问:
Given $AB=0$. Prove that every column of $B$ is a solution for $Ax=0$
I am studying for my test by going through some old tests from my university and in one of the tests there's the following question:

Given $AB=0$ prove that every column of $B$ is a solution for $Ax=0$

And I am very confused, isn't the solution to this problem is obvious? what are they expecting as an answer...?

Isn't it simply:
$AB=0 \rightarrow A[B]^c_i=[0]^c_i$ which shows that every column of $B$ is a solution for $Ax=0$

答:
Here is a formal proof.

Let $e_j$be the column vector with all entries zero, except for the $j$th one, which is $1$.
Then $AB=0$ implies $0=0e_j=(AB)e_i=A(Be_j)=AB_j$, where $B_j$ is the $j$th column of $B$.

The proof relies on the associativity of matrix multiplication.


问:
How many codes can be formed from three letters (A-Z) followed by two digits ($0$-$9$)?

Problem:

A code consists of 3 letters followed by 2 digits. How many codes can we create if a)repetition is allowed b)repetition is not allowed?

Attempt:

The code should looks like this LLLDD (L for letter and D for  digit) so if repetition is allowed, the number of possible codes is
$$(26^3)(10)^2$$
and if repetition is not allowed the number of possible  codes is
$$26 \mathrm{ P }3 \times 10 \mathrm{ P }2.$$

My concern:

How does $(26^3)(10)^2$ or $26 \mathrm{ P }3 \times 10 \mathrm{ P }2$ guarantee that the code has the correct pattern (LLLDD) and not other patterns like LDLDL, LLDDL, it's what permutations do after all (order is important), right?

答:
In both cases, you are counting the number of ways to create LLL and DD separately. Since there is only one way to put them together to create LLLDD, multiply and you're done.

If you wanted to count LDLDL and LLDDL, among other arrangements, you would need to choose the positions of the letters. Or of the numbers. Either way, it's an additional multiplication by ten.

ContentLicense: CC BY-SA 4.0


问:
Why are Haar measures finite on compact sets?
I'm working through the answer by t.b. to another user's question here:

A net version of dominated convergence?

because I am trying to work through a related problem and I think it will be illuminating.

From step two, "Then $KK'$ is compact and thus has finite Haar measure."

why is this true?

$\bf{\text{My attempt so far}}$:

Let $\mu$ be a left Haar measure on a locally compact group $G$.  Then  $\mu$ is non-zero, regular, and left $G$-invariant.  Let $K\subset G$ be compact.

$\color{red}{(!)}$ If I could construct set $U$ with the properties:

(i) $1\in U$

(ii) $0 < \mu(U) < \infty$

(iii) $U$ is open.

Then I could cover $K$ with $\{xU : x\in K\}$ and choose finitely many $k_{1}, ... , k_{n}\in K$ such that $K\subset \bigcup_{j=1}^{n}xU$ which forces $\mu(K)\leq \sum\limits_{j=1}^{n}\mu(xU) = n\mu(U) < \infty$.

I don't know how to construct such a $U$, though it seems innocent enough to ask for.

Does such a $U$ exist? or is the proof much more difficult?

答:
A Haar measure is finite on compact sets by definition: A Borel measure $\mu$ on a topological group $G$ is a (left) Haar measure if

$\mu$ is regular.

$\mu$ is left-invariant, i.e. $\mu(xU) = \mu(U)$ for any measurable $U \subset G$.

$\mu(U) > 0$ for all $U \subset G$ open.

$\mu(K) < \infty$ for all $K \subset H$ compact.

ContentLicense: CC BY-SA 3.0


问:
Using Regula-Falsi (false position) to solve a system of non-linear equations
I've written an algorithm implementing the Regula-Falsi ("false position") root finding method, and it works well for singular non-linear equations (where a single root exists). I'd now like to apply this to a localization problem, which can be described by a system of nonlinear equations.

Suppose we're given

a) The locations $x_i, y_i, z_i$ of three observers in two dimensions, or four observers in three dimensions, t

b) The velocity $v$ of some signal

c) The time of arrival, $t_i$, of the signal at each observer.

We want to determine the coordinates of the signal source, assuming straight-line travel, which is possible in general for any $n$ dimensions, given $n+1$ observers such that not all observers lie in an $n-1$ dimensional space.

I know that this problem can, of course, be represented in two dimensions by a non-linear system with equations of the form:

$$\sqrt{(X-x_i)^2+(Y-y_i)^2}-v(t_i-T)=0$$

...the system...

$\sqrt{(x-x_1)^2+(y-y_1)^2}+v(t_2-t_1) = \sqrt{(x-x_2)^2 + (y-y_2)^2}$

$\sqrt{(x-x_2)^2+(y-y_2)^2}+v(t_3-t_2) = \sqrt{(x-x_3)^2 + (y-y_3)^2}$

$\sqrt{(x-x_3)^2+(y-y_3)^2}+v(t_1-t_3) = \sqrt{(x-x_1)^2 + (y-y_1)^2}$

And in three dimensions:

$$\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2}-v(t_i-T)=0$$

How might Regula-Falsi be applied to such a problem?

I'm not sure if it's best to apply this method directly to the system, or if there's a 'better' way to apply Regula-Falsi to this problem. I know, in general, we'd want to 'bracket in' $F(x, y, z, ...)$, but I'm not sure where to go from here.

Note: I'm aware other methods exist (for example, I've solved this with the Newton Raphson method, thanks in large part to this answer on a previous, somewhat related post: ). I'm interested in using Regula Falsi here.

答:
You can't directly bracket a system of equations. You can instead bracket each equation individually, and then iteratively solve by back substitution.

Given $x(y,z)$ as a solution for $x$ in one equation, found by some method (e.g. regula-falsi).

Then find $y(x,z)$ as a solution for $y$ in another equation where $x$ is given by $x(y,z)$.

Then find $z(x,y)$ as a solution for $z$ in the last equation, where $x,y$ are given by the above.

This will involve a lot of solving however, since you will have to solve for $x$ on every iteration where you compute a new $y,z,$ and you will have to solve for $y$ on every iteration where you compute a new $z$.


问:
If $A$ is dense in $\Bbb R$, then $A∩Q$ is dense in $\Bbb R$ or $A∩(\Bbb R\setminus Q)$ is dense in $\Bbb R$
So I got this on an assignment and I can't find a way to prove or disprove the statement:

If $A$ is dense in $\Bbb R$, then $A∩Q$ is dense in $\Bbb R$ or $A∩(\Bbb R\setminus Q)$ is dense in $\Bbb R$.

I'm cetain it has something to do with the density of Rational/Irrational numbers, but I can't figure it out.

答:
The statement is wrong.

$A= \left((-\infty , 0) \cap \mathbb Q)\right) \cup (0, \infty) \cap \left( \mathbb R \setminus \mathbb Q)\right)$ is dense in $\mathbb R$. However none of $A \cap \mathbb Q$ nor $A \cap(\mathbb R \setminus \mathbb Q)$ is.


问:
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$
I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$

But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$

I ran out of imagination as what to try now

答:
This may be done using residue theory.  Consider

$$\oint_C \frac{dz}{1+z^4}$$

where $C$ is a closed contour that spans the perimeter of the quarter-circle in the 1st quadrant (i.e., the positive real and positive imaginary quarter plane), of radius $R$.  As $R \to \infty$, the integral over the circular arc vanishes, and we are left with this contour integral being equal to

$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4}$$

This integral is equal to $ i 2 \pi$ times the residue at the pole $z=e^{i \pi/4}$.  Thus

$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4} = \frac{i 2 \pi}{4 e^{i 3 \pi/4}}$$

which means that

$$\int_0^{\infty} \frac{dx}{1+x^4} =\frac{\pi}{2 \sqrt{2}}$$


问:
For what variable $a$ are vectors linearly independent?
I have vectors $(1, a, 1), (1, 1, 1), (2, 2, a)$. For which $a \in \mathbb{R}$ are vectors linearly independent?
I tried to calculate it like matrix $$ \left[
\begin{array}{ccc|c}
  1&1&2&0\\
  a&1&2&0\\
1&1&a&0
\end{array}
\right] $$
After substracting rows it is
$$ \left[
\begin{array}{ccc|c}
  1&1&2&0\\
  0&1-a&2a-2&0\\
0&0&a-2&0
\end{array}
\right] $$
Then $(a-2)z=0$ etc.
I will get some equations, but I don't know how should I continue. Without determinant.

答:
Analyze your reduced matrix.
Also, we can see that
$$\begin{pmatrix} 1 & a & 1 \\ 1 & 1 & 1 \\ 2 & 2 & a \end{pmatrix} \sim  \begin{pmatrix} 1 & 1 & 1\\ 0 & 1-a & 0 \\0 & 0 & a-2\end{pmatrix} $$
Since with the matrix totally reduced, no row vanishes for values $a\not=1,2$ the set is linearly independent for all $a \in \mathbb{R}\setminus\{1,2\}$.


问:
Calculating the integral is not giving the same simplified function
I calculated this integral below :
$$
\int_{-\pi/2}^{\pi/2} \cos(x)\cos(nx) dx
$$

And I got this result :
$\dfrac{\left(\left(n-1\right)\sin\left(\frac{{\pi}n+{\pi}}{2}\right)+\left(n+1\right)\sin\left(\frac{{\pi}n-{\pi}}{2}\right)\right)}{n^2-1}$

I had a check at the end at the solution :
$-\dfrac{2\cos\left(\frac{{\pi}n}{2}\right)}{n^2-1}$

So apparently there is some kind of simplification to do but I don't actually know how .

答:
Use Prosthaphaeresis Formulas,

$$n\left(\sin\dfrac{(n+1)\pi}2+\sin\dfrac{(n-1)\pi}2\right)-\left(\sin\dfrac{(n+1)\pi}2-\sin\dfrac{(n-1)\pi}2\right)$$

$$=2n\sin\dfrac{n\pi}2\cos\dfrac\pi2-2\cos\dfrac{n\pi}2\sin\dfrac\pi2$$

Alternatively,

like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,

as $\cos x\cdot\cos(nx)$ is an even function using Werner Formulas,

$$\int_{-a}^a\cos x\cdot\cos(nx)\ dx=2\int_0^a\dfrac{\cos(n-1)x+\cos(n+1)x}2\ dx$$


问:
Finding the domain of a function (that is described by a DE) such that the function is real
I have the following DE:

$$n\cdot y(x)^2=\sqrt{1+(y'(x))^2}$$

How can I find the domain of $x$ such that $y:\mathbb{R}\to\mathbb{R}$ (so that the solution is real-valued) in terms of $n$ (which is a real positive number) and the initial condition $y(a)=b$ (where $a$ and $b$ are real numbers)?

答:
Hint

Notice that if $y$ is a solution of the given ODE, so is $-y$. Therefore, we can without loss of generality suppose that $y(a) >0$.

Suppose that $y^\prime(a) >0$. You can deal with the other alternative in a similar way.

You have

$$y^\prime(x) =\sqrt{n^2 y^4(x) - 1}$$ which implies the condition $\vert y(a) \vert \gt 1/ \sqrt n$. Locally around $a$

$$x-a = \int_a^x \frac{y^\prime(x)}{\sqrt{n^2y^4(x) -1}} \ dx= \frac{1}{\sqrt{n}}\int_{y(a)}^{y(x)} \frac{dz}{\sqrt{z^4-1/n^2}}.$$

The integral

$$I(b,n)= \frac{1}{\sqrt{n}}\int_{b}^\infty \frac{dz}{\sqrt{z^4-1/n^2}}$$ is convergent. And the solution $y$ is defined  on the right on the interval
$(a, a + I(b,n))$. Not sure that there is a closed value for $I(b,n)$ though.


问:
Permutations on words

How many six-letter words can be formed by using the letters of the word ‘PRESSES’?

So my doubt comes to which proceedure is valid to solve this problem:

The book solution is:
We omit in turn each of the four letters ‘P’, ‘R’, ‘E’ and ‘S’.
This leaves six letters which we must then arrange in order.
1 If an S is omitted, there are then 2 Es and 2 Ss,
so number of words = 6!,
2! × 2! = 180
2 If an E is omitted, there are then 3 Ss,
so number of words = 6!,
3! = 120.
3 If P or R is omitted (2 cases), there are then 2 Es and 3 Ss,
so number of words = 6!,
3! × 2!
× 2
= 120
Hence the total number of words is $180 + 120 + 120 = 420$

While my logic works like $\frac{7P6}{3!\cdot2}$ , 7Permute6 because it involves permuting 7 letters into 6 spaces and there are 3 letters repeated (the S's) and 2 letters repeated (the E's). And this gives you 420 too.

Is my logic/approach valid? Or is it just a coincidence that both methods get to the same answer, but mine is ilogical?

答: Your logic gives the correct answer only because exactly one letter is omitted. The result is the same as the number of seven-letter words, $\frac{7!}{3!2!}$. There is a bijection between six- and seven-letter words: PRESSE <=> PRESSES

Title: Permutations on words


问:
Assuming $(\alpha x_{n-1}+\beta x_n)_{n\geq 1}$ converges and $\beta > \alpha$, show that $(x_n)_{n\geq 0}$ converges.
Let $\alpha,\beta$ be non-zero positive real numbers such that $\alpha < \beta$, and $(x_n)_{n\geq 0}$ a sequence of real numbers and $(y_n)_{n\geq 1}$ be the sequence defined for all $n\in \mathbb{N}^{\ast}$ :
$$ y_n=\alpha x_{n-1} +\beta x_n $$
Show that if $(y_n)_n$ converges, then $(x_n)_n$ converges as well.

I thought of using the definition of the limit and trying to show that $x_n\to \frac{\ell}{\alpha+\beta}$ where $\ell$ is the limit of $(y_n)_n$. However, in the process, I find that there is a symmetry between $\alpha$ and $\beta$, respectively between $x_{n-1}$ and $x_n$ which is not supposed to be the case due to the hypothesis. So I find myself stuck, I would really like some hints.

答:
Hint : Show inductively that for all $n \geq 0$, you have $$x_n = \left(\frac{-\alpha}{\beta}\right)^n x_0 + \frac{1}{\beta}\sum_{k=0}^{n-1} (-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k $$

The result follows.


问:
Show that the action of $\mathfrak{so}(3)$ on $\mathbb R^3$, $M_a(x)=a\wedge x$, satisfies $M_{ga}=g M_a g^{-1}$
In the notes on the Lie groups $SU(2)$ and $SO(3)$ found here, the authors state (page 75, before Proposition 2.2) the following.

Consider the $\mathbb R$-linear map $\mathbb R^3\ni a\mapsto M_a\in\mathfrak{so}(3)$ defined by $e_k\mapsto \eta_k$, where $e_k\in\mathbb R^3$ are the elements of the canonical basis of $\mathbb R^3$, $\eta_k\in\mathfrak{so}(3)$ are the corresponding generators: $\exp(i\eta_k)=\mathrm{Rot}(e_k,t)$, and $\mathrm{Rot}(a,t)$ denotes the rotation by angle $t$ around the axis $a\in\mathbb R^3$.

We can then write $M_a(x)=a\wedge x$. This is easy to see, for example writing $\eta_i=\varepsilon_{ijk} (e_j e_k^* - e_k e_j^*)$.
Then, they say, given $g\in SO(3)$ and $a,x\in\mathbb R^3$, we have
$$M_{ga}(x) = (ga)\wedge x = g(a\wedge (g^{-1}x)) = (g M_ag^{-1})(x). \tag1$$
What's a good way to show (1)?

By direct computation, using $x\wedge y=\sum_{ijk}\varepsilon_{ijk}e_i x_j y_k$, I get
$$[(ga)\wedge x]_i = \sum_{jk\ell}\varepsilon_{ijk} g_{j\ell}a_\ell x_k, \\
[g(a\wedge (g^{-1}x))]_i = \sum g_{ij}\varepsilon_{jk\ell} a_k g^{-1}_{\ell m}x_m.
$$
Because $g\in SO(3)$ I know that $g^{-1}_{\ell m}=g_{m\ell}$, but even with this I'm not sure how to simplify the expressions to show that they are equal.

I could probably just get the identity from geometrical intuition, observing that it must be that $g(x\wedge y)=(gx)\wedge (gy)$, but I wanted to show it explicitly algebraically here.

答:
I want to show that, for any $R\in SO(3)$ and $a,b\in\mathbb R^3$, we have
$R(a\wedge b) = (Ra)\wedge(Rb)$.

The condition $R\in SO(3)$ is equivalent to the rows of of $R$ being orthonormal and $\det(R)=1$. In terms of the Levi-Civita antisymmetric tensor, this condition reads $$\sum_{ijk}\varepsilon_{ijk}R_{1i}R_{2j}R_{3k}=1.$$
This can be interpreted as the dot product between $(R_{1i})_i$ and $(\sum_{jk}\varepsilon_{ijk}R_{2j}R_{3k})_i$, which can only equal $1$ if the vectors are equal. More generally, we have
$R_{i\ell} = \sum_{mn}\varepsilon_{\ell mn}R_{jm}R_{kn}$ where $\varepsilon_{ijk}=1$. We can also write this as
$$R_{i\ell} = \frac12 \sum_{jk mn}\varepsilon_{ijk}\varepsilon_{\ell mn} R_{jm}R_{kn}.$$
We therefore have
$$[R(a\wedge b)]_i = \sum_{\ell mn}R_{i\ell}\varepsilon_{\ell mn} a_m b_n
= \sum_{\ell mnpq} (\varepsilon_{\ell pq}R_{jp}R_{kq}) \varepsilon_{\ell mn} a_m b_n.$$
Using
$$\sum_\ell \varepsilon_{\ell mn}\varepsilon_{\ell pq} = \delta_{mp}\delta_{nq} - \delta_{mq}\delta_{np}$$
we then get
$$[R(a\wedge b)]_i = \sum_{pq}(R_{jp}R_{kq}a_p b_q-R_{jp}R_{kq}a_q b_b) \\
= (R_j a) (R_k b) - (R_k a) (R_j b)
= [(Ra)\wedge (Rb)]_i.$$

We can also perform this calculation in more concise notation writing $a\wedge b=\varepsilon(a\otimes b)$, where we are thinking here of $\varepsilon$ as a linear operator $\mathbb R^3\otimes\mathbb R^3\to \mathbb R^3$. Similarly, $R_i= \varepsilon(R_j\otimes R_k)$, and thus
$$[R(a\wedge b)]_i = (\varepsilon(R_j\otimes R_k)) \cdot (\varepsilon(a\otimes b)) = [(Ra)\wedge (Rb)]_i,$$
where we used $\mathrm{Tr}_1[\varepsilon\varepsilon^*] = \mathrm{Id}-\mathrm{Swap}$.


问:
Differential equation model and graph question
Consider the model

$dS/dt=f(S)= kS(1-S/N)((S/M)-1)$

Where S is the population of the squirrel, k and M are constant parameter, N is the carrying capacity. Know that the more people move in, the more N decrease.

Suppose M is less than or equal to N, and k,M are fixed value. Sketch the graph for f(S). What value of N does a bifurcation occur?

How can I do that? there are too many unknown value.

答:
You are correct that there are a lot of parameters, but here are some hints.

You are trying to see the general shape of the solution sets for various parameters.

Approach:

1) Let $M = N$, solve the model. Fix some values for k, M, N (with M = N) and plot the resulting f(S)

2) Let $M < N$, solve the model. Fix some values for k, M, N (with M = N) and plot the resulting f(S).

3) Play around with such things as $k = 0$, $M$ very small and very large, $N$ very small and very large. This will let you do a qualitative analysis.

From all of this, see if you can figure out the general shape to the system. You might even find such things as instabilities and you want to see if you can characterize those. It also helps to know general values for the fixed parameters in the model and use those if you know them. This is a qualitative analysis.

Example 1:

Let $M = N = k = 1$, we have:

$$c_1-t = -1/(s(t)-1)-\ln(s(t)-1)+\ln(s(t))$$

If we plot solution curves for various $c_1$, we have:

Notice the general shape of the solution over $t$?

Here is the same thing, except I plotted the direction field (green) alongside many solution curves (blue) for the model.

Example 2: $k = 0$, we get $s(t) = c_1$, a constant function over t.

Example 3: $k = -1$ (does that make sense?), $N = 1$, $M = 1/5$. Here is the phase portrait. Notice that for all those changes, the general behavior did not change that much (direction field changed, but the rest is very similar. What do these curves tell you about the model would be the next question.

You are basically doing a qualitative analysis of the model and seeing if the behaviors can be generalized for various sets of parameter selections (there may be one behavior or several (stable, unstable, indeterminate, for example).


问:
Differential geometry and curved space

To calculate the intrinsic curvature, we need to know only the metric gμν; we don’t need
to know the embedding functions.

I am studying curved space and differential geometry, the author said the quotation: The Theorema Egregium of Gauss. As is the first time i am seeing it, i still don't get what does the quotation really means. Could you say more about it? Particularly the meaning of embedding functions would help a lot, since the author do not said what them mean.

It is necessary to understand general relativity... And, to be honest, i like the subject too, because it is hard. I am not stuckling in the meaning of metric, and another things it can suggest. But, thinking better, i realized that i was assuming this theorem probably as true without even knowing the necessity of it. That is, i was always assuming that given the metric is enough to characterize our geometry. The problem is this embedding functions, intuitively, what i am interpretation is that if we use two different functions (surface) to measure the intrinsic curvature, it will give the same ans

答:
Hitsorically, the Theorema Egregium is a result on the geometry of surfaces in the three dimensionnal space.
If $S \subset \mathbb{R}^3$ is a surface, then one can define a metric on $S$ by restricting the scalar product on the tangent spaces, that is
$$g_p(u,v) = \langle u , v \rangle,~ u,v\in T_pS$$
This allow us to talk about $S$ as a metric space in itself. The geometric properties of $S$ are here inherited of the geometry of the ambiant space, $\mathbb{R}^3$. One can define the curvature of $S$ in terms of its second fundamental form, which is an extrinsic object: it closely depends on the embedding $S \subset \mathbb{R}^3$.

But what does "depend on the embedding" means? As said earlier, one has defined $(S,g)$ as an intrinsic metric space thanks to the induced metric. If $(S',g')$ is a riemannian intrinsic surface that is isometric to $(S,g)$, that is if there exists a diffeomorphism $f : S \to S'$ with $f^*g' = g$, then one can hope the two object $(S,g)$ and $(S',g')$ have the same geometric properties. For instance, the same curvature? But curvature is defined in terms of the embedding $S\subset \mathbb{R}^3$! So maybe, if $S'\subset \mathbb{R}^3$ is also a surface in $\mathbb{R}^3$, they do not have the same curvature because they are not the same surface as embedded in the three dimensional space.

The Theorema Egregium just says that, even if the definition of the curvature here looks like it depends on the embedding (and in the computations, it really looks like!), in reality it does not. Gauss made some horrible calculations to get a formula that does not involve the extrinsic property of the embedding but just the intrinsic property of $(S,g)$. That is, the curvature is a geometric notion that does make sense for a surface in itself, not just for a surface as a subset of the three dimensional space.

It has a natural generalisation for embedded submanifold in riemannian geometry.

With a modern point of view, the Theorema Egregium may look weird because we usually define the curvature in terms of tensor algebra and the Riemann tensor. But in fact, the reason we arrived to the definition of these intrinsics notions (Riemann tensor, sectionnal curvature, etc.) is because the Theorema Egregium has been proven, and one could have hoped for an intrinsic notion to defined it!

Title: Differential geometry and curved space


问:
Can $T\left( n\right) = 4T\left({n\over 2}\right) + 2^{n\over2}$ be solved using Akra-Bazzi method?
$$T\left( n\right) = 4T\left({n\over 2}\right) + 2^{n\over2}$$

Using Akra-Bazzi method

$a = 4, b = {1\over2}, g(n) = 2^{n\over2}, h(n) = 0$

$$g'(n) = {d\over dx}(2^{n\over2})=2^{{n\over2}-1}\ln 2$$

Since $\left|g'(n)\right|$ is not polynomially bound we cannot proceed further with Akra-Bazzi method.

Using Master theorem

$a = 4, b = 2, g(n) = 2^{n\over 2}$

$g(n) = \Omega(n^{\log_2 4+\epsilon})$ where $\epsilon=1$, also

$$ag\left( {n\over b}\right) \le cg(n)$$

$$4g\left( {n\over 2}\right) \le cg(n)$$

$$2^{{n\over 4}+2} \le c2^{n\over 2} \tag*{for $c={1\over 2}$ and $n\ge 12$ }$$

Therefore it follows from master theorem that $T(n) = \Theta\left( g(n)\right) = \Theta\left( 2^{n\over 2}\right)$.

How come the above problem can be solved using Master theorem which is just a corollary of Akra-Bazzi method but not using Akra-Bazzi method itself.

答:
Starting with $$T(2n)=4T(n)+2^n$$

In this case, a possibility is to try to get $U(p)=\sum f(k)$ by introducing a telescoping relation for $U$ : $U(p+1)=U(p)+f(p)$.

To do that we first set $n=2^p$ so as to deal with the $T(2n)$ .vs. $T(n)$ relation.

And to make the proportionality coefficient $4$ disappear we will introduce the coefficient $4^p$.

Let set $$T(n)=T(2^p)=4^p\,U(p)$$

Substitution in the equation gives:

$\require{cancel}T(2n)=T(2^{p+1})=\cancel{4^{p+1}}U(p+1)=4T(n)+2^n=\cancel{4\cdot 4^p}\,U(p)+2^{2^p}$

After regrouping $U$ terms on one side and $f$ terms on the other side:

$U(p+1)-U(p)=2^{2^p}/4^{p+1}=\underbrace{2^{(2^p-2p-2)}}_{f(p)}$

And you can sum it to $$U(p)=U(0)+\sum\limits_{k=0}^{p-1} 2^{2^k-2k-2}$$

Asymptotically this sum is equivalent to its last term $2^{(2^{p-1}-2(p-1)-2)}$ because $2^{2^k}$ is growing very fast (use Cesaro for instance).

In the end $$T(n)\sim 4^p\,2^{(2^{p-1}-2p)}=2^{(2^p/2-2p+2p)}=2^{n/2}$$

Therefore because $2^{n/2}$ grows very fast we have $4T(n/2)\ll T(n)$ and $T(n)$ is just asymptotically equivalent to this growing term.


问:
Proving identity in infinite series
I am trying to prove the equation

\begin{align*}
\sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i},
\end{align*}

where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:

\begin{align*}
\sum_{n=0}^{\infty} a^{c_1 + c_2 + \ldots + c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d}
&= \sum_{n=0}^{\infty} a^{c_1} \cdot a^{c_2} \cdot \cdot \cdot a^{c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdot \cdot \cdot b_d^{c_d} \\
&= \sum_{c_1}^{\infty} a^{c_1} \cdots \sum_{c_d}^{\infty} a^{c_d} \cdot \sum_{c_1}^{} b_1^{c_1} \cdots \sum_{c_d} b_d^{c_d} \\
&= \sum_{c_1}^{\infty} (a b_1)^{c_1} \cdots \sum_{c_d}^{\infty} (ab_d)^{c_d}.
\end{align*}

My doubts in here are whether it is even possible to split up the sum in this manner and then simply recombining it in the last line, since the product of two sums is not necessarily the sum of the product.

From here I would use

\begin{align*}
\sum_{k}^{\infty} x^k = \frac{1}{1-x},
\end{align*}

such that it would become

\begin{align}
\frac{1}{1-a b_1} \times \cdots \times\frac{1}{1 - a b_d} = \prod_{i = 1}^{d} \frac{1}{1-a b_i}.
\end{align}

I would like to ask where my errors are and if there is a better way to prove this identity. Thank you for your time.

答:
By induction on $d$.

$$\begin{aligned}
\frac{1}{1-ab_1}
\frac{1}{1-ab_2} &= \left(\sum_{n=0}^\infty (ab_1)^n\right)\left(\sum_{n=0}^\infty (ab_2)^n\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^nb_1^lb_2^{n-l}\right) a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+c_2=n}b_1^{c_1} b_2^{c_2}\right) a^n
\end{aligned}$$

according to Cauchy product formula.

Now apply again Cauchy product formula to pass from $d$ to $d+1$:

$$\begin{aligned}
\prod_{i=1}^{d+1} \frac{1}{1 - a b_i} &=\left(\prod_{i=1}^{d} \frac{1}{1 - a b_i}\right)\frac{1}{1 - a b_{d+1}}\\
&= \left(\sum_{n=0}^\infty \left(\sum_{c_1+ \dots +c_d=n}b_1^{c_1} \dots b_d^{c_d}\right) a^n\right)\left(\sum_{n=0}^\infty (ab_{d+1})^n\right)\\
&= \sum_{n=0}^\infty \left(\sum_{l=0}^n \left(\sum_{c_1+ \dots +c_d=l}b_1^{c_1} \dots b_d^{c_d}\right)b_{d+1}^{n-l}\right)a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+\dots + c_{d+1}=n}b_1^{c_1} \dots b_{d+1}^{c_{d+1}}\right) a^n
\end{aligned}$$

Title: Proving identity in infinite series


问:
A sequence of real numbers with limits as n $\to \infty$
Let $(x_n)_{n \in \mathbb{N}}$ be a sequence of real numbers:
Assume that $x_n \gt -1$, and $x_n \to 0$ as $n \to \infty$. Show that $log(1+x_n) \to 0$ as $n \to \infty$.

Okay so by inspection this seems pretty obvious since $x_n$ goes to 0 for large n. So $log(1+x_n)$ would just become $log(1)$ which is 0. However I know this wouldn't be a sufficient explanation.

I've been told to use the answer of a previous question as a hint. But I'm not sure how to solve the previous question. Which is as follows:

Let $x \in (-1, 1)$. Show that:
$\frac{x}{1+x} \leq log(1+x) \leq x$

Using the following result: $1+x \leq e^x \leq \frac{1}{1-x}$
I'm allowed to use this result without a proof. But am unsure how to apply it to the latter question asked which in turn means I can't answer the question that I want to. If anyone could offer some insight I'd greatly appreciate it.

答:
The intent of the hint is for you to use the squeeze theorem.

As $~x \to 0,~ \frac{x}{1+x}~$ and $x$ both go to $0$.


问:
How do I solve a nonlinear function with Newton-Raphson assuming that my variable is discrete?
Let's say I want to approximate $\sqrt 2$.

I, therefore, write my function
$$
f(x) = x^2-2 = 0
$$

I define $x$ to be a discrete variable which can take values $\color{brown}{0, 0.1, 0.2, 0.3, 0.4, 0.5}$, etc.

Is there a Newton-Raphson method for solving this?

Kindly also explain the solution procedure by working out the problem.

Regards.

答:
This is a way to approximate without knowing that $1.4 \approx \sqrt{2}$.

If we plug $1,2$ into $f(x)$, we'd get

$$ f(1) < f(2)  \\  -1 < 2$$

So for $1 < x_0 < 2$, $f(x_0) = 0$. So, choose $x_0 = \frac{1+2}{2}=1.5$. From here, you can approximate it normally (as shown by other answers).

Of course, you can take $1.1, 1.9$ and get a better $x_0$, but this is just a (very) crude way to choose an initial guess.


问:
Solving stochastic differential equation $ dX_t = \kappa(\theta - X_t)dt\ +\ \sigma dW_t $
How do I solve this stochastic differential equation ? I can't seem to get rid of $ X_t.$  Thank you

$ dX_t = \kappa(\theta - X_t)dt\ +\ \sigma dW_t $

答: Try the substitution: $Y_t = X_t  e^{kt}$.


问:
Given $AB=0$. Prove that every column of $B$ is a solution for $Ax=0$
I am studying for my test by going through some old tests from my university and in one of the tests there's the following question:

Given $AB=0$ prove that every column of $B$ is a solution for $Ax=0$

And I am very confused, isn't the solution to this problem is obvious? what are they expecting as an answer...?

Isn't it simply:
$AB=0 \rightarrow A[B]^c_i=[0]^c_i$ which shows that every column of $B$ is a solution for $Ax=0$

答:
An alternative approach, using the row-column rule for matrix multiplication:

$$AB = A [ b_1 \quad b_2 \quad \cdots \quad b_n ] = [Ab_1 \quad Ab_2 \cdots \quad Ab_n] = 0, $$

which shows that

$$Ab_j = 0$$

for all $ 1 \leq j \leq n$.


问:
prove this inequality $x^{n+1}\geq (n+1) x-n$
I tied to prove this by induction

$$\forall x\in\mathbb{R_+} ,\forall n \in\mathbb{N}:x^{n+1}\geq (n+1) x-n$$

I know it holds for $n=1$ but I get stacked after this step.

答:
Assume $x\geq 0, n\geq 0$.

Define $f(x)=x^{n+1}-(n+1)x+n$, then $f'(x)=(n+1)(x^n-1)$.

The function is monotone increasing in $[1,+\infty)$ and decreasing in $[0,1]$. So the minimal is $f(1)=0$.

Then, $f(x)\geq0$ for all $x\geq 0$, i.e., $x^{n+1}\geq(n+1)x-n$.

1.$n=1$ holds.

Assume $n=k$ holds. then, $x^{k+1}\geq(k+1)x-k, \forall x$.

Then, for $n=k+1$, we have
\begin{equation}
\begin{aligned}
x^{k+2}&=x*x^{k+1}\\
&\geq (k+1)x^2-kx\\
&=(k+2)x-(k+1)-(k+2)x+(k+1)x^2-kx+(k+1)\\
&=(k+2)x-(k+1)+(k+1)(x-1)^2\\
&\geq (k+2)x-(k+1).
\end{aligned}
\end{equation}
So, the inequality also holds for $k+1$

Therefore, the inequality holds for all positive integers.


问:
How many solutions are in equation over GF(16)
Can some one show me how to do multiplication over gf(16) step by step

I cannot determine how many solutions are in x^15 - 1

答: You don't have to know the multiplication table in $\mathbf F_{16}$: the multiplicative group  $\mathbf F_{16}^\times$ has order $15$, hence by Lagrange's theorem, each element in this group has order a divisor of $15$.


问:
Conformal map from $\mathbb{C}\setminus\{z\in\mathbb{C} : \Re(z)=0, \, |z| \geq 1 \}$ to unit disc
I want to find a conformal map from $U=\mathbb{C}\setminus\{z\in\mathbb{C} : \Re(z)=0, \, |z| \geq 1 \}$ to the unit disc. Thus far, this is what i have:

$z \mapsto iz+1$ to map $U$ to $\mathbb{C}\setminus \left( (-\infty,0] \cup [2,\infty)\right)$.

$z\mapsto \sqrt{z}$ to map to the right half plane minus $[\sqrt{2},\infty)$.

Can someone help me to find a conformal map from the latter to the unit disc?

Thank you in advance.

答:
I think it's best to imagine your set as a subset of the Riemann sphere. Visualizing it that way, it is the sphere with a straight slit going from $\mathrm i$ to $-\mathrm i$ via $\infty$. We can now rotate this slit sphere using a Möbius transformation such that the slit goes from $0$ to $\infty$ via $1$. This gives us a slit plane missing the positive real axis. To do so, use the Cayley transform
$$z\mapsto\frac{z-\mathrm i}{z+\mathrm i}.$$
Now apply $z\mapsto\sqrt z$ to transform the slit plane to the upper half plane. Follow up by applying the Cayley transform again, which transforms the upper half-plane to the unit disc.

The basic idea is to recognize that both the unit disc and the half-plane are hemispheres of the Riemann sphere, and that both your given set and any slit plane are slit Riemann spheres. Now all that remains is to transform a slit sphere to a hemisphere, which the square root does nicely if the slit starts or ends at $0$. So we just have to rotate the slit sphere so that the slit starts at $0$ (that's the first Cayley transform), transform it to a hemisphere (the upper half-plane in this case, that's the square root transformation), and then rotate back so we actually have the unit disc (that's what the second Cayley transform does).


问:
Curve fitting with an invertible function
I have some data that need to be transformed as shown in the picture given some function.

example of input data and desired result

The function in question can take whatever form but with the following requirements:

It needs to be able to fit the data shown (i.e f(orange data) = blue data as much as possible)

It needs to be invertible

I need to be able to adjust its parameters so that in certain conditions it becomes identity

I've played around with various functions and although I can find various options that give a perfect fit, either I can't invert them or I can't figure out how to adjust the parameters to get an identity.

To give a more concrete example, let's say that a function of the form f(x) = ax + b could work (it doesn't). In this case the inverse function would be finv(x) = (x-b)/a.
Now we could imagine a weight that is provided as an external parameter which goes between 0-1, in which case I could adjust parameters a and be so that

a'= w*a - (1-w) -> so that when the weight is at 1 we get the desired fitting behavior, when the weight is at 0 we get 1

b' = w*b -> so that b becomes 0 when the weight is 0
The above would give me f(x) = x when the weight is at 0 and ax+b when the weight is at 1.

What function could fulfil all 3 criteria given the form of the data I have?

Any ideas would be extremely welcome, I've been looking at this for days and I'm getting nowhere.

Thanks!

Edit to add some examples that give me a good fit:
f(x) = ax/(b-x)^c

good fit, I can get it to become identity, but I don't see how to invert
f(x) = (ax/(b-x))^c

good fit, invertible but can't get it to become identity

Throwing more parameters at it:
f(x) = (ax^b)/(c-x^d)

again good fit, identity if I set a=1, b=1, c=2, d=0 but can't see how to invert

答:
How to fit a function to the inverse of a non-invertible function ?

This seems firstly to be a contradiction. Of course this should be a contradiction if we were looking for the analytically exact inverse function. But the problem is a problem of fitting, thus not entirely accurate. We are not looking for the inverse function itself, but for a function as close as possible to the inverse function.

A numerical example is shown below, in order to facilitate understanding.

FIRST PART : Fitting a function f(x) to the given data (x,y).

Deliberately a non-invertible function is chosen for the example.

SECOND PART : Fitting a function g(y) to the inverted data (y,x).

Deliberately a non-invertible function is chosen for the example. Deliberately again the function g which involves logarithms is not on the same kind than f which involves powers. We could have chosen both functions on the same kind if we like. The choice of the functions is entirely open insofar the two independent fittings can be correctly carried out.

CHECKING :

Now we have two explicit functions f(x) and g(y) which are approximately the inverse one to the other.

We can check the discrepancy in comparing g(f(x)) to x and in comparing f(g(y)) to y :


Title: prove this inequality $x^{n+1}\geq (n+1) x-n$

Body:
I tied to prove this by induction

$$\forall x\in\mathbb{R_+} ,\forall n \in\mathbb{N}:x^{n+1}\geq (n+1) x-n$$

I know it holds for $n=1$ but I get stacked after this step.

Body:
Simple solution using AM-GM: $\frac{x^{n+1}+n}{n+1}=\frac{x^{n+1}+1+1+...+1}{n+1} \geq \sqrt[n+1]{x^{n+1}\cdot 1\cdot 1 \cdot ... \cdot1} = x \implies $ $$x^{n+1}\geq (n+1) x-n.$$

In addition, there is a solution where you don't need to know any AM-GM or use induction:
$x^{n+1}-x\ge n(x-1) \iff x(x^n-1)\geq n(x-1) \iff x(x-1)(x^{n-1}+x^{n-2}+...+x+1)\geq n(x-1) \iff$ $$(x-1)(x^n+x^{n-1}+...+x^2+x-n)\geq 0$$ If $x\ge1$, then $x^n+x^{n-1}+...+x^2+x\ge1+1+...+1=n$, and if $x \le 1$, then $x^n+x^{n-1}+...+x^2+x\le1+1+...+1=n$, so
$$(x-1)(x^n+x^{n-1}+...+x^2+x-n)\geq 0 \implies x^{n+1}-x\ge n(x-1)$$


问:
Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$

Prove that $|a| \leq \max\{|b|,|c|\}$ if $b\leq a \leq c$.

I showed that $a\leq c$ and therefore $-c\leq a \leq c$ so that $|a|\leq c$ but then I got stuck.

Is this the right approach?

答:
Hint:

Simply break the problem into 3 cases.

Case 1: $b < 0 \leq c.$

Case 2: $b < c < 0.$

Case 3: $0 \leq b < c.$

Then manually attack each case.


问:
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$
I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$

But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$

I ran out of imagination as what to try now

答:
make a change of variable $ x=y^{1/4} $ and use the identity

$$ \int_{0}^{\infty}\frac{t^{s-1}}{x+1}dt= \frac{\pi}{sin(\pi s)}$$


问:
Prove that if d(x,A) = 0 then x is an adherent point for A
Let $A \subset \mathbb{R}^n$ be a non empty set, and for $x \in \mathbb{R^n}$ let $$d(x,A) := inf\{ ||x-y||_2 : y \in A \}$$

be the distance between x and the set A.

Prove that:

$d(.,A): \mathbb{R^n} \rightarrow \mathbb{R} \ \ \ x \mapsto d(x,A)$ is (Lipschitz) continous.

$d(x,A) = 0 \iff x \in \bar{A}$ ( The set of adherent points for A )

I have already proven 1) using the triangle inequality but i'm struggling with the second question, so far the only idea i have is $d(x,A) = 0 \implies x \in A \implies x \in \bar{A}$ but it seems too easy to be correct and i still got no clue for the opposite way $\impliedby$

Thanks in advance!

答:
For the forward implication, note that as $d(.,A)$ is a continuous function and ${0}$ is a closed set then $d(.,A)^{-1} ({0})$ is closed. $d(.,A)^{-1} ({0})$ also certainly contains A, so we must have that it contains $\bar{A}$ by definition of closure.

For the reverse implication, note that $\bar{A}=A\cup A'$ where $A'$ is the set of limit points of A. If $x\in A$ then $d(x,A)=0$ certainly. If $x\in A'$ then for all $\epsilon >0$ we may choose some $a \in A$ with $\left\lVert x-a\right\rVert_{2}<\epsilon$, and so $d(x,A)=0$.


问:
Can anyone help me with statistics?
Question: Following is the distribution of marks obtained by 500 candidates in the statistics paper of a civil services examination:

```<code>+ marks more than:      0   10  20  30  40  50
+ number of candidates: 500 460 400 200 100 30
</code>```

Calculate the lower quartile marks. If $70\%$ of the candidates pass in the paper, find the minimum marks obtained by a passing candidate.

In the above problem, I am able to find the first part to calculate lower quartile marks but not able to find the second section.

Anyone, please help I am a newbie to statistics.

Thanks in advance, please ignore the bad English.

答:
Write out the cumulative distribution function $P(N \leq m)$ for number of candidates N and marks m, by subtracting the numbers given from 500 and dividing by 500.
You can compute what is required by simple interpolation.

Title: Can anyone help me with statistics?


问:
On what subintervals of $[0,\infty)$ is $f_n(x)=\frac{x^2}{1+x^n}$ uniformly convergent?
I have calculated the pointwise convergence which is $f(x)=x^2$ for $x<1\qquad$ and for $x=1$ we get $f(x)=\frac{1}{2} \quad$ and lastely for $f(x)=0$ for $x>1$. Now, I know that this sequence is not uniformly convergent for  the set $[0,\infty)$ but I know that it IS uniformly convergent for $[0,1)$ since we can do the following: $|f_n(x)-f(x)| =...= |\frac{x^n}{1+x^n}|<|x^n|$ and since we know that $x<1, \forall \epsilon >0,$ we can find a $N$ such that $x^n < \epsilon \forall n\ge N$. The problem I'm having now is how can I find all subintervals of $[0,\infty)$ where this sequence is uniformly convergent?

答:
The sequence is not uniformly convergent on $(a,1)$ for $0 < 1$. Indeed, we have $\frac1{\sqrt[n]{2}} \to 1$ from the left so for large enough $n\in\Bbb{N}$ we have $\frac1{\sqrt[n]{2}} \in (a,1)$ and

$$\left|f\left(\frac1{\sqrt[n]{2}}\right) - f_n\left(\frac1{\sqrt[n]{2}}\right)\right| = \frac1{\sqrt[n]{4}} -\frac{\frac1{\sqrt[n]{4}}}{1+\frac12} = \frac1{3\sqrt[n]4} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

The sequence is uniformly convergent on $[0,b]$ for $0 < b < 1$. Namely, for $x \in [0,b]$ we have
$$|f(x)-f_n(x)| = x^2-\frac{x^2}{1+x^n} = {x^2}\frac{x^n}{1+x^n} \le x^n \le b^n \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $b^n$ is independent of $x$.

The sequence is not uniformly convergent on $(1,b)$ for $b > 1$. Namely, we have $\sqrt[n]{2} \to 1$ from the right so for large enough $n\in\Bbb{N}$ we have $\sqrt[n]{2} \in (1,b)$ and
$$\left|f_n(\sqrt[n]{2}) - f(\sqrt[n]{2})\right| = \frac{\sqrt[n]{2^2}}{1+2}-0 = \frac{\sqrt[n]{4}}{3} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

The sequence is uniformly convergent on $[a,b]$ for $1 < a < b$. Namely, for any $x \in [a,b]$ we have
$$|f_n(x)-f(x)| = \frac{x^2}{1+x^n} \le \frac{b^2}{1+a^n} \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $\frac{b^2}{1+a^n}$ is independent of $x$.

The sequence is furthermore uniformly convergent on $[a,+\infty)$ for $a>1$. Namely, by calculating the derivative $$f_n'(x) = \frac{x(2-x^n(n-2))}{(1+x^n)^2}$$we see that $f_n$ is strictly decreasing for $x \ge \sqrt[n]{\frac{2}{n-2}}$. Since this bound converges to $1$, for large enough $n\in\Bbb{N}$ it will be less than $a$ so we will have that $f_n$ is strictly decreasing on $[a,+\infty)$. Therefore for $x \in [a,+\infty)$ we have
$$|f_n(x)-f(x)| = f_n(x) \le f_n(a) = \frac{a^2}{1+a^n} \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $\frac{a^2}{1+a^n}$ is independent of $x$.

Therefore we conclude that $f_n \to f$ uniformly on $[a,b]$ where $0 \le a < b$ if and only if
$$0\le a < b < 1 \quad \text{ or } \quad 1 < a \le b$$
and on the ray $[a,+\infty)$ for $1 < a$.

Body:
The sequence is not uniformly convergent on $(a,1)$ for $0 <a < 1$. Indeed, we have $\frac1{\sqrt[n]{2}} \to 1$ from the left so for large enough $n\in\Bbb{N}$ we have $\frac1{\sqrt[n]{2}} \in (a,1)$ and

$$\left|f\left(\frac1{\sqrt[n]{2}}\right) - f_n\left(\frac1{\sqrt[n]{2}}\right)\right| = \frac1{\sqrt[n]{4}} -\frac{\frac1{\sqrt[n]{4}}}{1+\frac12} = \frac1{3\sqrt[n]4} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

The sequence is uniformly convergent on $[0,b]$ for $0 < b < 1$. Namely, for $x \in [0,b]$ we have
$$|f(x)-f_n(x)| = x^2-\frac{x^2}{1+x^n} = {x^2}\frac{x^n}{1+x^n} \le x^n \le b^n \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $b^n$ is independent of $x$.

The sequence is not uniformly convergent on $(1,b)$ for $b > 1$. Namely, we have $\sqrt[n]{2} \to 1$ from the right so for large enough $n\in\Bbb{N}$ we have $\sqrt[n]{2} \in (1,b)$ and
$$\left|f_n(\sqrt[n]{2}) - f(\sqrt[n]{2})\right| = \frac{\sqrt[n]{2^2}}{1+2}-0 = \frac{\sqrt[n]{4}}{3} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

The sequence is uniformly convergent on $[a,b]$ for $1 < a < b$. Namely, for any $x \in [a,b]$ we have
$$|f_n(x)-f(x)| = \frac{x^2}{1+x^n} \le \frac{b^2}{1+a^n} \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $\frac{b^2}{1+a^n}$ is independent of $x$.

The sequence is furthermore uniformly convergent on $[a,+\infty)$ for $a>1$. Namely, by calculating the derivative $$f_n'(x) = \frac{x(2-x^n(n-2))}{(1+x^n)^2}$$we see that $f_n$ is strictly decreasing for $x \ge \sqrt[n]{\frac{2}{n-2}}$. Since this bound converges to $1$, for large enough $n\in\Bbb{N}$ it will be less than $a$ so we will have that $f_n$ is strictly decreasing on $[a,+\infty)$. Therefore for $x \in [a,+\infty)$ we have
$$|f_n(x)-f(x)| = f_n(x) \le f_n(a) = \frac{a^2}{1+a^n} \xrightarrow{n\to\infty} 0$$
uniformly in $x$ since the bound $\frac{a^2}{1+a^n}$ is independent of $x$.

Therefore we conclude that $f_n \to f$ uniformly on $[a,b]$ where $0 \le a < b$ if and only if
$$0\le a < b < 1 \quad \text{ or } \quad 1 < a \le b$$
and on the ray $[a,+\infty)$ for $1 < a$.


答:
It is not uniformly convergent on $[0, 1)$, you have to be cautious here. First, you made a little mistake:
$$
\lvert f_n(x) - f(x) \rvert = \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert
$$
Then:
$$
\sup_{x \in [0, 1)} \lvert f_n(x) - f(x) \rvert = \sup_{x \in [0, 1)} \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert = \frac{1^{n+2}}{1+1^n} = \frac{1}{2}
$$
This does not tend to zero, so we do not have uniform convergence on $[0, 1)$ to $f(x) = x^2$. But we do have uniform convergence on every interval $[a, b]$ (or $(a, b]$ or $[a, b)$ or $(a, b)$, it does not matter) where $a, b \in (0, 1)$, because:
$$
\sup_{x \in [a, b]} \lvert f_n(x) - f(x) \rvert = \sup_{x \in [a, b]} \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert = \frac{a^{n+2}}{1+{a^n}} \overset{n \rightarrow \infty}{\longrightarrow} 0
$$

Obviously, for $x = 1$, $f_n(x) = \frac{1}{2}$ so the pointwise convergence for $x=1$ is to the constant function $\frac{1}{2}$.

For $x>1$, the pointwise limit is $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$. So uniform convergence is not possible on intervals $J \subseteq \mathbb{R}$ where $1 \in \overline{J}$, because $f$ would be discontinuous which is not allowed since all the $f_n$ are in fact continuous. Now consider intervals of the form $[c, d]$ (or $(c, d)$ or $[c, d)$ or $(c, d]$) where $c, d>1$. Then:
$$
\sup_{x \in [c, d]} \lvert f_n(x) - 0\rvert = \sup_{x \in [c, d]} \left \lvert \frac{x^2}{1+x^n} \right \rvert \leq  \left \lvert \frac{d^2}{1+c^n} \right \rvert \overset{n \rightarrow \infty}{\longrightarrow} 0
$$

For the case $[a, \infty)$ where $a>1$ consider mechanodroid's answer.

Title: On what subintervals of $[0,\infty)$ is $f_n(x)=\frac{x^2}{1+x^n}$ uniformly convergent?

Body: I have calculated the pointwise convergence which is $f(x)=x^2$ for $x<1\qquad$ and for $x=1$ we get $f(x)=\frac{1}{2} \quad$ and lastely for $f(x)=0$ for $x>1$. Now, I know that this sequence is not uniformly convergent for  the set $[0,\infty)$ but I know that it IS uniformly convergent for $[0,1)$ since we can do the following: $|f_n(x)-f(x)| =...= |\frac{x^n}{1+x^n}|<|x^n|$ and since we know that $x<1, \forall \epsilon >0,$ we can find a $N$ such that $x^n < \epsilon \forall n\ge N$. The problem I'm having now is how can I find all subintervals of $[0,\infty)$ where this sequence is uniformly convergent?


问:
What will be the eigen vector of $A^t$
$Ax = ax$ and $x$ is an eigen vector of $A$ corresponding to the eigen value $a$ . What will be the eigen vector of $A^t$ corresponding to the eigen value $a$.

Can anyone please give me a hint ?

答:
Assuming that you actually want to construct the eigenvector, I suggest a solution using diagonalization. We know that there exists $g\in GL_n$ such that $$
gAg^{-1} = \begin{bmatrix}
a & 1 & 0 & 0 &\dots & \dots & 0\\
0 & a & 1 & 0 & \dots & \dots & 0\\
 &  & \ddots & & & \\
& & & a & & \\
& & & & J_1 & & \\
& & & & & \ddots &\\
& & & & & & J_k
\end{bmatrix}
$$ Here, the first block on the diagonal is the Jordan block corresponding to the eigenvalue $a$ and $J_i$ are the rest of Jordan blocks.

Observe that $gAg^{-1}e_1 = a e_1$ and so $gx = e_1$. Now, take the transpose. The observation is that if the Jordan block corresponding to the eigenvalue $a$ has size $k$, then its transpose has $e_k$ as an eigenvector with the eigenvalue $a$. This is because $$
\begin{bmatrix}
a & 0& \dots &\dots &0 \\
1 & a &0 &\dots & 0 \\
& 1& \ddots & &\\
& & \ddots & & \\
& & & 1 & a
\end{bmatrix}\cdot e_k = a e_k.
$$ We go back by multiplying $e_k$ with $g^{t}$, i.e. if $y=g^t e_k $, then $A^t y=ay$. This is true since $$
A^t y = A^t g^t e_k = g^{t} (g^{t})^{-1}A^t g^t e_k = g^t (ae_k)=ay.
$$

Title: What will be the eigen vector of $A^t$


答:
$|a| = a \text{ or } -a$

$a \leq c \leq |c|$

$- a \leq -b \leq |b|$

Title: Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$

Body:
Prove that $|a| \leq \max\{|b|,|c|\}$ if $b\leq a \leq c$.

I showed that $a\leq c$ and therefore $-c\leq a \leq c$ so that $|a|\leq c$ but then I got stuck.

Is this the right approach?


问:
Complexity and Big-O notation in mathematics
I have only rough idea about computational complexity and Big-O notation.

Is these idea of algorithm applicable in mathematics too, or only in computer science?

If applicable in mathematics too,

Why the complexity of an expression similar to $n(n-1)$ is $O(n^2)$, not something like $O(n^2)\land O(n)$? Is it because how machines work? I mean we calculate $13^2$ from our memory, and for $31^2$ we use rules of multiplication, but machines need to add $31$ to $0$ for $31$ consecutive times, and hence most of the computational time is spent there, is it the reason? Even if such reasons apply, we know that the complexity of an expression similar to $n+5$ is $O(n)$, not $1$. Hence $O(n)$ is not negligible. Perhaps an explanation of the complexity of $n^3+n^2+n$ would clear things up.

What is the complexity of $(41n)^2-1$? Why?

What is the complexity of $n!/(n-2)!$ By cancelling out $(n-2)!$, we can have $O(n^2)$. But if we do not cancel out, I found that the complexity of factorial is $O(n^n)$. (I suppose machines don't have the idea of cancelling out yet.)

答:
Yes, these are used in mathematics as well as computer science.  We may need to specify a "limit point" when talking about them.  In addition to $f(n) = O(g(n))$ as $n \to \infty$, we also see rates given when approaching some finite value: $f(x) = O(g(x))$ as $x \to 3$.

All of the following are correct (as $n \to \infty$):
$$
n(n-1) = O(n^2)
\\
n(n-1) = O(n^2+n)
\\
n(n-1) = O(4n^2-3n+7)
\\
n(n-1) = O(n^2+n^{3/2}+n+n^{1/2})
\\
n(n-1) = O(932n^2-12n+\pi)
\\
$$
Can you guess why the most often chosen version is the first one?

Proofs for these:
$$
\lim_n \frac{n(n-1)}{n^2} = 1 < +\infty,
\\
\lim_n \frac{n(n-1)}{n^2+n} = 1 < +\infty,
\\
\lim_n \frac{n(n-1)}{4n^2-3n+7} = \frac{1}{4} < +\infty,
\\
\lim_n \frac{n(n-1)}{n^2+n^{3/2}+n+n^{1/2}} = 1 < +\infty,
\\
\lim_n \frac{n(n-1)}{932n^2-12n+\pi} = \frac{1}{932} < +\infty,
$$

I would choose
$$
(41n)^2-1 = O(n^2)
$$
but all the ones above are also correct.  The proof for $O(n^2)$
looks like this:

As $n \to \infty$ we have
$$
\frac{(41n)^2-1}{n^2} = \frac{1681 n^2 - 1}{n^2} = 1681 + \frac{-1}{n^2}
 \to 1681
$$
Any finite limit (or even lim sup) will be OK.

It is true that
$$
\frac{n!}{(n-2)!} = O(n^2)
\tag1$$
since
$$
\limsup_n \frac{n!/(n-2)!}{n^2} = 1 < +\infty
$$
It is also true that
$$
\frac{n!}{(n-2)!} = O(n^n)
\tag2$$
since
$$
\limsup_n \frac{n!/(n-2)!}{n^n} = 0 < +\infty
$$
But of course $(1)$ tells us more information than $(2)$, so
(1) is a "better" choice.

Sometimes this notion of "better choice" is specified by saying
$$
\frac{n!}{(n-2)!} = \Theta(n^2)
$$
which means:
$$
\text{both}\quad \frac{n!}{(n-2)!} = O(n^2), \quad\text{and}\quad
n^2 = O\left(\frac{n!}{(n-2)!}\right)
$$


问:
Iteration Substitution, and $T(n)$ where $n = 0$

Iteration Substitution / Plug and Chug method

Hi, I'm currently working on recurrence relation. And I'm having trouble on iteration problems, such as:

$$T(n) = 1 + \frac 12\, T\left(\frac n2\right), \quad T(0) = 2$$

And if, for instance, I have already formulated a formula with respect to i, such as:
$$
\dfrac {n^2}i = 0
$$
I just don't know what to do, as when I try to separate $n$ and $i$, I would just get $n^2 = 0$ or the other way around. How would I go about these kind of problems?

Edit:

The only way I know right now is by substituting $0$ with a variable $x$. That way, I could get some formula. But I still don't know what to do afterwards.
$$
x = 0;\\
\frac{n^2}{i}\ = x \\
\frac{n^2}{x}\ = i
$$

答: The only solution that is bounded at the origin is the constant function $T(n) = 2$. This is because we can iterate the relation to obtain $$T(n) = 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k} + \frac{1}{2^{k+1}} T\Big(\frac{n}{2^{k+1}}\Big)$$ which tends to $2$ as $k \to \infty$, regardless of $n$.


问:
Is any norm on $\mathbb R^n$ invariant with respect to componentwise absolute value?
Given $\mathbf{x}=(x_1,...,x_n) \in \mathbb{R}^n$ , define  $ \mathbf{x}'=(|x_1|,...,|x_n|) $ .

Then, is it $||\mathbf{x}'|| = ||\mathbf{x}||$  for every norm on $ \mathbb{R}^n $  ?

NB: The answer is trivially yes for $p$-norms.

Body:
You may consider, for example,

$$\|(x_1,x_2)\|:=|x_1|+|x_1-x_2|.$$


答:
You can search for constant solutions $c$.

Substituting in the equation gives $c=1+\frac 12c\iff c=2$

Let shift the solution to get rid of the annoying constant $1$ : $$U(n)=T(n)-2$$

We get: $\require{cancel}\quad U(n)=T(n)-2=\left(1+\frac 12 T(\frac n2)\right)-2=\left(\cancel{1}+\frac 12(U(\frac n2)+\cancel{2})\right)-\cancel{2}$

We get the simplified relation:
$$U(n)=\frac 12 U(n/2)$$

Let set $n=2^p$ and $V(p)=U(2^p)$

$V(p)=U(2^p)=\frac 12U(2^p/2)=\frac 12U(2^{p-1})=\frac 12V(p-1)$

This induction relation is easy to solve:

$V(p)=\frac 12V(p-1)=\frac 14V(p-2)=\frac 18V(p-3)=\cdots=\frac 1{2^p}V(0)$

We can now express $T(n)$ while going back the chain of substitutions :

$V(0)=U(2^0)=U(1)=T(1)-2$

$V(p)=U(2^p)=U(n)=T(n)-2=\dfrac{V(0)}{2^p}=\dfrac{T(1)-2}{n}$

Final result $$T(n)=\frac{T(1)-2}{n}+2$$

Solving initial condition $T(0)=2$ requires $T(1)=2$ else this quantity would be infinite.

In this case $T(n)=2,\ \forall n\in\mathbb N$.

Title: Iteration Substitution, and $T(n)$ where $n = 0$

Body:
Iteration Substitution / Plug and Chug method

Hi, I'm currently working on recurrence relation. And I'm having trouble on iteration problems, such as:

$$T(n) = 1 + \frac 12\, T\left(\frac n2\right), \quad T(0) = 2$$

And if, for instance, I have already formulated a formula with respect to i, such as:
$$
\dfrac {n^2}i = 0
$$
I just don't know what to do, as when I try to separate $n$ and $i$, I would just get $n^2 = 0$ or the other way around. How would I go about these kind of problems?

Edit:

The only way I know right now is by substituting $0$ with a variable $x$. That way, I could get some formula. But I still don't know what to do afterwards.
$$
x = 0;\\
\frac{n^2}{i}\ = x \\
\frac{n^2}{x}\ = i
$$


答:
70% of 500 is 350. So the top 350 candidates passed the paper.

If the passing grade were 30 or more, this wouldn't be possible, since then only 200 candidates could have passed, at most. If the passing grade were 20 or less, it wouldn't be possible since then at least 400 students would have passed. So the passing grade is somewhere between 21 and 29 inclusive. However, it's impossible to know exactly what it is. Maybe by "the minimum" marks, they mean 21, the lowest the passing grade could possibly be?

Title: Can anyone help me with statistics?

Body:
Question: Following is the distribution of marks obtained by 500 candidates in the statistics paper of a civil services examination:

```<code>+ marks more than:      0   10  20  30  40  50
+ number of candidates: 500 460 400 200 100 30
</code>```

Calculate the lower quartile marks. If $70\%$ of the candidates pass in the paper, find the minimum marks obtained by a passing candidate.

In the above problem, I am able to find the first part to calculate lower quartile marks but not able to find the second section.

Anyone, please help I am a newbie to statistics.

Thanks in advance, please ignore the bad English.


问:
Prove : $\arctan(x)>\tanh(x)$ for $0< x\leq 1$
It's a classical inequality that I want to propose :

Claim:

Let $0< x\leq 1$ then we have :
$$\arctan(x)>\tanh(x)$$

My proof :

We introduce the function :

$$f(x)=\arctan(x)-\tanh(x)$$

We differentiate :

$$f'(x)=\frac{1}{(x^2+1)}+\frac{-4}{(e^{-x} + e^{x})^2}$$

Remains to show :

$$\frac{(e^{-x} + e^{x})^2}{4}\geq x^2+1\quad \quad (0)$$

Or :

$$-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2\geq 0$$

We introduce the function :

$$g(x)=-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2$$

We differentiate twice :

$$g''(x)=e^{-2x}+e^{2x}-2\geq 0$$

We deduce that the first derivative of $g(x)$ is increasing and $g'(0)=0$ so $g(x)$ is increasing but $g(0)=0$ so we establish the inequality $(0)$ . We conclude that the function $f(x)$ is increasing and $f(0)=0$ wich prove the claim .

My question :

Have you one or more alternative proof (using integration by example)?

Thanks in advance !

Erik

答:
Note $\sinh t-t>0$ and

\begin{align}
\arctan x-\tanh x & =\int_0^x \left(\frac{1}{t^2+1}-\frac{1}{\cosh^2 t}\right)dt\\
&= \int_0^x \frac{(\sinh t +t)(\sinh t-t)}{(t^2+1)\cosh^2 t}dt>0
\end{align}
Thus, $\arctan x>\tanh x$.


问:
Proof Using Mathematical Induction that $2^n \geq n + 1$
$2^n \geq  n + 1$, for all $n \geq  1$

So I know that first I should use 1, which I did to make sure step 1 holds.
$$2 ^ 1 \geq 1 + 1\\2 \geq 2\\ \mathsf{TRUE}$$

After I figure this out, I know I am then supposed to assume it true for $k$, so I get $2 ^ k \geq k + 1$, but its the whole $k+1$ step that completely looses me. Where would I do from here? I am assuming that I have to get $2 ^ k \geq k + 1$ to equal   $2 ^{k+1} \geq k+1 + 1$, but am not sure how to get there.

Body: By AM-GM inequality $\frac{2^n+(n-1)}{n}=\frac{2^n+1+1+...+1}{n}\ge \sqrt[n]{2^n}=2\implies 2^n \ge 2n-(n-1)=n+1$


问:
How many codes can be formed from three letters (A-Z) followed by two digits ($0$-$9$)?

Problem:

A code consists of 3 letters followed by 2 digits. How many codes can we create if a)repetition is allowed b)repetition is not allowed?

Attempt:

The code should looks like this LLLDD (L for letter and D for  digit) so if repetition is allowed, the number of possible codes is
$$(26^3)(10)^2$$
and if repetition is not allowed the number of possible  codes is
$$26 \mathrm{ P }3 \times 10 \mathrm{ P }2.$$

My concern:

How does $(26^3)(10)^2$ or $26 \mathrm{ P }3 \times 10 \mathrm{ P }2$ guarantee that the code has the correct pattern (LLLDD) and not other patterns like LDLDL, LLDDL, it's what permutations do after all (order is important), right?

答:
Alternative explanation.

how does $(26^3)(10)^2$ or ${}26 \mathrm{ P }3 \times 10 \mathrm{ P }2$ guarantee that the code
has the correct pattern (LLLDD) and not other patterns like LDLDL , LLDDL , it's what permutations do after all (order is important), right?

The formulas do not guarantee the pattern.  However, the formulas are still correct.  The original question is to enumerate the number of ways that the pattern LLLDD can occur.  The formulas are accurate, under the assumption that your attention is confined to the LLLDD pattern.

The fact that the formulas may also be correct for other patterns is totally irrelevant.


问:
Is $s_n$ a multiple of $2^n$ for all $n \geq 0$?
Let's define:
$$a_{0,0} = 1$$
$$\forall k \ne 0, a_{0,k} = 0$$
$$\forall n \geq 0, \forall k, a_{n+1,k} = -(4n-2k+3)a_{n,k-1} + (2k+1)a_{n,k}$$

```<code>    |   0    1    2    3    4  ...
----+-----------------------------
  0 |   1
  1 |   1   -1
  2 |   1   -8    3
  3 |   1  -33   71  -15
  4 |   1 -112  718 -744  105
...   ...  ...  ...  ...  ... ...
</code>```

Up to the alternated sign, this is OEIS' sequence number A214406.
Now let's define:
$$s_n = \sum_{k=0}^n a_{n,k}$$
$$ s_0 = 1, s_1 = 0, s_2 = -4, s_3 = 24, s_4=-32, \dots $$

It appears that for all $n$, $s_n$ is a multiple of $2^n$. Is it provable?

Context/own efforts: so far, I was able to prove connections with the sequence of functions $(R_n)_{n \in \mathbb{N}}$ defined by:

$$ R_0(x)=1 $$
$$ R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2} \right)' $$

The connection is:

$$ s_n = 2^n R_n(1) $$

We even have:

$$ R_n(x) = 2^n \frac{P_n(x^2)}{(1+x^2)^{2n}} $$

where
$$ P_n(x) = \sum_{k=0}^n a_{n,k} x^k $$

I don't know if it helps, but this also means that the exponential generating function

$$ \sum_{n \geq 1} \frac{s_{n-1}}{2^{n-1}} \frac{x^n}{n!} $$

is the series reversion of

$$ \frac{\ln(x+1)}{2} + \frac{x}{2} + \frac{x^2}{4} = 0 + 1x + 0 x^2
+ \frac{1}{6}x^3 - \frac{1}{8}x^4
+ \frac{1}{10}x^5 - \frac{1}{12}x^6
\dots $$

I also established that

$$ s_{n+1} = -4 \sum_{k=0}^n (n-k) a_{n,k} $$

but this appears just to push the problem forward, of proving that $ \sum_{k=0}^n k a_{n,k} $ is a multiple of $2^{n-1}$ ...

答:
Starting with $$s_n = 2^n R_n(1)$$

All we need to prove then is that $R_n(1)$ is an integer for all $n$

Clearly $s_n$ is a sequence of integers, implying that if $R_n(1)$ is not an integer, then it must be a fraction with denominator $2^m,m\leq n$.

Remember
$ R_0(x)=1, R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2} \right)' $

Note that since $R_n(x)$ can subsequently be written as $R_n(1)=\frac{a*2^b}{2^m}$ for integers $a,b$, we just need to prove $b\geq m$ to prove $R_n(1)$ is an integer

Proof by induction:

$R_0(1)$ clearly satisfies this condition.

Also note for $n>0$,
$R_{n+1}(x) = \cfrac{d}{dx} (R_n(x) \times (\frac{2x}{x^2+1}-1))=R^{'}_n(x)\times \frac{2x}{x^2+1}-R_n(x)\times \frac{2(x+1)(x-1)}{(x^2+1)^2}$

$R_{n+1}(1) = R^{'}_n(1)\times (1-0) = R^{'}_n(1)$

$R_{n+1}(1) = \frac{d^n}{dx^n}R_1(1) = \frac{d^{n+1}}{dx^{n+1}}\frac{2x}{x^2+1}$

By the quotient rule, if there are $m-b$ factors of $2$ in the simplified denominator of $R_n(1)$ (such that there are 0 factors of 2 in the numerator), then there is at most $2(m-b)$ factors of $2$ in the denominator of $R^{'}_n(1)$.

However, since for $R_0(1),m-b=0$, then for $R_1(1)$, the denominator can have at most $2(m-b) = 0$ factors of 2. Equivalently, $2(m-b)\leq 0,b\geq m$ By induction, this holds true for every $R_n(1)$, which is what we set out to prove.


问:
Proving identity in infinite series
I am trying to prove the equation

\begin{align*}
\sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i},
\end{align*}

where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:

\begin{align*}
\sum_{n=0}^{\infty} a^{c_1 + c_2 + \ldots + c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d}
&= \sum_{n=0}^{\infty} a^{c_1} \cdot a^{c_2} \cdot \cdot \cdot a^{c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdot \cdot \cdot b_d^{c_d} \\
&= \sum_{c_1}^{\infty} a^{c_1} \cdots \sum_{c_d}^{\infty} a^{c_d} \cdot \sum_{c_1}^{} b_1^{c_1} \cdots \sum_{c_d} b_d^{c_d} \\
&= \sum_{c_1}^{\infty} (a b_1)^{c_1} \cdots \sum_{c_d}^{\infty} (ab_d)^{c_d}.
\end{align*}

My doubts in here are whether it is even possible to split up the sum in this manner and then simply recombining it in the last line, since the product of two sums is not necessarily the sum of the product.

From here I would use

\begin{align*}
\sum_{k}^{\infty} x^k = \frac{1}{1-x},
\end{align*}

such that it would become

\begin{align}
\frac{1}{1-a b_1} \times \cdots \times\frac{1}{1 - a b_d} = \prod_{i = 1}^{d} \frac{1}{1-a b_i}.
\end{align}

I would like to ask where my errors are and if there is a better way to prove this identity. Thank you for your time.

答:
We introduce the indicator function notation defined by

$$ \mathbf{1}_{\{ \cdots \}} = \begin{cases}
1, & \text{if $\cdots$ holds}, \\
0, & \text{if $\cdots$ does not hold}.
\end{cases} $$

Then the sum is can be rewritten as

\begin{align*}
\sum_{n=0}^{\infty}a^n \sum_{\substack{c_1,\ldots,c_d\geq0\\c_1+\cdots+c_d=n}}
b_1^{c_1} \cdots b_d^{c_d}
&= \sum_{n=0}^{\infty} \sum_{\substack{c_1,\ldots,c_d\geq0\\c_1+\cdots+c_d=n}}
(a b_1)^{c_1} \cdots (a b_d)^{c_d} \\
&= \sum_{n=0}^{\infty} \sum_{c_1,\ldots,c_d\geq0}
\mathbf{1}_{\{ c_1+\cdots+c_d=n \}}(a b_1)^{c_1} \cdots (a b_d)^{c_d}.
\end{align*}

Since the two sigma-notations in the last line are independent of each other, we can freely interchange the order[1] to get

\begin{align*}
\sum_{c_1,\ldots,c_d\geq0}
\Biggl( \sum_{n=0}^{\infty}  \mathbf{1}_{\{ c_1+\cdots+c_d=n \}} \Biggr) (a b_1)^{c_1} \cdots (a b_d)^{c_d}.
\end{align*}

However, for each given $c_0, \ldots, c_d \geq 0$, there is exactly one $n \geq 0$ such that $c_1+\cdots+c_d=n$ holds, and so, the inner sum is always $1$. So the sum simplifies to

$$ \sum_{c_1,\ldots,c_d\geq0}
(a b_1)^{c_1} \cdots (a b_d)^{c_d}
= \Biggl( \sum_{c_1 \geq 0} (a b_1)^{c_1} \Biggr) \cdots \Biggl( \sum_{c_d \geq 0} (a b_d)^{c_d} \Biggr)
= \frac{1}{1 - ab_1} \cdots \frac{1}{1 - ab_d}. $$

Remark. The same conclusion can be derived without the use of indicator function. However, my personal impression is that this method provides great flexibility when working with multiple sums.

[1] Switching the order of summation is not always possible. Fubini's theorem provides a sufficient condition for this. In OP's case, this is possible if either $\left| ab_i \right| < 1$ for all $i = 1, \ldots, d$ or $ab_i \geq 0$ for all $i = 1, \ldots, d$. Alternatively, if we regard the sum as a formal power series, then no such assumption is required.

Title: Proving identity in infinite series


问:
Is $f(n)$ asymptotic to $g(n)$?
$\pi(n)$ is the prime counting function and $li(n)$ is the logarithmic integral. $$f(n)=e^{-\frac{1}{\pi\big(\frac{1}{n}\big)}}$$
and $$ g(n)=e^{-\frac{1}{li\big(\frac{1}{n}\big)}}$$

Is $f(n)\sim g(n)?$ (asymptotic).

$f(n)$ and $g(n)$ seem to grow "closer" to one another not "further" apart, as $n\to 0.$

答:
Let $x = \frac{1}{n}$ and so when $n \to 0^{+}$  then $x \to \infty$ and we need to prove that $ f(x) \sim g(x) $

Since e(exp) is continues function then $ \lim \limits_{x \to \infty} f(x) =\lim \limits_{x \to \infty} e^{-\frac{1}{\pi(x)}} = e^{\lim \limits_{x \to \infty} -\frac{1}{\pi(x)}} = e^0 =1$ ($\pi(x)$ is not bounded from above, there are infinitely many primes).

Also $ \lim \limits_{x \to \infty} g(x) = \lim \limits_{x \to \infty} e^{-\frac{1}{li(x)}} = e^{\lim \limits_{x \to \infty} -\frac{1}{li(x)}} = e^0=1$ ($ li(x) > \frac{x}{\ln x}$)

And so $f(x) \sim 1 \sim g(x) \implies f(x) \sim g(x)$ as $x \to \infty$ or $ n \to 0^+$

Title: Is $f(n)$ asymptotic to $g(n)$?


问:
Find the global extreme values of a the following function $f(x) = x|x^2-6|- \frac{3}{2} x^2 +2$ in [-2,4]
$f(x) = x|x^2-6|- \frac{3}{2} x^2 +2$

With this function I know that I need to differentiate it in order to find the local maxima and minima points. However it always confuses me a bit on what I should do when there's a modulus/absolute value within the equation. Once I've found the derivative I'm pretty certain on what to do.

Anyways what I'm thinking right now is to have three cases. Firstly when $x^2$ is greater than 6, secondly when $x^2$ is less than 6 and finally when it's equal to 6. Is this the right approach? I'm also going to test $f(-2)$ & $f(4)$ since they're the limits.

Body:
hint

Observe that
$$-\sqrt{6}<-2<\sqrt{6}<4$$
and
$$-\sqrt{6}<x\le \sqrt{6}\implies$$
$$ |x^2-6|=-x^2+6$$

$$\sqrt{6}\le x\le 4\implies |x^2-6|=x^2-6$$

Thus , at $ [\sqrt{6},4]$,
$$f'(x)=3x^2-6-3=3(x^2-3)>0$$
and at $ [-2,\sqrt{6}] $,
$$f'(x)=6-3x^2-3=3(1-x^2)$$

You have to check $ f(\pm 1) $.


问:
Wedge product of vector fields
Can somebody explain me step by step how can I compute the wedge product $X\wedge Y$ of two vector fields, $X,Y$, in $\mathbb{C}^2$?

We can consider
$$
X=X_1\partial_x+X_2\partial_y
$$
and
$$
Y=Y_1\partial_x+Y_2\partial_y
$$

Please!

答:
Expand out.  $ \ \ e_i \wedge e_i=0, \ \ e_i \wedge e_j=-e_j \wedge e_i$

$e_1=\partial_x, e_2=\partial_y
\\X\wedge Y=X_1Y_1e_1\wedge e_1+X_1Y_2e_1\wedge e_2+X_2Y_1e_2\wedge e_1+X_2Y_2e_2\wedge e_2\\
=(X_1Y_2-X_2Y_1)e_1\wedge e_2=(X_1Y_2-X_2Y_1) \partial_x \wedge \partial_y$

Title: Wedge product of vector fields


答:
Anyways what I'm thinking right now is to have three cases. Firstly when $x^2$ is greater than 6, secondly when $x^2$ is less than 6 and finally when it's equal to 6. Is this the right approach? I'm also going to test $f(-2)$ & $f(4)$ since they're the limits. Also I don't really need the case when $x^2$ is less than 6 do I? Since it would be outside of the range

First of all, you can collapse $x^2 > 6$ with $x^2 = 6.$

Second of all you do need the case of $x^2 < 6$ because (for example) $1$ is in the domain.

I would split the function into two separate functions, based on whether
$x^2 < 6$ or $x^2 \geq 6$.  Then you can differentiate each one, as long as you are careful to note that the separate functions each have a restricted domain.

Title: Find the global extreme values of a the following function $f(x) = x|x^2-6|- \frac{3}{2} x^2 +2$ in [-2,4]

Body:
$f(x) = x|x^2-6|- \frac{3}{2} x^2 +2$

With this function I know that I need to differentiate it in order to find the local maxima and minima points. However it always confuses me a bit on what I should do when there's a modulus/absolute value within the equation. Once I've found the derivative I'm pretty certain on what to do.

Anyways what I'm thinking right now is to have three cases. Firstly when $x^2$ is greater than 6, secondly when $x^2$ is less than 6 and finally when it's equal to 6. Is this the right approach? I'm also going to test $f(-2)$ & $f(4)$ since they're the limits.


问:
How to factorize equation of pair of lines?

Find two equations of lines
: $$6x^2 -xy -12y^2 -8x + 29y -14= 0$$

I tried doing it using quadratic equation but got some very much unusual value. How to solve this?

Body:
Hint:

$$6x^2-x(y+8)-(12y^2-29y+14)=0$$

The discriminant $(D)$ is $$(y+8)^2+24(12y^2-29y+14)=(17y-20)^2$$

$$\implies x=\dfrac{y+8\pm\sqrt D}{2\cdot6}$$


问:
Can $T\left( n\right) = 4T\left({n\over 2}\right) + 2^{n\over2}$ be solved using Akra-Bazzi method?
$$T\left( n\right) = 4T\left({n\over 2}\right) + 2^{n\over2}$$

Using Akra-Bazzi method

$a = 4, b = {1\over2}, g(n) = 2^{n\over2}, h(n) = 0$

$$g'(n) = {d\over dx}(2^{n\over2})=2^{{n\over2}-1}\ln 2$$

Since $\left|g'(n)\right|$ is not polynomially bound we cannot proceed further with Akra-Bazzi method.

Using Master theorem

$a = 4, b = 2, g(n) = 2^{n\over 2}$

$g(n) = \Omega(n^{\log_2 4+\epsilon})$ where $\epsilon=1$, also

$$ag\left( {n\over b}\right) \le cg(n)$$

$$4g\left( {n\over 2}\right) \le cg(n)$$

$$2^{{n\over 4}+2} \le c2^{n\over 2} \tag*{for $c={1\over 2}$ and $n\ge 12$ }$$

Therefore it follows from master theorem that $T(n) = \Theta\left( g(n)\right) = \Theta\left( 2^{n\over 2}\right)$.

How come the above problem can be solved using Master theorem which is just a corollary of Akra-Bazzi method but not using Akra-Bazzi method itself.

答:
From

$$
T\left(2n\right)=4T\left(n\right)+2^n\Rightarrow T\left(2^{\log_2(2n)}\right)=4T\left(2^{\log_2n}\right)+2^n
$$

now making

$$
\mathcal{T}(\cdot) = T\left(2^{(\cdot)}\right),\ \ \ z=\log_2 n
$$

we have

$$
\mathcal{T}(z+1)=4\mathcal{T}(z)+2^{2^z}
$$

solving this recurrence we obtain

$$
\mathcal{T}(z) = \frac 14 4^z\left(C_0+\sum_{k=0}^{z-1}2^{2^k-2k}\right)
$$

or as $4^{\log_2 n} = 2^{2\log_2 n} = 2^{\log_2 n^2} = n^2$

$$
T(n) = \frac 14 n^2\left(C_0+\sum_{k=0}^{\log_2 n-1}2^{2^k-2k}\right)
$$


问:
Recursively generating all partitions
Given a set $\{1,2,\ldots,n\}$, I would like to recursively generate all partition of size $r$ and $n-r$. For instance, we start with the partitions
$$\{1,2,\ldots, r\} \quad \{r+1,\ldots, n\}$$
and then iteratively swap elements.  We can write the solution in terms of all the collections of allowed swaps between the two sets. We do this by considering a swap operator $S_{ij}$, which swaps elements $i$, $j$ between these sets, for instance:

$$S_{1,r+1} \{\{1,2,\ldots, r\},\{r+1,\ldots, n\}\} = \{\{r+1,2,3,\ldots, r\},\{1, r+2,\ldots, n\}\}.$$

Then we can write the solution in terms of all $\binom{n}{r}$ strings of swap operators acting on the two sets.

I know that one way to do this is with Heap's algorithm, which would generate all the permutations, which we then just have to partition. However, this algorithm is inefficient for this problem since many permutations correspond to the same subsets. As such I am essentially looking for a simplification to Heap's algorithm for this specific case.

答: You need to generate all the $r$ sized subsets, of which there are $n \choose r$.  ${n-1 \choose r-1}$ of them contain $1$ and ${n-1 \choose r}$ do not contain $1$, so call your subset generator program with those parameters, add $1$ to each set in the output of the first instance, and put them together.

Title: Recursively generating all partitions


问:
Idempotents in a Quotient Ring
Let $R=\mathbb{Z}_p[x]/(x^p-x)$.  Show that $R$ has exactly $2^p$ elements satisfying $r^2=r$.

I know that for $f,g\in\mathbb{Z}_p[x]$, we have $f-g\in(x^p-x)$ if and only if $f(a)=g(a)$ for all $a\in\mathbb{Z}_p$.  It follows that $r^2=r\in R$ if and only if $r$ is represented by some $f\in\mathbb{Z}_p[x]$ such that $f(a)^2=f(a)$ for all $a\in\mathbb{Z}_p$.  Thus, it must follow that $f(a)=0$ or $f(a)=1$ for each $a$.

Now, for each of the $2^p$ $p$-tuples $(b_0,\ldots,b_{p-1})$ where each $b_i$ is either $0$ or $1$, does there exist a polynomial $f$ such that $f(a)=b_a$, and is this polynomial uniquely defined in $R$?

答:
Each of the $p$ elements of the field $\mathbb{Z}_p$ is a root of the polynomial $q(x) = x^p - x$.  Thus we have a factorization of $q(x)$ into $p$ distinct irreducible factors: $$q(x) = x (x-1) \cdots (x-(p-1)).$$ By the Chinese Remainder Theorem, we therefore have a ring isomorphism $$\mathbb{Z}_p[x]/(x^p - x) \cong \mathbb{Z}_p[x]/(x) \times \mathbb{Z}_p[x]/(x-1) \times \cdots \times \mathbb{Z}_p[x]/(x-(p-1)) \cong (\mathbb{Z}_p)^p.$$

It's now much easier to compute the number of idempotents of this ring.  I'll leave the rest of the fun to you!

Title: Idempotents in a Quotient Ring


答:
Hint 1: If $f: \mathbb{R}\to \mathbb{R}$ such that $$f(x)=x|x^{2}-6|-\frac{3}{2}x^{2}+1$$
So,
\begin{eqnarray*}
f'(x)&=&\frac{d}{dx}\left(x|x^{2}-6|-\frac{3}{2}x^{2}+1 \right)\\
&=&0-3x-\frac{d}{dx}\left(x|x^{2}-6\right)\\
&=&-3x-|x^{2}-6|-x\frac{d}{dx}\left(|x^{2}-6|\right)
\end{eqnarray*}
Hint 2: If we need to find $\displaystyle \frac{d}{dx}\left(|x^{2}-6|\right)$
we can use the chain rule $\displaystyle \frac{d}{dx}(|x^{2}-6|)=\frac{d|t|}{dt}\frac{dt}{dx}$ where $t=x^{2}-6$ and$\displaystyle \frac{d}{dt}(|t|)=\frac{t}{|t|}$.
Using that, you can find $$\frac{d}{dx}\left(|x^{2}-6|\right)=\frac{x^{2}-6}{|x^{2}-6|}\frac{d}{dx}(x^{2}-6)$$


问:
A thorough study of type III von Neumann algebras
In many refernce books and papers, type III von Neumann algebra factors were well studied.

If  $M$ is a type III von Neumann algebra has non-trivial centers, what is the structure of $M$?

答:
Given an abelian subalgebra $N$ of a von Neumann algebra $M$, such as the center of $M$, one can always write
$N=L^\infty (X)$, for some measure space $X$.   Furthermore
one can decompose $M$ as a "direct integral"
$$
M=\int_X^\oplus M_x\,dx
$$
(sort of a
continuous direct sum) of von Neumann algebras
indexed by $X$.   When $N$ is the center of $M$ the factors $M_x$ in this
decomposition are simple von Neumann algebras (called "factors" precisely for that reason) [1, 14.2.3]. The moral of the story is
that all there remains to be done  about von Neumann algebras is to study factors, so no one cares about non-simple von Neumann algebras
anymore!

[1] Kadison, Richard V.; Ringrose, John R., Fundamentals of the theory of operator algebras. Vol. 2, Pure and Applied Mathematics, 100-2. San Diego, CA: Academic Press, Inc.


答:
For all $u$,
$$
\sin^2(u)\le u^2\implies\cos^2(u)\ge1-u^2\implies\sec^2(u)\left(1-u^2\right)\le1\tag1
$$
Set $u=\tanh(x)$:
$$\newcommand{\sech}{\operatorname{sech}}
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\tan(\tanh(x))
&=\sec^2(\tanh(x))\sech^2(x)\tag{2a}\\
&=\sec^2(\tanh(x))\left(1-\tanh^2(x)\right)\tag{2b}\\[3pt]
&\le1\tag{2c}
\end{align}
$$
Thus, for $x\ge0$, the Mean Value Theorem says
$$
\tan(\tanh(x))\le x\tag3
$$
Therefore, because $\arctan(x)$ is increasing,
$$
\tanh(x)\le\arctan(x)\tag4
$$

Title: Prove : $\arctan(x)>\tanh(x)$ for $0< x\leq 1$

Body:
It's a classical inequality that I want to propose :

Claim:

Let $0< x\leq 1$ then we have :
$$\arctan(x)>\tanh(x)$$

My proof :

We introduce the function :

$$f(x)=\arctan(x)-\tanh(x)$$

We differentiate :

$$f'(x)=\frac{1}{(x^2+1)}+\frac{-4}{(e^{-x} + e^{x})^2}$$

Remains to show :

$$\frac{(e^{-x} + e^{x})^2}{4}\geq x^2+1\quad \quad (0)$$

Or :

$$-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2\geq 0$$

We introduce the function :

$$g(x)=-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2$$

We differentiate twice :

$$g''(x)=e^{-2x}+e^{2x}-2\geq 0$$

We deduce that the first derivative of $g(x)$ is increasing and $g'(0)=0$ so $g(x)$ is increasing but $g(0)=0$ so we establish the inequality $(0)$ . We conclude that the function $f(x)$ is increasing and $f(0)=0$ wich prove the claim .

My question :

Have you one or more alternative proof (using integration by example)?

Thanks in advance !

Erik


问:
Proving a parabola and polynomial intersect exactly 3 times
the equations are : $P(x)=5-8(x-1)^2$ for the parabola and the other equation is $E(x)=(x-1)^5$ we need to prove that they intersect exactly 3 times.

What I did was compare them at first  $5-8(x-1)^2 = (x-1)^5 $ and then make a new function $f(x)$ according to that we get $f(x)=(x-1)^5 -5+8(x-1)^2$and the function is continuous because its a polynomial and we can also have a derivative.
I tried applying some numbers and i got $f(-1)=-5$ , $f(0)=2$ , $f(-5)=-7493$ , $f(2)=4$  so according to the Intermediate value theorem we have at least 3 roots but what we need to prove is that it has exactly 3 roots. and this looked like rolle's theorem

What I tried then was we want to show that it has exactly 3 roots according to Rolle's theorem so if the original equation has 4 roots the first derivative needs to have exactly 3

$f'(x)=5(x-1)^4 +16(x-1)$

And I got stuck here I could not figure a way to continue from here ..

Thanks for any tips and help all is appreciated and I would also like to know if anyone can confirm that what I did was right until the last step.

Sorry for my english mistakes hope it is understandable.

答:
$$ Q = E-P = (x-1)^5 -5 + 8(x-1)^2$$

By descartes rule of signs, for the positive 'x' values this polynomial has a maximum of two real roots or no real roots.

Now shift $ x \to -x$

$$Q(-x) = -(x+1)^5 -5 + 8 (x+1)^2$$

By descartes rules of sign it has maximum of one roots for negative $x$ side

So, combining the two ideas, we get that it should have either one or three roots. Now, since you've identified existence of the three roots, this means the polynomial has only three real roots.


问:
Check my work on permutations of letters

In how many arrangements can the word "CORONA" be arranged in only 4 letters with no restriction?

I divided into 2 cases:

Case 1: No doubles, hence we have 1 C, 1 O, 1 R, 1 N and 1 A to arrange into 4 spaces, so 5P4, in other words 5!.

Case 2: one double, we have a double O, to arrange into 4 spaces with repetition, so $\frac{4P2}{2!}$ and the rest of the letters C, R, N and A, to be arranged in the 2 remaining spaces, hence $\frac{4P2}{2!}\cdot 4P2$

So the final answer will be adding both: $\frac{4P2}{2!}\cdot 4P2+5!=192$

Is this correct?

答:
In how many arrangements can the word "CORONA" be arranged in only 4 letters with no restriction?

Yes your work is both accurate and valid.  However, I found it a bit confusing.

I would have worded it this way.

$\underline{\text{case 1: 0 O's}}$

4! ways to permute the C, R, N, A.

4!

$\underline{\text{case 2: 1 O}}$

$\binom{4}{3} = 4$ ways to choose the other three letters.

Then, 4! ways to permute the O with the other 3 letters.

$4 \times 4!$

$\underline{\text{case 3: 2 O's}}$

$\binom{4}{2} = 6$ ways to choose the other two letters.

Then, $4 \times 3 = 12$ ways to place the other two letters.

$6 \times 12$

Final answer:

$$(4!) + (4 \times 4!) + (6 \times 12).$$


问:
Prove that if $x^TAx = 0 \Rightarrow Ax = 0$
A is $n \times n$ symmetric and positive semidefinite. So we can use $A^T = A$ and $x^TAx \geq 0$ for all x $\in \mathbb{R}^n$.

Proof by contradiction comes to mind, or $x^TAx = 0 \; \land \; Ax \neq 0$. This might seem like a given at first but I'm thinking what if there is some linear dependence between $x^T$ and the $Ax$ vector?

Edit: Ideally, I'm supposed to prove this using the rank of a matrix.

If I were to prove it by contradiction, $x^TAx = 0 \; \land \; Ax \neq 0$, I'd start off by saying that:

$x$ cannot be a zero vector in this case. The original statement $x^TAx = 0 \Rightarrow Ax = 0$ is trivial for a zero $x$ vector and needs no proof.

from $Ax \neq 0$ we have $rank(Ax) = 1$ since $Ax$ is also a non-zero vector.

problem: how to show that the $x^T$ vector and the $Ax$ vector cannot be orthogonal? (this would be the point of the contradiction proof)

problem 2: the only idea I have for a proof barely utilizes the rank of a matrix, which is my assignment.

答:
I think you have the best hint from Gerry Myerson in the link from @projectilemotion.

Symmetric $A \Rightarrow$ we can choose orthogonal eigenvectors $v$ for A. These form a basis for $\mathcal{R}^n$, so you can write $x=\sum b_jv_j$.

Positive semi-definite $A \Rightarrow$ its eigenvalues are non-negative. Express $x^TAx$ in terms of the eigenvector basis and use these two facts, you should be home.


问:
On first principles of differentiability
Question

Let

$$f(x) =
\begin{cases}
\frac {a} {x - 1} [3\sin(x - 1) - 2\tan(\ln x)]\ , & \text{if } \frac 1 2 < x < 1\ , \\
b\ , & \text{if } x = 1\ , \\
\int^{x^2}_{4(x - 1)} e^{x + [\ln (t + 1)]^c}\ \mathrm {d}t\ , & \text{if } x > 1\ .
\end{cases}$$

Given that $f$ is differentiable, find the values of $a$, $b$ and $c$.

My working

For $f$ to be differentiable, $f$ must first be continuous.

$$\implies \lim\limits_{x\to1^-}f(x) = \lim\limits_{x\to1}f(x) = \lim\limits_{x\to1^+}f(x)$$

\begin{align}
\lim\limits_{x\to1^-}f(x) & =
\lim\limits_{x\to1^-} \frac {a[3\sin(x - 1) - 2\tan(\ln x)]} {x - 1}
\\[5 mm] & =
a\lim\limits_{x\to1^-} [3\cos(x - 1) - \frac 2 x \sec^2(\ln x)]
\\[5 mm] & =
a
\end{align}

\begin{align}
\lim\limits_{x\to1}f(x) & =
b
\end{align}

$$\implies a = b $$

\begin{align}
\lim\limits_{x\to1^+}f(x) & =
\lim\limits_{x\to1^+} \int^{x^2}_{4(x - 1)} e^{x + [\ln (t + 1)]^c}\ \mathrm {d}t
\\[5 mm] & =
\int^1_0 e^{1 + [\ln (t + 1)]^c}\ \mathrm {d}t
\\[5 mm] & =
e\int^1_0 e^{[\ln (t + 1)]^c}\ \mathrm {d}t
\end{align}

Then, I am stuck, as evaluating $\lim\limits_{x\to1^+}f(x)$ does not seem to be that straightforward. I also know that if $f$ is differentiable at some $d$, then $\lim\limits_{x\to d^-} \frac {f(x) - f(d)} {x - d} = \lim\limits_{x\to d^+} \frac {f(x) - f(d)} {x - d}$ and I would usually find these limits to obtain some more equations, but in this case, I do not seem to be able to arrive with more "usable" relations between $a$ and/or $b$ and/or $c$.

Any help/hints/intuition on how I should proceed will be greatly appreciated :)

答:
Whoa. After 5 days, I think I have managed to figure it out. Posting my answer here for anyone who is interested :)

For $f$ to be differentiable, $f$ must first be continuous.

$$\implies \lim\limits_{x\to1^-}f(x) = \lim\limits_{x\to1}f(x) = \lim\limits_{x\to1^+}f(x)$$

\begin{align}
\lim\limits_{x\to1^-}f(x) & =
\lim\limits_{x\to1^-} \frac {a[3\sin(x - 1) - 2\tan(\ln x)]} {x - 1}
\\[5 mm] & =
a\lim\limits_{x\to1^-} [3\cos(x - 1) - \frac 2 x \sec^2(\ln x)]
\\[5 mm] & =
a
\end{align}

\begin{align}
\lim\limits_{x\to1}f(x) & =
b
\end{align}

$$\implies a = b $$

\begin{align}
\lim\limits_{x\to1^+}f(x) & =
\lim\limits_{x\to1^+} \int^{x^2}_{4(x - 1)} e^{x + [\ln (t + 1)]^c}\ \mathrm {d}t
\\[5 mm] & =
\int^1_0 e^{1 + [\ln (t + 1)]^c}\ \mathrm {d}t
\\[5 mm] & =
e\int^1_0 e^{[\ln (t + 1)]^c}\ \mathrm {d}t
\end{align}

$$\implies \frac a e = \int^1_0 e^{[\ln (t + 1)]^c}\ \mathrm {d}t$$

$$\because f\ \mathrm {is\ differentiable}$$

$$\therefore f'(1)\ \mathrm {must\ exist}$$

$$\implies \lim\limits_{x\to1^-} \frac {f(x) - f(1)} {x - 1} = \lim\limits_{x\to1^+} \frac {f(x) - f(1)} {x - 1}$$

\begin{align}
\lim\limits_{x\to1^-} \frac {f(x) - f(1)} {x - 1} & =
\lim\limits_{x\to1^-} \frac {\frac {a} {x - 1} [3\sin(x - 1) - 2\tan(\ln x)] - a} {x - 1}
\\[5 mm] & =
a \lim\limits_{x\to1^-} \frac {3\sin(x - 1) - 2\tan(\ln x) - (x - 1)} {(x - 1)^2}
\\[5 mm] & =
a \lim\limits_{x\to1^-} \frac {3\cos(x - 1) - \frac {2\sec^2(\ln x)} {x} - 1} {2(x - 1)}
\\[5 mm] & =
\frac a 2 \lim\limits_{x\to1^-} \left[-3\sin(x - 1) - \frac {4\sec^2(\ln x)\tan(\ln x) - 2\sec^2(\ln x)} {x^2}\right]
\\[5 mm] & =
a
\end{align}

\begin{align}
\lim\limits_{x\to1^+} \frac {f(x) - f(1)} {x - 1} & =
\lim\limits_{x\to1^+} \frac {\int^{x^2}_{4(x - 1)} e^{x + [\ln (t + 1)]^c}\ \mathrm {d}t - a} {x - 1}
\\[5 mm] & =
\lim\limits_{x\to1^+} \frac {e^x \int^{x^2}_{4(x - 1)} e^{[\ln (t + 1)]^c}\ \mathrm {d}t - a} {x - 1}
\\[5 mm] & =
\lim\limits_{x\to1^+} \frac {e^x \int^{x^2}_{4(x - 1)} e^{[\ln (t + 1)]^c}\ \mathrm {d}t - a} {x - 1}
\\[5 mm] & =
\lim\limits_{x\to1^+} \left[e^x \int^{x^2}_{4(x - 1)} e^{[\ln (t + 1)]^c}\ \mathrm {d}t + e^x(2xe^{[\ln (x^2 + 1)]^c} - 4e^{[\ln (4x - 3)]^c})\right]
\\[5 mm] & =
e \int^1_0 e^{[\ln (t + 1)]^c}\ \mathrm {d}t + e[2e^{(\ln 2)^c} - 4]
\end{align}

$$\because \frac a e = \int^1_0 e^{[\ln (t + 1)]^c}\ \mathrm {d}t$$

$$\therefore \lim\limits_{x\to1^+} \frac {f(x) - f(1)} {x - 1} = a + e[2e^{(\ln 2)^c} - 4]$$

$$\because \lim\limits_{x\to1^-} \frac {f(x) - f(1)} {x - 1} = \lim\limits_{x\to1^+} \frac {f(x) - f(1)} {x - 1}$$

$$\therefore a = a + e[2e^{(\ln 2)^c} - 4]$$

$$\implies c = 1$$

$$\because c = 1$$

\begin{align}
\therefore \frac a e & = \int^1_0 (t + 1)\ \mathrm {d}t
\\[5 mm] & =
\frac 1 2 [(t + 1)^2]^{t = 1}_{t = 0}
\\[5 mm] & =
\frac 3 2
\end{align}

$$\implies a = \frac 3 2 e$$

$$\therefore a = \frac 3 2 e = b\ \mathrm {and}\ c = 1$$


答:
Expand and collect:

$$(ax+by+c)(dx+ey+f)=(ad)\ x^2 + (ae+bd)\ xy + (be)\ y^2 + (af+cd)\ x + (bf+ce)\ y + cf$$

Identifying coefficients leads to solving this system:

$\begin{cases}
ad=6\\
ae+bd=-1\\
be=-12\\
af+cd=-8\\
bf+ce=29\\
cf=-14\end{cases}$

You can try to work out integer solutions, since this is equal to $0$ there will be a proportionality coefficient, for instance try to start with $a=2,d=3$ or $c=2,f=-7$.

This is what I found:
$$(3x+4y-7)(2x-3y+2)=0$$

Title: How to factorize equation of pair of lines?

Body:
Find two equations of lines
: $$6x^2 -xy -12y^2 -8x + 29y -14= 0$$

I tried doing it using quadratic equation but got some very much unusual value. How to solve this?


问:
why derivative of $\sin^{-1}(x)$ is the derivative of the inverse $\sin(x)$?
Why can't we directly apply the chain rule here?

Why $\sin^{-1}x$ is not the same as $1/\sin(x)$?

Is it the general notation as inverse only for trig function?

答:
$$ \sin^{-1} = \arcsin x \neq  \frac{1}{\sin x}$$

However there exists a relation between the derivative of inverse and regular function,I wrote an answer about it here


Title: Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$?

Body: How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.

Body:
$5^{6n}-2^{2n}=125^{2n}-2^{2n}$

$=(125-2)\left(125^{2n-1}+125^{2n-2}2+125^{2n-3}2^2+\cdots+125^22^{2n-3}+125\cdot2^{2n-2}+2^{2n-1}\right)$

is divisible by $3$, because $125-2$ is.

Now substitute $m=n+1$ to show that $5^{6m+6}-2^{2m+2}$ is divisible by $3$.


问:
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$
I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$

But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$

I ran out of imagination as what to try now

答:
Note that
$$\int_1^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{y^2dy}{1+y^4}$$
Hence,
$$I=\int_0^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{1+y^2}{1+y^4}dy$$
We have $y^4+1 = (y^2+1+y\sqrt2)(y^2+1-y\sqrt2)$. Hence,
$$1+y^2= \dfrac{(y^2+1+y\sqrt2) + (y^2+1-y\sqrt2)}2$$
Hence, we get that
\begin{align}
I & = \dfrac12\int_0^1\dfrac{dy}{1+y^2-y\sqrt2} + \dfrac12\int_0^1\dfrac{dy}{1+y^2+y\sqrt2}\\
& = \dfrac12\int_0^1\dfrac{dy}{\left(y-\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2} + \dfrac12\int_0^1\dfrac{dy}{\left(y+\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2}\\
& = \dfrac1{\sqrt2} \left(\arctan \left(y\sqrt2-1\right) + \arctan \left(y\sqrt2+1\right) \right)_{y=0}^1\\
& = \dfrac1{\sqrt2} \left(\arctan(\sqrt2 - 1)+\arctan(\sqrt2+1)\right) = \dfrac1{\sqrt2} \left(\arctan\left(\dfrac1{1+\sqrt2} \right)+\arctan(1+\sqrt2)\right)\\
& = \dfrac1{\sqrt2} \times \dfrac{\pi}2 = \dfrac{\pi}{2 \sqrt2}
\end{align}</p>


问:
Irreducibility over $\mathbb{Z}_3$

Suppose $R= \frac{\mathbb{Z}_3[x]}{x^5+2x^4+x^2+2x+1}$, then which of the following is true?

The number of ideals in $R$ is 4.

$R$ has exactly two prime ideals.

R is an integral domain.

R[x] is a PID.

I have tried to check if the polynomial $x^5+2x^4+x^2+2x+1$ is reducible. For that, I have checked for roots and then I assumed that the polynomial is reducible and tried to compare the left-hand side and right-hand side of
$x^5+2x^4+x^2+2x+1=(mx^3+ax^2+bx+d)(nx^2+ex+f)$
It seems that the polynomial is irreducible. Is there any general way of checking irreducibility in Finite fields? I know there is an Algorithm which is a bit lengthy for this type of problem.

答:
Yes, there is a general way of checking irreducibility of polynomials over finite fields. It is called the Berlekamp algorithm and is based on linear algebra (see your link, which it the Berlekamp–Welch algorithm). For the above polynomial is gives

$$x^5 + 2x^4 + x^2 + 2x + 1=(x^3 + x^2 + 2)(x^2 + x + 2).$$
Your computation is the right way to solve the problem without this algorithm by hand. The only problem seems that, maybe, you did not consider that we have a linear system of equations over $\Bbb F_3$ and not over a field of characteristic zero. So for example $4=1$ for the absolute term.

Title: Irreducibility over $\mathbb{Z}_3$


问:
F-test on two population variance where the F ratio is larger than 1
I am learning F-test for testing if two population variances are the same. I have two samples with sizes $n_1$ and $n_2$.

The null hypothesis is $H_0: \sigma_1 = \sigma_2$.

I am doing a two-tailed test, and so the alternative is $H_1: \sigma_1 \neq \sigma_2$.

So the F ratio is $F=\frac{s_1^2}{s_2^2}$. On various notes, it says that we have to put the highest sample variance on the nominator. So $s_1^2 > s_2^2$. Then it's clear that in such case, F is always larger than $1$.

Why does $F$ have a F-distribution with freedom $n_1-1$ and $n_2-1$? The F-distribution will have values less than $1$, but actually $F$ can never be less than 1.

答:
The convention of reordering to make the ratio at least $1$ comes from printed statistical tables, where making the ratio greater then $1$ save half the space.  Something similar happened with tables of the standard normal distribution where $\Phi(x)$ was only given for positive $x$.

If $F(x,d_1,d_2)$ is the cumulative distribution function of an $F$ distribution with $d_1$ and $d_2$ degrees of freedom then $$F\left(\frac1x,d_2,d_1\right)=1- F(x,d_1,d_2)$$ and for example in R you have pf(11,8,3) giving 0.963 while pf(1/11,3,8) gives 0.037.  One implies the other, so you not need both in printed tables.

The reason $d_1=n_1-1$ and $d_2=n_2-1$ is the same reason as the reduction of the degrees of freedom with the Student $t$-test.  In a hand-waving sense, you took the sample means to be the population means in the calculation of the sample variances; they are not and so you used up one degree of freedom in each case in those calculations.


问:
How long does it take to change $90\%$ of water in an aquarium?
I am thinking about buying an aquarium and thought about a system that automatically changes water. This led me to think about the following problem:

Given a $500$$\ell$ tank and a system that automatically changes $1$$\ell$ of the water in the tank with fresh water every day, how many days does it take to replace $90\%$ of the water that's in the tank on day $0$? We assume the old water and the fresh water are perfectly mixed every time fresh water is added.

My idea is that on the first day we simply replace $1/500$ of the water. On the second day again  we replace $1/500$ of the water in the tank. However, $1/500$ has already been replaced on the first day and we have to subtract it. So on the second day we replace $\frac{1}{500}-\frac{\frac{1}{500}}{500}=\frac{1}{500}-\frac{1}{500^2}$. Thus, in total we have replaced $\frac{2}{500}-\frac{1}{500^2}$ on the second day. On the third we do the same thing, replace $1/500$ and subtract the water we have already replaced, obtaining $\frac{1}{500}-\frac{\frac{2}{500}-\frac{1}{500^2}}{500}=\frac{1}{500}-\frac{2-\frac{1}{500}}{500^2}$ of fresh water on the third day.

However I can't find the pattern, but I think this is a series, isn't it? So all we would have to do is find an explicit form of the series and find the finite series that equals $450/500$, right?

I think I was posed a similar question in my calculus course once and thought that there's an easy solution to the problem. Is my reasoning above correct? How can I find the solution to my problem?

答: On day $2$, $.998$ of the original water remains.  On day $3$, $.998^2$ of the original water remains.  On day $n$, $.998^{ n+1}$ of the original water remains.  We want the smallest integer $n$ such that $.998^{n+1}<.1$ which is $n=1150.$


问:
Define a function f such that for x≠0 f(x) =cos(1/x) and f(0)=1/2.does it admits antiderivatives? (primitives)
Does the function f have antiderivatives? I know that because 1/2 is between - 1 and 1 it has Darboux 's property</p>

答:
Hint: See if you can prove this:  if $f$ is bounded, and for every $a$ with $0 < a < 1$, $f$ is Riemann integrable on $[a,1]$, then $f$ is Riemann integrable on $[0,1]$. Do the same thing for $[-1,b]$, $-1<b<0$.

If there is any anti-derivative of $f$, then this would be it:
$$
F(x) = \int_0^x f(t)\;dt,\qquad\text{for all real numbers } x
$$
If $f$ is continuous except at $0$, this gives us $F$ with $F'(x) = f(x)$ for all $x$ except possibly $x=0$.

In the particular case on the OP, we get $F'(0) = 0$,
but I will not prove that here.
So your answer is "no", your function has no anti-derivative on $[-1,1]$.

The Darboux property is necessary, but not sufficient, for the existence of an anti-derivative.


问:
Fixed points on functions mapping $[a,b]$ onto itself
A function $f:[a,b] \to [a,b]$ is said to have a fixed point $c \in [a,b]$ if $f(c)=c$. Show that every continuous function $f$ mapping $[a,b]$ onto itself has at least one fixed point.

I came up with a basic proof of this. I don't think it's quite correct, but let me know what you think.

Suppose $f:[a,b] \to [a,b]$ is continuous on $[a,b]$

Since the domain and codomain of $f$ are the same, $f(x)=x$

Let $\varepsilon>0, c \in[a,b]$

Let $\delta= \varepsilon$

Suppose $|x-c|<\delta$

$\Rightarrow |x-c|<\varepsilon$

$\Rightarrow |f(x)-f(c)|<\varepsilon$

Thus, $\lim_{x \to c} f(x)=f(c)$ and therefore $f(x)$ is continuous for all $c \in [a,b]$

Then $f(c)=c$ for all $c \in [a,b]$

Thus, every continuous $f$ mapping $[a,b]$ onto itself has at least one fixed point

答:
You can not assume $f(x)=x$. You need to prove this.

Hints: Let $g(x)=f(x)-x$. Can you see that $g$ is continuous? Use Intermediate value theorem. Good luck!


问:
Deriving the Dirac delta function sifting property from a top hat function
I am trying to derive the Dirac delta function from a top hat function whose width tends to $0$ and has an area of $1$. Here is my working below and let me know if there is anything wrong with it:

答: We want to prove $\lim_{a\to0^+}\int_{q-a}^{q+a}\frac{1}{2a}f(x)dx=f(q)$ (provided $f$ is continuous at $q$; the last $=$ in this answer uses that). Before trying, let's note how plausible this is; it says that if you average $f$ over $[q-a,\,q+a]$, as $a\to0^+$ the mean becomes $f(q)$. By L'Hôpital,$$\lim_{a\to0^+}\frac{\int_{q-a}^{q+a}f(x)dx}{2a}=\lim_{a\to0^+}\frac{f(q+a)+f(q-a)}{2}=f(q).$$


问:
Evaluate a limit of an integral
Here is another Prelim Question.

Evaluate:
$$
I = \lim_{\epsilon\to 0} \frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y)\frac{y}{\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}dxdy
$$
where $f$ and $\nabla f$ are continuous and bounded.

So I have tried to change the integration order and used integration by parts and then changing to polar coordinates to obtain
$$
I = \frac{\partial f}{\partial y}(0,0) + \lim_{\epsilon \to 0}\frac{1}{2 \pi}\int_0^{2\pi}\int_0^\infty\frac{\partial}{\partial r}\left(\frac{\partial f}{\partial y}(r\cos(\theta),r\sin(\theta) )\right) \exp\left(\frac{-r^2}{2\epsilon}\right) dr d\theta
$$
I hope the final answer is the first term and we can show the second term goes to zero.  Can we just apply DCT?  I am getting second order partials in the second term and we only know information for the first order derivatives.  Note that
$$
\frac{\partial }{\partial r} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}
$$

答:
Integrate w.r.t. $y$ first and use integration by parts to get $$
I =  \frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty f_y(x,y)\exp{\frac{-(x^2+y^2)}{2\epsilon}}dydx
$$
Here you have used the fact that $$\lim_{y\to\pm\infty} f(x,y)\exp{\frac{-(x^2+y^2)}{2\epsilon}}=0$$ for each fixed $x$ and that $$\frac{\partial}{\partial y} \exp{\frac{-(x^2+y^2)}{2\epsilon}}=-\frac{y}{\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}$$
Now $$\frac{1}{2\pi\epsilon}\int_{-\infty}^\infty\int_{-\infty}^\infty \exp{\frac{-(x^2+y^2)}{2\epsilon}}dydx=\frac{1}{2\pi\epsilon}\cdot (2\pi)\int_0^\infty\exp(\frac{-r^2}{2\epsilon})r\,dr=1$$
for every $\epsilon>0$. Since the function $\frac{1}{2\pi\epsilon}\exp{\frac{-(x^2+y^2)}{2\epsilon}}$  is nearly support at $(0,0)$ for small $\epsilon$,  you have an approximation to the delta function. Hence your required limit is $f_y(0,0)$.</p>

Title: Evaluate a limit of an integral


问:
$L^\infty([0,1])$ is dual space of space of Radon measure
In the lecture here (at 16:14s), he claims that the (continuous) dual space of space of Radon measures on $\Omega=[0,1]$ is $L^\infty(\Omega)$. So, I am wondering

Where I can cite such results?

Do we have similar result for (locally) compact set $\Omega$?

答:
I think the instructor made a mistake there.  It is not even true that $L^\infty(\Omega)$ is contained in the dual of the space $X$ of Radon measures (at least not in a natural way): for instance, it is impossible  to define
$$
  \int_\Omega f\,d\mu,
  $$
for $f\in L^\infty(\Omega)$, in case, say, $\mu=\delta_x$ is the Dirac measure supported on $x$, since $f(x)$ is not well defined.

If you replace $L^\infty(\Omega)$ by the space $B(\Omega)$ of bounded measurable functions on $\Omega$ (not moding out when functions agree a.e.) then at least you may embedd
$$B(\Omega) \hookrightarrow X'$$
and I think this is what the instructor had in mind.