问:
$\mathbb{R^2} $and $ \mathbb{R}\times\mathbb{R}$ difference?
I have read that you can 'identify' $\mathbb{R}\times\mathbb{R}$ with $\mathbb{R}^2$. My question is what does identify actually mean in linear algebra? And second question: isn't it the same? As an element $x\in \mathbb{R}^2 $ is defined as $(a,b)\in \mathbb{R^2} $ where $a,b \in \mathbb{R}$. Thanks

答:
There is a small notational difference here. Given two sets $A$ and $B$ the notation $A^B$ is the collection of all function from $B$ to $A$. When we write $A^n$ this refers to the set of all functions from a some set of cardinality $n$ to $A$. Say for $\mathbb{R}^3$ we might choose the set with three letters $\{x,y,z\}$ to be the set of three elements. An example of one of the functions in $\mathbb{R}^3$ would be $f(x)=1,f(y)=2,f(z)=3$.

Now instead of working over the set of functions we could instead write each function in coordinate form by creating the ordered tuple $(f(x),f(y),f(z))$ which for the example would correspond to the tuple $(1,2,3)$. This can be understood as the $x$ coordinate being $1$, the $y$ coordinate being $2$ and the $z$ coordinate being $3$. This gives us a bijection between the set of all functions from $\{x,y,z\}$ to $\mathbb{R}$ and the ordered triples of $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$. It's this bijection that allows us to identify $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ with $\mathbb{R}^3$.


问:
Modulo question with negative
$60-88  \equiv \,\,? \pmod 5$

$60-88 = -28$

Then what do I do?

Please tell me how to answer this question. Thanks.

答:
When you have a negative number, like $-28 \pmod 5$, all you really need to do keep adding the modulus, or an integer multiple of the modulus $m$, to the negative number until it is zero or in $\mathbb Z_m = \{0, 1, \cdots, m-1\}.\;$

In this case, we can add any multiple of the modulus $5$ to $-28$ until we obtain zero or a positive number, if you are looking to represent the equivalence class of $-28$ modulo $5$ with the least positive integer $x$, $\;x \in \{0, 1, 2, 3, 4\}$.

So given $$60 - 88 = -28 \equiv x\pmod 5\tag{1}$$ and wanting to find such a representative $x$, we know that $5k \equiv 0 \pmod 5$, and simply choose $k = 6$ as our multiple of $5$ so we have $$5 \cdot 6 = 30 \equiv 0 \pmod 5\tag{2}$$

Adding $(1), (2)$:

$$30 - 28 = 2 \equiv x \pmod 5$$

That is, $60 - 88 = - 28 \equiv 2 \pmod 5$.

What this all means is that any multiple $n$ of 5, added to $2$ is congruent modulo $5$ to $2$, including $-28$: The solutionS to all $$x \equiv 2 \pmod 5 = \{x\mid x = 2 \pm 5n, n \in \mathbb Z\}$$

ContentLicense: CC BY-SA 3.0

Title: Modulo question with negative


问:
Hi! I'm having a problem with the integral $\int_{n}^{n+1}x^2 \,dx$
Basically, I noticed that when $n\subset  \mathbb{N}$, $\displaystyle\int_{n}^{n+1}{x^2}\,dx$ gives a prime number divided by 3 or a multiplication of prime numbers.
Here's a table i made.

Is this something already known or a "function that generates prime numbers"?
I don't know much about math and I don't have a great academic background so I'm asking this here.

答:
Nice observation. But it does not hold for all $n$. There are plenty of counterexamples. For instance:

For $n=22$, three times the integral is $ 7 \cdot 7 \cdot 31$.

For $n=33$, three times the integral is $7 \cdot 13 \cdot 37$.

For $n=187$, three times the integral is $7 \cdot 13 \cdot 19 \cdot 61$.

ContentLicense: CC BY-SA 4.0


问:
Find two distinct, non-zero matrices, A and B, such that A*B+A+B = 0
this is a problem which is for homework in my math course. The problem states that you must find two distinct, non-zero matrices, (Size 2x2) such that A * B + A + B = 0.

I'm not really looking for an answer, but rather the methodology I should be using to come to this answer. It seems like the easiest way to do this would be through brute force, but I am somewhat slow when it comes to math and so I was hoping there might be an easier method out there. Thank you in advance.

答: Take any number $a\ne-1$. Then take $b=-\frac a{a+1}$. With this choice, $ab+a+b=0$. Now, take $A=\left[\begin{smallmatrix}a&0\\0&a\end{smallmatrix}\right]$ and $B=\left[\begin{smallmatrix}b&0\\0&b\end{smallmatrix}\right]$.


答:
$AB+A+B=0$ is equivalent to $(A+I)(B+I)=AB+A+B+I=I$ for the identity matrix $I=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$.

Thus, picking any invertible matrix $M$, the matrices $A=M-I$ and $B=M^{-1}-I$ are solutions. In fact, it is easy to show that these are the only solutions.

Title: Find two distinct, non-zero matrices, A and B, such that A*B+A+B = 0

Body:
this is a problem which is for homework in my math course. The problem states that you must find two distinct, non-zero matrices, (Size 2x2) such that A * B + A + B = 0.

I'm not really looking for an answer, but rather the methodology I should be using to come to this answer. It seems like the easiest way to do this would be through brute force, but I am somewhat slow when it comes to math and so I was hoping there might be an easier method out there. Thank you in advance.


问:
How to find the elapsed time for a medication when two people have to met a certain dose?
The problem is as follows:

In a children's hospital in Tokyo a doctor has prescribed to two kids
a medication to alleviate their sore throat infection. According to
this Hiroto and Midori must begin a treatment taking an amoxicillin
oral suspension. Hiroto must take $10$ mililiters each $8$ hours and
Midori must take $7$ mililiters each $6$ hours. They begin their first
dose together and they will finish the treatment when the two of them
have taken exactly two and a half bottles of this medicine. Assume
that each bottle of amoxicillin has a content of 120 militers. Find
how long the treatment of Midori lasted.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{84 hours}\\
2.&\textrm{108 hours}\\
3.&\textrm{96 hours}\\
4.&\textrm{114 hours}\\
\end{array}$

I'm confused exactly how to approach this problem, and this arises for the fact that when you take a medication which lasts for a certain time, you must account for each time interval. But the number of doses is those intervals increased by one. And this must be met in order to avoid falling into the telephone post error.

Then this means when the two kids took their dose as follows:

Hiroto:

$$10+\frac{10}{8}t$$

Midori:

$$7+\frac{7}{6}t$$

This is the part where it comes the right interpretation from the words, the two of them have taken the two and a half bottles of that medication.

I'm assuming that those two quanitities must be added in order to get the requested time. For both kids the elapsed time is the same.

Two and a half bottles of that medicine is:

$$120\times 2 + 120\times \frac{1}{2}=240+60=300$$

Then:

$$7+\frac{7}{6}t+10+\frac{10}{8}t=300$$

But this reduces to:

$$\frac{29}{12}t=283$$

Thus the time is:

$$t=\frac{283\cdot 12}{29}$$

But needless to say that this answer does not appear on any of the alternatives. What could be happening here?. Is it just me or what?. Can someone help me here to solve this problem?.

I think I did the right assumptions, but I could be wrong. It would help a lot that the answer could contain a wordy and highlighted explanation of the right interpretation so I can catch up where am I not getting this right.

答:
It does look like there is a problem with the question.

Your functions are continuous, but the dosing is discrete. In other words, your functions are not correct to give the amount received as a function of $t$. The correct functions are step functions, not linear functions.

Still, you get close to the right answer, as your functions are similar to the actual dosage step functions. Your answer is a bit over $117$ hours, so lets see what the kids have gotten after $114$ hours.

By $t=114$ hours, Hiroto has taken $10$ml at times $t=0,8,16,\dots112$, which is $15$ doses or $150$ml. By the same time, Midori has taken $7$ml doses at times $t=0,6,12,18,\dots,114$, which is $20$ doses, or $140$ml. At this point, $290$ml of medication has been given.

Six hours later (at $t=120$), it’s time for both kids’ next dose, which will take the amount of medication delivered over $2.5$ bottles, so it doesn’t seem like it’s possible for there to be a specific point where the children have gotten a total of $2.5$ bottles. If when they are scheduled to take doses at the same time, Hiroto goes first, and then the $2.5$ bottles run out before Midori’s $t=120$ dose, perhaps we could say that Midori's treatment ended after $114$ hours. (Hiroto’s treatment ended after $120$ hours, but the question did not ask about this.)

Added, in response to a comment: Using Midori as the example, you give this function for the amount of medication received after $t$ hours: $m(t)=10+\frac{10}{8}t$. As I noted, this is incorrect, because she isn’t given medication continuously. Instead, after $t$ hours, she has received $10$mg plus $10$ additional milligrams for each complete $8$-hour period. The correct function is $m(t)=10+8\displaystyle\left\lfloor{t\over10}\right\rfloor$, where $\lfloor\cdot\rfloor$ is the “floor” or “greatest integer” function.

However, there’s no easy way to solve an equation like
$$7+7\displaystyle\left\lfloor{t\over6}\right\rfloor+10+8\displaystyle\left\lfloor{t\over10}\right\rfloor=300$$
in a similar way to solving
$$7+\frac{7}{6}t+10+\frac{10}{8}t=300$$


问:
how to draw the graph of this function with rationals and irrationals
if someone can draw the graph of this function for me, thank you

I tried something like this but I'm not sure;

答:
The graph of your function will look like this:

Note that this function is (seriously) discontinuous at all points $x \neq 0$, because $\mathbb{Q}$ lies dense in $\mathbb{R}$, i.e. every interval $(a- \epsilon, a+\epsilon)$ contains a rational number for every $a \in \mathbb{R}$ and $\epsilon>0$. Because of this, the function will "jump" between the red and blue graph infinitely often in any (no matter how small) interval you choose.

If you like to read more about fuctions with this kind of property, I recommend https://en.wikipedia.org/wiki/Dirichlet_function.


答:
91 is also 13 times 7, so it seems to be coincidence that you are finding a few prime numbers.  All other numbers can always be written as a product of prime numbers, (but not necessarily just two) and there isn't a known formula for generating prime numbers.

Your formula for $3n$, which should be $3f(n)$, is $3n^2 + 3n + 1$, if n=26 you get 2107 which is 43 times 7 squared and there are probably higher ones which need 3 or more primes multiplied together, but nice try!

Title: Hi! I'm having a problem with the integral $\int_{n}^{n+1}x^2 \,dx$

Body:
Basically, I noticed that when $n\subset  \mathbb{N}$, $\displaystyle\int_{n}^{n+1}{x^2}\,dx$ gives a prime number divided by 3 or a multiplication of prime numbers.
Here's a table i made.

Is this something already known or a "function that generates prime numbers"?
I don't know much about math and I don't have a great academic background so I'm asking this here.


问:
How many epsilon numbers $<\omega_1$ are there?
An epsilon number is an ordinal $\epsilon$ such that $\epsilon=\omega^\epsilon.$ What is the cardinality of the set of all epsilon numbers less than $\omega_1$?

I'm asking this because of a proof I've just read that seems to presuppose that there are countably many such ordinals, and it seems to me intuitively that there should be uncountably many (although I don't know how to prove it).

Added. OK, I've just understood that the proof I mentioned is OK even if there are uncountably many such ordinals, but I still don't see how I can find their number.

答:
Note that $\varepsilon_0$ is countable. So if there are only countably many countable $\varepsilon$ numbers, they would have a countable supremum, $\alpha$. Consider now the same construction as $\varepsilon_0$, starting $\alpha+1$. That is:
$$\sup\{\omega^{\alpha+1},\omega^{\omega^{\alpha+1}},\ldots\}$$

The result is itself an $\varepsilon$ number, and it is countable (as the countable limit of countable ordinals). But all the countable $\varepsilon$ numbers were assumed to be below $\alpha$, which is a contradiction.


问:
Similarity by SSA (side-side-angle) in obtuse triangles
Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54:

"In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other acute angles are proportional. Prove that the triangles are similar."

I tried this first by cosine rule, and the question reduced to:

Assume $\triangle ABC$ and $\triangle DEF$ with $\angle A$ and $\angle D$ being obtuse, $\angle B=\angle E$ and $AC/BC=DF/EF$. We may take $DF=k\cdot AC$, $EF=k\cdot BC$ and apply $\cos B= \cos E$, obtaining $DE=k\cdot AB$ and the problem is solved. EDIT: Even this however is difficult to prove, as pointed out by cosmo5 in his answer.

What is interesting and challenging is to prove the same without the cosine law (probably the expected solution, as cosine law doesn't seem to help), as the law hasn't been introduced in the book until later chapters, while this problem is taken from chapter 3.

My attempt:

Fix $\triangle ABC$ as well as points $E$ and $F$ of $\triangle DEF$. Assuming $DF=k\cdot AC$, $EF=k\cdot BC$, the locus of $D$ will be a circle with centre $F$ and radius $k\cdot AC$.

After this however, I couldn't go further with the same concept. Either a hint or a solution related to my thought process, or any other thought process for that matter, would be greatly appreciated.

答: They are congruent if the known congruent angles are obtuse. If the known congruent angles are acute, then the triangles are congruent if the triangles are both acute or both obtuse.  Why this is not more widely distributed is above my paygrade!</p>


问:
Integral notation in Shifrin's *Multivariable Mathematics*
On page 348 of Shifrin's Multivariable Mathematics, for a 1-form $\omega=\sum F_{i}dx_{i}$ on $\mathbb{R}^{n}$ and a parameterized curve $C$ given by a function $\mathbf{g}:[a,b]\rightarrow\mathbb{R}^{n}$, he defines$$\intop_{C}\omega=\int_{[a,b]}\mathbf{g}^{*}\omega=\int_{a}^{b}F_{i}\left(\mathbf{g}\left(t\right)\right)g_{i}^{\prime}\left(t\right)dt.$$

Why does he write $$\int_{[a,b]}\mathbf{g}^{*}\omega$$ and not$$\int_{a}^{b}\mathbf{g}^{*}\omega\ ?$$

I vaguely recall the $\int_{[a,b]}$ notation as referring to an unsigned definite integral, but may have got that wrong.

答: Vaguely recall from where?! At this point, we've defined the integral of a differential form over a parametrized manifold (perhaps with boundary). The notation $\int_a^b f(t)\,dt$ is the classical notation for the Riemann integral of the function $f$.


问:
Analysis of series containing Zeta function:
Consider the given alternating series:

$$f(x) =2\sum_{n=0}^\infty \frac{a_n(x-1)^{2n+1}}{\zeta(-2n-1)}$$

Here, $$a_n= \left(1+\frac{1}{4n+1}\right)^6\left(1-\frac{1}{4n+3}\right)^6\left(\frac{4n²}{4n²-1}\right)^2.$$

How can we calculate the asymptotic of this series as $x\rightarrow\infty$?

Atleast how to compute values of this series for atleast upto x=100 accurately.?

Body:
To begin, using the formula for the zeta function for even integers, we have the following:$$\frac{a_n(2\pi)^{2n+2}(-1)^{n+1}}{\zeta(2n+2)\Gamma(2n+2)}=-a_n(2\pi)^{2n+2}\cdot\frac{2(2n+2)!}{(2\pi)^{2n+2}B_{2n+2}}\frac{1}{(2n+1)!}=-\frac{4n+4}{B_{2n+2}}a_n,$$

so that

$$f(x)=-4\sum_{n\ge0}\frac{n+1}{B_{2n+2}}a_n(x-1)^{2n+1}.$$


问:
How to prove that $\sum_{k=1}^m \cos^2(\frac{2 \pi k}{m}) = \frac{m}{2}$ for $m>2$

How to prove that for $m>2$ $$\sum_{k=1}^m \cos^2\left(\frac{2 \pi k}{m}\right) = \frac{m}{2}.$$

This could be useful for me solving a Differential Geometry problem. And Wolfram Alpha says it's true. But i can only be certain with a proof

答:
Let $ S = \sum_{k=1}^m \cos^2(\frac{2 \pi k}{m}) $

Multiply by 2 and subtract m

$2S - m = \sum_{k=1}^m 2\cos^2(\frac{2 \pi k}{m}) -1 $

Since $\cos(2x) = 2\cos^2(x)-1$

$2S - m = \sum_{k=1}^m \cos(\frac{4 \pi k}{m}) $

We have a formula when there is a summation of cosines where the angles are in AP

$Scn = \frac{\sin(\frac{md}{2})}{\sin(\frac{d}{2})} \cos(\frac{2a+(m-1)d}{2})$

We can see that $ d = \frac{4\pi}{m}$

Hence $\sin(\frac{md}{2}) = \sin(4\pi) = 0$

$2S - m = 0 $

$S = \frac{m}{2}$


答:
Let's come at it from first principles: by direct computation,
$$
\int_{n}^{n+1} x^2 dx = {1 \over 3}\left[ (n+1)^3 - n^3 \right].
$$
Are you testing the statement that the expression in the []'s is either a prime number or a multiple of 3?  If so, this can be tested directly, by expanding the first cube in []'s and collecting the like powers.


问:
Judge countable and uncountable set
Is $(1,2]$ uncountable? And is $\mathbb{Q} \cap [0,1]$ countable?
I'm confused about how to judge if it is countable, and I think the $(1,2]$ and $[0,1]$ are from $\mathbb{R}$?

答: First of all, $(a,b) , [a,b] , [a,b) , [a,b)$ have the same cardinality. And as any interval with $a\neq b$, $(1,2]$ is uncountable. The other set is a subtset of  $\mathbb{Q}$, so it can't be uncountable

Title: Judge countable and uncountable set


问:
$k$-regular even graph, still connected after removing k-2 edges has a 1$-factor
I have been asked to show that for a $k$-regular graph with even number of vertices has a $1$-factor if the graph stays connected after $k-2$ edges have been removed.
I do not know how to start this, but I think I could use Tutte's theorem somehow but I'm not sure. Does anyone have any hint or know how to solve it?

I have seen similar posts before but they have never had a complete answer, only hints.

Previous posts:

If a k-regular graph of even order remains connected when any k-2 edges are deleted, it has a 1-factor

Let $G$ be a $k$-regular graph, with $|V(G)|$ even, that remains connected when any $k − 2$ edges are deleted. Prove that $G$ has a $1$-factor.

答:
Lemma. In a $k$-regular graph $G$, if $S \subseteq V(G)$ with $|S|$ odd, there cannot be exactly $k-1$ edges between $S$ and $V(G)-S$.

Proof. Suppose there are $k-1$ such edges. Then $\sum_{v \in S}\deg(s) = k|S|$ counts each edge between two vertices in $S$ twice, and each edge from $S$ to $V(G)-S$ once, so there are $\frac{k|S|-(k-1)}{2}$ edges between two vertices in $S$. But this simplifies to $k \cdot \frac{|S|-1}{2} + \frac12$, which is not an integer. $\square$

Now we are ready to apply Tutte's theorem. Let $U$ be any nonempty subset of $V(G)$. Then there are at most $k|U|$ edges out of $U$. If $C$ is an odd component of $G-U$, it must receive at least $k$ of those edges: there cannot be $k-2$ or fewer by the connectivity condition, and there cannot be $k-1$ by the lemma. Therefore there can be at most $\frac{k|U|}{k}$ odd components.

If $U = \varnothing$, then the connectivity condition does not apply, because there might just be one odd component $C$ which is the whole graph. This is where we need the assumption that $|V(G)|$ is even.


问:
Question in Hatcher proposition 1.26 proof

(a) If $Y$ is obtained from $X$ by attaching 2-cells as described above, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to\pi_1(Y,x_0)$ whose kernel is $N$.

Rather than writing down the meaning 'describe above' and '$N$' in the statement, I think it's better to look p.49 of Hatcher's algebraic topology here.

The question is I can't understand the red line in the image. I understand why they're trying to choose such $\delta_\alpha$, but why such an element exists in $\pi_1(A\cap B,z_0)$? The map $\pi_1(A\cap B)\to \pi_1(A)$ is not surjective. Could you explain this?

答:
To describe the desired loop $\delta_\alpha$ in detail, let's look at the parts of the path $\gamma_\alpha \phi_\alpha \bar\gamma_\alpha$, and describe corresponding new parts which will be concatenated together to define $\delta_\alpha$.

The path $\gamma_\alpha$ goes along the bottom of the strip labelled $S_\alpha$ in that picture, starting from its initial point $x_0$. Define a correponding new path $\gamma'_\alpha$ which instead goes along the top of the strip, starting from its initial point $z_0$.

The closed path $\phi_\alpha$ goes around the boundary circle of the disc $e^2_\alpha$, which is where that disc is attached to $X$, and the base point of $\phi_\alpha$ coincides with the terminal endpoint of $\gamma_\alpha$.  Define a corresponding new path $\phi'_\alpha$ which instead goes around a circle in the disc $e^2_\alpha$ that is concentric to the outer circle, and such that the base point of $\phi'_\alpha$ meets the terminal endpoint of $\gamma'_\alpha$.

Thus the concatenation $\delta_\alpha = \gamma'_\alpha \, \phi'_\alpha \, \bar\gamma'_\alpha$ is a closed loop based at $z_0$, and $\delta_\alpha$ is entirely contained in $A \cap B$.

Furthermore, letting $h$ be "the line segment connecting $z_0$ to $x_0$ in the intersection of the $S_\alpha$'s", the concatenation
$$\bar h \, \delta_\alpha \, h = \bar h \, \gamma'_\alpha \, \phi'_\alpha \, \bar\gamma'_\alpha \, h
$$
is a closed loop in $A$ based at $x_0$ that is path homotopic to $\gamma_\alpha \, \phi_\alpha \, \bar\gamma_\alpha$. That path homotopy may be seen geometrically as moving through the strip $S_\alpha$ and through the annulus joining the outer circle $\phi_\alpha$ of $e^2_\alpha$ to the concentric circle $\phi'_\alpha$.

It follows that $\delta_\alpha$ is a loop "in $A \cap B$ based at $z_0$ representing the element of $\pi_1(A,z_0)$ corresponding to $[\gamma_\alpha\, \phi_\alpha\, \bar\gamma_\alpha]$ under the base-point change homomorphism $\beta_h$".


问:
Need help with proving that $(6n^2 -1/ (2n^2 + 5n))$ is a Cauchy Sequence by using definition

Using the definition of the Cauchy Sequence, show that $\forall\varepsilon\gt 0, \exists n_0\in\mathbb N$ s.t. $\forall m\gt n\ge n_0$, $\vert x_m - x_n\vert\le\varepsilon$.

I have tried to estimate it from above and got $\dfrac{12m^2}{4m^2}$, but in that case, the $m$ terms in the numerator and the denominator get canceled.

答:
well, the definition.

For any $\epsilon > 0$ there is an $N$ so that for any $n,m > N$ then

$|\frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}| < \epsilon$.

And $ \frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}=$

$\frac {6n^2 + 15n -15n-1}{2n^2 + 5n} - {6m^2 + 15m -15m-1}{2m^2 + 5m}=$

$3-\frac {15n+1}{2n^2 + 5n} - 3 + \frac {15m+1}{2m^2 + 5m}=$

$-\frac {15}{2n+5}-\frac 1{2n^2 + 5n}-\frac {15}{2m+5} - \frac 1{2n^2 + 5n}$.

Now we do know that each of those terms converge to $0$.

Or by definition for every $\epsilon_i$ there are $N_i$ so that $n > N_1$ $|-\frac {15}{2n + 5}| < \epsilon_1$ and so on....

If we let $\epsilon_i = \frac \epsilon 4$ and $N = \max (N_i)$ then if $n,m > M$ then

$|-\frac {15}{2n+5}-\frac 1{2n^2 + 5n}-\frac {15}{2m+5} - \frac 1{2n^2 + 5n}|=$

$|\frac {15}{2n+5}|+|\frac 1{2n^2 + 5n}|+|\frac {15}{2m+5}|+| + \frac 1{2n^2 + 5n}| <$

$4\cdot \frac \epsilon 4 < \epsilon$.

If need be we can caculate and find these $N_i$ individually.

we need $\frac {15}{2n + 5} < \frac \epsilon 4$ so

$\frac {\frac {15}{\frac \epsilon 4}-5}2 < n$ so we can have $N_1=N_3 = \frac {30}\epsilon$

we need $\frac 1{2n^2 + 5n} < \frac {\epsilon}4$ so $\frac 4{\epsilon} < 2n^2 + 5n$ so we can have $\frac 2{\epsilon} < n^2 < n^2 - \frac 52n$ if  $N_2 = N_4 =\sqrt{\frac 2{\epsilon}}$

So if $N > \max (\frac {30}\epsilon, \sqrt{\frac 2{\epsilon}})$ we will have $n,m > N$ would imply:

$|\frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}| =$

$|\frac {15}{2n+5}+\frac 1{2n^2 + 5n}+\frac {15}{2m+5} + \frac 1{2n^2 + 5n}|=$

$|\frac {15}{2n+5}|+|\frac 1{2n^2 + 5n}|+|\frac {15}{2m+5}|+| \frac 1{2n^2 + 5n}|<$

$|\frac {15}{2n}| + |\frac 1{2n^2}| + |\frac {15}{2m}| + |\frac 1{2n^2}| <$

$|\frac {15}{2\frac {30}\epsilon}| +|\frac 1{2\frac 2{\epsilon}}|+|\frac {15}{2\frac {30}\epsilon}| +|\frac 1{2\frac 2{\epsilon}}|=$

$|\frac \epsilon 4| +|\frac \epsilon 4| +|\frac \epsilon 4| +|\frac \epsilon 4| =\epsilon$


问:
Let $f(x)=(x+1)/x$ for all $x \in \mathbb{R}$ and $x\not= 0$ prove that this function is one-to-one correspondence
Let $f: R\setminus\{0\} \rightarrow R\setminus\{1\}$ be a function defined by the formula $f(x)=(x+1)/x$  for all $x \in \mathbb{R}$ and $x \not= 0$ prove that this function is one-to-one correspondence , I already proved that this function is one-to-one but I can't prove that it is onto.

答: Hint: Solve $(x+1)/x=y$ for $x$ when $y\ne1$.


答:
By the fundamental theorem of calculus,
\begin{align}
f(n)=\int_{n}^{n+1}x^2 \, dx &= \left[\frac{x^3}{3}\right]_{n}^{n+1} \\[4pt]
&= \frac{(n+1)^3}{3}-\frac{n^3}{3} \\[4pt]
&= \frac{3n^2+3n+1}{3} \, .\\
\end{align}
This means that $3f(n)=3n^2+3n+1$. So what we are dealing with is a polynomial that, when you input an integer into it, you get another integer out. Some of the time we will get a prime number. E.g. when $n=1$, we get $3f(n)=7$. But, as Ihf's answer has already noted, some of the time we won't.

Since $3n^2+3n+1=3n(n+1)+1$, and $3n(n+1)$ is even, we know that $3n(n+1)+1$ is odd. If an integer is relatively small and also odd, then it is fairly likely that it is prime. Hence, it shouldn't be too suprising that the polynomial outputs prime numbers for small values of $n$.


问:
Definite integral - closed form: $\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{\rm d}x$
I'm struggling with this definite integral:

$$
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{\rm d}x.
$$

Any help will be greatly appreciated.

答:
Note

\begin{align}
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{d}x
&= \frac12 \int_{-\infty}^{\infty}\cos\left((x-\frac1x)^2+2\right)\>dx= \frac12 \int_{-\infty}^{\infty}\cos\left(x^2+2\right)\>dx\\
&= \cos2\int_{0}^{\infty}\cos x^2\>dx - \sin2\int_{0}^{\infty}\sin x^2\>dx\\
&
= \frac{\sqrt\pi}2\cos(2+\frac\pi4)
\end{align}
where
$$\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx= \int_{-\infty}^{\infty}f(x)dx, \>\>\>
 \int_{0}^{\infty}\cos x^2dx = \int_{0}^{\infty}\sin x^2dx = \frac{\sqrt\pi}{2\sqrt2} $$
are used.


问:
Show that $\|x\|_1 = 1/2 \inf_{y > 0}(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$

Show that $\|x\|_1 = 1/2 \inf_{y > 0}(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$ for $x \in R^{m}$.

I think an equivalent problem is:
minimize $f(y) = 1/2 (\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$.

$1/y$ is a convex function on $y > 0$, $1^Ty$ is a convex function so, $1/y + 1^Ty$ is a convex function. I remember something like that the $\inf$ of a convex function over a convex set is convex.

$df(y)/dy_i = 1/2(-\frac{x_i^2}{y_i^2} + 1) = 0, y_i^2 = x_i^2, y_i = |x_i|$

The function value at $y_i$ becomes $m/2 + 1^Ty = m/2 + 1^T|x|$ which is not equal to the $\|x\|_1 = 1^T|x|$.

答:
Your derivation is almost correct until you claim that the function value becomes $m/2 + 1^T|x|$. When $y_i = |x_i|$,

$$\frac{1}{2}\left(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty\right) = \frac{1}{2}\left(\sum_{i=1}^{m}\frac{x_i^2}{|x_i|} + 1^T|x| \right)=\frac{1}{2}\left(\sum_{i=1}^{m}|x_i|+||x||_1\right)=||x||_1.$$


问:
Using Calculus to Solve for a Differential Equation
The question I have is: $x''(t)+yx'(t)+z^2x(t)=0$, where y and z $\in \mathbb{R}$ without $0$. $t \geq 0$ by the way. The total energy of the system is $$\frac{1}{2}(x'(t))^2+ \frac{1}{2}z^2x(t)^2$$ The question is, for t > 0, under what conditions is the total energy of the system always increasing? Since the only things that can change are y and z, it certainly has to do with these constants. I'm just not sure what it is, though.

答:
hint

Let $ E $ be the total energy of your system.

$$E=\frac 12\Bigl((x'(t))^2+z^2(x(t))^2\Bigr)$$

$ E $ is always increasing in time if
$$\frac{dE}{dt}\ge 0$$
with
$$\frac{dE}{dt}=x'(t)\Bigl(x''(t)+z^2x(t)\Bigr)$$

$$= -y(x'(t))^2$$

So, we need the condition
$$y<0$$


问:
The 'compactness cardinal' of a space
I'm looking for references (and a name!) for the following invariant of a topological space $X$:

The least (infinite) cardinal $\kappa$ such that any open cover of $X$ has a subcover of cardinality less than $\kappa$.

For compact spaces, for example, this cardinal is $\aleph_0$.

答: Such spaces have been called finally $\kappa$-compact. More generally, $X$ is $[\kappa,\lambda]$-compact if every open cover of $X$ of cardinality at most $\lambda$ has a subcover of cardinality less than $\kappa$. An $[\omega,\kappa]$-compact space is said to be initially $\kappa$-compact. Thus, Lindelöf spaces are finally $\omega_1$-compact, and countably compact spaces are initially $\omega$-compact.

Title: The 'compactness cardinal' of a space


问:
How might one prove, by contradiction, that $n$ is prime if $\mathrm{gcd}((n-k),(n-2k))=1$ for for all $k$ s.t. $1\leq k\leq(n-3)/2$?
How might I go about proving this by contradiction.  I came about this on my own, but I know it is known.  It's rather obvious if one looks at a binary representation of Euclid's Orchard.  Does that structure itself perhaps constitute, by way of explanation of it's construction, a visual proof?</p>

答:
Let $d$ be a divisor of both $n-k,n-2k$. Then $d$ is also a divisor of $2(n-k)=2n-2k$. Finally, $d$ is also a divisor of $(2n-2k)-(n-2k)=n$

But by hypothesis, $n$ is prime, so $d=1,n$

Since $d$ is a divisor of $n-k$, either it is true that $1$ divides $k$ or $n$ divides $k$, or both.

But $n$ divides $k$ is a contradiction, because $k$ is smaller than $n$.

Thus, the only possibility remaining is $d=1$, which means the same as $\gcd ((n-k),(n-2k))=1$


问:
Equation $\frac{dy}{dx}=y-x$
Consider an equation $\frac{dy}{dx}=y - x$. Is there a solution to this? If so, will it have a local mininum along $y=x$? Sorry if the question is not specific, this is the best I can phrase it, and I don't really understand how to approach the problem.

答:
If we write it as
$$
\frac{dy}{dx}-y=-x \, ,
$$
then we are dealing with a linear first-order differential equation of the form
$$
\frac{dy}{dx}+p(x)y=q(x) \, .
$$
To solve an equation of this form, you have to multiply both sides by $e^{P(x)}$, where $P$ is an antiderivative of $p$. Here $p(x)=-1$, and so we can take $P(x)=-x$. The equation becomes
$$
e^{-x}\frac{dy}{dx}-ye^{-x}=-xe^{-x} \\
\frac{d}{dx}(e^{-x}y)=-xe^{-x} \, .
$$
Integrating both sides and rearranging gives you the answer.

Title: Equation $\frac{dy}{dx}=y-x$


问:
How to compute probability of a bootstrap sample

The Question

Consider the  samples $\{1, 3, 4, 6\}$ from some distribution.

a) For one random bootstrap sample, find the probability that the mean
is $1$.

b) For one random bootstrap sample, find the probability that the
maximum is $6$.

c) For one random bootstrap sample, find the probability that exactly
two elements in the sample are less than $2$.

My Understanding

We just started to learn the bootstrap in class and I came across this question. I'm a little confused, as I feel like this question is too easy, as the mean of any sample with those numbers is always $3.5$ so a) is $0$. The maximum will always be $6$ so b) is $1$. And $2$ of the numbers cannot be less than $2$. So c) is $0.$

Is there something major that I'm missing?

答: My understanding is that you should take a random sample, presumably of size 4, of those four values with replacement. This means that they need not all appear, in which case the probabilities you are asked to find will not be those got from the given sample.</p>


问:
Expanding a PDE in powers of a small parameter?
I'm working on an assignment for my quantum mechanics class and I've arrived at a nonlinear inhomogeneous partial differential equation for a complex function $S:\mathbb{R}^2\to\mathbb{C}~;~S:(x,t)\mapsto S(x,t)$ (This is actually the Quantum Hamilton-Jacobi equation in one spatial dimension)
$$\partial _{t} S+\frac{1}{2m} |\partial _{x} S|^{2}+U =\frac{\mathrm{i} \hbar }{2m}\partial ^{2}_{x}S\tag{1}$$
Here, $\mathrm{i}$ is the imaginary unit, $\hbar$ and $m$ are positive real constants (reduced Planck's constant and the mass of the particle), the $|~|$ denotes complex norm, and $U:\mathbb{R}^2\to\mathbb{R}$ is a real function (potential energy). My professor has given me the following instruction:

Expand $S(x,t)$ in terms of the small parameter $\hbar$
$$S(x,t)=S_0(x,t)+\hbar S_1(x,t)+\hbar^2S_2(x,t)+\dots$$
Show that
$$\partial_tS_0+\frac{|\partial_x S_0|^2}{2m}+U=0\tag{2}$$
$$\partial_t S_1+\frac{1}{m}(\partial_x S_0)(\partial_x S_1)=\frac{\mathrm{i}\hbar}{2m}\partial_x^2S_1\tag{3}$$

I'm not really sure how he wants us to approach the "expand in terms of the small parameter" part. It looks kind of like a Taylor series, but I don't know how they are related. I've done small parameter expansions on single variable Fredholm integral eigenvalue problems, but I really don't know how I can apply that here. I think my question is kind of similar to this one and this one, but as for the first I'm not sure how to state my equation as an eigenvalue problem and as for the second the only answer on that post is very unclear, and I can't see a way to use geometric series as Ted Shifrin mentioned in the comments. This is from a quantum mechanics class, so we haven't been given a general treatment of the mathematical methods needed for this. "Perturbation theory" is covered later in the course, but only really on linear eigenvalue problems. I have another course that deals with this sort of thing, but its focused mostly on integral equations and on linear single variable differential equations, so I don't know how to apply those methods here.

How should I approach this??

CORRECTIONS:

My professor has issued a revised assignment (still containing errors I might add) with some corrections to equations (1),(2), and (3). They now agree with @Sal 's work. They read
$$\partial _{t} S+\frac{1}{2m} (\partial _{x} S)^{2}+U =\frac{\mathrm{i} \hbar }{2m}\partial ^{2}_{x}S\tag{1*}$$
$$\partial_tS_0+\frac{(\partial_x S_0)^2}{2m}+U=0\tag{2*}$$
$$\partial_t S_1+\frac{1}{m}(\partial_x S_0)(\partial_x S_1)=\frac{\mathrm{i}}{2m}\partial_x^2S_1\tag{3*}$$

答:
Obligatory first remark (apologies in advance) $\hbar$ is a dimensionful constant, and so is not 'small' or 'large'. I may choose to work in units where $\hbar=10$, where it is certainly not true that $\hbar^n \to 0$ for large $n$. Let me rewrite the equation, setting $m=1/2$ and $\hbar=1$ for my convenience$^\dagger$

$$
\partial_t S +(\partial_x S)^2 +U =i\partial_{xx}S
$$

We wish to solve this equation with the perturbative ansatz

$$
S=\sum_{n=0}^\infty S_n \varepsilon^n
$$

That is, we claim the solution may be written as a power series in the small parameter $\varepsilon$: a dimensionless parameter that we have introduced. We wish to solve for the $S_n$ recursively. The next step is to write an $\varepsilon$ in the original equation. Following the spirit of the question, we choose

$$
\partial_t S +(\partial_x S)^2 +U =i\varepsilon \partial_{xx}S
$$

The strategy is now straightforward. We substitute our ansatz for $S$ into this equation, and equate powers of $\varepsilon$

$$
\sum_{n=0}^\infty \partial_tS_n \varepsilon^n + \left( \sum_{n=0}^\infty \partial_xS _n \varepsilon^n \right)^2+U=i\sum_{n=0}^\infty \partial_{xx}S_n \varepsilon^{n+1}
$$

Products such at the second term on the left are expanded using Cauchy's product, however, we will not need that to get the two terms required in your question. Note the lowest power of $\varepsilon$ on the right is one. The coefficients of $\varepsilon^0$ in the equation are

$$
\partial_t S_0+(\partial_xS_0)^2+U=0
$$

This is your first equation. In principle, you solve this for $S_0$. So presume $S_0$ is known. Look at coefficients of $\varepsilon^1$

$$
\partial_t S_1 + \partial_x S_0 \partial_x S_1 + \partial_x S_1 \partial_x S_0 + 0 = i \partial_{xx}S_0
$$
$$
\partial_t S_1 + 2 ( \partial_x S_0 )(\partial_x S_1) = i \partial_{xx}S_0
$$

Which is your second equation. The middle terms arise from picking the coefficients of $\varepsilon^1$ from the product

$$
(\partial_x S)^2=\left(\sum_{n=0}^\infty \partial_xS _n \varepsilon^n \right)^2 =(\partial_x S_0 + \varepsilon \partial_x S_1 + \varepsilon^2 \partial_x S_2+\dots) (\partial_x S_0 + \varepsilon \partial_x S_1 + \varepsilon^2 \partial_x S_2 +\dots)
$$

In principle, since $S_0$ is known, you can now solve the equation for $S_1$, and so on.

$\dagger$ The ansatz $\psi=e^S$ occurs frequently in perturbative analysis. With this ansatz, we get the equation in your post from the Schrodinger equation, without the complex norm. I assume this was a typo, or that what was meant is: $\nabla S \cdot \nabla S$.


问:
Terminology to describe that, eg, $ax^2+bx+c=0$ cannot be solved by simply moving terms around to isolate $x$
For example, a quadratic equation in the form
$$ax^2+bx+c=0$$
cannot be solved by simply "moving" the terms around; we wouldn't be able to reduce $x$ to just one term without introducing new terms.

Is there a way to describe this? Say I'm trying to teach the derivation of the quadratic formula. How would I explain that we must introduce new terms, construct a perfect square trinomial, factor it, etc. rather than simply attempt to solve it via basic algebraic manipulation?

答:
If you were teaching students about this, then perhaps you could mention strategies that ultimately don't work. If we rewrite
$$
ax^2+bx+c=0
$$
as
$$
x=-\frac{ax^2+c}{b}
$$
then $x$ is still written in terms of $x^2$, meaning that we can't get anywhere. Similarly, if we make $x^2$ the subject of the equation, then we don't get anywhere. The crux of the derivation of the quadratic formula is realising that
\begin{align}
ax^2+bx+c &= a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right) \\[4pt]
&= a\left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right) \, .
\end{align}
From here, there is only a single term containing $x$, and the rest that follows is the basic algebraic manipulation you mentioned earlier. You might want to make things look cleaner by writing the equation
$$
x^2+\frac{b}{a}x+\frac{c}{a}=0
$$
as
$$
x^2 + 2Bx + C = 0 \, ,
$$
with $2B=b/a$ and $C=c/a$. Then, the factorisation is
$$
(x+B)^2-B^2+C=0 \, ,
$$
and here it is abundantly clear why completing the square is such a powerful method.


问:
A hint to the differential equation $y^2(1-y'^2)=C$ for some real constant $C$
As the title suggests, I am seeking for some kind of hint to the above differential equation, just to get started. I've been sitting all afternoon without any kind of progress.

My approach so far has been substituting $y$ for some various trigonometric functions; however no luck.

Edit: I don't know how much of critical information this is, but $y(0)=0$ is to be satisfied.

答:
As @Cameron_Williams said try solving for $y'$

$$y^2(1-y'^2)=C$$

$$1-y'^2=\dfrac{C}{y^2}$$

$$y'^2=1-\dfrac{C}{y^2}$$

$$y'=\;\;?$$

Can you find it now?


问:
To need a way of thinking about Ji_Chen's nice result $(a- c)^{2}+ (b- d)^{2}\geq\frac{7}{9}ab- \frac{7}{20}(c^{2}+ 4d^{2})$
given four real numbers $a, b, c, d$

Ji Chen gave a nice result on.AoPS
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{7}{9}ab- \frac{7}{20}\left ( c^{2}+ 4d^{2} \right )$$

The original problem consists a condition $ab= c^{2}+ 4d^{2}= 4$ and hides the rightside of the earlier inequality, I wondered that how he could find this.. of course I searched in AoPS because of him but I also found another interesting work, which made me sense.

He (AoPS/@ye109) claimed that
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}ab+ \frac{k^{4}- 4}{k^{4}- 16}\left ( c^{2}+ 4d^{2} \right )$$
with the best $k\cong 1.672955253$ is a real root of $k^{12}- 28k^{8}- 36k^{6}- 128k^{4}+ 1024$

That guy is very good at making second solution, we just need explain his solution from Ji Chen's 1st one, if Ji Chen used discriminant to find $7/9, -7/20$ then how could he use discriminant to find $\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}, \frac{k^{4}- 4}{k^{4}- 16},$ maybe there could exist discriminant for polynomials not only for constants. What's your idea ? Who can give me a way of thinking ? Thanks a real lot !

答:
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq Uab-V \left ( c^{2}+ 4d^{2} \right )$$
(for all reals $a,b,c,d$)

precisely when $$  V \geq 0$$  and
$$ |U| \leq \frac{4V}{\sqrt{(1+V)(1+4V)}}  $$

In turn, the largest $U$ is rational when
$$  V = \frac{m^2 - n^2}{4n^2 - m^2} \geq 0$$
where we get
$$ |U| \leq 4\frac{|m^2-n^2|}{3mn} $$

In particular, when we take $n=3,m=4$ we get $V= 7/20$ and the largest $|U|$ that still gives positive semidefinite is $7/9$

The positive semidefinite matrix with extremal $|U|$

using $2n > m > n,$
$$
\left(
\begin{array}{cccc}
1&\frac{2m^2-2n^2}{3mn}&-1&0\\
\frac{2m^2-2n^2}{3mn}&1&0&-1\\
-1&0&\frac{3n^2}{4n^2-m^2}&0\\
0&-1&0&\frac{3m^2}{4n^2-m^2}\\
\end{array}
\right)
$$


问:
Solving differential equations with signum
Is there a general solution to:

$A x''+ B x' + C \mathop{\rm sgn}(x')+ D x=0$  where $\mathop{\rm sgn}(x')$ is the sign of $x'$

答:
The solution will satisfy $A x'' + B x' + C + D x = 0$ in intervals where $x' > 0$ and $A x'' + B x' - C + D x = 0$ in intervals where $x' < 0$.  If $B^2 - 4 A D > 0$, for example (the overdamped case), the solution of $A x'' + B x' + C + D x = 0$ are of the form $x = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} - C/D$ for arbitrary constants $c_1$ and $c_2$, where $\lambda_i$ are the roots of the quadratic $A \lambda^2 + B \lambda + D$.  Then $x' = c_1 \lambda_1 e^{\lambda_1 t} + c_2 \lambda_2 e^{\lambda_2 t} = 0\ $ for
$t = \ln(-c_1 \lambda_1/(c_2 \lambda_2))/(\lambda_2 - \lambda_1)\ $ in the case where
$c_1 \lambda_1/(c_2 \lambda_2) < 0$, or never if $c_1 \lambda_1/(c_2 \lambda_2) > 0$.  Thus we have solutions which are either monotonic solutions of $A x'' + B x' + C + D x = 0$ or of $A x'' + B x' - C + D x = 0$, or piecewise solutions (a solution of one of these equations on one interval and a solution of the other equation on its complement, joined at a point where $x' = 0$ (on both sides).
In the underdamped case, things can be more complicated because we will have infinitely many intervals.</p>


问:
How to prove $\big[\text{Frac}\left(\mathbb{F_p}[x]\right):\mathbb{F} \big]=\infty$ for a prime number $p$?
Let $p$ be a prime number and consider the ring $\mathbb{F}_P=\{0,1,\ldots,p-1\}$ of integers modulo $p$, which is a field. It follows that the polynomal ring $\mathbb{F}_p[x]$ is an integral domain, and so we can consider the field of fractions $\text{Frac}\left(\mathbb{F_p}[x]\right)$ of $\mathbb{F}_p[x]$. Now, since $\mathbb{F}_p \subseteq \text{Frac}\left(\mathbb{F_p}[x]\right)$, $\text{Frac}\left(\mathbb{F_p}[x]\right)$ becomes a field extension of $\mathbb{F}_p$.

Problem: Prove that $\big[\text{Frac}\left(\mathbb{F_p}[x]\right):\mathbb{F}_p \big]=\infty$ is true.

I don't know where to start to prove this statement. I have a feeling that a proof by contradiction, utilizing the fact that the polynomial ring as a vector space is of infinite dimension, might be a course to take.

答: Any finite extension of $\Bbb{F}_p$ is finite. The field of fractions contains $\Bbb{F}_p[x]$ as a subring, so it is infinite.


问:
Analysis of series containing Zeta function:
Consider the given alternating series:

$$f(x) =2\sum_{n=0}^\infty \frac{a_n(x-1)^{2n+1}}{\zeta(-2n-1)}$$

Here, $$a_n= \left(1+\frac{1}{4n+1}\right)^6\left(1-\frac{1}{4n+3}\right)^6\left(\frac{4n²}{4n²-1}\right)^2.$$

How can we calculate the asymptotic of this series as $x\rightarrow\infty$?

Atleast how to compute values of this series for atleast upto x=100 accurately.?

答:
Not a full answer. Define
$$
F(x): =  - \frac{{f(\sqrt x  + 1)}}{{2\sqrt {\pi x} }} = \sum\limits_{n = 0}^\infty  {\frac{{a_n \pi ^{2n + 2} }}{{\zeta (2n + 2)\Gamma \left( {n + \frac{3}{2}} \right)}}\frac{{( - x)^n }}{{n!}}} .
$$
Then, by Ramanujan's master theorem,
$$
\int_0^{ + \infty } {x^{s - 1} F(x)dx}  = \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma \left( {\frac{3}{2} - s} \right)}}.
$$
Using the Mellin inversion formula, we then obtain
$$
F(x) = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {x^{ - s} \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma \left( {\frac{3}{2} - s} \right)}}ds}
$$
with any $0 < c < \frac{1}{2}$. Thus,
$$
f(x) = \frac{i}{{\sqrt \pi  }}\int_{c - i\infty }^{c + i\infty } {(x - 1)^{1 - 2s} \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma \left( {\frac{3}{2} - s} \right)}}ds} .
$$
You may then deform the contour and, depending on the location of the non-trivial zeros of the Riemann zeta function, obtain some estimates for $f(x)$. A crude esimate under the Riemann hypothesis would be $f(x) = \mathcal{O}(x^{ - 1/2 + \varepsilon } )$.


问:
Is $\Gamma(x+y+1)\ge\Gamma(x+1)\Gamma(y+1)$?
I apologize if this has already been asked, but I could not find it. Given $x,y\ge 0$, is it true that $$\Gamma(x+y+1)\ge \Gamma(x+1)\Gamma(y+1),$$ where $\Gamma$ is the gamma function? This certainly holds for natural numbers $x,y$, as then $$\Gamma(x+y+1) = (x+y)! \ge x!y! = \Gamma(x+1)\Gamma(y+1),$$ and some plots in Mathematica seem to show that this relation also holds true for positive reals, but I have so far been unable to find a proof myself.

答:
$f(x) = \log \Gamma(x + 1)$ is convex for $x \ge 0$ (compare Bohr–Mollerup theorem) with $f(0) = 0$, and therefore superadditive:
$$
 f(x+y) \ge f(x) + f(y) \, ,
$$
see for example properties of a convex function.


问:
Contour integral of $f(z)=z$ from $z_1 = 5i$ to $z_2 = 2-2i$
I have to make two different contour integrals of $f(z)=z$ from $z_1 = 5i$ to $z_2 = 2-2i$. I did the first contour to be $\gamma_1 (t) = 5i + t(2-2i)$ with $0\le t \leq 1$. But my question is how can I make a second contour different from the first one I have? Is it valid to do $\gamma_2 = (2-2i) + t(5i)$ with $0\le t \leq 1$? I was trying to make a contour with a half circle but I don't really know how to make it with parameters. I get that we will have $e^{it}$ with $0 \leq t \leq \pi$ but I don't know how to include the details of $z_1$ and $z_2$. Or is it easier to make another contour with just some random point in-between $z_1$ and $z_2$?

Thanks for the help.

Body: You can take, say, $\gamma(t)=5i+t^2(2-2i)$ ($t\in[0,1]$).


问:
how to draw the graph of this function with rationals and irrationals
if someone can draw the graph of this function for me, thank you

I tried something like this but I'm not sure;

答: When drawing graphs such as this, there is a convention that the points on the irrational part of the curve are drawn more closely together than the points on the rational part of curve. This is to emphasise that the cardinality of the irrationals is greater than that of the rationals. This might not seem like an 'accurate' sketch, but, when you think about it, the graphs of functions are never drawn accurately anyway.</p>


问:
Triple Integral in Cylindrical Coordinates using x axis instead of z axis
Today in my Calculus class my teacher made an example using change of variables using this problem:

Find the volume of the solid inside the cylinder $y^{2} + z^{2} = 2y$ bounded by $x=0$ and the surface of the cone $x^{2}=y^{2} + z^{2}$. With $x \geq 0$ .

He wrote directly the change of variables like this: from (x, y, z) to $(\theta, r, x)$ which is, I think, the same as cylindrical coordinates but with rotated axis (I don't know how to call it)? because instead of $z$ he used $x$ for the height because it was convenient for the problem (if you make a graph, the cylinder lies in the $x$ axis).

The thing is, I didn't understand the bounds for $(\theta, r, x)$. He wrote the answer and it should be this:

$$-\frac{\pi}2 \leq \theta\leq \frac{\pi}2,$$
$$0 \leq r\leq 2cos(\theta),$$
$$0 \leq x \leq r,$$

After the integration, the result should be 32/9.

I've been trying to think all day but I still have troubles. Could you explain to me the change of variables, please? The  $\theta$ angle bounds are the most confusing. Please.

I'm sorry if I made a mistake with the equations format.

答: Note that$$y^2+z^2\leqslant2y\iff(y-1)^2+z^2\leqslant1.$$But then $|y-1|\leqslant1$, which means that $0\leqslant y\leqslant2$. But $y=r\cos\theta$ and so, since $y\geqslant0$, $\theta\in\left[-\frac\pi2,\frac\pi2\right]$. Besides,$$y^2+z^2\leqslant2y\iff r^2\leqslant2r\cos\theta\iff r\leqslant2\cos\theta.$$Furthermore,$$x^2\leqslant y^2+z^2\iff x^2\leqslant r^2\iff0\leqslant x\leqslant r,$$since we are assuming that $x\geqslant0$. So, the computation will be\begin{align}\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}\int_0^rr\,\mathrm dx\,\mathrm dr\,\mathrm d\theta&=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}r^2\,\mathrm dr\,\mathrm d\theta\\&=\int_{-\pi/2}^{\pi/2}\frac83\cos^3(\theta)\,\mathrm d\theta\\&=\frac{32}9.\end{align}


问:
How to prove this propositional logic equivalence?
Problem : $\lnot[(\lnot p∧q)∨\lnot q]∨[\lnot(p∧\lnot q)]≡\lnot p∨q$  Prove this equility without using truth table.

I have managed to solve this question until this part :

$$\begin{align}\lnot[(\lnot p∧q)∨[\lnot(p∧\lnot q)] &≡ [\lnot(\lnot p∧q)∧q]∨(\lnot p∨q)\\& ≡ [(p∧q)∨(\lnot q∧q)]∨(\lnot p∨q)]\\&  ≡ (p∧q)∨(\lnot p∨q)\end{align}$$

I stucked here. I tried to $(p∧q)∨\lnot(p∧\lnot q)$ like this for de morgan rule but it doesnt work. What do I need to do?

答:
Observe that $((p \land q) \lor (\lnot p \lor q)) \leftrightarrow (\lnot p \lor q)$ by the fact that
$$((\phi \rightarrow \psi) \land (\psi \rightarrow \phi)) \rightarrow (\phi \leftrightarrow \psi)$$ where $\phi$ and $\psi$ are arbitrary propositions.

Thus, all you have to show that

\begin{align}
(\lnot p \lor q) &\rightarrow ((p \land q) \lor (\lnot p \lor q)) && (1) \\
((p \land q) \lor (\lnot p \lor q)) &\rightarrow (\lnot p \lor q) && (2)
\end{align}

The first statement is the definition of $\lor \text{ Introduction}$, and the second is proved trivially by $\lor \text{ Elimination}$.


问:
Fitting a derivative of a curve to data
Some derivatives cannot be integrated to an analytic function but have the function embedded in the derivative itself.  I need to know how to fit those derivative functions to a set of data.  Here is an example:

I have a time series data, lets call it yData.

I have a non-integrable derivative, in this example:
$$ \frac{dy}{dt} = \frac{(ay(t)(K-y(t)))}{K - y(t) +cy(t)}$$

I need to come up with an objective function (figure of merit, $\chi^2$) to optimize, but don't know what to use for the "y" terms in the derivative.

What do I do?  The only thing I can think of is to use the data itself:

$$ \chi^2 = \sum \frac{\left(yData_i - \frac{ayData_i(K-yData_i)}{K - yData_i +c*yData_i}\right)^2}{\sigma_i}$$

Does this make any sense?

答:
The ODE is separable thus

$$
\frac{A dy}{y}+\frac{B dy}{K-y} = \frac{K+(c-1)y}{y(K-y)}dy = adt
$$

with $A=1$ and $B=c$ then

$$
\ln y - c\ln(y-K) = a t + c_0\Rightarrow \frac{y}{(y-K)^c}=c_1 e^{a t}
$$

The determination of $c_1$ is done as $\frac{y(0)}{(K-y(0))^c}=c_1 $ and then

$$
\frac{y}{(y-K)^c}=\frac{y(0)}{(y(0)-K)^c} e^{a t}
$$

now if $\{t_j, y_j\},\ \ j=1,\cdots,n$ is the data to adjust we can proceed defining

$$
\mathcal{O}(a,c,K,y_0)=\sum_{j=1}^n\|\frac{y_j}{(y_j-K)^c}-\frac{y_0}{(y_0-K)^c} e^{a t_j}\|
$$

and then follow with

$$
\min_{a,c,K,y_0}\mathcal{O}(a,c,K,y_0)
$$

to determine the best parameters.

Title: Fitting a derivative of a curve to data


问:
SDE with constant coefficients
So the simplest possible SDE would be
\begin{equation}
\begin{cases}
dX_t = a dt + b dW_t, \text{ with } a,b\in \mathbb{R}   \\
X_0 = x_0
\end{cases}
\end{equation}

describing a 1-dimensional stochastic process $X_t$ on a probability space $(\Omega, \mathcal{F}, P)$ and $t \in [0,\infty)$.

I know that in this case, the distribution of the solution is a Gaussian. So it is enough to characterize $\mathbb{E}[X_t]$ and $\text{Var}(X_t)$. Taking the expectation and variance of the equation, I get that $X_t \sim \mathcal{N}(x_0 + at, b^2t)$.

The problem I am interested in is a piecewise constant SDE, where
$$ a = \begin{cases}
a_1, \text{ if } X_t >0\\
a_2, \text{ if }X_t \leq 0 \end{cases}$$

$$ b = \begin{cases}
b_1, \text{ if } X_t >0\\
b_2, \text{ if }X_t \leq 0 \end{cases}.$$

How would one "piece together" the two solutions on either side of $0$? It would be simple to solve the equation on either side, but I think there would have to be some compatibility equation at 0.

One problem I know is that a Brownian motion will hit $0$ infinitely many times in a time interval $[0,\epsilon)$, which means that each time $X_t=0$, there is a small interval where the process may not be well-defined. I was thinking to fix this by stopping the process everytime it hits $0$, and restarting it after it "escapes" some neighborhood of 0. Perhaps this could be reframed as a reflected diffusion.

Note: there are two relevant solution methods to similar problems that I've seen in the literature, one involving an analytic solution to the constant-coefficients problem (Karatzas and Shreve), and a numerical solution to the reflecting boundary problem (Skorokhod), not sure which is appropriate here.

答:
You could replace the constant coefficients $a$ and $b$ with the coefficient functions in terms of indicator functions:
$$a(x)=a_1 \cdot \mathbb{1}_{(0, \infty)}(x)+a_2 \cdot  \mathbb{1}_{(-\infty, 0]}(x)$$
and a similar expression for $b$ but with the constants $b_1, b_2$. Then the SDE is
$$dX_t=a(X_t)dt+b(X_t)dB_t$$

Here is a simulation of a sample-path via the Euler-Maruyama scheme implemented in R with initial point $x_0=0$, total time $T=1$, and parameters $a_1=-0.05$, $a_2=0.1$, $b_1=0.1$ and $b_2=0.3$ with $n=1000$ time-subintervals:

and here is one more simulation that crosses the line $x=0$ with all the same parameters except starting below zero at $x_0=-1$ and running the path for $T=10$. If I can derive any analytic information, I will update this post.

Title: SDE with constant coefficients


问:
Given an ODE then regard it as an Euler-Lagrange equation, how to find a functional relative to it?
Recently, I have been studied about Lp Minkowski problem. I met some confusion. The equation is an ODE like that
$$u_{xx}+u=\frac{g(x)}{u^{p+1}},   \quad x\in\mathbb{R} , \quad p\geq 0.$$
$g(x)$ is a positive function and $u>0, u(k)=u(2\pi+k),k\in\mathbb{R}$ . Then we regard it as an Euler-lagrange equation of a functional. The papers gave this one:
$$J[u]=(\int_0^{2\pi}\frac{g(x)}{u^p}dx)^{\frac{2}{p}}(\int_0^{2\pi}(u^2-u_x^2)dx).$$
In other word, a solution of this ODE is essentially a critical point of the functional. My question is how to find this functional? I think it is strange and maybe relative to eigenvalue of ODE? But why it has the power $\frac{2}{p}$? Thanks in advance.

答:
OP's functional is of the form
$$ J[u]~= F[u]^r G[u]^s, \qquad r,s~\in~\mathbb{R},\tag{1}$$
where
$$ F[u]~:=~ \int_0^{2\pi}\!\mathrm{d}x(u^2-u_x^2)
\qquad\text{and}\qquad
G[u]~:=~ \int_0^{2\pi}\!\mathrm{d}x\frac{g(x)}{u^p}. \tag{2}$$

The functional derivatives are
$$ \frac{\delta F[u]}{\delta u(x)}~=~2u(x)+2u_{xx}(x)
\qquad\text{and}\qquad
\frac{\delta G[u]}{\delta u(x)}~=~ -p \frac{g(x)}{u^{p+1}(x)}.\tag{3} $$

The sought-for functional derivative becomes$^1$
$$\begin{align} \frac{1}{J[u]}\frac{\delta J[u]}{\delta u(x)}
~=~& \frac{r}{F[u]}\frac{\delta F[u]}{\delta u(x)}+ \frac{s}{G[u]}\frac{\delta G[u]}{\delta u(x)}\cr
~=~&\frac{2r}{F[u]}(u(x)+u_{xx}(x))- \frac{sp}{G[u]}\frac{g(x)}{u^{p+1}(x)}\end{align}\tag{4}$$

The stationary condition for $J$ is hence an ODE of the form
$$ u(x)+u_{xx}(x)~= k\frac{g(x)}{u^{p+1}(x)}, \qquad k~\in~\mathbb{R}\backslash\{0\}.\tag{5}$$

Multiplying the ODE (5) with u(x) and integrating over $x$ implies that
$$F[u]~=~kG[u]\tag{6}$$
on-shell.

Moreover, if the powers satisfy
$$ 2r~=~sp,\tag{7}$$
then the ODE (5) is a stationary path for $J$. This essentially answers OP's question about the power $\frac{2}{p}$.

--

$^1$ Strictly speaking $J[u]$ could be zero, so the factor $\frac{1}{J[u]}$ on the LHS of eq. (4) is better kept as a factor $J[u]$ on the RHS of eq. (4). We leave it to the reader to improve this.


问:
Is $x\sin\left(\frac{1}{x}\right)$ uniformly continuous in $\mathbb{R}$?
Let $f:\mathbb{R}\xrightarrow{}\mathbb{R}$ with rule defined as
$$
f(x) = \left\{\begin{array}  xx\operatorname{sin}\left(\frac{1}{x}\right) &\text{  if }x\neq0 \\
                      0  &\text{ if } x=0
\end{array}\right.
$$
I would like to show that this function is uniformly continous over $\mathbb{R}$.
I started out with the $\epsilon-\delta$ definition and tried to manipulate the following expression without success,
$$
\left|x \operatorname{sin}\left(\frac{1}{x}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right |\\
=\left|x \operatorname{sin}\left(\frac{1}{x}\right) -x \operatorname{sin}\left(\frac{1}{y}\right) +x \operatorname{sin}\left(\frac{1}{y}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right | \\
\leq \left|x \left(\operatorname{sin}\left(\frac{1}{x}\right) - \operatorname{sin}\left(\frac{1}{y}\right)\right) + (x - y) \operatorname{sin}\left(\frac{1}{y}\right) \right |
$$
Also, the function isn't Lipschitz, so I can't go this way too.
I've seen some other proofs at mathSE however they try to show things for smallers domains as $[0,1]$.

答:
Hints:

(1) The function is continuous on any compact interval $[-a,a]$.

(2) The derivative is bounded on semi-infinite intervals $(-\infty,-a]$ and $[a,\infty)$ where $a > 0$.


问:
Determine whether the given set is bounded and find its supremum and its infimum.
Let $A = \{x \mid \Bbb R \mid x^2 \lt 2\}$. Is the set $A$ bounded?
If yes, find its supremum and its infimum.

Attempt:
Notice that
\begin{align*}
A &= \{x \in \Bbb R \mid x^2 \lt 2 \} \\
&= \{x \in \Bbb R \mid x^2 - 2 \lt 0 \} \\
&= \{x \in \Bbb R \mid (x - \sqrt{2})(x + \sqrt{2}) \lt 0 \} \\
&= \{x \in \Bbb R \mid -\sqrt{2} \lt x \lt \sqrt{2} \} \\
&= (-\sqrt{2},\sqrt{2}).
\end{align*}
Hence, by definition, $A$ is bounded. Now, I claim $\sup A = \sqrt{2}$ and $\inf A = -\sqrt{2}$ which is true, indeed. For the supremum, let $M$ be the another upper bound of $A$ and suppose $M \lt \sqrt{2}$.
Then, since $A$ is bounded, there exist an element of $A$, say $r$, such that
$M \lt r$. A contradiction with the definition of an upper bound, that's $M$. Hence, we must have
$\sqrt{2} \le M$ i.e. $\sup A = \sqrt{2}$.

Now, for the infimum, let $m$ be the another lower bound of $A$ and suppose
$-\sqrt{2} \lt m$.
Then, since $A$ is bounded, there exist an element of $A$, say $s$, such that
$s \lt m$. A contradiction with the definition of a lower bound, that's $m$. Hence, we must have
$m \le -\sqrt{2}$ i.e. $\inf A = -\sqrt{2}$.

Am I true ?

答: This is mostly right, but I think you should be a little more explicit about where you get the contradiction from. In particular, how do you know that there is an element $r \in A$ such that $M<r$? This doesn't come from the fact that $A$ is bounded, it's because $M < \sqrt{2}$.


问:
Is this integral $\int_0^\infty\frac{\cos(a x+ 2b \arctan x)}{x^2+1}dx$ exactly zero when $b\in\mathbb{N}$?
I recently encountered this integral
$$\int_0^\infty\frac{\cos(a x+ 2b \arctan x)}{x^2+1}dx$$
which suspiciously close to 0 for nonzero integer values of $b$, as indicated by numerical calculations. When $b$ is not an integer it is not 0. Based on my experiments I conjecture that
$$\int_0^\infty\frac{\cos(a x+ 2b \arctan x)}{x^2+1}dx=0,~~~~b\in \mathbb{Z}\setminus \{0\}.$$

Question. Is the above conjecture true?

答:
Assume that $a > 0$ and that $n$ is a positive integer.

Then we have

$$ \begin{align}\int_{0}^{\infty} \frac{\cos \left(ax  + 2 n \arctan x \right)}{1+x^{2}} \, \mathrm dx &= \frac{1}{2} \, \Re \int_{-\infty}^{\infty} \frac{\exp \left(iax + i2n \arctan x \right)}{1+x^{2}} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\infty}^{\infty} e^{iax} \, \frac{ \exp \left(n \log(1+ix) - n\log(1-ix)  \right)}{1+x^{2}} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\infty}^{\infty} e^{iax} \, \frac{(1+ix)^{n}}{(1-ix)^{n}(1+x^{2})} \, \mathrm dx \\ &=  \frac{1}{2} \, \Re \int_{-\infty}^{\infty} e^{iax} \,  \frac{(1+ix)^{n-1}}{(1-ix)^{n+1}} \, \mathrm dx.  \end{align}$$

But the integrand is analytic in the upper half of the complex plane, and by Jordan's lemma (or the estimation lemma) the integral $$\int_{R\exp(i[0, \pi])} e^{iaz} \, \frac{(1+iz)^{n-1}}{(1-iz)^{n+1}} \, \mathrm dz$$  vanishes as  $R \to \infty$.

Therefore, $$\int_{0}^{\infty} \frac{\cos \left(ax  + 2 n \arctan x \right)}{1+x^{2}} \, \mathrm dx =0. $$


问:
Uniqueness of the connecting morphism in the snake lemma
Consider the following commutative diagram in an abelian category:$$\require{AMScd}
\begin{CD}
@. A @>{f}>> B @>{g}>> C @>>> 0\\
@. @VV{a}V @VV{b}V @VV{c}V  \\
0@>>> A' @>>{f'}> B' @>>{g'}> C'
\end{CD}$$

The snake lemma gives a morphism $\delta:\ker c\to \operatorname{coker} a$ making the sequence
$$\ker a \to \ker b\to \ker c\to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c$$
exact. I wonder if the morphism $\delta$ is unique such that the induced sequence above is exact.

The proof of the snake lemma using elements makes me believe that the answer is yes, but I'm not really sure.

(This is basically the same as Uniqueness of the connecting homomorphism in theory of abelian categories but my question adds a condition, so I think that it was worth making a new question. If someone is against this, I can delete my question.)

答:
In order for the sequence to be exact, all you need is an isomorphism between $\def\coker{\operatorname{coker}}\ker c/\operatorname{im}(\ker b\to \ker c)$ and $\ker(\coker a\to\coker b)$. By the snake lemma, these objects are isomorphic, but they may well have non-trivial automorpshims that you can sneak into $\delta$.
For abelian groups, you may for example always replace $\delta$ with $-\delta$ (which is the same as $\delta$ only when the quotients in question are of exponent 2). Incidentally, $-\delta$ is also natural.


答:
Since $$\lim_{x\to\ 0} f(x) = f(0)$$
we know that $f$ is continuous particularly on $[-1,1]$. Hence, it is uniform continuous on $[-1,1]$.

Now, assume $x \in \mathbb R \setminus [-1,1]$. Then, by Lagrange's theorem (or MVT), we have
$$\exists \xi \in (x,y): f(x)-f(y) = f'(\xi)(x-y) \tag 1$$
Also,
$$|f'(x)| = \left|\sin\frac 1x - \frac 1x\cos \frac 1x\right| \le 1 + 1 = 2 \tag 2$$
Therefore,
$$(1)\land (2) \implies |f(x) - f(y)| = |f'(x)||x-y| \le 2|x-y|$$
Now, apply $\varepsilon$-$\delta$ definition.

Title: Is $x\sin\left(\frac{1}{x}\right)$ uniformly continuous in $\mathbb{R}$?

Body:
Let $f:\mathbb{R}\xrightarrow{}\mathbb{R}$ with rule defined as
$$
f(x) = \left\{\begin{array}  xx\operatorname{sin}\left(\frac{1}{x}\right) &\text{  if }x\neq0 \\
                      0  &\text{ if } x=0
\end{array}\right.
$$
I would like to show that this function is uniformly continous over $\mathbb{R}$.
I started out with the $\epsilon-\delta$ definition and tried to manipulate the following expression without success,
$$
\left|x \operatorname{sin}\left(\frac{1}{x}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right |\\
=\left|x \operatorname{sin}\left(\frac{1}{x}\right) -x \operatorname{sin}\left(\frac{1}{y}\right) +x \operatorname{sin}\left(\frac{1}{y}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right | \\
\leq \left|x \left(\operatorname{sin}\left(\frac{1}{x}\right) - \operatorname{sin}\left(\frac{1}{y}\right)\right) + (x - y) \operatorname{sin}\left(\frac{1}{y}\right) \right |
$$
Also, the function isn't Lipschitz, so I can't go this way too.
I've seen some other proofs at mathSE however they try to show things for smallers domains as $[0,1]$.


问:
Show that the series $\sum_{n=0}^{\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$ is divergent
In this thread, I use the "C+iS" summation method for trigonometric series to sum it:

lab bhattacharjee pointed out that:

$\left|\dfrac{e^{ix}}{\cos(x)}\right|=|\sec(x)|\geq1$

What I don't understand is why $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, I thought it is $e^{ix}\sec(x)$. By looking at the graph of $\sec(x)$, it is clear that $\sec(x)\geq1 $ on the interval $(0,\pi/2)$ and $(0,5\pi/2)$ and so on. $\sec(x)\leq1$ on the interval $(\pi, 3\pi/2)$ and $(3\pi/2, \pi)$. So I get the part if $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, then the series diverge.

Also, how can I show that the series $\left(\dfrac{e^{-ix}}{\cos(x)}\right)^n$ is a divergent series?

答:
Note that
$$\left|\dfrac{e^{ix}}{\cos x}\right| = \left|\frac{\cos x + i\sin x}{\cos x}\right| = \left| 1 + i\tan x\right| = \sqrt{1+\tan^2 x} = \left|\sec x\right|$$


问:
Convergence of non-negative $(a_n)^p$ series if the series $a_n$ converges
I tried to prove the following:

Suppose that $a_n$ is a sequence of non-negative numbers and that the series $\sum a_n$ converges. Prove that the series $\sum a_n^p$ converges for p>1.

How should I start this prove? Should I break down $a_n$ into three cases: $a_n$ equals 1, less then 1, greater than 1? or Should I use comparison test?

答:
Hint:

There exists $\;N\in\Bbb N\;$ s.t. $\;n>N\implies a_n<1\implies a_n^p<a_n\;$ for $\;p>1\;$ ...


答:
hint

$ f $ is even and continuous at $ \Bbb R $ .

Given $ \epsilon>0$.

$$\lim_{|x|\to+\infty}f(x)=1$$

So, for $ |x| $ and $ |y| $ large enough,

Let us say, $$x>y>A>0\text{ or  } x<y<B<0$$

$$|f(x)-f(y)|\le|f(x)-1|+|f(y)-1|<\epsilon$$

On the other hand, $ f $ is uniformly continuous at the compact $ [A,B] $,

So, There exists $ \eta>0 $, such that

for $ x,y\in[A,B] $
$$|x-y|<\eta\implies |f(x)-f(y)|<\epsilon$$

Now, use continuity at $ A $ and $ B$ for the cases
$$x<A<y \text{ or } x<B<y$$


答:
Well, you already gave the answer in your question.

If $\sum a_n$ converges, $\{a_n\}$ converges to zero. As we’re dealing with non negative series and $p \gt 1$, we have for all $n $ large enough

$$0\le a_n^p \le a_n$$ and therefore $\sum a_n^p$ converges.

Title: Convergence of non-negative $(a_n)^p$ series if the series $a_n$ converges

Body:
I tried to prove the following:

Suppose that $a_n$ is a sequence of non-negative numbers and that the series $\sum a_n$ converges. Prove that the series $\sum a_n^p$ converges for p>1.

How should I start this prove? Should I break down $a_n$ into three cases: $a_n$ equals 1, less then 1, greater than 1? or Should I use comparison test?


问:
Need a good method for computing $\int _{\infty }^a\:\frac{1}{\left(x^2-1\right)^p}dx$
I need a good method for computing $$\int _{\infty }^a\:\frac{1}{\left(x^2-1\right)^p}dx$$
Where $p>0$ and $a>1$ are real. Maybe you have a good numerical algorithm for such a computation? That would really help me, thank you.

答:
Some useful information

The integral converges if $a>1$ and $p>1/2$.

The integrand is decreasing rapidly. We will only have to look at a few small values.

We define the integral
$$I(a,p)=\int\limits_{a}^{\infty}\frac{1}{(x^2-1)^p}\,\mathrm{d}x.$$

For big $p$

In this graph we set $p=50$. You can see that the values are very small. The spike at $3$ might seem confusing, but note that that the $y$-values only go up to $2.5\cdot 10^{-40}$. Also observe that if we increase $a$, the integral area under the integrand get's smaller and smaller. I will only work out some values for $p\le3$.

For $p\le3$

Here are the values for $p/2$, where $p\in\mathbb{N}$ and $p<6$.

Input
Output
Logical $a$

$I(a,1)$
$-\dfrac{\ln\left(a+1\right)-\ln\left(a-1\right)}{2}$
-

$I(a,1.5)$
$1-\dfrac{a}{\sqrt{a^2-1}}  $
$a<50$

$I(a,2)$
$\dfrac{\left(a^2-1\right)\ln\left(a+1\right)-\ln\left(a-1\right)a^2-2a+\ln\left(a-1\right)}{4a^2-4}$
$a<10$

$I(a,2.5)$
$\frac{\sqrt{a^2-1}\left(2a^3-3a\right)}{3a^4-6a^2+3}-\dfrac{2}{3}$
$a<5$

$I(a,3)$
$-\dfrac{\left(3a^4-6a^2+3\right)\ln\left(a+1\right)-3\ln\left(a-1\right)a^4-6a^3+6\ln\left(a-1\right)a^2+10a-3\ln\left(a-1\right)}{16a^4-32a^2+16}$
$a<4$

What I mean by "Logical $a$" is a (nice) value, where the output $I(a,p)$ is $>0.0001$, except for $p=1$. So as you can see, for $p=1$ it really makes sense to get a closed form. To get values (from hand) between $p$ and $p+0.5$, you can use the values in the table for upper and lower bounds. If you can use a computer to evaluate these values, I don't see a problem to do this with arbitrary $p$. Also if you are looking for a good calculator, I recommend Mathematica or Matlab.


问:
Differences Between Row Echelon and Reduced Row Echelon
Let $$A =\begin{bmatrix} 1&3&-6&2\\5&12&-24&7\\9&15&-30&6\end{bmatrix}$$

$$RREF(A) =\begin{bmatrix} 1&0&0&-1\\0&1&-2&-1\\0&0&0&0\end{bmatrix}$$
$$REF(A) =\begin{bmatrix} 1&3&-6&2\\0&1&-2&1\\0&0&0&0\\\end{bmatrix}$$

Now, provided this information, it can be found that the basis for the nullspace of $A$ using the $RREF(A)$ is: $$\begin{pmatrix} \begin{bmatrix} 0\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} 1\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$

Using the $REF(A)$ the "basis" for the nullspace is:

$$\begin{pmatrix} \begin{bmatrix} 6\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} -2\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$

Is the "basis" that I provided for the $REF(A)$ actually a basis? How does the  $REF$ and $RREF$ of a matrix differ?

答:
Row Echelon form(REF) requires backward  substitution while
Row Reduce Echelon form (RREF) requires no backward substitution.</p>


问:
Using L'Hopital's Rule of Multivariable Limits Converted into Single-Variable Limits
I've been given a couple of multivariable limit questions to solve but I don't fully understand if L'Hopital's rule can apply.  Some other people I have talked to say we can split up the limits and calculate them separately e.g.

$\lim\limits_{(x,y) \to (0,0)} \frac{x^2y + sin(y)}{y} = \lim\limits_{(x) \to (0)}x^2 + \lim\limits_{(y) \to (0)} \frac{sin(y)}{y}$

Then applying L'Hopital's Rule to get the limit to be 1, however, some other people are saying we can't use L'Hopital's Rule on multivariable limits.  My understanding is that we have now separated this limit into two single variable limits so we should be able to use L'Hopital's Rule.

My second question about using L'Hopital's Rule is, if we are trying to prove that a limit doesn't exist by for example making $y = \alpha x$.  If we replace all instances of $y$ with $\alpha x$, can we now use L'Hopital's Rule to simplify down the limit and show that if we have different values of $\alpha$ we would get a different limit hence the limit doesn't exist?

答:
When you convert a multivariable limit to a single-variable limit all the single-variable limit rules apply.  This means that if you have:

$\lim\limits_{(x,y) \to (0,0)} \frac{x^2y + sin(y)}{y} = \lim\limits_{(x) \to (0)}x^2 + \lim\limits_{(y) \to (0)} \frac{sin(y)}{y}$

You can use any of the rules you would a single-variable limit.  This means L'Hopital's Rule is allowed.  This is also true after you have substituted an equation e.g. $y = \alpha x$.


问:
Do Cauchy-Riemann Equations Hold?
Consider $f(z) = \frac{(\overline{z})^2}{z}$ if $z \neq 0$ and $f(z) = 0$ if $z = 0$

Do Cauchy-Riemann Equations only hold at (0,0)?

I'm confused as to how to start this problem. I have tried converting the function $\frac{(\overline{z})^2}{z}$ into the form U(x,y) + iV(x,y) by letting $z = x+yi$ and $\overline{z} = x-yi$ but that just makes things over complicated and I end up with division by zero at the point (0,0).

Can anyone point me in the right direction?

答:
First, using the form $u+iv$ this problem can be solved after a lengthy calculation in which it is show that the function is differenciable in the real sense. You obtain that

$$f(x+iy)=\frac{x^3-3xy^2}{x^2+y^2}+i\frac{3x^2y-y^3}{x^2+y^2}$$

And here, differentiating, using the definition of derivative, we obtain at $(0,0)$ that

$$\frac{\partial u}{\partial x}(0,0)=\lim_{(x,y)\to (0,0)}\frac{\frac{x^3-3xy^2}{x^2+y^2}-0}{x}=\lim_{(x,y)\to (0,0)}\frac{x^2-3y^2}{x^2+y^2}$$

Where we see that this limit does not exist, since by taking the paths $t\mapsto(\sqrt{3}t,t)$ and $t\mapsto(t,t)$ we obtain different limits when $t\to 0$.

Second, you can do this directly by computing

$$\lim_{z\to 0}\frac{f(z)-f(0)}{z}=\lim_{z\to 0}\left(\frac{\overline{z}}{z}\right)^2$$

And then use polar form to see that the limit does not exist, since it depends of the path of approximation to zero.

Title: Do Cauchy-Riemann Equations Hold?


问:
Show a function is bounded below by $\sqrt{a}$
I have to show a function $$f(x)= \frac{1}{2}\left(\frac{a}{x}+x\right),$$ where $x>0$, is bounded from below by $\sqrt a$.

The first thing I think of is: if a converges, then it is bounded. So my question: is looking if the limit of $f(x)$ from left and right is $\sqrt a$ the right way to go?

答:
hint

$ f $ is an odd function.

Assume $ x\ne 0 $, then

$$2x(f(x)-\sqrt{a})=(\sqrt{a})^2+x^2-2x\sqrt{a}$$

$$=(\sqrt{a}-x)^2$$


问:
Algebraic derivatives of a given function
I was studying about fractional derivatives and after that did some own work on imaginary order derivatives. Now I'm curious to know are algebraic order derivatives possible?
Like the xth derivative of x.</p>

答:
Continuing my comment:

You should notice how this is an entirely new structure. As when we differentiate we usually do it a $\mathbb{N_+}$ - [integer] amount of times. Now people generalized this idea to the $\mathbb{Q}$ - [rational numbers], $\mathbb{R}$ - [reals] or even $\mathbb{C}$ - [complex numbers]. The common idea is numbers. But the core concept stems from a discrete amount of operations.

But [x] is not a number. It can be a function or an indexing letter or a parameter. Now when you differentiate a function $f(x)$ you don't just take it at one point (like $x_0$) . You try to find its derivative on an as big domain as possible (often $\mathbb{R}$ and its varieties).

If you want to to differentiate $[x]$ amount of times while $x$ is going from $-\infty$ to $\infty$ (so the domain being $\mathbb{R}$ presumably) then you'll have to find a different kind of derivative for every single point and you'll run into some problems.

Using the original idea, you'd need to calculate a lot.

Namely $$\text{card}(\mathbb{R})=\mathfrak{c}=2^{\aleph_0}$$

amount of times. You can imagine $\aleph_0$ as how many rationals are there. So you'd need to calculate an exponential amount.

Of course this logic is rather the computer science side of the problem. You could just design a new structure where the derivative-operator continously changes as we shift through the reals. But figuring out that is rather your job.

Specify your question so we can give you a better answer!


问:
Evaluating a definite integral with e
Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

答:
I'm not quite certain what you did - usually when you substitute you are looking for something in the integrand to substitute out (or more advanced, to reduce the integrand into something "nicer", which can lead to some creative substitutions).

A good rule of thumb for intro calculus is, look for something whose derivative is already in the problem or almost in the problem, and pick that for your u. Here, I would make the substitution $u=2x^3$ since $u'=6x^2$ differs only by a constant from $x^2$.

Your integral becomes $\frac{1}{6}\int_0^2 e^{-u} du $. Can you work out the rest?

NB to OP (taking into consideration some of the other answers): Remember you're doing a definite integral so you shouldn't have $+C$ in your answer. Some of the other answers that have it aren't wrong but they've only taken you up to the antiderivative step; you still have to evaluate it at your bounds of integration.

Title: Evaluating a definite integral with e


问:
When $\frac{z-1}{z+1}$ is a real number?
I wrote it this way: $\frac{z-1}{z+1} = \frac{(z-1)(\overline{z}+1)}{(z+1)(\overline{z}+1)}=\frac{|z|^2 + z - \overline{z} -1}{|z+1|^2} = \frac{|z|^2 -1+2iImz}{|z+1|^2}$ and
I calculated that for $|z|=1 $ this quotient $ \frac{z-1}{z+1}$ is a purely imaginary number, but I don't know for which $ z $ it is a real number.

答:
You've shown that $$\frac{z-1}{z+1} = \frac{|z|^2 -1}{|z+1|^2} +i \frac{2 \operatorname{Im}z}{|z+1|^2}$$
so that the fraction is purely real if and only if $\operatorname{Im} z = 0$ and $z \not= -1$.


问:
Calculate $\lim_{n \to \infty}\frac{a^n}{n!}$
Calculate $$\lim_{n \to \infty}\frac{a^n}{n!}$$

Attempt

Consider $$\lim_{n \to \infty}\exp(\log(\frac{a^n}{n!}))=\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)$$

for $a>1$
$$ \lim_{n \to \infty}\left(n\log(a)\right)=\infty $$
$$ \sum_{n \geq 1}\log(n)\leq \int_{1}^{\infty}{\log(x)dx}=\infty$$

therefore $$\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)=\exp(-\infty)=0$$
therefore

$$\lim_{n \to \infty}\frac{a^n}{n!}=0$$
Is my proof right?

I think that is not general, since $a>1$ and in the problem $a$ is arbitrary

答: Hint : Starting from your idea, try to find the limit of $n log(a) - \sum_{k=1}^n log(k)$. Then apply what you know about the $exp$ function.


问:
Create non-principal filter contained in a principal ultrafilter
Please, consruct a non-principal filter $\mathscr F$, which is contained in a principal ultrafilter $\mathscr G$.</p>

答:
I’m assuming for this answer that you translated as substantive and chief some word whose correct English translation in this context is principal.

Let $E=\{2n:n\in\Bbb N\}$, let $\mathscr{U}$ be a non-principal ultrafilter on $\Bbb N\setminus E$, and let $$\mathscr{F}=\{E\cup U:U\in\mathscr{U}\}\;;$$ then $\bigcap\mathscr{F}=E$, but $E\notin\mathscr{F}$, so $\mathscr{F}$ is non-principal. However, it’s easy to extend $\mathscr{F}$ to a non-principal filter $\mathscr{G}$ on $\Bbb N$: if $\mathscr{V}$ is any non-principal filter on $E$, $\mathscr{V}$ is a base for such a $\mathscr{G}$.


问:
Solivng differential equation by subtracting Y.
I have the equation $$ \frac{dy}{dx} = y + 2$$ And normally u would isolate the y's and the x's but what if I subtract y and then solve the equation. That would like the following. $$ \frac{dy}{dx} - y =2 $$ $$
 \int(\frac{dy}{dx} - y)dx = \int2dx$$ $$ \int\frac{dy}{dx}dx - \int ydx = \int 2dx $$ $$ \int1dy - \int ydx = \int2dx $$ $$ y - yx = 2x + C$$ $$ y = \frac{2x+C}{1-x}$$ But when you check this u get an equation where your only solution that works is when x=0. What am is wrong with the math.

答:
$$ \frac{dy}{dx} = y + 2$$
you should  do
$$\frac{dy}{y+2}=dx$$
and then integrate
$$\log(y+2)=x+C$$
$$y=e^{x+C}-2$$


答: Alternatively, if you know the fact that a Möbius transformations sends lines in $\mathbb{C}\cup\{\infty\}$ to lines or circles in $\mathbb{C}\cup\{\infty\}$, then it suffices to check the image of the points $-1,0,1$ under $\phi(z)\equiv(z-1)/(z+1)$. We have $\phi(0)=-1$, $\phi(1)=0$, and $\phi(-1)=\infty$, so it follows that $\phi(\mathbb{R}\cup\{\infty\})=\mathbb{R}\cup\{\infty\}$. Since $\phi$ is a homeomorphism of $\mathbb{C}\cup\{\infty\}$ (and hence a bijection), it follows that $\phi(z)$ is real if and only if $z$ is real.

Title: When $\frac{z-1}{z+1}$ is a real number?

Body:
I wrote it this way: $\frac{z-1}{z+1} = \frac{(z-1)(\overline{z}+1)}{(z+1)(\overline{z}+1)}=\frac{|z|^2 + z - \overline{z} -1}{|z+1|^2} = \frac{|z|^2 -1+2iImz}{|z+1|^2}$ and
I calculated that for $|z|=1 $ this quotient $ \frac{z-1}{z+1}$ is a purely imaginary number, but I don't know for which $ z $ it is a real number.


答:
Replying to your last comment: Yes there is a faster way. Show that for sufficiently large $n$ the following inequality holds:
$$n!>(a+1)^n.$$
We now have
$$\lim_{n\to\infty} \frac{a^n}{n!} \leq \lim_{n\to\infty} \frac{a^n}{(a+1)^n} = \lim_{n\to\infty} \left(\frac{a}{a+1}\right)^n \to 0.$$

Title: Calculate $\lim_{n \to \infty}\frac{a^n}{n!}$

Body:
Calculate $$\lim_{n \to \infty}\frac{a^n}{n!}$$

Attempt

Consider $$\lim_{n \to \infty}\exp(\log(\frac{a^n}{n!}))=\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)$$

for $a>1$
$$ \lim_{n \to \infty}\left(n\log(a)\right)=\infty $$
$$ \sum_{n \geq 1}\log(n)\leq \int_{1}^{\infty}{\log(x)dx}=\infty$$

therefore $$\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)=\exp(-\infty)=0$$
therefore

$$\lim_{n \to \infty}\frac{a^n}{n!}=0$$
Is my proof right?

I think that is not general, since $a>1$ and in the problem $a$ is arbitrary


问:
Number of functions - combinatorics
I have a problem with a question in combinatorics even though it seems like a simple one unfortunately I am still having trouble to see the solution...

The question is:

Let $A = \{1,2,3,4,5,6\}$
Find the number of functions $f: A \rightarrow A$ such that $\{1,2,3,4\} \subseteq f[A]$

I tried to solve it with Inclusion-exclusion principle:

We will denote $Ai$ = {Group of all functions where i isn't in the image of $f[A]$}

Then I looked for of $A1^\mathsf{c} \cap A2^\mathsf{c} \cap A3^\mathsf{c} \cap A4^\mathsf{c}$
which is by De-Morgan : $(A1 \cup A2 \cup A3 \cup A4)^\mathsf{c}$ and then I looked for $U - A1 \cup A2 \cup A3 \cup A4$

Calculations:
$U = 6^6$

$|A1| = |A2| = |A3| = |A4| = 6^5$

$|Ai \cap Aj| = 6^4$

$|Ai \cap Aj \cap Ak| = 6^3$

$|Ai \cap Aj \cap Ak \cap At| = 6^2$

$|U - A1 \cup A2 \cup A3 \cup A4| = 6^6 - {4 \choose 1}6^5 + {4 \choose 2}6^4 - {4 \choose 3} 6^3 + {4 \choose 4}6^2 = 22,500$

Then used inclusion-exclusion 'formula' to solve it.
But it seems to give me the wrong answer (Correct answer is 5880)

I'll appreciate any help, Thanks !

答: You mistake is the cardinality $|A_i|$. For example $A_1$ maps a set of 6 elements onto the set of 5 elements, therefore the number is $5^6$ not $6^5$. The same story is with other cardinalities.

Title: Number of functions - combinatorics


答:
It is real if & only if it is equal to its conjugate, namely, since complex conjugation is a field automorphism,
\begin{align}
\frac{z-1}{z+1}=\frac{\bar z-1}{\bar z+1}&\iff(z-1)(\bar z+1)=(z+1)(\bar z-1) \\
&\iff z\bar z+z-\bar z-1= z\bar z-z+\bar z-1 \\&\iff 2z=2\bar z,\quad\text{ i.e. }\enspace z=\bar z,
\end{align}
in other words, if & only if $z$ is real.


问:
How to prove that an irrational number to an irrational number could be irrational. - SOLVED
There exists $a,b \in \mathbb{R}-\mathbb{Q}$ such that $a^b \in \mathbb{R}-\mathbb{Q}$.

I am lost as to where to begin. In my class, the only irrational numbers we've really covered is $\sqrt{2}$ and cubed root of two. Any help would be appreciated!

Edit : How my professor did it.

Proof: Consider $\sqrt{2}^{\sqrt{2}}\cdot\sqrt{2}^{1-\sqrt{2}} = \sqrt{2}$. Since a rational multiplied by an irrational is irrational, either $\sqrt{2}^{\sqrt{2}}$ is irrational or $\sqrt{2}^{1-\sqrt{2}}$ is irrational. We will proceed by cases.

Case 1: $\sqrt{2}^{\sqrt{2}}$ is irrational. Then $a=\sqrt{2},b=\sqrt2$.

Case 2: $\sqrt{2}^{1-\sqrt{2}}$ is irrational. Then $a=\sqrt2,b=1-\sqrt2$.

Thus there exists irrationals a,b where $a^b$ is irrational.

答: There are several ways to approach this problem at an elementary level. But it depends a lot on the context and the material you are expected to know. Here is a suggestion. Consider the function $f(x)=e^x$. I assume you know that $e$ is irrational and that $f$ is injective on the reals. If it were the case that for all irrational $x$ the value $f(x)$ were rational, then you would have an injection from the irrationals to the rationals, thus violating cardinality.


问:
Convex hull of an open set is an open set
I have to prove this. Actually, I have the proof, but I don't understand one part.

It says:

"Since $\operatorname{co}A$ is intersect of all convex sets that contain set A, it follows that $\operatorname{co}A\subset (\operatorname{co}A)^\circ$".

答:
I assume that an earlier part of the proof showed that the interior of a convex set is convex, so that $(\operatorname{co}A)^\circ$ is convex. Also, $A$ is open so it is contained in its own interior, and therefore in the interior of the larger set $\operatorname{co}A$.

So now $(\operatorname{co}A)^\circ$ is a convex set containing $A$, and therefore it also contains $\operatorname{co}A$.


问:
How do continue/end this infinite limit expression? Shall I use the Reciprocal Rule? - Almost finished
The picture of my work is here

I am almost done with this, but I do not know how to end it? Shall I use the Reciprocal rule?

答:
You are missing the $\lim_{x \to \infty}$ on the final step; you haven't applied the limit yet.

When you do apply it, the terms $\frac 4x$, $\frac 2{x^2}$, etc., will disappear since they approach $0$. You will be left with $\frac25$.


问:
Cauchy-Schwarz Inequality Proof Question
I'm trying to follow proof 2 of the Cauchy-Schwarz Inequality from Wikipedia https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Statement_of_the_inequality

I've gotten to the point where they let $\mathbf {u} ={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} +\mathbf {z}$. Then they say this gives $ \|\mathbf {u} \|^{2}=\left|{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\right|^{2}\|\mathbf {v} \|^{2}+\|\mathbf {z} \|^{2}$

How do they get this equivalence? I started by taking the norm of both sides, and and get $\|\mathbf{u}\| = ||{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} \|+\|\mathbf {z} \|$. How do you go further from here?

Really, I'm very confused as to how you can start with norms and ever end up with an absolute value?

答:
If $\mathbf{x}$ and $\mathbf{y}$ are orthogonal, then
$$\begin{align*}
\lVert \mathbf{x}+\mathbf{y}\rVert^2 &= \langle \mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle\\
&= \langle \mathbf{x},\mathbf{x}\rangle + \langle\mathbf{x},\mathbf{y}\rangle + \langle \mathbf{y},\mathbf{x}\rangle + \langle \mathbf{y},\mathbf{y}\rangle\\
&= \lVert \mathbf{x}\rVert^2 + \lVert \mathbf{y}\rVert^2.
\end{align*}$$
This is the "generalized Pythagorean Theorem" (think about vectors $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{x}+\mathbf{y}$ on the plane, forming a triangle).

In your expression, $\mathbf{v}$ and $\mathbf{z}$ have been shown to be orthogonal to each other, hence so are $\alpha\mathbf{v}$ and $\mathbf{z}$ for any scalar $\alpha$. The expression now follows.


问:
Necessity and utility of infinite dimensional spaces
I have a group of similar questions about infinite dimensional spaces. I think these are a little discursive and trivial, but I can't find any answer.

-Why is the concept of infinite dimensional spaces so important and useful?

-From which field did the concept arise and for which necessity?

-Is this concept in some cases related to describing objects with a non finite number of degrees of freedom? If so, does this connection hold for every case?

答:
There are entire books answering these questions. Here is a nutshell answer to your three questions.

The concept is important since it arises in applications, primarily (and the origin of the research into it) in quantum mechanics as well as in differential equations. It is useful since the theory of infinite dimensional spaces is very well behaved and thus offers practical techniques for solving hard problems.

Similar to the above. The field arose from formulations in physics around the turn of the 20th century that used spaces of functions.

Yes. No.


答:
Here is an alternative:

It seems you are working with $a>1$, although it is not stated at the beginning but rather in the middle. I will go with that assumption:

$a^n>0$ and $n! > 0, \forall n \in \mathbb{N}$.

Now,

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n} $

At some point we will have $n > a$, say this occurs at $N$, take $n > N$ then

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n}  \le \frac{a \cdots a}{1 \cdots (N-1)}\cdot \frac{a \cdots a}{N \cdots n} \le  \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n}$

Now, since we have $m > a, \forall m \in \{N,\ldots,n\}$ we know $\displaystyle \frac{a}{m} < 1$

Hence

$\displaystyle \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n} \le \frac{a^{N-1}}{(N-1)!}\cdot \frac{a}{n} \to 0$ as $n \to \infty$


问:
Find a value of $\;\lim\limits_{n\rightarrow\infty}n\left ( e- e^{\frac{1}{e}}\uparrow\uparrow n \right )$

Find a value of
$$\lim_{n\rightarrow\infty}n\left ( e- e^{\frac{1}{e}}\uparrow\uparrow n \right )$$
For your information$,\quad\uparrow\uparrow$ is a tetration defined as
$$a\uparrow\uparrow n:=\underbrace{a^{a^{\cdot^{\cdot^{\cdot^{a}}}}}}_{n}$$

I think we must use the function $\ln x$ here then use the Laurent series of $\ln x.$ But I couldn't find an extension formula of $\ln e^{\frac{1}{e}}\uparrow\uparrow n,$ I need to the help, thanks a real lot !

答:
To put that into a more reasonable (though less prone to enthusiastic upvotes) form: We have $x_1=e^{1/e}$ and $x_{n+1}=e^{x_n/e}$, and we're looking for $\lim_{n\to\infty}n\,(e-x_n)$. Letting $y_n=1-x_n/e$, some elementary algebra gives $$y_{n+1}=1-e^{-y_n}.$$ The precise value of $y_1$ is not so important, as long as it is positive. Then, $y_n$ is monotone decreasing, and positive, so it must have a limit $y$ satisfying $y=1-e^{-y}$, i.e. we must have $\lim_{n\to\infty}y_n=0$. Now some more elementary algebra gives
$$\frac1{y_{n+1}}-\frac1{y_n}=\frac{e^{-y_n}+y_n-1}{y_n\,(1-e^{-y_n})}\to\frac12$$ as $n\to\infty$, and by the   Stolz–Cesàro theorem, we have $$\lim_{n\to\infty}\frac1{n\,y_n}=\frac12.$$ Substituting the definition of $y_n$, we obtain $$\lim_{n\to\infty}n\,(e-x_n)=2\,e.$$ Convergence is rather poor, the error is $O\left(\frac{\log n}n\right)$.


答: @IttayWeiss's answer is good, but we don't even need to know irrationals are uncountable. Since $x:=\sqrt{2}^{\sqrt{2}}>1$, even if $x$ is rational its prime factorization ensures all sufficiently large $n\in\Bbb N$ satisfy $x^{1/n}\notin\Bbb Q$, whence we can take $a=\sqrt{2},\,b=\sqrt{2}/n$. (The Gelfond-Schneider theorem shows $n=1$ works anyway.)

Title: How to prove that an irrational number to an irrational number could be irrational. - SOLVED

Body:
There exists $a,b \in \mathbb{R}-\mathbb{Q}$ such that $a^b \in \mathbb{R}-\mathbb{Q}$.

I am lost as to where to begin. In my class, the only irrational numbers we've really covered is $\sqrt{2}$ and cubed root of two. Any help would be appreciated!

Edit : How my professor did it.

Proof: Consider $\sqrt{2}^{\sqrt{2}}\cdot\sqrt{2}^{1-\sqrt{2}} = \sqrt{2}$. Since a rational multiplied by an irrational is irrational, either $\sqrt{2}^{\sqrt{2}}$ is irrational or $\sqrt{2}^{1-\sqrt{2}}$ is irrational. We will proceed by cases.

Case 1: $\sqrt{2}^{\sqrt{2}}$ is irrational. Then $a=\sqrt{2},b=\sqrt2$.

Case 2: $\sqrt{2}^{1-\sqrt{2}}$ is irrational. Then $a=\sqrt2,b=1-\sqrt2$.

Thus there exists irrationals a,b where $a^b$ is irrational.


Title: Evaluating a definite integral with e

Body:
Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

Body:
$$u=2x^3$$
$$du=6x^2\space{dx}$$
$$\int\frac{x^2}{e^{2x^3}}dx$$
$$\frac{1}{6}\int{e^{-u}}du$$
$$-\frac{1}{6}e^{-u}+C=-\frac{1}{6}e^{-2x^3}+C$$


答:
Consider $x=\sqrt{2}^\sqrt{2}$. It's either rational or irrational. If it's irrational, you have your answer. If it's rational $x=\frac{p}{q}$, then using the decompositions of $p$ and $q$ into prime factors it's easy to show that there exist $n\in\mathbb{N}$ such that
$$ \sqrt[n]{x} = \sqrt{2}^{\sqrt{2}/n} $$
is irrational, and then your answer is the pair $(\sqrt{2},\sqrt{2}/n)$.


问:
How to compute probability of a bootstrap sample

The Question

Consider the  samples $\{1, 3, 4, 6\}$ from some distribution.

a) For one random bootstrap sample, find the probability that the mean
is $1$.

b) For one random bootstrap sample, find the probability that the
maximum is $6$.

c) For one random bootstrap sample, find the probability that exactly
two elements in the sample are less than $2$.

My Understanding

We just started to learn the bootstrap in class and I came across this question. I'm a little confused, as I feel like this question is too easy, as the mean of any sample with those numbers is always $3.5$ so a) is $0$. The maximum will always be $6$ so b) is $1$. And $2$ of the numbers cannot be less than $2$. So c) is $0.$

Is there something major that I'm missing?

答:
I think this is a very thoughtful question, leading up to an understanding of how bootstrapping works. Your attempted answers are not exactly on target, so here is my attempt
to clarify. Because bootstrapping is a simulation-based, I begin with simulation results before showing exact binomial probabilities for each part.

Simulation: One bootstrap sample will take four values at random with replacement from the set $\{1,3,4,6\}.$ Let's simulate many re-samples and
see what happens. With a million bootstrap samples simulated
probabilities should be accurate to two or three places.

(a) Average of four is $1;$ (b) max is $6;$ (c) need exactly two $1$'s.

```<code>set.seed(1)
x = c(1,3,4,6)

a = replicate(10^6, mean(sample(x, 4, rep=T)))
mean(a == 1)
[1] 0.003916  # aprx  0.00390625

w = replicate(10^6, max(sample(x, 4, rep=T)))
mean(w == 6)
[1] 0.683426  # aprx 0.6835938

nr.ones = replicate(10^6, sum(sample(x, 4, rep=T)==1))
mean(nr.ones==2)
[1] 0.210837  # aprx 0.2109375
</code>```

So the respective probabilities for parts (a)-(c) are
approximately $0.004, 0.684,$ and $0.211.$

Exact binomial probabilities: Exact probabilities can be found using the binomial distribution. Exact probabilities computed using R, where dbinom is a binomial PDF and pbinom is a binomial CDF. You can easily use
the appropriate binomial PDF formula to do the computations.

(a) Ones are successes. The number of ones in four draws
is $X_1 = \mathsf{Binom}(n=4, p = 1.4).$ In order for the average to be $1,$ we need all ones. $P(X_1 = 4) = 0.0039.$

```<code>dbinom(4, 4, 1/4)
[1] 0.00390625
</code>```

(b) Sixes are successes. In order for the max to be $6,$
we need at least one six. $X_2 = \mathsf{Binom}(4,1/4),$
$P(X_2 \ge 1) = 0.6836.$

```<code>sum(dbinom(1:4, 4, 1/4))
[1] 0.6835938

1 - dbinom(0, 4, 1/4)
[1] 0.6835938
</code>```

(c) Values below $2$ are successes. Only ones are smaller.
So we need exactly two Ones:  Probability is $0.2109.$

```<code>dbinom(2, 4, 1/4)
[1] 0.2109375
</code>```

Body:
I think this is a very thoughtful question, leading up to an understanding of how bootstrapping works. Your attempted answers are not exactly on target, so here is my attempt
to clarify. Because bootstrapping is a simulation-based, I begin with simulation results before showing exact binomial probabilities for each part.

Simulation: One bootstrap sample will take four values at random with replacement from the set $\{1,3,4,6\}.$ Let's simulate many re-samples and
see what happens. With a million bootstrap samples simulated
probabilities should be accurate to two or three places.

(a) Average of four is $1;$ (b) max is $6;$ (c) need exactly two $1$'s.

set.seed(1)
x = c(1,3,4,6)

a = replicate(10^6, mean(sample(x, 4, rep=T)))
mean(a == 1)
[1] 0.003916  # aprx  0.00390625

w = replicate(10^6, max(sample(x, 4, rep=T)))
mean(w == 6)
[1] 0.683426  # aprx 0.6835938

nr.ones = replicate(10^6, sum(sample(x, 4, rep=T)==1))
mean(nr.ones==2)
[1] 0.210837  # aprx 0.2109375

So the respective probabilities for parts (a)-(c) are
approximately $0.004, 0.684,$ and $0.211.$

Exact binomial probabilities: Exact probabilities can be found using the binomial distribution. Exact probabilities computed using R, where dbinom is a binomial PDF and pbinom is a binomial CDF. You can easily use
the appropriate binomial PDF formula to do the computations.

(a) Ones are successes. The number of ones in four draws
is $X_1 = \mathsf{Binom}(n=4, p = 1.4).$ In order for the average to be $1,$ we need all ones. $P(X_1 = 4) = 0.0039.$

dbinom(4, 4, 1/4)
[1] 0.00390625

(b) Sixes are successes. In order for the max to be $6,$
we need at least one six. $X_2 = \mathsf{Binom}(4,1/4),$
$P(X_2 \ge 1) = 0.6836.$

sum(dbinom(1:4, 4, 1/4))
[1] 0.6835938

1 - dbinom(0, 4, 1/4)
[1] 0.6835938

(c) Values below $2$ are successes. Only ones are smaller.
So we need exactly two Ones:  Probability is $0.2109.$

dbinom(2, 4, 1/4)
[1] 0.2109375


问:
Why expected value from a tries example $n=3$ is equal to expected value $n=654$. Why the population mean is still the same?
From a sample we have the expected value:
$E[x] = μ$.

And if we have differents trials $n=654$ the mean still is
$E[x] = μ$.

What is the reason of this relation? What causes this relationship to be true?

答:
It depends what you mean when you say from the different trials the mean is the same. Presumably you're describing the statistic that corresponds to the mean of the trials (since I can make any function of the trials and have it not equal the mean of a single trial!)

Consider $S_n = \frac{1}{n} \sum_{j=1}^n X_j$, where the $X_j$ are the results of your trial, each independent and with mean $E(X_j) = \mu$. Then one can show (by linearity of the expectation operator):
$$ E(S_n) = E \left(\frac{1}{n} \sum_{j=1}^n X_j \right) = \frac{1}{n} \sum_{j=1}^n E(X_j) = \frac{1}{n} \sum_{j=1}^n \mu = \mu. $$