问:
Help understanding the proof: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$
In Spivak's Calculus, one is asked to prove that $(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}$

I will write his proof (from the solution manual) verbatim, but then I have some pedantic questions at the end, which are the primary purpose of this post.

The only extra piece of information Spivak provides is:

$\text{P1}$. $a+(b+c) = (a+b)+c \quad$ Associative Law for Addition

and the following definition:

$a_1+a_2+a_3+...+a_{n-2}+a_{n-1}+a_n:= a_1+(a_2+(a_3+...+(a_{n-2}+(a_{n-1}+(a_n)))...)$

However, it seems to me that he should have also mentioned that his bracket symbols $[$ and $]$ are presumably meant to represent different layers of the parentheses symbols $($ and $)$.

Base Case:

$a_1+a_2 = a_1 + a_2 $

Assume:

$(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$

Prove the $k+1$ case: $(a_1+...+a_k + a_{k+1})+a_{k+2}=a_1+...+a_k+a_{k+1}+a_{k+2}$

\begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
& \quad \quad \text{ by P1} \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
&=a_1+...+a_{k+2} \\
\end{align}

First Question

I do not understand Spivak's base case. Shouldn't his base case start as $(a_1)+a_2$? It seems to me the only way of going beyond this sentence is to first make the definition that $(x):=x$. However, in making this definition, doesn't the base case already solve everything? Because any finite expression of $k$ terms can be lumped together, called $x$, and then you'd have $(x)+a_{k+1}=x+a_{k+1}$...which is the whole point of this proof.

Edit: A work around for this seems to be to dictate that this proof only operate on $k\geq 2$. Because if so:

We could say $(a_1+a_2) +a_3=a_1+a_2+a_3$ because by definition,  $a_1+a_2+a_3=a_1+(a_2+a_3)$ and then we can invoke the associative property to PROVE .$(a_1+a_2) +a_3=a_1+a_2+a_3$.

Second Question

The offending transition that has confused me is when Spivak writes:
\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}

Now, I imagine what Spivak is trying to convey is that we can think of $a_{k+1}+a_{k+2}$ as being a single term...say $\alpha$...and that our assumption in our Assume step can be reframed as $(a_1+...+a_k) + \alpha=a_1+...+a_k+\alpha$.

From what I've learned in set theory, $a$ is actually a function with a domain of some subset in $\mathbb N$ and, in this context, a range that is a subset of $\mathbb R$. Given this, I do not see why at all our initial assumption case of $(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$ can be applied to $(a_1+...+a_k)+(a_{k+1}+a_{k+2})$. In general, the quantity $(a_{k+1}+a_{k+2}) \neq a_{k+1}$.

So how exactly is this step actually being applied?

Edit: The only thing I can think to do here is invoke ordinals; I will explain. Originally, the formal definition of the proof I thought was: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$. However, to get around the issue of $(a_{k+1}+a_{k+2}) \neq a_{k+1}$, it seems like I should instead write the proof as:

$\forall k \in \mathbb N \Big [ \forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big) \Big]$

In doing this, when I arrive at my assumption case, I assume for some $k$ that:

$\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$

WITH THIS, when we arrive at the situation of:

\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}

we can use our updated formula to tackle this. Specifically, because $\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$, there definitely exists one function, call it $a'$, that has the first $k$ terms in common with $a$ but, as its $k+1$ term, has $a_{k+1} + a_{k+2}$. That is to say: $a'_{k+1}=a_{k+1} + a_{k+2}$. Then we can absolutely apply our assumption to produce: $a_1+...+a_k+(a_{k+1}+a_{k+2})$

答:
First, the proof you've written contains some incorrect justifications:

$$ \begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
& \quad \quad \text{ by P1} \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
&=a_1+...+a_{k+2} \\
\end{align}$$

Here is a corrected version:
$$\begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
&\quad \quad \text{since the equation holds for } k \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
& \quad \quad \text{ by P1} \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_{k+2} \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
\end{align}$$

Spivak defines addition as an operation performed on a pair of numbers. If $a$ and $b$ are numbers, $a + b$ is a single number formed by the sum. It's probably best to put aside for now everything you know about set theory and functions. Spivak's talking only about the addition of pairs of (real) numbers.

If we have a sum involving more than two numbers, we can use brackets to specify the order in which terms are to be added together, but the end result of a sum is always a single number. $b + c$ is a single number. $a + (b + c)$ is a single number.

I'm not sure which edition of the book you're using, but in the 3rd edition version of this exercise, Spivak never uses brackets around individual terms. Expressions like $(a_n)$ don't clarify anything, since addition works only on pairs of numbers.

In the inductive argument, you can prove base cases up to n = 3 (via P1). I did this because the theorem doesn't really mean anything for less than 3 terms. (Complete induction works also.)

For the $k$-case you have:
$$(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$$

The statement says the sum of $k$ terms (bracketed as specified by Spivak's convention) added to a single term produces the number equal to the sum of all $k+1$ terms, bracketed as specified.

If the last term $a_{k+1}$ is itself a sum of several terms, that's ok, because the end result of those sums is a single number: $a_{k+1}$.

It's easy to mistake indexes for the number of terms here.

Sometimes it can be helpful to step through a specific case to see how the inductive step works. Just for overkill, let's do that here for the $n = 5$ case.

Assuming it's true for 4 terms, lets prove the 5 term case. Let's show
$$(a_1 + (a_2 + (a_3 + a_4))) + a_5 = a_1 + (a_2 + (a_3  + (a_4 + a_5)))$$

Now we know from the $n = 4$ case that
$$a_1 + (a_2 + (a_3 + a_4)) = (a_1 + (a_2 + a_3)) + a_4$$
(By the $n = 4$ case assumption, we can "pop" out the innermost term $a_4$)

Thus
$$(a_1 + (a_2 + (a_3 + a_4))) + a_5 = ((a_1 + (a_2 + a_3)) + a_4) + a_5$$

$$= (a_1 + (a_2 + a_3)) + (a_4 + a_5)  \text{ (by P1)}$$

$$= a_1 + (a_2 + (a_3 + (a_4 + a_5))) \text{ (again by the $n=4$ case)}$$

Above, we've "popped back in" the last term, which is allowed because of the $n=4$ case assumption. $a_4 + a_5$ is just a number.

Finally, be warned: the last step in this exercise (not mentioned here) is kinda hairy.


问:
find a upper bound of a product of gcd
Let $n,k \in \mathbb{N}$.

Find a upper bound for

$\gcd(n,2^n) \gcd(n+1, 2^{n+1}) \cdots \gcd(n+2^k, 2^{n+2^k})$.

Note that in the upper bound, there should be no $n$, but having $k$ is fine.

Attempt:

$\gcd(n,2^n) \gcd(n+1, 2^{n+1}) \cdots \gcd(n+2^k, 2^{n+2^k})
\leq \gcd(n,  2^{n+2^k}) \gcd(n+1, 2^{n+2^k}) \cdots \gcd(n+2^k, 2^{n+2^k})
= \frac{\gcd(1,  2^{n+2^k}) \gcd(2, 2^{n+2^k}) \cdots \gcd(n+2^k, 2^{n+2^k})}{\gcd(1,  2^{n+2^k}) \gcd(2, 2^{n+2^k}) \cdots \gcd(n-1, 2^{n+2^k})}$

Then I tried to eliminate the $n's$ by using How to calculate "gcd product" $\operatorname{gcdp}(n,m)=\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m)$.

Then it just got really messy.

What I'm really trying to do is to find a bound without the $n$. But leaving $k$
is fine.

It would be even better to find the actually value...

Thank you.

答:
The important thing to note is that there is no upper bound in terms of $k$ only.

For any $k \in \mathbb{N}$, let $n=2^N$ where we can choose $N$ to be as large as we like. Then the quantity that you are trying to bound contains the factor $$2^N=\gcd(2^N,2^{2^N})$$ independently of whatever $k$ is.

ContentLicense: CC BY-SA 4.0

Title: find a upper bound of a product of gcd


问:
What type of input does trigonometric functions take in
I see in my Book that 45 deg is equivalent of π/4 . Ι also do the conversion if I simply convert degrees into radians like this

45* π/180 = π/4 radians

and back again

π/4 * 180/π = 45 deg

.
So I think that I grasp the Idea finally that π/4 is another way of saying 45 degrees. Now if I use my calculator and I say sin(45) witch I assume that 45 is degrees I get the number 0.7071067812. If I do the calculation sqrt(2)/2 I will get the same number hence sin(45)=sqrt(2)/2 . But when I input sin(π/4) I get the number 0.0137073546. So I say to my self that probably that function does not get that type of input. But then again I see in my book that sin(π/4) = sqrt(2)/2 . So this forces me to ask the question, "What type of input do trigonometric functions take in?" Is the calculator not working properly ?

Excuse me if this looks simple to you but I am so bad in math and the calculator is the only validator I have at the moment and this confuses me allot.

答:
If you ask your calculator for $\tan ^{-1}(1)$, there are four possible answers:

If you get $45$. your calculator is set to use (receive and report) angles in degrees;

If you get $50$, your calculator is set to use grads;

If you get $0.785398163$, your calculator is set to use radians;

If you get anything else, your calculator is broken.

The grad is a unit of angle such that a right-angle contains $100$ grads

ContentLicense: CC BY-SA 3.0


问:
What formulas are required to calculate a 3d transformation?
Considering the point of view of square A and B, what math tranformations must be applied (either to the 3d camera or world) to transition from A to B?

I can tell that for the B viewpoint I had to move right and up, but I lack the math background to know what formulas can give me accurate values. Thanks

答:
I would recommend an approach where you first determine the initial basis unit vectors $\hat{u}$, $\hat{v}$, $\hat{w}$ and a fixed point $\vec{t}$, and corresponding final basis unit vectors $\hat{U}$, $\hat{V}$, $\hat{W}$ and a fixed point $\vec{T}$, such that the basis vectors are orthonormal,
$$\begin{array}{rcr}
\hat{u} \cdot \hat{u} = 1 & ~ & \hat{U} \cdot \hat{U} = 1 \\
\hat{u} \cdot \hat{v} = 0 & ~ & \hat{U} \cdot \hat{V} = 0 \\
\hat{u} \cdot \hat{w} = 0 & ~ & \hat{U} \cdot \hat{W} = 0 \\
\hat{v} \cdot \hat{u} = 0 & ~ & \hat{V} \cdot \hat{U} = 0 \\
\hat{v} \cdot \hat{v} = 1 & ~ & \hat{V} \cdot \hat{V} = 1 \\
\hat{v} \cdot \hat{w} = 0 & ~ & \hat{V} \cdot \hat{W} = 0 \\
\hat{w} \cdot \hat{u} = 0 & ~ & \hat{W} \cdot \hat{U} = 0 \\
\hat{w} \cdot \hat{v} = 0 & ~ & \hat{W} \cdot \hat{V} = 0 \\
\hat{w} \cdot \hat{w} = 1 & ~ & \hat{W} \cdot \hat{W} = 1 \\
\hat{u} \times \hat{v} = \hat{w} & ~ & \hat{U} \times \hat{V} = \hat{W} \\
\hat{w} \times \hat{u} = \hat{v} & ~ & \hat{W} \times \hat{U} = \hat{V} \\
\hat{v} \times \hat{w} = \hat{u} & ~ & \hat{V} \times \hat{W} = \hat{U} \\
\end{array}$$

Next, form two $4 \times 4$ transformation matrices:
$$\mathbf{M}_\text{before} = \left[ \begin{matrix}
u_x & v_x & w_x & t_x \\
u_y & v_y & w_y & t_y \\
u_z & v_z & w_z & t_z \\
  0 &   0 &   0 &   1 \\
\end{matrix} \right ], \quad \mathbf{M}_\text{after} = \left[ \begin{matrix}
U_x & V_x & W_x & T_x \\
U_y & V_y & W_y & T_y \\
U_z & V_z & W_z & T_z \\
  0 &   0 &   0 &   1 \\
\end{matrix} \right ]$$
Then, you can calculate the transformation matrix $\mathbf{M}$ using
$$\mathbf{M} = \mathbf{M}_\text{after} \mathbf{M}_\text{before}^{-1}$$
noting that because of orthonormality,
$$\mathbf{M}_\text{before}^{-1} = \left[ \begin{matrix}
v_y w_z - v_z w_y & v_z w_x - v_x w_z & v_x w_y - v_y w_x & t_x (v_z w_y - v_y w_z) + t_y (v_x w_z - v_z w_x) + t_z (v_y w_x - v_x w_y) \\
u_z w_y - u_y w_z & u_x w_z - u_z w_x & u_y w_x - u_x w_y & t_x (u_y w_z - u_z w_y) + t_y (u_z w_x - u_x w_z) + t_z (u_x w_y - u_y w_x) \\
u_y v_z - u_z v_y & u_z v_x - u_x v_z & u_x v_y - u_y v_x & t_x (u_z v_y - u_y v_z) + t_y (u_x v_z - u_z v_x) + t_z (u_y v_x - u_x v_y) \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$$
but this also applies to non-orthonormal basis vector sets, as long as the initial vector $\vec{u}$ corresponds to final vector $\vec{U}$, initial vector $\vec{v}$ to final $\vec{V}$, initial $\vec{w}$ to final $\vec{W}$, they are all nonzero, and initial vector $\vec{t}$ corresponds to final vector $\vec{T}$.

The resulting transformation matrix $\mathbf{M}$ is applied to point $\vec{p} = ( x, y, z )$, transforming it to $\vec{q} = ( \chi , \gamma, \zeta )$, via
$$\left[\begin{matrix} \chi \\ \gamma \\ \zeta \\ 1 \end{matrix} \right] = \mathbf{M} \left[ \begin{matrix} x \\ y \\ z \\ 1 \end{matrix} \right]$$
and represents both rotation and translation: the upper left $3\times 3$ matrix of $\mathbf{M}$ is a rotation matrix, and upper right $3 \times 1$ vector is the translation after rotation.

These $4\times 4$ matrices are extremely common in 3D graphics, in e.g. OpenGL.


问:
Proof that if: a^2=b^2 then a=b or a=-b
I have to prove that if a^2=b^2 then a=b or a=-b, and I came up with the following and I am wondering if it is valid use of absolute values?:
a^2=b^2 |square root of both sides

|a|=|b| |particularly interested in whether or not this next step is valid:

a=b or a=-b

Thank you in advance.

Edit: I do understand that perhaps easier way to prove is the following:
a^2=b^2

a^2-b^2=0

(a+b)=0 or (a-b)=0

a=-b or a=b

But I am interested in whether or not my prior proof was valid.

答: That is correct, from the logical point of view. But it is incomplete. You proved correctly that, if $a^2=b^2$, then $|a|=|b|$. But now you still have to prove that if $|a|=|b|$, then $a=b$ or $a=-b$.

Title: Proof that if: a^2=b^2 then a=b or a=-b


Title: Proof that if: a^2=b^2 then a=b or a=-b

Body:
I have to prove that if a^2=b^2 then a=b or a=-b, and I came up with the following and I am wondering if it is valid use of absolute values?:
a^2=b^2 |square root of both sides

|a|=|b| |particularly interested in whether or not this next step is valid:

a=b or a=-b

Thank you in advance.

Edit: I do understand that perhaps easier way to prove is the following:
a^2=b^2

a^2-b^2=0

(a+b)=0 or (a-b)=0

a=-b or a=b

But I am interested in whether or not my prior proof was valid.

Body: $a^2=b^2$ means $a^2-b^2 = 0$ and so $(a-b)(a+b)=0$. Thus $a=b$ or $a=-b$.


问:
Inner product in hyperbolic space
I am trying to learn about Hyperbolic Spaces. I can't find information about the inner product in hyperbolic spaces. So the paper "Multi-relational Poincaré Graph Embeddings" from Balaževic et al. states: "... rely on inner product as a similarity measure and there is no clear correspondence to the Euclidean inner product in hyperbolic space". Whereas the paper "Neural Embeddings of Graphs in Hyperbolic Space" from Chamberlain et al. gives the following definition of an inner product of two vectors in polar coordinates in the Poincaré disk model: $$x = (r_x,\theta)\\y=(r_y,\theta)\\ \langle x,y \rangle = \lVert x \rVert \lVert y \rVert cos(\theta_x - \theta_y) = 4 \text{arctanh } r_x \text{arctanh } r_y  \cos(\theta_x - \theta_y) $$

I am confused whether there exists an inner product in hyperbolic space and if so, how is it defined for the Poincaré ball? I already tried to read introductions in Hyperbolic space but I did not find an answer. I would really appreciate your help! I am studying CS, so an intuitive explanation would be nice ;)

Thank you!

答:
The answer is "yes" and "no."

Why "no": The notion of an inner product is reserved for vector spaces and hyperbolic plane

${\mathbb H}^2$ does not have a natural vector space structure. This is especially clear if one uses an axiomatic definition of the hyperbolic plane, which you can find, for instance, in Greenberg's book
"Euclidean and Non-Euclidean Geometries."

Why "yes": Nevertheless, one can construct an "identification" (a mathematician would call it a "diffeomorphism") between the hyperbolic plane and a 2-dimensional vector space $V$. Such an identification requires a choice of the "origin" $o$ in the hyperbolic plane (which will correspond to zero in the vector space). In the case of the unit disk model, the convenient choice of the origin is the Euclidean center of the disk. In the paper you are reading, the identification is given by
$$
f: z= (x,y)\mapsto (d(o, z), \theta),
$$
where $d$ is the hyperbolic distance, $\theta$ is the angle and $(\rho, \theta)$ is understood as the polar coordinates of a vector. Now, the vector space $V$ has a natural inner product (called the "Riemannian metric" in differential geometry) and this is the one described in the paper you are reading (except they do it via an ad hoc method which obscures the procedure). A Riemannian geometer would call the inverse $\exp_o$ of the map $f: {\mathbb H}^2\to V$ the exponential map (the map $f$ itself is sometimes called the Riemannian logarithmic map).

Using $f$ one then "transfers" the inner product $V$ to  ${\mathbb H}^2$:
$$
\langle \exp_o(v), \exp_o(w)\rangle := \langle v, w\rangle.
$$
This is what the authors of the paper are doing.

What is this good for? I do not know, maybe the authors of the paper you are reading found some use for this construction.

In the literature you will likely encounter other notions of "products" on hyperbolic spaces, most importantly, Gromov product
$$
(z, w)_p= \frac{1}{2}(d(p,z)+ d(p,w)- d(z,w)).
$$
This will be also different from the one defined in your question.

Title: Inner product in hyperbolic space


问:
Probabilities with/without ordering are equal
During my math classes, the teacher always highlighted the importance of distinguishing between combinations and permutations. However, I recently noticed that when calculating probabilities with a sample space consisting of equally likely outcomes these two approaches yield the same result.

For instance, let the sample space $U$ consist of $N$ outcomes and consider some subset $A$ of $U$ consisting of $n$ outcomes. We draw $r$ outcomes without replacement from $U$.

Without ordering the probability of all $r$ elements being in $A$ is given by,
$$P(A_\text{no ordering}) = \frac{{n\choose r}}{{N \choose r}} = \frac{n!}{r!(n-r)!}\cdot \frac{r!(N-r)!}{N!} = \frac{n!}{(n-r)!}\cdot \frac{(N-r)!}{N!}$$

With ordering the probability of all $r$ elements being in $A$ is given by,
$$P(A_\text{ordering}) =\frac{n!}{(n-r)!}\cdot \frac{(N-r)!}{N!}$$

Thus the two probabilities are equal.

I am surprised that my teachers haven't pointed this out when studying counting principles and probability. Therefore, I am concerned whether I have misunderstood something or simply made a mistake. I am, of course, aware of the conceptional difference between number of combinations/permutations and that they are related to distinct questions.

I hope someone would take the time to give me their opinion on the matter.

Best regards,

Michael

答:
Yes, this is true. If you're taking $r$ samples from $U$ without replacement, and the event you want to know about doesn't care about their order, then you can use either ordered or unordered sampling.

In general, one or the other might be more straightforward. For example:

Say you are drawing two cards from a deck and you want to know the probability that the first is a spade, and the second is a heart. This talks about a "first card" or "second card", so we have to use ordered sampling: the probability is $\frac{13 \cdot 13}{P_2^{52}}$, where $P^n_k = \frac{n!}{(n-k)!}$.

Say you are drawing five cards from a deck and you want exactly two out of five to be hearts. You don't care which two are hearts, and don't want to care, so it's easier to use unordered sampling: the probability is $\frac{\binom{13}{2} \binom{39}{3}}{\binom{52}{5}}$.

You can always make a problem into an ordered sampling problem if it isn't already, but you might make the problem more complicated that way. In the second example I gave, we'd have to write $\frac{\binom 52 P^{13}_2 P^{39}_3}{P^{52}_5}$: the extra factor of $\binom 52$ counts the number of ways to pick which two cards are hearts.

But by far the most common mistake students make is to be inconsistent and use ordered sampling for part of the calculation and unordered sampling for the rest. This is always wrong.


问:
Prove that the covariance of 2 linear combinations of random variables is the linear combination of the covariances
How would you prove the following property about covariances?

I found it here:

https://www.probabilitycourse.com/chapter5/5_3_1_covariance_correlation.php

答:
It is enough to prove for suitable random variables $X,Y,Z$ and constant $a$ that:

$\mathsf{Cov}(X+Y,Z)=\mathsf{Cov}(X,Z)+\mathsf{Cov}(Y,Z)$

$\mathsf{Cov}(aX,Y)=a\mathsf{Cov}(X,Y)$

$\mathsf{Cov}(X,Y)=\mathsf{Cov}(Y,X)$

which can be done on base of definition of $\mathsf{Cov}$.

The first two bullets show that $\mathsf{Cov}$ is linear on first argument.

The third bullet makes it possible to show that this is also the case for the second argument.


问:
How can $n\log n +n $ is less than less than $n\log n$?
How can we show, methodically, that $n\log n + n \ll n\log n$ ?

![Image][1]

[1]: https://imgur.com/a/c0SeNQf “problem image”

答: You can apply the definition of $\ll$ for $C = 2$, as eventually $n \log n + n \leq 2(n \log n)$.


问:
Showing that an ideal is not a free $R-$module.
Here is the question I want to answer:

Let $R = \mathbb Z[\sqrt{-5}]$

Show that $I = (2, 1 + \sqrt{-5})$ is not a free $R-$module.

Hint: It helps to view $R$ as $\mathbb Z[x]/(x^2 + 5).$

My questions are::

1- How can I use the hint?

I have found here How to show that an $\mathbb{Z}[\sqrt{-5}]$-module is not free? how to show that it is not free as it says in the answer in that link the following:

It is generated by two elements so if it were free, it would be free of rank $1$ or $2$.

If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.

If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent.

2- for showing this "If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.", should I use the norm function?

Could someone push me in the right direction to answer those questions please?

答:
Hint: Any two elements of $R$ are linearly dependent. (There is an obvious relation. Can you see this?)

From this, can you conclude that an ideal is free only if it is principal?

Edit: To see the first part, for distinct $a, b \in R$, note that $$(-b)\cdot a + a \cdot b = 0$$
and not both $a$ and $-b$ are $0$.

The second part now follows because any $R$-basis of $I \subset R$ must now either be $\varnothing$ or a singleton.


问:
Natural measure on space of continuous functions
Let $C(I)$ to be the space of continuous functions on the interval $I=[0,1]$, we equip $C(I)$ with the norm $||f||=\sup_{x\in I}|f(x)|$. Is there is a "natural" Borel probability measure on $C(I)$ such that we have the following statement:

Almost every continous function $f\in C(I)$ is not differentiable.

答:
The Wiener measure should give you this, but it is unfortunately a priori cumbersome to define.
You first need Brownian motion. A stochastic process $(X_t)_{t \geq 0}$ is a family of random variables. We say that $X_t$ is a brownian motion if

$t \mapsto X_t(\omega)$ is continuous for every $\omega \in \Omega$

$B_0(\omega) = 0$ for almost all $\omega \in \Omega$

for all $0 = t_0 < t_1 < ... < t_n$, the random variables $B_{t_{i+1}} - B_{t_i}$ are idependent and Gaussian with expected value $0$ and variance $t_{i+1} - t_i$.

We then define the Wiener measure on $C([0,1])$ by
$$\mathcal{W}(M) = \mathbb{P}([t \mapsto X_t]\vert_{[0,1]} \in M) \qquad \forall M \in \mathcal{B}(C([0,1])).$$
What happens here intuitively is that the Brownian motion gives us a bunch of random paths, and we can measure the size of a set of paths by how likely it is that a given realization of the Brownian motion is in this set.
Since all paths generated by Brownian motion are nowhere differentiable, under this measure paths are indeed almost surely nowhere differentiable.


答:
Despite the notation, "$\ll$" is not asymmetric: we can have two functions $f,g$ such that $f\ll g$ and $g\ll f$ at the same time.

That's exactly what's happening here:

$n\log n\ll n\log n+n$ trivially (take $C=1$).

$n\log n+n\ll n\log n$ as well, however: taking $C=2$ we have for large enough $n$ that $$n\log n + n\le n\log n+n\log n=2n\log n=Cn\log n.$$

Note that the example right after the definition of $\ll$ also has this behavior: the text explains why $$7x^3-12x+9\ll x^3,$$ even though we obviously have $x^3<7x^3-12x+9$ for large enough $x$ (in this case consider $C=7$).

Title: How can $n\log n +n $ is less than less than $n\log n$?

Body:
How can we show, methodically, that $n\log n + n \ll n\log n$ ?

![Image][1]

[1]: https://imgur.com/a/c0SeNQf “problem image”


答:
You can prove the result $Cov(aX+bY,cW+dZ)=abCov(X,Y)+acCov(X,W)+bcCov(Y,W)+bdCov(Y,Z)$ as follows.

$$\begin{split}Cov(aX+bY,cW+dZ)&=E[(aX+bY-E(aX+bY))(cW+dZ-E(cW+dZ))]\\
&=E[(a(X-E(X)+b(Y-E(Y))(c(W-E(W)+d(Z-E(Z))]\\
&=E[ac(X-E(X))(W-E(W))]+E[ad(X-E(X))(Z-E(Z))+E[bc(Y-E(Y))(W-E(W))]+E[bd(Y-E(Y))(Z-E(Z))]]\\
&=acCov(X,W)+adCov(X,Z)+bcCov(Y,W)+bdCov(Y,Z)\end{split}$$

Repeated application of this gives the desired result.

Title: Prove that the covariance of 2 linear combinations of random variables is the linear combination of the covariances

Body:
How would you prove the following property about covariances?

I found it here:

https://www.probabilitycourse.com/chapter5/5_3_1_covariance_correlation.php


问:
Spheres in different dimension are not homotopy equivalent
Is there a way to prove that $\textbf{S}^n$ and $\textbf{S}^m$ are not homotopy equivalent if $n\neq m$ without using the machinery of homology or higher homotopy groups?</p>

答:
Just for fun, here's an answer that doesn't use the machinery you're talking about, at the expense of using crazier machinery.  Suppose you had a homotopy equivalence $S^n \stackrel{\sim}{\to} S^{n+k}$ for $k > 0$.  By smashing this with $S^{km}$, you get homotopy equivalences $S^{n+km} \stackrel{\sim}{\to} S^{n + (k+1)m}$ for each $k$.  Composing these gives you a sequence
$$ S^n \stackrel{\sim}{\to} S^{n+k} \stackrel{\sim}{\to} S^{n+2k} \stackrel{\sim}{\to} \dotsb $$
and taking the colimit gives you a weak equivalence $S^n \stackrel{\sim}{\to} S^\infty$.  But $S^\infty$ is contractible.  (And as with Andreas's answer, you'd then have to prove that spheres aren't contractible.)</p>


答:
IMO the use of the relational operator $\ll$ is awkward, as it usually means "much smaller than". The common notation would be

$$n\log n+n=O(n\log n).$$

This is true because

$$n\log n+n\le 2n\log n$$
or
$$1\le \log n$$

for $n\ge e$.


问:
Show for the centroid $M$ of a triangle $ABC$ that $\vec{OM}=\frac13(\vec{OA}+\vec{OB}+\vec{OC})$
Show for the centroid $M$ of a triangle $ABC$ that $$\vec{OM}=\dfrac13\left(\vec{OA}+\vec{OB}+\vec{OC}\right)$$
where $O$ is an arbitrary point.

I haven't studied position vectors and am very new to vectors so I would like a simple solution. Any help would be appreciated. I don't even know how to start. Maybe we can use the fact that the centroid divides each median in a ratio of $2:1$? Thank you!

答:
As $A_1$ is the midpoint of $BC$,

$2\vec{OA_1} = \vec{OB} + \vec{OC}$

As $M$ is the centroid of the triangle, it divides median $AA_1$ into ratio of $AM:MA_1 = 2:1$

$\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3} = \frac{\vec{OA}+\vec{OB} + \vec{OC}}{3}$

That is all the proof is.

But if you need to further show how $2\vec{OA_1} = \vec{OB} + \vec{OC}$,

$\vec{OA_1} = \vec{OB} + \vec{BA_1}$

But also, $\vec{OA_1} = \vec{OC} + \vec{CA_1} = \vec{OC} - \vec{A_1C}$

As $\vec{BA_1} = \vec{A_1C}$, adding both you have

$2\vec{OA_1} = \vec{OB} + \vec{OC}$

Similarly you can show $\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3}$.

$\vec{OM} = \vec{OA}+\vec{AM}$

As $\vec{AM} = \frac{2}{3} \vec{AA_1}$,

$\vec{OM} = \frac{3\vec{OA}+2\vec{AA_1}}{3} = \frac{\vec{OA}+2(\vec{OA}+\vec{AA_1})}{3} = \frac{\vec{OA}+2\vec{OA_1}}{3}$


问:
Number of repetitions of modular operation
I am looking into a problem that is about threaded pins on a circle. See image bellow for an example:

I approached it as follows:

Let $P$ be the number of pins and $G$ the number of gaps.

Number the pins on the circle from $0$ to $(P-1)$. The process of moving the thread from pin to pin is just a modulo operation on $P$. I.e. for $5$ pins and gap of $2$ we connect:

$0$ with $2$ i.e. $(0+2)\space mod \space 5$

$2$ with $4$ i.e. $(2 + 2)\space mod\space 5$

$4$ with $1$ i.e. $(4 + 2)\space mod\space5$

$1$ with $3$ i.e. $(1 + 2)\space mod\space5$

$3$ with $0$ i.e. $(3 + 2)\space mod\space 5$

That requires 1 thread since we reached 0. This process can be applied to all the rest of the configurations.

For 6 pins and 2 gaps we connect:

$0$ with $2$ i.e. $(0 + 2)\space mod \space 6$

$2$ with $4$ i.e. $(2 + 2)\space mod \space 6$

$4$ with $0$ i.e. $(4 + 2)\space mod \space 6$

That was thread $1$ and we start again with pin $1$ since we got back to pin $0$.

We connect:

$1$ with $3$ i.e. $(1 + 2)\space mod \space 6$

$3$ with $5$ i.e. $(3 + 2)\space mod \space 6$

$5$ with $1$ i.e. $(5 + 2)\space mod \space 6$

That was thread $2$. I.e. we need $2$ threads as $2$ repeats of the process happened.

So each time we just wrap around the modulo $P$ and increment by $1$ and start over. The number of repetitions equals the number of threads

My problem is how to formalize the process in a way to shows that the number of threads increments based on this modulo operation.

Note: There is a post about the same problem here: Threaded pins but my question is different.

I do understand the solution about using the greatest common divisor. What I am looking for is how could I extend the modulo process I have described to reach the solution. Or is gcd also related in this approach too?

答:
The answer to your final question is no - it is a nicely presented question.

Your modulo method returns to $0$ when $$0,G,2G,3G,...,kG$$ reaches $0$ i.e. $$kG\equiv 0\pmod P.$$

This requires $$k=\frac{P}{\gcd(P,G)}.$$

Therefore the number of threads you require is $$\frac{P}{k}=\gcd(P,G).$$

Thus the methods are the same.


问:
Question about vector spaces involving polynomials
Consider the set $V$ of real polynomials $f$ of degree at most
3 which satisfy $f(1) = f(2)$. Define addition and scalar multiplication
of such polynomials by $(f +g)(x) = f(x)+g(x)$ and $(αf)(x) = α·f(x).$

Show that the following polynomials $f_1(x) = 1$, $f_2(x) = x^2 − 3x$ and $f_3(x) = x^3 − 7x$ form a basis of $V$, and is the polynomial $g(x) = 3 + x + 2x^2 − x^3$
in $V$ ?

If yes, what
are the coordinates of g with respect to the basis in above?
What is the dimension of V ?

Progress:
I am thinking of finding the RREF of the matrix associated with the 3 polynomials, and if it gives the identity matrix, then it is a basis. To check if g is in $V$, should I be writing $g(x)=\alpha f_1 + \beta f_2 + \gamma f_3$ then solve for the scalars? But how do I find the coordinates of $g$ with respect to the basis above and also the dimension of V?
Is $dim_v$= $4$?

Can anyone confirm my method in doing this question? Thanks!

答:
Consider the defining properties of a basis: all the proposed basis vectors are orthogonal to each other, i.e. are not scalar multiples, and there are the same number of them as there is dimension to the space.

By observation, the three functions you've been given are all orthogonal - they each have distinct powers of $x$ so it's impossible to scale any of them to form any of the others. You have three of them, but it's not obvious that $V$ is only three dimensional.

To demonstrate that $V$ is actually three dimensional rather than four, you can interpret the constraint as defining one of the four coefficients in terms of the other three (freely variable) ones, meaning you only have three choices. Specifically, we can imagine expanding out $f(x)=\alpha x^3 + \beta x^2 + \gamma x + \delta$:

$$\alpha 1^3+\beta 1^2+\delta 1+\gamma = \alpha 2^3+\beta 2^2+\delta 2+\gamma$$
$$\alpha +\beta + \delta + \gamma = 8\alpha +4\beta +2\delta +\gamma$$
$$\delta = -7\alpha -3\beta $$

(The choice of solving for $\delta$ was arbitrary, I just chose that because I expected the result to be neater; $\gamma$ disappearing entirely isn't a problem, it just means it is both unconstrained by and has no influence on the value of the other three)

The "coordinates of g with respect to the basis" means, what scalars do we have to multiply the polynomials $f$ by so that the sum of them is $g$? Constructing $g = Af_1 + Bf_2 + Cf_2 $ and solving for the coefficients should produce the right answer and immediately means e.g. $A=3$


问:
Permutation involving 4 cups and 4 saucers.
I study maths as a hobby.  I have come across this problem:

On a shelf there are 4 saucers of different colours and 4 matching cups. In how many ways can the cups be arranged on the saucer so that no cup is on a matching saucer?

I start off by saying the first cup can be placed on any of 3 saucers.  For the second cup, there are 3 choices, unless the first cup was placed on the second cup’s matching saucer, in which case there are only 2 choices.  That gives 8 outcomes so far.  But the answer in the book is 9.  So I know my method is wrong.

I have seen similar problems posted on here but the solutions were too complex for me.

答:
There are $3$ cups on which you can place on saucer $4$.  For concreteness, let's say you put cup $3$ there.  Now, you might place cup $4$ on saucer $3$, and then only cups $1$ and $2$ and saucers $1$ and $2$ remain, and there's only one way to place the cups on non-matching saucers, so in this case we have $3$ possibilities.

Alternatively, cup $4$ might be placed on either saucer $1$ or saucer $2$.  Now we still have to place cups $1$ and $2$ and saucers $2$ and $3$ are available.  Again, there is only one way to place the cups on non-matching saucers, so we have $2\cdot3=6$ possibilities in this case, and $9$ overall.

The generalization of this line of reasoning to $n$ cups and saucers is given here.


问:
Strong operator topology limit of projections
Let $\mathcal{H}$ be a Hilbert space and $r$ be a fixed natural number. Suppose $\{P_n\}_{n=1}^\infty$ be a sequence of projections on $\mathcal{H}$ with $\text{dim ran }P_n<r$ for all $n$. Let $P$ be another projection on $\mathcal{H}$ such that $P_n\rightarrow P$ in SOT. Then show that $\text{dim ran }P<r$.

Comments: My attempts are mainly in the following two ways so far:

Suppose $\mathcal{H}$ be separable Hilbert space with orthonormal basis $\{e_n:n\geq 1\}$. Let $f\in\text{ran}P$. Now $f=\sum\limits_{n\geq 1}\langle f,e_n\rangle e_n$. Using the SOT convergence, I was trying show $\langle f,e_n\rangle=0$ for all but finitely many $n$ and those finitely many could be atmost $r$. But I could not be successful by this approach.

I assumed that $\text{dim ran}P>r$ and trying to get a contradiction using the SOT limit of $P_n$ but could not be successful here also.

Any comment is highly appreciated. Thanks in advance.

答:
Arguing by contradiction,  if $\text{dim ran } P\geq r$, there exists  an orthonormal set
$$
  \{e_i:1\leq i\leq s\}\subseteq  \text{ran } P.
  $$
Observing that
$$
  \lim_{n\to \infty }P_n(e_i) = P(e_i) = e_i,
  $$
we see that  the coeficients of the matrix
$$
  A_n=\Big (\langle P_n(e_i), P_n(e_j) \rangle \Big)_{i, j=1}^r
  $$
can be made as close as desired to the coefficients of   $\big (\langle e_i, e_j \rangle \big )_{i, j=1}^r$, a.k.a. the identity
$r\times r$ matrix.

Since the invertible matrices form an open set,  we may choose $n$ such that $A_n$ is invertible,  and this implies that
$\{P_n(e_i):1\leq i\leq r\}$ is a linearly independent set,  contradicting the assumption that $\text{dim ran } P_n<r$.


问:
Push-through rule for square root of specific projection matrix
Is the following a valid identity?

$$
(I - XX^T)^{1/2} X = X (I - X^T X)^{1/2}
$$

Without the square root, it is quite easy to see that

$$
(I - XX^T) X = X (I - X^T X)
$$

holds but I have trouble verifying that the upper equation holds.

答:
A very general rule exists for these things.

Let $A,B\,$ be any two matrices such that the product $AB$ is a square matrix, and let $f(z)$ be an analytic function of a scalar argument.

Then the following is true
$$f(AB)\cdot A = A\cdot f(BA)$$

The proof consists of a series expansion of the function coupled with the associativity of matrix multiplication.

In this particular case, $\,A=X,\,B=X^T\,$ and  $\,f(z) = \sqrt{1-z}$


问:
limit of a sequence of integral functions
I have to study punctual and uniform convergence of the following succession
$$f_n(x)=n\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt.$$
I showed that the functions $f_n$ are uniformly convergent in $\mathbb R$ but I'm not able to calculate explicitly the limit function $f(x):=\underset{n\to +\infty}{\lim}f_n(x)$.

My idea was to split the integral in half and use the characteristic function of the domain of integration but this seems to be inconclusive.

Thank you for your help and your time.

Body:
$$f_n(x)=\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt=\frac{1}{2} \sqrt{\pi }
   \left(\text{erf}\left(\frac{1}{n}-x\right)+\text{erf}\left(\frac{1}{n}+x\right)\right)$$ Using the infinite series representation of $\text{erf}(t)$ and continuing with Taylor series, when $n$ is large
$$f_n(x)\sim g_n(x)=\frac{2 e^{-x^2}}{n}\Bigg[1+\frac{2 x^2-1}{3 n^2}+\frac{4 x^4-12 x^2+3}{30
   n^4}+O\left(\frac{1}{n^6}\right) \Bigg]$$

For example, using $x=\pi$ and $n=2^k$
$$\left(
\begin{array}{cc}
 k & \frac {f_n(x)} {g_n(x)} \\
 0 & 1.28419669254952350166175610011 \\
 1 & 1.02205560798133497330954918511 \\
 2 & 1.00074377793925422180746341333 \\
 3 & 1.00001499797854822520136495734 \\
 4 & 1.00000025126261683720744981856 \\
 5 & 1.00000000399669102804908420208 \\
 6 & 1.00000000006272933330307433397 \\
 7 & 1.00000000000098124840982741539 \\
 8 & 1.00000000000001533631803233576 \\
 9 & 1.00000000000000023964681613758 \\
 10 & 1.00000000000000000374454731476 \\
 11 & 1.00000000000000000005850880888 \\
 12 & 1.00000000000000000000091420114 \\
 13 & 1.00000000000000000000001428440 \\
 14 & 1.00000000000000000000000022319 \\
 15 & 1.00000000000000000000000000349 \\
 16 & 1.00000000000000000000000000005 \\
 17 & 1.00000000000000000000000000000
\end{array}
\right)$$


问:
Basic computation in Steenrod algebra
In Homology operations for $H_\infty$ and $H_n$ spectra (pdf), Steinberger makes the computation of the Dyer-Lashof operations in $H\mathbb F_p$, and at some point uses the following "basic fact"

Lemma 6.1. The following equalities hold in $\mathcal A_*$. For $p\geq2$ and $i\geq0$,
$$P^r_*\chi\xi_i=\left\{\begin{array}{ll}-\chi\xi_{i-k}^{p^k}&\text{if }r=\frac{p^k-1}{p-1}\\0&\text{otherwise}\end{array}\right.$$
[...]

I have tried my best to show it by hands but I couldn't manage to make the computation myself. Also I've been looking for a reference without success...

Would someone know a reference where this is made explicit ?

答:
Let me give this a shot.  I'm bound to mess up signs, so assume $p = 2$.  Write $\langle \cdot, \cdot \rangle$ for the pairing between $\mathcal{A}_*$ and $\mathcal{A}^*$.

Lemma. $\newcommand{\Sq}{\mathrm{Sq}}\langle \chi \xi_k, \Sq^r \rangle = \delta_{r = 2^k-1}$.

Proof. We have
\begin{align}
\langle \chi \xi_k, \Sq^r \rangle &= \langle \xi_1^{2^k-1} \bmod{(\xi_2, \xi_3, \ldots)}, \Sq^r \rangle \\
&= \langle \xi_1^{2^k-1}, \Sq^r \rangle \\
&= \delta_{r = 2^k-1}.
\end{align}

Proposition. $\Sq^r_* \chi \xi_i = \begin{cases} \chi \xi_{i-k}^{2^k}, & \text{if } r = 2^k - 1 \\ 0, & \text{otherwise}. \end{cases}$

Proof. For any $x \in \mathcal{A}^*$, we have
\begin{align}
\langle \Sq^r_* \chi \xi_i, x \rangle &= \langle \chi \xi_i, \Sq^r x \rangle \\
&= \langle \chi \xi_i, \mu(\Sq^r \otimes x) \rangle \\
&= \langle \xi_i, \chi \mu(\Sq^r \otimes x) \rangle \\
&= \langle \xi_i, \mu(\chi x \otimes \chi \Sq^r) \rangle \\
&= \langle \Delta \xi_i, \chi x \otimes \chi \Sq^r \rangle \\
&= \langle \sum_k \xi_{i-k}^{2^k} \otimes \xi_k, \chi x \otimes \chi \Sq^r \rangle \\
&= \sum_k \langle \xi_{i-k}^{2^k}, \chi x \rangle \cdot \langle \xi_k, \chi \Sq^r \rangle \\
&= \sum_k \langle \chi \xi_{i-k}^{2^k}, x \rangle \cdot \langle \chi \xi_k, \Sq^r \rangle \\
&= \sum_k \langle \chi \xi_{i-k}^{2^k}, x \rangle \cdot \delta_{r=2^k-1} \\
&= \begin{cases} \langle \chi \xi_{i-k}^{2^k}, x \rangle, & \text{if } r = 2^k - 1 \\ 0, & \text{otherwise}. \end{cases}
\end{align}

Title: Basic computation in Steenrod algebra


问:
Isomorphism vs equality of graphs
I have just started studying graph theory and having trouble with understanding the difference b/w isomorphism and equality of two graphs.According what I have studied so far, I am able to conclude that isomorphic graphs can have same diagrams when represented on paper, but equal graphs also have same diagram on paper, if that is so, then what is the difference b/w equality and isomorphism.   </p>

<p>It would be really helpful, if a person could throw some light on what are the difference b/w the above two terms. Specifically, any example of an isomorphic graph which is not equal. </p>

答:
I am comming back to this old question for a reason and I would like to add some thoughts which might be helpful for this discussion.

This answer is inspired by a comment in response to this question.

Let me eleborate some context first.

All considered graphs are finite, simple and undirected.

Def. 1: Let $G=(V,E)$ and $G'=(V',E')$ be two graphs. We call $G$ and $G'$ isomorphic, and write $G \simeq G'$, if there exists a bijection $\varphi: V \to V'$ with $(x,y) \in E \iff (\varphi(x), \varphi(y)) \in E', \forall x,y \in V$. Such a map $\varphi$ is called an isomorphism from $G$ to $G'$.

Let us call a graph which has a fully labelled vertex set, a concrete graph.

Now, in most graph theoretic paper, something like the following is stated all the time and it seems very clear at first.

Given two graphs $G$ and $G'$.

A graph $G$ has property $x$ iff $G$ is isomorphic to $G'$.

What we mean is that $G'$ is an abstract representative of some graph for which we ignore concrete (vertex) labeling. $G$ on the other hand can be considered as fully labelled. In that sense: we plug in a concrete graph $G$ and prove isomorphism to $G'$ by creating a concrete labeling for $G'$ a posteriori and we can do that in general for any concrete graph $G$ having property $x$. Usually in the proof the labelling of $G'$ becomes equal to the one in $G$ (because if not, we must define and use the isomorphism map all the time, which is usually omitted). This is in fact found very often. Assume $G'$ is isomorphic to a cycle. You see statements like "Let $G$ be such a graph having property $x$ [... show some stuff ...] then $v_1, \cdots, v_n$ form a cycle, which concludes the proof". But what we mean is: $v_1,\cdots,v_n$ induce a graph isomorphic to a cycle. But what cycle? What isomorphism? The obvious answer is that $v_1,\cdots,v_n$ form the cycle itself.

We might ask "what is graph equality in the first place"? Consider a graph $G=(V,E)$ and $G'=(V',E')$ and assume that $G \simeq G'$ with given isomorphism map $\varphi: V \to V'$, then we can say that $G$ and $G'$ are equal if and only if $V=V'$. The isomorphism map $\varphi:V \to V'$ becomes $\varphi: V \to V$ and is obviously the identity mapping. We can write $G=G'$. In that sense graph equality is just some special isomorphism.

But in fact this scheme is found very inconsistently and often language is abused. Let us say again that $G'$ is a cycle (or should I say "is isomorphic to a cycle"?). Then, it will not take a long time to find a similar statement:

A graph $G$ has property $x$ iff $G$ is a cycle.

What the author really means is: iff $G$ is isomorphic to a cycle. This kind of abuse of language is found very often and is wideley accepted. However, we never say the following (and now for obvious reasons)

A graph $G$ has property $x$ iff $G$ is equal to a cycle.

Let me provide another example which could make the reasoning clear. Given a graph $H$. We call a graph $G$ $H$-free if no induced subgraph of $G$ is isomorphic to $H$. Again, I consider $H$ to be some kind of abstract and in that sense unlabelled graph. We write isomorphic in this setting to underline the fact that we could "plug in" any concrete graph $G$ and define "$H$-freeness". But why is it different than writing: $G$ is $H$-free if no induced subgraph of $G$ is equal to $H$? Because this would be drastically false. Each induced  subgraph $X$ of $G$ could only be equal to $H$ if $X \simeq H$ and $V(H)=V(X)$. In that sense we fixed the vertex set of $H$ to be equal to $V(X)$. But each induced subgraph of $G$ differs in at least one vertex, hence $V(H) \neq V(X')$ for an induced subgraph $X'$ of $G$ with $V(X) \neq V(X')$. We can not define $H$-freeness without the notion of isomorphism.

Title: Isomorphism vs equality of graphs


答: Place cup A on saucer B say, there are $3$ ways to do this. Place cup B on any remaining saucer, there are again $3$ ways to do this. The last $2$ cups are forced onto specific saucers, hence $9$ ways in total.

Title: Permutation involving 4 cups and 4 saucers.

Body:
I study maths as a hobby.  I have come across this problem:

On a shelf there are 4 saucers of different colours and 4 matching cups. In how many ways can the cups be arranged on the saucer so that no cup is on a matching saucer?

I start off by saying the first cup can be placed on any of 3 saucers.  For the second cup, there are 3 choices, unless the first cup was placed on the second cup’s matching saucer, in which case there are only 2 choices.  That gives 8 outcomes so far.  But the answer in the book is 9.  So I know my method is wrong.

I have seen similar problems posted on here but the solutions were too complex for me.


问:
Probability to have at least two kings when drawing 13 cards from a standard deck. What is wrong with reasoning?
I know there are $\binom{52}{13}$ ways to pick 13 cards from a standard deck.

There are $\binom{4}{2}$ was to pick the two kings. Then I reasoned there are 50 cards remaining (52 cards minus the two kings) to pick 11 cards. This results in $\frac{\binom{4}{2}\binom{50}{11}}{\binom{52}{13}}$. However, this answer is wrong according to my notes. But what is wrong with my reasoning?

The correct answer should be $\sum_{i=2}^{4} \frac{\binom{4}{i} \binom{48}{13-i}}{\binom{52}{13}}$. I understand the reasoning behind this solution, i.e. the probability to have at least 2 kings is equal to the sum of the probability of having 2, 3 or 4 kings.

Thanks in advance.

Edit: I also simulated this problem and the simulation agreed with the answer in the notes.

答:
The product $\binom 42 \binom{50}{11}$ counts the outcomes in two stages: first, you pick the two kings, and then, you pick the other cards. This would be a correct way to count if every outcome was obtained exactly once when we do this.

But an outcome with three or more kings is obtained multiple times. If you draw the king of hearts, king of diamonds, and king of spades (and ten more cards), then we can obtain that by:

picking the king of hearts and king of diamonds in the first stage, then the other cards in the second stage;

picking the king of hearts and king of spades in the first stage, then the other cards in the second stage;

picking the king of diamonds and king of spades in the first stage, then the other cards in the second stage.

So it is triple-counted. An outcome with all four kings is counted six times.


问:
In a reflexive Banach space, weak and weak* convergences are equivalent

Let $X$ be are flexive Banach Space. Then weak and weak* convergences are equivalent.

Let $(f_n) \subset X'$, $f\in X'$ and $$f_n \xrightarrow{w}f,$$ then $$f_n  \rightarrow{w*} f.$$ I'm trying to show that the convergence of $(f_n)$ to $f$ in a weak* topology implies a convergence in a weak topology.

I've tried this:

Let $g \in X''$. X is reflexive so $X'' = X $. Then $(g(f_n)) \subset X$ is Cauchy. Since $X$ is Banach space, $(g(f_n))$ converges.

But I wasn't able to show that $g(f_n)\rightarrow g(f)$.

答: If $X$ is reflexsive $X''=X$ that means $\forall x \in X \Rightarrow x\in X''$.Also $f_n \rightarrow^{w*}  f$ then $\forall x \in X$ ; $f_n(x)\rightarrow f(x)$.$X$ reflexsive then $x:X' \rightarrow \Bbb{F}$ $\forall h\in X'$ ; $h(x)=x(h)$ is a linear functional$X''$. Hence $\forall x\in X''$; $x(f_n) \rightarrow x(f) $ because we known $x(f_n)=f_n(x)\rightarrow x(f)=f(x)$ so that says $(f_n)$weak convargence $f$


问:
How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $

$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$

I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.

答:
Note that $7x-10-x^2 = \frac94-(\frac72-x)^2$. So, substitute  $\frac72-x=\frac32\sin t$ to integrate as follows
\begin{align}
& \int \frac{dx}{(x-2)(1+\sqrt{7x-10-x^2})}\\
= & - \int \frac{2\cos t}{(1-\sin t)(2+3\cos t)}dt\\
= & - \int \left ( \frac{\cos t}{1-\sin t}-\frac {3\sin t}{2+3\cos t} - \frac {3}{2+3\cos t} \right)dt\\
=& \>\ln(1-\sin t) -\ln(2+3\cos t)+\frac3{\sqrt5}\tanh^{-1}\frac{\tan\frac t2}{\sqrt5}+C
\end{align}


问:
The unit interval homomorphic to the unit circle
Is $[0,1]$ homomorphic to $S^1=\{x^2+y^2=1:x,y \in \mathbb{R}\}$?

I've seen a function $g:[0,1) \rightarrow S^1$ defined by
$g(x)=(\cos 2x\pi,\sin 2x \pi)$, with the question, is $g$ is homomorphism?

$g$ is clearly continuous, but I can not see weather it is a bijection and then has continuous inverse.

If the answer of the above question is No, I wont need to work with the above $g$ function. Thanks for any help

答:
It is a bijection but the inverse isn't continous. At $1 \in S^1$ you have the only discontinuity.

By the way. At first you say $[0,1]$, later $[0,1)$. Only the latter one makes the function injective.


问:
Poincaré theorem about the unit ball and the unit polydisc
In Rudin's book "Function theory in the unit ball of $\mathbb{C}^n$", it is proven that the open unit ball $\mathbb{B}^n$ and the open unit polydisc $\Delta^n$ are not biholomorphically equivalent, because otherwise there would be a linear biholomorphism between them, and invertible linear transformations map balls onto ellipsoids.

Now, my question is whether there is a simple way to show that a cartesian product of discs cannot be the image of the unit ball by means of an invertible linear transformation. In particular, I am looking for something that does not involve the fact that the ball has a smooth boundary, while the polydisc does not. The whole point of Rudin's proof is to use only linear maps.

答:
It seems like I have found a satisfactory answer. As Rob and Giuseppe suggested, it involves convexity. The proof I give is taken from Kaup's book "Holomorphic functions of several variables", page 10.

If $F:\mathbb{B}^n\rightarrow\mathbb{\Delta}^n$ is a linear bijection, then it is defined on all of $\mathbb{C}^n$, hence $F(\partial\mathbb{B}^n)=\partial\Delta^n$ (this follows in fact from a quite elementary argument). But $\partial\Delta^n$ includes the real segment $\{(1,t,0, \dots,0):t\in[-1,1]\}$. Since $F^{-1}$ is linear, $\partial\mathbb{B}^n=F^{-1}(\partial\Delta^n)$ must contain a real segment as well, which is clearly impossible.

Thanks for all the kind suggestions!


问:
Modify the rules of Gomoku (Five-in-a-row) or Connect Four type games to enforce the fairness among players
One colleague and me were discussing this problem during lunch today, and I did a little bit digging for several hours after returning to my office.

Fact: For an $(m,n,k)$-game, there does not exist a strategy that assures that the second player will win. For example, in Five-in-a-row, first player wins with perfect play.

Now our question is: How to modify the rules to make the game fair for both players?

I googled a bit more, found an MSE question that addresses the fairness of this kind of games.

I think the fairness in that question can be strenghthen by the following definition: A game is fair if and only if that two players play with perfect strategy, the game will always be a draw. This is fair in that this game favors player with less mistakes. For example, Tic-tac-toe is fair.

Several proposals to modify the rules to enforce the fairness:

Impose more restrictions on the player who plays first. For example, Five-in-a-row bans black to play "three and three", "four and four", and "overlines".

Change the $(m,n,k)$-game to an $(m,n,k,p,q)$-game: $k$-in-a-row on an $(m\times n)$-board, first player put $p$ stones on board, in subsequent moves, players put $q$ stones on board. For example, [Connect 6].

Go to higher dimension. For example, Five-in-a-row played in a $19\times 19\times 19$ "board".

I am no expert in combinatorial game theory or computational complexity in board games, but it is always nice to learn some new things in summer time. My question is: Any papers or treatises dealing the fairness of this kind of games? or more specifically, any proof of the fairness using above three modifications of rules? especially, for Five-in-a-row, do those additional rules make the game fair?

Any analysis of fairness on simpler cases like Connect 4 is welcome too.

Lastly just out of curiousity, as an avid Go player myself, I wonder is there any mathematical analysis on addressing the fairness using komi in the Go game? Or Go is just too complex to analyze...

答:
Here's one option of a type you didn't suggest.

Take any symmetric game.  Modify the game as follows:  After the first player's first move, give the second player the option to continue the game as usual, or to switch places, taking the first player's position and giving the (former) first player the next move as second player.

Assuming that ties are possible, best play on both sides must produce a tie.  The first player will not make a first move that guarantees him a win, since if he did, the second player would just switch places and take the winning position.  And nor will the first player hand the second player a win by making a bad first move.  Best play for the first player is to play into a position that he can tie but not win, and then play for the tie. If the second player switches with him, he can still play for the tie.


问:
If $f\,:\,A\to A$ is surjective and not injective then what can I say about $A$'s size?
Studying for my finals in Logic. Although the question is about Set Theory. In one of the previous exams, there was the following statement:

If $f\,:\,A\to A$ is injective and not surjective then the size of $A$ is infinite.

It's been a while sense I last had to deal with Set Theory. Then I saw this post If $f:A\rightarrow A$ is injective but not surjective then $A$ is infinite. which cleared some things. But I could not find explanations of the following questions:

If $f\,:\,A\to A$ is surjective and not injective then what can I say about $A$'s size? is it finite?

If $f\,:\,A\to A$ is surjective and injective then what can I say about $A$'s size? is it infinite? I guess it's more general case the the explained one so $A$ has to be infinite. Is it true?

答:
If $f: A \rightarrow A$ is surjective and not injective then $A$ must be infinite. To see this let take for example the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. If you construct $f$ in such a way that it assigns even two (or more) elements to one element there is no way to make it surjective. It can be shown more formally using induction, I think.

But if $A$ is infinite then it is possible to construct $f$ so that it would be surjective and not injective. Simple example: let us consider $A = N_+$. Then let's define $f$ in this way:

$f(1) = 1$

$f(2) = 1$

$f(3) = 2$

$f(4) = 3$

$f(5) = 4$

$.....$

Nothing can be said from the fact that $f: A \rightarrow A$ is surjective and injective. For identity is always surjective and injective and you can define identity on any set.


答:
If $f: A \to A$ is surjective and not injective then $A$ must be infinite. This is a direct consequence of the 'pigeonhole principle', which says that if $A$ and $B$ are two finite sets with the same number of elements and $f:A \to B$, then $f$ is an injection $\iff$ $f$ is a surjection. Thus, if you suppose that $A$ is finite, you get an immediate contradiction through the existence of this function $f$.

As for your second question if $f$ is both surjective and injective you can tell nothing about the cardinality of $A$. For any set $A$ there is such a function (the identity for instance), and this holds irrespective of the cardinality of $A$.

Body:
Studying for my finals in Logic. Although the question is about Set Theory. In one of the previous exams, there was the following statement:

If $f\,:\,A\to A$ is injective and not surjective then the size of $A$ is infinite.

It's been a while sense I last had to deal with Set Theory. Then I saw this post If $f:A\rightarrow A$ is injective but not surjective then $A$ is infinite. which cleared some things. But I could not find explanations of the following questions:

If $f\,:\,A\to A$ is surjective and not injective then what can I say about $A$'s size? is it finite?

If $f\,:\,A\to A$ is surjective and injective then what can I say about $A$'s size? is it infinite? I guess it's more general case the the explained one so $A$ has to be infinite. Is it true?


答: The median passing through $A$ has equation$$r=a+s((b+c)/2-a)=(1-s)a+(s/2)b+(s/2)c.$$The median through $B$ is$$r=(t/2)a+(1-t)b+(t/2)c.$$These meet at$$s=t=2/3,\,r=(a+b+c)/3.$$This point lies at $u=2/3$ on the third median,$$r=(u/2)a+(u/2)b+(1-u)c.$$

Title: Show for the centroid $M$ of a triangle $ABC$ that $\vec{OM}=\frac13(\vec{OA}+\vec{OB}+\vec{OC})$

Body:
Show for the centroid $M$ of a triangle $ABC$ that $$\vec{OM}=\dfrac13\left(\vec{OA}+\vec{OB}+\vec{OC}\right)$$
where $O$ is an arbitrary point.

I haven't studied position vectors and am very new to vectors so I would like a simple solution. Any help would be appreciated. I don't even know how to start. Maybe we can use the fact that the centroid divides each median in a ratio of $2:1$? Thank you!


答: As @principal-ideal-domain explains, your map $g$ is not a homeomorphism. In fact there cannot be a homeomorphism (for both cases). If you are interested in the case $[0,1)$, then we can argue that $[0,1)$ is not compact while $S^1$ is. For the case $[0,1]$ we could for example say that if we remove a point from the interior of $[0,1]$ then the interval becomes disconnected but this doesn't happen when we do that to $S^1$ (this argument also works for $[0,1)$.

Title: The unit interval homomorphic to the unit circle

Body:
Is $[0,1]$ homomorphic to $S^1=\{x^2+y^2=1:x,y \in \mathbb{R}\}$?

I've seen a function $g:[0,1) \rightarrow S^1$ defined by
$g(x)=(\cos 2x\pi,\sin 2x \pi)$, with the question, is $g$ is homomorphism?

$g$ is clearly continuous, but I can not see weather it is a bijection and then has continuous inverse.

If the answer of the above question is No, I wont need to work with the above $g$ function. Thanks for any help


问:
Searching for a counter example for this statement.
Let $A\in \mathbb{C}^{n\times n}$ be similar to $A^2$, then $A=A^2$.

This was in a multiple choice question, I'm trying to construct a counter example, but it's getting complicated for me since I'm trying to use upper triangular or diagonal matrices, I'm trying to fill in the diagonal with different numbers so I can get different eigen values which makes my life easier to decide similarity between $A$ and $A^2$. but It's not really working for me.

I would appreciate any help, and it would mean alot to explain to me your steps of constructing the counter example.

Thanks in advance.

答:
First notice, that the Eigenvalues of $A^2$ are the squares of the Eigenvalues of $A$. Also, since $A$ and $A^2$ are similar, they have the same Eigenvalues. Or, to rephrase, if we take $\Lambda \subseteq \mathbb{C}$ to be the set of Eigenvalues of $A$, we have $\{\lambda^2 | \lambda \in \Lambda\} = \Lambda$.

Now, try to find a finite set $\Lambda$ with this property and containing at least two values:
First, since the set is finite, all elements need to have norm $1$. Visualizing these as points on the unit circle, you can find $\Lambda = \{ e^{\frac{2\pi i}3}, e^{\frac{4\pi i}3} \}$.

Thus you get $A = \begin{pmatrix}
e^{\frac{2\pi i}3} & 0 \\
0 & e^{\frac{4\pi i}3}
\end{pmatrix}$ as a counterexample.


问:
Interchange of set theoretical limit
Find for what type of real valued measurable functions g, the equality holds:
$$ \lim_{n \to \infty} \{x: |g(x)|>a-1/n\}= \{x: |g(x)| \geq a\}$$ where $a \geq 0$.

答:
Let $a \geq 0$ and $g: \Omega \to \mathbb{R}$ any function. I don't think we need measurability here since we do not measure anything.

Set $A_{g,n} := \{x: |g(x)| > a - 1/n\}$ and $A_g := \{x: |g(x)| \geq a\}$. For the sequence of sets $(A_{g,n})_{n \geq 1}$, we have two notions of limits:

$\overline{A}:= \limsup_{n \to \infty} A_{g,n} := \bigcap_{n \geq 1} \bigcup_{k \geq n}A_{g,k}\quad$ and

$\underline{A} := \liminf_{n \to \infty} A_{g,n} := \bigcup_{n \geq 1} \bigcap_{k \geq n}A_{g,k}$

If $\overline{A} = \underline{A}$, we call this set $A$ and write $A = \lim_{n \to \infty} A_{g,n}$.

For all $n \geq 1$, it holds that $A_{g,n+1} \subset A_{g, n}\,$, i.e.  $(A_{g,n})_{n \geq 1}$ is nonincreasing. In this case, we have (why?)

$$
{\displaystyle \liminf _{n\rightarrow \infty }A_{g,n}=\bigcup _{n\geq 1}\bigcap _{k\geq n}A_{g,k}=\bigcap _{k\geq 1}A_{g,k}=\bigcap _{n\geq 1}\bigcup _{k\geq n}A_{g,k}=\limsup _{n\rightarrow \infty }A_{g,n}.}
$$

So, $\overline{A} = \underline{A}$ and  $A = \lim_{n \to \infty} A_{g,n} = \bigcap_{k \geq 1} A_{g,k}$. But this intersection exactly yields $A_g$.

Title: Interchange of set theoretical limit


问:
Application of Colorful Caratheodory for Directed Cycles
I'm working through Irme Barany's convexity lectures (http://wiki-math.univ-mlv.fr/gemecod/lib/exe/fetch.php/barany_lecture_2.pdf) and I am struggling to prove an exercise the author left to the reader. I'll include the context and my thoughts on it below, though you are also welcome to just read the proof of Theorem 5 in the linked document.

Having just introduced the Colorful Caratheodory's Theorem, Barany presents the following theorem as an application:

Let $G(V,A)$ be a directed graph with $|V|=n$ and let $C_1, \ldots, C_n$ be directed cycles in $G$. Then there exist arcs $a_i \in C_i$ such that the set $\{a_1, \ldots, a_n\}$ contains a directed cycle.

The proof goes as follows (paraphrased):

Let $V = \{v_1, \ldots, v_n\}$ and for an arc $a\in A$ from $v_i$ to $v_j$ consider the point $p_a\in \mathbb R^n$ given by $p_a = e_j -e_i$, where $e_1, \ldots, e_n$ is the standard basis. Note that $p_a \in H = \{x\in \mathbb R^n: 1^Tx = 0\}$ for all $a\in A$. This implies that the points $p_a$ lie in an affine space of dimension $(n-1)$.

Additionally, note that for any directed cycle $C$ in $G$ we have that $\sum_{a\in C} p_a =0$.
Let $K_i = \text{Conv}\{p_a : a\in C_i\}$, and note that $0 \in K_i$ for every $i\in [n]$ since,
$$
0 = \sum_{a\in C} p_a = \sum_{a\in C} \frac{1}{|C|} p_a.
$$
It follows from the Colorful Caratheodory Theorem that there exist points $p_{a_1}, \ldots, p_{a_n}$ such that $a_i \in C_i$ for each $i$ and $0 \in \text{Conv}\{p_{a_1}, \ldots, p_{a_n}\}$. WLOG let $k$ be the number of non-zero terms, so there exist constants $\alpha_i >0$ such that
$$
0 = \sum_{i=1}^k \alpha_i p_{a_i}, \text{ and } \sum_{i=1}^k \alpha_i =1.
$$

It is at this point that I get stuck on an assertion made in the document: it follows from the above that the non-zero constants are all equal, i.e. that $\alpha_1 = \alpha_2 = \ldots = \alpha_k =\frac{1}{k}$.

From there it is easy to show that the corresponding arc set $\{a_1 , \ldots, a_k\}$ has a directed cycle. I did this by a proof by contradiction. Start at the first arc, $a_1 = v_i \to v_j$. By the results above, there must be an another arc in our set with start node $v_j$ to cancel the contribution of the first arc. Proceeding in this way we show that eventually we must return to an arc we already visited, giving us the directed cycle.

Any help understanding the assertion in bold would be appreciated. I figure there must be some counting argument, since among the points $p_a$ we have exactly $k$ 1's and $k$ -1's, but I am not seeing it.

答:
First note, be very careful in how you restate the proof. Saying "$k$ is the number of non-zero terms" is very different from letting $k$ be minimal so that $\mathbf{0} \in \text{conv}(\{p_{a_i}\}_{i=1}^k)$. After all, there are no zero terms!

Every vector $p_{a_j}$ is the same distance from $\mathbf{0}$. Indeed, for each $i$ we have $d(\mathbf{0},p_{a_i}) = 2^{1/n}$. It follows that the unique expression of $\mathbf{0}$ as a convex combination of the $p_{a_i}$ has some constant scalar whose $k$-fold sum is $1$. The only candidate for such a scalar is $1/k$.


问:
Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}?
Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?

Obviously the set $L$ depends on where $C$ is, which is why I've written $L$ as a function of $C.$

If $C$ is the midpoint of $AB$, or on the perpendicular bisector of $AB$, it is not clear to me that $L= \{$ the perpendicular bisector of $AB\},$ because maybe $L$ contains more points. [Edit: Karan Elangovan has answered this part of the question.]

答:
Let $C(a,b),s(x,y)$ where $y\not=0$.

Also, let $U(u,0)$ be the intersection point of the angle bisector with $x$-axis.

Then, we have
$$\begin{align}&sA:sB=AU:BU
\\\\&\iff (1-u)\sqrt{(x+1)^2+y^2}=(1+u)\sqrt{(x-1)^2+y^2}
\\\\&\iff u=\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}\end{align}$$

So, $C$ is on the angle bisector ($sU$) if and only if
$$\begin{align}&b(x-u)=y(a-u)
\\\\&\iff (y-b)u=ay-bx
\\\\&\iff (y-b)\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}=ay-bx
\\\\&\iff (y-b)\bigg(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\bigg)
\\&\qquad\qquad=(ay-bx)\bigg(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\bigg)
\\\\&\iff (y-b-ay+bx)\sqrt{(x+1)^2+y^2}=(y-b+ay-bx)\sqrt{(x-1)^2+y^2}\end{align}$$

So, we have to have
$$(y-b-ay+bx)(y-b+ay-bx)\ge 0$$
under which we have
$$(y-b-ay+bx)^2\bigg((x+1)^2+y^2\bigg)=(y-b+ay-bx)^2\bigg((x-1)^2+y^2\bigg)$$
i.e.
$$\color{red}{bx^3-  a b x^2-  b x+(a^2 -b^2 +  1) x y -  a x^2 y +  b x y^2  -  a y^3+  a b y^2   -  a y     +  a b =0}$$

Examples :

If $a=b=0$, then we get a line $x=0$.

If $b=0$ with $a\not=0$, then we get a circle$$  \bigg(x-\frac{a^2+1}{2a}\bigg)^2 +  y^2   =\bigg(\frac{1-a^2}{2a}\bigg)^2$$under the condition that $-1\lt a\lt 1$ as jlammy commented.

If $a=0$ with $b\not=0$, then we get a line $x=0$ or a circle $$x^2  +\bigg(y-\frac{b^2-1}{2b}\bigg)^2  =\bigg(\frac{b^2+1}{2b}\bigg)^2$$under the condition that $(y+bx-b)(y-bx-b)\ge 0$.


答:
$$0\le f_n(x)=\int_{x-1/n}^{x+1/n}e^{-t^2}dt\le\int_{x-1/n}^{x+1/n}dt=\left(x+\frac 1n\right) - \left(x-\frac 1n\right)=\frac2n.
$$
So $f_n(x)$ congerves uniformly to the null function over $\Bbb R$.

Title: limit of a sequence of integral functions

Body:
I have to study punctual and uniform convergence of the following succession
$$f_n(x)=n\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt.$$
I showed that the functions $f_n$ are uniformly convergent in $\mathbb R$ but I'm not able to calculate explicitly the limit function $f(x):=\underset{n\to +\infty}{\lim}f_n(x)$.

My idea was to split the integral in half and use the characteristic function of the domain of integration but this seems to be inconclusive.

Thank you for your help and your time.


问:
Describe by set notation that a subset of events happen
I struggle with finding an efficient way to notate the following tasks. From $n$ events $A_1, A_2, \dots, A_n$

exactly one event happens

exactly $n-1$ happens

There is a post on finding the notation for two events but not on the general case of $n$ (to the best of my knowledge).

If possible, please elaborate a bit on your solution. Thanks for any advice!

答:
Let $\overline{A_k}$ be the complement of $A_k$, that is the event that $A_k$ does not happen.  To reduce the notational clutter, notate intersection by juxtaposition.  That is, $AB$ means $A\cap B$.  Then the probability that exactly one event happens is $$\bigcup_{k=1}^n\overline{A_1}\overline{A_2}\cdots\overline{A_{k-1}}A_k\overline{A_{k+1}}\cdots\overline{A_n}$$

The second case may be written similarly, by reversing the complemented and non-complemented symbols.

I don't know what kind of elaboration would be appropriate.  Please ask me if you have questions.


问:
Determine the variance of $X$ given the probabilty distributive function $f(x) = (1-x^2)$ for $0\le x\le 2$.
I tried to solve the same using $E(x^2)-[E(x)]^2$, but the answers I found come out to be negative.

答: $f$ is not a valid probability density function - one requirement is that for any event $E$, $P(E)\in[0,1]$. However, $P(1.5)<0$ for example and so the question is impossible.


问:
Got $e^{-x}=\cos(ix)$ from Euler's formula. Where is my mistake?
When I was messing around with Euler's formula, I came across this: $$e^{ix}=\cos(x)+\sin(x)i$$ Then, let $x$ be an imaginary value, $ix$, so then: $$e^{i(ix)}=\cos(ix)+\sin(ix)i$$ which we can simplify to $$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$. so that means that $$e^{-x}=\cos(ix)$$ This doesn't seem right, so could someone please point out where I made a mistake?

答: If $a+bi=\lambda\in\Bbb R$, with $a,b\in\Bbb R$ then, indeed, you must have $a=\lambda$ and $b=0$. But, in the equality$$e^{-x}=\cos(ix)+\sin(ix)i,$$you have no reason to assume that $\cos(ix),\sin(ix)\in\Bbb R$.


问:
Expressing a projective conic in standard form
I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:

$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$

into the form:

$$X^2 + Y^2 + Z^2 = 0$$

Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:

$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end {array} \right]
$$

Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).

In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:

$$  \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\,
\sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3}
\\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array}
 \right]
$$

I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?

答:
I found some notes online, the task is to begin with symmetric $H$ and solve $P^T HP = D$ diagonal, which can be done over the reals. Then forcing the identity over the complexes requires multiplying by an additional diagonal complex matrix on both sides.

Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 -  \frac{ 1 }{ 2 }  & 1 & 0 \\
 - 2 &  \frac{ 11 }{ 2 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
4 & 2 &  - 3 \\
2 & 0 & 4 \\
 - 3 & 4 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  -  \frac{ 1 }{ 2 }  &  - 2 \\
0 & 1 &  \frac{ 11 }{ 2 }  \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
4 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 30 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 \frac{ 1 }{ 2 }  & 1 & 0 \\
 -  \frac{ 3 }{ 4 }  &  -  \frac{ 11 }{ 2 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
4 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 30 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  \frac{ 1 }{ 2 }  &  -  \frac{ 3 }{ 4 }  \\
0 & 1 &  -  \frac{ 11 }{ 2 }  \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
4 & 2 &  - 3 \\
2 & 0 & 4 \\
 - 3 & 4 & 2 \\
\end{array}
\right)
$$

Alright, given this $P$ and the resulting $D,$  we multiply on the right and left by
$$
P_1 = \left(
\begin{array}{rrr}
\frac{1}{2} & 0 &  0 \\
0 & i & 0 \\
 0 & 0 & \frac{1}{\sqrt {30}} \\
\end{array}
\right)
$$

to reach $I.$ I began with double your symmetric matrix, so you may need to multiply throughout by $\sqrt 2$ to take your actual matrix to $I.$ Note how completely the answer changes based on choices made; mine looks nothing like your given answer.


问:
Show that for every triangle $ABC$ $\frac{1}{h_b}+\frac{1}{h_c}-\frac{1}{h_a}=\frac{1}{r_a}$
Show that for every triangle $ABC$ $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{1}{r_a}.$$
Let $S_{ABC}=S$. Then $$S=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}\Rightarrow h_a=\dfrac{2S}{a};h_b=\dfrac{2S}{b};h_c=\dfrac{2S}{c}.$$ On the other hand, $$S=r_a(p-a)\Rightarrow r_a=\dfrac{S}{p-a}.$$ where $r_a$ is the exradii.
So $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{b+c-a}{2S}.$$ I can't see that this is equal to $\dfrac{1}{r_a}=\dfrac{p-a}{S}.$ Thank you in advance!

答:
If $p$ is the semiperimeter $\frac {a+b+c}2$:

$$\frac1{h_b}+\frac1{h_c}-\frac1{h_a}=\frac {b+c-a}{2\color{red}S} = \frac {a+b+c-2a}{2\color{red}S} = \frac{\frac {a+b+c}2-\frac {2a}2}{\color{red}S} = \frac {p-a}{\color{red}S} = \frac1{r_a}$$


答:
You assumed $\sin(ix)=0$, but according to the formula
\begin{equation}
\sin(x)=\mathrm{i}\frac{e^{-\mathrm{i}x}-e^{\mathrm{i}x}}{2},
\end{equation}
($\sin(x)$ is sometimes defined this way for all $x\in\mathbf{C}$),
\begin{equation}
\sin(\mathrm{i}x)=\mathrm{i}\frac{e^x-e^{-x}}{2},
\end{equation}
and thus this is true if and only if $x=0$ (if $x=0$, $\cos(ix)=1$ and there is no contradiction).

I guess you assumed this because you know that a complex number is real if and only if its imaginary part is zero, but $\sin(\mathrm{i}x)$ is not the imaginary part of $\cos(\mathrm{i}x)+\mathrm{i}\sin(\mathrm{i}x)$.

Title: Got $e^{-x}=\cos(ix)$ from Euler's formula. Where is my mistake?

Body: When I was messing around with Euler's formula, I came across this: $$e^{ix}=\cos(x)+\sin(x)i$$ Then, let $x$ be an imaginary value, $ix$, so then: $$e^{i(ix)}=\cos(ix)+\sin(ix)i$$ which we can simplify to $$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$. so that means that $$e^{-x}=\cos(ix)$$ This doesn't seem right, so could someone please point out where I made a mistake?


问:
What type of input does trigonometric functions take in
I see in my Book that 45 deg is equivalent of π/4 . Ι also do the conversion if I simply convert degrees into radians like this

45* π/180 = π/4 radians

and back again

π/4 * 180/π = 45 deg

.
So I think that I grasp the Idea finally that π/4 is another way of saying 45 degrees. Now if I use my calculator and I say sin(45) witch I assume that 45 is degrees I get the number 0.7071067812. If I do the calculation sqrt(2)/2 I will get the same number hence sin(45)=sqrt(2)/2 . But when I input sin(π/4) I get the number 0.0137073546. So I say to my self that probably that function does not get that type of input. But then again I see in my book that sin(π/4) = sqrt(2)/2 . So this forces me to ask the question, "What type of input do trigonometric functions take in?" Is the calculator not working properly ?

Excuse me if this looks simple to you but I am so bad in math and the calculator is the only validator I have at the moment and this confuses me allot.

答:
There is a unit-less sine function (call it, very originally, $\sin (x)$) for which the input is a number and the output a number.  The computation of this in a calculator might be something like taking the first few terms of the formula $\hskip5pt \sin x = x + \frac{x^3}{6} + \frac{x^5}{120} + \cdots = \sum \frac{x^{2n-1}}{(2n-1)!}$.  Notice that $x$ should be dimensionless for addition of different powers to be meaningful, or we would be adding degrees to cubes of degrees and higher powers.  In any case, the input and output are (real) numbers.

The sine of "$x$ radians" is $\sin (x)$.

The sine of "$x$ degrees" is $\sin (\frac{2 \pi}{360} x)$, in which you can see the conversion factor that equates $2 \pi$ radians to $360$ degrees.

The same remarks apply for all the other trigonometric functions.  They are unitless mathematical functions of period $2\pi$, the functions with units are rules for calling on the unitless function, and radians are a name for the units in which the conversion factor (to the standard $2\pi$ normalization of the mathematical functions) is $1$.

To harmonize this with the other answers:

there are three sine functions, a primary one (mathematicians' sine, if you will) and the other two derived from it by detaching the input from its "unit of measurement", multiplying the input by a (unitless) conversion constant, and applying to the result the primary $\sin(x)$ function

setting a calculator to degrees or radians is a choice of which sine function will be called when you press the $\sin x$ button.


问:
Computing limit involving integration (application of FTC)
The question asks:

Let $f(x)$ be continuous with domain $\mathbb{R}$. Compute the following limit:
$$\lim_{x \to 0} \int_{0}^{x} \frac{(x-t)f(t)}{x^2} \,dt$$

I sense I have to use Fundamental Theorem of Calculus somehow but I don't know how. Could anyone help? Thank you so much.

答:
Let $ x\in\mathbb{R}^{*} $, notice that using the change of variable $ \left\lbrace\begin{aligned}t&=xy\\ \mathrm{d}t&=x\,\mathrm{d}y\end{aligned}\right. $, we get that : \begin{aligned} \frac{1}{x^{2}}\int_{0}^{x}{\left(x-t\right)f\left(t\right)\mathrm{d}t}&=\int_{0}^{1}{\left(1-y\right)f\left(xy\right)\mathrm{d}y}\\&=\frac{f\left(0\right)}{2}+\int_{0}^{1}{\left(1-y\right)\left(f\left(xy\right)-f\left(0\right)\right)\mathrm{d}y} \end{aligned}

Now let $ \varepsilon >0 $, since $ f $ is $ \mathcal{C}^{0} $ on $ \mathbb{R} $, then there exists some $ \eta >0 $ such that, for all $ x\in\left(-\eta,\eta\right) $, we have : $$ \left|f\left(x\right)-f\left(0\right)\right|<\varepsilon $$

That means if $ x $ was in $ \left(-\eta,\eta\right) $, then, since $ \left(\forall y\in\left[0,1\right]\right),\ xy\in\left(-\eta, \eta\right) $, we would have that $ \left|f\left(xy\right)-f\left(0\right)\right|<\varepsilon $.

Thus, if we let $ x\in\left(-\eta,\eta\right)\setminus\left\lbrace 0\right\rbrace $, we'll get : \begin{aligned} \left\vert\int_{0}^{1}{\left(1-y\right)\left(f\left(xy\right)-f\left(0\right)\right)\mathrm{d}y}\right\vert&\leq\int_{0}^{1}{\left(1-y\right)\left\vert f\left(xy\right)-f\left(0\right)\right\vert\mathrm{d}y}\\ &\leq\varepsilon\int_{0}^{1}{\left(1-y\right)\mathrm{d}y}=\frac{\varepsilon}{2} \end{aligned}

Hence : $$ \left\vert\frac{1}{x^{2}}\int_{0}^{x}{\left(x-t\right)f\left(t\right)\mathrm{d}t}-\frac{f\left(0\right)}{2}\right\vert\leq\frac{\varepsilon}{2} $$

Which proves that : $$ \frac{1}{x^{2}}\int_{0}^{x}{\left(x-t\right)f\left(t\right)\mathrm{d}t}\underset{x\to 0}{\longrightarrow}\frac{f\left(0\right)}{2} $$


问:
Trying to understand a proof from math overflow regarding direct limit

Direct limit of $\mathbb {C^*}$ behaves well with quotients.

The following solution is from mathoverflow:

Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $\phi_n: A_n \to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $\phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $\tilde{\phi}_n: A_n/I_n \to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $\tilde{\phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.

I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?

答:
In general, if $A_1\xrightarrow{\phi_1}A_2\to\dots$ is an inductive sequence and $(A, \{\mu_n\})$ is a $C^*$-algebra with $*$-homomorphisms $\mu_n:A_n\to A$ so that $\mu_{n+1}\circ\phi_n=\mu_n$ for all $n$, then $(A,\{\mu_n\})$ is the inductive limit of the sequence if and only if

$A=\overline{\bigcup_{k=1}^\infty\mu_k(A_k)}$

$\ker(\mu_n)=\{x\in A_n: \lim_{m\to\infty}\|\phi_{m,n}(x)\|=0\}$ for all $n$, where $\phi_{m,n}=\phi_{m-1}\circ\phi_{m-2}\circ\dots\circ\phi_n$.

Now if $I$ is a (closed, two-sided) ideal in $A$, then set $I_n:=\mu_n^{-1}(I)$. Note that $I_n$ is a closed, two-sided ideal in $A_n$. We now set $\psi_n:A_n/I_n\to A_{n+1}/I_{n+1}$ by $\psi_n(x+I_n)=\phi_n(x)+I_{n+1}$. These are well-defined, since if $\phi_n(x)-\phi_n(x')\in I_{n+1}$ then $\mu_{n+1}(\phi_n(x-x'))\in I$, so $\mu_n(x-x')\in I$ so $x-x'\in I_n$. It is easy to see that each $\psi_n$ is a *-homomorphism. Moreover, if $\psi_n(x+I_n)=0$, then $\varphi_n(x)\in I_{n+1}$, so $\mu_{n+1}(\varphi_n(x))\in I$, so $\mu_n(x)\in I$ so $x\in I_n$, so $x+I_n=0$, thus the maps $\psi_n$ are injective. So we have an inductive sequence
$$A_1/I_1\xrightarrow{\psi_1}A_2/I_2\to\dots $$
where the maps $\psi_n$ are injective. Note that if we set $\pi_n:A_n/I_n\to A/I$ as $\pi_n(x+I_n)=\mu_n(x)+I$, then the $\pi_n$'s are well-defined, injective $*$-homomorphisms (verified as we did for $\psi_n$). Moreover,
$$\pi_{n+1}\circ\psi_n(x+I_n)=\pi_{n+1}(\phi_n(x)+I_{n+1})=\mu_{n+1}(\phi_n(x))+I=\mu_n(x)+I=\pi_n(x+I_n)$$
so $\pi_{n+1}\circ\psi_n=\pi_n$. If we verify the conditions (1) and (2) for the system of quotiens, then we can conclude that $\varinjlim(A_n/I_n, \psi_n)=(A/I,\{\pi_n\})$. But indeed, (2) is trivial since $\ker(\pi_n)=0$ and $\|\psi_{m,n}(x)\|=\|x\|$ for all $m$, since each $\psi_k$ is injective (thus isometric) so $\{x\in A_n: \lim_{m\to\infty}\|\psi_{m,n}(x)\|=0\}=\{0\}$.

Condition (1) is easily verified: let $x+I\in A/I$, $\varepsilon>0$. Since $x\in A$, we may find $k\geq1$ and $y\in A_k$ so that $\|x-\mu_k(y)\|<\varepsilon$, because $A=\overline{\bigcup_{k=1}^\infty\mu_k(A_k)}$ (since $A$ is the inductive limit of $A_n$). But now
$$\|(x+I)-\pi_k(y+I_k)\|=\|(x+I)-(\mu_k(y)+I)\|=\|(x-\mu_k(y))+I\|\leq\|x-\mu_k(y)\|<\varepsilon $$
and we are done. Does this answer your question?


答:
Let $g: X \to \mathbb{R}$ be any function. Given any $a \geqslant 0$, we have that, for all $n \in \Bbb N$, $n \geqslant 1$,

$$ \left \{x: |g(x)|>a-1/(n+1) \right \} \subseteq \left \{x: |g(x)|>a-1/n \right \}   $$

So $$\lim_{n \to \infty} \{x: |g(x)|>a-1/n\}= \bigcap_{n \geqslant 1} \{x: |g(x)|>a-1/n\} = \{x: |g(x)| \geqslant a\}$$

Title: Interchange of set theoretical limit

Body:
Find for what type of real valued measurable functions g, the equality holds:
$$ \lim_{n \to \infty} \{x: |g(x)|>a-1/n\}= \{x: |g(x)| \geq a\}$$ where $a \geq 0$.


问:
Integrate in Mathematica takes forever
I'm trying to calculate the length of a curve from a polynomial in Mathematica and for some reason Integrate doesn't complete, it just runs until I abort the execution.

The polynom:

```<code>58.1735 - 36.4175 x + 7.10945 x^2 - 0.538311 x^3 + 0.0138115 x^4
</code>```

Is a part of an integration like this:

where "koordfitprim" is the polynom above (i.e. what is integrated is in fact the square root of 1 + the above polynom^2).

Any ideas why this wouldn't execute?

答:
This is a very tough integral and you asked Mathematica to evaluate it symbolically i.e. analytically. I am not sure whether it can be calculated at all.

What you wanted was a numerical integral

NIntegrate[Sqrt[1 + ko[x]^2], {x, 3, 18}]

The result is, after a fraction of a second,

59.211531

Title: Integrate in Mathematica takes forever


问:
Explanation for Proof of $[T(u)]_{\gamma}=[T]_{\beta}^{\gamma}[u]_{\beta}$
I am struggling with understanding the proof of the following theorem:

Theorem. Let $$ and $$ be finite-dimensional vector spaces over a field $$, having ordered basis $\beta$ and $\gamma$, respectively. Let $: \rightarrow $ be linear. Then for each $u∈$ we have:

$$[T(u)]_{\gamma}=[T]_{\gamma}^{\beta}[u]_{\beta}$$

The proof for this theorem is given as:

Proof: Fix $u∈V$, and define the linear transformations $:F \rightarrow V$ by $(a)=au$ and $g:F \rightarrow W$ by $g(a)=aT(u)$ for all $a∈F$. Let $\alpha=\{1\}$ be the standard ordered basis for $F$. Notice that $g=T$. Identifying column vectors as matrices and using Theorem 2.11, we obtain

$$[T(u)]_{\gamma}=[g(1)]_{\gamma}=[g]_{\alpha}^{\gamma}=[Tf]_{\alpha}^{\gamma}=[T]_{\beta}^{\gamma}[f]_{\alpha}^{\beta}=[T]_{\beta}^{\gamma}[f(1)]_{\beta}=[T]_{\beta}^{\gamma}[u]_{\beta}$$

I am having trouble understanding the logic behind this proof. It seems like it just plugs in 2 linear transformations ($f$ and $g$) that make this proof work. How does it show that there isn't an example of $f$ and $g$ that won't make this theorem work? How does this proof prove the theorem for all possible $T$ and $u$?

答: It's not that all $f$ and $g$ must work. The logic here is that the authors pick some special $f$ and $g$ to make the notation easier. The claim needs to hold for all $V$, $W$, $F$, $T$, $\beta$, $\gamma$, and $u\in V$. The proof precisely demonstrates this holds. The claim does not claim it must hold for all $f$ and $g$.


问:
How to simplify this combinatorial summation?
Expression: ${m}\choose{1}$${n-m}\choose{k-1}$ + ${m}\choose{3}$${n-m}\choose{k-3}$ + ... + ${m}\choose{k}$${n-m}\choose{0}$ (assuming k is odd)

We could write it as ${n}\choose{k}$ if the even terms were also present, but I am clueless as they are not.

答:
I think there is no simple formula for the first expression, let alone the second one.

We have
$$\sum_{j=0}^k \binom{m}{j}\binom{n-m}{k-j} x^j = [y^k]\ (1+xy)^m(1+y)^{n-m},$$
where $[y^k]$ is the operator taking coefficient of $y^k$.
Then the sum of coefficients at odd $k$ equals halved difference of the above expression evaluated at $x=1$ and at $x=-1$, that is
\begin{split}
\sum_{j=0\atop j\text{ is odd}}^k \binom{m}{j}\binom{n-m}{k-j} &= [y^k]\ \frac{(1+y)^m(1+y)^{n-m} - (1-y)^m(1+y)^{n-m}}2 \\
&=\frac{1}{2}\binom{n}{k} - \frac{1}{2} \sum_{i=0}^k (-1)^i\binom{m}i\binom{n-m}{k-i}.
\end{split}
The last sum here simplifies is some cases -- e.g., if $n=2m$, then it's zero.

For the second expression,
$$\sum_{j=0}^k \binom{m}{j} x^j\sum_{i=0}^{k-j}\binom{n-m}{i} = [y^k]\ (1+xy)^m\frac{(1+y)^{n-m}}{1-y}.$$
Hence, we similarly have
\begin{split}
\sum_{j=0\atop j\text{ is odd}}^k \binom{m}{j}\sum_{i=0}^{k-j}\binom{n-m}{i}
&= \frac{1}{2} [y^k]\ \big(\frac{(1+y)^n}{1-y} - (1-y)^{m-1}(1+y)^{n-m}\big) \\
&=
\frac{1}{2}\sum_{i=0}^k\big(\binom{n}{i} - (-1)^i\binom{m-1}i \binom{n-m}{k-i}\big).
\end{split}


问:
On linear, continuous, injective map $T:V \to W$ with closed/ dense image
Let $V,W$ be normed linear spaces and $T:V \to W$ be a linear continuous, injective map .

Then is $T^{**}: V^{**} \to W^{**}$ is injective ? If this is not true in general, what if we also assume $Im T$ is closed/ dense in $W$ ?

My try: $g\in \ker T^{**} \implies  g\circ T^*=T^{**}(g)=0 \implies g(T^*(f))=0, \forall f \in W^*$.

So $Im (T^*) \subseteq \ker g$.

Now $\{0\}=\ker T^{**} $ if and only if $g\circ T^*=0,\forall g\in \ker T^{**}$ if and only if $\ker g=V^*, \forall g \in \ker T^{**}$.

Now if $Im(T^*)$ is dense in $V^*$, then $Im (T^*) \subseteq \ker g$ and $\ker g$ is closed would readily imply $\ker g=V^*$, but unfortunately we don't know that $Im (T^*)$ is dense in $V^*$.

I don't know how to proceed further.

答:
I am almost sure that this is not true in general without assuming that $T$ has closed image. I cannot think of a counter example now, but I am almost sure.

However, if one assumes that $T$ has closed image then the result follows; we prove the following two claims:

If $S:X\to Y$ is a bounded operator that is injective and has closed range, then the adjoint operator $S^*:Y^*\to X^*$ is surjective.

If $S: X\to Y$ is a bounded operator that is surjective, then the adjoint operator $S^*:Y^*\to X^*$ is injective.

Proof of the claims:

For (1), since $S$ is injective and has closed range, we conclude (from the open mapping theorem) that $S$ is bounded below, so there exists a constant $\delta>0$ so that $\delta\|x\|\leq\|Sx\|$ for all $x\in X$. Let $\varphi\in X^*$ be a bounded linear functional on $X$. We define a functional $\psi_o:\text{Im}(S)\to\mathbb{C}$ as $\psi_o(Sx)=\varphi(x)$ for all $x\in X$. Note that this is well-defined, since $S$ is injective. Moreover, this is bounded, since $S$ is bounded below: we have that
$$|\psi_o(Sx)|=|\varphi(x)|\leq\|\varphi\|\cdot\|x\|\leq\|\varphi\|\cdot\delta^{-1}\cdot\|Sx\|$$
for all $x\in X$. By the Hahn-Banach theorem we may extend $\psi_o$ to a bounded functional (of the same norm) $\psi\in Y^*$. Now we have that $S^*(\psi)=\psi\circ S=\psi_o\circ S=\varphi$, so $S^*$ is indeed surjective.

For (2), Let $\varphi\in Y^*$ be a functional so that $S^*(\varphi)=0$, i.e. $\varphi\circ S=0$. Since $S$ is surjective, we have that $\varphi(z)=0$ for all $z\in Y$, so $\varphi=0$.

Combining these two: we have $T:X\to Y$ an injective operator with closed range. By (1) the adjoint operator $T^*:Y^*\to X^*$ is surjective. Applying (2) to $T^*:Y^*\to X^*$ we conclude that $T^{**}:X^{**}\to Y^{**}$ is injective. I hope this answers your question.


问:
Show that the function $f(x)=x/(x^2-1)$ is bijective.
I have to show that the function $f:(-1,1)\to \mathbb{R}$
$$f(x)=\frac{x}{x^2-1}$$
is bijective. I have shown that it is injective which is pretty simple. I'll write it out here for future reference.

To show that $f$ is injective, let $x_1,x_2\in(-1,1)$. Assume that $f(x_1)=f(x_2)$. Then
$$\frac{x_1}{x_1^2-1}=\frac{x_2}{x_2^2-1}$$
Multiplying both sides by $(x_1^2-1)(x_2^2-1)$ which we know can't be $0$. On simplifying the equation further, we get
$$(x_1x_2+1)(x_2-x_1)=0$$
This means either $x_2x_1=-1$ or $x_2=x_1$. It isn't possible that $x_2x_1=-1$ because if it were true then $x_2=\cfrac{-1}{x_1}$ and for all the value of $x_1$ in the domain, $x_2$ will go out of the domain. Thus, it must be that $x_2=x_1$. This concludes the injectivity part.

What I am confused about is the surjectivity part. I know that you usually invert the function and then show that for any value $y$ in the codomain, there exists a value $x$ in the domain of the function such that $f(x)=y$. I can't seem to invert this function appropriately. Any hints?

答:
Let $y\in \mathbb{R}$. If $y=0$ then $f(0)=y$ with $0\in(-1,1)$. Suppose  $y\neq 0$. Let $x_1,x_2$ be defined by

$$x_1=\frac{1+\sqrt{1+4y^2}}{2y}\hspace{1cm}x_2=\frac{1-\sqrt{1+4y^2}}{2y}$$

Then $x_1 x_2=-1$, so one of them must lie in the interval $[-1,1]$. Since $x_1, x_2$ are roots of $yx^2-x-y$, we see that none of them can be equal to $1$. Hence none of them can be equal to $-1$ either. Therefore one of them must lie in $(-1,1)$. Let $z$ be this one. Then $yz^2-z-y=0$ and $z\in(-1,1)$. This implies that $f(z)=\frac{z}{z^2-1}=y$.


问:
What programs can calculate (this) series (get expression in closed form)?
What programs can calculate this (type of) series?

$$
\sum_{m,n=0}^\infty \frac{(-1)^{-2n}2^{-2m-n}\,(1+m)}{2+m+n}
$$

One program I know of is XSummer math-ph/0508008, XSummer.

The result is:

$$ -16\cdot \left(\frac{1}{3}-\log\frac{3}{2}\right) $$

Because I will need to calculate more complicated series (4+ fold) of the same form (also resulting in log2 and log3 terms) I'd like to get a decent overview. Common CAS like maple or mathematica can perform just one series, resulting in hypergeometric expressions, for which they cannot perform the second series summation.

答:
I just typed that into mathematica and got $\frac{8}{3} \left(\log \left(\frac{729}{64}\right)-2\right)$.

Image as follows:


答: You are right to say that $e^{-x}$ is a real value for all real inputs. But you seem to assume that $\sin(ix)$ is also a real value for all real inputs. Maybe this assumption is hasty...


问:
To prove that it is possible to find a prime number q with a primitive root r such that ind(p) is a prime, for a particular prime p.
Question: Let $p$ be a prime number. Prove that there exist a prime number $q$ such that for every integer $n$, the number $n^{p}-p$ is not divisible by $q$.
I was trying to solve the above question with an original method using indices and I couldn't proceed after a certain point. My approach is shown below

$\textbf{Approach:}$  The congruence $n^{p}\equiv p\pmod{q}$ can be reduced to
$$p\cdot ind(n)\equiv ind(p) \pmod{q-1}$$
We can do this since every prime has a primitive root. A solution to this congruence exists iff $gcd(ind(n), q-1)|ind(p)$. Thus, if the congruence has no solution for all integers $n$, then it is logical to conclude that $ind(p)$ could be a prime. Hence, I present the following lemma

$\textbf{Lemma:}$ It is possible to find a prime $q$ with a primitive root $r$ such that $ind_{r}(p)$ is a prime number.

It would be a great help if someone could prove or disprove the above Lemma. Thanks a lot for your help!! :)

答:
If $ind_r(p)$ is coprime with $q-1$, then $p \equiv r^{ind_r(p)}$ is a primitive root $\bmod q$ (as $ind_r(p)$ is invertible $\bmod q-1$). Conversely, given a prime $q$ for which $p$ is a primitive root, we may pick any prime $i$ coprime with $q - 1$ and obtain a primitive root $r$ for which $ind_r(p) = i$ by $r \equiv p^{i^{-1}}$.

We therefore need to prove that all primes $p$ are primitive roots modulo some prime $q > p$.


问:
The limitations of the middle square method in generating pseudo-random numbers.
I have been trying to research the middle square method as a way to generate pseudo-random numbers for a school project. Everywhere I look claims that this method is limited by its short period and will either loop or repeat if used enough times. However, I don't really understand how this happens or even why it happens. I wrote a Python script that takes a seed of 6 digits and I have yet to witness any looping or repetition. Is this because my seed is too large? If so, what digit seed would be appropriate to show the limitations of this method? Lastly, why does this method apparently repeat or loop if done enough times?
Thank you!</p>

答:
Here's a simple example - let's say we are taking the middle two digits of a four digit number and start with a seed of 99.

99 x 99 = 9801

Next seed is 80

80 x 80 = 6400

Next seed is 40

40 x 40 = 1600

Next seed is 60

60 x 60 = 3600 - so the seed is now stuck at 60 for ever.


问:
Intuition for formula family of circles passing through two given points involving determinants

$$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + \lambda| \begin{bmatrix} x&y&1\\x_1 & y_1 & 1 \\ x_2 &y_2  &1 \end{bmatrix}| = 0$$

The above equation is supposed to represent family of all circles passing through two points, but I can't understand why it should be. I understand so far that the first two terms i.e: the $(x-x_1)(x-x_2) + (y-y_1)(y-y_2)$ part represents the diametric form of circle and the determinant is the equation of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ but I don't understand why we needed to add line to get family of all circles passing through the point. Isn't the first two terms enough?

答:
Given two points $A = (x_1, y_1)$ and $B = (x_2, y_2)$, we want to find a circle for which $AB$ is a chord, not necessarily a diameter. Choose $C = (x, y)$ on this circle. Because the inscribed angle $\angle ACB$ is independent of the choice of $C$ (it depends only on the subtended arc), we get an equation
$$\frac{CA \cdot CB}{|CA||CB|} = k$$
for some constant $k$.

(When $AB$ is a diameter, $\angle ACB$ is a right angle. Then $k= 0$ and you get the diametric equation for a circle.)

In any case, we get
$$(x_1 - x)(x_2 - x) + (y_1 - y)(y_2 - y) = k|CA||CB|.$$

Here's where my argument gets somewhat inelegant, but I think it still works:

Note that $\frac12|CA||CB|\sin(\angle ACB)$ is the area of $\Delta ABC$. Moreover, the determinant in the equation is the volume of a tetrahedron with base $\Delta ABC$ and height $1$, i.e. $1/3$ of the area of $\Delta ABC$. Since $\sin(\angle ACB)$ is a constant depending only on the circle, you can now solve for the $\lambda$ to get the third parameter.