mathbf{d} - R d_z) \\
  c & =\mathbf{x}_0 \cdot \mathbf{x}_0 - 2 R {x_0}_z
\end{align}$$

In some cases $R$ can be very large (for a barely curved lens); or large and negative (concave vs convex); and we may want to explore the effect of changing from a slighly convex to slighly concave surface, including the intermediate 'flat' situation. And it's much better if the calculations are continuous through that space. So, we can define 'curvature' $C=1/R$, and then divide all three coefficients by $R$:

$$\begin{align}
  a & = C\mathbf{d} \cdot \mathbf{d} \\
  b & = 2(C \mathbf{x}_0 \cdot \mathbf{d} - d_z) \\
  c & = C \mathbf{x}_0 \cdot \mathbf{x}_0 - 2 {x_0}_z
\end{align}$$

Now we have a situation where $a$ becomes small as $C$ tends to zero; and it's essential to use the 'alternate' form to find $s$; the quadratic equation we are solving tends to linear as $a$ tends to zero; at which point the solution is just $s=-c/b$; and the 'alternate' formula clearly gives this result at that limit (with proper choice of the sign of the radical) while the 'conventional' form tends to 0/0.
So $C=0$ is no longer a special case, it's continuous with slightly negative and slightly positive $C$ (and we can, for instance, evaluate an analytic derivative for $\frac{ds}{dC}$ even at $C=0$, if needed for non-linear optimization).

For small $C$, the sign of $b$ is determined by the sign of the $z$ component of $\mathbf{d}$ -- is the ray going up or down? -- something which is known to the solving framework, and the 'near' root is the one obtained when the radical has the same sign as $d_z$, even as $C$ changes sign.

Title: Alternative Quadratic Formula


问:
Why is it that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$?
Unfortunately I am stuck on one step of a proof for an algebraic limit theorem, specifically:

Why is it exactly that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$ ?

If this doesn't make sense without more context, please let me know. Otherwise, thank you for your help!

答:
Rewriting the antecedent, we want to show $\frac{|b|}{2} > |b_n-b| \Rightarrow |b_n| > \frac{|b|}{2}$.
\begin{align*}
\frac{|b|}{2} \color{red}{+|b_n| - \frac{|b|}{2}} = \color{LimeGreen}{|b_n|} &> |b_n-b| \color{red}{+|b_n| - \frac{|b|}{2}} \\
&= |b-b_n| \color{red}{+|b_n| - \frac{|b|}{2}} \\
&> |b-b_n+b_n| - \frac{|b|}{2} \\
&= |b| - \frac{|b|}{2} \\
&= \color{LimeGreen}{\frac{|b|}{2}}
\end{align*}


问:
Homology group of $X$
Can I say that the homology group  $ H_n(X)$  it gives us the information of how many holes the topological space $X$ has ? And for an example to say that the $H_n(\bigtriangleup^{n} )=0$ because n simplexis dont have holes and compare it to $H_n(X )=1$ where $X$ here is a torus which has one hole? Meaning if $ H_n(X)\neq H_n(Y) $ then $X$ can't be isomorphic to $Y$ ?

答:
Intuitively, you can think of $H_n(X)$ as measuring the $n$-dimensional holes of your space $X$. This observation is motivated by the homology groups we encounter on simple spaces such as spheres and tori etc.

Saying $H_n(\Delta^n) = 0 $ would mean there are no $n$-dimensional holes in your space $\Delta^n$. If $\Delta^n$ is supposed to be the $n$-simplex, note that $\Delta^n$ is homeomorphic to the $n$-dimensional ball $D^n$ which is homotopy equivalent to a point $*$ and homotopy equivalent spaces have isomorphic homology groups, therefore

$$H_n(\Delta^n) \cong H_n(D^n) \cong H_n(*) = 0\ \ \text{for}\ \ n > 0$$

A bit more algebraic background: What homology does actually measure is the failure of exactness of a given chain complex $(C_*,\partial_*)$:

$$\cdots \to C_n \xrightarrow{\partial_n}  C_{n-1} \xrightarrow{\partial_{n-1}} C_{n-2} \xrightarrow{\partial_{n-2}}  \cdots $$

Usually one consideres the chain complex of the (free abelian) chain groups $C_k(X)$ that are generated by the (singular) $k$-chains in $X$.

Now by definition, the $n$th homology group $H_n(X)$ is defined as the quotient $$H_n(X) := \ker \partial_{n}/\operatorname{im}\partial_{n+1}$$

Therefore, the "bigger" the homology group $H_n(X)$, the further your chain complex is away from being exact at $C_n(X)$, the "smaller" $H_n(X)$, the closer you are to exactness at $C_n(X)$. Consequently, if $H_n(X) = 0$, your chain complex is exact at degree $n$, i.e.

$$\ker \partial_{n} = \operatorname{im}\partial_{n+1}.$$

That's what homology groups measure algebraically.

Title: Homology group of $X$