答:
Assuming the question is:

A's arrows hit the target so that the distribution of the distance from the center is uniform, i.e.,
$f_A(r,\theta )= \begin{cases}
 \frac{1}{2\pi R} & 0<r<R, 0<\theta<2\pi \\
 0, &\text{otherwise}
\end{cases}
$,

D's arrows hit the target uniformly distributed over the target, i.e.,
$f_D(r,\theta )= \begin{cases}
 \frac{r}{\pi R^2} & 0<r<R, 0<\theta<2\pi \\
 0, &\text{otherwise}
\end{cases}
$.

Further, if the question asks for the expected distance of the arrow from the center for each of them, then:

$E_A\{r\} = \int\limits_{r=0}^{R} r\frac{1}{R}dr = \frac{1}{2}R$, and

$E_D\{r\} = \int\limits_{r=0}^{R} r\frac{2r}{R^2}dr=\frac{2}{3}R$.

Title: A problem with uniform distribution?

Body:
In a game Alis and Daniel shoot arrows on a circular target with a
radius of R.

Alis throws arrows such that their distance from the center have
Continuous uniform distribution (0,R) - o,R are the parameters for
uniform distribution.

While Daniel throws arrows such that their distance from the center have
random Continuous uniform distribution (From the
target)

I was asked to calculate the expected value for the distance of the arrows for both participant.

For Alis it's R/2 but what about Daniel? I didn't understand what's given about him, what are the parameters for him...

I claimed: "For Alis it's R/2" Proof:

The expected value for random variable with uniform distribution (a,b) (with parameters a and b) is (a+b)/2


问:
Christoffel Symbols: when did the bracket notation first appear?
An early notation to Christoffel symbols is:

\begin{equation}
\begin{Bmatrix}
  m  \\
  \hspace{1.5mm}b\hspace{5mm} c\hspace{1.5mm}
\end{Bmatrix} = \frac{1}{2}g^{ma}\{ \partial_{b}g_{ca} + \partial_{c}g_{ab} - \partial_{a}g_{bc}\} \tag{1}
\end{equation}

with these curly brackets.

Now, in Einstein's notebook and in the seminal paper of Einstein Field Equations we see a clear usage of this notation:

and

I'm not sure if was Einstein indeed the first one to use this curly notation (or the german publishers). So, It was Elwin Christoffel the one who introduced this notation? If no,  would like to know the one who first have used this notation.

答:
The bracket notation and brace notation was the first notation, due to Christoffel and appearing in his 1869 paper. See both the 3rd and 4th pages.

The $\Gamma$-notation is due to Einstein in his work on general relativity.

This is discussed, with links to a bunch of papers by different authors (Christoffel, Einstein, Ricci, Weyl, et al.), in the question I asked about this issue a couple of years ago here.  It is suggested in comments there that the use of $\Gamma$ by Einstein might be an allusion to "gravitation" and the $\Gamma$-notation should be called Einstein symbols instead of Christoffel symbols.

ContentLicense: CC BY-SA 4.0


问:
Linear polynomials of finite fields
I have a final tomorrow, and I was looking over some exercises in my textbook. However, I can't seem to work this problem out.

Let $F$ be a field of $p^n$ elements and let $\alpha \in F^*$, where $F^*=F-\{0\}$.
Show $(x- \alpha)(x- \alpha^p)(x- \alpha^{p^2})\cdots(x- \alpha^{p^{n-1}}) \in \mathbb{Z}_p[x]$. Also, my text has some weird notation, $\mathbb{Z}_p$ is shorthand for $\mathbb{Z}/p\mathbb{Z}$.

It hints that I need to show both $\alpha+\alpha^{p}+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\in\mathbb{Z}_p$ and $\alpha\alpha^{p}\alpha^{p^2}\cdots\alpha^{p^{n-1}}\in\mathbb{Z}_p$, but I'm having trouble with it. Help would be much appreciated.

答:
In a field of characteristic $p$, $(a-b)^p = a^p-b^p$.

Now,
$$\left[\prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right)\right]^p
= \prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right)^p
= \prod_{i=0}^{n-1}\left(x^p-\alpha^{p^{i+1}}\right)
= \prod_{i=1}^{n}\left(x^p-\alpha^{p^i}\right)
= \prod_{i=0}^{n-1}\left(x^p-\alpha^{p^i}\right)$$
where the rightmost equality follows from the fact that $\alpha^{p^n} = \alpha$ for all $\alpha \in \mathbb F_{p^n}$.  Thus we have that
$$f(x) = \prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right) = \sum_{j=0}^n f_jx^j$$
enjoys the property that $[f(x)]^p = f(x^p)$.  But,
$\displaystyle [f(x)]^p = \left[\sum_{j=0}^n f_jx^j\right]^p = \sum_{j=0}^n f_j^px^{jp}$
equals
$\displaystyle f(x^p) = \sum_{j=0}^n f_jx^{jp}$ if and only if $f_j^p = f_j, 0 \leq j \leq n$, that is, $f_j \in \mathbb F_p$ and hence $f(x) \in \mathbb F_p[x]$.

ContentLicense: CC BY-SA 3.0

Title: Linear polynomials of finite fields


问:
Number of prime knots for crossing number greater than 16
How many prime knots are there for crossing numbers greater than or equal to 17? I can't seem to find any useful resources indicating that they have been computed or that algorithms exists to compute the number of prime knots.</p>

答:
Using his program Regina, Benjamin Burton has computed the number of prime knots with 17, 18, and 19 crossings (see this link for reference). The table below summarizes the data from the link:

Crossings
Alternating
Non-alternating
Total

17
1,769,978
6,283,385
8,053,363

18
8,400,285
39,866,095
48,266,380

19
40,619,384
253,510,828
294,130,212


问:
partial sum of $\sum^h_{i=1}\sum^i_{j=1}ij$
I am trying to come up with a fast partial sum for a big series in a program that I am writing in order to simplify some loops. However there is an internal sum inside that to me should be able to be simplified. What would be the partial sum of this be:
$$\sum^h_{i=1}\sum^i_{j=1}ij$$

Any help would be much appreciated! Thanks in advance

答:
$$
\begin{align*}
\sum_{i = 1}^h \sum_{j = 1}^i ij
&= \sum_{i = 1}^h i \sum_{j=1}^i j\\
&= \sum_{i=1}^h i \left ( \frac{i(i+1)}{2} \right )\\
&= \frac{1}{2} \sum_{i=1}^h i^3 + i^2\\
&= \frac{1}{2} \left ( \frac{h^2 (h+1)^2}{4} + \frac{h(h+1)(2h+1)}{6} \right )
\end{align*}
$$

Of course, you can keep simplifying from here.

I hope this helps ^_^


问:
Computing the intersection of all (minimal) vertex covers
Let $t<n$ be two integers and $G$ an undirected, irreflexive graph without multi-edges of size $n$.

How hard is it to compute the intersection of all (minimal) vertex covers of cardinality $ \leq t$?
Or alternatively, how hard is it to compute the intersection's cardinality?

Note: Obviously, the intersection of all vertex covers ($\leq t$) is equal to the intersection of all minimal vertex covers ($\leq t$), hence the parenthesis.

I was wondering whether the hardness of this particular problem has been studied in the literature.
Is it efficiently computable, maybe reducible to the MVC-problem or completely unknown?

For $t \leq n-1$ the naive approach would be to consider all subsets of cardinality $t$ and intersect all those which are vertex covers.
However this approach would require the consideration of $\binom{n}{t}$ subsets;
which is efficient (polynomial in $n$) if and only if $t$ or $n-t$ is bounded by a constant.

Edit:
To see that it suffices to intersect only all subset of size equal $t$ instead of all subsets at most $t$, one can write:
$$
\begin{equation}\begin{aligned}
\bigcap_{i} VC_i^{t} = \bigcap_{i} MVC_i^{t} \cap \bigcap_{i} NMVC_i^{t}
\end{aligned}\end{equation}
$$
where $MVC^{t}$ denotes a minimal VC of size $t$ and $NMVC^{t}$ denotes a non-minimal VC.
Further a NMVC can be split into an MVC of size $t'< t$ and an extension $E^{t-t'}$,
hence

$$
\begin{equation}\begin{aligned}
\bigcap_{i} VC_i^{t} &= \bigcap_{i} MVC_i^{t} \cap \bigcap_{i} \left(MVC_{i'}^{t'} \cup E_{i'j'}^{t-t'}\right)
\\
&=
\bigcap_{i} MVC_i^{t} \cap \bigcap_{i'} \underbrace{\bigcap_{j'} \left(MVC_{i'}^{t'} \cup E_{i'j'}^{t-t'}\right)}_{MVC_{i'}^{t'}}
\\
&=
\bigcap_{i} MVC_i^{t} \cap \bigcap_{i'} MVC_{i'}^{t'}
\end{aligned}\end{equation}
$$
which is the intersection of all MVCs of size at most $t$.
Note that the last equality only holds if there exist at least two disjoint extensions of $MVC_{i'}^{t'}$.
This is true for $t' < n-2$ and hence for $t \leq n - 1$ because $t' < t$.

答:
In turns out, there is a reduction from the decision Minimum VC (MinVC) problem, i.e. deciding whether a graph G has a MinVC of size $\leq k$, to the intersecting MVC problem (IMVC) as follows:

```<code>hasMVC(G, k):
G' := copy(G);
X := {};
while !isVC(G,X) do
    removeIsolatedNodes(G');
    X' := {};
    t := 0;
    while X' = {} do
        if t < |G'| then
            X' := findIMVC(G', t);
        else
            X' := X' ∪ {P_next};
        endif
        t := t + 1;
    endwhile
    X := X ∪ X';
    G' := G' \ X';
    if |X| > k then
        return false;
    endif
endwhile
return true;
</code>```

where findIMVC is an oracle function to solve the IMVC problem, $P_{\mathrm{next}}$ is the first vertex of $G'$ in some arbitrary canonical order (e.g. lexicographical or even random) and removeIsolatedNodes is the obvious subroutine for removing vertices without neighbors from $G'$.

Correctness:
Intuitively, the outer loop checks whether a candidate vertex set $X$ is a vertex cover of admissible size (<= t), if so, it terminates.
Otherwise, it tries to find new vertices $X'$ to add to the set $X$ to make it into a MinVC. The inner loop has two cases:

If $t \geq |G'|$, then $G'$ does not contain any IMVC of any size, hence there is a set of MinVCs, say $\{MinVC_1, ..., MinVC_m\}$, such that their intersection is empty.
Because the graph $G'$ is connected (via removeIsolatedNode) for each vertex $v$ there is a MinVC that contains $v$.
Otherwise, if there was a vertex in no $MinVC_i$, then the neighbors of $v$, which must exists by connectivity, would have to be in all $MinVC_i$;
in contradiction to the case condition that the IMVC is empty.
Therefore adding any vertex $v$ to $X'$ is valid.

If $t < |G'|$, then $t$ is incremented and $G'$ checked for IMVCs of size at most $t$.
If one exists, it is added to $X'$ since it is in all MVCs of $G'$.
Otherwise repeat the inner loop.

After finding new vertices $X'$ for $X$, the counter $t$ is reset.
This fashion of adding only vertices to the candidate set $X$ ensures that the candidates set eventually only contains vertices from a single MinVC.
If the size of the (minimal) candidate set $X$ exceeds $k$, then return $false$ since any MinVC of $G$ must then have size at least $|X| > k$.

Runtime:
In each outer loop the size of $X$ is increased by at least one because $X' \neq \emptyset$.
Thus there are at most $n$ (number of vertices) outer iterations.
The inner loop increments $t$ at most until $t = |G'| \leq n$,
hence there are at most $n$ inner iterations.
The subroutines isVC and removeIsolatedNodes have linear runtime in $n$.
So overall, we get a crude runtime bound of $\mathcal{O}(n^2)$.

Body:
In turns out, there is a reduction from the decision Minimum VC (MinVC) problem, i.e. deciding whether a graph G has a MinVC of size $\leq k$, to the intersecting MVC problem (IMVC) as follows:

hasMVC(G, k):
G' := copy(G);
X := {};
while !isVC(G,X) do
    removeIsolatedNodes(G');
    X' := {};
    t := 0;
    while X' = {} do
        if t < |G'| then
            X' := findIMVC(G', t);
        else
            X' := X' ∪ {P_next};
        endif
        t := t + 1;
    endwhile
    X := X ∪ X';
    G' := G' \ X';
    if |X| > k then
        return false;
    endif
endwhile
return true;

where findIMVC is an oracle function to solve the IMVC problem, $P_{\mathrm{next}}$ is the first vertex of $G'$ in some arbitrary canonical order (e.g. lexicographical or even random) and removeIsolatedNodes is the obvious subroutine for removing vertices without neighbors from $G'$.

Correctness:
Intuitively, the outer loop checks whether a candidate vertex set $X$ is a vertex cover of admissible size (<= t), if so, it terminates.
Otherwise, it tries to find new vertices $X'$ to add to the set $X$ to make it into a MinVC. The inner loop has two cases:

If $t \geq |G'|$, then $G'$ does not contain any IMVC of any size, hence there is a set of MinVCs, say $\{MinVC_1, ..., MinVC_m\}$, such that their intersection is empty.
Because the graph $G'$ is connected (via removeIsolatedNode) for each vertex $v$ there is a MinVC that contains $v$.
Otherwise, if there was a vertex in no $MinVC_i$, then the neighbors of $v$, which must exists by connectivity, would have to be in all $MinVC_i$;
in contradiction to the case condition that the IMVC is empty.
Therefore adding any vertex $v$ to $X'$ is valid.

If $t < |G'|$, then $t$ is incremented and $G'$ checked for IMVCs of size at most $t$.
If one exists, it is added to $X'$ since it is in all MVCs of $G'$.
Otherwise repeat the inner loop.

After finding new vertices $X'$ for $X$, the counter $t$ is reset.
This fashion of adding only vertices to the candidate set $X$ ensures that the candidates set eventually only contains vertices from a single MinVC.
If the size of the (minimal) candidate set $X$ exceeds $k$, then return $false$ since any MinVC of $G$ must then have size at least $|X| > k$.

Runtime:
In each outer loop the size of $X$ is increased by at least one because $X' \neq \emptyset$.
Thus there are at most $n$ (number of vertices) outer iterations.
The inner loop increments $t$ at most until $t = |G'| \leq n$,
hence there are at most $n$ inner iterations.
The subroutines isVC and removeIsolatedNodes have linear runtime in $n$.
So overall, we get a crude runtime bound of $\mathcal{O}(n^2)$.


答:
First note that

$$
\begin{align}
\left(\sum^h_{i=1}i\right)^2 &= \left(\sum^h_{i=1}i\right) \left(\sum^h_{j=1}j \right) \\
&= \sum^h_{i=1}\sum^h_{j=1}ij \\
&= \sum^h_{i=1}\sum^i_{j=1}ij + \sum^h_{i=1}\sum^h_{j=i}ij \\
&= 2 \sum^h_{i=1}\sum^i_{j=1}ij - \sum^h_{i=1}i^2.\tag1
\end{align}
$$

On the other hand (see arithmetic progression)

$$
\sum^h_{i=1}i = \frac{h(h+1)}{2}
$$

and (see square pyramidal numbers)

$$
\sum^h_{i=1}i^2 = \frac{h(h+1)(2h+1)}{6}.
$$

Substituting into $(1)$

$$
\sum^h_{i=1}\sum^i_{j=1}ij = \frac{h^2(h+1)^2}{8} + \frac{h(h+1)(2h+1)}{12}.
$$

Title: partial sum of $\sum^h_{i=1}\sum^i_{j=1}ij$

Body:
I am trying to come up with a fast partial sum for a big series in a program that I am writing in order to simplify some loops. However there is an internal sum inside that to me should be able to be simplified. What would be the partial sum of this be:
$$\sum^h_{i=1}\sum^i_{j=1}ij$$

Any help would be much appreciated! Thanks in advance


问:
Why are the order-of-operations conventions good?
Children are sometimes taught silly mnemonics like "PEMDAS" to remember conventions on order of operations.  (I never heard of "PEMDAS" until long after graduating from college, as far as I can recall.  I think it means (1) parentheses, (2) exponentiation, (3) multiplication and division, and (4) addition and subtraction.)

I think it would be better to help them understand why those particular conventions, rather than some others, are a good thing.  Maybe even demonstrably optimal by some precisely definable desiderata?

How would one make the case for the usual conventions against possible alternatives?

I think I might say that if one operation distributes over another, it should come first, and we go from left to right because we normally read that way.  Maybe I'll post my own answer to this if I'm so inspired at some point.

答:
Order of operations conventions are rooted in reality. At the heart of it all, mathematics tries to model our everyday experience, and our order of operations reflects that.

First of all, let's agree that addition and subtraction are really the same thing, and multiplication and division (when defined) are also really the same thing, so I can just say "addition" when I mean "addition and subtraction" and "multiplication" when I mean "multiplication and division."

Why should multiplication take precedence over addition? Suppose Alice were to give me $5$ bags of $8$ apples, and Bob were to give me $3$ bags of $4$ apples. How many apples do I have?

A good guess is that the number of apples I have is equal to the number that Alice gave me plus the number that Bob gave me. Alice gave me $5 \cdot 8=40$ apples, and Bob gave me $3\cdot 4=12$ apples; therefore I should have received $40+12=52$ apples. This is an observation from everyday experience: if I get $x$ things from Alice and $y$ things from Bob, I received $x+y$ things in total.

Let's look at how varying the order of operations plays out! The number is given by the expression $5\cdot 8+3\cdot 4$. If multiplication comes first, we get $52$ apples; if addition comes first, then we get $220$ apples; and if neither takes precedence and we proceed left to right, we get $172$ apples. Via experiment, we have obtained evidence suggesting that doing multiplication first is wiser than the other options.

The reasoning for exponentiation is similar. Why should $3\cdot 2^3$ be $3\cdot 8$ instead of $6^3$? Imagine I had $3$ identical cubes of side length $2$ cm. What is the total volume of the cubes? Volume is additive, so I should add together the volumes of each cube. Each cube has volume $2^3~\text{cm}^3$, and there are 3 cubes; thus the total volume is $24~\text{cm}^3$. I would be very surprised if you told me these 3 cubes took up $6^3=216$ cubic centimeters - they certainly didn't look that big when I first got them!

I'll let you imagine a reason in the same vein for exponentiation before addition. For parentheses fgp has given the essential answer - the entire purpose of parentheses is to group things and make sure operations inside them happen independently of the rest of the expression. And you can make a similar real-life analogy here too.

Of course, once we start talking about real numbers things get a little more delicate. However, historically these operations started out defined for just the natural numbers (no zero or negative numbers yet), then became generalized. And when they were generalized, the generalization happened in a way that preserved these properties. (Michael Spivak gives a charming explanation of this sort when he defines the exponential function in Calculus, 4th edition.) And this is why the group and field axioms are what they are!

Moral: Mathematics is an experimental science at heart.


问:
Is it possible to solve for x in sin^2(x) + tanx - 3/2 =0
original question: Is it possible to solve for x in   $$\sin^2x + \tan x  + \frac{3}{2} = 0 .$$

please show steps

edit:
sorry for typo mistake, the equation should be
$$\sin^2x + \tan x  - \frac{3}{2} = 0 .$$

答:
$$\begin{align*}
\sin^2x + \tan x - \frac 32 &= 0\\
\frac{\tan^2x}{1+\tan^2x} + \tan x - \frac32 &= 0\\
2\tan^2x + (2\tan x) (1+\tan^2x) - 3(1+\tan^2x) &= 0 &&(1+\tan^2x \ne 0)
\end{align*}$$

This will be a cubic equation in $\tan x$.


答:
If you are asking for an existence proof, try using the Intermediate Value Theorem. As Ron Jensen suggested, since the function $\sin^2(x) + 3/2$ has range $[0,5/2]$, you need the following two things:

A value $x$ such that $\tan(x)< 0$.

A value $y$ such that $\tan(y)>5/2$.

If this is true, then there must be a value $z$ between $x$ and $y$ at which your function is equal to $0$.

For example, $x = -\pi/4$ and $y=1.2$ should suffice for the proof.

For example, you could

Title: Is it possible to solve for x in sin^2(x) + tanx - 3/2 =0

Body:
original question: Is it possible to solve for x in   $$\sin^2x + \tan x  + \frac{3}{2} = 0 .$$

please show steps

edit:
sorry for typo mistake, the equation should be
$$\sin^2x + \tan x  - \frac{3}{2} = 0 .$$


问:
Limits of Vector Valued Functions
For vector-valued functions of the form $\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}=(x(t),y(t))$, I learned that the derivative of this function is another vector valued function, $\vec{r'}(t)=\lim_{h\to0}\frac{(\vec{r}(t+h)-\vec{r}(t))}{h}=\lim_{h\to0}\frac{x(t+h)-x(t)}{h}\hat{i}+\lim_{h\to0}\frac{y(t+h)-y(t)}{h}\hat{j}=(x'(t),y'(t))$.

Is the limit of a vector-valued function as the parameter approaches some value defined to be the limit of each of its components as the parameter approaches that value, or is there a definition for these limits that's similar to the epsilon-delta definition of limits that's used for single variable functions?

答:
You can use the Euclidean norm.

Definition of $\lim_{(x,y) \to (x_0, y_0)} (f(x), g(y)) = (a,b)$:

For any $\epsilon > 0$ there exists $\delta$ such that if $(x,y)$ satisfies $\sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta$, then $\sqrt{(f(x)-a)^2 + (g(y)-b)^2} < \epsilon$.

In finite dimensions, the choice of Euclidean norm is arbitrary, since all norms are equivalent.

Title: Limits of Vector Valued Functions


问:
Drawing Lines between Rectangles - Vector Network/Geometry Math Algorithm
I need help coming up with an algorithm that draws lines between rectangles. Why? I'm a UX Designer working on a plugin for the design tool I use. The plugin should be able to connect two mockup screens with arrows in an efficient way.

Here's the prompt I'm giving myself for this problem:

Imagine two standard (not-rotated) rectangles on an infinite plane. On each rectangle, a side has been chosen. You must come up with an algorithm to describe a series of connected lines (a vector network) that connects the midpoints of each selected rectangle sides (starting and ending nodes).

The segments cannot intersect the rectangles' sides, the segments (vector edges) must be perfectly vertical or horizontal (meaning all turns are right angles), and it must produce the minimal number of segments.

Here's an image to help: two rectangles connected by segments of lines that start and end at various points of the rectangles

I'm looking for something that will always describe the vertices (aka nodes) between the rectangles regardless of their position in the plane or the side chosen. For example, the output would look like: {3, 5} {10, 5} {10, 20} {32, 20}.

Does anyone have ideas or literature that might help me get started?

答:
It seems this is partly described here:
https://stackoverflow.com/questions/4647454/drawing-a-line-connecting-two-rectangles

easy to extend to other sides


问:
Does every category have a functor?
Is there any one (or more) categories that doesn't have a functor? Functors go between categories, so is there any category that only has an identity functor but no other functor that maps it to another category?</p>

答:
For every category $C$ there is a functor from $C$ to the category $1$ consisting of exactly one object and one arrow.

So you can map every category $C$ to the category $1$.

Conversely, $1$ can be embedded into any other category.

So every category "has a functor" other than identity functor.

Title: Does every category have a functor?


问:
what do you mean by the length of an integer k in base b?
I have a problem where there is a statement that says there is an ID 'n' which is non-negative integer of length 'k' in base 'b'?
I dont understand what they are trying to say by this?
Is it something to do with log? like log with base 'b' or something?
can someone please explain with an example?</p>

答:
It means it has $k$ digits. Nothing more nothing less.

You can use logs and exponents to estimate the size of a base $b$ number with $k$ digits by the following reasoning.

A $1$ digit number Can be as small as $1$ or as large as $b-1$.  So if your value is between $1$ and $b-1$ it has $1$ digit.

A $2$ digit number can be as small as $10_b = b$ or large as $[b-1][b-1]_b = (b-1)\cdot b + (b-1) = b^2 -1$.  so if you value is between $b$ and $b^2-1$ it will have $2$ digits.

A $3$ digit number can be as  small as $100_b = b^2$ or as large as $[b-1][b-1][b-1]_b = (b-1)\cdot b^2 + (b-1)\cdot b + (b-1) = b^3 - 1$.  so if your value is between $b^2$ and $b^3-1$ it will have $3$ digits.

ANd generalizing. A $k$ digit number can be as small as $100....0_b = b^{k-1}$ or as large as $[b-1][b-1]....[b-1] = b^{k} -1$.

so if you want to know how many digits a number will have in base $b$ we do the following.

If we want to know how many digits $M$ has, we say. Let $M$ have $k$ digits.  To solve for $K$ we figur that $b^{k-1} \le M \le b^{k+1} -1< b^{k}$.

so $\log_b b^{k-1} \le \log_b M < \log_b b^{k}$ so

$k - 1\le \log M < k$.  And that is how we solve for $k$ and find out how many digits $M$ will have in base $b$.

For example if we wante to figure out how many digits $3456$ will have in base $6$ we figure if it has $k$ digits then $6^{k-1} \le 3456 < 6^{k}$  so $k-1 \le \log_6 3456 = \frac {\ln 3456}{\ln 6} < k$. so $k-1 \le 4.35 < k$ so $k = 5$.  It will have $5$ digits.

And indeed $6^4 = 1296 < 3456 < 6^5= 7776$

so $3456 = 2*1296 + 864$.  And $6^3 = 216$ and $3456 = 2*1296 + 4*216 + 0$.  (That was a complete coincidence!  I didn't plan it.)  So $3456_{10} = 24000$ has $5$ digits.

=======

Alternatively we can go the other way.

If you have $243253242_6$ how many digits will it have in base $10$.

well,  $M = 243253242$ has $9$ digits so $6^8 \le  M \le 6^9 -1$.

$\log_{10} 6^8 = 8\log_{10} 6 \approx  6.2$ and $\log_{10} 6^9 = 9\log_{10} 6\approx 7$  so if $243253242_6$ has $k$ digits then $k- 1=6$ and it will have $7$ digits.

And indeed $2*6^8 =3359232_{10} < M < 2*6^8 + 5*6^7 = 5318784_{10}$ which has $7$ digits.


问:
Find an improper fraction equal to $9$
Simple premise but I can't get anywhere with this problem!

The problem involves a fraction of the form $\frac{ABCDE}{FGHIJ}$ where $A,B,C,D,E,F,G,H,I,J$ are a number from $0-9$, each being used only once. This improper fraction must be equal to $9$ and I believe there are 3 possible answers.

I began working out the number of possible combinations, but there were thousands so I decided that was a dead end! Then I started looking into excel solutions but again I hit a wall. Any help would be appreciated.

Edit: I also believe that the $0$ is not allowed at the start of the 5 digit numbers i.e $A,F\neq 0$

答:
Here are some reductions that should hopefully be able to get you started. They are motivated by pretending the order of $A,B,C,D,E,F,G,H,I,J$ chosen is "random," and figuring out what sorts of things could stop the quotient from being $9$.

You should expect the quotient to be roughly $1$, and, since it's the quotient of two $5$-digit numbers (two numbers of roughly the same size), it really shouldn't be that much larger than $1$. In particular, if $F\neq 1$, then $FGHIJ\geq 2\cdot 10^4$ and then $ABCDE \geq 18\cdot 10^4$, which can't occur. So, $F$ must be $1$, and $A$ must be $9$.

In addition, for the numerator to be $<10^5$, we need the denominator to be $<\frac{10^5}{9}<12000$; since $G$ cannot also be $1$, $G$ must be $0$.

$9$ must certainly divide the numerator. This means that $9|A+B+C+D+E$, and since $$9|A+B+C+D+E+F+G+H+I+J=45,$$ we also know that $9|F+G+H+I+J$, and thus the denominator is divisible by $9$ as well. So, the numerator is divisible by $81$. The fact that it lies between $9\cdot 10^4$ and $10^5$, has all distinct digits, and can't have a digit of $0$ or $1$ should narrow it down a fair bit.

Title: Find an improper fraction equal to $9$


问:
Is the 2-norm of a matrix bounded by the maximum of its 1-norm and Infinity-norm?
I am implementing the algorithm in "Approximating the Logarithm of a Matrix to Specified Accuracy" by Sheung Hun Cheng, Nicholas J. Higham, Charles S. Kenny, Alan J. Laub, 2001.

In this algorithm, I would to avoid computing the 2-norm of a real-valued square matrix $A\in\mathbb{R}^{n\times n}$. Numerical experiments suggest to me that the following upper bound holds

$\|A\|_2 \leq \max ( \|A\|_1, \|A\|_\infty )$

Can anybody confirm whether this inequality always holds? Thank you and happy new year!

One user remarked that Cauchy-Schwarz implies

$\|A\|_2 \leq \sqrt n \min ( \|A\|_1, \|A\|_\infty )$

which in some cases improves the bound, but not always. So I hope my initial question is still of relevance. A counter example to the suggested inequality would also be appreciated, if it exists.

答:
Indeed:

$\|A\|_2 \leq \max ( \|A\|_1, \|A\|_\infty )$

follows from

$\|A\|_2 \leq \sqrt { \|A\|_1  \|A\|_\infty } \leq \max ( \|A\|_1, \|A\|_\infty )$

which - according to Wikipedia - is a special case of Hölder's inequality.


答: Well I think you can guarantee that $A=9$ and $F=1$, just because otherwise would make it impossible to have it end up reducing to $9$ (it would be too small), unless you can say that $F=0$. For the rest, you can narrow it down and see that the numerator has to be AT LEAST divisible by $9$ (because it ends up reducing to $9$), but otherwise, I don't have anything for you.

Title: Find an improper fraction equal to $9$

Body:
Simple premise but I can't get anywhere with this problem!

The problem involves a fraction of the form $\frac{ABCDE}{FGHIJ}$ where $A,B,C,D,E,F,G,H,I,J$ are a number from $0-9$, each being used only once. This improper fraction must be equal to $9$ and I believe there are 3 possible answers.

I began working out the number of possible combinations, but there were thousands so I decided that was a dead end! Then I started looking into excel solutions but again I hit a wall. Any help would be appreciated.

Edit: I also believe that the $0$ is not allowed at the start of the 5 digit numbers i.e $A,F\neq 0$


答:
Since $F$ cannot be $0$, the denominator is at least $10,000$.  Therefore, the numerator is at least $90,000$.  Since the numerator is a $5$-digit number which is at least $90,000$, we know that $A=9$.

Since the numerator is a $5$-digit number, it is less than $100,000$, dividing this by $9$ shows that the denominator is at most $11,111$.  Therefore, $F=1$ and $G=0$ or $1$.

Since numbers cannot be repeated, $G\not=1$ and so $G=0$.

At this point, we know the values of $A$, $F$, and $G$ and the remaining digits are $2,3,4,5,6,7,8$.

The smallest that the denominator can be is $10,234$ as all other digits are taken and the largest the denominator can be is $10,876$ since the larger digits are taken.  Therefore, the numerator varies between $92,106$ and $97,884$.  In particular, $B$ is between $2$ and $7$ and $H$ is between $2$ and $8$.

Now, let's do a bit of a case-by-case analysis.

If $H$ is $2$, then the largest that the denominator could be is $10,287$.  In this case, the numerator is $92,583$.  Therefore, $B$ is at most $2$, but we already know that $B$ is at least $2$, so $B=2$ in this case.  This is not possible since $H$ was already assumed to be $2$.

Therefore, $H$ cannot be $2$ and the smallest that the denominator can be is $10,324$, so $H$ is between $3$ and $8$.

If $H$ is $3$, then the largest that the denominator could be is $10,387$.  In this case, the numerator is $93,483$.  Therefore, $B$ is at most $3$, but since $H$ is $3$ and $B$ is at least $2$, it must be that $B=2$.  Therefore, the largest that the numerator could be is $92,876$.  Dividing this by $9$, we find that the denominator is at most $10,319$, which contradicts the conclusion that we have that the denominator is at least $10,324$.

Therefore, $H$ cannot be $3$, so $H$ is between $4$ and $8$ and the smallest that the denominator can be is $10,423$.  Multiplying by $9$ gives us that the numerator is at least $93,807$, so $B$ is between $3$ and $7$.

If $H=4$, then the largest that the denominator could be is $10,487$.  Therefore the largest that the numerator could be is $94,383$.  Hence, we find that $B$ is at most $4$, but we already know that $H=4$, so that is not possible.  Since $B$ is at least $3$, it must be that $B=3$.  The largest that the numerator could be when $B=3$ is $93,876$.  So, by dividing by $9$, the largest that the denominator could be is $10,430$.  By checking all the values between $10,423$ and $10,430$ for the denominator, we see that none of them satisfy the requirements.

Therefore, $H$ cannot be $4$ and the smallest the denominator can be is $10,523$.  In this case, the numerator is at least $94,788$.  Therefore, $H$ is between $5$ and $8$ and $B$ is between $4$ and $7$.

If $H=5$, then the largest the denominator could be is $10,587$.  Therefore, the largest the numerator could be is $95,283$.  Therefore, $B$ is at most $5$, but since $5$ is already taken, it must be that $B=4$.  The largest that the numerator could be is then $94,876$.  This means that the denominator is at most $10,541$.  Therefore, we know that the denominator is between $10,523$ and $10,541$, which leaves only a few cases to check.

At this point, it is helpful to note that the units digits of either number can't be $0$ or $5$ because then the units digit of the other number is also $0$ or $5$.  In addition, the units digit of either number can't be $1$ or $9$ since those have already been used.  By careful inspection, there are only ten denominators to check when $H=5$.

Let's assume that you've dealt with the case where $H=5$ and checked all ten denominators.  Therefore, we're assuming that $H$ is at least $6$, so the smallest the denominator could be is $10,623$.  Then the numerator is at least $95,607$, so $B$ is between $5$ and $7$.

Now, you can continue with this pattern on up, you would next assume that $H=6$ and see if you can determine $B$.  Then, find the largest that the numerator can be and use that to restrict the denominator.  You'll end up with a few more numbers to check for each value of $H$, but by looking closely, you can cut out a lot of options for the denominator and make each calculation reasonable (I'll leave the rest to you).

You can also use @Carl Schildkraut's answer to really cut down the things that you need to check, to just a few for each possibility for $H$.

Title: Find an improper fraction equal to $9$


问:
"Differential of" as an operator
I am reading the paper Learning Dynamical Systems from Partial Observations and I am having a hard time following the calculations in Appendix B. Let $F: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a multivariate function. The paper defines the differential $\partial_u F(u_0)$ of $F$ with respect to $u$ as the operator such that
$$F(u_0 + \delta u) = F(u_0) + \partial_u F(u_0) \delta u + o(\delta u).$$

Now, let $F_{\theta}$ to mean that $F$ is a function of $\theta$. Likewise, let $X^{\theta}$ mean that $X$ is a function of $\theta$.

The paper states that

$$\partial_{\theta} F_{\theta} (X_t^{\theta} + \partial_{\theta} X_t^{\theta} \cdot \delta \theta + o(\delta \theta)) = \partial_{\theta} F_{\theta}(X_t^{\theta}) + \partial_{X} \partial_{\theta} F_{\theta}(X_t^{\theta}) \cdot \partial_{\theta} X_t^{\theta} \cdot \delta \theta + o(\delta \theta).$$

Here are my questions:

Is the definition of the differential accurate? Shouldn't it be
$$F(u_0 + \delta u) = F(u_0) + \partial_u F(u_0) \delta u + o(\|\delta u\|^2)$$
instead?

Can anyone guide me on the calculation of $$\partial_{\theta} F_{\theta} (X_t^{\theta} + \partial_{\theta} X_t^{\theta} \cdot \delta \theta + o(\delta \theta))?$$ I can't figure out the calculation. Just looking at the resulting expression above, I wonder why the argument $(X_t^{\theta} + \partial_{\theta} X_t^{\theta} \cdot \delta \theta + o(\delta \theta)$ is not present at all in the $F_{\theta}$'s that appear on the right hand side if the chain rule was indeed used.

Suggestions appreciated. Thanks!

答:
Question 1

No, it should not be $o\left(\Vert\delta u\Vert^2\right)$, but some mathematicians would likely prefer $o\left(\Vert\delta u\Vert\right)$ over $o(\delta u)$. See English Wikipedia on "the total derivative as a linear map" for details, or search the internet for things like little o definition of derivative.

Question 2

The paper writes $\partial_\theta F_\theta\left(X_t^\theta+\partial_\theta X_t^\theta\cdot\delta\theta+o(\delta\theta)\right)=\partial_\theta F_\theta\left(X_t^\theta\right)+\partial_X\partial_\theta F_\theta\left(X_t^\theta\right)\cdot\partial_\theta X_t^\theta\cdot\delta\theta+o(\delta\theta)$, and we must justify this.

For simplicity, use "$\delta X$" to denote $\partial_\theta X_t^\theta\cdot\delta\theta+o(\delta\theta)$. Then we have $\partial_\theta F_\theta\left(X_t^\theta+\delta X\right)$ on the left. By the twice differentiability of $F$, we can apply the earlier $X$-differentiability equation to write \begin{align}\partial_\theta F_\theta\left(X_t^\theta+\delta X\right)&=\partial_\theta F_\theta\left(X_t^\theta\right)+\partial_X\partial_\theta F_\theta\left(X_t^\theta\right)\cdot\delta X+o(\delta X)\\
&=\partial_\theta F_\theta\left(X_t^\theta\right)+\partial_X\partial_\theta F_\theta\left(X_t^\theta\right)\cdot\left(\partial_\theta X_t^\theta\cdot\delta\theta+o(\delta\theta)\right)+o\left(\partial_\theta X_t^\theta\cdot\delta\theta+o(\delta\theta)\right)\\&=\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)+\partial_{X}\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)\cdot\partial_{\theta}X_{t}^{\theta}\cdot\delta\theta+\partial_{X}\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)\cdot o(\delta\theta)\\&\phantom{=}+o\left(\partial_{\theta}X_{t}^{\theta}\cdot\delta\theta+o(\delta\theta)\right)\text{ by linearity of }\partial_{X}\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)\\
&=\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)+\partial_{X}\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)\cdot\partial_{\theta}X_{t}^{\theta}\cdot\delta\theta+o(\delta\theta)\checkmark\end{align}

The last step in the above uses the fact that the linear operators $\partial_{X}\partial_{\theta}F_{\theta}\left(X_{t}^{\theta}\right)$ and $\partial_{\theta}X_{t}^{\theta}$ are continuous/bounded, so they won't change whether something is $o(\delta\theta)$.

Title: "Differential of" as an operator


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Algebraic Number theory you have the Kronecker-Weber theorem.

Title: Landmarks of subjects of mathematics


答: A very strong no! For every two categories $C,D$ and an object $d\in ob(D)$ there is a functor $F:C\to D$ mapping every thing in $C$ to $d$ and $id_d$.</p>

Title: Does every category have a functor?

Body: Is there any one (or more) categories that doesn't have a functor? Functors go between categories, so is there any category that only has an identity functor but no other functor that maps it to another category?</p>


问:
Is number of knot strands an invariant?

Question: Does the number of components in a knotwork depend on the particular planar embedding?

I've been investigating how to compute the number of components ("separate strands") in a Celtic knot based on the underlying planar graph structure. (See relationship between knots/links and planar graphs here).

Apparently the calculation for general graphs is a little complicated; for example, the reference in this question points out that for a uniform $m\times n$ grid of squares, the number of components is $\mathrm{lcd}(m,n)$.

It would satisfy me to find a formula for computing the number of components ("strands"), or a relationship between the number of strands and various graph properies such as its degree, spectrum, etc., even if those properties were hard to calculate.

One approach I've taken is in terms of connected components: each separate strand  follows a particular trajectory, and the connected components of those trajectories correspond exactly to the strands. You can define the trajectory as a transition function mapping (some additional structure plus) each edge to its successor; this is a permutation on (structured) edges whose cycles are the components.

The transition function can be encoded as its own, derived, directed graph (similar to a graph-encoded map), whose connected components are the components of the knotwork. From linear algebra, we know that the number of connected components can be recovered as the multiplicity of the zero eigenvalue of the adjacency matrix's Laplacian.

However, I know that the same graph $G$ can have multiple non-isomorphic planar embeddings (i.e. whose duals are non-isomorphic). So far in my experience, this has changed some of the knotting properties (such as number of twists in each component) but not the number of components:

My question is this:

Question: Does the number of components in a knotwork depend on the particular planar embedding? How do we prove it?

My intuition says that the number of components is an invariant, but I haven't been able to produce a counterexample or proof using my approach above.

Conjecture: If $G$ is a graph, then the corresponding knotwork has $c$ components, where

$$T_G(-1,-1) = (-1)^{|E(G)|}\cdot (-2)^{c - 1}$$

and $T_G$ is the Tutte polynomial, and $|E(G)|$ is the number of edges in the graph. (?)

答:
Let $D$ be the diagram of a link. For example, $D$ could be the diagram of the Celtic knot or link pictured in your post. Let $G$ be the checkerboard graph of $D$. The graph $G$ is the graph described in your first bullet point.

Answer: The number of components of $D$ is determined by the abstract graph $G$ and does not depend on how $G$ is embedded in the plane.

To the best of my knowledge, this was first proved by Michel Las Vergnas in 1979. He showed that the number of components of $D$ is determined by the Tutte polynomial evaluation $T_G(-1,-1)$. Since the Tutte polynomial doesn't depend on a particular embedding of $G$, the result follows. The reference for this paper is

Las Vergnas, Michel. On Eulerian partitions of graphs. Graph theory and combinatorics (Proc. Conf., Open Univ., Milton Keynes, 1978), pp. 62–75,
Res. Notes in Math., 34, Pitman, Boston, Mass.-London, 1979.

I could not easily find a copy of the above paper, so here's another way to get the solution, due to Dan Silver and Susan Williams (arXiv link). They define a matrix $Q_2(G)$ whose entries are in the field with two elements $\mathbb{F}_2$ as follows. Both the rows and columns of the matrix are indexed by the vertices $v_1,\dots,v_n$ of $G$. If $i\neq j$, then the $ij$ entry of $Q_2(G)$ is the number of edges between vertices $v_i$ and $v_j$ (taken$\mod 2$). The $ii$ entry of $Q_2(G)$ is the sum of the other entries in the row $i$ (again taken$\mod 2$). Equivalently, we could say the $ii$ entry in $Q_2(G)$ is the sum of the other entries in the column $i$.

In Theorem 1.1 of the linked paper, they prove that the number of components of $D$ equals the nullity of $Q_2(G)$. They note in Remark 1.2 that this implies the number of components of $D$ is independent of the plane embedding of $G$.

Edit: I don't have access to the Las Vergnas paper, but I can give another explanation of the result using the Tutte polynomial and the Jones polynomial.

Let $L$ be an alternating link, let $D$ be an alternating diagram of the link, and let $G$ be the checkerboard graph of $D$. Then the Tutte polynomial $T_G(x,y)$ of $G$ and the Jones polynomial $V_L(t)$ of $L$ are related as follows:
$$V_L(t) = f_D(t) T_G(-t,-t^{-1})$$
for the function $f_D(T)$ defined by
$$f_D(t) = (-1)^{w(D)}t^{\frac{1}{4}(|E| - 2(|V|-1)+3w(D))}$$
where $w(D)$ is the writhe of $D$, $|E|$ is the number of edges in $G$, and $|V|$ is the number of vertices of $D$. Notice that $|f_D(1)|=1$, and thus $|V_L(1)| = |T_G(-1,-1)|$.

The Jones polynomial satisfies the skein relation
$$(t^{\frac{1}{2}}-t^{-\frac{1}{2}})V_{L_0}(t) = t^{-1}V_{L_+}(t) - tV_{L_-}(t)$$
where $L_+,L_-,$ and $L_0$ are as below.

Setting $t=1$ in the above skein relation yields $V_{L_+}(1)=V_{L_-}(1)$. In other words the Jones polynomial evaluated at $t=1$ does not change under crossing changes, and thus $V_L(1)=V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(1)$ where $\bigcirc\sqcup\dots\sqcup\bigcirc$ is the trivial link with the same number of components as $L$. The Jones polynomial of $\bigcirc\sqcup\dots\sqcup\bigcirc$ is $V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(t) = (-t^{\frac{1}{2}}-t^{-\frac{1}{2}})^{m-1}$ where $m$ is the number of components of $\bigcirc\sqcup\dots\sqcup\bigcirc$. Thus
$$|T_G(-1,-1)|=|V_L(1)|=|V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(1)| = 2^{m-1}.$$

The above case handles when $L$ is alternating. If $L$ is non-alternating, then proceed as follows. Let $D$ be any diagram of $L$. Define $D_{\text{alt}}$ to be a diagram with the same shadow as $D$ but whose crossings are changed to be alternating, and define $L_{\text{alt}}$ to be the link whose diagram is $D_{\text{alt}}$. Note that $D$ and $D_{\text{alt}}$ have the same checkerboard graph $G$. The above argument implies that $|T_G(-1,-1)|=2^{m-1}$ where $m$ is the number of components of $L_{\text{alt}}$. Since $L_{\text{alt}}$ and $L$ have the same number of components, the result follows for $L$ as well.

Title: Is number of knot strands an invariant?


问:
Numbers to base $i$
Does it make sense to think of numbers to base $i$, i.e. $\sqrt{-1}$?

For conventional bases e.g. $N = 2, 8, 10, 16$ etc., we talk of using digits from $0$ through the base $N-1$ in the numeric representation. What digits would be permitted for base $i$?

What about other unconventional bases like $\pi$? $e$? Irrational numbers?

Just need a "frame of reference" to think through these things logically. Would appreciate pointers to papers or websites. I am not a Math PhD, just a hobbyist, so please keep explanations simple. Thank you.

答: As far as base $i$ goes, see Donald Knuth's Quarter-Imaginary System: https://en.wikipedia.org/wiki/Quater-imaginary_base. It's pretty cool, and I think almost all complex numbers can be expressed uniquely (some may have two expressions). You could have base e or base pi, but they don't have much practical use imo (not just for scientific application, but in a pure mathematical sense, it would also have like no use). For something like base $\sqrt{2}$, it would alternate between just being binary and being a weird root 2 thing, which is interesting but again, not really practical.

Title: Numbers to base $i$


问:
Using Dominance Convergence Theorem in derivation of Leibniz's rule
I really need help understanding how Dominance Convergence Theorem applies to the following limit:

$$
\underset{\Delta t\rightarrow 0}{lim}\int_{a(t)}^{b(t)}\frac{f(x, t+\Delta t)-f(x,t)}{\Delta t}dx
$$

So that we can change the limit to the following integral:

$$
\int_{a(t)}^{b(t)}\underset{\Delta t\rightarrow 0}{lim}\left ( \frac{f(x, t+\Delta t)-f(x,t)}{\Delta t} \right )dx
$$

For clarification as to where I got this limit from, I was watching a video on the derivation of Leibniz's rule (link) and this was the only portion that I was unable to understand. The person doing the derivation just said we can switch the integral and limit, but I would like to understand why. Looking into it on my own, I understand that Dominance Convergence Theorem is used to switch a limit and integral, I just don't understand why this theorem can be applied in this specific case.

答:
Fix $t$. Let $g_h(x) = \frac{f(x, t+h) - f(x,t)}{h}$. If there exists a function $g^*$ that

is integrable on the interval $[a(t), b(t)]$, and

is such that $|g_h(x)| \le g^*(x)$ for all $x \in [a(t), b(t)]$,

Then $\lim_{h \to 0} \int_{a(t)}^{b(t)} g_h(x) \, dx = \int_{a(t)}^{b(t)} \lim_{h \to 0} g_h(x) \, dx$ (i.e. you can switch the order of limit and integration.)

I didn't watch the video, but I assume there are some assumptions on $f$ such that the above conditions hold.


答: The problem is that the powers of $i$ don't grow because $i^4=1$.  Negative integers work fine for bases.  If you work in base $-2$ the position values are $1,-2,4,-8,16\ldots$.  You can represent all positive and negative integers with just $0$ and $1$.  $9_{10}=16-8+1=11001_{-2}$.  With a bit of work you can find algorithms for normal arithmetic.  Irrational bases can be made to work using digits up to the largest natural less than the base.  Base $\pi$ would use $0,1,2,3$.  You don't get a unique representation for some numbers and integers have infinite decimals.  That makes it such a mess that I haven't seen anyone find a use for it.

Title: Numbers to base $i$

Body:
Does it make sense to think of numbers to base $i$, i.e. $\sqrt{-1}$?

For conventional bases e.g. $N = 2, 8, 10, 16$ etc., we talk of using digits from $0$ through the base $N-1$ in the numeric representation. What digits would be permitted for base $i$?

What about other unconventional bases like $\pi$? $e$? Irrational numbers?

Just need a "frame of reference" to think through these things logically. Would appreciate pointers to papers or websites. I am not a Math PhD, just a hobbyist, so please keep explanations simple. Thank you.


问:
The Baer-McCoy (a.k.a. prime) radical of $A$

Let $B$ a ring and let $A$ a subring of $B$.

Show that $A\cap \mathrm{Nil}_{*}(B)\subset \mathrm{Nil}_{*}(A)$.

If $A$ is contained in the center of $B$, show that $A\cap \mathrm{Nil}_{*}(B)=\mathrm{Nil}_{*}(A)$.

The notation $\mathrm{Nil}_{*}(A)$ being the lower nilradical, radical of Baer-McCoy, or prime radical of $A$.
$$\mathrm{Nil}_{*}(A)=\sqrt{(0)}$$

I know that if $A$ is conmutative then equality in 2) trivially holds.

I ask for your help in section 1.

Could be helpful that $Nil_{*}(B)$ is the set of nilpotent elements?

答:
(Hints along the path of $m$-systems, ala Manny Reyes's suggestion.)

A set $S\subset R$ is called an $m$-system if whenever $a,b\in S$, there exists $r\in R$ such that $arb\in S$.

You can check that $P$ is a prime ideal iff $R\setminus P$ is an $m$-system.

Now also along the lines of the commutative version of multiplicative systems, you can show (by the routine Zorn's Lemma argument) that an ideal maximal with respect to being disjoint from an $m$-system is a prime ideal.

So: suppose that $a\in A$ but there is a prime ideal $P\lhd A$ such that $a\notin P$. Then $A\setminus P$ is an $m$-system in $A$, and also in $B$. Deduce that there is a prime ideal of $B$ not containing $a$. From this, conclude that the containment you desire holds.


问:
Why is the non-symmetric function, the Jacobian used in changes of variables?
For example, if we take the polar coordinates $x(r, \theta) = r\cos \theta$ and $y(r, \theta) = r \sin \theta$ the Jacobian is

$$\frac{\partial(x, y)}{\partial(r,\theta)} = \frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial x}{\partial\theta}{\frac{\partial y}{\partial r}} = r\cos^2(x) + r\sin^2(x) = r$$

Hence $dxdy = r\space drd\theta$ This is not symmetric with respect to the variables like one would expect though. Expressing $x$ and $y$ as functions of $r$ and $\theta$ in that order is an arbitrary convention. We could just as easily write $x(\theta,r) = r\cos\theta$ and $y(\theta, r) = r\sin\theta$ in which case the Jacobian would be $-r$ and $dx dy$ would transform into $-r\space drd\theta$.

答:
The change-of-variables formula is with the absolute value of the determinant of the Jacobian, not with just the Jacobian itself. So in either case we have
$$ dx dy = | \det J |dr d\theta = r dr d\theta.$$

See, for example, https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables


问:
An unbounded sequence s.t. $\forall t>0 |[a_nt]-a_nt|\rightarrow 0$
While studying probability I accidentally proved the result below by what I think is an overly complicated proof (a sketch of this proof below). Is there an easy way to prove it?

The result is the following.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals such that for all $t>0$ we have $\lim_{n} |[a_nt]-a_nt|=0$, where we denote with $[x]$ the nearest integer to x.

Then $\{a_n\}_{n\in\mathbb{N}}$ converges to $0$.

This is equivalent to another result.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals and $a$ a real number such that for all $t>0$ we have $\lim_{n} \text{e}^{\text{i}a_nt}=e^{\text{i}at}$

Then $\{a_n\}_{n\in\mathbb{N}}$ converges.

Proof. Let us suppose that $\{a_n\}_{n\in\mathbb{N}}$ doesn't converge. We will study two possible scenarios, both leading to contradiction.

First case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded.
Then, choosing a small enough $t$, we get that $\forall n |a_nt|<\pi$, and we conclude, using the continuity of the complex logarithm, that the sequence converges to $a+2k\pi$ for some $k\in\mathbb{Z}$.

Second case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is unbounded. Then, we consider a sequence $\{X_n\}_{n\in\mathbb{N}}$ of random variables such that $\forall n X_n\sim N(a_n,1)$ and their respective characteristic functions $\{\text{exp}(\text{i}a_nt-t^2/2)\}_{n\in\mathbb{N}}$. Our initial assumptions imply that the sequence of characteristic functions is pointwise convergent to $\text{exp}(\text{i}at-t^2/2)$, so, by Paul-Levy's theorem, the sequence of random variables $\{X_n\}_{n\in\mathbb{N}}$ converges in distribution to a random variable $X$ such that $X\sim N(a,1)$. However, it can be proved via Chebishev inequality that if the sequence of numbers $\{a_n\}_{n\in\mathbb{N}}$ is unbounded then the sequence of random variables doesn't converge in distribution to any random variable.

答:
Suppose $a_n$ does not converge to zero, so there is some $\varepsilon > 0$ for which infinitely many $a_n$ have $|a_n| > \varepsilon$. Let $\{a_{n_k}\}$ be the subsequence consisting of those elements with absolute value exceeding $\varepsilon$, and let $b_k = a_{n_k}/\varepsilon$, so all $|b_k| > 1$ and $|[b_k t] - b_k t| \to 0$ for all $t > 0$. Define $f(x) = |[x] - x|$, and note that $f$ is nonnegative, bounded above by $1$, periodic with period $1$, and has $\int_0^1 f(x) \,dx = \frac{1}{4}$. Then since $f(b_n x)$ is periodic with period $\frac{1}{|b_n|}$, $f(b_n x)$ achieves $\lfloor |b_n| \rfloor \geq \frac{|b_n|}{2}$ full periods on the interval $[0, 1]$, and thus
$$\int_0^1 f(b_n x) \,dx \geq \frac{\lfloor |b_n| \rfloor}{4|b_n|} \geq \frac{1}{8}.$$

Now, for each $N$, let $B_N$ be the set of $t \in (0, 1)$ for which $f(b_n t) < \frac{1}{20}$ for all $n \geq N$. Then $\{B_N\}$ is an ascending sequence of (measurable) sets, and by assumption, each $t$ is in some $B_N$, so $\bigcup_N B_N = (0, 1)$. This means $\lim_{N \to \infty} \mu(B_N) = 1$, so in particular there is some $N$ for which $B_N$ has measure at least $1 - \frac{1}{20}$. But then for this $N$ we have
$$\int_0^1 f(b_N x) \,dx \leq \int_{B_N} f(b_N x) \,dx + \int_{(0, 1) - B_N} f(b_N x) \, dx \leq \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$$
contradicting our lower bound on the integral. Thus $\{a_n\}$ must converge to zero.


问:
The limit points of the disk $\{z \mid|z| < 1\}$
I'm having troubles proving that the complex numbers $x$ such that $|x|=1$ are the limit points of the complex open disk  $\{z \in \mathbb{C} \mid |z| < 1\}$. I tried to consider the complex numbers $(x_{1}- \frac{1}{n}, x_{2}- \frac{1}{n})$ for such a $x$ but I'm not sure that these complex are in the disk in question.

Could you help me?

答:
I'll denote by $\mathbb{D}$ the open unit disc and $\partial\mathbb{D}$ its boundary (i.e. the unit circle).

Clearly every point of $\mathbb{D}$ is a limit point. Let $z'\in\partial\mathbb{D}$, and consider the sequence $(z_n)_{n\in\mathbb{N}}$ where $z_n=(1-1/n)z'$ for each $n\in\mathbb{N}$. It's easy to check that this sequence converges to $z'$, and that each $z_n$ has $|z_n|<1$, so each $z_n\in\mathbb{D}$. This shows that $z'$ is a limit point of $\mathbb{D}$, and hence that $\partial\mathbb{D}$ is a subset of the limit points of $\mathbb{D}$.

Now for any $z'$ with $|z'|>1$, consider the open ball $B(z',|z'|-1)$. Check that this open ball has empty intersection with $\mathbb{D}$ - this suffices to show that the limit points of $\mathbb{D}$ are precisely $\mathbb{D}\cup\partial\mathbb{D}$.


问:
$\ker \varphi_p \subset (\ker \varphi)_p$ where $(\cdot)_p$ is taking the stalk of sheaves at the point $p$ (Diagram inside!).
I've already proven that $(\ker \varphi)_p \subset \ker \varphi_p$ using a commutative diagram and the definition $F_p = \lim\limits_{\longrightarrow \\ U \ni p} F(U) = \bigsqcup\limits_{U \ni p} F(U)/ \sim$ where $[x] = [y]$ in the equivalence relation $\iff \rho_{UW}(x) = \rho_{VW}(y)$ for some $W \subset U \cap V$ open sets in the topology $X$.  I got that from Wikipedia article for direct limit.  Where $\rho_{UV}$ are the restriction maps of the sheaf $F$.

So I have a morphism of sheaves $\varphi : F \to G$, which induces $\varphi_p : F_p \to G_p$.  Intuitively it seems like if we assume $[x] \in \ker \varphi_p$ it's going to take a "subtraction trick" to show that $y \sim x$ for some $y \in \ker \varphi(U)$.  I'm just not sure where to deploy this trick.

The sheaves are abelian group-valued.

Any ideas?

The red and the blue represent two different diagram chases.  First we assume (blue) $[x] \in \ker \varphi_p$.  That's as far as I've gotten and shows that I did some work. :)

Diagram drawn using: https://www.mathcha.io/

$j$ is the inclusion map I proved in the converse direction on paper.

答: Suppose $x \in F_p$ is in $\ker(\varphi_p)$.  Then there exist a neighborhood $U$ of $p$ and $\bar x \in F(U)$ such that $(\bar x)_p = x$.  Furthermore, since $(\bar x)_p \in \ker(\varphi_p)$, we have that $[\varphi_U(\bar x)]_p = 0 \in G_p$.  Therefore, there exists some neighborhood $V \subseteq U$ of $p$ such that $[\varphi_U(\bar x)] |_V = 0$.  But then $[\varphi_U(\bar x)]|_V = \varphi_V(\bar x |_V)$.  So, $\bar x |_V \in [\ker(\varphi)](V)$, and $[\bar x |_V]_p = x$.


问:
Surface Area of Cap below, Top Part of Sphere
How do I find the Surface Area of an oval cap below (Top of sphere)?

Oval Cape

Internet is stating: 2πRh

However, a simple 8 radius circle is π * 8^2 = 201

So with 8 radius with height of 1: 2π * 8 * 1 = 50.24 (This is less than above, and doesn't make sense)

Not sure, what the real answer is.

答:
Hint:

Express $R$ in the area formula $A=2\pi hR$ in term of $h$ and $r$, which comes out as
$$R= \frac12h\left(1+\frac{r^2}{h^2}\right)$$


问:
Fourier transform in $\mathbb{L}^2(\mathbb{R})\setminus\mathbb{L}^1(\mathbb{R})$
In a book I've found  the following definition for the Fourier transform in  $\mathbb{L}^2(\mathbb{R})\\$:

$g(k)$=$\frac{1}{\sqrt{2\pi}}\frac{d}{dk}\int_{-\infty}^{+\infty}\frac{e^{-ikx}-1}{-ix}f(x)dx\\$
Now I don't understand why this definition is equivalent to the usual one that is:
$g(k)=\lim_{N \to \infty} \frac{1}{\sqrt{2\pi}}\int_{-N}^{+N}e^{-ikx}f(x)dx\\$  where the limit is in the sense of the norm $||\cdot||_2$. Where can I find a proof of the equivalence of the two definitions?

答:
\begin{align}
    \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{e^{-ikx}-1}{-ix}f(x)dx
    &=\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\int_0^k e^{-isx}ds f(x)dx \\
    &=\lim_{R\rightarrow\infty}\int_0^k\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-isx}f(x)dx ds \\
    &=\int_0^k \hat{f}(s)ds
\end{align}


问:
Find all positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$
All I have got so far is that $a$ must divide $2002$... Could anyone share some ideas? I mean is there any method rather than trial and error?

答:
At most one of $a,a+1$ is a multiple of $3$.

If exactly one of $a,a+1$ is a multiple of $3$, then
$$
a^{n+1}-(a+1)^n
$$
would not be a multiple of $3$, contradiction, since $2001$ is a multiple of $3$,

Thus neither of $a,a+1$ is a multiple of $3$, hence $a\equiv 1\;(\text{mod}\;3)$,

If $n$ is odd, then

\begin{align*}
&
a^{n+1}-(a+1)^n=2001
\\[4pt]
\implies\;&
a^{n+1}-(a+1)^n\equiv 2001\;(\text{mod}\;3)
\\[4pt]
\implies\;&
1^{n+1}-2^n\equiv 0\;(\text{mod}\;3)
\\[4pt]
\implies\;&
1-2\equiv 0\;(\text{mod}\;3)
\end{align*}
contradiction.

Hence $n$ is even.$\;$Then

\begin{align*}
&
a^{n+1}-(a+1)^n=2001
\\[4pt]
\implies\;&
a^{n+1}-(a+1)^n\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
(-1)^{n+1}-0^n\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
-1\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
2002\equiv 0\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
(a+1){\,\mid\,}2002
\\[4pt]
\end{align*}

and as you noted, we also have $a{\,\mid\,}2002$, thus both $a$ and $a+1$ are divisors of $2002$.

Hence since $\gcd(a,a+1)=1$, it follows that $\bigl(a(a+1)\bigr){\,\mid\,}2002$.

Then $a(a+1)\le 2002$, so $a < 45$.

From the prime factorization of $2002=(2)(7)(11)(13)$, we get that the only positive integer divisors of $2002$ which are less than $45$ are $1,2,7,11,13,14,22,26$, and of those potential values of $a$, only $a=1$ and $a=13$ are such that $a+1$ is also a divisor of $2002$.

But if $a=1$, then $a^{n+1}-(a+1)^n=1-2^n < 0$, contradiction.

It remains to analyze the case $a=13$.

Suppose $13^{n+1}-14^n=2001$.

If $n > 2$, then

\begin{align*}
&
13^{n+1}-14^n=2001
\\[4pt]
\implies\;&
13^{n+1}-14^n\equiv 2001\;(\text{mod}\;8)
\\[4pt]
\implies\;&
13^{n+1}\equiv 1\;(\text{mod}\;8)
\\[4pt]
\implies\;&
(13^n)(13)\equiv 1\;(\text{mod}\;8)
\\[4pt]
\implies\;&
(1)(5)\equiv 1\;(\text{mod}\;8)
\\[4pt]
\end{align*}
contradiction.

Hence since $n$ is even, the only remaining possibility is $n=2$.

By direct evaluation, it can be verified that $13^3-14^2=2001$.

Therefore $a=13, n=2$ is the only solution in positive integers to the given equation.


答: According to answerer's textual diagram chase.

Title: $\ker \varphi_p \subset (\ker \varphi)_p$ where $(\cdot)_p$ is taking the stalk of sheaves at the point $p$ (Diagram inside!).

Body:
I've already proven that $(\ker \varphi)_p \subset \ker \varphi_p$ using a commutative diagram and the definition $F_p = \lim\limits_{\longrightarrow \\ U \ni p} F(U) = \bigsqcup\limits_{U \ni p} F(U)/ \sim$ where $[x] = [y]$ in the equivalence relation $\iff \rho_{UW}(x) = \rho_{VW}(y)$ for some $W \subset U \cap V$ open sets in the topology $X$.  I got that from Wikipedia article for direct limit.  Where $\rho_{UV}$ are the restriction maps of the sheaf $F$.

So I have a morphism of sheaves $\varphi : F \to G$, which induces $\varphi_p : F_p \to G_p$.  Intuitively it seems like if we assume $[x] \in \ker \varphi_p$ it's going to take a "subtraction trick" to show that $y \sim x$ for some $y \in \ker \varphi(U)$.  I'm just not sure where to deploy this trick.

The sheaves are abelian group-valued.

Any ideas?

The red and the blue represent two different diagram chases.  First we assume (blue) $[x] \in \ker \varphi_p$.  That's as far as I've gotten and shows that I did some work. :)

Diagram drawn using: https://www.mathcha.io/

$j$ is the inclusion map I proved in the converse direction on paper.


问:
Does $\lim\limits_{n\to\infty}\frac{\sum_{i=1}^{n^2} 1}{\sum_{i=1}^n i}$ result into 2?
How does the following limit $$\lim\limits_{n\to\infty}\frac{\displaystyle\sum_{i=1}^{n^2} 1}{\displaystyle\sum_{i=1}^n i}$$ result into 2?
Something like:

$$\frac{1+3+5+7+9+11+13+15+\dots}{1+2+3+4+5+6+7+8+\dots}$$

Can I plot this in WolframAlpha?

Body:
The notation $$\sum_{i=1}^{M}1$$ means you sum $1$, $M$ times, so in your case you sum $1$, $n^2$ times, so

$$\sum_{i=1}^{n^2} 1=\underbrace{1+1+\cdots+1}_{n^2\text{ times} }=n^2$$

while

$$\sum_{i=1}^{n} i=\frac 1 2n(n+1)$$

which means your expression is $$=\frac{2n^2}{n(n+1)}$$ $$=\frac{2n}{n+1}$$ and this goes to $2$ as $n\to\infty$.

Let

$$S_n=1+2+\cdots n$$

Then $$S_n=1+2+\cdots+(n-1) +n\\S_n=n+(n-1)+\cdots + 2 +1$$

and summing column-wise gives:
 $$2S_n=(n+1)+(n-1+2)+(n-2+3)+\cdots +(3+n-2)+(2+n-1)+(n+1)$$

$$2S_n=(n+1)+(n+1)+\cdots+(n+1)=n(n+1)$$ Thus $$S_n=\frac{n(n+1)}2$$

In $\Sigma$ notation, $$S_n=\sum_{k=1}^n k=\sum_{k=1}^n (n-k+1)$$

so $$S_n+S_n=2S_n=\sum_{k=1}^n (n+1)=n(n+1)$$


答:
You may continue with

$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$

knowing that $ \cos x = \frac1{\sqrt2} $ is a solution from inspection, as commented. So, factorize  the equation as

$$ (\cos x - \frac1{\sqrt2})(\cos^3x +(2+\frac1{\sqrt2})\cos^2x+ (\frac12+\sqrt2)\cos x-1+\frac1{2\sqrt2})=0
$$
where the cubic polynomial has one real root that can be calculated analytically with the Cardano formula.

Title: Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $

Body:
How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$

My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$  (\sin x + \sin x \cos x)^2 = (\frac{1}{\sqrt{2}} + \frac{1}{2})^2
$$
$$ \sin^2x + 2\sin^2 x \cos x + \sin^2x \cos^2x - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 =  0
$$

$$ \sin^2 x( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$

$$ (1 - \cos^2x)( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$

$$ 1 + 2 \cos x + \cos^2 x - \cos^2 x - 2 \cos^3 x - \cos^4 x- (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$

$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$

thx for all ur help


问:
A metrizable Lindelöf space has a countable basis
$X$ is called metrizable Lindelöf space if $X$ is a metrizable space and every open covering of $X$ contains a countable subcovering. Would you help me prove that $X$ has a countable basis? Thanks

答:
In the metrizable space $X$, the following conditions are equivalent, cf. General Topology by Engelking Page 255:

$X$ is separable

$X$ is second countable

$X$ is Lindelöf

$X$ has countable extent

$X$ is star countable

$X$ is  CCC


问:
Complex chain rule
Suppose $V_1$ and $V_2$ are open sets in $\mathbb{C}^m$ and $\mathbb{C}^n$, repsectively, $f:V_1\mapsto V_2$ and $g:V_2\mapsto \mathbb{C}^k$ are $\mathbb{R}-$ differentiable. We know that the real chain rule for the differential of $g\circ f$ at a point $z$ gives
$$D_z(g\circ f)=D_{f(z)}g\circ D_zf.$$ Can someone explain how can we prove that
$$\partial_z(g\circ f)=\partial_{f(z)}g\circ \partial _z f+\overline{\partial}_{f(z)}g\circ \overline{\partial}_zf,  $$and
$$\overline{\partial}_z(g\circ f)=\overline{\partial}_{f(z)}g\circ \partial _z f+\partial_{f(z)}g\circ \overline{\partial}_zf?  $$
We use the following notation:
$$\partial _z\varphi=\sum_p\frac{\partial \varphi}{\partial z_p}dz_p\quad \& \quad \overline{\partial}_z\varphi=\sum_p\frac{\partial \varphi}{\partial \overline{z}_p}d\overline{z}_p .  $$ Also that would be great if someone knows a reference for that kind of stuff. Thanks.

答:
You are searching for details about the Wirtinger chain rule: the differential of a function of complex variable(s) is the ordinary differential plus some extra terms which turn out to vanish if such a function is holomorphic.
You can find a detailed explanation in the book by Michael Range "Holomorphic Functions and Integral Representations in Several Complex Variables": 1st chapter in the proof of Lemma 2.2 (page 20 in my edition).</p>

Title: Complex chain rule


问:
Surface integrals in spherical coordinates
If I am given a surface in spherical coordinates $(r,\theta,\varphi)$, such that it is parametrised as:

$$
\begin{align}
r&=r(\theta,\varphi)\\
\theta&=\theta\\
\varphi&=\varphi
\end{align}
$$

What is the area $S$ of such surface? Or more specifically, can you show how to get the result:
$$
S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{r^2+\left(\frac{\partial r}{\partial \theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\frac{\partial r}{\partial \varphi}\right)^2}\;r\sin\theta\;{\rm d}\theta\,{\rm d}\varphi
$$

Some definitions that I am using:

$k$-surface:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is called a $k$-surface, if there exists a non-empty open set $E\subset \mathbb{R}^k$ and a map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, such that: (i) $\varphi(E)=M$, (ii) $\varphi\in C^1(E;\mathbb{R}^N)$, and (iii) the rank of Jacobi matrix of $\varphi$ is equal $k$ everywhere on $E$.
The surface is called simple if $\varphi$ is also injective on $E$ and $\varphi^{-1}$ is continuous of $\varphi(E)$.

Surface integral of the first kind:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is a simple $k$-surface parametrized by the map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, from the open set $E\subset \mathbb{R}^k$ and $f:\mathbb{R}^N\to\mathbb{R}$ is defined on $M$. The surface integral of the first kind is defined by:
$$
\int_M f\,\mathrm{d}S:=\int_E f(\varphi(t))\sqrt{\det{G(D_\varphi(t))}}\,\mathrm{d}t\,,
$$
if the integral on the right exists in the Lebesgue sense and is finite. Here, $G(A)$ denotes the Gramm matrix made from columns of $A$ and $D_\varphi$ is the Jacobi matrix of the map $\varphi$. The numeric value of:
$$
S_k(M):=\int_M f\,\mathrm{d}S\,,
$$
is called the $k$-dimensional surface area of the $k$-surface $M$.

Motivation for the question

Now these definitions can be used to calculate e.g. the surface of a unit sphere. When one describes the sphere by the map (omitting one longitudinal line): $\varphi: (\eta,\psi)\mapsto(\cos\psi\cos\eta,\cos\psi\sin\eta,\sin\psi)$, where $(\eta,\psi)\in E=(0,2\pi)\times(-\frac{\pi}{2},\frac{\pi}{2})$. The Gramm matrix looks like:
$$
\begin{pmatrix}
\cos^2\psi & 0 \\
0 & 1
\end{pmatrix}
$$

one ends up with the (here $\lambda_2$ denotes the Lebesgue measure):
$$
S_2(M) = \int_E 1\sqrt{\det{G(D_\varphi(\eta,\psi))}}\,\mathrm{d}\lambda_2(\eta,\psi)=\int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\psi\,\mathrm{d}\psi\,\mathrm{d}{\eta}=4\pi
$$

I did not do anything else but blindly followed the definitions.

The wrong result:

If I do the same approach to the above problem at hand. My map is $\varphi:\,(\theta,\phi)\mapsto (r(\theta,\phi),\theta,\phi)$, the Jacobian is then:
$$
\begin{pmatrix}
\frac{\partial r(\theta,\phi)}{\partial \theta} & \frac{\partial r(\theta,\phi)}{\partial \phi} \\
1 & 0 \\
0 & 1
\end{pmatrix}
$$

The rank of this matrix is 2 as needed for a 2-surface. The Gramm matrix is then:
$$G(D_\varphi)=
\begin{pmatrix}
1+ \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 & \frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} \\
\frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} & 1+ \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
\end{pmatrix}
$$
with determinant evaluated to:
$$
\det(G(D_\varphi)) = 1 + \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 + \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
$$

which leads to the surface area:

$$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{1+\left(\frac{\partial r}{\partial \theta}\right)^2 + \left(\frac{\partial r}{\partial \varphi}\right)^2}\;{\rm d}\theta\,{\rm d}\varphi$$

Which is incorrect. The question is why, and please show the correct way with explanation because as suggested by the example with unit sphere, I did not have to do any transformations and it did work with curvilinear coordinates right away. So the answer provided by Quanto does not address this at all.

Note to the edit:

I have added quite a lot of details since the answer by Quanto, but the root problem is the same - to see by calculation explicitly how one arrives to the correct result and to understand why the my calculation for a sphere works (where i am not using Cartesian coordinates) but fails here.

答:
There is an elementary way to get the areal element, but it's kind of tedious. Along the surface let
$$\vec r=\langle x,y,z\rangle=\langle r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta\rangle$$
Then
$$\begin{align}d\vec r&=\left\langle\frac{\partial r}{\partial\theta}\sin\theta\cos\phi+r\cos\theta\cos\phi,\frac{\partial r}{\partial\theta}\sin\theta\sin\phi+r\cos\theta\sin\phi,\frac{\partial r}{\partial\theta}\cos\theta-r\sin\theta\right\rangle\,d\theta\\
&\quad+\left\langle\frac{\partial r}{\partial\phi}\sin\theta\cos\phi-r\sin\theta\sin\phi,\frac{\partial r}{\partial\phi}\sin\theta\sin\phi+r\sin\theta\cos\phi,\frac{\partial r}{\partial\phi}\cos\theta\right\rangle\,d\phi\end{align}$$
This writes an infinitesimal displacement along the surface in terms of changes in the independent variables. We find a vector normal to the surface whose magnitude is the area of a parallelogram two of whose sides are the two infinitesimal vectors above:
$$\begin{align}d^2\vec A&=\pm\left\langle\frac{\partial r}{\partial\theta}\sin\theta\cos\phi+r\cos\theta\cos\phi,\frac{\partial r}{\partial\theta}\sin\theta\sin\phi+r\cos\theta\sin\phi,\frac{\partial r}{\partial\theta}\cos\theta-r\sin\theta\right\rangle\,d\theta\\
&\quad\times\left\langle\frac{\partial r}{\partial\phi}\sin\theta\cos\phi-r\sin\theta\sin\phi,\frac{\partial r}{\partial\phi}\sin\theta\sin\phi+r\sin\theta\cos\phi,\frac{\partial r}{\partial\phi}\cos\theta\right\rangle\,d\phi\\
&=\pm\left\langle\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\sin\phi+r\frac{\partial r}{\partial\phi}\cos^2\theta\sin\phi\right.\\
&\quad-\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\sin\phi+r\frac{\partial r}{\partial\phi}\sin^2\theta\sin\phi-r\frac{\partial r}{\partial\theta}\sin\theta\cos\theta\cos\phi+r^2\sin^2\theta\cos\phi,\\
&\quad\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\cos\phi-r\frac{\partial r}{\partial\phi}\sin^2\theta\cos\phi-r\frac{\partial r}{\partial\theta}\sin\theta\cos\theta\sin\phi+r^2\sin^2\theta\sin\phi\\
&\quad-\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\cos\phi-r\frac{\partial r}{\partial\phi}\cos^2\theta\cos\phi,\\
&\quad\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin^2\theta\sin\phi\cos\phi+r\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\sin\phi\cos\phi+r\frac{\partial r}{\partial\theta}\sin^2\theta\cos^2\phi\\
&\quad+r^2\sin\theta\cos\theta\cos^2\phi-\frac{\partial r}{\partial\theta}\frac{\partial r}{\partial\phi}\sin^2\theta\sin\phi\cos\phi-r\frac{\partial r}{\partial\phi}\sin\theta\cos\theta\sin\phi\cos\phi\\
&\quad\left.+r\frac{\partial r}{\partial\theta}\sin^2\theta\sin^2\phi+r^2\sin\theta\cos\theta\sin^2\phi\right\rangle\,d\theta\,d\phi\\
&=\pm\left\langle r\frac{\partial r}{\partial\phi}\sin\phi-r\frac{\partial r}{\partial\theta}\sin\theta\cos\theta\cos\phi+r^2\sin^2\theta\cos\phi,\right.\\
&\quad-r\frac{\partial r}{\partial\phi}\cos\phi-r\frac{\partial r}{\partial\theta}\sin\theta\cos\theta\sin\phi+r^2\sin^2\theta\sin\phi,\\
&\quad\left.r\frac{\partial r}{\partial\theta}\sin^2\theta+r^2\sin\theta\cos\theta\right\rangle\,d\theta\,d\phi\end{align}$$
So now
$$\begin{align}d^2A&=\left\lVert d^2\vec A\right\rVert=\left\{r^2\left(\frac{\partial r}{\partial\phi}\right)^2\sin^2\phi+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^2\theta\cos^2\theta\cos^2\phi+r^4\sin^4\theta\cos^2\phi\right.\\
&\quad+r^2\left(\frac{\partial r}{\partial\phi}\right)^2\cos^2\phi+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^2\theta\cos^2\theta\sin^2\phi+r^4\sin^4\theta\sin^2\phi\\
&\quad-2r^2\left(\frac{\partial r}{\partial\phi}\right)\left(\frac{\partial r}{\partial\theta}\right)\sin\theta\cos\theta\sin\phi\cos\phi+2r^3\left(\frac{\partial r}{\partial\phi}\right)\sin^2\theta\sin\phi\cos\phi\\
&\quad-2r^3\left(\frac{\partial r}{\partial\theta}\right)\sin^3\theta\cos\theta\cos^2\phi+2r^2\left(\frac{\partial r}{\partial\phi}\right)\left(\frac{\partial r}{\partial\theta}\right)\sin\theta\cos\theta\sin\phi\cos\phi\\
&\quad-2r^3\left(\frac{\partial r}{\partial\phi}\right)\sin^2\theta\sin\phi\cos\phi-2r^3\left(\frac{\partial r}{\partial\theta}\right)\sin^3\theta\cos\theta\sin^2\phi\\
&\quad\left.+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^4\theta+2r^3\left(\frac{\partial r}{\partial\theta}\right)\sin^3\theta\cos\theta+r^4\sin^2\theta\cos^2\theta\right\}^{1/2}\,d\theta\,d\phi\\
&=\left\{r^2\left(\frac{\partial r}{\partial\phi}\right)^2+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^2\theta\cos^2\theta+r^4\sin^4\theta-2r^3\left(\frac{\partial r}{\partial\theta}\right)\sin^3\theta\cos\theta\right.\\
&\quad\left.+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^4\theta+2r^3\left(\frac{\partial r}{\partial\theta}\right)\sin^3\theta\cos\theta+r^4\sin^2\theta\cos^2\theta\right\}^{1/2}\,d\theta\,d\phi\\
&=\left\{r^2\left(\frac{\partial r}{\partial\phi}\right)^2+r^2\left(\frac{\partial r}{\partial\theta}\right)^2\sin^2\theta+r^4\sin^2\theta\right\}^{1/2}\,d\theta\,d\phi\\
&=\sqrt{\frac1{\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2+\left(\frac{\partial r}{\partial\theta}\right)^2+r^2}\,r\sin\theta\,d\theta\,d\phi\end{align}$$
A lot of typing but we have the correct answer if the typo will. The areal element given in the original question was wrong at least because quantities with different units ($\text{length}^2$ and no units) were being added under the square root.

EDIT: There is a much more efficient way to organize the arithmetic. Since
$$\vec r=\langle r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta\rangle$$
we have
$$\begin{align}d\vec r&=\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle\,dr+\langle r\cos\theta\cos\phi,r\cos\theta\sin\phi,-r\sin\theta\rangle\,d\theta\\
&\quad+\langle-r\sin\theta\sin\phi,r\sin\theta\cos\phi,0\rangle\,d\phi\\
&=\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle\,dr+\langle \cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta\rangle\,rd\theta\\
&\quad+\langle-\sin\phi,\cos\phi,0\rangle\,r\sin\theta d\phi\\
&=\hat rdr+\hat{\theta}rd\theta+\hat{\phi}r\sin\theta d\phi\end{align}$$
Since we are considering $\theta$ and $\phi$ to be the independent variables,
$$dr=\left(\frac{\partial r}{\partial\theta}\right)d\theta+\left(\frac{\partial r}{\partial\phi}\right)d\phi=\frac1r\left(\frac{\partial r}{\partial\theta}\right)rd\theta+\frac1{r\sin\theta}\left(\frac{\partial r}{\partial\phi}\right)r\sin\theta d\phi$$
So along the surface
$$d\vec r=\left(\frac1r\left(\frac{\partial r}{\partial\theta}\right)\hat r+\hat{\theta}\right)rd\theta+\left(\frac1{r\sin\theta}\left(\frac{\partial r}{\partial\phi}\right)\hat r+\hat{\phi}\right)r\sin\theta d\phi$$
Then since we can show that $\hat r\times\hat{\theta}=\hat{\phi}$, $\hat{\theta}\times\hat{\phi}=\hat r$, and $\hat{\phi}\times\hat r=\hat{\theta}$,
$$\begin{align}d^2\vec A&=\pm\left(\frac1r\left(\frac{\partial r}{\partial\theta}\right)\hat r+\hat{\theta}\right)rd\theta\times\left(\frac1{r\sin\theta}\left(\frac{\partial r}{\partial\phi}\right)\hat r+\hat{\phi}\right)r\sin\theta d\phi\\
&=\pm\left(-\frac1r\frac{\partial r}{\partial\theta}\hat{\theta}-\frac1{r\sin\theta}\frac{\partial r}{\partial\phi}\hat{\phi}+\hat r\right)r^2\sin\theta d\theta d\phi\end{align}$$
Since we can also show that $\hat r\cdot\hat r=\hat{\theta}\cdot\hat{\theta}=\hat{\phi}\cdot\hat{\phi}=1$ and that $\hat r\cdot\hat{\theta}=\hat{\theta}\cdot\hat{\phi}=\hat{\phi}\cdot\hat r=0$ we have
$$\begin{align}d^2A&=\left\lVert d^2\vec A\right\lVert\\
&=\sqrt{\left(-\frac1r\frac{\partial r}{\partial\theta}\hat{\theta}-\frac1{r\sin\theta}\frac{\partial r}{\partial\phi}\hat{\phi}+\hat r\right)\cdot\left(-\frac1r\frac{\partial r}{\partial\theta}\hat{\theta}-\frac1{r\sin\theta}\frac{\partial r}{\partial\phi}\hat{\phi}+\hat r\right)}\,r^2\sin\theta d\theta d\phi\\
&=\sqrt{\frac1{r^2}\left(\frac{\partial r}{\partial\theta}\right)^2+\frac1{r^2\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2+1}\,r^2\sin\theta d\theta d\phi\end{align}$$
Much nicer :)


问:
Prove that if $E \simeq E'$, $E \bigoplus F \simeq E' \bigoplus F'$ then $F \simeq F'$.
All the spaces in this question are Banach Spaces, and "$\simeq$" means they are linear homeomorphic. The question is to prove the following:

Prove that if $E \simeq E'$, $E \bigoplus F \simeq E' \bigoplus F'$ then $F \simeq F'$.

Any hints on how to do this? I'm kind lost.

Thanks!

答: A counterexample: $\ell_2\oplus\Bbb R$ is linearly homeomorphic to $\ell_2\oplus\Bbb R^2$, but $\Bbb R$ is not linearly homeomorphic to $\Bbb R^2$.


答:
Using the tangent half-angle substitution, you end with
$$\left(1+\sqrt{2}\right) t^4+2 \left(1+\sqrt{2}\right) t^2-8 t+(1+\sqrt{2})=0$$By inspection (with all these $1$'s and $\sqrt{2}$'s),  $\color{red}{t=(\sqrt{2}-1)}$ is a root.

Factoring, we are left with
$$\left(1+\sqrt{2}\right) t^3+t^2+\left(1+3 \sqrt{2}\right) t-(3+2 \sqrt{2})=0$$ for which $$\Delta=-64 \left(63+44 \sqrt{2}\right) \quad <0$$ Using the hyperbolic method for only one real root which is (!!)
$$\color{red}{t=\frac{1}{3} \left(1-\sqrt{2}+4 \sqrt{3-\sqrt{2}} \sinh \left(\frac{1}{3}
   \sinh ^{-1}\left(\frac{5}{14} \sqrt{\frac{1}{7} \left(173+81
   \sqrt{2}\right)}\right)\right)\right)}$$ I shall not try to find what is this number which is not recognized by inverse symbolic calculators.

To the first root corresponds
$$x_1=2 \tan ^{-1}\left(\sqrt{2}-1\right)=\frac \pi 4+2k\pi$$
and the second is
$$x_2\sim 2\tan ^{-1}(0.778669) \sim 1.323197$$


问:
Compute $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$

Compute$$\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$$.

The primitive function of each one seems not elementary function.

I tried like this
$$
\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} + \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\int_0^1\dfrac{\mathrm{d}(x-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-x^2}},
$$
but it still does not work.

Actually, I can compute it by Gamma and Beta function. But I wonder is there other approach? Appreciate any help!

答:
Yes, there is another approach – elliptic integrals.

By Byrd and Friedman 214.05 and 318.02, the two integrals evaluate as follows:
$$\left(\frac1{\sqrt2}K(m=1/2)\right)\left(\sqrt2E(1/2)-\frac{\sqrt2}2K(1/2)\right)$$
This can be rewritten as
$$\frac12(K(1-1/2)E(1/2)+K(1/2)E(1-1/2)-K(1/2)K(1-1/2))=\frac\pi4$$
where we have used Legendre's relation.


问:
$a^b = c$, is it possible to express $b$ without logarithms?
$ a^b = c $

is it possible to express b without logarithms?

Body:
It is possible, certainly not the most practical though.

$5^{3x}=\dfrac{1}{125}$

$5^{3x}=\dfrac{1}{5^3}$

$5^{3x}=5^{-3}$

$3x=-3$

$\dfrac{3x}{3}=\dfrac{-3}{3}$

$\boxed{x=-1}$


问:
Computing the resolvent of a rank one projector
I'm reading through this paper in dynamical systems, that's using a bit of perturbation theory and operator theory. The authors make the following claim:

Now, since $P_0$ is a projection of rank one, it is easy to compute explicitly its resolvent, namely
$$ R_0(ζ) = (P_0 − ζ)^{−1} = \frac{P_0}{1−ζ} − \frac{I−P_0}{ζ} $$

I have no background in operator theory, but I've skimmed through Kato's Perturbation Theory for Linear Operators and done a quick Google search and it's not at all obvious to me where this formula comes from. Does anyone have a source for the resolvent of a rank one projection?

答:
A geometric series expansions gives you what you want, at least for $|\zeta| > 1$:
\begin{align}
    (P_0-\zeta I)^{-1}
    &=\left(\zeta\left(\frac{1}{\zeta}P_0-I\right)\right)^{-1} \\
   &= -\frac{1}{\zeta}\left(I-\frac{1}{\zeta}P_0\right)^{-1} \\
   &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}P_0+\frac{1}{\zeta^2}P_0^2+\frac{1}{\zeta^3}P_0^3+\cdots\right) \\
   &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}\left(1+\frac{1}{\zeta}+\frac{1}{\zeta^2}+\cdots \right)P_0\right) \\
   &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}\frac{1}{1-1/\zeta}P_0\right) \\
   &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta-1}P_0\right) \\
   &=-\frac{1}{\zeta}I-\left(\frac{1}{\zeta-1}-\frac{1}{\zeta}\right)P_0 \\
   &=-\frac{1}{\zeta}(I-P_0)-\frac{1}{\zeta-1}P_0.
\end{align}
Then you can check that the right-hand side is the inverse for all $\zeta\ne 0,1$.


Body:
One solution, $\pi/4$, is immediately extracted from the form of the given equation, as Bumblebee commented after a few seconds as the question was posted:
$$
\sin x (1+\cos x)=\frac1{\sqrt2}\left(1+\frac1{\sqrt2} \right)\ .
$$
There are two solutions in $[0,2\pi]$, as the following pari/gp-raw plot shows it:

? plot(x=0, 2*Pi, sin(x) + sin(x)*cos(x) - (1/sqrt(2) + 1/2) )

 0.088734 |'''''''_x""x_'''''''''''''''''''''''''''''''''''''''''''''''''|
          ```````x``````x`````````````````````````````````````````````````
          |     x        "_                                              |
          |    x           _                                             |
          |   _             x                                            |
          |                  x                                           |
          |  "                x                                          |
          | _                  x                                         |
          |                     "_                                       |
          |x                      x_                                     |
          _                         xx____                               _
          |                               """"xx                         |
          |                                     "x                      x|
          |                                       "_                     |
          |                                         x                  " |
          |                                          x                _  |
          |                                           x                  |
          |                                            x             "   |
          |                                             "           x    |
          |                                              "_        x     |
          |                                                x      x      |
-2.502948 |................................................."x__x".......|
          0                                                       6.283185
?

Both solutions appear in the interval $[0,\pi/2]$.

To see that the second solution is not a "beautiful solution", i.e. a(n easy) rational multiple of $\pi$, we have to solve explicitly.

Numerically, this is simple. In sage, we have for instance:

? \p 80
   realprecision = 96 significant digits (80 digits displayed)
? c1 = solve(c=0, 0.5, 1 + 2*c - 2*c^3 - c^4 - (1/sqrt(2) + 1/2)^2)
%13 = 0.24507719212504432088553369017245407857420915894751612868529986637012882459682178
? c2 = solve(c=0.5, 1, 1 + 2*c - 2*c^3 - c^4 - (1/sqrt(2) + 1/2)^2)
%14 = 0.70710678118654752440084436210484903928483593768847403658833986899536623923105352
? acos(c1)/Pi
%15 = 0.42118669118081648201964584654917700002378936841121635895529014330812148926140359
? acos(c2)/Pi
%16 = 0.25000000000000000000000000000000000000000000000000000000000000000000000000000000

Here, we have used the equation of degree four from the OP in the unknown $c$, extracted to correspond to $\cos x$. The solution $c_2=1/\sqrt2$ corresponds to $x=\arccos c_2=\pi/4$. But $c_1$ "looks ugly".

To get it explicitly, we can factorize in sage, using the letter $a$ for $\sqrt 2$ and $C$ as unknown (playing the rolw of $\cos x$):

sage: K.<a> = QuadraticField(2)
sage: R.<C> = PolynomialRing(K)
sage: factor(1 + 2*C - 2*C^3 - C^4 - (1/a + 1/2)^2)
(-1) * (C - 1/2*a) * (C^3 + (1/2*a + 2)*C^2 + (a + 1/2)*C + 1/4*a - 1)

The first factor corresponds to the root $c_2=a/2=\sqrt 2/2$.
The second factor has three complex roots, one of them only is real:

sage: (C^3 + (1/2*a + 2)*C^2 + (a + 1/2)*C + 1/4*a - 1).roots(ring=QQbar, multiplicities=False)
[0.2450771921250444?,
 -1.476091986655796? - 0.6774060585198529?*I,
 -1.476091986655796? + 0.6774060585198529?*I]

And the real value above corresponds to the (numerical) one delivered by pari/gp.

Its exact expression is as follows.
Let us introduce:
$$
\begin{aligned}
S &= \frac 1{54}\Big(23 + \sqrt 2 + 3\sqrt{3(13+6\sqrt 2)}\Big)\ ,\\
T &= \frac 1{54}\Big(23 + \sqrt 2 - 3\sqrt{3(13+6\sqrt 2)}\Big)\ ,\\
&\qquad\text{Then the needed unknown root $c_1$ is:} \\
c_1 &= -\frac 16(4+\sqrt 2) + S^{1/3}+ T^{1/3}\ .
\end{aligned}
$$
And the corresponding $x$-solution is obtained by taking arccos of $c_1$.

Let us check this (numerically):

sage: a = sqrt(2); var('C');
sage: pol = (C^3 + (1/2*a + 2)*C^2 + (a + 1/2)*C + 1/4*a - 1)
sage: S = (+3*sqrt(3)*sqrt(6*sqrt(2) + 13) + sqrt(2) + 23)/54
sage: T = (-3*sqrt(3)*sqrt(6*sqrt(2) + 13) + sqrt(2) + 23)/54
sage: c1 = -(4+sqrt(2))/6 + S^(1/3) + T^(1/3)

sage: c1.n(300)
0.245077192125044320885533690172454078574209158947516128685299866370128824596821778908696124

sage: pol.roots(ring=RealField(300), multiplicities=False)
[0.245077192125044320885533690172454078574209158947516128685299866370128824596821778908696122]

sage: x1 = acos(c1)
sage: x1.n(250)
1.3231970148034460170431250253951876668350360601328894523066983942144393866

This corresponds also to the numerical solution given by pari/gp when asked for...

? solve(x=Pi/3, Pi/2, sin(x) + sin(x)*cos(x) - (1/sqrt(2) + 1/2))
%17 = 1.3231970148034460170431250253951876668350360601328894523066983942144393866408301


问:
Finding the angle between $x-5=0$ and $2x+2y-7=0$
I have 2 line equations: $x-5=0$ and $2x+2y-7=0$

I believe the angle is $45^\circ$, but I don't know how to calculate the second equation when there's a $y$ as well.

Thanks!

答: The angle $\theta$ between any two lines is related to their slopes as follows, $tan\theta=|\frac{m_1-m_2}{1+m_1m_2}|$, where the slope $m$ can be calculated as $m=\frac{-a}{b}$, where $a$ is the coefficient of $x$ and $b$ is the coefficient of $y$. Now in this case, $m_1=\frac{1}{0}$ (I know $\frac{1}{0}$ is undefined, but then it works. I appreciate any comments as to why it works. I think it has to do with limits.) and $m_2=-1$. Putting these values in the equation, we get $tan\theta=|\frac{\frac{1}{0}+1}{1+\frac{-1}{0}}|$, or $tan\theta=|-1|=1$, hence $\theta=45^\circ$.


问:
Is the matrix used to prove independence of $\bar Y $ and $S^2$ for a random normal sample orthogonal?
I'm looking at a proof that $\bar Y $ and $S^2$ for a sample of $n$ random normal variables are independent using two different sources. Both sources makes use of identical matrices linked in the images below.

In the first source, the matrix is defined as $A$ but the author states that $A'A = A^TA =I$. In the second source, the matrix is defined as $B^T$ (so $B^T = A$), but the author also states that $B^T B = I$.

If I denote this matrix as $A$, is $A$ orthogonal or does $A^T A = AA^T = I$? Otherwise the two sources seem to contradict.

From Statistical Models by Davison

From Mathematical Statistics by Wackerly

答:
The matrix is orthogonal as it consists of orthogonal unit row and column vectors. The norm of the first row is$$\sum_{i=1}^n\left(\frac1{\sqrt n}\right)^2=1$$The norm of $j^\text{th}(j>1)$ row is$$\sum_{i=1}^{j-1}\left(\frac1{\sqrt{j(j-1)}}\right)^2+\left[\frac{1-j}{\sqrt{j(j-1)}}\right]^2=1$$The norm of the $i^\text{th}$ column is given by$$\left(\frac1{\sqrt{n}}\right)^2+\left[\frac{1-i}{\sqrt{i(i-1)}}\right]^2+\sum_{j=i+1}^n\left(\frac1{\sqrt{j(j-1)}}\right)^2=1$$

 Use the fact $\frac1{j(j-1)}=\frac1j-\frac1{j-1}$.

The dot product of $i^\text{th},j^\text{th}(j>i)$ columns is$$\frac1{\sqrt n}\cdot\frac1{\sqrt n}+\frac{1-j}{\sqrt{j(j-1)}}\cdot\frac1{\sqrt{j(j-1)}}+\sum_{k=j+1}^n\left(\frac1{\sqrt{k(k-1)}}\right)^2=0$$

The dot product of the first and $j^\text{th}(j\ne1)$ rows is$$\frac1{\sqrt n}\left[\sum_{i=1}^{j-1}\frac1{\sqrt{j(j-1)}}+\frac{1-j}{\sqrt{j(j-1)}}\right]=0$$

The dot product of $i^\text{th},j^\text{th}(i\ne j,j>i>1)$ rows is$$\frac1{\sqrt{j(j-1)}}\left[\sum_{k=1}^{i-1}\frac1{\sqrt{i(i-1)}}+\frac{1-i}{\sqrt{i(i-1)}}\right]=0$$


答:
For the antiderivatives, make $x=\sin(t)$
$$I_1=\int \dfrac{dx}{\sqrt{1-x^4}}=\int \dfrac{dt}{\sqrt{1+\sin^2(t)}}=F\left(\left.t\right|-1\right) $$
$$I_2=\int \dfrac{x^2}{\sqrt{1-x^4}}dx=\int \dfrac{1+\sin^2(t)-1}{\sqrt{1+\sin^2(t)}}dt$$ that is to say
$$I_2=\int \sqrt{1+\sin^2(t)}\,dt -I_1=E(t|-1)-F(t|-1)$$

Integrated between $0$ and $\frac \pi 2$
$$I_1=K(-1)\qquad I_2=E(-1)-K(-1)$$ $$I_1\,I_2=K(-1)\Big[E(-1)-K(-1) \Big]=\frac \pi 4$$

Title: Compute $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$

Body:
Compute$$\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$$.

The primitive function of each one seems not elementary function.

I tried like this
$$
\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} + \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\int_0^1\dfrac{\mathrm{d}(x-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-x^2}},
$$
but it still does not work.

Actually, I can compute it by Gamma and Beta function. But I wonder is there other approach? Appreciate any help!


问:
Happy new year 2021
Happy new year to all mathematics lover.
I am looking for some mathematical way to say Happy new year  2021.
If possible, then please show your ideas.

$$\color{red}{\binom {20}{19}}
+\color{orange}{\binom {19}{16}
+{\binom  {16}{14}}}+
\color{green}{\binom {14}{12}
+\binom {12}{7}}+
\color{blue}{\binom {7}{6}+
\binom {6}{5}}+
\color{purple}{\binom {5}{3}+\binom {3}{1}}+
\color{magenta}{\binom {2}{0}}+\color{magenta}{\binom {1}{1}}+\color{magenta}{\binom {1}{0}}
=\color{red}{2021}$$

Above idea is copied from here.

答: A product of two consecutive primes is a four digit number $N$, written as (the string concatenation of) two consecutive natural numbers. Well, $N$ is...

Title: Happy new year 2021


问:
The equivalent of multiplying by $-1$ in $GF(4)$?
In $\mathbb{R}$, we have the notion of multiplying an element by $-1$. For example:

$12 \in \mathbb{R}$, then $12 \times (-1) = -12 \in \mathbb{R}$,

$7 \in \mathbb{R}$, then $7 \times (-1) = -7 \in \mathbb{R}$.

Now, assume we are in $GF(4)$ and its definition is a standard definition with elements $\{0, 1, \omega, \bar{\omega}\}$. We will label them as $\{0, 1, 2, 3\}$, respectively:

$$
\begin{array}{|c|cccc|}\hline
    + & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 1 & 2 & 3\\
    1 & 1 & 0 & 3 & 2\\
    2 & 2 & 3 & 0 & 1\\
    3 & 3 & 2 & 1 & 0\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    - & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 1 & 2 & 3\\
    1 & 1 & 0 & 3 & 2\\
    2 & 2 & 3 & 0 & 1\\
    3 & 3 & 2 & 1 & 0\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    \times & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 0 & 0 & 0\\
    1 & 0 & 1 & 2 & 3\\
    2 & 0 & 2 & 3 & 1\\
    3 & 0 & 3 & 1 & 2\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    \div& 0 & 1 & 2 & 3\\\hline
    0 & - & 0 & 0 & 0\\
    1 & - & 1 & 3 & 2\\
    2 & - & 2 & 1 & 3\\
    3 & - & 3 & 2 & 1\\\hline
\end{array}\qquad
$$

It is usually understood that $-x$ is the  additive inverse so that for any element $x \in GF(4)\,\,$, $x + (-x) = 0$.

I am interested in multiplying by $-1$ in $GF(4)$ defined as the same mathematical meaning as ordinary multiplying by $-1$ in $\mathbb{R}$. How would I go about doing something like this?

In the case where you would like more context, this deals with finding determinant of a matrix with elements in $GF(4)$ and multiplying by $-1$ is a necessary condition.

答:
$${GF(4)= \{0,1,w,w^2\}}$$  Now, ${w+1=w^2}$, so ${w^3=w \cdot w^2=w \cdot(1+w)=w+w^2=w + (1+w)=1}$, so ${w^2}$ is multiplicative inverse of ${w}$.

Now if you create the same field with ${Z_2[x]}$ with irreducible polynomial ${X^2+X+1}$, then you will have the set ${\{0,1,X,X+1\}}$

in ${GF(2^n)}$ we consider ${1=-1}$ , Hence ${X-1}$ is same as ${X+1}$  as mentioned by @Stinking Bishop


问:
Trouble understanding the question regarding dual spaces

If $(\alpha_0,\alpha_1,\alpha_2,\dots)$ is an arbitrary sequence of
complex numbers, and if $x$ is an element of $\mathcal{P}$,
$x(t)=\sum\limits^n_{t=0}\xi_i t^i$, write
$y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $y$ is an element of
$\mathcal{P'}$ and that every element of $\mathcal{P'}$ can be
obtained in this manner by a suitable choice of the $\alpha's$.

Here $\mathcal{P'}$ is a dual space of $\mathcal{P}$. Can you explain what the second question means?

答:
I’m not sure whether you only wanted clarification on the (second) question or a demonstration that it’s true. To answer them below, let’s write $\alpha=(\alpha_1,\alpha_2,\ldots)$ for the stated arbitrary infinite sequence. For any polynomial $x$, let $x^{(k)}$ be the $k$th derivative and write $Dx:=(x(0),x^{(1)}(0), \frac{x^{(2)}(0)}{2!},\ldots,\frac{x^{(n)}(0)}{n!},\ldots)$, which is simply the sequence coefficients of $x$ (consisting of an eventually trailing sequence of $0$s). Also, for a sequence $\delta=(\delta_1,\delta_2,\ldots)$, write $\delta\cdot\alpha:=\delta_1\alpha_1+\delta_2\alpha_2+\cdots$ for the dot product.

Now, observe that your space $\mathcal{P}$ is linearly isomorphic to the space $c_{00}$ of eventually zero sequences (consisting of all sequences which have only finitely many nonzero elements) by the canonical map
$$x\mapsto Dx\,.$$
It is not difficult to prove (I will supply details upon request, but you can mimic the situation for finite-dimensional spaces) that a linear functional $L$ in $c_{00}’$ is specifically of the form $$L(\delta)=\delta\cdot\alpha$$ for some infinite sequence $\alpha$ and all $\delta\in c_{00}$.

Thus via the isomorphism $\mathcal{P}\sim c_{00}$, then for the first question, the map
$$f\colon\mathcal{P}\to\mathbb{C}\,,~\,~\,~ x\mapsto Dx\cdot\alpha\,,$$
which gives you $fx=y$ in your notation, is a linear functional; and for the second question, every linear functional in $\mathcal{P}’$ is therefore of the form as $f$ above.


问:
$\inf_Y[\sup(2x+2y:0<x<1)]$ How to approach this?
$\inf_Y\sup_X f(x,y)$
Let $Y=(-1,0)$ and $X=(0,1)$ and $f(x,y)=2x+2y$
Let's start simply with $\inf_Y[\sup(2x+2y:0<x<1)]$
now here occurs the first problem for me, can I use the properties of $Y$ in this step, or would it be $\inf_Y[2+2y]$ or since i know that Y is negative i could finish here and say $\inf_Y[2]=2$
but the first approach would tell me it is: $\inf_Y[2+2y]=2-2=0$ since $Y=(-1,0) =>\inf_Y(2y)=-2$

So do I have to take sup of f(x,y) regardless of what Y is, which doesn't make sense. Maybe i need a proper definition for multivaraible sup and inf

Please give a comprehensive answer which approach is correct. Any explanation is appreciated

答:
The notation here is sloppy. It is ambiguous and forces the reader to make assumptions to be able to put any meaning to it. That is the source of your confusion. If this is exactly the notation you yourself are receiving, and it did not come with additional explanations, then you have my condolences. If this is the result of you short-cutting on the notation, then please stop. Don't take shortcuts on notation you do not thoroughly understand.

But the only reasonable interpretation of it that I see is
$$\inf\{\sup\{2x + 2y \mid 0<x<1\} \mid -1 < y < 0\}$$
Or to break it down, let $A_y := \{2x + 2y\mid 0 < x < 1\}$, then you are looking for
$$\inf\{\sup A_y\mid -1 < y < 0\}$$

Clearly, for each $y, \sup A_y = 2 + 2y$, so this becomes
$$\inf\{ 2 + 2y \mid -1 < y < 0\}$$
and so the answer is $0$.