答:
Let's assume that $x \in I = [1, +\infty).$ It is easy to show that the image of the function $f(x) = 0.5\left(x + \frac{1}{x}\right)$ is $I$ too.

This means that for any $a_n \in I$, then also $a_{n+1} = f(a_{n}) \in I.$
In other words, since $a_1 = 2 \in I$, then $a_{n} \in I$ for all $n > 1.$

Notice that:

$$f'(x) = \frac{d f(x)}{dx} = \frac{x^2-1}{2x^2}.$$

Using this fact, it is also easy to show that $0 \leq f'(x) < 0.5$ for all $x \in I$. As a consequence $f(x)$ is Lipschitz continuous in $I$. This means that:

$$\frac{|f(x)-f(y)|}{|x-y|} \leq k ,$$

for all $x, y \in I$, for $k = \sup_{x\in I} |f'(x)| = 0.5$.

Since $k = 0.5 < 1$, then $a_{n+1} = f(a_{n})$ is a contraction mapping. For the Banach fixed-point theorem, we have that:

$$\lim_{n \to + \infty} a_{n} = a^* \in I,$$

provided that $a_1 \in I$. Here, $a^*$ is the unique fixed point of the map, at it can be evaluated as follows:

$$a_{n+1} = a_n = a^* \Rightarrow 0.5\left(a^* + \frac{1}{a^*}\right) = a^* \Rightarrow a^* = 1.$$

At this point, we know that for $a_1= 2 \in I$, then

$$\lim_{n \to + \infty} a_{n} = 1.$$

Now, observe that:

$$e_{n+1} = a_{n+1}-a_{n} = 0.5\left(\frac{1}{a_n} - a_n\right) = \frac{1-a_n^2}{2a_n^2} $$

is negative for $a_n \in I.$ Hence:

$$a_{n+1} = f(a_n) < a_n,$$

for all $n$. Since $a_{n+1} < a_{n}$ for $a_n \in I$, this prove that your sequence, which starts from $a_n =2 \in I$ is decreasing, and it will converge to $1$.


问:
Law of Total Probability with Conditionals
I would greatly appreciate some help on this problem, thank you! It is a past problem, so an answer is acceptable with steps.

Suppose we have a sample space $S$ and two events $A$ and $B$ such that

$$\begin{array}{lcl}S=A\cup B & \text{and} & A\cap B=\emptyset\end{array}$$

and an event $E$,

Suppose we have the following results:

$P\left(A\right)=1/3$

$P\left(B\right)=2/3$

$P\left(E\mid A\right)=1/2$

$P\left(E\mid B\right)=1/4$

What is $P\left(E\right)\thinspace$?

This is all I know:

$\begin{array}{rcl}
P\left(E\mid A\right) & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\
 & =\\
 & = & 1/2
\end{array}$

$\begin{array}{rcl}
P\left(E\mid B\right) & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\
 & = & 1/4
\end{array}$

EDIT:

I figured it out. I was completely overthinking it.

$\begin{array}{rcccl}
P\left(E\mid A\right) & = & 1/2 & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\
 & = &  & = & \frac{P\left(A\cap E\right)}{1/3}\\
 & = & P\left(A\cap E\right) & = & 1/6
\end{array}$

$\begin{array}{rcccl}
P\left(E\mid B\right) & = & 1/4 & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\
 & = &  & = & \frac{P\left(B\cap E\right)}{2/3}\\
 &  & P\left(B\cap E\right) & = & 1/6
\end{array}$

Using the law of total probability,

$P\left(E\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$

答:
Since $S$ is the sample space, then use what you have been told about $S,A,B$ to express this in terms of the provided probabilities.

$$\begin{align}\mathsf P(E)&=\mathsf P(S\cap E)\\&=\mathsf P((A\cup B)\cap E)\\&~~\vdots\end{align}$$

$$\begin{align}\mathsf P(E)&=\mathsf P(S\cap E)\\&=\mathsf P((A\cup B)\cap E)\\&=\mathsf P((A\cap E)\cup(B\cap E))\\&=\mathsf P(A\cap E)+\mathsf P(B\cap E)\\&=\mathsf P(A)\mathsf P(E\mid A)+\mathsf P(B)\mathsf P(E\mid B)\end{align}$$

ContentLicense: CC BY-SA 4.0


问:
How to calculate $\text{Var}(2X)$ where $X$ is a random variable?
Roll a fair die and let $X$ be the number of dots showing on top.
What are $\text{E}(2X)$ and $\text{Var}(2X)$?

答:
Notice that:

$$\text{Var}[X] = \mathbb{E}[(X-\mathbb{E}[X])^2].$$

Consider $Y = 2X.$ Then:

$$\mathbb{E}[Y] = \mathbb{E}[2X] = 2 \mathbb{E}[X].$$

Moreover:

$$\text{Var}[Y] = \mathbb{E}[(Y-\mathbb{E}[Y])^2] = \mathbb{E}[(2X-2\mathbb{E}[X])^2] = \mathbb{E}[4(X-\mathbb{E}[X])^2] = 4 \text{Var}[X].$$

In general:

$$\text{Var}[aX] = a^2 \text{Var}[X].$$

Addedum

Recall that $\mathbb{E}[aX] = a \mathbb{E}[X]$.


问:
Is it possible to define a bijection from nonnegative to positive numbers?
Let $\mathbb{R}_{\geq 0}$ be the set of nonnegative numbers and $\mathbb{R}_{>0}$ the set of positive numbers, that is

$$
\mathbb{R}_{\geq 0} = \{\,x \geq 0 \mid x \in \mathbb{R} \,\}
$$

and

$$
\mathbb{R}_{> 0} = \{\,x > 0 \mid x \in \mathbb{R} \,\}
$$

Is it possible to define a bijection $f$ between these two sets?

答: The have the same cardinality, so a bijection exists. You mentioned in a comment that you're also interested in a function that doesn't have any fixed points. You can adapt roddick's answer by just mixing up the intervals. For instance, you could send $[0,1)$ to $(10,11]$, $[1,2)$ to $(11,12]$, $[2,3)$ to $(0,1]$, and so on.


问:
Finding the subgroup of $(\mathbb Z_{56},+)$ which is isomorphic with $(\mathbb Z_{14},+)$ by GAP
I am sorting some easy questions for the students in Group Theory I. One of them is:

Is $(\mathbb Z_{14},+)$ isomorphic to a subgroup of $(\mathbb Z_{35},+)$? What about $(\mathbb Z_{56},+)$?

I know the first claim is false because if it is true then I have $14\nmid 35$ which is a contradiction. For the second one, I see the function $f:(\mathbb Z_{14},+)\to(\mathbb Z_{56},+),  f(g)=4g$ is a nice one-one homomorphism.

My problem is to find $f(\mathbb Z_{14})$ by using GAP. I know how to define Cyclic Groups or Free Groups in GAP's environment also. Thanks for the ANY help.

答:
GAP prefers multiplicative groups. However, you can make some limited use of additive groups if you are willing to inform GAP of a few things it doesn't calculate automatically.</p>

```
gap> g:=Integers mod 14;
(Integers mod 14)
gap> h:=Integers mod 56;
(Integers mod 56)
gap> f:=MappingByFunction( g, h, x -> 4*Int(x)*One(h) );
MappingByFunction( (Integers mod 14), (Integers mod 56), function( x ) ... end )
gap> SetRespectsZero( f, Image( f, Zero(Source(f))) = Zero(Range(f)) );
gap> SetRespectsAddition( f, ForAll( Tuples(Source(f),2),
     xy -> Image(f,xy[1]+xy[2]) = Image(f,xy[1])+Image(f,xy[2]) ));
gap> IsAdditiveGroupHomomorphism(f);
true
gap> Image(f);
[ ZmodnZObj( 0, 56 ), ZmodnZObj( 4, 56 ), ZmodnZObj( 8, 56 ),
  ZmodnZObj( 12, 56 ), ZmodnZObj( 16, 56 ), ZmodnZObj( 20, 56 ),
  ZmodnZObj( 24, 56 ), ZmodnZObj( 28, 56 ), ZmodnZObj( 32, 56 ),
  ZmodnZObj( 36, 56 ), ZmodnZObj( 40, 56 ), ZmodnZObj( 44, 56 ),
  ZmodnZObj( 48, 56 ), ZmodnZObj( 52, 56 ) ]
gap> Size(Kernel(f));
1
gap> IsInjective(f);
true
```

ContentLicense: CC BY-SA 3.0

Body:
GAP prefers multiplicative groups. However, you can make some limited use of additive groups if you are willing to inform GAP of a few things it doesn't calculate automatically.

gap> g:=Integers mod 14;
(Integers mod 14)
gap> h:=Integers mod 56;
(Integers mod 56)
gap> f:=MappingByFunction( g, h, x -> 4*Int(x)*One(h) );
MappingByFunction( (Integers mod 14), (Integers mod 56), function( x ) ... end )
gap> SetRespectsZero( f, Image( f, Zero(Source(f))) = Zero(Range(f)) );
gap> SetRespectsAddition( f, ForAll( Tuples(Source(f),2),
     xy -> Image(f,xy[1]+xy[2]) = Image(f,xy[1])+Image(f,xy[2]) ));
gap> IsAdditiveGroupHomomorphism(f);
true
gap> Image(f);
[ ZmodnZObj( 0, 56 ), ZmodnZObj( 4, 56 ), ZmodnZObj( 8, 56 ),
  ZmodnZObj( 12, 56 ), ZmodnZObj( 16, 56 ), ZmodnZObj( 20, 56 ),
  ZmodnZObj( 24, 56 ), ZmodnZObj( 28, 56 ), ZmodnZObj( 32, 56 ),
  ZmodnZObj( 36, 56 ), ZmodnZObj( 40, 56 ), ZmodnZObj( 44, 56 ),
  ZmodnZObj( 48, 56 ), ZmodnZObj( 52, 56 ) ]
gap> Size(Kernel(f));
1
gap> IsInjective(f);
true


答:
We know that $S = A \cup B$ and $A \cap B = \emptyset$. So $A, B$ are disjoint events whose union is the entire sample space.

Hence we can apply the law of total probability:

$$P(E) = P(A)P(E \mid A) + P(B)P(E \mid B)$$

and all the probabilities on the RHS are known.

Title: Law of Total Probability with Conditionals

Body:
I would greatly appreciate some help on this problem, thank you! It is a past problem, so an answer is acceptable with steps.

Suppose we have a sample space $S$ and two events $A$ and $B$ such that

$$\begin{array}{lcl}S=A\cup B & \text{and} & A\cap B=\emptyset\end{array}$$

and an event $E$,

Suppose we have the following results:

$P\left(A\right)=1/3$

$P\left(B\right)=2/3$

$P\left(E\mid A\right)=1/2$

$P\left(E\mid B\right)=1/4$

What is $P\left(E\right)\thinspace$?

This is all I know:

$\begin{array}{rcl}
P\left(E\mid A\right) & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\
 & =\\
 & = & 1/2
\end{array}$

$\begin{array}{rcl}
P\left(E\mid B\right) & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\
 & = & 1/4
\end{array}$

EDIT:

I figured it out. I was completely overthinking it.

$\begin{array}{rcccl}
P\left(E\mid A\right) & = & 1/2 & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\
 & = &  & = & \frac{P\left(A\cap E\right)}{1/3}\\
 & = & P\left(A\cap E\right) & = & 1/6
\end{array}$

$\begin{array}{rcccl}
P\left(E\mid B\right) & = & 1/4 & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\
 & = &  & = & \frac{P\left(B\cap E\right)}{2/3}\\
 &  & P\left(B\cap E\right) & = & 1/6
\end{array}$

Using the law of total probability,

$P\left(E\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$


问:
Proof: not a perfect square
Let $y$ be an integer.
Prove that

$$(2y-1)^2 -4$$

is not a perfect square.

I Found this question in a discrete math book and tried solving it by dividing the question into two parts:

$$y = 2k , y = 2k + 1$$

But that got me nowhere.

答:
$(2y-1)^2-4=4(y^2-y)-3$  If it were a perfect square it would be $=c^2$, where c is an integer.  Solve for $y$ in $4(y^2-y)-3-c^2=0$ and get $y=\frac{4\pm \sqrt{16+16(3+c^2)}}{8}=\frac{1\pm \sqrt{4+c^2}}{2}$.

However $c^2+4$ cannot be a square, unless $c=0$ (where $y$ is not an integer). Assume $c^2+4=b^2$ so $b=c+a$ with $(c+a)^2=c^2+2ac+a^2$.    $2ac+a^2=4$ has no possible integer solutions.   ($a=1$ LHS is odd, $a\gt 1$ LHS $\gt 4$).

Therefore no possible integer $y$.

Title: Proof: not a perfect square


问:
What is the difference between isometric deformation and conformal defomation?
What is the difference between an isometric deformation and a conformal deformation? In fact, even the definition of both deformations is still unclear clear for me.

Is it possible to define an isometric deformation from $\mathrm{R}^2$ to $\mathrm{R}^3$ ?

So I would like to ask if anyone has a reference in mind that can help me understand each concept. Any help will be appreciated, but please note that I am not a geometry student, but a mathematics student working in applied mathematics. Honestly, I am working in computer vision, and precisely in  the 3D reconstruction domain. There is an idea saying that one can reconstruct a 3D object after being isometricaly defomed, just using its initial 3D form and a 2D image of the deformed object. So, in order to comprehensively understand the problem  it is important for me to understand what "deformation" (isometric) refers to.

I already searched for some references, but isometry is usually considered  from an algebraic point of view, as isometric transfomation.

答:
In the context of 3D shapes:

An isometric deformation is a deformation of the shape the preserves geodesic distances on the shape. https://en.wikipedia.org/wiki/Isometry

A conformal deformation is a deformation of the shape that preserves angles on the surface of the shape. https://en.wikipedia.org/wiki/Conformal_map

These definitions can be used on smooth manifolds or discrete manifold approximations like triangle meshes.


问:
Eigenvalues of $\sum_{i=1}^n \frac{(x_i - x_{i-1})^2}{\lambda_i}$
Consider the cuadratic form
$$
\mathbf{x}^{\intercal}Q\mathbf{x} = \frac{x_1^2}{\lambda_1} + \sum_{i=2}^n \frac{(x_i - x_{i-1})^2}{\lambda_i}\ .
$$
Is it true that the eigenvalues of $Q$ are $\lambda_i^{-1}$ ?</p>

答:
The answer to your question is no. $Q$ can be written in the form
$$\begin{bmatrix}1 & -1 \\ & 1 & -1 \\ & & 1 & \ddots \\  & & & \ddots & -1 \\ & & & & 1 \end{bmatrix}
\begin{bmatrix}\lambda_1^{-1} \\ & \lambda_2^{-1} \\ & & \lambda_3^{-1} \\ & & & \ddots \\ & & & & \lambda_n^{-1} \end{bmatrix}
\begin{bmatrix}1 \\ -1 & 1 \\ & -1 & 1 \\  & & \ddots & \ddots \\ & & & -1 & 1 \end{bmatrix}
$$
Call that right-hand matrix $Z$. The eigenvalues would be $\lambda_i^{-1}$ if $Z^TZ=ZZ^T=I$. But $Z^TZ\neq I$. The eigenvalues are not $\lambda_i^{-1}$, and it's not hard to see this by trying a couple of numerical examples.


答: odd squares are $1 \pmod 4,$ but it is more specific than that. Odd squares are $1 \pmod 8.$  You can check this by squaring, say, $1,3,5,7$ and find the remainder when divided by $8$. In particular, squares are never $5 \pmod 8.$ Your $(2y-1)^2 - 4 \equiv 5 \pmod 8$ and cannot be a square

Title: Proof: not a perfect square

Body:
Let $y$ be an integer.
Prove that

$$(2y-1)^2 -4$$

is not a perfect square.

I Found this question in a discrete math book and tried solving it by dividing the question into two parts:

$$y = 2k , y = 2k + 1$$

But that got me nowhere.


答:
Suppose:

$$(2 y - 1)^2 - 4 = a^2$$

for some $a$.

Then

$$(2 y - 1 + 2)(2 y - 1 - 2) = (2 y + 1)(2 y - 3) = a^2$$

Can you take it from here?

Think about the prime factorization of each side.

Title: Proof: not a perfect square


问:
Let $T:V \rightarrow V$ such that $T^2 = \frac{1}{2}T$. Find its characteristic polynomial.
I am used to dealing with transforms such as $T:P_2(\mathbb{R}) \rightarrow P_2(\mathbb{R})$ where $T(ax^2 + bx + c) = (a+b)x^2+cx$. In this case you would just use a basis of $P_2(\mathbb{R})$ and create a matrix representation of the transform. But in this case you cannot do that since you don't really have a lot of information about the transformation.

Can this even be done? If so how?

答: If $T^2=\frac{1}{2}T$, then $T^2-\frac{1}{2}T = 0$, so polynomial $x^2 - \frac{1}{2}{x}$ divides the characteristic polynomial.  But this is a $2 \times 2$ matrix, so the characteristic polynomial has degree $2$, and thus $x^2 - \frac{1}{2}x$ is the characteristic polynomial.


问:
Find tangent plane parallel to $x+y+z=1$

Given surface defined by $2x^2+3y^2+z^2=\frac{13}{6}$ parallel to the plane $x+y+z=1$. Find the tangent plane corresponding to it. The intersection of the plane and the surface happens in the first quadrant.

I found the normal of the surface as <4x,6y,2z> and equated it to the normal of the plane $x+y+z=1$ and got value of x in terms of a constant k and plugged it in the equation of the surface to get k and thus values x,y and z.

But here I am getting k as irrational and the calculation is becoming complex and lengthy.

Can you please let me know if my approach to the above formulation is correct or suggest a better way for the same?

答:
Matching the two normals
$$4x:6y:2z=1:1:1$$
or, $y=\frac23x$ and $z=2x$. Then, substitute them into
$2x^2+3y^2+z^2=\frac{13}{6}$ to get
$$2x^2 +3(\frac23 x)^2+(2x)^2=\frac{13}6$$
which leads to the tangent point $(
\frac12\sqrt{\frac{13}{11}},\frac13\sqrt{\frac{13}{11}}, \sqrt{\frac{13}{11}})$. Then, the tangent plane is
$$(1,1,1)\cdot\left(x-\frac12\sqrt{\frac{13}{11}},
y-\frac13\sqrt{\frac{13}{11}}, z-\sqrt{\frac{13}{11}}\right)=0
$$
which is
$$x+y+z= \frac{\sqrt{143}}6$$


问:
Avoiding Extended Kalman Filter linearization with a mixture of linear and non-linear observation model
I have a doubt about a possible alternative to EKF that i have applied successfully but that seems not to be theoretically correct.
Suppose to have a system
\begin{equation}
x_{k+1} = Ax_k  + w_k\\
y_k = h(x_k)+Cx_k + v_k
\end{equation}
where $w,v$ are i.i.d guassian vectors and the measurement equation can be divided into a linear term plus a nonlinear function $h(x)$.

The standard EKF formulation needs to compute the correction term based on the measurement equation jacobian H, which results in this case
\begin{equation}
 H = \frac{\partial h(\hat{x}_{k+1|k})}{\partial x}+C
\end{equation}
The corrected estimate is thus obtained by
\begin{equation}
\hat{x}_{k+1|k+1} = \hat{x}_{k+1|k} + PH^T(HPH^T+R)^{-1}(y_{k+1}-h(\hat{x}_{k+1|k})-C\hat{x}_{k+1|k})
\end{equation}

What i tried was to create a fictitious measurement defined as
\begin{equation}
\tilde{y}_{k+1} =   y_{k+1}-h(\hat{x}_{k+1|k})
\end{equation}
In this way, the new measurement equation becomes
\begin{equation}
\tilde{y}_{k+1} = Cx_{k+1}
\end{equation}
and it is possible to apply the Linear Kalman Filter using the modified measurement in the correction term obtaining
\begin{equation}
\hat{x}_{k+1|k+1} = \hat{x}_{k+1|k} + PC^T(CPC^T+R)^{-1}(\tilde{y}_{k+1}-C\hat{x}_{k+1|k})
\end{equation}
without resorting to any linearization.

I would like to know if it can be a reasonable  approximation in order to avoid linearization, maybe in systems such that $h(x)$ is highly non-linear.

Thank you everyone

答:
Unfortunately this won't work, and might in some cases actually become unstable. In order to show this by using a first order Taylor approximation of $h(\hat{x}_{k+1|k})$

$$
h(\hat{x}_{k+1|k}) = h(x_{k+1}) + \left.\frac{\partial\,h(x)}{\partial x}\right|_{x=x_{k+1}} (\hat{x}_{k+1|k} - x_{k+1}) + O(\|x_{k+1} - \hat{x}_{k+1|k}\|^2). \tag{1}
$$

Using $(1)$ in the measurement error yields

\begin{align}
y_{k+1} - \hat{y}_{k+1} &= h(x_{k+1}) + C\,x_{k+1} - h(\hat{x}_{k+1|k}) - C\,\hat{x}_{k+1|k},  \tag{2a} \\
&\approx \left(C + \left.\frac{\partial\,h(x)}{\partial x}\right|_{x=x_{k+1}}\right) (x_{k+1} - \hat{x}_{k+1|k}). \tag{2b}
\end{align}

So the partial derivative is still indirectly there, but now you are using and incorrect $H$ to update the covariance, from which you calculate the correction gain.

Another way of seeing that your approach can't work reliable is the extreme case of redefine the problem with a new nonlinear function $h'(x_k) = h(x_k) + C\,x_k$ and $C' = 0$, which would make the Kalman gain always zero.

An example of where this can become unstable would be

$$
A =
\begin{bmatrix}
\sqrt{3}/2 & 1/2 \\ -1/2 & \sqrt{3}/2
\end{bmatrix}, \quad
C =
\begin{bmatrix}
1 & 0
\end{bmatrix}, \quad
h(x_k) = -2\tanh(\begin{bmatrix}
0 & 1
\end{bmatrix} x_k),
$$

where the Kalman gain is obtained from the stationary solution using $R=1$ and $Q = I$. Simulation this even without noise and an initial state estimate close to the true state yields the following plots, where the blue lines are the true states and the red lines are the estimated states.


问:
Perimeter or Calculus Word problem

A rectangular plot of farmland will be bounded on one side by a river
  and on the other three sides by a single strand of electric fence.
  With 1400m of wire at your disposal, what is the largest area you can
  enclose, and what are its dimensions?

Is there a way to solve this without Calculus? This sounds like a simple perimeter problem. If you have to use Calculus to solve this, how would I start? Could someone show me the steps?

答:
Let $r$ be the length of the rectangle along the river and $s$ be the length of the other side. Thus, we have to:

Max $rs$

Subject to:

$r + 2s = 1400$

Substitute for $r$ back into the objective function and we get the problem reformulated as:

Max $(1400-2s) s$

You can find the value of $s$ either via calculus or by completing the square.

Title: Perimeter or Calculus Word problem


问:
Munkres' Section 67 Exercise 6 ( Rank of the subgroups of free abelian group with finite rank)

Hi. I am studying this topic and I cannot see the statement "If this subgroups is nontrivial, choose $x_m\in B_m$ so that $\pi(x_m)$ is a generator of this subgroup.

I want proves that $\langle \pi(x_m)\rangle=\pi_m(B_m)$.
Let $y\in \pi_m(B_m)$ then $y=\pi_m(x)$ some $x\in B_m$ but this $x$ depens of $y$...

Why exists that $x_m$?

答: $\Bbb Z$ is cyclic.  So all its subgroups are cyclic.


问:
Showing that $\mathbf I-\mathbf L(\mathbf L'\mathbf L)^{-1}\mathbf L'\mathbf I\mathbf L(\mathbf L'\mathbf L)^{-1'}\mathbf L'$ is an idempotent matrix
I am trying to show that $\mathbf I-\mathbf L(\mathbf L'\mathbf L)^{-1}\mathbf L'\mathbf I\mathbf L(\mathbf L'\mathbf L)^{-1'}\mathbf L'$ is an idempotent matrix where $\mathbf L$ is a $p\times r$ matrix, $\mathbf I$ is a $p\times p$ identity matrix, and $\mathbf L'\mathbf L$ is assumed to be invertible.

My try:

Let $\mathbf m=\mathbf L(\mathbf L'\mathbf L)^{-1}\mathbf L'$ so that

$$\mathbf Q=\mathbf I-\mathbf L(\mathbf L'\mathbf L)^{-1}\mathbf L'\mathbf I\mathbf L(\mathbf L'\mathbf L)^{-1'}\mathbf L'=\mathbf I-\mathbf{mIm}'$$

We have that

\begin{align*}
\mathbf{QQ}
&=(\mathbf I-\mathbf{mm}')(\mathbf I-\mathbf{mm}')\\\\
&=\mathbf I-\mathbf{mm}'-\mathbf{mm}'+\mathbf{m}\mathbf{m}'\mathbf{m}\mathbf{m}'\\\\
&=\mathbf Q-\mathbf{mm}'+\mathbf{m}\mathbf{m}'\mathbf{m}\mathbf{m}'\\\\
&=\mathbf Q-\mathbf{m}'\mathbf{m}+\mathbf{L}\underbrace{(\mathbf{L}'\mathbf{L})^{-1}\mathbf{L}'\mathbf{L}}_{=\mathbf{I}}(\mathbf{L}'\mathbf{L})^{-1'}\mathbf{L}'\mathbf{L}\underbrace{(\mathbf{L}'\mathbf{L})^{-1}\mathbf{L}'\mathbf{L}}_{=\mathbf I}(\mathbf{L}'\mathbf{L})^{-1'}\mathbf{L}'\\\\
&=\mathbf Q-\mathbf{m}'\mathbf{m}+\mathbf{L}(\mathbf{L}'\mathbf{L})^{-1'}\mathbf{L}'\mathbf{L}(\mathbf{L}'\mathbf{L})^{-1'}\mathbf{L}'\\\\
&=\mathbf Q-\mathbf{m}'\mathbf{m}+\mathbf{m}'\mathbf{m}'\\\\
\end{align*}

I'm not sure how I can proceed in showing that $\mathbf Q^2=\mathbf Q$ or where I went wrong. Any help would be appreciated!

答: Your first step should be $(I-mm')(I-mm')$; for some reason you wrote $m'm$ in the second term. If you propagate this correction through the rest of your work, you will get $Q-m'm + m'm = Q$ at the last step.


答:
$(2y-1)^2 - 4 = (2y-1)^2 - 2^2 = (2y-1+2)(2y-1-2) = (2y+1)(2y-3)$

Note that $2y+1$ and $2y-3$ are always distinct integers. Hence proving their product cannot be a square is accomplished by showing they're coprime (no prime factors in common) and that they're not both squares at the same time.

$\mathrm{gcd}(2y+1, 2y-3) =\mathrm{gcd}(2y+1, (2y+1)-(2y-3)) = \mathrm{gcd}(2y+1, 4) = 1$ (the last part is trivially observing that one is odd, the other even). Hence $2y+1$ and $2y-3$ are coprime.

Now note that both $2y+1$ and $2y-3$ are odd with a difference of $4$. The minimum difference between two odd squares is $3^2 - 1^2 = 8$. So they cannot both be squares.

Therefore $(2y+1)(2y-3) = (2y-1)^2 - 4$ cannot be a square.

Title: Proof: not a perfect square


问:
Prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$ with $a_1=2$ is decreasing
I need to prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$, $a_1=2$ is decreasing.

I tried to use induction: I assumed that $a_n<a_n-1$, but couldn't prove that $a_{n+1}<a_n$.

How can I prove it?

答:
Update: Inspired by Daniel Schepler, I found another way to solve directly:

$$\frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n+\frac{1}{a_n}+2}{a_n+\frac{1}{a_n}-2}=\left(\frac{a_n+1}{a_n-1}\right)^2 \\\implies \frac{a_n+1}{a_n-1}=\left(\frac{a_1+1}{a_1-1}\right)^{2^{n-1}}=3^{2^{n-1}} \\ \implies
a_n =\frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1} = 1+\frac{2}{3^{2^{n-1}}-1}$$
and it's straightforward to prove the sequence is decreasing.

Original answer
$$2a_{n+1} a_n = a_n^2+1, 2a_n a_{n-1} = a_{n-1}^2+1.\\
\implies a_n^2-a_{n-1}^2=2a_n(a_{n+1}-a_{n-1})=2a_n(a_{n+1}-a_n)+2a_n(a_n-a_{n-1})\\
\implies 2a_n (a_{n+1}-a_n)=(a_n+a_{n-1})(a_n-a_{n-1}) - 2a_n(a_n-a_{n-1})= -(a_n-a_{n-1})^2.
$$

Note that $a_n>0, \forall n$. Now you can complete your induction.


问:
Why is it so that we consider multiplying a vector by a negative scalar as reversing the direction of it.
I know that negative numbers reside in the opposite direction of positive numbers in the number line. I want a solid argument regarding what might the intuition be behind reversing the direction of a vector when multiplied by a negative scalar in any dimension (1, 2 or 3)</p>

答:
I think at first at least it's best to motivate the algebra by thinking geometrically. From a geometric perspective, we start by thinking of directed line segments in a reasonably simple space like $2$- or $3$-dimensional Euclidean space. Vector addition then is just "placing end-to-end:" to add two vectors, we translate one so that its tail rests at the head of the other, and then connect the tail of the first to the head of the second.

This gives us an obvious notion of the "additive inverse" of a vector $v$: it's just the same vector but with head and tail switched (the "backwards" version of the original vector). Call this "$v_{back}$." The point, then, is the following:

In order to satisfy the basic arithmetic properties we want, it needs to be the case that $$(-1)v=v_{back}.$$

This amounts exactly to the idea that multiplying by a negative reverses direction.

So how do we argue this? Well, we first observe that $$v+(-1)v=(1)v+(-1)v=(1-1)v=0v=0.$$ But now since we know geometrically that $v+v_{back}=0$ as well, we have $$v+(-1)v=v+v_{back},$$ and so by subtracting $v$ from both sides of the above (or if you prefer, adding - say - $v_{back}$ to both sides) we get $$(-1)v=v_{back}$$ as desired. All the above reasoning relies on only very basic assumptions about how vectors and scalars interact; essentially, we have that "multiplying by a negative reverses direction" is the only possible way to wind up with the basic algebraic properties we want.


问:
Prime number generator, how to make
Can anybody point me an algorithm to generate prime numbers, I know of a few ones (Mersenne, Euclides, etc.) but they fail to generate much primes...

The objective is:
    given a first prime, generate the 'n' next primes. But thanks for the link ;-)

```<code>for example : primes( 17, 50 ) -> Generate 50 consecutive primes starting at 17
</code>```

and do not fail any prime in this 50... no holes!

答:
Here you go - give any random number as an input, and it will give you the next prime after it. You can feed its output into itself to make a list. It basically uses Fermat's Little Theorem to heuristically check if each number is a prime, and keeps checking successive odd numbers until you get to something that works.

It has a very small chance of failure (eg. nextprime(560) = 561, but 561=3*11*17), but if you go high enough this becomes negligible in practice.

```<code>def modexp(b,e,n):
  if e == 0: return 1
  elif e%2 == 0: return modexp(b,e/2,n)**2 % n
  else: return b*modexp(b,e/2,n)**2 % n

def nextprime(inp):
  inp += inp%2 + 1
  while modexp(2,inp,inp) != 2 or modexp(3,inp,inp) != 3: inp += 2
  return inp
</code>```

Title: Prime number generator, how to make

Body:
Here you go - give any random number as an input, and it will give you the next prime after it. You can feed its output into itself to make a list. It basically uses Fermat's Little Theorem to heuristically check if each number is a prime, and keeps checking successive odd numbers until you get to something that works.

It has a very small chance of failure (eg. nextprime(560) = 561, but 561=3*11*17), but if you go high enough this becomes negligible in practice.

def modexp(b,e,n):
  if e == 0: return 1
  elif e%2 == 0: return modexp(b,e/2,n)**2 % n
  else: return b*modexp(b,e/2,n)**2 % n

def nextprime(inp):
  inp += inp%2 + 1
  while modexp(2,inp,inp) != 2 or modexp(3,inp,inp) != 3: inp += 2
  return inp


答: For the sake of contradiction write $(2y-1)^2-4=n^2$ where $n$ is an integer. Equivalently $$4=(2y-1-n)(2y-1+n).$$ Difference between the two factors is $2n$, i.e. even. Only ways to factor $4$ with factors that differ by even number are $(-2)\cdot(-2)$ and $2 \cdot 2$, both cases are impossible as they imply $n=0$ and $(2y-1)^2=4$.

Title: Proof: not a perfect square


问:
If $f_{n}\to f$ in $L^{2}$ and $f_{n}$ are absolutely continuous does $f_{n}\to f$ a.e?
Suppose a sequence of absolutely continuous functions $f_{n}\to f$ in $L^{2}$ on $[0,1]$. Then does $f_{n}\to f$ almost everywhere?

I know the statement is not true for arbitrary sequences $f_{n}$ that converge in $L^{2}$ because one can choose a sequence of moving characteristic (indicator) functions where the width of the interval goes to $0$, but the sequence converges almost nowhere.

However, does imposing the absolutely continuous (or even continuous) restriction remove all such possibilities? How would one use the properties of absolutely continuous functions to prove this statement?

答:
The claim is false even if the $f_n$ are infinitely differentiable on $(0,1)$. Here is a sequence $(f_n)$ for which $f_n\to0$ in $L^2$ but $f_n$ converges nowhere to $0$.

Let $(g_n)$ be the typewriter sequence. We will construct $f_n$ to dominate $g_n$. If the $n$th member $g_n$ is a rectangle of height $1$ over the interval $[a_n,b_n]$, define $f_n$ to be
$$
f_n(x):=\exp-\frac14\left[\left(\frac{x-\mu_n}{\sigma_n}\right)^2-1\right]
$$
with $\mu_n:=\frac{a_n+b_n}2$ the midpoint of $[a_n,b_n]$ and $\sigma_n:=\frac{b_n-a_n}2$ the half-width. By comparison with the density for a Gaussian distribution with mean $\mu_n$ and variance $\sigma_n^2$, we have $$\int_0^1 f_n^2\le \int_{-\infty}^\infty f_n^2=\sigma_n\sqrt{2\pi e},$$
and, since $\sigma_n\to0$, we conclude $f_n\to0$ in $L^2$.

On the other hand,
observe that $f_n(x)\ge1$ whenever $|x-\mu_n|\le\sigma_n$, hence $f_n\ge g_n$ for all $n$. Since $g_n(x)$ does not converge to zero for any $x$, conclude the same is true for $f_n$.


答: Another proof: WLOG assume $y>0$. Look at the differences between square of two consecutive numbers: $1, 3, 5, 7$, etc.. Therefore the only way to get a difference of 4 is 2^0-0=1+3, which is impossible because $2y-1$ is odd.

Title: Proof: not a perfect square


问:
Prove that there isn't a polynomial with $\text {f(x)}^{13} = {(x-1)}^{143}+(x+1)^{2002}$
Prove that there isn't a polynomial with  $\text {f(x)}^{13} = {(x-1)}^{143}+(x+1)^{2002}$

We can easily find out that  $\text {deg}(f) = 154$

Then?

答: Hint: Consider the coefficients of $x^0,x^1$ on both sides of the equation. This suffices.


答:
The difference between any two squares $a^2$ and $b^2$ with $a^2< b^2$ is at least 5 if $|b|$ is at least 3.

So all you are left with is to check directly for $(2y-1)^2 =0,1,4$. And as $2y-1$ is odd, infact only $2y-1=1$.

Title: Proof: not a perfect square


答:
Note: This is a half tongue-in-cheek answer, in that it's probably not something we would expect students to come up with on their own; but hopefully it's still a useful viewpoint on the problem.

Suppose that $a_n = \coth\theta$.  Then
$$a_{n+1} = \frac{1}{2} \left( \frac{\cosh\theta}{\sinh\theta} + \frac{\sinh\theta}{\cosh\theta}\right) = \frac{\cosh^2\theta + \sinh^2\theta}{2\sinh\theta \cosh\theta} = \frac{\cosh(2\theta)}{\sinh(2\theta)} = \coth(2\theta).$$
Therefore, by induction it should be easy to show that $a_n = \coth(2^{n-1} \theta_1)$ where $\theta_1 := \operatorname{arccoth}(2)$.

Now, since $\coth$ is a decreasing function on $[0, \infty)$, and the sequence $2^{n-1} \theta_1$ is increasing, we see that $a_n$ is decreasing.

(With a bit of algebraic manipulation, the solution above could also be written as: $a_n = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1}$.)

Title: Prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$ with $a_1=2$ is decreasing

Body:
I need to prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$, $a_1=2$ is decreasing.

I tried to use induction: I assumed that $a_n<a_n-1$, but couldn't prove that $a_{n+1}<a_n$.

How can I prove it?


问:
Question about derivative and limits
$f(x): [0,1]\to[0,1]$ - continuous function.

Let $f'(x_0) = \lim_{x\to x_0} \frac{f(x_0) - f(x)}{ x_0 - x }$
exists.

Consider the sequence of intervals $G_n = [ a(n), b(n) ]$,
where $ \forall n\in\mathbb{N}$ we have  $a(n)<b(n)<x_0$
and
$a(n) \to x_0$, $b(n) \to x_0$ at $n\to\infty$.

Is it true that
$\lim_{n\to\infty} \frac{f(b) - f(a)}{ b - a } = f'(x_0)$ ?

答:
As it turns out, the answer is no. Define
$$
f(x) = \begin{cases}
x^2\sin(\frac1x), &\text{if } x\ne0, \\
0, & \text{if } x=0.
\end{cases}
$$
It can be shown that
$$
f'(x) = \begin{cases}
2x\sin(\frac1x)-\cos(\tfrac1x), &\text{if } x\ne0, \\
0, & \text{if } x=0
\end{cases}
$$
($x=0$ being the only nontrivial point). In particular, at every point of the form $x(n)=-\frac{1}{2n\pi}$ we have $f'(x(n)) = -1$. We then obtain a counterexample (with $x_0=0$) by choosing $a(n)=x(n)$ and $b(n)$ so close to $x(n)$ that $\frac{f(b(n))-f(x(n))}{b(n)-x(n)}$ is very close to $-1$, which is possible since $\lim_{b(n)\to x(n)} \frac{f(b(n))-f(x(n))}{b(n)-x(n)} = f'(x(n))$.

This counterexample essentially comes from the fact that $f'(x)$ exists at all points but is not continuous. It's less clear whether the OP's question can be answered in the affirmative if one assumes that $f'(x)$ is continuous.

Title: Question about derivative and limits


答:
For $y\le-1$, $(2y-1)^2-4$ is between consecutive squares $(2y)^2$ and $(2y-1)^2$.

For $y\in\{0,1\}$, $(2y-1)^2-4$ is negative, so not a   square.

For $y\ge2$, $(2y-1)^2-4$ is between consecutive squares $(2y-2)^2$ and $(2y-1)^2$.

Title: Proof: not a perfect square


问:
Calculating upper eccentricity in a graph
I was going through a paper. There calculating upper eccentricity was mentioned.
Can anybody help me in finding out how it was done? I tried hard but was unable to get it.
A little hint or explanation of only one vertex is welcome. That would help me a lot. Thanks a lot.

Centers of $n$-Fold Tensor Products of Graphs, page $495$, Fig. $1$

答:
The figure in the paper is the following:

The vertices are marked by an ordered pair.  In the first coordinate is the usual eccentricity of the vertex, namely $\mathrm{ecc}(v)=\max_u \mathrm{dist}(u,v)$, where $\mathrm{dist}(u,v)$ is the number of steps in the shortest walk from $u$ to $v$.

The second coordinate contains the upper eccentricity, which is defined similarly, except we ignore walks that have the same parity as $\mathrm{ecc}(v)$ (meaning walks of length congruent to $\mathrm{ecc}(v) \pmod 2$).

So, for example, the upper eccentricity of $b$ above is $4$ since we cannot get to $a$ in $0$ or $2$ steps, but can reach $a$ in $4$ steps (we're not allowed to use an odd number of steps since $\mathrm{ecc}(b) \equiv 1 \pmod 2$).  We can also reach either $c$ or $d$ in $2$ steps.

The upper eccentricity of $a$ is $5$, since we cannot get from $a$ to $a$ in $1$ or $3$ steps, but can reach $a$ in $5$ steps.


问:
Show that $f \vartriangle g$ is not a function.
This is an exercise from Velleman's "How To Prove It".

Suppose that $f$ and $g$ are functions from $A$ to $B$ and $f \neq g$. Show that $f \vartriangle g$ is not a function.

I was trying to reach a contradiction by supposing that $f \vartriangle g$ is a function. Since $f \neq g$, I broke the proof into cases based on $(a,b) \in f$ but $(a,b) \notin g$ or $(a,b) \in g$ but $(a,b) \notin f$. I'm failing to find a contradiction though. Any hints?

答:
Hint The condition $f \neq g$ is precisely that there is an $a \in A$ such that $f(a) \neq g(a)$.

Additional hint For such an element $a \in A$, what can you say about the elements of the form $(a, \,\cdot\,)$ in $f \bigtriangleup g$?


答:
I'm giving a concrete example, but this should illustrate the general case.
Let $A = \{0,1\} = B$, $f = \{(0,1),(1,0)\}$ and $g = \{(0,0), (1,0)\}.$ Then
$A\vartriangle B = \{(0,1),(0,0)\}$, but this is not a function, since it is not well-defined.

UPDATE: Travis Willse gave an excellent hint as well. Between his answer and mine, you should have a clear approach.

Body:
This is an exercise from Velleman's "How To Prove It".

Suppose that $f$ and $g$ are functions from $A$ to $B$ and $f \neq g$. Show that $f \vartriangle g$ is not a function.

I was trying to reach a contradiction by supposing that $f \vartriangle g$ is a function. Since $f \neq g$, I broke the proof into cases based on $(a,b) \in f$ but $(a,b) \notin g$ or $(a,b) \in g$ but $(a,b) \notin f$. I'm failing to find a contradiction though. Any hints?


问:
Norm convergence of partial sum operator on $L^p$, $1 < p < \infty$.
I am working through Mark Pinsky's "Introduction to Fourier Analysis and Wavelets" textbook and have been working through the following problems.  Note that here $S_R$ is the partial sum operator defined by
$$
(\widehat{S_R f}) = \chi_{B_R} \widehat{f}.
$$

For any $f \in \mathcal{S}(\mathbb{R})$, the Schwartz space, we have the bound $ \|S_Rf\|_p \leq C_p \|f\|_p $, where $C_p$ is independent of $R$.

If $2 < p < \infty$ and $f \in \mathcal{S}(\mathbb{R})$, then $\|S_R f - f\|_p \rightarrow 0$ as $R \rightarrow \infty$.

If $f \in L^p(\mathbb{R}), 2 < p < \infty$, then $\|S_R f - f\|_p \rightarrow 0$ as $R \rightarrow \infty$.

Use the duality of $L^p$ and $L^{p^\prime}$ to prove that for any $f \in L^p(\mathbb{R}), 1 < p < 2$, then $\|S_R f - f\|_p \rightarrow 0$ as $R \rightarrow \infty$.

As you can see these problems build upon one another.  I have figured out the first 3 problems, but find myself struggling to solve 4. using duality.  I'm not exactly sure what the author means by duality, but to me I take it that for $f \in L^p, 1 < p < 2$, then $\widehat{f} \in L^{p^\prime}$, where $p^\prime$ is the conjugate exponent of $p$.  This aided me in solving 2. using the Hausdorff-Young inequality, however the inequality doesn't seem to help me when considering $f \in L^p$ for $1 < p < 2$.  Duality also brings up the following definition:
$$
\|S_R f\|_p = \sup \left\{ \left| \int S_R f \cdot g \right|\, : \, \|g\|_{p^\prime} \leq 1 \right\},
$$
however I seem unable to "swap" the $S_R$ operator between $f$ and $g$ the way you can with the Hilbert transform.  Thank you for the guidance!

答:
By the definition you noted, it's enough to show that
$|\langle S_Rf-f,g\rangle|\rightarrow 0$ for all $g\in L^{p'}(\mathbb{R})$ where $\langle S_Rf-f,g\rangle$ is the pairing $\int_\mathbb{R} (S_Rf(x)-f(x))g(x)\,dx$. We know that $||S_Rg -g||_{L^{p'}(\mathbb{R})}\rightarrow 0$ so the idea is to try to rewrite the above pairing in terms of $S_Rg$. Thus, you should try to rewrite $\int_\mathbb{R}S_Rf(x)g(x)\,dx$ as $\int_\mathbb{R}f(x)T_Rg(x)\,dx$ where $T_R$ is some operator very closely related to $S_R$. In fact, $T_R$ is what is called the adjoint of $S_R$ meaning that for all $f\in L^{p}(\mathbb{R}),g\in L^{p'}(\mathbb{R})$, $\langle  S_Rf,g\rangle=\langle f,T_Rg\rangle$. You should see that $\langle f,T_Rg\rangle\rightarrow \langle f,g\rangle $ because of $T_R$'s relationship to $S_R$ and that fact that $S_Rg\rightarrow g$ in $L^{p'}(\mathbb{R})$.

More abstractly, what is meant by "using duality" is proving something about an operator or sequence of operators (usually that it is bounded or that the sequence converges to some limit) by knowing something about the adjoint(s). In the setting of $L^p$ spaces, if $T^*$ is similar to $T$, or even the same, and we know something about $T^*$ on $L^p$ for $2<p<\infty$, we can learn something about $T$ for $1<p'<2$, $1/p'+1/p=1$ since these are the dual spaces.


问:
Good books to Learn General topology
I am quite interested in learning topology from the basics, I would like to find a book that is quite detailed and has a large number of examples.

What are the best books to easily understand topology?

Thanks.

答:
My recommendations: Elementary Topology: Problem Textbook and Topology without Tears. (Both are available online.)

Stephen Willard's General Topology is also a nice choice.

Title: Good books to Learn General topology


答:
This, I think, is a nice problem, and deserves a hint rather than an answer.

See what happens with some values of $x$ which are easy to calculate. What does this tell you about the form of $f(x)$?

Body:
Prove that there isn't a polynomial with  $\text {f(x)}^{13} = {(x-1)}^{143}+(x+1)^{2002}$

We can easily find out that  $\text {deg}(f) = 154$

Then?


问:
In what sense does Martin's axiom say that cardinals between $\aleph_0$ and $2^{\aleph_0}$ are "similar" to $\aleph_0$?
Continuum Hypothesis says that every infinite cardinal smaller than $2^{\aleph_0}$ is equal to $\aleph_0$. In the other hand I've heard that Martin Axiom says that these cardinals are "similar" in some sense (and that's a quote from Polish Wikipedia). What exactly does it mean? I couldn't find anything about it.

答:
Roughly speaking, Martin's axiom says that many statements that can be proven about countable sets by simple diagonalization arguments also hold for sets of all cardinalities below $2^{\aleph_0}$.  Here is a typical example.

Two subsets of $\omega$ are called almost disjoint if their intersection is finite.  Consider the following simple theorem.

Theorem: Let $S$ be a countable set of subsets of $\omega$ which are pairwise almost disjoint and such that the union of any finitely many elements of $S$ is coinfinite.  Then there exists an infinite subset $A\subseteq\omega$ that is almost disjoint from every element of $S$.

Proof: Enumerate the elements of $S$ as $(A_n)_{n\in\omega}$.  Define a sequence $(n_k)$ of distinct elements of $\omega$ as follows.  Having defined $n_j$ for all $j<k$, let $n_k$ be some element of $\omega$ that is different than every such $n_j$ and is not in $A_j$ for all $j<k$.  Such an element exists since $\bigcup_{j<k}A_j$ is coinfinite by hypothesis.

Now let $A=\{n_k\}$; since the $n_k$ are all distinct, $A$ is infinite.  But for any $j$, $n_k\not\in A_j$ for all $k>j$, so $A\cap A_j$ is finite.  Thus $A$ has the desired properties.

Now you might ask, what if $S$ is not countable?  Well, clearly the theorem cannot remain true without some additional constraint on $S$, since you could take $S$ to be a maximal infinitely family of pairwise almost disjoint subsets of $\omega$ (which exists by Zorn's lemma, say).  In particular, the theorem cannot be true for all $S$ of cardinality $2^{\aleph_0}$.  But what if $S$ is uncountable, but $|S|<2^{\aleph_0}$?  The proof of the theorem certainly wouldn't apply to $S$; trying to continue the construction of $(n_k)$ transfinitely runs into trouble already at $k=\omega$, since the $n_j$ you have chosen already could exhaust the entire complement of $\bigcup_{j<k} A_j$.

However, Martin's axiom implies that the theorem does apply to such $S$.  In this sense, Martin's axiom says that sets of size less than $2^{\aleph_0}$ behave "as though they were countable" for these purposes.

Of course, this raises the question of exactly which "purposes" Martin's axiom applies to.  The exact statement is quite technical (and probably won't mean much to you if you aren't familiar with forcing).  It says that if $P$ is a poset satisfying the countable chain condition (given any uncountable subset $A\subseteq P$, there exist distinct $a,b\in A$ and $c\in P$ such that $c\leq a$ and $c\leq b$) and $S$ is a collection of fewer than $2^{\aleph_0}$ dense subsets of $P$ (a subset $D$ of $P$ is called dense if for each $a\in P$, there exists $b\in D$ such that $b\leq a$), then there exists a filter on $P$ which intersects every element of $S$.  In the case that $S$ is countable, it is easy to construct such a filter by a diagonalization argument (pick a descending sequence of elements of each set in $S$ one by one, and take the filter they generate).  So, this says that in a wide variety of situations (anything that can be encoded by finding a filter in a ccc poset meeting dense sets), sets of fewer than $2^{\aleph_0}$ conditions can be simultaneously satisfied in the same way that countable sets of conditions can be simultaneously satisfied by building an object to meet the conditions one by one.

Here's another example that may be more familiar.  By the Baire category theorem, any intersection of countably many open dense subsets of $\mathbb{R}$ is dense.  A consequence of Martin's axiom is that this is also true for intersections of fewer than $2^{\aleph_0}$ open dense subsets of $\mathbb{R}$.


问:
Show that the sequence $x_n = \sqrt{n^2 + 1} - \sqrt{n + 1}$ is increasing
already tried to do $\frac{x_{n+1}}{x_n}<1$ or $x_{n+1}-x_n>0$ but I didn't get anything. any tips?

I want to use this to show the divergence. Only this step is missing from my demonstration

答:
Hint: let
$$ f(x)=\sqrt{x^2+1}-\sqrt{x+1} $$
and then show $f'(x)>0$. It is not necessary to use this to show $\{x_n\}$ diverges. In fact $\sqrt{n+1}\le\frac12n$ for $n\ge3$ and hence
$$ x_n=\sqrt{n^2+1}-\sqrt{n+1}\ge n-\frac12n =\frac12n$$
and hence $\{x_n\}$ diverges.


问:
understanding construction and definition of classifying space BG
Let $G$ be a discrete group. $EG$ is defined as the $\triangle$-complex (Hatcher p.102) whose $n$-simplices are given by  $[g_0,g_1,...,g_n]$ glued together in the obvious way. Then define $BG=EG/G$. I am trying to understand this definition by a simple example. If I choose $G= \mathbb{Z}/2\mathbb{Z}$, then I think $EG$ should be a path with endpoints $0$ and $1$, and then $BG$ will identify the endpoints and give us $S^1$. This does not seem right since $S^1 = K(\mathbb{Z},1)$. What am I doing wrong here?

答:
In your construction, $E(\mathbb{Z}/2\mathbb{Z})$ is $S^\infty$ with the antipodal $\mathbb{Z}/2\mathbb{Z}$-action.

You can think of $n$-simplices here as binary strings of length $n+1$.  Nondegenerate simplices are given by binary strings that alternate strictly between $0$ and $1$.

There are two $0$-simplices $[0]$ and $[1]$.  So far this is $S^0$.

There are two (nondegenerate) $1$-simplices $[01]$ and $[10]$ which are attached to their boundaries: $[0]$ at one end, $[1]$ at the other.  This is $S^1$.

There are two (nondegenerate) $2$-simplices $[010]$ and $[101]$, which are attached to $[01]$ and $[10]$ (and other degenerate simplices).  This is $S^2$.

This pattern continues, and you may recognize the whole thing as the usual presentation of $S^\infty$ as a $\mathbb{Z}/2\mathbb{Z}$-CW complex.

The quotient of $S^\infty$ by the antipodal action is $S^\infty / (\mathbb{Z}/2\mathbb{Z}) \simeq \mathbb{R}P^\infty$, which is indeed $K(\mathbb{Z}/2\mathbb{Z}, 1)$.


问:
prove the function $w=\sqrt{z}$ is continuous for all $z \in \mathbb{C}$
I search a lot for the answear and many topics here about this questions says the function is continuous on $[0,\infty)$. But the book doesn't says anything about this interval. It requests to prove for all set of complex. Well, what I know is for any point $z$ and $z_0 \neq 0$ we have

$|f(z) - f(z_0)| = |\sqrt{z}- \sqrt{z_0}|= | \frac{z-z_0}{\sqrt{z}+\sqrt{z_0}} | < \epsilon$.

Now what should I do to prove this using $\sqrt{z} = r^{\frac{1}{2}}e^\frac{i\theta}{2}$ and $\sqrt{z_0} = r_0^{\frac{1}{2}}e^\frac{i\theta_0}{2}$ ?

The book says to consider the fact $|\sqrt{z}+\sqrt{z_0}|^2= r + r_0 + 2\sqrt{rr_0}cos(\frac{\theta + \theta_0}{2}) > r_0 $

答: It isn't continuous on all of $\Bbb C$.  You need a branch of the square root function, and the largest subset you can define it on is $\Bbb C$ minus a slit, like $(-\infty,0)$.


答:
An alternative approach: note that
$$
\begin{align}
x_n &=(\sqrt{n^2 + 1} - \sqrt{n+1}) \cdot \frac{\sqrt{n^2 + 1} + \sqrt{n+1}}{\sqrt{n^2 + 1} + \sqrt{n+1}}
\\ & = \frac{n^2 + 1 - (n + 1)}{\sqrt{n^2 + 1} + \sqrt{n+1}}
\\ & = \frac{n^2 - n}{\sqrt{n^2 + 1} + \sqrt{n+1}} \cdot \frac{1/n}{1/n}
\\ & = \frac{n - 1}{\sqrt{1 + \frac 1{n^2}} + \sqrt{\frac 1n + \frac 1{n^2}}}.
\end{align}
$$
From there, we can easily see that $x_n \to \infty$ as $n \to \infty$.

Title: Show that the sequence $x_n = \sqrt{n^2 + 1} - \sqrt{n + 1}$ is increasing

Body:
already tried to do $\frac{x_{n+1}}{x_n}<1$ or $x_{n+1}-x_n>0$ but I didn't get anything. any tips?

I want to use this to show the divergence. Only this step is missing from my demonstration


问:
Prove that the number of solutions to $x^a = 1 \pmod p$ in $\mathbb Z_p^\times$ is $\gcd(a,p-1)$
Let $p$ be a prime number and $a\geq 1$ an integer.  Show that the number of solutions to $x^a = 1 \pmod p$ in $\mathbb  Z_p^\times$ is $\gcd(a,p-1)$.</p>

答:
Applying Discrete Logarithm wrt an arbitrary primitive root $g\pmod p,$

$$a\cdot ind_gx\equiv0\pmod {p-1}$$

Now using Linear Congruence Theorem, as $(a,p-1)$ divides $0,$ the congruence will have exactly $(a,p-1)$ solutions.

This can be generalized to any number $n$ having at least one primitive root where $p-1$ will be replaced with $\phi(n)$


问:
Perpendicularity of real and imaginary part of a complex function
For a couple days now, I have been thinking about the graph of the real and imaginary parts of complex functions, for example: $$f(z)=z^2=(x+iy)^2=x^2-y^2+i(2xy)$$

I found that if I set, for any constants c$$x^2-y^2=c_1, 2xy=c_2$$
Then if I zoom in more and more on the intersections of these two curves, the angle at which they intersect appears to always approach a right angle. I have tried this also for sines, cosines, exponentials, and other elementary functions and found the same result, but I am not sure what is actually going on here. Why is this happening? There must be something more fundamental about the nature of these functions that I am missing.

I have done some research into conformal maps that show this phenomenon for many, many values of c, but I don't believe I have quite developed the mathematical maturity for complex analysis quite yet. I am coming at this from the perspective of a motivated high school senior with knowledge of multivariable calc and some linear algebra/differential equations.

If anyone could provide me some insight or direction into, more generally, why this happening and what I am missing, I would greatly appreciate it!

答:
This happens specifically with differentiable complex functions. I have a kind of intuitive explanation of this phenomenon.

First, as I hope you know, every complex number can be written in polar form, which looks like this
$$r(\cos(\theta) + i\sin(\theta))$$
for some real $r \ge 0$ and $\theta \in \Bbb{R}$. Multiplying complex numbers in polar form is nice and intuitive. It's easy to verify that:
\begin{align*}
&r_1(\cos(\theta_1) + i\sin(\theta_1)) \times r_2(\cos(\theta_2) + i\sin(\theta_2)) \\
= \, &r_1 r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)).
\end{align*}
What this means is, if we are multiplying a complex number $z$ in general by a specific complex number $r(\cos(\theta) + i\sin(\theta))$, then all we have to do is:

Scale $z$ by a factor of $r$, and

Rotate $z$ (counter-clockwise) by $\theta$ radians.

Note that both of these operations preserve lines and don't change angles. If we start with two lines in the complex plane intersecting at an angle $\alpha$, then multiplying every complex number in these lines by $r(\cos(\theta) + i\sin(\theta))$ will result in two (rotated) lines, still intersecting at an angle of $\alpha$.

Now, let's say we have a complex function $f$ which is differentiable at some complex number $z_0$. Essentially what this means is that the function $f$ behaves like its linearisation:
$$L(z) = f'(z_0)z + f(z_0)$$
for $z$ close to $z_0$. All that $L(z)$ does is multiply $z$ by a fixed complex number $f'(z_0)$ (which just scales and rotates), then adds a fixed complex number $f(z_0)$, which just shifts the picture without changing angles. So, $L$ will also not affect angles!

And since $f(z) \approx L(z)$ for $z$ around $z_0$, this means that, if two smooth curves meet at $z_0$ and their tangent lines meet at an angle $\alpha$, then whatever curves $f$ maps them to will have their tangents meet at angle $\alpha$.

So, what about your particular question? You have a complex function $f(a + ib) = u(a, b) + iv(a, b)$, which we will want to be differentiable. You are looking at the curves $u(a, b) = c_1$ and $v(a, b) = c_2$. The curve $u(a, b) = c_1$ is the set of points which will map to the vertical line $\operatorname{Re} z = c_1$. Similarly, $v(a, b) = c_2$ is the curve that will map to the horizontal line $\operatorname{Im} z = c_2$.

So, if the two curves intersect at $z_0$, and their tangents form an angle $\alpha$, then because $f$ is differentiable at $z_0$ (as it is everywhere else), the image of these curves must also form an angle $\alpha$. But, as we discussed, the image of these curves are horizontal and vertical lines, which are perpendicular! So, we must have $\alpha = \pi/2$. That is, if we look very closely at intersections between the curves, they must occur at right-angles.


问:
Prove that this is a sufficient condition for function continuity
Prove that this is a condition for continuity of $f(x)$:

For any sequence $x_n$ $\rightarrow$ $x$ and for any $w$, if $f(x_n) \geq w$ $\forall n$ then $f(x) \geq w$ and if $f(x_n) \leq w$ $\forall n$ then $f(x) \leq w$.

I was trying to prove by contradiction but didn't really know how to proceed. I thought that if there exists an $\epsilon$ such that $\forall n$ $|f(x_n) - f(x)| > \epsilon$, there exists no $N$ such that $\forall n > N, |x_n - x| > \epsilon$. I couldn't show this, though.

答:
Use the negation of continuity. If $f$ is discontinuous at $x$ then there is a specific $\epsilon>0$ such that every neighborhood of $x$ contains points $t$ for which $|f(t) - f(x) |\geq \epsilon$.

Thus we have an infinity of points $t$ near $x$ for which $f(t) \geq f(x) +\epsilon $ or $f(t) \leq f(x) - \epsilon$. If there are infinitely many points to satisfy first inequality then we have a sequence $t_n\to x$ with $t_n\neq x$ such that $f(t_n) \geq f(x) +\epsilon$. But by given hypothesis this means that $f(x) \geq f(x) +\epsilon $ an obvious contradiction. A similar contradiction is achieved if there are infinitely many points near $x$ to satisfy the inequality $f(t) \leq f(x) - \epsilon $.

It follows that $f$ must be continuous at $x$.


问:
Convert a recursive formula to explicit
Given sequence:

$A_{1} = 0$ $A_{2} = 3$ $A_{3} = \frac{3}7$
$A_{4} = \frac{21}{13}$  $A_{5} =  \frac{39}{55}$

And the recursive formula is given as $_{n+1}=\frac{3}{2_{n} + 1}$

How can one find this sequence's explicit formula?

答:
It's a Möbius transformation.

check my answer at Recursive sequence depending on the parameter for a shortcut method. A few others have also posted such approach in the past.

Title: Convert a recursive formula to explicit


问:
Inequality proof by induction
I'm supposed to prove that for any integer $n\geq 2$, if $x_{1},\ldots,x_{n}$ are real numbers in $]0,1[$, then

$(1-x_{1})\cdot\ldots\cdot(1-x_{n}) > 1 - x_{1} - \ldots - x_{n}$

I am trying the induction method so I first tried to find if it's true for n=2 :
$(1-x_{1})(1-x_{2}) > 1- x_{1} - x_{2}$

$1-x_{1}-x_{2}+x_{1}x_{2} > 1-x_{1}-x_{2}$ and this is true because $x_{1}x_{2}$ is positive

for n+1 : $(1-x_{1})(1-x_{2})...(1-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

I am stuck here.. what should I do next?

Edit : I then did

$(1-x_{1}-...-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

How do I compare $(1-x_{n+1})$ and $-x_{n+1}$ to find out that the left side is bigger?

Title: Inequality proof by induction

Body:
Suppose it is true for some $n$ as you've shown.

Then $(1-x_{1})(1-x_{2})\cdots(1-x_{n})(1-x_{n+1}) > (1-x_{1}-\cdots-x_{n})(1-x_{n+1})$

$=(1-x_{1}-\cdots-x_{n})-x_{n+1}+x_{1}x_{n+1}+\cdots+x_{n}x_{n+1}>1-x_{1}-\cdots-x_{n}-x_{n+1}$

as $x_{i}x_{n+1} > 0$.


答:
To simplify this step, let $y = x_1+x_2+\dots +x_n$. Then

$$(1-x_1-\dots-x_n)(1-x_{n+1})=(1-y)(1-x_{n+1}) = 1-y-x_{n+1} + yx_{n+1} > 1-y- x_{n+1}$$

Finally substitute in $y$ and finish the proof.

Title: Inequality proof by induction

Body:
I'm supposed to prove that for any integer $n\geq 2$, if $x_{1},\ldots,x_{n}$ are real numbers in $]0,1[$, then

$(1-x_{1})\cdot\ldots\cdot(1-x_{n}) > 1 - x_{1} - \ldots - x_{n}$

I am trying the induction method so I first tried to find if it's true for n=2 :
$(1-x_{1})(1-x_{2}) > 1- x_{1} - x_{2}$

$1-x_{1}-x_{2}+x_{1}x_{2} > 1-x_{1}-x_{2}$ and this is true because $x_{1}x_{2}$ is positive

for n+1 : $(1-x_{1})(1-x_{2})...(1-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

I am stuck here.. what should I do next?

Edit : I then did

$(1-x_{1}-...-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

How do I compare $(1-x_{n+1})$ and $-x_{n+1}$ to find out that the left side is bigger?


Title: Inequality proof by induction

Body:
Assume $(1-x_1)(1-x_2)...(1-x_n) > 1 - x_1 -x_2 ... -x_n$ as the indutive hypothesis.

Then certainly $x_{n+1}x_1 + ... + x_{n+1}x_n > 0$ since each number is positive.

Thus $1- x_1 -... - x_n -x_{n+1} + x_{n+1}x_1 + ... +x_{n+1}x_n > 1 - x_1 ... - x_n - x_{n+1}$.

The LHS can be rephrased as follows:

$1- x_1 -... - x_n -x_{n+1} + x_{n+1}x_1 + ... +x_{n+1}x_n = (1- x_1...-x_n)(1 - x_{n+1})$, so that $(1...-x_n)(1-x_{n+1}) > 1 -x_1 -...-x_n - x_{n+1}$.

Now by the inductive assumption that $(1-x_1)(1-x_2)...(1-x_n) > 1 - x_1 -x_2 ... -x_n$, multiply both sides by $(1-x_{n+1})$ which is positive, to get

$(1-x_1)(1-x_2)...(1-x_n)(1-x_{n+1}) > (1 - x_1 -x_2 ... -x_n)(1-x_{n+1}) > 1 - x_1 ... - x_n - x_{n+1}$, and thus $(1-x_1)(1-x_2)...(1-x_n)(1-x_{n+1})  > 1 - x_1 ... - x_n - x_{n+1}$ as desired.


问:
Why is the double quotient $B\backslash G /B$ the same as $G\backslash (G/B \times G/B)$?
Why is the double quotient  $B\backslash G /B$ the same as $G\backslash (G/B \times G/B)$?

The particulars don't matter, as I'm really just trying to understand this on a set level. That being said, $G$ is a Lie group and $B$ is a Borel subgroup.

For a little context, i'm currently learning about the Bruhat decomposition. See Emerton's post here: https://mathoverflow.net/questions/15438/a-slick-proof-of-the-bruhat-decomposition-for-gl-nk/15554#15554

答:
$G$ acts on the set $S=G/B \times G/B$ through $$g.(aB,cB)=(gaB,gcB)$$

Let $$T=B/B\times G/B \subset S$$ We have a bijection $$G\setminus S \to B\setminus T\to B\backslash G /B$$ given by $$G.(aB,cB)=G.(B,a^{-1}cB)\to B.(B,a^{-1}cB)\to Ba^{-1}cB$$


问:
Reorganising alternating sum of products of binomial coefficients
The summation
$$
\sum_{k=0}^{\lfloor\frac{p}{s}\rfloor}(-1)^k {n\choose k}{p-ks+n-1\choose n-1}\quad;n > 0, s > 0 \text{ and } 0\le p\le ns,
$$
with ${n\choose k}$ denoting the binomial coefficient, has the limitation that implemented fixed sized integers, there are cases where individual terms exceed the integer size even though the result does not. (Even assuming ideal calculation of binomial coefficients.) Is there a way to reorganise that summation to avoid that? The integer size bound can be assumed to be $s^n$, the sum of that result for all $p$. A stronger result would that all terms summed do not exceed their sum in absolute value. A yet stronger result would be that all terms summed are non-negative. But even the weakest case is fine.

答: I cannot answer your question exactly, but assuming you are trying to compute this sum while avoiding integer overflow, here is a trick which might work. Let $a_k$ be the absolute value of the $k^{th}$ summand in your sum. You can prove that the numbers $a_k$ increase up to a point, and then decrease afterwards. This implies that your alternating sum is bounded above by the largest $a_k$. Letting $m=\max_k a_k$, you can compute $m$ in advance, and then you instead compute the sum with all additions performed modulo $m$. This ensures you never have any partial computations larger than $m$, while still giving the correct answer.


问:
Expected maximum deviation of lattice path
Suppose there is an election with two candidates, Alice and Bob. Each candidate receives $n$ votes. As the votes are counted in a random order, what is the expected value of the maximum difference between Alice and Bob's partial totals? Note that this is process is equivalent to choosing a random lattice path from $(0,0)$ to $(n,n)$, equally likely from all $\binom{2n}n$ possibilities, and looking at the maximum distance of this path from the diagonal $y=x$ (up to a factor of $\sqrt{2}$).

I heard the problem from this blog post, where it is phrased in terms of sock matching. I ask this question because I expect it has a nice approximate answer; namely, the author of that post crunched the numbers and found that the expected value is well approximated by $\sqrt{\frac{3}2n}$, as shown in the data below. I wonder if anyone knows if this problem appears in the literature, and if they can prove or refute this claim?

Here is what I know:

Let $M$ be the maximum difference in votes. In this other answer, I showed that for any $m\ge 1$, we have
$$
P(M<m) = \frac1{\binom{2n}n}\sum_{k=-\lfloor n/m\rfloor}^{\lfloor n/m\rfloor}(-1)^{k}\binom{2n}{n+km}\tag1
$$
You can then use this to compute $E[M]$, via $E[M]=\sum_{m=1}^n \big(1-P(M<m)\big)$. Furthermore, the expression in $(1)$ direcctly implies
$$
P(M<m)=\frac1{\binom{2n}n}\cdot\frac2m\cdot 2^{2n}\sum_{k=1}^{\lfloor m/2\rfloor} \bigg[\cos\left(\frac{(2k-1)\pi}{2m}\right)\bigg]^{2n}
$$
by using a roots-of-unity filter on the generating function $(x^{1/2}+x^{-1/2})^{2n}$. As $n\to \infty$, the main contribution is from the $k=1$ term, so $P(M<m)$ grows like $\frac1{\binom{2n}n}\cdot\frac2m\bigg[2\cos\left(\frac{\pi}{2m}\right)\bigg]^{2n}$.

A somewhat related problem is finding the expected height of a random Dyck path of order $n$, which is just a random lattice path conditioned on staying above the diagonal $y=x$. In $[1]$, de Bruijn, Knuth, and Rice show that the expected height of a Dyck path is asymptotically $\sqrt{\pi n}-\frac12$. Perhaps something similar to their methods works here.

Bruijn, de, N. G., Knuth, D. E., & Rice, S. O. (1972). The average height of planted plane trees. In R. C. Read
(Ed.), Graph Theory and Computing (pp. 15-22). Academic Press Inc. [PDF]

答:
Not an answer, but perhaps a step in the right direction.

Wikipedia has a nice approximate formula for $C(n,k)$ for when $n$ is large and $k$ is close to $n/2$. Using $n\to 2n$ and $k\to n-ku$ we get
$$C(2n,n-ku)\approx\frac{2^n}{\sqrt{n\pi/2}}\exp\left(\frac{-(n-2(n-ku))^2}{2n}\right)$$
$$=\frac{2^n}{\sqrt{n\pi/2}}\exp\left(\frac{-(n-ku)^2}{2n}\right)$$
$$=\frac{2^n}{\sqrt{n\pi/2}}e^{-n/2}\cdot e^{ku/n}\exp\left(\frac{-k^2u^2}{2n}\right)$$
So then going to the sum mentioned in the blog post we have
$$\mathrm{E}[U]\approx\frac{2^{n+1}e^{-n/2}}{C(2n,n)\sqrt{n\pi/2}}\sum_{u=1}^n\sum_{k=1}^{\lfloor n/u\rfloor}(-1)^{k-1}e^{ku/n}\exp\left(\frac{-k^2u^2}{2n}\right)$$

Perhaps this will submit more readily to analysis, but I'm a bit out of my depth here.


问:
Clarification on the notion of isometry between surfaces.
I seek to clarify some confusion I have with regards to the notion of an isometry with the use of an example. This is related to a question I had asked previously. Consider a helicoid with an axis of length $l$ and pitch $p$ parametrised as $$x(s,v) = (v\cos{s}, v\sin{s}, ps) ,$$
where $ s\in [0,l]$ and $v \in [-a,a]$. In general I have the relation $ \frac{2\pi}{p}= \frac{l}{n}$, where $n$ is the number of turns of the helicoid around the axis.

The first fundamental form coefficients of this helicoid will be $$E = 1, F =0 , G = p^2+v^2.$$

Now I take another helicoid with same parametrization as before but this time, I take $ s \in [0,  p]$ and $v \in [-a,a]$. Essentially the pitch is same as before, so the first fundamental coefficients dont change but the number of turns is reduced to 1. This seems to be a local isometry to me but the length of the axis is not preserved, so this violates the isometry condition that lengths of curves on the surfaces are preserved.

Alternatively one can say I am attempting to isometrically unscrew a helicoid of n turns to a helicoid with just one turn. But although it seems to be an isometry due to the matching first fundamental coefficients, it doesnt seem to preserve distances. The attached figure is meant to sort of convey what I am saying. Can anyone please clarify what is happening here??

答:
The parametrization of the second is $$\big(v\cos(s/n), v\sin(s/n),ps\big).$$
Now compute the first fundamental form. Certainly no local isometry.


问:
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$
I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks</p>

答:
Welcome to Math Stack Exchange!

With lab bhattacharjee's helpful hint for the integral, we can rewrite the expression as:

$$\int \left(1 + \frac{(-2x - 5)}{2(x^2 + 1)} + \frac{7}{4(x - 1)} - \frac{3}{4(x + 1)}\right)dx$$

$$= \int 1 dx + \int \frac{(-2x - 5)}{2(x^2 + 1)} dx + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$

$$= \int 1 dx - \int \frac{2x}{2(x^2 + 1)} dx - \int \frac{5}{2(x^2 + 1)} + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$

$$= \int 1 dx - \int \frac{x}{x^2 + 1} dx - \frac{5}{2} \int \frac{1}{x^2 + 1} dx + \frac{7}{4} \int \frac{1}{x - 1} dx - \frac{3}{4} \int \frac{1}{x + 1} dx$$

Next steps should be "direct-cut", meaning that it's really possible to work out the computation one way! :D

For the first integrand, use power rule to obtain $x$.

For the second integrand, let $u = x^2 + 1 \rightarrow du = 2x dx$.

For the third integrand, you get $\arctan(x)$.  You can find the special integral here.

For the fourth integrand, let $v = x - 1 \rightarrow dv = dx$.

For the fifth integrand, let $w = x + 1 \rightarrow dw = dx$.

So we are left off with:

$$x - \frac{5}{2}arctan(x) - \frac{1}{2} \int \frac{du}{u} + \frac{7}{4} \int \frac{dv}{v} - \frac{3}{4}  \int \frac{dw}{w}$$

Finally, the expression becomes...

$$x - \frac{5}{2}arctan(x) - \frac{ln(u)}{2} + \frac{7ln(v)}{4} - \frac{3ln(w)}{4} + K \text{ where } K \text{ is an arbitrary constant}$$
$$= x - \frac{5}{2}arctan(x) - \frac{ln(x^2 + 1)}{2} + \frac{7ln(x - 1)}{4} - \frac{3ln(x + 1)}{4} + K$$

The solution for the integral is equivalent to the one in Wolfram Alpha.


问:
Counting problem concerning Generalized Pigeonhole Principle
In a game, all points in the (x, y) plane with coordinates obeying $ x, y \in \mathbb{Z} $ are labelled as
belonging to one of three players, i.e., either to Alice, Bob or Carol.

Show that one of the players will possess four points whose vertices form a rectangle.

Here is the problem I've been thinking for days. It seems easy since the coordinates are unlimited.
But it is also hard to find pigeonholes I want to divide.

I've encountered several problems like finding a parallelogram on a $n \times n$ chessboard with $2n$ pawns. It is relatively simpler to me.

答:
HINT: Consider the points $\langle 0,n\rangle$, $\langle 1,n\rangle$, $\langle 2,n\rangle$, and $\langle 3,n\rangle$ for $n=0,1,2,\ldots\;$. Imagine that each point is labelled $A$, $B$, or $C$. Thus, one row might be labelled $ABCA$, another might be labelled $BCCB$, and so on.

How many different labellings of a row are possible?

Why would you like to find two rows with the same labelling?

How many rows does it take to guarantee that you have two rows with the same labelling?


问:
How do I calculate where a ballistic missile is going to land?
I've been trying to design a homemade rocket, and I'm stumped on how I will calculate where the rocket will land. I can't figure out how to calculate the velocity (in m/s) a rocket will be at when its motor burns out. Is there an equation (or equations) I can use to determine a rockets velocity at burnout? I would prefer if the equations ignored air drag.a</p>

答:
If you ignore drag, the equations are simple but the error will be large.  You need the dry mass of the rocket, the mass of the propellant, the thrust/time curve and the launch angle.  If we define $m_0$ as the dry mass of the rocket, $m_p$ as the mass of the propellant, $F(t)$ as the thrust as a function of time, $\theta$ as the angle off the vertical of the rocket, and $g$ the acceleration of gravity our dynamic variables are $x(t),y(t), \theta(t), m(t)$
First compute the specific impulse as $I_{sp}=\frac 1{m_p}\int F(t)dt$  Then
$$\dot m=-\frac {F(t)}{I_sp}\\
\ddot x=\frac{F(t)}{m(t)}\sin \theta(t)\\
\ddot y=\frac{F(t)}{m(t)}\cos\theta(t)-g\\
\theta(t)=\operatorname{arccot} \frac {\dot y}{\dot x}$$
and you can integrate these numerically.


问:
Perimeter or Calculus Word problem

A rectangular plot of farmland will be bounded on one side by a river
  and on the other three sides by a single strand of electric fence.
  With 1400m of wire at your disposal, what is the largest area you can
  enclose, and what are its dimensions?

Is there a way to solve this without Calculus? This sounds like a simple perimeter problem. If you have to use Calculus to solve this, how would I start? Could someone show me the steps?

答:
Yes, you can solve this without calculus by a process called quadratic optimisation (though the math behind it really uses calculus, you can do it without taking a single derivative or integral).

The plot is rectangular, so we have four sides which sum to 1400: $x + 2z = 1400$

Area $A = x \cdot z$

We'll want to state $A$ in terms of x only, so we use a simultaneous equation. Rearrange $x + 2z = 1400$ to form $2z = 1400 - x$, simplify to $z = 700 - \frac{x}{2}$

Plug that into the area equation:

$A = x \cdot (700 - \frac{x}{2})$

$A = 700x - \frac{x^{2}}{2}$

Now we have a quadratic, so we can optimise it. (If we wanted to use calculus, we could see where the derivative is equal to 0, but we won't do that.) We will do this by finding the vertex of the quadratic.

We'll want the $a,b,c$ values of our quadratic in form ($ax^{2} + bx + c$):

$a = \frac{-1}{2}$

$b = 700$

$c = 0$

To find the x-value of the vertex, we use the formula $x=\frac{-b}{2a}$:

 $x=\frac{-700}{2\frac{-1}{2}}$

 $x=\frac{-700}{\frac{-2}{2}}$

 $x=\frac{-700}{-1}$

 $x=700$

We just plug that in to find the A-value:

$A = 700x - \frac{x^{2}}{2}$

$x=700$

$A = 700(700) - \frac{700^{2}}{2}$

$A = 245000$

This only leaves the question of what the length of the other 2 sides is. If $x + 2z = 1400$, and $x = 700$, then $z = 350.$

So there's the answer: The area is maximised (to 245000) when one side is 700 units long, and the other two sides are 350 each.

Title: Perimeter or Calculus Word problem


问:
Can someone explain the relation between "Achilles chasing turtle" paradox and monotonic, bounded sequence to me?
Recently I have read a book called "Caculus - Basic Concepts for High School"

The possibility of infinite but bounded sets
was not known, for example, to ancient Greeks. Suffice
it to recall the famous paradox about Achilles chasing
a turtle.
Let us assume that Achilles and the turtle are initially
separated by a distance of 1 km. Achilles moves 10 times
faster than the turtle. Ancient Greeks reasoned like this:
during the time Achilles covers 1 km the turtle covers 100 m.
By the time Achilles has covered these 100 m, the turtle
will have made another 10 m, and before Achilles has covered these 10 m, the turtle will have made 1 m more, and
so on. Out of these considerations a paradoxical conclusion
was derived that Achilles could never catch up with the
turtle.
This "paradox" shows that ancient Greeks failed to grasp
the fact that a monotonic sequence may be bounded.

I know how to prove the conclusion of this paradox is incorrect, and understand that the described movement of Achilles and the turtle each is a monotonic, bounded sequence. However, I don't understand author's claim that

This "paradox" shows that ancient Greeks failed to grasp
the fact that a monotonic sequence may be bounded.

Can someone help to explain this part? Thank you very much.

EDIT:
Thank you all so much for your instant replies. However, I still don't get how these two movements being strictly monotonic, bounded sequences, has anything to do with Greek ancient people's conclusion.

For Achilles' movement:
$$ x = \sum_{i=0}^\infty \frac{1000}{10^i}$$

For the turtle's movement:
$$ y = 1000 + \sum_{i=0}^\infty \frac{100}{10^i} $$

So the Greek ancient people are trying to find an $n$ where:
$$ x_n = y_n $$

However, since $ y_n = x_{n+1} $ and $ x_n < x_{n+1} $ so there is no such $n$ exist.

With everything mentioned above, I still don't find anything to do with bounded sequence part.

答:
I think it means that they did not know that there could be sequences where every value is bigger than the last that themselves have a limit, for example the sequence:
1/2, 2/3, 3/4, 4,5, 5/6, etc
it has a limit of 1 but it always increses, the statement says that the greek did not have the understanding that there where such sequences, where the numbers always increse never get to infinity.</p>


答:
(For fun)

Greek philosopher:

Now you see, by the time Achille reaches the place where the turtle is right row, the turtle will have gone to another place. So Achill will have to run an extra time to reach turtle again. Clearly, that pattern repeats infinitely to Achille, thus Achille has to run forever.

Greek student

But Philosopher, why does AChille have to run forever?

Greek Philosopher:

Stupid! Because all the instants -at which that pattern happens to Achille - is an infinite increasing sequence. And an infinite increasing sequence cannot be bounded.

Count your cows! Count the houses! Do you see that even though those numbers may be big but you still can count them all. That is because they are bounded. Finite is bounded! Infinite is very very big!

You potato head surely have wasted all the potatoes you have eaten.

Title: Can someone explain the relation between "Achilles chasing turtle" paradox and monotonic, bounded sequence to me?

Body:
Recently I have read a book called "Caculus - Basic Concepts for High School"

The possibility of infinite but bounded sets
was not known, for example, to ancient Greeks. Suffice
it to recall the famous paradox about Achilles chasing
a turtle.
Let us assume that Achilles and the turtle are initially
separated by a distance of 1 km. Achilles moves 10 times
faster than the turtle. Ancient Greeks reasoned like this:
during the time Achilles covers 1 km the turtle covers 100 m.
By the time Achilles has covered these 100 m, the turtle
will have made another 10 m, and before Achilles has covered these 10 m, the turtle will have made 1 m more, and
so on. Out of these considerations a paradoxical conclusion
was derived that Achilles could never catch up with the
turtle.
This "paradox" shows that ancient Greeks failed to grasp
the fact that a monotonic sequence may be bounded.

I know how to prove the conclusion of this paradox is incorrect, and understand that the described movement of Achilles and the turtle each is a monotonic, bounded sequence. However, I don't understand author's claim that

This "paradox" shows that ancient Greeks failed to grasp
the fact that a monotonic sequence may be bounded.

Can someone help to explain this part? Thank you very much.

EDIT:
Thank you all so much for your instant replies. However, I still don't get how these two movements being strictly monotonic, bounded sequences, has anything to do with Greek ancient people's conclusion.

For Achilles' movement:
$$ x = \sum_{i=0}^\infty \frac{1000}{10^i}$$

For the turtle's movement:
$$ y = 1000 + \sum_{i=0}^\infty \frac{100}{10^i} $$

So the Greek ancient people are trying to find an $n$ where:
$$ x_n = y_n $$

However, since $ y_n = x_{n+1} $ and $ x_n < x_{n+1} $ so there is no such $n$ exist.

With everything mentioned above, I still don't find anything to do with bounded sequence part.


问:
Proving $a\equiv b \pmod m$ if $a \bmod m = b\bmod m$
I want to prove that for integers $a, b$ and positive integers $m$, that $a \equiv b\pmod m$ iff $a \bmod m = b\bmod m$

I’m on my tablet currently so I cannot type up my latex well. But I’ve been trying to do a direct proof using the fact that $b=mq+r$. I also tried using propositional logic, since this is a biconditional I tried proving the implications in both directions, but I’m still getting confused

答:
Here are the first steps to get you started.

Let $a,b,m \in \mathbb{Z}$ and assume $a \equiv k \pmod{m}$ and $b \equiv k \pmod{m}$. Then, there exist $n_a, n_b \in \mathbb{Z}$ such that
$$
\begin{split}
a &= m n_a &+ k \\
b &= m n_b &+ k \\
\end{split}
$$
Can you please subtract the two equations?

HINT 2

Since $n_a,n_b$ are integers, so is $n_b - n_a$. Note that
$$
a = mn_a+k = m(n_a-n_b) + mn_b + k = m(n_a-n_b) + b.
$$


答:
Say $a=qm + r$, $0\leq r\lt m$, and $b=pm + s$, $0\leq s\lt m$, so that $a\bmod m = r$ and $b\bmod m = s$. Without loss of generality, say $r\geq s$.

You want to show that $a\equiv b\pmod{m}$ if and only if $r=s$. Presumably you know that $a\equiv b\pmod{m}$ if and only if $m|a-b$, if and only if $m$ divides $(qm+r)-(pm+s) = (q-p)m + (r-s)$.

Can you take it from there?

Title: Proving $a\equiv b \pmod m$ if $a \bmod m = b\bmod m$

Body:
I want to prove that for integers $a, b$ and positive integers $m$, that $a \equiv b\pmod m$ iff $a \bmod m = b\bmod m$

I’m on my tablet currently so I cannot type up my latex well. But I’ve been trying to do a direct proof using the fact that $b=mq+r$. I also tried using propositional logic, since this is a biconditional I tried proving the implications in both directions, but I’m still getting confused


问:
Partial fraction method in Integration
My question is can we proceed with the method without even factorising the denominator.
For eg take a question.
$$\int \frac{2x-3}{(x^2-1)(2x+3)}dx$$
Can we write directly
$$\frac{2x-3}{(x^2-1)(2x+3)}=\frac{Ax+B}{x^2-1}+\frac{C}{2x-3}$$
without simplifying denominator term $(x^2-1)=(x-1)(x+1)$?

答:
There is no problem in this; it's just a matter of choice. HOWEVER, if you resolve the integrand as
$$\frac{2x-3}{(x^2-1)(2x+3)}=\frac{Ax+B}{x^2-1}+\frac{C}{2x-3}$$
then how would you go about integrating $(Ax+B)/(x^2-1)$?

Unless you are lucky that $B=0$, (in the above case you get $B=1$), in all other cases, you will have no choice but again resolving it into two more fractions, so why not do it directly?

Edit

Suppose we get $A=5$ and $B=0$ in the above one, then integrating is trivial.
$$\int \frac{5x}{x^2-1}\ dx=\int\frac 5 2 \frac{2x}{x^2-1}\ dx$$
$$ = \frac{5}{2}\ln |x^2-1|$$
However, say we get $B=1$, then how would you solve
$$\int\frac{5x+1}{x^2-1}$$
You have no other choice but to use partial fraction decomposition.

Hope this is clear :)

Title: Partial fraction method in Integration


问:
Logic - Proof of a resolution rule
I am going to show the following resolution rule:
$$
(p \lor q) \land (\lnot p \lor r) \vdash (q \lor r)
$$
The proof I tried:
\begin{align}
&\quad \ (p \lor q) \land (\lnot p \lor r) \\
&\equiv ((p \lor q) \land \lnot p) \lor ((p \lor q) \land r) \\
&\equiv ((p \land \lnot p) \lor (\lnot p \land q)) \lor ((p \lor q) \land r) \\
&\equiv (\lnot p \land q) \lor ((p \lor q) \land r) \\
&\equiv ((\lnot p \land q) \lor(p \lor q)) \land ((\lnot p \land q) \lor r) \\
&\equiv (((\lnot p \land q) \lor p) \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (((\lnot p \lor p) \land (p \lor q)) \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (p \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (p \lor q) \land (\lnot p \lor r) \land (q \lor r) \\
&\vdash (q \lor r)
\end{align}

It is strange that $(p \lor q) \land (\lnot p \lor r)$ appeared again with $q \lor r$ being connected by conjunction. Is this correct?

答:
Your sentenceial logic proof is valid.

$$\begin{array}{cll}
\quad & (p \lor q) \land (\lnot p \lor r) \\
\equiv& ((p \lor q) \land \lnot p) \lor ((p \lor q) \land r) &\text{distribution} \\
\equiv& (\lnot p \land q) \lor ((p \lor q) \land r) &\text{absorption} \\
\equiv& ((\lnot p \land q) \lor(p \lor q)) \land ((\lnot p \land q) \lor r) &\text{distribution} \\
\equiv& (((\lnot p \land q) \lor p) \lor q) \land ((\lnot p \land q) \lor r) &\text{association} \\
\equiv& ((p\lor q)\lor q)  \land ((\lnot p \land q) \lor r) &\text{absorption}\\
\equiv& (p \lor q) \land ((\lnot p \land q) \lor r) &\text{idemptoptence} \\
\equiv& (p \lor q) \land (\lnot p \lor r) \land (q \lor r) &\text{distribution} \\
\vdash& (q \lor r) &\text{simplification}
\end{array}
$$

It is strange that $(p \lor q) \land (\lnot p \lor r)$ appeared again with $q \lor r$ being connected by conjunction.

It is interesting, but not disturbing.  Since $q\lor r$ is entailed by $(p\lor q)\land(\lnot p\lor r)$, this is nothing more than a case of $$\begin{split}A~~\vdash&~ B\\\hline A\dashv\vdash&~ A\land B\end{split}$$

Title: Logic - Proof of a resolution rule


问:
Using goal seek in excel.
Hi I don't know if this is an appropriate question for this page but I thought it was worth a shot and if nothing else a learning experience.

So I am studying actuarial mathematics and I am currently working on problem involving calculating the yield of an investment.

I have been working through these questions alongside prerecorded lectures and have been fairly understanding the concept and how to use it however my lecturer constantly says that these answers are not that exact or precise but would be a lot better when worked out using Goal seek in excel.

However my problem is that he has never shown us goal seek or mentioned any further on the topic.

Hence I was wondering if anyone could explain this too me and/or how to use it as I'm getting answers to questions wrong due to the situation.

Thanks in advance.

Adding from comments:

The question I was given was: You invest £10,000 in a project and receive back £700.87 per annum payable half-yearly in arrears for the next 25 years. Calculate the yield on this investment.

So from this we worked out that the money coming back at the end of this question is £17521.75 and the average time for this question is 12.75 $(last time period-first time period) /2$ which is $(25.5)/(2)$

And from this we do the calculation

$(17521.75/10000)^(1/12.75)$

And we get 4.5% which my lecturer then states in his lecture that this isn't a very good answer as the real answer is 4.98% but doesn't proceed to tell us how to obtain this answer.

答:
As at least one of the comments suggested, it is reasonable to understand the math.

The following approach will require a calculator that can (for example) readily compute $v^{(50)}$, for $0 < v < 1.$

The easiest way to determine the annual interest rate $i$ is to first solve for the semi annual interest rate $j$.

Then you will have that

$$1 \to (1 + j) \to (1 + j)^2 = (1 + i) \implies i = 2j + j^2.$$

Therefore, the problem reduces to solving for $j$.

Let $v = \frac{1}{1 + j}.$

Then $v^k$ represents the present value of a payment of $1$ that will be made $[k \times (6 ~\text{months})]$ from now.

Therefore, 50 payments of $1$, every 6 months, starting 6 months from now have a present value of

$$\sum_{k=1}^{50} ~v^k ~~=~~ \frac{v \times [1 - v^{50}]}{1 - v}.$$

Therefore

$$10000 = 350.435 \times \frac{v \times [1 - v^{50}]}{1 - v}.$$

Let $m = \frac{10000}{350.435} \approx 28.536.$

Then $$m ~\text{is slightly less than} ~\frac{v}{1 - v}.$$

If you were to presume that

$$m = \frac{v}{1-v} ~~~\text{then} ~~~m(1 - v) = v \implies v = \frac{m}{m+1} \approx 0.96614.$$

Proceeding iteratively, and using $0.96614$ as the first estimate of $v$ gives

$$\left[1 - v^{(50)}\right] \approx 0.8213.$$

Refining the first estimate:

$$m \approx \frac{v(0.8213)}{1-v} \implies m(1 - v) = v(0.8213) \implies v = \frac{m}{m+(0.8213)} \approx 0.97202.$$

Using the second estimate of $v = 0.97202$ gives

$$1 - v^{(50)} \approx 0.758.$$

Refining the second estimate:

$$m \approx \frac{v(0.758)}{1-v} \implies m(1 - v) = v(0.758) \implies v = \frac{m}{m+(0.758)} \approx 0.9741.$$

Using the third estimate of $v = 0.9741$ gives

$$1 - v^{(50)} \approx 0.7304.$$

Refining the third estimate:

$$m \approx \frac{v(0.7304)}{1-v} \implies m(1 - v) = v(0.7304) \implies v = \frac{m}{m+(0.7304)} \approx 0.9750.$$

Obviously, you could continue this process with a hand calculator and get an estimate for $v$ accurate to within 7 or 8 decimal places in less than an hour.  I suspect that the spreadsheet is doing something very similar.

Anyway, if you presume that $~v = 0.9750~$, then

$$1 + j = \frac{1}{v} \approx \frac{1}{0.9750} \implies
j \approx 0.256 \implies i = 2j + j^2 \approx 0.519.$$

For an intuitive explanation of why this estimate is on the high side, notice that each refinement of $v$ yielded a slightly higher estimate than the previous refinement.  The refinements stopped after the fourth estimate of $v$.

Title: Using goal seek in excel.


问:
How many distinct multiplications?
I have a multiset $\left\{ {2,3,3,3,4,4,5,5,6,7} \right\}$. How many distinct multiplications of three numbers chosen from the multiset?

e.g. $2×3×3$; $2×3×4$; $4×4×3$;... not $2×3×3$; $3×3×2$;... and $4×3×3$; $6×2×3$; ... (equal multiplication) are considered as the same.

答: How many distinct products are there when all numbers are different? With two $3$'s, $4$'s or $5$'s? With three $3$'s?</p>

Title: How many distinct multiplications?


问:
What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$?
We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$

What is the limit of this function as $n \rightarrow \infty$?

My idea:

$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$

Is this correct?

答:
HINT:

Such application of limit (esp. to $\infty$) to individual summand is applicable only when the number of summand is finite.

$$\dfrac{1+2+3+...+ \space n}{n^2} =\frac{\frac{n(n+1)}2}{n^2}=\frac12\cdot\left(1+\frac1n\right)$$


答:
Let $a = mx + r_1$ and let $b = my + r_2$

Given $a \mod{m} = b\mod{m} \implies r_1 = r_2 = k$

$ \implies a = mx + k, b = my + k$

$a - b = m(x - y)$

$m \mid m(x - y) \implies m \mid (a - b) \implies a \equiv b (\mod{m}) $


问:
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$

Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.

How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.

答:
Guide:

Notice that $|x+y|+|x-y|=2$ describes the boundary of a  square.

$x^2-6x+y^2=(x-3)^2+y^2-9$

The largest value of $y^2$ is $1$. Given that $-1 \le x \le 1$. Try to solve the $-1 \le x \le 1$, how would you optimize $(x-3)^2$.


问:
A question regarding the Poisson Process
I have the following question: Buses arrive at a city according to a Poisson process with a rate of 5 per hour. What is the probability that the fifth bus of the day arrives after midday given they start arriving at 9 a.m.

What I think is the correct way to go about this question is that I calculate:
$ P(N(0,3)\le4) $ The probability that less than or equal to four buses occur between 9am to midday. Because we have to reach the 4th bus in that time before the 5th bus can arrive. Provided I am correct with that notion the formula would look like: $   \sum_i_=_0^4 {15^i e^{-15}\over i!} $.

Thank you

答: a short answer: yes your answer is correct. the number of arrival before 12 is distributed as Poisson (15), so you need this to be strictly smaller than 5.</p>


Body:
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.

How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.

Body:
Hint:

If $f(x,y)=|x+y|+|x-y|-2,$

observe that  $f(x,y)=f(-x,y)=f(x,-y)=f(-x,-y)$

WLOG $x+y, x-y\ge0;$

$$x=1, -x\le y\le x, y^2\le x^2=1$$

Now,
$$x^2+y^2+10y=1+(y+5)^2-25$$

$-1\le y\le1\iff5-1\le5+y\le5+1\implies4^2\le(y+5)^2\le6^2$


问:
How is it correct:$\lim_{n\rightarrow{\infty}}n[(n+1)\sqrt[n+1]{n+1}-n\sqrt[n]{n}-1)] = 1$
It is known that:
$$\lim_{n\rightarrow{\infty}}[(n+1)\sqrt[n+1]{n+1}-n\sqrt[n]{n}] =1$$
I tried to calculate the limit:
$$\lim_{n\rightarrow{\infty}}n[(n+1)\sqrt[n+1]{n+1}-n\sqrt[n]{n}-1)]. $$
I managed to show that:
$$\lim_{n\rightarrow{\infty}}\frac{n}{\ln{n}}[(n+1)\sqrt[n+1]{n+1}-n\sqrt[n]{n}-1] = 0.$$
Using: $$\lim_{n\rightarrow{\infty}}n(\sqrt[n]{n}-1) = \lim_{x\rightarrow{\infty}}{\ln{n}}    (*)$$
and
$$\lim_{n\rightarrow{\infty}}\frac{\sqrt[n]{n}-1}{\ln{\sqrt[n]{n}}} =1$$
I found:
$$\lim_{n\rightarrow{\infty}}n[(n+1)\sqrt[n+1]{n+1}-n\sqrt[n]{n}-1)] =$$

$$=\lim_{n\rightarrow{\infty}}n[(n+1)(\sqrt[n+1]{n+1}-1)-n(\sqrt[n]{n}-1)] =$$

$$=\lim_{n\rightarrow{\infty}}n[\frac{\sqrt[n+1]{n+1}-1}{\ln{\sqrt[n+1]{n+1}}}\cdot\ln{(n+1)}-\frac{\sqrt[n]{n}-1}{\ln{\sqrt[n]{n}}}\cdot \ln{n}]=$$

$$?=?\lim_{n\rightarrow{\infty}}n[\ln{(n+1)}-\ln{n}]= \lim_{n\rightarrow{\infty}}\ln(1 + \frac{1}{n})^n= 1$$

Is the result correct?

Is the application of equality (*) correct?

Is there another method of calculating, without derivatives, the respective limit?

答:
Consider the expression $$f(n)=\frac{n^{1/n}-1}{n^{1/n}}$$ which can be rewritten as $$\dfrac{\exp\dfrac{\log n} {n} - 1}{\dfrac{\log n} {n} }=1+\frac{\log n} {2n}+\dots$$ and then $$f(n) \log(n+1)-f(n)\log n=\log(1+1/n)+\frac{(\log(n+1)) ^2}{2(n+1)}-\frac{(\log n) ^2}{2n} +\dots$$ By mean value theorem the second term equals $$\frac{2\log\xi-(\log\xi)^2}{2\xi^2}$$ where $n<\xi<n+1$ and this multiplied by $n$ clearly tends to $0$ as $n\to\infty $.

It follows that $n(f(n+1)\log (n+1)-f(n)\log n) $ has same limit as that of $n\log(1+n^{-1})$ which is $1$.

The above uses derivatives and Taylor expansions and I don't see a clear way to avoid these tools right now.


问:
Reorganising alternating sum of products of binomial coefficients
The summation
$$
\sum_{k=0}^{\lfloor\frac{p}{s}\rfloor}(-1)^k {n\choose k}{p-ks+n-1\choose n-1}\quad;n > 0, s > 0 \text{ and } 0\le p\le ns,
$$
with ${n\choose k}$ denoting the binomial coefficient, has the limitation that implemented fixed sized integers, there are cases where individual terms exceed the integer size even though the result does not. (Even assuming ideal calculation of binomial coefficients.) Is there a way to reorganise that summation to avoid that? The integer size bound can be assumed to be $s^n$, the sum of that result for all $p$. A stronger result would that all terms summed do not exceed their sum in absolute value. A yet stronger result would be that all terms summed are non-negative. But even the weakest case is fine.

答:
If the summation index $k$ went up to $n$ instead of $p/s \le n$, it would be the inclusion-exclusion formula for the number of $(n-1)$-subsets of $\{1,\dots,p+n-1\}$ that contain at least one element from each of the following $n$ disjoint intervals:
$$
\{1,\dots,s\} \\
\{s+1,\dots,2s\} \\
\vdots \\
\{(n-1)s+1,\dots,ns\}
$$
The number of such subsets is clearly $0$ because $n > n-1$.  As in the linked answer, you can generalize by replacing the $n-1$ with any number smaller than $n$ and still obtain the same $0$ result.

So your sum is
$$
\sum_{k=0}^{\lfloor p/s \rfloor}(-1)^k \binom{n}{k}\binom{p-ks+n-1}{n-1}
=
-\sum_{k=\lfloor p/s \rfloor+1}^n(-1)^k \binom{n}{k}\binom{p-ks+n-1}{n-1}
$$


答:
Recall that $$1 + 2 + \cdots + n =\sum_{i = 1}^n i = \frac{n(n+1)}{2}$$

So you want $$\lim_{n\to \infty}\dfrac{n(n+1)}{2n^2} = \lim_{n\to\infty}\frac{n^2+ n}{2n^2} = \lim_{n\to \infty} \left(\frac 12 + \frac 1{2n}\right)$$

Title: What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$?

Body:
We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$

What is the limit of this function as $n \rightarrow \infty$?

My idea:

$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$

Is this correct?


问:
Multivariate Inverse Transformation Sampling

Summary

Given a multivariate density distribution, I use inverse transformation sampling to sample points from this distribution. While the first dimension exhibits the correct distribution, all other dimensions contain a slight, stable error.

Details

My density distribution is given as a bilinear interpolation on the $([0-1], [0-1])$ rectangle. In the rest of this question, I use the following example:

$$
d(x,y)=\frac{4}{11}(2+x+2y-3xy)
$$

This results in the following density plot (brighter colors represent higher density):

The cumulative density function is

$$
cum(x,y)=\int_{0}^{x} \int_{0}^{y} d(px,py) dpy\ dpx = \frac{1}{11}(8xy+2x^2y+4xy^2-3x^2y^2)
$$

In order to sample from this distribution, I draw two samples $ux$ and $uy$ from the uniform $[0, 1)$ distribution and transform them to $x$ and $y$ as follows:

$$
cum(x, 1)=ux \\
x = 6-\sqrt{36-11ux}
$$

and
$$
cum(x, y)=ux \cdot uy \\
y = \frac{4+x-\sqrt{16+48uy+8x-40uy x+x^2+3 uy x^2}}{3x-4}
$$

The samples resulting from this transformation look reasonable (5000 samples in the following figure):

I tried to verify the result by approximating the cumulative density from the samples by simply counting how many of the samples have a smaller or equal x and y coordinate. Here are some results for 10 million samples. I report the analytic expected value and the results of two samplings. Digits are truncated:

```<code>                   i : 0.0   0.1   0.2   0.3   0.4   0.5   0.6   0.7   0.8   0.9   1.0
---------------------------------------------------------------------------------------
  analytic cum(i, 1) : 0.0   0.108 0.214 0.319 0.421 0.522 0.621 0.719 0.814 0.908 1.0
sampling 1 cum(i, 1) : 0.0   0.108 0.214 0.319 0.421 0.522 0.621 0.718 0.814 0.908 1.0
sampling 2 cum(i, 1) : 0.0   0.108 0.214 0.319 0.421 0.522 0.621 0.719 0.814 0.908 1.0
---------------------------------------------------------------------------------------
  analytic cum(1, i) : 0.0   0.091 0.185 0.280 0.378 0.477 0.578 0.680 0.785 0.891 1.0
sampling 1 cum(1, i) : 0.0   0.080 0.164 0.253 0.347 0.445 0.547 0.653 0.764 0.880 1.0
sampling 2 cum(1, i) : 0.0   0.080 0.164 0.253 0.347 0.445 0.547 0.653 0.764 0.880 1.0
</code>```

Obviously, the cumulative density of the x-coordinates ($cum(i, 1)$) is almost exactly the analytic expression. On the other hand, there is a clear error in the y-coordinates ($cum(1, i)$). This error is stable across different samplings.

I cannot explain this slight error. Both the theoretical fundamentals and the implementation (with Mathematica) look sound.

Is there something I might have missed? Univariate sampling works perfectly as can be seen from the x-coordinates. However, multivariate sampling exhibits a slight error.

答:
The accepted answer to this question is good! However, I thought I would supplement it with some R code showing how to do the bivariate inverse sampling correctly with the distribution provided in the original question. I hope this helps future readers of this question.

```<code># Bivariate Inverse Transformation Sampling

# Provided Bivariate Distribution
dis<-function(x,y){(4/11)*(2+x+2*y-3*x*y)}
dis<-Vectorize(dis)

# Heatmap of the true bivariate distribution
library(ggplot2)
data<-expand.grid(X=seq(0,1,0.01),Y=seq(0,1,0.01))
data<span class="math-container">$Z<-dis(data$</span>X,data$Y)
ggplot(data, aes(X, Y, fill= Z)) +
  geom_tile() +
  scale_fill_distiller(palette = "RdYlGn") + coord_fixed() +
  theme_bw()

# Now, note that we can find the marginal of X by just integrating y from 0 to 1
margX<-function(x){(-2/11)*(-6 + x)}
# Then, find the marginal cdf of X by integrating from 0 to x
margxcdf<-function(x){-1/11 * (-12 + x) * x}

# Then, using the marginal of X we can get the conditional distribution of Y|X by dividing the joint by the marginal
# Then integrate that with respect to y to get the conditional CDF of Y|X. (I did this in wolfram alpha)
condcdfY_X<-function(y,x){(y*(-2*(2 + y) + x*(-2 + 3*y)))/(-6 + x)}

# Now we can use inverse sampling to get the desired result. First, find the inverse of margXcdf and condcdfY_X
# I used wolfram alpha to solve for the inverse of the conditional (note: it has to be explicitly defined at x=2/3. See wolfram alpha for details)
margXcdf_inv<-Vectorize(function(u){6 - sqrt(36 - 11*u)})
condcdfY_X_Inv<-Vectorize(function(z,x){ifelse(x!=2/3,(2 + x - sqrt(4 + x*(4 - 20*z) + 12*z + (x^2)*(1 + 3*z)))/(-2 + 3*x),z)})

# Now to get a random sample from a uniform[0,1] for X & transform it
Xs<-runif(1000000)
Xs<-margXcdf_inv(Xs)

# Now to get a random sample from a uniform[0,1] for y
Ys<-runif(1000000)
Ys<-condcdfY_X_Inv(Ys,Xs)

samples<-data.frame(X=Xs,Y=Ys)

# Now to plot the estimated bivariate distribution (using kernel density estimation) using the samples
library(MASS)
library(ggplotify)
library(gridExtra)
test<-kde2d(Xs,Ys,n=100)
hm_col_scale<-colorRampPalette(c("black","darkgreen","green","yellow","red"))(1000)
p1<-as.ggplot(~image(test<span class="math-container">$z,
      col = hm_col_scale,
      zlim=c(min(test$</span>z), max(test$z))))

p2<-ggplot(data, aes(X, Y, fill= Z)) +
  geom_tile() +
  scale_fill_distiller(palette = "RdYlGn") +
  theme_bw()

grid.arrange(p1 + coord_fixed(),p2 + coord_fixed(),nrow=1)
</code>```

Here are the final images produced by the code above:
R code Final Images comparing sample distribution to true distribution

As you can see, the suggested solution by the accepted answer works correctly.

Body:
The accepted answer to this question is good! However, I thought I would supplement it with some R code showing how to do the bivariate inverse sampling correctly with the distribution provided in the original question. I hope this helps future readers of this question.

# Bivariate Inverse Transformation Sampling

# Provided Bivariate Distribution
dis<-function(x,y){(4/11)*(2+x+2*y-3*x*y)}
dis<-Vectorize(dis)

# Heatmap of the true bivariate distribution
library(ggplot2)
data<-expand.grid(X=seq(0,1,0.01),Y=seq(0,1,0.01))
data$Z<-dis(data$X,data$Y)
ggplot(data, aes(X, Y, fill= Z)) +
  geom_tile() +
  scale_fill_distiller(palette = "RdYlGn") + coord_fixed() +
  theme_bw()

# Now, note that we can find the marginal of X by just integrating y from 0 to 1
margX<-function(x){(-2/11)*(-6 + x)}
# Then, find the marginal cdf of X by integrating from 0 to x
margxcdf<-function(x){-1/11 * (-12 + x) * x}

# Then, using the marginal of X we can get the conditional distribution of Y|X by dividing the joint by the marginal
# Then integrate that with respect to y to get the conditional CDF of Y|X. (I did this in wolfram alpha)
condcdfY_X<-function(y,x){(y*(-2*(2 + y) + x*(-2 + 3*y)))/(-6 + x)}

# Now we can use inverse sampling to get the desired result. First, find the inverse of margXcdf and condcdfY_X
# I used wolfram alpha to solve for the inverse of the conditional (note: it has to be explicitly defined at x=2/3. See wolfram alpha for details)
margXcdf_inv<-Vectorize(function(u){6 - sqrt(36 - 11*u)})
condcdfY_X_Inv<-Vectorize(function(z,x){ifelse(x!=2/3,(2 + x - sqrt(4 + x*(4 - 20*z) + 12*z + (x^2)*(1 + 3*z)))/(-2 + 3*x),z)})

# Now to get a random sample from a uniform[0,1] for X & transform it
Xs<-runif(1000000)
Xs<-margXcdf_inv(Xs)

# Now to get a random sample from a uniform[0,1] for y
Ys<-runif(1000000)
Ys<-condcdfY_X_Inv(Ys,Xs)

samples<-data.frame(X=Xs,Y=Ys)

# Now to plot the estimated bivariate distribution (using kernel density estimation) using the samples
library(MASS)
library(ggplotify)
library(gridExtra)
test<-kde2d(Xs,Ys,n=100)
hm_col_scale<-colorRampPalette(c("black","darkgreen","green","yellow","red"))(1000)
p1<-as.ggplot(~image(test$z,
      col = hm_col_scale,
      zlim=c(min(test$z), max(test$z))))

p2<-ggplot(data, aes(X, Y, fill= Z)) +
  geom_tile() +
  scale_fill_distiller(palette = "RdYlGn") +
  theme_bw()

grid.arrange(p1 + coord_fixed(),p2 + coord_fixed(),nrow=1)

Here are the final images produced by the code above:
R code Final Images comparing sample distribution to true distribution

As you can see, the suggested solution by the accepted answer works correctly.