问:
A weird contour integral calculation
I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as  $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because when the poles are at $|z|=1$. It's outside the boundary. So I have no idea about how I can calculate this integral

Here's my Residues

$\operatorname{Res}_{z=i}[f(z)]=\lim_{z\to i}\frac{\sqrt{1-z^2}}{i(1-zi)}dz=\frac{\sqrt{2}}{2i}$

$\operatorname{Res}_{z=-i}[f(z)]=\lim_{z\to -i}\frac{\sqrt{1-z^2}}{i(1+zi)}dz=\frac{\sqrt{2}}{2i}$

But I guess these are wrong

Edit: since I need a closed contour i replaced $x=cos\theta$ and $dx=-sin\theta d\theta$

And my integral is now $-\frac{1}{2}\oint \frac{\sqrt{1-cos^2\theta}}{cos^20+cos^2\theta}(-sin\theta) d\theta$=
$-\frac{1}{2}\oint \frac{-sin^2\theta}{cos^20+cos^2\theta}d\theta$

答:
Time for the dog-bone contour (dumbbell contour), which  works perfectly.
Indeed, consider the contour $\mathcal{C}$ given as follows:

$\hspace{12em}$

With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,

\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^2+1)} \, dz
&\longrightarrow
\int_{-1-0^+i}^{1-0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \, dz
- \int_{-1+0^+i}^{1+0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \, dz \\\\
&\quad = 2\int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^2+1} \, dx.
\end{align*}

On the other hand, Residue Theorem tells that

\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
&= - 2\pi i \sum_{z_k \ : \ z_k^2 + 1 = 0} \underset{z=z_k}{\mathrm{Res}} \, \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \\\\
&= 2\pi (\sqrt{2}-1)
\end{align*}

(In order to use Residue Theorem, consider a large circle, apply Residue Theorem to the region enclosed by this circle and $\mathcal{C}$, and then let the radius of the circle go to $\infty$.) Therefore

$$ \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^2+1} \, dx = \pi (\sqrt{2}-1) $$

Title: A weird contour integral calculation


问:
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.

Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$

My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$

I stucked here.

答:
A Physicist's Point of View.

The closed form solution to the integral is very impressive, of course.

Sometimes when needed a quick numerical estimate for such a hard-to-calculate integral it is useful to take a closer look at the integrand.

Consider the denominator of the integrand

$$\sin^2x + \cos^3x$$

What happens if we replace here $\cos^3x$ with $\cos^2x$?

Probably not much because $\cos x$ varies within $\left [0,1  \right ]$

But we get a considerable simplification

$$I_{approx}\approx\int_0^{\pi/2} \cos^3x dx.=\frac{2}{3}$$

Absolute error from the exact value is about $0.05$

ContentLicense: CC BY-SA 4.0


问:
If I chop my apple into smaller pieces, can I fit more apple in my lunchbox?
Well, that's how my little sister asked it, and I couldn't provide a quick answer.

If we ignore any changes in shape from the cutting process, then my first thought was that the scaling wouldn't make any difference. But then I started thinking that more efficient packings might only become possible after we cut the apple pieces down to a certain size.

I realise this is almost unanswerable without multiple assumptions (spherical pieces of apple anyone? :) ) but at the end of the day I'm interested in what is known (if anything) about the packing of irregularly shaped solids in finite containers of similar dimension (within an order of magnitude) to the solids themselves.

Thanks

答:
This restates some of the discussion in the comments, but here is a more rigorous proof of what I take this question to be asking:

Say that a 3D shape is "reasonable" if it has positive volume and the limit as $\epsilon\to0^+$ of the volume of the points within distance $\epsilon$ of its boundary is $0$.

In apple terms, we're requiring the apple is not all peel - if you take a fine enough apple-peeler, you can remove the peel without taking out any significant part of the apple. (This is to avoid pathological shapes that are arbitrarily close to their boundary and otherwise hard to dice up nicely.)

Claim: Given a reasonably-shaped apple and a reasonably-shaped lunchbox with more volume than the apple, it is possible to cut the apple into finitely many pieces with a knife so that they can be rearranged to fit inside the lunchbox.

Proof: Suppose that the ratio $\frac{\text{Volume}(\text{box})}{\text{Volume}(\text{apple})}$ is equal to some $r>1$.

Let $\epsilon_1>0$, and cut the apple along planes normal to the $x$, $y$, and $z$ axes so that it is cut along a grid of cubes of side length $\epsilon_1$. Extend every nonempty piece until it fills up its corresponding cube. By our reasonableness property, this will give us a set of cubes with total volume approaching $\text{Vol}(\text{apple})$ from above as $\epsilon_1\to0$. (This is because any cube which contains some of the apple has no points more than $\epsilon_1\sqrt{3}$ outside the boundary.)

Conversely, if we cut the box along planes normal to the $x$, $y$, and $z$ axes so that it is cut along a grid of cubes of side length $\epsilon_2>0$, our reasonableness property tells us that the cubes which fit entirely inside the box (without touching the boundary) will have total volume approaching $\text{Vol}(\text{box})$ from below as $\epsilon_2\to0$.

So then we can choose $\epsilon$ small enough that cutting the box into cubes of side length $\epsilon$ gives us "inside cubes" with total volume at least $\frac{\text{Vol}(\text{box})}{\sqrt{r}}$, and cutting the apple into cubes of side length $\epsilon$ gives us "outside cubes" with total volume at most $\text{Vol}(\text{apple})\cdot\sqrt{r}$.

But then we have at least as many empty lunchbox cubes to fill as we have partially-filled apple cubes to place inside, so we can just pair up the (identically-shaped) cubes with each other however we like. This completes the proof. (In fact, for most nice apple and lunchbox shapes it will do so in a relatively small number of knife cuts, because we only need to make $O(1/\epsilon)$ cuts in order to produce $O(1/\epsilon^3)$ pieces.)

If you place any absolute bound on the number of pieces, then it is not necessarily possible to do this, even for apples and lunchboxes of similar dimensions. Consider a worm-eaten apple with many tiny holes inside it, so that it is more like a collection of soap bubble films - even though its volume is very small, it takes a large number of cuts to allow these holes to be filled nicely and to pack it into a tighter space.

ContentLicense: CC BY-SA 3.0


问:
Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule?

I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..

Edited

COSINE RULE

$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)

After finding the angle, I then use trigonometry to find

$$Cos{A} = \frac{PS}{60}$$

答:
Three equations three unknowns. (and one assumption that all terms are positive)

$PS^2 + RS^2 = 60^2$

$RS^2 + QS^2 = 102^2$

$PS+QS = 126$

They aren't linear so they may not satisfy the simple "$k$ (linear and independant) equations; $k$ unknowns = solvable" mantra, but they probably do.

Substitute for one variable and combine to one equation with one unknown.

$x = PS$ and so $x^2 + (102^2 - (126-x)^2)=60^2$

or in other words $102^2 -126^2 +2\cdot 126x = 60^2$ so

$PS = x = \frac {60^2 + 126^2 -102^2}{2\cdot 126}$

$QS = 126 - \frac {60^2 + 126^2 -102^2}{2\cdot 126}$

$RS =\sqrt{60^2 -(\frac {60^2 + 126^2 -102^2}{2\cdot 126})^2 }$

.... of course you can make things easier by dividing out common factors and noting the difference of square rule that $126^2 - 102^2 = (126-102)(126+102)$ which may be things easier to calculate (especially if you divided out the common factor of $6$).....

=====

Alternatively

$PS^2 + RS^2 =60^2$ and $RS^2 + QS^2 = 102^2$ so

$QS^2 - PS^2 = 102^2 - 60^2$ and

$(QS + PS)(QS-PS) = (102+60)(102-60)$

$126(QS-PS) = 162*42$

$(QS-PS) = \frac {162*42}{126}\frac {6*27*6*7}{6*21} = 9*6  =54$.

$OS-PS = 54$ and $QS+PS = 126$ so $QS=\frac {54+126}2=\frac 62(9+21)=3*30=90$

And $PS = \frac {126-54}2 = \frac 62{21-9} =3*12 = 36$

And $RS = \sqrt{60^2 - 36^2} = \sqrt{102^2 - 90^2}=$

$6\sqrt{10^2 -6^2} = 6\sqrt{17^2-15^2}=$

$6\sqrt{(10-6)(10+6)} = 6\sqrt{(17-15)(17+15)}=$

$6\sqrt{4*16} = 6\sqrt {2*32}=$

$6*2*4 = 6*8 = 48$.


问:
Applying Gauss's formula to find flux
Using divergence theorem, find flux of vector field
${\bf F}=\left(x\left(z - 1\right),y\left(z - 1\right),xy\right)$ on the area determined with inequalities
$$
\left(z-1\right)^2\le 4-x^2-y^2,\quad
\left(z-1\right)^2\ge x^2+y^2,\quad z\ge1
$$

This comes down to calculating the integral $2\displaystyle\iiint_{V}\left(z-1\right)\,\mathrm{d}V$. Then, what is left is to find bounds.

When I set $z=0$, in the $OXY$ plane I get
$x^2+y^2\le 1$ and $x^2+y^2\le3$, so it is enough to watch circle with radius $\sqrt3$ and center at origin.

After cylindrical coordinates I have
$$
\left\{\begin{array}{rcccl}
{\displaystyle 0} & {\displaystyle \le} & {\displaystyle \phi} & {\displaystyle \le} & {\displaystyle 2\pi}
\\[0.5mm]
{\displaystyle 0} & {\displaystyle \le} & {\displaystyle r} & {\displaystyle \le} & {\displaystyle \sqrt{3}}
\\[0.5mm]
{\displaystyle r + 1} & {\displaystyle \le} &  {\displaystyle z} & {\displaystyle \le}  & {\displaystyle 1 + \sqrt{4 - r^{2}}}
\end{array}\right.
$$
And then I would integrate in order ${\rm d}z, {\rm d}r, {\rm d}\phi$.

Is this correct?

答:
Everything seems correct except bound of $r$.

This is a closed surface enclosed by an inverted cone along $z-$axis with vertex at $(0,0,1)$ and a sphere of radius $2$ with center at $(0,0,1)$. Radius of resulting surface is maximum where they intersect which we can find by equating them,

Sphere: $(z-1)^2\le 4-r^2$, Cone: $(z-1)^2\ge r^2$.

At intersection, $4-r^2 = r^2 \implies r = \sqrt2$

So bounds of $r$ is $0 \leq r \leq \sqrt2$.

Title: Applying Gauss's formula to find flux


问:
Prove the following alternative group definition.
Motivation: If operation is change, and there is no change, then there is no operation. Then the meaning of  1+0 =1 is not in the equality itself but in the comparison with other equalities as  1+2=3. The following definition does not use the meaningless equality  1+0 =1, or generally a*e=a. I guess that trying to prove statements using the next definition, matters/problems will arise that do not arise with the standard definition.

Let be the non empty set G = {a,b,c,d,.....z} with the operation *, satisfy the following 3 conditions:

a(bc) = (ab)c, for every a, b, c ∈ G (associative law).

For every a,b there is c (unique) such that ac = b. (c  is the "path" from a to b).

For every a,b there is d (unique) such that bd = a. (d is the "path" back from b to a).

P.S.: the meaning of 2) and 3) is the possibility to "go"/"move" from any member to any member, through the same path (if a->b->c then c->b->a as in the Klein group) or through other path (if a->b->c then c->X->a).

Prove: G,*  is a group.

Proof:

Associativity:

From 1)

Identity element:

Let's prove for every a there is e such that ae=ea=a.

Proof: According to 2) For every a,a there is e such that ae = a. (3)

According to 2) for every e,a there is x such that ex=a, that means ae =ex =a (4)

Let's proof x=a.

Multiply by a on the left, such that the result is some g ∈ G:

ae = ex = a (4) <---> (a)ae = (a)ex = aa =g <---> aae = (ae)x = aa =g (associative) <---> <---> aae = ax = aa =g (3). (5)

According to 2) for every a,g there is x (unique) such that aa=g and ax=g , then a and x are the same, and it's proved x=a. Then replacing in (5): ae = ea = a.

Inverse element: Let's prove that for every a, there is b such that a ⋅ b = e and b ⋅ a = e

According to 2) for every a,e there is b such that ab = e. (6)

According to 2) for every b,e there is X such that bX=e. (7)

Let's proof that X=a

ab = bX = e (6),(7)

Multiply be a on the right:

aba =ea

a(ba) = ea; Associativity

ea = a; Identity element

a(ba) = a; then ba = e

bX=e according to (7)

ba = bX = e; according to 2) for every b,e there is x unique such that bX = e then x=a and it's proved

that for every a, there is b such that a ⋅ b = e and b ⋅ a = e.

答: Your axioms are satisfied if one takes any nonempty set $G$ and defines the operation $*$ by setting $xy=y$ for all $x$ and $y$ in $G$. If $G$ has at least two elements, this is not a group.


问:
There are pairs of rational numbers $(a,b)$
There are pairs of rational numbers $(a,b)$, $a<b$, such that
$\int^b_a \frac{(2x-5)^{2020}}{x+1} \,dx \in \mathbb Q$
I do not know what to do here. I really have no idea and I am sorry for that. Could you help me, please?

答:
$$\frac{(2 x-5)^{2020}}{x+1}=p_{2019}(x)+(-7)^{2020}\cdot \frac{1}{x+1}$$
integrating on $[a,b]$ we get
$$P_{2020}(b)-P_{2020}(a)+(-7)^{2020}\log\frac{b+1}{a+1}$$
the unique values that make this result a rational number are $a=b$. For any rationals $a,b$ such that $a\ne b$ we have $\log\frac{b+1}{a+1}$ irrational,  transcendental to be more precise.

Therefore there is no rationals $a<b$ such that the given integral is rational


问:
Numbers on clock represented by polyhedra

The "1" is replaced with a four-sided object, then the next one could be a five sided object, then six (the cube), but then after that, it is either a five-sided pyramid, or a eight-sided die.

Would you please give me an explanation or a rule for this image? Enlighten me, please!

答:
I do not think there is any particular pattern to these shapes other than increasing "complexity" in some sense (as mentioned in the comments), but to describe exactly which polyhedra are pictured:

$n$
Polyhedron
Vertices
Edges
Faces

$1$
Tetrahedron
4
6
4

$2$
Square pyramid
5
8
5

$3$
Cube
8
12
6

$4$
Octahedron
6
12
8

$5$
Pentagonal trapezohedron
12
20
10

$6$
Cuboctahedron
12
24
14

$7$
Dodecahedron
20
30
12

$8$
Icosahedron
12
30
20

$9$
Icosidodecahedron
30
60
32

$10$
Snub cube
24
60
38

$11$
Rhombicosidodecahedron
60
120
62

$12$
Small stellated dodecahedron
12
30
12

If one considers the small stellated dodecahedron as a simple polyhedron rather than a self-intersecting Kepler-Poinsot polyhedron, it has $32$ vertices, $90$ edges, and $60$ faces.


问:
Finding Area of a shape
I'm doing some revision and I'm a bit stuck. How do you find out the area of this shape?

I know you have to do 10cm x 2cm = 20cm devided by 2 = 10cm - 2cm x ? (something) = area.

I'm not sure what the cm of the line next to 10cm is (the question mark bit). I mean would you do 13cm - 10cm = 3cm to find the missing length. Or would it be 2cm like the other side?

Would it be 6cm squared or 4cm squared.

Btw that corner bit is not part of the shape, it's only the 2 sides traingle without the right angle... that's just added on so I can work it out.

P.S. I know this is an easy question but it's bugging me.

答:
Your triangle is not one, as $2 + 10 \lt 13$.  But let's change the $2$ to $4$ so it makes a triangle.  If you know the three sides, you can find the area of a triangle from Heron's formula.  Define $s=\frac{a+b+c}2$, then the area $A=\sqrt{s(s-a)(s-b)(s-c)}$

To do what you are drawing, let the extension of the 10 side to the right angle be $x$, the side from the right angle to the corner near "2 cm" be $y$.  We have $x^2+y^2=4^2=16$ (because I made that side $4$ long).  We also have $y^2+(10+x)^2=y^2+x^2+20x+100=13^2=169$.  This gives $116+20x=169$ or $x=\frac {53}{20}, y=\sqrt {16-(\frac {53}{20})^2}$ and finally the area of your original triangle is half the base times the height, $\frac 12 (10)y$

Title: Finding Area of a shape


问:
If $f:A \to B$ is a real function, and $y\in B, $ Is Rudin's definition of $f^{-1}(y)$ commonly accepted as convention or not?
Let $f:A \to B\ $ be a function.

In Rudin's PMA, at the bottom of page 24 and top of page 25, he states:

If $y \in B, f^{-1}(y)\ $ is the set of all $x \in A\ $ such that
$f(x) = y.\ $

This notation could be confused with the function $f^{-1}:B \to A.$

Is Rudin's use of $f^{-1}(y)$ as a set still conventional today?

答: Yes, $f^{-1}(y)$ is a common (I dare say the most common) notation for the set $f^{-1}(\{y\})$, when it is clear by context that $f^{-1}$ does not indicate a purported inverse map of $f$.


答:
The notation $f^{-1}(\{y\}) = \{x\in A : f(x) \in \{y\}\}$ or more generally for a subset $C\subseteq A$
$$
f^{-1}(C) = \{x\in A : f(x) \in C\}
$$
is pretty standard and known as the preimage.

When it is clear by context $f^{-1}(y)$ would also work but i would always make sure it is clear [as it might lead to confusion with the inverse function $f^{-1}$ which not necessarily exists]

Title: If $f:A \to B$ is a real function, and $y\in B, $ Is Rudin's definition of $f^{-1}(y)$ commonly accepted as convention or not?

Body:
Let $f:A \to B\ $ be a function.

In Rudin's PMA, at the bottom of page 24 and top of page 25, he states:

If $y \in B, f^{-1}(y)\ $ is the set of all $x \in A\ $ such that
$f(x) = y.\ $

This notation could be confused with the function $f^{-1}:B \to A.$

Is Rudin's use of $f^{-1}(y)$ as a set still conventional today?


问:
Relations between partial derivatives used in the Maxwell relations
In the context of Maxwell's relations, there are two frequently used expressions relating the partial derivatives of different thermodynamic quantities:

$$\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y}= -1 \tag{1}$$

$$\left(\frac{\partial y}{\partial x}\right)_{z}=1 \bigg{/} \left(\frac{\partial x}{\partial y}\right)_{z} \tag{2}$$

What is the mathematical background of these two expressions? Is there a simple demonstration for them?

答:
They follow from the mathematical definition of the derivative, which is:

$\left(\frac{\partial x}{\partial y} \right)_z=\lim_{h \to 0}\frac{x(y+h,z)-x(y,z)}{h}$,

$\left(\frac{\partial y}{\partial z} \right)_x=\lim_{h \to 0}\frac{y(z+h,x)-y(z,x)}{h}$,

$\left(\frac{\partial z}{\partial x} \right)_y=\lim_{h \to 0}\frac{z(x+h,y)-z(x,y)}{h}$,

$dx=\left(\frac{\partial x}{\partial y} \right)_z dy+\left(\frac{\partial x}{\partial z} \right)_y dz$.

OwnerDisplayName: Peter


问:
Statistics Question
This is probably super simple to most of you on here, but I was chatted by a friend earlier with a question.  It reads just like this:

```<code>A sample of n =7 scores has a mean of M =5. After one new score is added to the sample,
the new mean is found to be M =6. What is the value of the new
score? (Hint: Compare the values for E X before and after the score was added.)
</code>```

She said the E X looks like the greek symbols.  Can someone please explain this problem to me and provide a solution or tell me how to solve it?  I guess I don't understand the question or how to set up an equation to solve it.  Thanks!

答:
The sum of the first seven scores is $5 \cdot 7$. It's $7$ times the mean.

Similarly, you can find the sum of all eight scores - to find the last one, all you need to do is subtract.

OwnerDisplayName: AndyWarren

Title: Statistics Question


问:
arbitrary product of the closed unit interval is Hausdorff
In showing that $\Pi_{\alpha \in A} [0,1]_\alpha$ is Hausdorff, is the proof pretty much the same as showing :

"$\Pi_\alpha X_\alpha$ is Hausdorff iff $X_\alpha$ is Hausdorff for each $\alpha \in A$ some arbitrary indexing set."

Or? You take two points and since they are distinct there exists an index element on which their component do not agree and everywhere else take the open sets to be everything? Does similar argument work here since $[0,1]$ is Hausdorff? (It is a subspace of a Hausdorff space, $\mathbb{R}$.

答: Yes, the product of Hausdorff spaces is Hausdorff: if $x \neq y$ are points in $X:= \prod_{i \in I} X_i$, then for some $i_0 \in I$ we have $x_{i_0} \neq y_{i_0}$; we can find disjoint open subsets $U$ and $V$ in $X_{i_0}$ (which is Hausdorff) such that $x_{i_0} \in U, y_{i_0} \in V$ and note that $\pi_{i_0}^{-1}[U]$ and $\pi_{i_0}^{-1}[V]$ are (sub)basic open sets in $X$ that are still disjoint and contain $x$, and $y$ respectively.


问:
Solution to 5 equations in 5 variables
I have the following 5 equations in 5 variables(A, B, C, D, E):

$2 \times E \times A + D = 1    $

$2\times E\times B + D = 0    $

$D + E = 2$

$A + B + C = 1$

$A^2   + B^2 + C = \frac 7 4$

I know the solution is: $ (0 , \frac {-1} 2 , \frac 3 2 ,  1, 1) $. Just don't know how to get there.

答:
From the first two, $E\ne0$. After elimination* of $A, B, C, D$, you get

$$ \frac{E-1}{2E} \left( \frac{E-1}{2E} -1\right)+\frac{E-2}{2E} \left(\frac{E-2}{2E} -1\right)=\frac34.$$

or simply

$$E^2=1.$$

The rest follows.

*Use $D=2-E$, express $A, B$ in terms of $E$ and plug them in the difference of the last two.

Title: Solution to 5 equations in 5 variables


问:
Solve the system by graphing. y = -3x + 1 y = x -7.
Solve the system by graphing.

$y = -3x + 1$

$y = x -7.$

Which of these coordinates point fits this exact equation?

Step by step process and an explanation to show if it is the correct answer.

$$(-2, 5);~~
(-2, -5);~~
(2, -5);~~
(2, 5)
$$

答: This may help you get started:

Body: This may help you get started:


问:
Verify $f: \mathbb{Z} ⟶ \mathbb{Z}_{r} \times \mathbb{Z}_{s}$ given by $a ⟶ (ā_r, ā_s)$ is an epimorphism
Let $r,s \in \mathbb{Z}^+$ such that $(r,s)=1.$  Consider the ring $\mathbb{Z}_{r} \times \mathbb{Z}_{s}$ (with respect to ordinary multiplication and addition, component by component), prove that the application $f: \mathbb{Z} ⟶ \mathbb{Z}_{r} \times \mathbb{Z}_{s}$ given by $a ⟶ (ā_r, ā_s)$ is an epimorphism of rings.

I've already proved that it's a homomorphism since

$(a+b)$ maps to $([a+b]_r, [a+b]_s)$, which is $([a]_r + [b]_r) + ([a]_s + [b]_s)$.

$(ab)$ maps to $([ab]_r, [ab]_s)$, which is $([a]_r [b]_r) ([a]_s [b]_s$)$.

These proposition are true owning to the operations defined in modular arithmetic. So it's a homomorphism.

If the map is surjective then for $(x,y) \in ℤ_r \times ℤ_s$ there's an $a \in ℤ$. This means that for any $(x,y)$, the equations $[a]_r = x$, and $[a]_s = y$ are simultaneously solvable. But I'm failing at showing the latter proposition.

答:
Given any relatively prime integers $r$ and $s,$ consider the ring homomorphism $\pi : \mathbb Z \to (\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z)$ defined by $\pi(n) = (n + r \mathbb Z, n + s \mathbb Z).$ By hypothesis that $r$ and $s$ are relatively prime, Bézout's Lemma implies that there exist integers $a$ and $b$ such that $ra + sb = 1.$ Consequently, for any element $(x + r \mathbb Z, y + s \mathbb Z)$ of $(\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z),$ we have that $$(x + r \mathbb Z, y + s \mathbb Z) = (xra + xsb + r \mathbb Z, yra + ysb + s \mathbb Z) = (xsb + r \mathbb Z, yra + s \mathbb Z).$$ Can you find an element $n$ of $\mathbb Z$ such that $\pi(n) = (xsb + r \mathbb Z, yra + s \mathbb Z)$ to finish the proof?

Like I indicated in my comment above, the original question is part of the proof of the Chinese Remainder Theorem for arbitrary rings; the only thing to do is to replace "relatively prime integers" with "comaximal ideals $I$ and $J.$" From here, it follows that there exist elements $i \in I$ and $j \in J$ such that $i + j = 1_R,$ and the rest follows similarly to the above proof.


答: The mean is the sum $S$ of the scores divided by the number of scores.  Here we have $M = S/n$.  Plugging in our values gives $S = 7 \cdot 5 = 35$.  Now the new mean is $6 = (S + X)/(n+1) = (35 + X)/8$.  Then just solve for $X$.</p>

Title: Statistics Question

Body:
This is probably super simple to most of you on here, but I was chatted by a friend earlier with a question.  It reads just like this:

```<code>A sample of n =7 scores has a mean of M =5. After one new score is added to the sample,
the new mean is found to be M =6. What is the value of the new
score? (Hint: Compare the values for E X before and after the score was added.)
</code>```

She said the E X looks like the greek symbols.  Can someone please explain this problem to me and provide a solution or tell me how to solve it?  I guess I don't understand the question or how to set up an equation to solve it.  Thanks!


答:
These are known as the reciprocal and reciprocity theorems. I feel these are perhaps not stated as clearly as they should be in introductory courses. I always had trouble with them until I read Andrew Steane's fantastic book on Thermodynamics (pretty much the only book on the subject that I like), which I will be following in this answer.

The idea is to start with the well known theorem of partial derivatives. If $f$ is a function of two variables $x$ and $y$, then we can relate a change in $f$ to changes in $x$ and $y$ using the partial derivatives: $$\text{d}f = \left(\frac{\partial f}{\partial x}\right)_{y} \text{d}x + \left(\frac{\partial f}{\partial y}\right)_{x} \text{d}y.\label{1}\tag{1}$$

In the above formula, the notation $(\cdot)_x$ means that the partial derivative is taken while keeping the subscript (here, $x$) constant. Note that in general (I encourage you to try to come up with an example for this)$$\left(\frac{\partial f}{\partial x}\right)_{y} \neq \left(\frac{\partial f}{\partial x}\right)_{z}.$$

We can now use the above relation carefully, while realising that we can't quite be as cavalier with partial derivatives as we could with total derivatives. For example, we can't talk about quantities like $\text{d}f/\text{d}x$, since $f$ is a function of two variables, $x$ and $y$. You might naively call the quantity in question the "rate of change of $f$ with $x$", but since it depends on $y$ as well, we need to specify along which path we choose to describe this variation completely.

The Reciprocal Theorem:

let's start by dividing Equation (\ref{1}) by $\text{d}x$, while keeping $y$ constant. The second term in the equation would trivially be zero, since $\text{d}y=0$, and we'd get the trivial result: $$\left(\frac{\partial f}{\partial x}\right)_{y}  = \left(\frac{\partial f}{\partial x}\right)_{y}.$$

A more interesting result appears if we carefully divide Equation (\ref{1}) by $\text{d}f$, but we still keep $y$ constant. In this case, it reduces to: $$1 = \left(\frac{\partial f}{\partial x}\right)_{y} \left(\frac{\partial x}{\partial f}\right)_{y},$$ from which you get the relation

$$\boxed{\left(\frac{\partial f}{\partial x}\right)_{y} = \left(\frac{\partial x}{\partial f}\right)_{y}^{-1}}.$$

The Reciprocity Theorem:

Let's go back to Equation (\ref{1}) again, and this time we divide throughout by $\text{d}x$ again, except we don't keep $y$ constant, but rather we do it while keeping $f$ constant. It should be easy to show that $$0 = \left(\frac{\partial f}{\partial x}\right)_{y} + \left(\frac{\partial f}{\partial y}\right)_{x} \left(\frac{\partial y}{\partial x}\right)_{f}.$$

Using the reciprocal theorem above, we can easily see that this means that $$\boxed{\left(\frac{\partial f}{\partial y}\right)_{x} \left(\frac{\partial y}{\partial x}\right)_{f}  \left(\frac{\partial x}{\partial f}\right)_{y} = -1.}$$

Title: Relations between partial derivatives used in the Maxwell relations

Body:
In the context of Maxwell's relations, there are two frequently used expressions relating the partial derivatives of different thermodynamic quantities:

$$\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y}= -1 \tag{1}$$

$$\left(\frac{\partial y}{\partial x}\right)_{z}=1 \bigg{/} \left(\frac{\partial x}{\partial y}\right)_{z} \tag{2}$$

What is the mathematical background of these two expressions? Is there a simple demonstration for them?


问:
How to integrate $\int_{0}^{2\pi}\cos x\times \exp\left [ \frac{a}{\cos x}\right]\,dx$
$$\int_{0}^{2\pi}\cos x\times \exp\left [ \frac{a}{\cos x}\right]\,dx$$

where $a$ is complex parameter. (Integral does not converge if $a$ is Real)

I tried to solve it by two methods.

The first is direct replacement $z=e^{ix}$. After that the integral goes to $$-\frac{i}{2}\oint_{\left| z \right| < 1}\frac{1+z^2}{z^2}\exp{\left[\frac{2az}{z^2+1}\right]}\,dz$$

Before switching to complex numbers, you can use substitution $t=1/\cos x$ to reduce this integral to the form: $$2\int_{0}^{+\infty }\frac{e^{at}}{t^2\sqrt{t^2-1}}\,dt$$
And after that we can go to complex plane.

But because of the complex contours, integrating in both cases is a difficult task for me.

答:
As noted in comments, the integral converges only if $a$ is purely imaginary (otherwise zeros of $\cos x$ represent non-integrable singularities). The substitution $t=1/\cos x$, done carefully, leads to $$f(a)=2\int_1^\infty\frac{e^{at}-e^{-at}}{t^2\sqrt{t^2-1}}\,dt,$$ with $f''(i\lambda)=2\pi i J_0(\lambda)$ for $\lambda>0$, according to this integral for the Bessel function.

This may give a way to confirm (or refute) the CAS result given in comments.


问:
Why isn't the growth of $\mathrm {Iso}(\mathbb R ^2)$ sub-exponential?
Given a finite subset $S$ of a group, denote by $\gamma_S(n)$ the number of distinct elements that are words of length $n$ in $S\cup S^{-1}$.

A group is of sub-exponential growth if for every finite subset $S$, the number $\gamma_S(n)$ satisfies $\sqrt[n]{\gamma_S(n)} \to 1$.

I have proved the false claim that the isometries of the plane are a group of sub-exponential growth.

This must be false, because a group of sub-exponential growth does not admit a paradox, while the group of isometries of the plane does admit a paraox. (
The Sierpinski-Mazurkiewicz paradox.)

Which of the following parts is my mistake?

A semidirect product of two groups of sub-exponential growth is again sub-exponential.

(Proof: if $G=H \rtimes Q$, take a word $h_1q_1h_2q_2\ldots $ and make it into a word of the form $q_1\ldots q_n\tilde h_1\ldots \tilde h_n $ using the conjugations $h_iq_j=q_j\tilde h_i$.)

The group of isometries is $\mathrm {Iso}(\mathbb R ^2) = O(2) \ltimes \mathbb R^2$.

$O(2)=SO(2) \rtimes {\pm 1}$.

Alternative formulation
It is known that the group $\mathrm {Iso}(\mathbb R ^2) = O(2) \ltimes \mathbb R^2$ contains a free semigroup of order $2$. Why don't claims 1,2,3 disprove this?

答: The mistake is that $\tilde h_i$ does not depend only on $h_i$, but also on $q_j$. (This is unlike the case of isometries of $\mathbb R^1$.) Thus the set of possible $\tilde h_i$ is no longer finite.


问:
${\{n\sqrt{2}\}>\frac{1}{2n\sqrt{2}}}$ for all $n\in \mathbb{N}^*$
For $x\in \mathbb{R}$, we define $\{x\}=x-\lfloor x\rfloor$.

Prove that for all $n\in \mathbb{N}^*$, $\displaystyle{\{n\sqrt{2}\}>\frac{1}{2n\sqrt{2}}}$.

I really don't see how to start. The fact that $\sqrt{2}\notin \mathbb{Q}$ is surely important but I don't know how to use it.

答:
If  $p=\lfloor n\sqrt{2}\rfloor $, we have:

$\{n\sqrt{2}\}= n\sqrt{2}-p=\dfrac{2n^2-p^2}{n\sqrt{2}+p}\geqslant \dfrac{1}{n\sqrt{2}+p} >\dfrac{1}{2n\sqrt{2}} $


答:
The sample mean is given by:

$M=\dfrac{\sum_{i=1}^n x_i}{n}$

where $x$ is your individual observations and $n$ the number of observations.

Before you added the observation you have:

$5=\dfrac{\sum_{i=1}^7 x_i}{7}$

And after you add the observation you have:

$6=\dfrac{\sum_{i=1}^8 x_i}{8}$

Given these you can solve for the sums of $x_i$

These turn out to be:

$\sum_{i=1}^7 x_i=5 \times 7 = 35$

$\sum_{i=1}^8 x_i=6 \times 8 = 48$

We know that the above summations share the first $7$ observations. Therefore we can find the value of the new observation by taking the difference.

$x_8=\sum_{i=1}^8 x_i- \sum_{i=1}^7 x_i=48 - 35 = 13$

Therefore your new observation is $13$

Some further explanation:

We may also write $\sum_{i=1}^7 x_i=x_1 + x_2 + x_3 +x_4 + x_5 +x_6 +x_7$

As we, we may write: $x_8=\sum_{i=1}^8 x_i = x_1 + x_2 + x_3 +x_4 + x_5 +x_6 +x_7 +x_8$

We had shown that $\sum_{i=1}^7 x_i=x_1 + x_2 + x_3 +x_4 + x_5 +x_6 +x_7=35$

Therefore, we may substitute this into the equation for $\sum_{i=1}^8 x_i$:

$\sum_{i=1}^8 x_i = x_1 + x_2 + x_3 +x_4 + x_5 +x_6 +x_7 +x_8 = 35 +x_8$

Since using the formula for the mean we also shown:

$\sum_{i=1}^8 x_i= 45$

Then we have:

$45 = 35 + x_8$

and simply solving for $x_8$:

$x_8 = 13$

Title: Statistics Question


问:
If $R$ is a commutative ring, can we show that the set of nilpotent elements of $R$ form an ideal without using the Binomial Theorem?
All answers I have seen regarding this question (showing that the set of nilpotent elements of $R$ form an ideal) make use of the Binomial theorem in order to prove that $(N,+)$ is an additive subgroup of $(R,+)$. My book (Abstract Algebra A First Course by Dan Saracino) makes no reference to the Binomial theorem, and so I am unsure as to whether I can use it (perhaps that is a silly question).

答:
Given $\ \color{#c00}{b^k} = 0 = \color{#0a0}{a^n},\,$ Factor Theorem $\Rightarrow (a\!+\!b)^n-a^n$ has factor $\,(a\!+\!b)-a = b,\,$ therefore $\, bc = (a\!+\!b)^n - \color{#0a0}{a^n}= (a\!+\!b)^n\,$ $ \overset{\large (\,\ )^{\Large k}}\Longrightarrow\, (a\!+\!b)^{nk} = (bc)^k = \color{#c00}{b^k}c^k = 0$.

Note $ $ The proof only needs $\,(a+b)^n = a^n + bc\,$ for some $\,c.\,$ There are many elementary ways to prove that besides the Factor Theorem (or Binomial Theorem), including an obvious simple inductive proof. These allow  us to replace the Factor Theorem by various closely related well-known facts, e.g. the Congruence Power Rule  $\Rightarrow \bmod b\!:\ a\!+\!b\equiv a\,\Rightarrow\, (a\!+\!b)^n\equiv a^n\, $ (or equivalent arithmetic in the quotient ring $R/b\,).\, $ Alternatively we can employ the Polynomial Remainder Theorem, or Taylor's Theorem (to first order), etc.


问:
Interactions (MIMO systems)
Page

According to the page above, an LTI MIMO system could be affected by many interactions.
I have two questions about this page:

Is it described the system using differential equations (as well as it would be done with a spring-mass system)?

Why does it say that the MIMO system is described by differential equations if the set of equations (8.11) is in the complex variable 's'? I guess here it was applied a Laplace transformation.

答:
Yes the system is described with differential equations, i.e:

$$
\begin{align}
\dot{x} &= A x + B u \\
y &= C x + D u
\end{align}
$$

MIMO for the LTI system just mean that $B$ has $\geq 2$ columns (multiple inputs) and $C$ has $\geq 2$ rows (multiple outputs). And yes the 8.11 equation is the Laplace transformation.

For example:

$$
A = \begin{pmatrix}
0&0&1 \\0&-1&0 \\-5&-10&-15
\end{pmatrix},B = \begin{pmatrix}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{pmatrix},C =\begin{pmatrix}1&0&0\\0&1&0\\0&0&1  \end{pmatrix},D=\begin{pmatrix}0&0\\0&0\\0&0\end{pmatrix}
$$

Laplace transform:

$$
\begin{pmatrix} y_1(s)\\y_2(s)\\y_3(s) \end{pmatrix} = \begin{pmatrix}\frac{-10}{s^3 + 16 s^2 + 20 s + 5} & \frac{1}{s^2 + 15 s + 5} \\
\frac{1}{s + 1}  & 0  \\
\frac{-10s}{s^3 + 16 s^2 + 20 s + 5} & \frac{s}{s^2 + 15 s + 5} \end{pmatrix} \begin{pmatrix} u_1(s) \\ u_2(s) \end{pmatrix}
$$

You can see that $u_1$ influences all outputs but $u_2$ influences only $y_1$ and $y_3$ but not $y_2$.

Title: Interactions (MIMO systems)


问:
Comparing the size of the sets
Let $A, B, C, D$ be sets. Prove that if $|A|=|B|$ and $|C|=|D|$, then $|A^C|=|B^D|$

I am asking for help in solving the task.

答:
You have that $|A^C| = |A|^{|C|}$ and $|B^D| = |B|^{|D|}$ so, because $|A| = |B|$ and $|C| = |D|$, is trivial that $|A|^{|C|} = |B|^{|D|}$, meaning that $|A^C| = |B^D|$

Check: https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_exponentiation

Title: Comparing the size of the sets


问:
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.

First I can make an assumption that $a+b+c$ will be the smallest if we can obtain the highest power of b, then c and later a.

Let me go in the general way, $12^a\cdot 324^b\cdot48^c$ = $ 3^a \cdot  2^{2a} \cdot 3^{4b} \cdot 2^{2b} \cdot3^c \cdot 2^{4c}$ = $3^{a+4b+c} \cdot 2^{2a+2b+4c}$

And $6^{70}$ = $2^{70}\cdot 3^{70}$

Then I need to find the solution for $a+4b+c = 70$ and $a+b+2c=35$ where a+b+c need to be the samllest.

Please anyone can show me the solution how to approach this type of questions.

答:
$$\left(2^2 3\right)^a \cdot \left(2^2 3^4\right)^b \cdot \left(2^4 3\right)^c=3^{a+4 b+c} \cdot2^{2 a+2 b+4 c}$$
$$
\begin{cases}
a+b+c \text{ is minimum}\\
a+4b+c=70\\
2a+2b+4c=70\\
\end{cases}
$$
From second and third equations we get
$$
\begin{cases}
4b+c=70-a&\to -4b-c=-70+a\\
2b+4c=70-2a&\to 4b+8c=140-4a\\
\end{cases}
$$
$$7c=70-3a\to c=\frac{70-3a}{7}$$
and finally
$$b= 15-\frac{a}{7},c= 10-\frac{3 a}{7}$$

edit

```<code>(thanks to @TonyK for correcting)
</code>```

which are integer for $a=0$, therefore the minimum value is $a+b+c=25$ when
$$a=0,b=15,c=10.$$

Body:
$$\left(2^2 3\right)^a \cdot \left(2^2 3^4\right)^b \cdot \left(2^4 3\right)^c=3^{a+4 b+c} \cdot2^{2 a+2 b+4 c}$$
$$
\begin{cases}
a+b+c \text{ is minimum}\\
a+4b+c=70\\
2a+2b+4c=70\\
\end{cases}
$$
From second and third equations we get
$$
\begin{cases}
4b+c=70-a&\to -4b-c=-70+a\\
2b+4c=70-2a&\to 4b+8c=140-4a\\
\end{cases}
$$
$$7c=70-3a\to c=\frac{70-3a}{7}$$
and finally
$$b= 15-\frac{a}{7},c= 10-\frac{3 a}{7}$$

edit

(thanks to @TonyK for correcting)

which are integer for $a=0$, therefore the minimum value is $a+b+c=25$ when
$$a=0,b=15,c=10.$$


答:
\begin{align}
2e a+d&=1
\tag{1}\label{1}
,\\
2 e b+d&=0
\tag{2}\label{2}
,\\
d+e&=2
\tag{3}\label{3}
,\\
a+b+c&=1
\tag{4}\label{4}
,\\
a^2+b^2+c&=\tfrac74
\tag{5}\label{5}
.
\end{align}

Substitution of $e=2-d$ from \eqref{3} into \eqref{1} and \eqref{2}
leads to expression of $a$ and $b$ in terms of $d$:

\begin{align}
a &= \tfrac12\frac{d-1}{d-2}
\tag{6}\label{6}
,\\
b &= \tfrac12\frac{d}{d-2}
\tag{7}\label{7}
.
\end{align}

Then,
substitution of these expressions for $a,b$
into \eqref{4}
gives the expression of $c$ in terms of $d$:

\begin{align}
c&=-\tfrac32\frac1{d-2}
\tag{8}\label{8}
.
\end{align}

substitution expressions for $a,b,c$ in terms of $d$ into \eqref{5}
after simple rearrangements,
results in equation is $d$:

\begin{align}
-\tfrac54\frac{(d-1)(d-3)}{(d-2)^2}
&=0
,
\end{align}

so there are two solutions for $d$, $1$ and $3$,
consequently, there are two solutions of \eqref{1}-\eqref{5}:

\begin{align}
a &= 0
,\quad
b = -\tfrac12
,\quad
c = \tfrac32
,\quad
d=1
,\quad
e=1
,\\
\text{and}\quad
a &= 1
,\quad
b = \tfrac32
,\quad
c = -\tfrac32
,\quad
d=3
,\quad
e = -1
.
\end{align}

Title: Solution to 5 equations in 5 variables

Body:
I have the following 5 equations in 5 variables(A, B, C, D, E):

$2 \times E \times A + D = 1    $

$2\times E\times B + D = 0    $

$D + E = 2$

$A + B + C = 1$

$A^2   + B^2 + C = \frac 7 4$

I know the solution is: $ (0 , \frac {-1} 2 , \frac 3 2 ,  1, 1) $. Just don't know how to get there.


问:
How to find the third root of a complex number by transforming the complex number into a root
I'd like to find all 3rd roots of this number z = i - 1. Now I've found formulas on how to do it; First we transform the complex number into this form

$$ \sqrt[n]{r} * e^{i\frac {\phi + 2k\pi}{n}} $$ Where n should be 3 (because of 3rd root) and k should be k = n - 1 (Including 0)

Now what is not really clear to me is what to do with this -1? If I transform this I'd pressume I get this.

$$\sqrt[3]{-1} * e^{i\frac {\pi/2 + 2k\pi}{3}}$$

But than what? If it was a 2nd root I could say that it is simply i (the root of a negative number) but since it is a 3rd root I'm not quite sure what to do next?

Now for the phi (angle) I just assumed this;  $$i = e^{i \frac {\pi}{2}}$$ And I'm not too sure about this

Thanks!

EDIT: With great help from the people in the comments; I think I've solved it.So as already stated in the comments; $$ r = \sqrt 2, \phi = \frac{3\pi}{4} $$ Now If i would to input that into my formula we should  get this;

$$ \sqrt[6]{2} * e^{i\frac{3pi/4 + 2k\pi}{3}} $$

Now as stated before k = 0... n-1;

k = 0 --> $$ \sqrt[6]{2} * e^{i(\pi/4)} $$
k = 1 --> $$ \sqrt[6]{2} * e^{i(11\pi/12)} $$
k = 2 --> $$ \sqrt[6]{2} * e^{i(19\pi/12)} $$

You reckon I did this correctly?

答:
$$\sqrt[3]{re^{i\theta+2k\pi}}=\sqrt[3]re^{i\theta/3}e^{i2k\pi/3}.$$

If we let $\omega=e^{i2\pi/3}$, the three roots are

$$\sqrt[3]re^{i\theta/3}, \sqrt[3]re^{i\theta/3}\omega,\sqrt[3]re^{i\theta/3}\omega^2.$$

This is true for any complex number.

Notice that

$$\omega=-\frac12+i\frac{\sqrt3}2,\omega^2=-\frac12-i\frac{\sqrt3}2.$$ The next powers repeat periodically (this is why there are only three distinct roots). The three roots are obtained from any root, multipliying once and twice by $\omega$.


问:
Compute $\oint_\gamma\sin z\, dz$
Compute $$\oint_\gamma\sin z\, dz$$ where $\gamma  =z_0 +e^{i\theta}$

My attempt : I was thinking about residue theorem , but i don't know  how to apply residue theorem and compute

答: The integral of any entire function along any closed path is $0$. In the specific case of the sine function, you use the fact that you know a primitive of it (which is $-\cos$) to reach the same conclusion.

Title: Compute $\oint_\gamma\sin z\, dz$


问:
Does Euler totient function gives exactly one value(answer) or LEAST calculated value(answer is NOT below this value)?
I was studying RSA when came across Euler totient function. The definition states that- it gives the number of positive values less than $n$ which are relatively prime to $n$.

I thought I had it, until I came across this property:-

Euler Totient function is multiplicative function, that is: $\varphi(mn) = \varphi(m)\varphi(n)$

Now, if $p$ is a prime number, $\varphi(p)=p-1$.

Putting values of $p$ as 11 and 13 one by one,

$$\varphi(11)=10$$

$$\varphi(13)=12$$

Applying above stated property,

$$\varphi(11\cdot 13)=\varphi(11)\varphi(13)$$

$$\varphi(143)=12 \cdot 10$$

$$\varphi(143)=120$$

Is it correct? Does that mean we have $23$ values between $1$ and $143$ which are not relatively prime to $143$? Sorry if its something basic I'm missing. I'm not some genius at maths and came across this during study of RSA Algo. Thanks.

答: You are correct that there are $23$ numbers between $1$ and $143$ that share a factor with $143$.  They are $11,22,33,\ldots 132$ ($13$ of them) and $13,26,39,\ldots 130$ ($10$ of them.)</p>


问:
What morphisms must exist in a category if the morphism between two hom-sets relative to that category's objects exist?
I'm asking this and other questions in an attempt to tackle something I'm having a hard time to understand from another point of view.

Let's consider a category $C$ with 4 objects, $a$, $b$, $c$, and $d$, and some morphisms such that

$|Hom(a,c)| \ge 1$,

$|Hom(b,d)| \ge 1$,

there is a morphism $m$ in $Set$ such that $m: Hom(a,c) \to Hom(b,d)$.

What do the 3 hypothesis above imply as regards the other hom-sets in $C$?

My understanding is that, at the minumum, $|Hom(b,a)| \ge 1\ \wedge\ |Hom(c,d)| \ge 1$ must be true, but it's not required that the strict inequality holds, nor that the other hom-sets are not empty.

Furthermore, I believe that the concept of dual/opposite category need not be brought in to answer this question.

Is all the above correct?

答:
Let's consider a category $C$ with $4$ objects, $a$, $b$, $c$, and $d$,
and some morphisms such that

$|\mathrm{Hom}(a,c)| \ge 1$,

$|\mathrm{Hom}(b,d)| \ge 1$,

there is a morphism $m$ in $\mathsf{Set}$ such that $m: \mathrm{Hom}(a,c) \to \mathrm{Hom}(b,d)$.

Let's pause and observe that the third hypothesis follows from the second hypothesis (so it doesn't add anything). Assuming $\mathrm{Hom}(b,d)$ is non-empty, we can pick some $f\in \mathrm{Hom}(b,d)$ and define $m: \mathrm{Hom}(a,c) \to \mathrm{Hom}(b,d)$ to be the constant function with value $f$, $m(x) = f$.

What do the 3 hypothesis above imply as regards the other hom-sets in $C$?

Absolutely nothing.

My understanding is that, at the minumum, $|\mathrm{Hom}(b,a)| \ge 1\ \wedge\ |\mathrm{Hom}(c,d)| \ge 1$ must be true.

Nope, that's not the case. And I'm very curious what your reasoning is - this seems like a  total non sequitur!

For example, $C$ could have exactly six arrows: $\mathrm{id}_a$, $\mathrm{id}_b$, $\mathrm{id}_c$, $\mathrm{id}_d$, $f\colon a\to c$, and $g\colon b\to d$. Then $\mathrm{Hom}(b,a)$ and $\mathrm{Hom}(c,d)$ are both empty.


问:
Relations between partial derivatives used in the Maxwell relations
In the context of Maxwell's relations, there are two frequently used expressions relating the partial derivatives of different thermodynamic quantities:

$$\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y}= -1 \tag{1}$$

$$\left(\frac{\partial y}{\partial x}\right)_{z}=1 \bigg{/} \left(\frac{\partial x}{\partial y}\right)_{z} \tag{2}$$

What is the mathematical background of these two expressions? Is there a simple demonstration for them?

答:
The second one, which mathematicians usually call the "inverse function theorem", should actually be rather geometrically obvious: if you have the same set of constraints, then $\frac{\Delta y}{\Delta x}$ will be just the reciprocal of $\frac{\Delta x}{\Delta y}$ even for finite changes. It is only nontrivial when you write it down in terms of functions instead of relations between quantities. Indeed, thinking in terms of functions, the equation seems rather strange, since the two sides don't appear to depend on the same thing (the left side seems to depend on $x$ and not on $y$ while the right side seems to depend on $y$ and not on $x$). The fix at the level of functions is that you have to evaluate $\frac{\partial x}{\partial y}$ at $y^{-1}(x)$, or vice versa.

The first one is more mysterious, especially when written in this nice symmetrical "triple product rule" form. It is a lot more intuitive to me when you distinguish one quantity from the other two by rewriting it in the form

$$\left ( \frac{\partial y}{\partial x} \right )_f = -\frac{\left ( \frac{\partial f}{\partial x} \right )_y}{\left ( \frac{\partial f}{\partial y} \right )_x}.$$

This obviously has the right units, so the only remaining mysterious element is why the coefficient is just $-1$ all the time. The intuition for why the coefficient is negative is that if $\left ( \frac{\partial f}{\partial x} \right )_y$ and $\left ( \frac{\partial f}{\partial y} \right )_x$ have the same sign, then $f$ will change if you change both $x$ and $y$ in the same direction. As for why it is actually $-1$, the intuitive explanation is to write out the total differential

$$df=\left ( \frac{\partial f}{\partial x} \right )_y dx + \left ( \frac{\partial f}{\partial y} \right )_x dy = 0$$

since $f$ is held constant, and then formally rearrange to get $\frac{dy}{dx}$, which gets interpreted as $\left ( \frac{\partial y}{\partial x} \right )_f$ since you held $f$ constant in the derivation. In math notation, we formalize this by considering an arbitrary path $(x(t),y(t))$ along which $f$ is constant, and then

$$\frac{df}{dt} = \left ( \frac{\partial f}{\partial x} \right )_y \frac{dx}{dt} + \left ( \frac{\partial f}{\partial y} \right )_x \frac{dy}{dt} = 0$$

from the chain rule. Then you apply the property $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to conclude.


问:
A question about two rules in isomorphism and cardinality
In class I have encountered the next two rules:

the next models - $\langle \mathcal{P}(A),\subseteq \rangle$ and $\langle \mathcal{P}(B),\subseteq \rangle$ are isomorphic, if $A$ and $B$, have the same cardinality.

I had a chance to use this rule in a certain question, but I wondered to find out that I was wrong. Now, I am not sure when I can apply this rule. The question was finding isomorphism between $\langle \mathcal{P}(\mathbb{N})\cup\{\mathbb{Z}\},\subseteq \rangle$ and $\langle \mathcal{P}(\mathbb{N}),\subseteq \rangle$, which are not isomorphic. But on the next question, I have found that: $\langle \mathcal{P}(\mathbb{N}\cup\{\mathbb{Z}\}
),\subseteq \rangle$ and $\langle \mathcal{P}(\mathbb{N}),\subseteq \rangle$ are indeed isomorphic, and I cannot see the differences at all, because in both scenarios we have the same cardinalities of the models. Therefore, I need an extended explanation of when I can apply this rule, etc.

if $B=\{X\in\mathcal{P}(\mathbb{Z}):x\ is\ finite\}$ then $\left|B\right|=\left|\mathcal{P}(\mathbb{Z})\right|=\left|\mathbb{R}\right|$.

Can you give me an explanation for why is true?

Thanks!

答:
Note that $\mathcal P(\Bbb N)\cup\{\Bbb Z\}$ is not of the form $\mathcal P(\text{something})$, and therefore the rule that is only about posets ofa that form cannot be applied.

We can see that $\langle \mathcal P(\Bbb N)\cup\{\Bbb Z\},\subseteq \rangle$ and $\langle \mathcal P(\Bbb N),\subseteq \rangle$ are not isomorphic because only the first has a unique second-largest element, i.e.,
$$ \exists a\,\exists b\,\forall c\colon (c=a\lor c\subseteq b)\land a\ne b$$
holds in only one of the two posets.

Moreover, in your second question, the set $B$ of finite subsets of $\Bbb Z$ is only countably infinite (do you see how you can write it as countable union of countable sets?), hence strictly smaller than $\Bbb R$ in cardinality.


问:
Limit point clarification
Proposition 3.1.6:
A is closed iff A contains its limit points

and the def of a limit point.
https://en.wikipedia.org/wiki/Limit_point

I usually think closed sets have limit points
because of 3.1.6.
So l recently thought of open sets, and
found they do too.

If both open and closed sets have limit
points why is 3.1.6 useful?

Where is the defining difference in using
it

Ok . I thought more. An open set could have more outside the
set

答: There is a difference between "having" limit points and "containing" them. An open interval $(0,1)$ has $0$ and $1$ as limit points, but they do not belong to the interval. For closed sets, their limit points belong to them.

Title: Limit point clarification


答: Because while, indeed, usually open sets have limit poits, they do not contain all limit points.

Title: Limit point clarification

Body:
Proposition 3.1.6:
A is closed iff A contains its limit points

and the def of a limit point.
https://en.wikipedia.org/wiki/Limit_point

I usually think closed sets have limit points
because of 3.1.6.
So l recently thought of open sets, and
found they do too.

If both open and closed sets have limit
points why is 3.1.6 useful?

Where is the defining difference in using
it

Ok . I thought more. An open set could have more outside the
set


问:
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$
Find the limit without the use of L'Hôpital's rule or Taylor series

$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

答:
sorry about to answer my Question but Zarrax's way and her comment lead me to the answer
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}$$

$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}$$

$$L=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}$$

$$L=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}=\frac{3L}{27}-\frac{4}{27}$$

$$L=\frac{3L}{27}-\frac{4}{27}$$

$$L=-\frac{1}{6}$$

$$2L=-\frac{1}{3}=\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

thanks alot Zarrax
thanks for every one


答:
Note that the given function is holomorphic in the whole complex plane, and so it has no singularities at all. If you then look at the statement of the residue theorem, you see that the terms determining the value of the integral are only the values of the residue of the function at its singularities, which are none. So, in this case, the integral has value zero.

Another way to look at it is the following: When speaking of holomorphic functions (for example, $f \colon D \subset \mathbb{C} \longrightarrow \mathbb{C}$), if you can find a primitive of the function (that is, a holomorphic function $F$ with $F' = f$ in $D$), then the integral of $f$ along any path $\gamma \colon [0,1] \longrightarrow D$ with $\gamma (0) = a $ and $\gamma (1) = b$ will be exactly $F(b) - F(a)$ (In this case, since it is a closed path, it would then be zero). The implication I just referred to is actually an equivalence.

Title: Compute $\oint_\gamma\sin z\, dz$

Body:
Compute $$\oint_\gamma\sin z\, dz$$ where $\gamma  =z_0 +e^{i\theta}$

My attempt : I was thinking about residue theorem , but i don't know  how to apply residue theorem and compute


答:
Proposition 3.1.6 is a characterization of closed sets. The original statement in the book (bold font mine):

Let $A$ be a subset of a topological space $(X,\tau)$. Then $A$ is closed in $(X,\tau)$ if and only if $A$ contains all of its limit points.

Your book comment on the previous page of the proposition that

"The next proposition provides a useful way of testing whether a set is closed or not."

and right after Proposition 3.1.6, it gives an example on how it is useful:

Title: Limit point clarification


问:
Recurrence relations for strings of $0$'s and $1$'s of length $n$ where no $k$ consecutive $1$'s appear.
This question is related to this one, but the answers there do not explain my confusion.

This is exercise 20 in page 28 in the book generatingfunctionology:

Let $f(n,m,k)$ be the number of strings of $n$ $0$'s and 1's that contain exactly $m$ $1$'s, no $k$ of which are consecutive. Then

(a) Find a recurrence formula for $f$. It should have $f(n,m,k)$ on the left side, and exactly three terms on the right.

(b) Find, in simple closed form, the generating functions

$$ F_k(x,y)=\sum_{n,m\geq 0}f(n,m,k)x^ny^m\quad(k=1,2,...)$$

Here's what I have:

I didn't get exactly three terms in the recurrence, but I did come up with one plausible. Well, we want to use $n-m$ $0$'s to separate $m$ $1$'s. That's the same as putting $m$ balls into $n-m+1$ ordered boxes, and each box can at most contain $k-1$ balls. We divide the cases by how many balls the last box has: it could be $0,1,...,k-1$. Hence we have the recurrence relation:

$$f(n,m,k)=\sum_{i=1}^{k}f(n-i,m-i+1,k)$$

This recurrence should hold for arbitrary $n,m,k$ as long as we specify the appropriate initial values, or at least it should hold for $k$ large enough. (This might be where I ran into problems but I can not see it immediately).

However, when I carry this to the next part I am confused, because

\begin{aligned}
 F_k(x,y)& = \sum_{n,m\geq 0}f(n,m,k)x^ny^m \\
& = \sum_{n,m\geq 0}\sum_{i=1}^{k}f(n-i,m-i+1,k)x^ny^m \\
& = \sum_{i=1}^k\sum_{n,m\geq 0}f(n-i,m-i+1,k)x^ny^m \\
& = \sum_{i=1}^kx^iy^{i-1}\sum_{n,m\geq 0}f(n-i,m-i+1,k)x^{n-i}y^{m-i+1}\\
& = \sum_{i=1}^kx^iy^{i-1}F_k(x,y)
\end{aligned}
The second to bottom equality holds in my mind because $f(m,n,k)=0$ if either of $m$ or $n$ is $0$. However, it confuses me since at the end the equation implies $F_k(x,y)=0$?

Where did I go wrong? Any kind of help would be appreciated.

答:
Based on @Mike Earnest's help, I revised my solution and include it here in case it helps anyone. It also coincides with @robjohn's answer in this post, which I didn't get at the beginning.

\begin{aligned}
f(n,m,k) = \begin{cases}
\sum_{i=1}^k f(n-i,m-i+1,k),\quad n>m \\
0\;\quad\qquad\qquad\qquad\qquad ,\quad n<m \\
[n<k]\qquad\qquad\qquad\quad,\quad n=m
\end{cases}
\end{aligned}
Hence,
\begin{aligned}
F_k(x,y) & =\sum_{n,m\geq 0}f(n,m,k)x^ny^m \\
 & = \sum_{ n > m\geq 0}f(n,m,k)x^ny^m + \sum_{ n = m\geq 0}[n<k]x^ny^m\\
& = \sum_{n>m}\sum_{i=0}^kf(n-i,m-i+1,k)x^ny^m+\sum_{ n = 0}^{k-1}x^ny^n \\
& = \sum_{i=1}^kx^iy^{i-1}\sum_{n>m\geq 0}f(n-i,m-i+1,k)x^{n-i}y^{m-i+1}+\sum_{ n = 0}^{k-1}x^ny^n \\
& = \sum_{i=1}^kx^iy^{i-1}F_k(x,y) + \frac{x^ky^k-1}{xy-1} \\
& = \frac{x(x^ky^k-1)}{xy-1}F_k(x,y) + \frac{x^ky^k-1}{xy-1}
\end{aligned}

where
$$[n<k]=\begin{cases}1\quad, n<k \\ 0\quad, n\geq k \end{cases}$$

Finally we get
$$F_k(x,y)=\frac{x^ky^k-1}{xy-1-x(x^ky^k-1)}$$


问:
Exercise 2.13 Atiyah-Macdonald Introduction to commutative algebra

Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B\otimes_A N$. Show that the homomorphism $g:N\to N_B$ which maps $y$ to $1\otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$. [Define $p:N_B\to N$ by $p(b\otimes y) = by$ and show that $N_B = \text{Im}(g)\oplus\text{Ker}(p)$.]

First, as a $A$-module homomorphism, $p\circ g =1_{N}$ so $g$ is injective.

I want to prove the summand part in two ways first by finding isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$ and next by splitting lemma.

1). How can I define a $A$-module isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$?

I first tried this by defining $b\otimes n\mapsto (1\otimes p(b\otimes n),b\otimes n-(g\circ p)(b\otimes n))$ as $b\otimes n = 1\otimes p(b\otimes n)+b\otimes n-(g\circ p)(b\otimes n)$. But I think this map is not surjective.

2). This is a verification question. For splitting lemma, I considered the exact sequence $0\to\text{Ker}(p)\to N_B\xrightarrow{p} N\to 0$. Since $p\circ g = 1_N$, by splitting lemma, $N_B\simeq N\oplus \text{Ker}(p)\simeq \text{Im}(g)\oplus\text{Ker}(p)$ as $g$ is injective. Is this correct?

I saw some related post to this question(defining explicit isomorphism) but I think the defined map in the post is wrong. Anyway, any comment will be appreciated.

Edit: Both $A,B$ are commutative rings with unity

答:
1)

Take an $x\in N_B$, and decompose it as
$$
\Big(g\circ p(x),x-g\circ p(x)\Big)
$$
Clearly the first component is in the image of $g$. And since $g\circ p$ (which maps $b\otimes_A n$ to $1\otimes_A bn$) is idempotent, the second component is clearly in the kernel of $g\circ p$, which by injectivity of $g$ means the kernel of $p$.

As for mapping in the other direction, we have simply $(x,y)\mapsto x+y$.

2)

Yes, that looks good.


问:
Let $|G|$ be infinite. Use normal cores (not Poincaré's Theorem) to prove that if $H, K$ have finite index in $G$, then so does $H\cap K$.
This is Exercise 7.4 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE. This Google search yields this question, which is close but does not have the restriction to normal cores nor the prohibition of Poincaré's Theorem.

The Details:

On page 62 of Roman's book, there is

Poincaré's Theorem: Let $H_1, \dots, H_n\le G$ and $[G: H_i]<\infty$ for all $i$ and let $I_k=H_1\cap\dots\cap H_k$. Then Poincaré's Inequality holds:

$$[G: H_1\cap \dots\cap H_n]\le [G: H_1]\dots[G: H_n]$$

and so, in particular, $[G: H_1\cap \dots\cap H_n]$ is also finite.

The inequality above can be replaced can be replaced by division if $$I_kH_k\le G\tag{1}$$ for all $k\in\{1,\dots, n-1\}$.

Equality holds in Poincaré's Inequality if and only if $$[I_kH_{k+1}:H_{k+1}]=[G:H_{k+1}]$$ for all $k\in\{1,\dots, n-1\}$. Hence, if $G$ is finite or $(1)$ holds for all $k$, then equality holds in Poincaré's Inequality if and only if $$I_kH_{k+1}=G$$ for all $k\in\{1,\dots, n-1\}$.

On page 125 ibid., we have, within a theorem, the

Definition: Let $H\le G$. The largest subgroup of $H$ that is normal in $G$ is called the normal interior or core of $H$, which we denote by $H^\circ$.

The Question:

This question is from the chapter on group actions and the structure of $p$-groups, although I think the latter topic is not relevant here.

Exercise 7.4: Let $G$ be an infinite group. Use normal interiors (not Poincaré's Theorem) to prove that if $H$ and $K$ have finite index in $G$, then so does $H\cap K$.

Context:

As I said above, I think group actions are relevant here. Indeed, the theorem the definition above is in goes like this:

Theorem 4.19: Let $H\le G$. [. . .] The core of $H$ is the kernel $$H^\circ=\bigcap_{a\in G}H^a$$ of the action $\sigma$ of $G$ on $G/H$ by left translation.

Proof (p.125): We have

$$\begin{align}
\ker(\sigma)&=\{x\in G\mid \sigma_x={\rm id}\}\\
&=\{x\in G\mid xaH=aH, \forall a\in G\}\\
&=\{x\in G\mid xa\in aH, \forall a\in G\}\\
&=\{x\in G\mid x\in a^{-1}Ha, \forall a\in G\}\\
&=\bigcap_{a\in G} H^a.
\end{align}$$ $\square$

Thus, if I compose/compare somehow the actions $\sigma_L$ of $G$ on $G/L$ for $L\in\{H,K\}$ in question if possible, then I might get somewhere. I'm not sure how to flesh out this vague idea though.

Do I need to calculate $$(H\cap K)^\circ=\bigcap_{g\in G}(H\cap K)^g?$$ If so, I'm not sure how it helps.

(Here $L^g$ is $g^{-1}Lg$.)

Please help :)

答:
Since $G/H$ is finite, $G/H^\circ$ is isomorphic to a subgroup of a finite symmetric group, i.e., $H^\circ$ is also of finite index. Likewise, $K^\circ$ is of finite index.
Then the kernel of the obvious homomorphism $G\to G/H^\circ \times G/K^\circ$ is also of finite index and is clearly contained in $H\cap K$.


问:
Is the map $\ g:S \otimes_R A \to T \otimes_R A \ $ also faithfully flat?
Let $R, S, T$ are commutative rings with $R \subset S \subset T$ and  $f:S \to T$ is faithfully flat.

Assume $A$ be a finite flat $R$-algebra. $$ \text{Is the map $\ g:S \otimes_R A \to  T \otimes_R A \ $ also faithfully flat ?}$$
If $R \to S$ is also faithfully flat, case then in that  case, I think the above map $g$ will be faithfully flat.

Does the result follows without assuming $R \to S$ faithfully flat?

Any help  please

答:
Let $u: M \rightarrow N$ be a morphism of $S \otimes_R A$-modules. Then $M \otimes_{S \otimes_R A} (T \otimes_R A) \cong M \otimes_S T$ with the $A$-algebra structure on $M$ directly (same for $N$); and $u'=u \otimes_{S \otimes_R A} (T \otimes_R A)$ acts under these isomorphisms as $u \otimes_S T: M \otimes_S T \rightarrow N \otimes_S T$. So $u$ is injective iff $u \otimes_S T$ is injective, iff $u'$ is injective.

So $S \otimes_R A \rightarrow T \otimes_R A$ is faithfully flat.

More directly, you can check that $T \otimes_R A = T \otimes_S (S \otimes_R A)$ so your map is a "base change" of the faithfully flat map $S \rightarrow T$.


答:
Denote $A=12^a \cdot 324^b \cdot 48^c.$

From $$12^a \cdot 324^b \cdot 48^c=6^a2^a6^{2b}3^{2b}6^c2^{3c}$$ we see that the best strategy is to take $b$ as large as possible, because the steps of $b$ are twice those of $a$ or $c.$

Further, note that each $324^b$ offers $3^2$ extra that we need to join to some remaining $2s$ from $2^a$ or $2^{3c}.$

Clearly, $2^{3c}$ is the best choice, because a single $c$ offers $2^3.$

Let us start with $(a,b,c)=(0,35,0),$ here $A=6^{70}3^{70}$ and $a+b+c=35.$

Note that $(0,33,1)$ gives $A=6^{70}3^{63}$ and $a+b+c=34.$

In general, $(a,b,c)=(0,35-2k,k)$ gives $A=6^{70-4k}3^{70-4k}6^k2^{3k}$ or
$$A=6^{70-3k}3^{70-4k}2^{3k}$$

To obtain $A$ divisible by $6^{70},$ solve $$70-4k=3k.$$

We obtain $k=10,$ the triple is $(a,b,c)=(0,15,10)$ and the sum $$a+b+c=0+15+10=25.$$

Body:
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.

First I can make an assumption that $a+b+c$ will be the smallest if we can obtain the highest power of b, then c and later a.

Let me go in the general way, $12^a\cdot 324^b\cdot48^c$ = $ 3^a \cdot  2^{2a} \cdot 3^{4b} \cdot 2^{2b} \cdot3^c \cdot 2^{4c}$ = $3^{a+4b+c} \cdot 2^{2a+2b+4c}$

And $6^{70}$ = $2^{70}\cdot 3^{70}$

Then I need to find the solution for $a+4b+c = 70$ and $a+b+2c=35$ where a+b+c need to be the samllest.

Please anyone can show me the solution how to approach this type of questions.


问:
Given conditions on differentiability, what can we say about second derivative?
Let $f:[0,10] \rightarrow [10,30]$ be a continuous and twice differentiable function such that $f(0)=10$ and $f(10) =20$. Suppose $\mid f'(x) \mid \leqslant 1 \; \forall x \in [0,10]$. Then the value of $f"(5)$ is

$(a) 0$

$(b) 1/2$

$(c) 1$

$(d) Can't be determined$

My trial:

We need the function to be twice differentiable. So let $f"(x)=c$ where $c$ is a constant.
Then $f(x) =\frac{cx^{2}}{2} + dx + e$ where $c,d,e \in \mathbb{R}$

Also, $f(0) =10$ implies $e=10$

and $f(10) = 20 \;$ implies $ \; 50c + 10d =10$ or $5c + d=1$

Now, $f'(5) = 5c +d = 1$ (suprise, shawty!)

therefore $c = \frac{1-d}{5} , d \in \mathbb{R}$

How to proceed? what values of $d$ to chooses such that $\mid f'(x) \mid \leqslant 1$

Any other easier approach is also welcome. Thanks. Can we somehow use MVT or Brower's Fixed Point here?

答:
Hint

One can prove that $f'(x)=1$ for all $x\in [0,10]$.


问:
CDF and PDF of non-monotonic transformations
$y = x(1-x)$. Distribution of $x$ is uniform on $[0,1]$ for $x$ in $(0,1)$.

Find $p(y)$ and $P(Y)$

Can you guys give me hints on how to proceed?

I think the general way for this problem is to partition the function into monotone parts. But when I proceed, I got stuck with the function $y = x(1-x)$ as I cannot express $x$ in terms of $y$.

Thank you

答:
With $y=x(1-x)$ you have

$y$ increasing from $0$ to $\frac14$ as $x$ increases from $0$ to $\frac12$

$y$ decreasing from $\frac14$ to $0$ as $x$ increases from $\frac12$ to $0$

$x=\frac12 \pm \frac12\sqrt{1-4y}$

$\mathbb P(Y \le y) = \mathbb P\left(X \le \frac12 -\frac12\sqrt{1-4y} \right)+\mathbb P\left(X \ge \frac12 +\frac12\sqrt{1-4y} \right) = 1-\sqrt{1-4y}$ when $0 \le y \le \frac14$

$p_Y^{\,}(y) = \frac{d}{dy}\mathbb P(Y \le y) = \frac{2}{\sqrt{1-4y}}$ when $0 \le y \le \frac14$


问:
Show that this set is convex
Let $K\subset \mathbb{R}^p$ be convex with nonempty interior, i.e. $K^{\circ}\neq \emptyset$. I want to show that the set

$$A:=\{x\in \bar{K}:d(x,\mathbb{R}^p\setminus K^{\circ}) \geq 1/n\}$$

is convex, where $n\geq 1$, $\bar{K}$ denotes the closure of $K$, and $d(x,\mathbb{R}^p\setminus K^{\circ}):=\inf\{||x-y||: y\in\mathbb{R}^p\setminus K^{\circ}\}.$ Note that $A=\bar{K}$ if $K=\mathbb{R}^p$.

Using continuity of the distance function I was able to show that $A$ is closed (using the sequence criterion), but I don't know how to proceed next.

Any help is greatly appreciated.

答:
Let $x,y \in A$ and $\lambda \in [0,1]$ and suppose $\lambda x + (1-\lambda)y \not\in A$. Then for all $a \in \mathbb{R}^p\setminus K^{\circ}$ we have $d(x,a)\geq 1/n$ and $d(y,a) \geq 1/n$, but there exists a $b \in\mathbb{R}^p\setminus K^{\circ}$ such that $d(\lambda x + (1-\lambda)y,b)<1/n$. I will show that this leads to a contradiction.

Note that $\{a : d(x,a)< 1/n\} \subset K^{\circ}$ and $\{a : d(y,a)< 1/n\} \subset K^{\circ}$. Since $d(x, x+b-\lambda x - (1-\lambda)y)<1/n$, $x+b-\lambda x + (1-\lambda)y \in K^{\circ}$ . Similarly, $y+b-\lambda x - (1-\lambda)y \in K^{\circ}$. Since $K^\circ$ is convex,
$$\lambda\left[x+b-\lambda x - (1-\lambda)y\right] + (1-\lambda)\left[y+b-\lambda x - (1-\lambda)y\right]=b\in K^\circ,$$
which is a contradiction with $b \in\mathbb{R}^p\setminus K^{\circ}$.

Title: Show that this set is convex


答:
You cannot assume that $f''$ is constant (though it will turn out that it is).
Assume $f'(a)<1$ for some $a\in[0,1]$. Then by continuity(!) of $f'$, there exists an interval $[a,b]$ and a number $q<1$ with $f'(x)<q$ for all $x\in[a,b]$.
Then
$$ \begin{align}10&=f(10)-f(0)\\
&=\int_0^{10}f'(x)\,\mathrm dx \\
&=\int_0^{a}f'(x)\,\mathrm dx+\int_a^bf'(x)\,\mathrm dx+\int_b^{10}f'(x)\,\mathrm dx\\
&\le \int_0^{a}1\,\mathrm dx+\int_a^bq\,\mathrm dx+\int_b^{10}1\,\mathrm dx\\
&=a+(b-a)q+10-b\\&=10-(b-a)(1-q)\\
&<10,\end{align}$$
contradiction.

Therefore $f'(x)=1$ for all $x$.

In fact, let's forget about the second derivative, i.e., start with once differentiable $f\colon [0,10]\to [20,30]$ with $f(0)=20$, $f(10)=30$, and $f'(x)\le 1$ for all $x$. Then this is enough to show $f(x)=x+20$ (and consequently, $f''$ exists and is identically $0$):
Consider $x$ with $0<x<10$

Assume  $f(x)>x+20$. Then by the Mean Value Theorem, there exists $\xi$ between $0$ and $x$ with
$$f'(\xi)=\frac{f(x)-f(0)}{x-0}>\frac{(x+20)-(0+20)}{x} =1,$$
contradiction.

Assume $f(x)<x+20$. Again by the mean value theorem, there eixts $\xi$ betwewn $x$ and $10$ with
$$f'(\xi)=\frac{f(10)-f(x)}{10-x}>\frac{30-(x+20)}{10-x} =1,$$
contradiction.

Therefore, the only remaining possibility $f(x)=x+20$ remains.

Title: Given conditions on differentiability, what can we say about second derivative?

Body:
Let $f:[0,10] \rightarrow [10,30]$ be a continuous and twice differentiable function such that $f(0)=10$ and $f(10) =20$. Suppose $\mid f'(x) \mid \leqslant 1 \; \forall x \in [0,10]$. Then the value of $f"(5)$ is

$(a) 0$

$(b) 1/2$

$(c) 1$

$(d) Can't be determined$

My trial:

We need the function to be twice differentiable. So let $f"(x)=c$ where $c$ is a constant.
Then $f(x) =\frac{cx^{2}}{2} + dx + e$ where $c,d,e \in \mathbb{R}$

Also, $f(0) =10$ implies $e=10$

and $f(10) = 20 \;$ implies $ \; 50c + 10d =10$ or $5c + d=1$

Now, $f'(5) = 5c +d = 1$ (suprise, shawty!)

therefore $c = \frac{1-d}{5} , d \in \mathbb{R}$

How to proceed? what values of $d$ to chooses such that $\mid f'(x) \mid \leqslant 1$

Any other easier approach is also welcome. Thanks. Can we somehow use MVT or Brower's Fixed Point here?


问:
Construct an affine transformation given the image of 2 points without skewing
I am attempting to create a mobile app in which the user can interact with content on the screen by using two fingers to translate, scale and rotate the content - very much like the interaction with the map in the Google Maps mobile app.

Intuitively, I am convinced that the added constraint that no skewing operations are used to construct the transformation should result in a unique transformation. I compare it to using two fingers to move a piece of a paper on a flat surface - if one can imagine a piece of paper that can be stretched as well, the comparison is complete.

Given the transformation matrix

$$T = \begin{bmatrix} a & b & 0 \\ c & d & 0 \\ t_x & t_y & 1 \end{bmatrix}$$

In the general case, this results in 6 unknowns ($a, b, c, d, t_x, t_y$) and only 4 equations (one for each $x$ and $y$ for each of the two known points). My current attempt is adding the additional constraints that $a = d$ and $b = c$, since the scaling operations I am interested in scale $a$ and $d$ by the same amount, and rotation operations result in $b = c$ or $b = -c$. This gives me 4 equations in 4 unknowns ($a = d, b = c, t_x, t_y$) which I can solve.

Specifically, given an unknown desired transformation $T$, and known points $A$, $A'$, $B$ and $B'$:

$$\begin{align}A = (a_x, a_y) \\ T(A) = A' = (a'_x, a'_y) \\ B = (b_x, b_y) \\ T(B) = B' = (b'_x, b'_y)\end{align}$$

I can evaluate the system of equations

$$\begin{bmatrix}a_x & a_y & 1 & 0 \\ b_x & b_y & 1 & 0 \\ a_y & a_x & 0 & 1 \\ b_y & b_x & 0 & 1 \end{bmatrix} \begin{bmatrix} a = d \\ b = c \\ t_x \\ t_y \end{bmatrix} = \begin{bmatrix} a'_x \\ b'_x \\ a'_y \\ b'_y \end{bmatrix}$$

The result of this is scaling and translations work as expected. However, instead of a rotation, I end up with a symmetrical skewing effect that you can see in the below images. The 6 orange numbers at the top of each image represent the first two columns of the transformation matrix; namely, $\begin{bmatrix} a & c & t_x \\ b & d & t_y \end{bmatrix}$:

Initial identity transformation:

After rotating two fingers clockwise a small amount:

After rotating two fingers clockwise a larger amount

I am sure this is simply a result of my lack of understanding of how to calculate this new transformation matrix correctly.

How can I calculate this new transformation matrix?

Thanks in advance for any help!

答:
In case anyone finds this in the future, I have solved my problem with a different approach. Instead of attempting to calculate the entries in the transformation matrix using a system of equations, I now construct a transformation as the concatenation of four simple transformations.

Given an unknown desired transformation $T$, and known points $A$, $A'$, $B$ and $B'$:

$$\begin{align}A = (a_x, a_y) \\ T(A) = A' = (a'_x, a'_y) \\ B = (b_x, b_y) \\ T(B) = B' = (b'_x, b'_y)\end{align}$$

I construct $T$ as follows:

Translating the identity transformation by $-A$

Scaling the resulting transformation by the ratio $\overline{B'A'}\over\overline{BA}$

Rotating the resulting transformation by $\angle B'A' - \angle BA$

Translating the resulting transformation by $A'$

Intuitively, 1 translates the line segment $AB$ moving $A$ to the origin, 2 and 3 scale and rotate this segment to be the same length and angle as the line segment $A'B'$, and 4 translates this segment moving the origin to $A'$. In this way, we know $T(A) = A'$ and $T(B) = B'$. Further, since it is a composition of translations, rotations and scaling, we can be assured that the resulting transformation does not contain any shearing.

With a few small caveats surrounding implementation, this has exactly the effect I am looking for - two fingers can be used to move, scale and rotate the content on the screen.


答:
You used it correctly, but left something out when stating multiplicativity: A number-theoretic function such as $\phi$ is called multiplicative if $\phi(nm)=\phi(n)\phi(m)$ holds if $n,m$ are relatively prime. For example $\phi(4)=2\ne\phi(2)\phi(2)$.
Otherwise, your result is correct: Precisely the multiples of $11$ and the multiples of $13$ are not relatively prime to $143$, so thats $11,22,33,\ldots , 143$ and $13,26,39,\ldots,143$ (with $143$ occuring in both exception lists).

Title: Does Euler totient function gives exactly one value(answer) or LEAST calculated value(answer is NOT below this value)?

Body:
I was studying RSA when came across Euler totient function. The definition states that- it gives the number of positive values less than $n$ which are relatively prime to $n$.

I thought I had it, until I came across this property:-

Euler Totient function is multiplicative function, that is: $\varphi(mn) = \varphi(m)\varphi(n)$

Now, if $p$ is a prime number, $\varphi(p)=p-1$.

Putting values of $p$ as 11 and 13 one by one,

$$\varphi(11)=10$$

$$\varphi(13)=12$$

Applying above stated property,

$$\varphi(11\cdot 13)=\varphi(11)\varphi(13)$$

$$\varphi(143)=12 \cdot 10$$

$$\varphi(143)=120$$

Is it correct? Does that mean we have $23$ values between $1$ and $143$ which are not relatively prime to $143$? Sorry if its something basic I'm missing. I'm not some genius at maths and came across this during study of RSA Algo. Thanks.


问:
Graphing $x^2+1$ over the complex plane
I was watching this video: https://youtu.be/T647CGsuOVU?t=113

and I wanted to be able to graph the function $x^2+1$ as they have in the video (the link takes you to the moment when the graph is shown). The graph was what I found an interesting way of what the function would look like over the complex plane.

Essentially, I'd like to try and get some visual representations of the complex roots of the function. How could I go about this with matlab or geogebra (preferably the latter)?

答:
Following @user856's comment, I tried to plot the real part of y=z^2.

$
\begin{aligned}
y &= z^2 \\
&= (a + ib)^2 \\
&= (a^2 - b^2) + i(2ab)
\end{aligned}
$

Using Python:

```<code>import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

A = np.linspace(-5, 5, 20)
B = np.linspace(-5, 5, 20)
A, B = np.meshgrid(A, B)
Y = A**2 - B**2

C = np.linspace(-5, 5, 10)
D = np.linspace(-25, 25, 10)
C, D = np.meshgrid(C, D)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(A, B, Y, zorder=2)
ax.plot_surface(A, B, np.zeros_like(A), color='red', alpha=0.5, zorder=1)
ax.plot_surface(C, np.zeros_like(C), D, color='green', alpha=0.5, zorder=1)
ax.set_xlabel('a')
ax.set_ylabel('b')
ax.set_zlabel('y')
plt.show()
</code>```

I have drawn two planes:

b=0, which will show the plot of $z^2$ in the real plane.

y=0, so that you can see how $z^2$ in the real plane never crosses that plane, but outside b=0 it does.

EDIT:

After a second thought, you might get a glimpse of what the function looks like in its entirety, by coloring the plot for instance.

```<code>import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

A = np.linspace(-10, 10, 40)
B = np.linspace(-10, 10, 40)
A, B = np.meshgrid(A, B)
Re = A**2 - B**2
Im = 2 * A * B

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
img = ax.scatter(A, B, Re, c=Im)
ax.set_xlabel('a')
ax.set_ylabel('b')
ax.set_zlabel('Re(z^2)')
bar = fig.colorbar(img)
bar.ax.set_ylabel('Im(z^2)', rotation=0)
plt.show()
</code>```

Title: Graphing $x^2+1$ over the complex plane

Body:
Following @user856's comment, I tried to plot the real part of y=z^2.

$
\begin{aligned}
y &= z^2 \\
&= (a + ib)^2 \\
&= (a^2 - b^2) + i(2ab)
\end{aligned}
$

Using Python:

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

A = np.linspace(-5, 5, 20)
B = np.linspace(-5, 5, 20)
A, B = np.meshgrid(A, B)
Y = A**2 - B**2

C = np.linspace(-5, 5, 10)
D = np.linspace(-25, 25, 10)
C, D = np.meshgrid(C, D)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(A, B, Y, zorder=2)
ax.plot_surface(A, B, np.zeros_like(A), color='red', alpha=0.5, zorder=1)
ax.plot_surface(C, np.zeros_like(C), D, color='green', alpha=0.5, zorder=1)
ax.set_xlabel('a')
ax.set_ylabel('b')
ax.set_zlabel('y')
plt.show()

I have drawn two planes:

b=0, which will show the plot of $z^2$ in the real plane.

y=0, so that you can see how $z^2$ in the real plane never crosses that plane, but outside b=0 it does.

EDIT:

After a second thought, you might get a glimpse of what the function looks like in its entirety, by coloring the plot for instance.

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

A = np.linspace(-10, 10, 40)
B = np.linspace(-10, 10, 40)
A, B = np.meshgrid(A, B)
Re = A**2 - B**2
Im = 2 * A * B

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
img = ax.scatter(A, B, Re, c=Im)
ax.set_xlabel('a')
ax.set_ylabel('b')
ax.set_zlabel('Re(z^2)')
bar = fig.colorbar(img)
bar.ax.set_ylabel('Im(z^2)', rotation=0)
plt.show()


问:
Find all units, zero divisors and nilpotent elements of $\mathbb{Z}_4[X]$ and $\mathbb{Z}_6[X]$
I've been solving some problems from my abstract algebra course as training for the final exam, and I want to check if my solution to this one is correct:

Describe the units, the nilpotent elements and the zero divisors of
the rings $\mathbb{Z}_4[X]$ and $\mathbb{Z}_6[X].$

This is the work I did:

For $\mathbb{Z}_4[X]$:

Units: this one's easy, since units in a polynomial ring $R[X]$ are the same as the units in $R$, so I conclude that the units of $\mathbb{Z}_4[X]$ are $\bar1,\bar3$ (the coprime elements to $4$).

Zero divisors: First, it's obvious that $\bar 2$ is zero divisor. Then, in order for a polynomial to get cancelled by multiply6ing it by a non-zero element, it must have all of it's coefficients $\bar0$ or $\bar2$. So I conclude that the zero divisors of $\mathbb{Z}_4[X]$ are all the polynomials with $\bar 0$ and $\bar 2$ as coefficients (not including $\bar 0$).

Nilpotent elements: in order for the power of a polynomial in $\mathbb{Z}_4[X]$ to be cancelled, the same reasoning from the zero divisors gives us that the nilpotent elements of $\mathbb{Z}_4[X]$ are the polynomials with coefficients $\bar0$ and $\bar 2$ (this time including $\bar 0$).

For $\mathbb{Z}_6[X]$:

Units: The units in $\mathbb{Z}_6[X]$ are the same of $\mathbb{Z}_6$, so they are $\bar 1$ and $\bar 5$ (the coprimes to $6$)

Zero divisors: using a similar reasoning to the one I used for $\mathbb{Z}_4[X]$, I end up concluding that the zero divisors in $\mathbb{Z}_6[X]$ are the polynomial with coefficients $\bar0$, $\bar2$, $\bar 4$; and the polynomials with coefficients $\bar0$, $\bar 3$ (not including $\bar 0$)(also not mixing coefficients, either it has $\bar2$, $\bar4$ or either it has $\bar3$).

Nilpotent elements: in order for a power of a polynomial to be cancelled, it's clear that it's leading coefficient must be multiple of $6$, as well as it's independent coefficient, but $6$ is $\bar 0$ in $\mathbb{Z}_6[X]$, so I end up concluding that the only nilpotent element in $\mathbb{Z}_6[X]$ is $\bar 0$.

Is my solution correct? If not, where did I go wrong? Any help will be appreciated, thanks in advance.

答: I think that parts with units and nilpotent elements are correct, but I'm not convinced about the part with zero divisors. It looks like you only included the case of multiplying our polynomial by something from $R$, not from $R[X]$. It is not clear why for polynomial like $2x^2+x+2$ it's not possible that multiplying it by some strange element of $R[X]$ won't result in some fancy cancelling and product equal to zero. (However I suppose the answer should be correct.)


问:
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.

First I can make an assumption that $a+b+c$ will be the smallest if we can obtain the highest power of b, then c and later a.

Let me go in the general way, $12^a\cdot 324^b\cdot48^c$ = $ 3^a \cdot  2^{2a} \cdot 3^{4b} \cdot 2^{2b} \cdot3^c \cdot 2^{4c}$ = $3^{a+4b+c} \cdot 2^{2a+2b+4c}$

And $6^{70}$ = $2^{70}\cdot 3^{70}$

Then I need to find the solution for $a+4b+c = 70$ and $a+b+2c=35$ where a+b+c need to be the samllest.

Please anyone can show me the solution how to approach this type of questions.

答:
We need $a+4b+c \ge 70$ and $a+b+2c \ge 35$.

Let's suppose we have $\alpha + 4\beta + \gamma \ge 70$ and $\alpha + \beta + 2\gamma \ge 35$ and $\alpha +\beta + \gamma = k$.

But then $\alpha - 1 + 4(\beta +1) +\gamma\ge  73>70$ and $\alpha -1 +(\beta + 1) + 2\gamma \ge 35$ annd $(\alpha -1) + (\beta + 1) + \gamma = k$.

So wolog we may assume $a = 0$.  (If we have a small value for $a,b,c$ with $a \ge 1$ it can't hurt to replace decrease $a$ and increase $b$ if necessary)

So if $a = 0$ we need $b \ge \frac {70-c}4$ and $c \ge \frac {35-b}2$.

If we replace $b = 17-x$ and $c=17-y$ we get

$68 - 4x + 17 -y \ge 70$ or $4x+y \le 15$ and $17-x + 34-2y\ge 35$ or $x+2y\le 16$.

To get as close to these values as possible we can, if we solve for equality we get $x=2; y=7$ so

$a = 0; b= 15; c =10$ is as solution with $a+4b + c =70$ and $a+b+2c = 35$.  And $a+b +c =25$ is pretty small.

And it's easy to see that is an absolute minimum.  We can't decrease $a$ at all.  We can't decrease $b$ without increasing $a$ or $c$ four times as much to maintain $a + 4b + c \ge 70$.  And we can't decrease $c$ without increasing $a$ or $b$ twice as much to maintain $a + b = 2c \ge 35$.


问:
Given a line segment, a line parallel to it, and a straightedge, divide the segment into $n$ equal segments

Given a line segment, a line parallel to it, and a straightedge, how to divide the segment into $n$ equal segments?

(With a straightedge, you are allowed only to draw straight lines. You are not allowed to mark off distances on the straightedge.)

答:
This took a lot of time to finish up. I hope it conveys the interesting nature of this problem. Although the problem is stated quite clearly, it seems to have caused some confusion in the comments and answers, so I will restate it differently:

Given a line segment and a line parallel to it, how can the line segment be partitioned into n equal segments only using a
straightedge?

Let's start with the $n=2$ case. Let the given line segment be AB, and let the given parallel line be defined with the points $\textbf{F}$ and $\textbf{E}$. Let's name the intersection point of the rays $\textbf{AF}$ and $\textbf{BD}$ as $\textbf{D}$. Let the intersection point of $\textbf{AE}$ and $\textbf{FB}$ be $\textbf{O}$. Let the intersection point of the ray $\mathbf{DO}$ with the line $\textbf{FE}$ be $\mathbf{H_{0}}$. Finally, let the point intersection of  the ray $\mathbf{DO}$ and $\textbf{AB}$ be $\mathbf{I_{1}}$. I created a sketch to work on in GeoGebra in the figure below.

Theorem 1

In the following figure, point $\mathbf{I_{1}}$ is the midpoint of $\mathbf{AB}$, ie.:
\begin{equation}\label{theorem1}
\mathbf{AI_{1}}=\mathbf{I_{1}B}.
\end{equation}

The proof of this theorem will use two well-known theorems; namely Thales' Theorem and Ceva's Theorem.
Proof.
Since $\textbf{AB} \parallel \textbf{FE}$, from Thales' Theorem:
$$\tag{1}\label{1}
    \frac{\textbf{AF}}{\textbf{FD}}=\frac{\textbf{BE}}{\textbf{ED}}$$

From Ceva's Theorem,
$$\frac{\textbf{AF}}{\textbf{FD}}\cdot\frac{\textbf{DE}}{\textbf{EB}}\cdot\frac{\mathbf{BI_{1}}}{\mathbf{I_{1}A}}=1$$

Substituting from Eq. $\ref{1}$,
$$\mathbf{AI_{1}}=\mathbf{I_{1}B}$$
we obtain the required result. $\blacksquare$

Corollary 1.1

\begin{equation}\tag{2}\label{2}
       \mathbf{FH_{0}}=\mathbf{H_{0}E}
\end{equation}
Proof.
This can be proved in several different ways. Since Eq.($\ref{1}$) and Eq.($\ref{2}$) were established before, by considering a homothety centered at $\textbf{D}$ it is clear that $\mathbf{FH_{0}}=\mathbf{H_{0}E}$. Alternatively, using the following similar triangles: $\bigtriangleup\mathbf{AI_{1}O} \sim \bigtriangleup\mathbf{EH_{0}O}$ and $\bigtriangleup\mathbf{BI_{1}O} \sim \bigtriangleup\mathbf{FH_{0}O}$, the corollary can be proven.
$\blacksquare$

Next, for the case $n=3$ (as in the Figure  above}, let's connect the points $\mathbf{I_{1}}$ and $\textbf{E}$ and name the intersection of $\mathbf{I_{1}E}$ and $\textbf{FB}$ as $\mathbf{X_{1}}$. Let the point intersection of  the ray $\mathbf{DX_{1}}$ and $\textbf{AB}$ be $\mathbf{I_{2}}$. Let the intersection point of the ray $\mathbf{DX_{1}}$ and line segment $\textbf{AE}$ be $\mathbf{Y_{1}}$. Let the intersection point of the ray $\mathbf{DX_{1}}$ with the line $\textbf{FE}$ be $\mathbf{H_{1}}$. Finally, let the intersection points of the line segment $\textbf{AB}$ and the rays $\mathbf{H_{1}O}$, $\mathbf{H_{1}Y_{1}}$ be points $\mathbf{W_{1}}$, $\mathbf{W_{2}}$ respectively. Notice that the ray $\mathbf{H_{1}Y_{1}}$ is contained by the ray $\mathbf{DX_{1}}$, and hence, $\mathbf{I_{2}}$ is coincident with $\mathbf{W_{2}}$. So, the point is labeled as  $\mathbf{I_{2}(W_{2})}$.
(Alternatively, we could've connected the points $\mathbf{I_{1}}$ and $\textbf{F}$ and name the intersection of $\mathbf{I_{1}F}$ and $\textbf{AE}$ as $\textbf{P}$. Then, we could obtain $\mathbf{W_{1}}$ by finding the intersection of ray $\textbf{DP}$ and $\textbf{AB}$. However, we will stick to the previous method as it is easier to generalize.)

Theorem 2

In the figure above,
\begin{equation}\tag{3}\label{3}
\mathbf{AW_{1}}=\mathbf{W_{1}W_{2}}=\mathbf{W_{2}B}
\end{equation}

Proof.
Using Eq.(3) and similar triangles in Figure 2 , Theorem 2 can be proven.
\begin{equation}\tag{4}\label{4}
\bigtriangleup\mathbf{I_{1}X_{1}B} \sim \bigtriangleup\mathbf{EX_{1}F} \;and \bigtriangleup\mathbf{I_{1}X_{1}W_{2}} \sim \bigtriangleup\mathbf{EX_{1}H_{1}} \implies \frac{\mathbf{I_{1}B}}{\mathbf{2FH_{O}}}=\frac{\mathbf{I_{1}W_{2}}}{\mathbf{H_{1}E}}
\end{equation}

\begin{equation}\tag{5}\label{5}
\bigtriangleup\mathbf{AOW_{1}} \sim \bigtriangleup\mathbf{EOH_{1}} \;and \bigtriangleup\mathbf{W_{1}OI_{1}} \sim \bigtriangleup\mathbf{H_{1}OH_{0}} \implies
\frac{\mathbf{AW_{1}}}{\mathbf{H_{1}E}}=\frac{\mathbf{W_{1}I_{1}}}{\mathbf{H_{0}H_{1}}}
\end{equation}

\begin{equation}\tag{6}\label{6}
\bigtriangleup\mathbf{I_{1}OB} \sim \bigtriangleup\mathbf{H_{0}OF} \;and \bigtriangleup\mathbf{W_{1}OI_{1}} \sim \bigtriangleup\mathbf{H_{1}OH_{0}} \implies
\frac{\mathbf{I_{1}B}}{\mathbf{FH_{O}}}=\frac{\mathbf{W_{1}I_{1}}}{\mathbf{H_{0}H_{1}}}
\end{equation}

Combining Eq.(5),  Eq.(6), and Eq.(7), we get
$\mathbf{AW_{1}}=\mathbf{2{I_{1}W_{2}}}$.
By the symmetry of the figure, it is also true that,
$\mathbf{W_{2}B}=\mathbf{2{W_{1}I_{1}}}$.
Combining the two results with Eq. (1),
$\mathbf{AW_{1}}=\mathbf{W_{1}W_{2}}=\mathbf{W_{2}B}. \;  \blacksquare$

The proof for Theorem Theorem 2 using similar triangles can be extended to construct proofs for the rest of the theorems that will be deduced (for Theorem 3 and Theorem 4). Before trying to generalize, let's also analyze the case for $n=4$ in Figure below. Similar to what we did for the $n=3$ case let's connect the points $\mathbf{I_{2}}$ and $\textbf{E}$ and name the intersection of $\mathbf{I_{2}E}$ and $\textbf{FB}$ as $\mathbf{X_{2}}$. Let the point intersection of the ray $\mathbf{DX_{2}}$ and $\textbf{AE}$ be $\mathbf{Y_{2}}$. Let the intersection point of the ray $\mathbf{DX_{2}}$ with the line $\textbf{FE}$ be $\mathbf{H_{2}}$. Let the intersection points of the line segment $\textbf{AB}$ and the rays (purple) $\mathbf{H_{2}O}$, $\mathbf{H_{2}Y_{1}}$, and $\mathbf{H_{2}Y_{2}}$ be points $\mathbf{W_{1}}$, $\mathbf{W_{2}}$, $\mathbf{W_{3}}$ respectively. Note that $\mathbf{I_{1}}$ is coincident to $\mathbf{W_{2}}$ and $\mathbf{I_{3}}$ is coincident to $\mathbf{W_{3}}$.

Theorem 3

In the figure above,
\begin{equation}\tag{7}\label{7}
\mathbf{AW_{1}}=\mathbf{W_{1}W_{2}}=\mathbf{W_{2}W_{3}}=\mathbf{W_{4}B}
\end{equation}

The same methodology can be applied repeatedly to obtain solutions for all values of ${\textit{n}}$. Let the intersection points of the line segment $\textbf{BF}$ with the rays $\mathbf{DH_{1}, DH_{2}, DH_{3},...,DH_{n-2}}$ be called as $\mathbf{X_{1}, X_{2}, X_{3}, ..., X_{n-2}}$. Let the intersection points of the rays $\mathbf{DO, DX_{1}, DX_{2}, ..., DX_{n-2}}$ with the line $\textbf{FE}$ be $\mathbf{H_{0}, H_{1}, H_{2}, ..., H_{n-2}}$. Then, let's name the intersection points of the rays $\mathbf{DO, DX_{1}, DX_{2}, ..., DX_{n-2}}$ with the line segment $\textbf{AB}$ as $\mathbf{I_{1}, I_{2}, I_{3}, ..., I_{n-1}}$. Let's form the rays $\mathbf{H_{n-2}O, H_{n-2}Y_{1}, H_{n-2}Y_{2}, ... , H_{n-2}Y_{n-2}}$ and name the intersection points of these rays with $\textbf{AB}$ as $\mathbf{W_{1}, W_{2}, W_{3}, ..., W_{n-1}}$. Voil'a, the line segment $\textbf{AB}$ is partitioned into $\textit{n}$ equal segments. As a final note, $\forall \; (k+1)|n$, $\mathbf{I_{k}}$ is coincident to $\mathbf{W_{m}}$ for certain values of $\textbf{m}$.

Theorem 4

In general,
\begin{equation}\tag{8}\label{8}
\mathbf{AW_{1}}=\mathbf{W_{1}W_{2}}=\mathbf{W_{2}W_{3}}=\mathbf{W_{3}W_{4}}=...=\mathbf{W_{n-1}B}
\end{equation}

Let's illustrate it with an example. Suppose, we are trying to divide $\textbf{AB}$ into six equal segments. Thus, in this case $n=6$, and we can use the general method we derived. Applying the method discussed above, the figure below can be obtained. Notice the labeling of the following points: $\mathbf{I_{1}(W_{3})}, \mathbf{I_{2}(W_{4})}, \mathbf{I_{5}(W_{5})}$. As stated above, this coincidence occurs because 2, 3, and 6 are factors of 6.


问:
A weird contour integral calculation
I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as  $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because when the poles are at $|z|=1$. It's outside the boundary. So I have no idea about how I can calculate this integral

Here's my Residues

$\operatorname{Res}_{z=i}[f(z)]=\lim_{z\to i}\frac{\sqrt{1-z^2}}{i(1-zi)}dz=\frac{\sqrt{2}}{2i}$

$\operatorname{Res}_{z=-i}[f(z)]=\lim_{z\to -i}\frac{\sqrt{1-z^2}}{i(1+zi)}dz=\frac{\sqrt{2}}{2i}$

But I guess these are wrong

Edit: since I need a closed contour i replaced $x=cos\theta$ and $dx=-sin\theta d\theta$

And my integral is now $-\frac{1}{2}\oint \frac{\sqrt{1-cos^2\theta}}{cos^20+cos^2\theta}(-sin\theta) d\theta$=
$-\frac{1}{2}\oint \frac{-sin^2\theta}{cos^20+cos^2\theta}d\theta$

答:
By your comments in my previous question, we can also deal in another way (but I am about to tell you that is more tedious, even if apparently at the beginning it looks easier).

With $x = \cos\theta$ we get

$$-\int_{\pi}^0 \frac{\sin^2(\theta)}{1 + \cos^2(\theta)}\ \text{d}\theta$$

We can close the contour by doubling the integration range:

$$\frac{1}{2}\int_0^{2\pi} \frac{\sin^2(\theta)}{1 + \cos^2(\theta)}\ \text{d}\theta$$

From here we are ready to Complex again:

$$\sin(\theta) = \frac{1}{2i}\left(z - \frac{1}{z}\right) ~~~~~~~ \cos(\theta) = \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}\theta = \frac{\text{d}z}{iz}$$

Whence

$$\frac{1}{2}\oint_{|z| = 1} \frac{\left(\frac{1}{2i}\left(z - \frac{1}{z}\right)\right)^2}{\left(1 + \frac{1}{4}\left(z + \frac{1}{z}\right)^2\right)}\frac{\text{d}z}{iz} \longrightarrow \frac{1}{2}\oint_{|z| = 1} \frac{i \left(z^2-1\right)^2}{z^5+6 z^3+z}$$

Here there are five poles:

$$\left\{\{z\to 0\},\left\{z\to -i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to -i \sqrt{2 \sqrt{2}+3}\right\},\left\{z\to i \sqrt{2 \sqrt{2}+3}\right\}\right\}$$

But only three of them lie inside the unit circle:

$$\{z\to 0\},\left\{z\to -i \sqrt{3-2 \sqrt{2}}\right\},\left\{z\to i \sqrt{3-2 \sqrt{2}}\right\}$$

Now it's about Residues time again.

Whereas the residue at $z = 0$ is rather easy, not easy (just tedious) are the calculations of the two others. But eventually, by doing that calculation, you will get the identical final result!

Title: A weird contour integral calculation


问:
Group as the Union of Subgroups
We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4.

In this paper http://www.jstor.org/stable/2695649 by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$.

I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?

答:
One way to ensure this happens is to have every maximal subgroup be characteristic.  To get every maximal subgroup normal, it is a good idea to check p-groups first.  To make sure the maximal subgroups are characteristic, it makes sense to make sure they are simply not isomorphic.  To make sure there are not too many maximal subgroups, it makes sense to take p=2 and choose a rank 2 group.

In fact the quasi-dihedral groups have this property.  Their three maximal subgroups are cyclic, dihedral, and quaternion, so each must be fixed by any automorphism.

So a specific example is QD16, the Sylow 2-subgroup of GL(2,3).

Another small example is 4×S3.  It has three subgroups of index 2, a cyclic, a dihedral, and a 4 acting on a 3 with kernel 2.  Since these are pairwise non-isomorphic, they are characteristic too.  It also just so happens (not surprisingly, by looking in the quotient 2×2) that every element is contained in one of these maximal subgroups.

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Title: Group as the Union of Subgroups

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问:
Prove a property of when simple modules are isomorphic
Let $\mathfrak{m}_1$ and $\mathfrak{m}_2$ be left maximal ideals of a unital ring $A$. Show that the simple modules $A/\mathfrak{m}_1$ and$A/\mathfrak{m}_2$ are isomorphic if and only if there exist $a\in A\setminus \mathfrak{m}_2$ so that $\mathfrak{m}_1 a \subseteq \mathfrak{m}_2$.

I don't really know how to begin, I'd love some help.

答:
If such an $a$ exists you can consider the $A$-module map defined by multiplication by $a$ on the right from $A/m_1$ to $A/m_2$, you must use the fact that this modules are simple and the assumption on $a$ to show that it is an isomorphism.

If there exists an isomorphism $\varphi$ between the modules, then take $a$ to be the image of $1$ in $A/m_2$ (actually pick someone in the equivalence class). Now how does $\varphi$ look like? Deduce that it is the $a$ you are looking for


问:
Does $\int_{I}f(t)g(t)dt = 0$ for every $g|_{\partial I} = 0$ implies $f\equiv 0$?
Let $I=[t_{0},t_{1}]$ be a bounded interval of $\mathbb{R}$ and suppose $f: I \to \mathbb{R}$ is $C^{k}(I)$ for some $k \ge 2$. Suppose that:
$$\int_{t_{0}}^{t_{1}}f(t)g(t) dt = 0$$
for every $C^{k}(I)$ function $g:I\to \mathbb{R}$ such that $g(t_{0})=g(t_{1}) = 0$. Does it follow that $f(t) \equiv 0$?

答:
Suppose $f(t_{0}) > 0$ for some $t_{0} \in I$. Then, once $f$ is continuous, there exists $\delta>0$ such that if $t \in (t_{0}-\delta, t_{0}+\delta)\subset I$, then $f(t) \ge c > 0$. Define:
\begin{eqnarray}
\varphi(t) = \begin{cases}
\displaystyle Ke^{-\frac{1}{\frac{\delta^{2}}{4}-(t-t_{0})^{2}}} \quad \mbox{if $t \in (t_{0}-\frac{\delta}{2},t_{0}+\frac{\delta}{2})$} \\
\displaystyle 0 \quad \mbox{otherwise}
\end{cases}
\end{eqnarray}
so that $\varphi \in C_{c}^{\infty}(I)$, $\varphi \ge 0$ and choose $K$ so that:
$$\int_{t_{0}}^{t_{1}}\varphi(t)dt = \int_{t_{0}-\delta/2}^{t_{0}+\delta/2}\varphi(t)dt = 1 $$
Then:
$$\int_{t_{0}}^{t_{1}}f(t)\varphi(t)dt = \int_{t_{0}-\delta/2}^{t_{0}+\delta/2}f(t)\varphi(t)dt \ge c \int_{t_{0}-\delta/2}^{t_{0}+\delta/2}\varphi(t)dt = c > 0 $$
and this contradicts the hypothesis.


问:
Derivation of equation of the straight line through two points $x_1$ and $x_2$ in $\mathbb{R}^N$ (in convex function definition).
I am getting started on reading convex optimization. One equation that is being used to represent traveling from one point to another in a straight line in a convex set is:

$$y = (1-\theta) x_1 + \theta x_2$$ for two points $ x_1 \neq x_2$ where  $x_1, x_2 \in \mathbb{R}^N$.

I think I have an intuitive understanding of how this is a straight line, but I am trying to derive is from the usual equation of straight line going through two points. $y = y_1 + m(x-x_1)$ where $m = \frac{y_1 - y_2}{x_1-x_2}$. Could anyone convince me how the first equation was derived and what the parameter $\theta$ means or represents?

Thanks a lot.

答:
If $x, y\in \mathbb{R}^n$, then
$$
L := \{ x + \theta(y-x) \mid \theta \in \mathbb{R} \}
$$
represents the line passing through $x$ (when $\theta = 0$) and $y$ (when $\theta = 1$) in the direction of the vector $y-x$.

Notice that $x + \theta(y-x) = (1-\theta)x + \theta y$ and
$$
L(x,y) := \{ (1-\theta)x + \theta y \mid \theta \in [0,1] \}
$$
is the line-segment containing $x$ and $y$.

This can be visualized in $\mathbb{R}^2$ using the parallelogram law: