问:
Prove that if a subset of $\mathbb{R}^2 $ is convex closed and has non empty interior they are equal to the closure of their interior
This question was asked here 5 years ago Here but there was a typo in the body that resulted in the question not being answered. There was a comment that linked to another different question but I didn't get the solution for the one that I am stuck on

The question is as follows:

Suppose that $A \subset \mathbb{R}^2$ is convex, closed and has nonempty interior. Prove that A is the closure of it's interior

I drew a picture and was able to clearly able to see that by convexity and closure if $x \in \text{int} A$ there is a line segment that connects it to every point in the closure of A but I'm not sure if that's relevant or if I'm on the right track. Can someone give me a nudge in the right direction?

答:
Let $A\subset {\mathbb R}^2$ be a convex, closed set with nonempty interior. Let $y\in A$ and let $x \in {\rm int}\,A$. We will show that $y$ is in the closure of $\rm{int}\, A$.

Let the ball of radius $\epsilon$ centered at $x$ be in $A$, $B(x, \epsilon)\subset A$. Such $\epsilon > 0$ exists since $x$ in the interior.

Now consider $z_t = (1-t)y + tx$ for $t\in (0,1)$. We will show show that $z_t\in {\rm int}\, A$ and thus conclude that $y$ is in the closure of the interior of $A$.

To see that $z_t\in {\rm int}\,A$, note that $B(z_t, \epsilon t)\subset A$. Why? Any arbitrary point in $B(z_t, \epsilon t)$ can be written as $z_t + \epsilon t v$ where $v$ is a vector with magnitude smaller than one. Thus,
$$
z_t + \epsilon t v = (1-t) y + t (x + \epsilon v)
$$

Since $x+\epsilon v \in B(x, \epsilon) \subset A$, convexity implies that $z_t + \epsilon t v \in A$. So $B(z_t, \epsilon t)\subset A$ and thus $z_t$ is in the interior of $A$ for all $t\in(0,1)$. Thus $y$ (which was an arbitrary point in $A$) is in the closure of the interior of $A$.


问:
Is it true that $a \wedge b \wedge c \wedge d = (a \wedge b) \wedge (c \wedge d)$ for the exterior product?
Let us suppose we have a (finite-dimensional) vector space $X$ and let $\bigwedge^k X$ be its exterior product of order $k$. Then $a_1 \wedge a_2 \wedge \dots \wedge a_k \in \bigwedge^k X$ denotes the exterior product of $a_1, a_2, \dots, a_k \in X$.

Question: What is the relation between $\bigwedge^4 X$ and $\bigwedge^2 ( \bigwedge^2 X )$?

For example, suppose we have two exterior products $a \wedge b$ and $c \wedge d$. What is the relation between $(a \wedge b) \wedge (c \wedge d)$ and $a \wedge b \wedge c \wedge d$?

It seems that in $(a \wedge b) \wedge (c \wedge d)$ obeys the (anti)symmetries

$(a \wedge b) \wedge (c \wedge d) = -(b \wedge a) \wedge (c \wedge d)$

$(a \wedge b) \wedge (c \wedge d) = -(a \wedge b) \wedge (d \wedge c)$

$(a \wedge b) \wedge (c \wedge d) =  (c \wedge d) \wedge (a \wedge b)$

but I do not see any antisymmetry such as

$(a \wedge b) \wedge (c \wedge d) =  -(a \wedge c) \wedge (b \wedge d)$

I suspect I am getting confused about the notation. Are there two different types of wedge products in the play?

答:
The exterior product defines a bilinear map $\wedge^i(V) \otimes \wedge^j(V) \to \wedge^{i+j}(V)$ for any $i, j$, and so in particular defines a bilinear map

$$\wedge^i(V) \otimes \wedge^i(V) \to \wedge^{2i}(V).$$

This map is alternating when $i$ is odd and symmetric when $i$ is even, so induces maps

$$S^2(\wedge^{2i}(V)) \to \wedge^{4i}(V)$$
$$\wedge^2(\wedge^{2i+1}(V)) \to \wedge^{4i+2}(V).$$

In particular, we do not get a natural map $\wedge^2(\wedge^2(V)) \to \wedge^4(V)$, but instead a natural map $S^2(\wedge^2(V)) \to \wedge^4(V)$. ("Natural" here means "natural in $V$" and so in particular these are all morphisms of $GL(V)$-representations.)

The phrasing of your question in the title is ambiguous because $(a \wedge b) \wedge (c \wedge d)$ would usually be interpreted in terms of the usual exterior product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^4(V)$ (which is symmetric), whereas you are asking about a different exterior product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^2(\wedge^2(V))$ (which is alternating). To be less ambiguous the two different exterior products involved should be notated differently.

ContentLicense: CC BY-SA 4.0


问:
explicit value of the integral.
If $I= \int_1^4(1+x)^{1-m}x^{-1} \, dx$ where $0<m<1.$ Is it possible to compute the explicit value of the convergent integral $I$.

Title: explicit value of the integral.

Body:
Incomplete Beta function is
$$B(x;a,b)=\int _0^xt^{a-1} (1-t)^{b-1}\,dt$$
In this case we change variable $x+1=t$ and get
$$\int_1^4 (x+1)^{1-m}x^{-1}\, dx=\int_2^5 t^{1-m}(t-1)^{-1} \, dt=\int_5^2 t^{1-m}(1-t)^{-1} \, dt=$$

$$=B_2(2-m,0)-B_5(2-m,0)$$


问:
Kernel decomposition as direct sum, related to minimal polynomial of a linear operator
How to show the following:

Let $T: V_F \to V_F$ be a linear operator and $f(x)$ be the minimal polynomial of $T$ over $F$. Let $$f(x)=g_1(x)g_2(x)\cdots g_n(x)$$ where the $g_i$'s are monic and pairwise relatively prime. Then $g_i(x)$ is the minimal polynomial of the linear operator $T|_{\ker g_i(T)}: \ker g_i(T) \to \ker g_i(T)$ and $V=\oplus_{i=1}^n \ker g_i(T).$

Added: Using the result for $n=2$ is permissible.

i.e. you may assume that

Let $T: V_F \to V_F$ be a linear operator and $f(x)$ be the minimal polynomial of $T$ over $F$. Let $$f(x)=g_1(x)g_2(x)$$ where $g_1$ and $g_2$ are monic and relatively prime. Then $g_i(x)$ is the minimal polynomial of the linear operator $T|_{\ker g_i(T)}: \ker g_i(T) \to \ker g_i(T)$ and $V=\ker g_1(T)\oplus\ker g_2(T).$

答:
For an induction, the cases $n\leq2$ are either obvious or given. For the inductive case $n>2$, the essential observation is that $g_1g_2\ldots g_{n-1}$ and $g_n$ are relatively prime; this is the reason that it is essential that the hypothesis be that the $g_i$ are pairwise relatively prime (no two have a common irreducible factor, as opposed to the weaker condition of having no common irreducible factor to all of them at once). To obtain from that hypothesis that $g_1g_2\ldots g_{n-1}$ and $g_n$ are relatively prime, one could for contradiction assume a common irreducible factor, and by the repeated form of Euclids lemma see that it is a common factor of $g_n$ and at least one of $g_1,\ldots,g_{n-1}$. With this established, one gets by the admitted case $n=2$:
$$
 V=\ker(g_1g_2\ldots g_{n-1}(T))\oplus\ker g_n(T).
$$
Finally apply the induction hypothesis to further decompose the first summand.

ContentLicense: CC BY-SA 3.0


问:
$f$ linear map such that $\ker(f)=V$ and $\text{Im}(f)=W$
In an exercise I'm asked the following:

Let $W,V$ be subspaces of $\mathbb R^3$: $W=\{(x,y,z):x+y+z=0\}$ and $V=\{(x,y,z):x=y=z\}.$ Show that it exists a linar map $f:\mathbb R^3 \to \mathbb R^3$ such that $\ker(f)=V$ and $\text{Im}(f)=W$.

I not sure how I'm supposed to do this. It's trivial that $\dim(\mathbb R^3)=\dim\ker(f)+\dim\text{Im}(f)$, but this does not prove that that $f$ exists right? Is there a way to prove that it exists without explicitly finding $f$? And, how can I find a linear map $f$ that satisfies this condition?

答:
The space $V$ is $1$-dimensional and it is spanned by $(1,1,1)$. On the other hand, $W$ is $2$-dimensional it is spanned by $\{(1,-1,0),(0,1,-1)\}$. So let $f$ be, for instance, the only linear map from $\Bbb R^3$ into itself such that:

$f(1,1,1)=(0,0,0)$;

$f(1,0,0)=(1,-1,0)$;

$f(0,1,0)=(0,1-1)$.

Such a map exists, since $\{(1,1,1),(1,0,0),(0,1,0)\}$ is a basis of $\Bbb R^3$.


问:
How to calculate the derivative of the angular momentum vector $ d\vec L = d(\hat I \vec \omega)?$
How to calculate the derivative of the angular momentum vector?
$$ d\vec L = d(\hat I \vec \omega)$$
or
$$ d\vec L = d(\vec r \times m \vec v)$$

I would like to note that the change of the moment of inertia $d \hat I$ must result in change vector $\vec L$ and there must be a second element $d \vec \omega$ that compensates for this change that the angular momentum vector remains unchanged.

答:
When I asked this question, I didn't know the full answer but now I know the full answer in several ways.

These calculations are performed only for reference systems where the center of the reference system is aligned with the center of mass of the system

The angular momentum vector $\vec L$ is defined as the vector product of the position vector $\vec r$ over the derivative of the position vector $\frac {d \vec r} {dt}$ times the mass $ m $ , i.e. the momentum vector $m \vec v$

$$\vec r \times m \vec v = \vec L \tag 1$$

The position vector can be changed in two directions:

the perpendicular component of change to the position vector $\vec v_{\bot}$

the parallel component of the change to the position vector $\vec v_{{}={}}$

The vector product of two parallel vectors is zero, which gives us

$$\vec r \times m \vec v_{\bot} + \vec r \times m \vec v_{{}={}} = \vec L = \vec r \times m \vec v_{\bot}  \tag {1.a}$$

To be more precise, the angular momentum vector is the vector product of the position vector on the perpendicular component of the velocity vector times the mass.

The parallel component of the velocity vector $\vec v_{{}={}}$ is not part of the angular momentum vector.

The perpendicular component of the velocity vector can be calculated as the vector product of the angular velocity vector times the position vector

$$  \vec \omega \times  \vec r  = \vec v_{\bot}  \tag 2$$

We can develop a pattern (1.a)

$$ \vec L = \vec r \times m \vec v_{\bot} = m(\vec r \times (\vec \omega \times  \vec r))  \tag 3$$

Derivative of the formula (3)

$$ \frac {d \vec L} {dt}  = m(\frac {d \vec r} {dt} \times (\vec \omega \times  \vec r) + \vec r \times (\frac {d \vec \omega} {dt} \times  \vec r) + \vec r \times (\vec \omega \times  \frac {d \vec r} {dt}))  \tag {3.a}$$

simpler version

$$ \frac {d \vec L} {dt}  = m( \vec v \times (\vec \omega \times  \vec r) + \vec r \times (\vec \omega \times  \vec v)+ \vec r \times (\vec \varepsilon \times  \vec r) )  \tag {3.b}$$

Let us also derive the derivative of the formula (1)

$$ \frac { d (\vec r \times m \vec v)} {dt} = \vec r \times m \vec a \tag 4$$

The angular momentum vector can also be written as the product of the moment of inertia tensor on the angular velocity vector

$$ \vec L = \hat I \vec \omega  \tag 5$$

and its derivative

$$ \frac {d \vec L} {dt}  = \frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon  \tag {5.a}$$

I have an additional formula which is used to derive Euler's formulas for the rotation of a rigid body

$$  \hat I \frac {d \vec \omega} {dt} + \vec \omega \times \hat I \vec \omega = \vec M  \tag 6$$

This formula is based on the formula for transforming the vector derivative to a non-inertial frame of reference

$$   \frac {d \vec A} {dt}= \frac {d \vec A `} {dt} + \vec \omega_o \times \vec  A   \tag {6.a}$$

When transforming the derivative of a vector to a non-inertial frame of reference (6.a), we use the angular velocity of the non-inertial frame of reference $ \vec \omega_o $.
In deriving Euler's formulas (6), however, we use the inertial angular velocity $ \vec \omega$ of a rigid body.

Details of the calculation of the proof of equality (3.a)=(4) for the point can be found on my blog in Polish

https://www.salon24.pl/u/przestrz/1192364,mechanika-obrotu-punktu-pochodna-pedu

Since this proof is very extensive, I will only present the results here.

The acceleration vector resulting from the equation (2)

$$ \frac {d \vec v_{\bot}} {dt} = \vec \omega \times  \vec v  + \vec \varepsilon \times  \vec r = \vec a_{\omega \varepsilon } \tag 7$$

Because there are such situations when the angular acceleration vector is zero, therefore

$$ \exists  \vec \varepsilon (0,0,0) ; [ \vec \varepsilon \times  \vec r = (0,0,0) \to \vec \omega \times  \vec v = \vec a_{\omega } ] \tag {7.a}$$

this is why

$$\vec \omega \times  \vec v = \vec a_{\omega }  \tag {7.b}$$

so the angular acceleration vector will give the following acceleration vector

$$  \vec \varepsilon \times  \vec r = \vec a_{\varepsilon \omega } - \vec a_{\omega }  = \vec a_{\varepsilon  }  \tag   {7.c}$$

The angular velocity vector of a point

$$  \frac {\vec r_n \times \vec v_n} { \vert \vec r_n \vert ^2 } = \vec \omega_n \tag {8}$$

the derivative of the vector of the angular velocity of a point

$$  \frac {(\vec r_n \times \vec a_n)  \vert \vec r_n \vert ^2 - (\vec r_n \times \vec v_n)(2 r_x v_x + 2 r_y v_y  + 2 r_z v_z) } { \vert \vec r_n \vert ^4 } =  \frac {d\vec \omega_n} {dt} \tag {8.a} $$

I split the angular acceleration vector into two components

$$  \vec \varepsilon _1 = \frac {(\vec r_n \times \vec a_n) }  { \vert \vec r_n \vert ^2 } \tag {8.b.1}$$

$$\vec \varepsilon _2 = - 2 \frac { (\vec r_n \times \vec v_n)(\vec r_n \circ \vec v_n) } { \vert \vec r_n \vert ^4}  \tag {8.b.2} $$

After substituting the angular velocity vector of the point (8) into the formula (3.b) and calculating the value, I obtained the following equations

$$  \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec r) = \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec v) = 2 \frac { (\vec r_n \times \vec v_n)(\vec r_n \circ \vec v_n) } { \vert \vec r_n \vert ^4}  \tag {9.a}$$

and I was able to reset part of the angular velocity vector to zero

$$  \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec r) + \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec v) + \vec r \times (\vec \varepsilon _2 \times  \vec r) =0 \tag {9.b}$$

and

$$   \vec r \times (\vec \varepsilon _1 \times  \vec r) =  \vec r \times  \vec a \tag {9.b}$$

by which I obtained equation between the equations (3.b)=(4)

$$ m( \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec r) + \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times  \vec v) + \vec r \times (\vec \varepsilon _2 \times  \vec r) +   \vec r \times (\vec \varepsilon _1 \times  \vec r) )=\vec r \times  m \vec a  \tag {9.e}$$

Now let's count (5.a).

The first moment of inertia tensor

$$     \left [ \begin{matrix}
I_{xx} & I_{xy} & I_{xz} \\
I_{yx} & I_{yy} & I_{yz} \\
I_{zx} & I_{zy} & I_{zz} \\
   \end{matrix} \right ]  \tag {10}  $$

elements of the moment of inertia tensor

$$
I_{xx}= \sum _n   {m_n (r_{ny} ^2 + r_{nz} ^2)} ;\qquad
I_{yy} = \sum _n   {m_n (r_{nz} ^2 + r_{nx} ^2)}; \qquad
I_{zz}  = \sum _n  {m_n (r_{nx} ^2 + r_{ny} ^2)};
\tag {10.a}  $$

$$
I_{xy}= I_{yx}= - \sum _n   {m_n r_{nx} r_{ny}} ;\qquad
I_{yz}= I_{zy}= - \sum _n   {m_n r_{ny} r_{nz}} ; \qquad
I_{zx}= I_{xz}= - \sum _n   {m_n r_{nz} r_{nx}} ;
\tag {10.a}  $$

derivative of the moment of inertia tensor

$$     \left [ \begin{matrix}
\frac {d I_{xx}} {dt} & \frac {d I_{xy}} {dt} & \frac {d I_{xz}} {dt} \\
\frac {d I_{yx}} {dt} & \frac {d I_{yy}} {dt} & \frac {d I_{yz}} {dt} \\
\frac {d I_{zx}} {dt} & \frac {d I_{zy}} {dt} & \frac {d I_{zz}} {dt} \\
   \end{matrix} \right ]  \tag {10.b}  $$

derivative of the elements of the moment of inertia tensor

$$
\frac {d I_{xx}} {dt}= \sum _n {2 m_n (v_{ny} r_{ny} + v_{nz} r_{nz} )} ;\quad
\frac {d I_{yy}} {dt}= \sum _n {2 m_n (v_{nz} r_{nz} + v_{nx} r_{nx})}; \quad
\frac {d I_{zz}} {dt}= \sum _n {2 m_n (v_{nx} r_{nx} + v_{ny} r_{ny} )};
\tag {10.c}  $$

$$
\frac {d I_{xy}} {dt}= \frac {d I_{yx}} {dt}= - \sum _n {m_n (v_{nx} r_{ny} + r_{nx} v_{ny})} ;\qquad
\frac {d I_{yz}} {dt}= \frac {d I_{zy}} {dt}= - \sum _n {m_n (v_{ny} r_{nz} + r_{ny} v_{nz})} ; \qquad
\frac {d I_{zx}} {dt}= \frac {d I_{xz}} {dt}= - \sum _n {m_n (v_{nz} r_{nx} + r_{nz} v_{nx})} ;
\tag {10.c}  $$

We count the first element (5.a) $ \frac {d \hat I} {dt} \vec \omega $

$$ \frac {d \hat I} {dt} \vec \omega =(\omega_x \frac {d I_{xx}} {dt}+ \omega_y \frac {d I_{xy}} {dt}+ \omega_z \frac {d I_{xz}} {dt}, \omega_x \frac {d I_{yx}} {dt}+ \omega_y \frac {d I_{yy}} {dt}+ \omega_z \frac {d I_{yz}} {dt}, \omega_x \frac {d I_{zx}} {dt}+ \omega_y \frac {d I_{zy}} {dt}+ \omega_z \frac {d I_{zz}} {dt})  \tag {11}$$

where

$$ \omega_x \frac {d I_{xx}} {dt}+ \omega_y \frac {d I_{xy}} {dt}+ \omega_z \frac {d I_{xz}} {dt} = \sum _n {2 m_n \omega_x (v_{ny} r_{ny} + v_{nz} r_{nz} )}- \sum _n {m_n \omega_y (v_{nx} r_{ny} + r_{nx} v_{ny})} - \sum _n {m_n \omega_z (v_{nz} r_{nx} + r_{nz} v_{nx})} =
\sum _n { m_n( 2 \omega_x v_{ny} r_{ny} + 2 \omega_x v_{nz} r_{nz} -  \omega_y v_{nx} r_{ny} - \omega_y r_{nx} v_{ny} - \omega_z v_{nz} r_{nx} - \omega_z r_{nz} v_{nx})} =
\sum _n { m_n( r_{ny} (\omega_x v_{ny} - \omega_y v_{nx}) + r_{nz} (\omega_x v_{nz} - \omega_z v_{nx}) + v_{ny} (\omega_x r_{ny} - \omega_y r_{nx}) + v_{nz} (\omega_x r_{nz} - \omega_z r_{nx} ))} \tag {11.a}
$$

$$ \omega_x \frac {d I_{yx}} {dt}+ \omega_y \frac {d I_{yy}} {dt}+ \omega_z \frac {d I_{yz}} {dt} =
 \sum _n { m_n (- \omega_x v_{nx} r_{ny} - \omega_x v_{ny} r_{nx} + 2 \omega_y v_{nz} r_{nz} + 2 \omega_y r_{nx} v_{nx} - \omega_z v_{ny} r_{nz} + r_{nz} v_{ny})} =
\sum _n { m_n( r_{nz} (\omega_y v_{nz} - \omega_z v_{ny}) + r_{nx} (\omega_y v_{nx} - \omega_x v_{ny}) + v_{nz} (\omega_y r_{nz} - \omega_z r_{ny}) + v_{nx} (\omega_y r_{nx} - \omega_x r_{ny} ))} \tag {11.b}
$$

$$ \omega_x \frac {d I_{zx}} {dt}+ \omega_y \frac {d I_{zy}} {dt}+ \omega_z \frac {d I_{zz}} {dt} =
 \sum _n { m_n (- \omega_x v_{nx} r_{nz} - \omega_x v_{nz} r_{nx} - \omega_y v_{ny} r_{nz} - \omega_y r_{nz} v_{ny} + 2 \omega_z v_{nx} r_{nx} + 2 \omega_z r_{ny} v_{ny})} =
\sum _n { m_n( r_{nx} (\omega_z v_{nx} - \omega_x v_{nz}) + r_{ny} (\omega_z v_{ny} - \omega_y v_{nz}) + v_{nx} (\omega_z r_{nx} - \omega_x r_{nz}) + v_{ny} (\omega_z r_{ny} - \omega_y r_{n} ))} \tag {11.c}
$$

let's write out the details of the formula (7.b)

$$\vec \omega \times  \vec v = (\omega_y v_z - \omega_z v_y , \omega_z v_x - \omega_x v_z , \omega_x v_y - \omega_y v_x ) =  \vec a_{\omega }  \tag {12.a}$$

and the details of the formula (2)

$$\vec \omega \times  \vec r = (\omega_y r_z - \omega_z r_y , \omega_z r_x - \omega_x r_z , \omega_x r_y - \omega_y r_x ) =  \vec v_{\bot}  \tag {12.b}$$

Now we can simplify the formulas (11)

$$ \frac {d \hat I} {dt} \vec \omega = \sum _n { m_n (\vec v_n \times (\vec \omega \times  \vec r_n) + \vec r_n \times (\vec \omega \times  \vec v_n))} = \sum _n { m_n (\vec v_n \times \vec v_{n \bot} + \vec r_n \times  \vec a_{n \omega })}\tag {13}$$

So, we have a fragment of the pattern (3.b) $ m (\vec v \times (\vec \omega \times  \vec r) + \vec r \times (\vec \omega \times  \vec v))$. And naw let's check out the rest of this pattern $m (\vec r \times (\vec \varepsilon \times  \vec r)) $

$$\sum _n { m_n (\vec r_n \times (\vec \varepsilon \times  \vec r_n))} = \sum _n {m_n (\varepsilon_x (r_{ny}^2 + r_{nz}^2) - \varepsilon_y r_{nx} r_{ny} - \varepsilon_z r_{nx} r_{nz},- \varepsilon_x r_{ny} r_{nx} + \varepsilon_y (r_{nz}^2 + r_{nx}^2) - \varepsilon_z r_{ny} r_{nz}, - \varepsilon_x r_{nz} r_{nx} - \varepsilon_y r_{nz} r_{ny} + \varepsilon_x (r_{nx}^2 + r_{ny}^2))} \tag {14} $$

that is

$$\sum _n { m_n (\vec r_n \times (\vec \varepsilon \times  \vec r_n))}  = \hat I \vec \varepsilon  \tag {14.a} $$

On the basis of equality (13) and (14.a) we got another equality (3.b)=(5.a)

$$\frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon= \sum _n { m_n( \vec v_n \times (\vec \omega \times  \vec r_n) + \vec r_n \times (\vec \omega \times  \vec v_n)+ \vec r_n \times (\vec \varepsilon \times  \vec r_n) )}  \tag {15}$$

The properties of a rigid body show that the distance of a point from the center of mass is constant over time that is, for a rotating rigid body, a point dont has velocity vector along its position vector

$$ \vec v_{{}={}}=0 \land \vec v = \vec v_{\bot}   \tag {16}$$

that is, always for a rigid body

$$ \vec v_n \times (\vec \omega \times  \vec r_n)=0 \tag{16.a}$$

and the formula for the velocity vector is ours (2). So the formula (3.b) for a rigid body has the following form

$$ \frac {d \vec L} {dt}  = \sum _n { m_n(  \vec r_n \times (\vec \omega \times (\vec \omega \times  \vec r_n))+ \vec r_n \times (\vec \varepsilon \times  \vec r_n) )}  \tag {16.b}$$

so let's count $m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times  \vec r_n)))$

$$ \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times  \vec r_n))) =(M_{nx},M_{ny},M_{nz})}  \tag {16.c}$$

$$M_{nx} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_z r_{ny} - \omega_y r_{nz}))}$$

$$M_{ny} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_x r_{nz} - \omega_z r_{nx}))}$$

$$M_{nz} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_y r_{nx} - \omega_x r_{ny}))}$$

after simplification

$$ \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times  \vec r_n))) }=- \sum _n {m_n (\vec \omega \circ \vec r) (\vec \omega \times \vec r)} \tag {16.d}$$

we check the missing element of the formula (6) $ \vec \omega \times \hat I \vec \omega$

$$  \vec \omega \times \hat I \vec \omega = \vec \omega \times (\omega_x I_{xx} + \omega_y I_{xy} + \omega_z I_{xz}, \omega_x I_{yx} + \omega_y I_{yy}+ \omega_z I_{yz}, \omega_x I_{zx} + \omega_y I_{zy} + \omega_z  I_{zz}) = (\sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_z r_{ny} - \omega_y r_{nz}))},\sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_x r_{nz} - \omega_z r_{nx}))}, \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_y r_{nx} - \omega_x r_{ny}))}) \tag {16.e}$$

that is

$$\vec \omega \times \hat I \vec \omega =- \sum _n {m_n (\vec \omega \circ \vec r) (\vec \omega \times \vec r)}= \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times  \vec r_n))) }\tag {16.f}$$

So for a rigid body with assumptions (16) and (16.a) we obtain equality (5.a)=(6)

$$ \frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon  =  \hat I \frac {d \vec \omega} {dt} + \vec \omega \times \hat I \vec \omega  \tag {16.g}$$

In this answer I presented four different forms of the angular momentum vector derivative (3.b), (4), (5.a) and (6) which, as I showed, are contable with each other.


问:
Why does the short exact sequence for projective module split?
I want to understand the definition of surjective module in terms of splitting sequence. The definition says for a projective $R$-module $P$, the following short exact sequence $$0 \to A \xrightarrow{f} B \xrightarrow{g} P \to 0$$ splits, where $A,B$ are also $R$-modules.

I want to see why $B \cong A+P$ ?

The Wikipedia, says, then there is a section map $h:P \to B$ such that $gh=1_P$.

$(1)$ Why is so ?

This then says $B=\text{Im}(h)\oplus \text{ker}(g)$.

$(2)$ why is so ?

I am trying in the following way:

Since the given sequence is short exact sequence, the map $g$ is surjective. This means that $\text{Im}(g)=P$.

Now as $P$ is surjective module any module homomorphism factors through an epimorphism to $B$   i.e.,  for any $R$-module $C$  there exists an epimorphism (surjective module homomorphism) $i: C \twoheadrightarrow  B$ and a module homomorphism $j: P \to C$  such that $$ij=g.$$
I can't go further. What is the section map $h$ here ?

Any explanation of above two questions ?

答:
$\DeclareMathOperator{\im}{im}\DeclareMathOperator{\id}{id}$(2): Suppose $y = s(x) \in \im(s) \cap \ker(g)$. Then on the one hand $g(y) = 0$ and on the other hand $g(y) = g(s(x)) = \id_P(x) = x$ so $x = 0$. Therefore $\im(s) \cap \ker(g) = \{0\}$.

Next, suppose $y \in B$ and let $x = g(y)$, $z = s(x)$ and $z' = y - z$. Then

$y = z + z'$

$z \in \im(s)$

$g(z') = g(y) - g(s(x)) = x - \id_P(x) = 0$ so $z' \in \ker(g)$

Thus $B = \im(s) \oplus \ker(g)$.


问:
How to find the following limit?

Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.

答:
MVT for integrals:

$\displaystyle{\int_{0} ^{x}} \cos (t^2) dx=$

$\cos (s^2)\displaystyle{\int_{0}^{x}} dx$ where s \in [0,x];

Note :

$\lim x \rightarrow 0$  implies $\lim t^2 \rightarrow 0.$

$\lim_{x \rightarrow 0} \cos (t^2)=1.$

Title: How to find the following limit?


答:
A first and simple way to do it is by observing that : $$\lim_{x\to 0}{\frac{1}{x}\int_{0}^{x}{\cos{\left(y^{2}\right)}\,\mathrm{d}y}}=\lim_{x\to 0}{\frac{F\left(x\right)-F\left(0\right)}{x-0}}=F'\left(0\right)=\cos{0}=1 $$

Where $ F $ is the function $ F:x\mapsto\int_{0}^{x}{\cos{\left(y^{2}\right)}\,\mathrm{d}y} $.

Another way to do it would be squeezing, using the inequality $ \left(\forall x\in\mathbb{R}\right),\ \left|1-\cos{x}\right|\leq\frac{x^{2}}{2} $.

Let $ x\in\mathbb{R}^{*} $, we have :

\begin{aligned}\frac{1}{x}\int_{0}^{x}{\cos{\left(y^{2}\right)}\,\mathrm{d}y}=1-\frac{1}{x}\int_{0}^{x}{\left(1-\cos{\left(y^{2}\right)}\right)\mathrm{d}y}\end{aligned}

And since : $$ \left|\frac{1}{x}\int_{0}^{x}{\left(1-\cos{\left(y^{2}\right)}\right)\mathrm{d}y} \right|\leq\frac{1}{x}\int_{0}^{x}{\left|1-\cos{\left(y^{2}\right)}\right|\mathrm{d}y}\leq\frac{1}{2x}\int_{0}^{x}{y^{2}\,\mathrm{d}y}=\frac{x^{2}}{6}\underset{x\to 0}{\longrightarrow}0$$

Then : $$ \frac{1}{x}\int_{0}^{x}{\cos{\left(y^{2}\right)}\,\mathrm{d}y}\underset{x\to 0}{\longrightarrow}1 $$

Title: How to find the following limit?

Body:
Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.


问:
Dividing by an almost-surely positive random variable
I am reading Shreve's "Stochastic Calculus for Finance II".  In it, he states (Theorem 1.6.1) that if $Z$ is an almost-surely strictly positive random variable on a probability space $(\Omega, \mathcal F, P)$ with $E(Z) = 1$, then the probability measure defined as $\widetilde P(A) := \int_A Z d P$ satisfies
$$
E(Y) = \widetilde E(Y/Z)
$$
for any random variable Y (here E and $\widetilde E$ are the expected values in the respective measures).  I don't understand the meaning of this expression; $Y/Z$ is not a random variable since there may be $\omega \in \Omega$ upon which $Z$ vanishes.

Is $Y/Z$ a valid expression, or is it shorthand for something more rigorous?

答:
You can think of it as the following shorthand:
$$(Y/Z)(\omega) := \begin{cases}
Y(\omega)/Z(\omega), & Z(\omega) \ne 0 \\
43.7, & Z(\omega) = 0
\end{cases}
$$

(Exercise: check that $Y/Z$ thusly defined is measurable.)

The point being: you can consider $Y/Z$ to be whatever you want on the event $\{Z=0\}$.  Since the event has probability zero, it doesn't matter what you pick.


问:
Solution of cubic equations in terms of quadratic equations
Is there a general way in which a cubic equation of $3$rd degree can be represented by a quadratic equation of $2$nd degree such that $2$ solutions of a cubic equation is equal to $2$ solutions of corresponding quadratic equation.

For example $2$ solutions of equation $Ax^3+Bx^2+Cx-(A+B+C)=0$ is equal to $2$ solutions of quadratic equation $Ax^2+(A+B)x+(A+B+C)=0$.

But here the last term of cubic equation is $-(A+B+C)$.

If the last term is arbitrarily some constant number not equal to $-(A+B+C)$ then what quadratic equation will give the $2$ solution of cubic equation $Ax^3+Bx^2+Cx+D=0$ such that $D$ is not equal to $-(A+B+C)$

答:
$$a x^3+b x^2+c x-(a+b+c)=0\tag{1}$$
has the integer solution $x=1$ thus can be factored as
$$(x-1)\left[a x^2+(a+b) x+(a+b+c)\right]=0$$
So two soolutions of $(1)$ are solutions of
$$a x^2+(a+b) x+(a+b+c)=0$$


问:
What is Qing Liu's rule of index and reference in "Algebraic Geometry and Arithemetic Curves "
I am just so confused about Qing Liu's way of referencing definition/lemma/theorem. For example, on Page 166, he said Definition 2.3.11 but I couldn't find this in his book. Is there a rule of his referencing or it's just a typo? What does 2.3.11 means? Sometimes there is one definition in one chapter and a lemma in the other chapter that has the same number, x.x.xx.</p>

答: That would be definition 3.11 from chapter 2, which you will find on page 44. Besides, take a look at the Numbering style section from the preface.


问:
Geometric problem concerning relation of equal segments and angles
Triangle ABC

Suppose that $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $△ABC$. The perpendiculars to the sides $AB$ and $AC$ at $P$ and $Q$ respectively meet at $D$, an interior point of $△ABC$. If $M$ is the midpoint of $BC$, prove that $PM = QM$ if and only if $∠BDP=∠CDQ$.

What i don't get is that how can i relate the sides with the angles. I tried to extend some segments and find similar triangles, but to no avail.

答:
First of all continue segments $CD,BD$ to intersect with $AB,AC$ in two points. Then there will be a cyclic quadrilateral and the problem will be something like the figure and we have to prove that $NF=NE$
For proving this we will name midpoints of $DG,CG$ as $T,S$ respectively. Then we'll prove that $ETN,FSN$ are equal triangles:
$i) SN=GD/2=ET$
$ii) TN=GC/2=FS$
$ETN=ETG+NTG=180-2×EGD+180-CGD=360-2×EGD-CGD$
$FSN=FSG+NSG=180-2×CGF+180-CGD=360-2×CGF-CGD$
And because $CGF=DGE$,
$iii) ETN=FSN$
So the two triangles are equal and thus $NF=NE$.


问:
integral $\frac {\overline{f(z)}}{z-a}$ and Cauchy formula
Problem:Let $f(z)$ be analytic on some domain containing the closed unit disk $|z|\leq1$. Calculate
$$\frac{1}{2 \pi i}\int_D\frac{\overline{f(z)}dz}{z-a}$$ if $|a|<1$, where D is unit circle.

Solution:
Let$$f(z)=u(z)+iv(z)$$
where, $u,v$ are real functions.
By Cauchy integral formula we have:
$$\frac{1}{2 \pi i}\int_D\frac{f(z)dz}{z-a}=f(a)$$
so,

$$\int_D\frac{(u(z)+iv(z))dz}{z-a}=2 \pi if(a)$$
$\int_D \frac {u(z)}{z-a}dz=-2 \pi Im(f(a))$ and $\int_D \frac {u(z)}{z-a}dz=2 \pi Re(f(a))$.
So,
$$\frac{1}{2 \pi i}\int_D\frac{\overline{f(z)}dz}{z-a}=\frac{1}{2 \pi i}(-2 \pi Im(f(a))-i2 \pi Re(f(a))=- \overline{f(a)}$$
What is wrong with this solution( in book solution is $\overline{f(a)}$)?

答:
If $g$ is holomorphic on $D(0,R), R>1,$ then

$$\tag 1 g(0)=\frac{1}{2\pi}\int_0^{2\pi} g(e^{it})\,dt.$$

This falls right out of Cauchy's theorem (or the mean value property of harmonic functions if you like).

Do a little work to see our expression equals

$$\tag 2 \frac{1}{2\pi}\int_0^{2\pi} \frac{ \overline{f(e^{it})}}{1-ae^{-it}}\,dt = \overline{ \frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{it})}{1-\bar{a}e^{it}}  }\,dt.$$

If we let $g(z)=f(z)/(1-\bar{a}z)$ and use $(1),$ we see that $(2) $ equals $\overline{f(0)}.$


问:
Matrix similarity and diagonalizable matrices
I understand that two matrices $A$ and $B$ (in $\mathbb{R}^{n \times n})$ are similar matrices if there exists an invertible matrix $P \in \mathbb{R}^{n \times n}$ (a change of basis matrix) such that $B = P^{-1}AP$.

My question is whether;

does a matrix $P$ always exists for any choice of $A$ and $B$;

and as such doest this imply that the $A$ is always diagonalizable;

is $B$ always a diagonal matrix;

in which cases does $P$ not exists?

I would really appreciate detailed answer to help me structure my understanding.

Many thanks

答:
No. For instance, if $A$ is the identity matrix and if $B$ is the null matrix, no such matrix $P$ exists.

There are non-diagonalizable matrices, such as $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$.

No.

When $A$ and $B$ are not similar (by definition).


问:
question about Riemann zeta $\zeta (0)$
i know that

$$\zeta (m)=\sum_{n=1}^\infty n^{-m}$$

so

$$\zeta (0)=\sum_{n=1}^\infty n^0=1+1+1+1+1+1+\cdots=\infty $$

but actually

$$\zeta (0)=-0.5$$

where is the wrong

please help

thanks for all

答:
I also struggled with this for a while. Your definition of the Riemann zeta function is only its definition when the real part of $ m $ is greater than $ 1 $.

The domain of $ \zeta $ though is $ \mathbb{C} $, so the question is: how do we move from $ \{z \mid \Re(z) > 1 \} $ to $ \mathbb{C} $? The answer is analytic continuation.

Using the functional equation for $ \sum\limits_{n = 1}^\infty \frac{1}{n^z} $, we can extend the domain of $ \zeta $ to the complex numbers.

If you want to know the details, I suggest looking at the Wikipedia page on the function.

Title: question about Riemann zeta $\zeta (0)$


问:
Prove that, for any two positive integers $x$ and $y$, the value of $x \pmod y$ equals $x$ or is less than $ \frac{x}{2}$.
I started by taking three cases viz $x<y$, $x=y$ and $x>y$. The proofs for the former two cases were pretty easy but I can't figure the same out for the remaining case. I have an intuition but it isn't too solid for formal proof.

答:
For $x>y$: you can say $x=yq+r$, where $q$ is the quotient in division of $x$ by $y$.

We can see that $q\ge 1$ (as $x\gt y$), and $r=x\mod y$ is the remainder in that division, so $0\le r\lt y$.

Now we have $x=yq+r\ge y+r\gt 2r$, therefore $r\lt x/2$.


答:
If $y\geqslant x$ then clearly the claim is true.
Otherwise assume that $x=ay+b$ form some non-negative integers $a,b$ such that $y\geq b$. Then for the sake of the contradiction assume that $b\geqslant x/2$. Then, $ay$ is less than or equal to $x/2$ and so is $y$ itself. Then we will have $b\geqslant (x/2)\geqslant y$ but this contradicts $y\geq b$.

Title: Prove that, for any two positive integers $x$ and $y$, the value of $x \pmod y$ equals $x$ or is less than $ \frac{x}{2}$.

Body: I started by taking three cases viz $x<y$, $x=y$ and $x>y$. The proofs for the former two cases were pretty easy but I can't figure the same out for the remaining case. I have an intuition but it isn't too solid for formal proof.


问:
How to establish the truth values of a set of statements which depend on $\lnot p \rightarrow (q\lor \lnot r)$ when its false?
The problem is as follows:

First find the truth values of $p$, $q$ and $r$ such as the complex statement from below is false.

$$\lnot p \rightarrow (q\lor \lnot r)$$

Then using this information find the truth values of each of the following statements,

I. $\lnot (p \lor q) \rightarrow (p \leftrightarrow \not q)$

II. $(r \lor \lnot p) \bigtriangleup r$

III. $(\lnot p \bigtriangleup r)\lor (\lnot p \rightarrow q)$

Assuming you answer correctly all the statements I, II and III which would be your answer?.

The answers given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{FFF}\\
2.&\textrm{FFT}\\
3.&\textrm{TFF}\\
4.&\textrm{TFT}\\
\end{array}$

My book defines the $\bigtriangleup$ operator as a strong disjunction as follows from these equivalences:

$p \bigtriangleup q \equiv \lnot (p \leftrightarrow q)$

$p \bigtriangleup q \equiv  (\lnot p \leftrightarrow q)$

$p \bigtriangleup q \equiv (p \lor q) \lor \lnot (p \land q)$

$p \bigtriangleup T \equiv \lnot p$

From then on I don't know exactly how to "guess" the adequate values which would make the statement to pinpoint the answer for the each of the statement.

Can someone help me here?. I don't know exactly what to do. The only thing which I do recall is that the definition of the $\bigtriangleup$ for a strong disjunction is given by this truth table.

$\begin{array}{|c|c|c|}
\hline
p& q & p \bigtriangleup q \\ \hline
T&T&F\\ \hline
T&F&T\\ \hline
F&T&T\\ \hline
F&F&F\\ \hline
\end{array}$

But I don't know if this would help into the solution. Can this be used to find the values for getting the truth values for each of the statements given?. Can someone help me?.

答:
$$\lnot p \rightarrow (q\lor \lnot r)$$ is false if and only if $\lnot p$ is true, and $(q\lor \lnot r)$ is false.

If $\lnot p$ is true, then $\lnot \lnot p \equiv p$ is not true.

$q\lor \lnot r$ is false only when both $q$ is false and $\lnot r$ is false.

$(p, q, r) = (F, F, T)$.


答:
does a matrix $P$ always exists for any choice of $A$ and $B$?

No. For example, if $A$ is the zero matrix (all entries are zero), then $PAP^{-1} = 0$ for any $P$. So for any non-zero choice of $B$, there cannot be a $P$ which satisfies this equation. There are less trivial examples, but this gets the point across.

and as such doest this imply that the A is always diagonalizable?

This is (sort of) answered by question 1. Since not every two matrices are conjugate, it is not necessarily true that all matrices are diagonalizable. For example,
$$ \begin{pmatrix} 1&1\\0&1 \end{pmatrix} $$
is not diagonalizable (over $\Bbb{R}$). Its characteristic polynomial is $(1-\lambda)^2$, and so its only eigenvalue is $\lambda=1$. But you can check that it does not have two linearly independent eigenvectors (the only eigen-direction is $(1,0)$). If it were diagonalizable, it would need two linearly independent eigenvectors with eigenvalue $1$.

is B always a diagonal matrix?

No. The equation $B = PAP^{-1}$ is often used when talking about diagonalization, but the equation just by itself is just saying two matrices are conjugate.

In which cases does $P$ not exist?

This was answered in question #1.

Title: Matrix similarity and diagonalizable matrices

Body:
I understand that two matrices $A$ and $B$ (in $\mathbb{R}^{n \times n})$ are similar matrices if there exists an invertible matrix $P \in \mathbb{R}^{n \times n}$ (a change of basis matrix) such that $B = P^{-1}AP$.

My question is whether;

does a matrix $P$ always exists for any choice of $A$ and $B$;

and as such doest this imply that the $A$ is always diagonalizable;

is $B$ always a diagonal matrix;

in which cases does $P$ not exists?

I would really appreciate detailed answer to help me structure my understanding.

Many thanks


问:
Differential structures on $\mathbb{R}$
Let $\mathbb{R}$ be the real line with the differential structure from the maximal atlas given by the identity map. And let $\mathbb{R}^{\prime}$ be the real line with the differential structure given by the map $\phi: x \rightarrow x^{\frac{1}{3}}$

(i)- Show that these two differential structures are distinct.

(ii)- Show these smooth manifolds are diffeomorphic. (Hint: try not to use the identity map as a diffeomorphism)

for (i) let $(\mathbb{R},\phi=\operatorname{Id}_\mathbb{R}:\mathbb{R}\to\mathbb{R})$ and $(\mathbb{R},\psi:\mathbb{R}\to\mathbb{R})$ where $\psi(x)=x^{1/3}$. now $(\psi \circ \operatorname{id}_\Bbb R \circ\ \phi^{-1})(x) = x^{1/3}$. This map is not smooth at $0$. Hence the identity map is not a diffeomorphism.

how we can solve (ii) ?

答:
For (i) you are right: these two charts are not compatible, hence belong to two distinct differential structure.

For (ii), just notice that the map
\begin{align}
f : \mathbb{R} & \to \mathbb{R}' \\
x & \mapsto x^3
\end{align}
is a diffeomorphism. To see this, consider $\varphi$ the identity chart on $\mathbb{R}$ and $\psi$ the cube map on $\mathbb{R}'$, and look at $\psi^{-1}\circ f \circ \varphi$.

This shows that even if these two differentiable structures are not compatible, they are diffeomorphic.

Title: Differential structures on $\mathbb{R}$


答:
Let$$r \equiv x \pmod y$$
If $y > x$, clearly $r = x$. Now, assume $y < x$. Then
$$r \equiv x \equiv \underbrace{x - \left[\frac xy \right]y}_{<y} \pmod y$$  So, what we need to prove is $$r=x - \left[\frac xy \right]y \le \frac x2 $$or$$ \left[\frac xy \right]y \ge \frac x2$$
Since $x - 1 < [x] \le x$, it is enough to show that
$$\left(\frac xy - 1\right) y \ge \frac x2 $$or$$ xy - y \ge \frac x2 $$or$$ y \ge \frac {x}{2(x-1)}$$ which is obvious since $$\mathbb N \ni y > \frac 12 \ge \frac {x}{2(x-1)}$$


答:
Once you find one root of a cubic, you can take out a linear factor by the factor theorem, leaving a quadratic to solve for the other roots. It's not pretty, but it works. (The weird part is some other quadratics are of interest along the way.)

In analogy with completing the square, divide out the leading coefficient and shifting the unknown to lose the second highest-order term. We want a root of $x^3+px+q=0$. Given any root $x$, some $u,\,v\in\Bbb C$ satisfy $u+v=x,\,uv=-p/3$, namely the roots of $t^2-xt-p/3$. Then$$u^3+v^3=x(x^2-3uv)=x^3+px=-q,\,u^3v^3=(-p/3)^3,$$so $u^3,\,v^3$ are roots of $t^2+qt-p^3/27$. The root we'll take out as aforementioned is $\sum_\pm\sqrt[3]{\frac{-q\pm\sqrt{q^2+4p^3/27}}{2}}$. (Note you have to define these cube roots in just the right way, so their product is $-p/3$; we don't want to be off by a factor of $\exp\tfrac{\pm2\pi i}{3}$.)

Title: Solution of cubic equations in terms of quadratic equations

Body:
Is there a general way in which a cubic equation of $3$rd degree can be represented by a quadratic equation of $2$nd degree such that $2$ solutions of a cubic equation is equal to $2$ solutions of corresponding quadratic equation.

For example $2$ solutions of equation $Ax^3+Bx^2+Cx-(A+B+C)=0$ is equal to $2$ solutions of quadratic equation $Ax^2+(A+B)x+(A+B+C)=0$.

But here the last term of cubic equation is $-(A+B+C)$.

If the last term is arbitrarily some constant number not equal to $-(A+B+C)$ then what quadratic equation will give the $2$ solution of cubic equation $Ax^3+Bx^2+Cx+D=0$ such that $D$ is not equal to $-(A+B+C)$


问:
Prove that there exists an integer $a$ such that $\frac{a^2-x}{p}\in \mathbb{Z}$
Let $x$ be an odd number, $y$ a positive integer, and $p$ an odd prime number such that $x^2+y^2=p$. Show that there exists an integer $a$ such that  $\dfrac{a^2-x}{p}\in \mathbb{Z}.$

This problem is one that my friend asked me, and I've considered it for some time, but I can't prove it. Thank you everyone.

答:
Using Jocobi symbol form of quadratic reciprocity,
$$
\left(\frac{x}{p}\right)
= \left(\frac{p}{x}\right)(-1)^{\tfrac{x-1}{2}\tfrac{p-1}{2}}=1.
$$</p>


答:
First case $ : x<y$

$$x\mod y =x$$

Second case $ : x=y $

$$x\mod y =0$$

Third case $ : x>y $

$$x=ny+r \; with\; 0\le r <y \; and\; n\ge1$$

$$\implies x< (n+1)y$$
$$\implies x<2ny $$
$$\implies \frac x2 < ny$$
$$\implies \frac x2 < x-r$$
$$\implies r < \frac x2$$
$$\implies x\mod y < \frac x2$$


问:
Solutions that oscillate more rapidly
Which one of the following two equations has solutions that oscillate more rapidly?

$$x''+\sqrt{t^6+3t^5+1} x =0$$

$$ x''+2t^3 x =0$$

I was trying to use sturm separation theorem but not getting exact approach towards it!!

Is there any direct way to do it?

Edit: The book from where I have given this question has answer as first equation in the hints provided at the back of book.
Book name: "A Textbook  on Ordinary  Differential  Equations" by Shair Ahmad, Antonio Ambrosetti(Second Edition)-Chapter5-Exercise-D2

答: If we assume that $t$ is confined in an interval with small amplitude we may assume that $t^3$ is approximately constant. By recalling that a solution of $x''+ K x=0$ (with $K>0$) has a frequency of $\frac{\sqrt{K}}{2\pi}$ we immediately get that a solution of $x''(t)+2t^3 x(t)=0$ oscillates faster than a solution of $x''(t)+\sqrt{t^6+3t^5+1}\,x(t)=0$.

Title: Solutions that oscillate more rapidly


问:
Sentences into LTL
Is this the correct conversion to LTL formula

Lions and zebras never drink at the waterhole together.
(Lions -> Drink) U (zebra -> Drink)

Even if you ask her infinitely often, she will never tell you.
GFask -> ¬Ftell

Once red, the light becomes green eventually after being yellow for a while. FRed -> Xyellow -> Fgreen

答:
I will use the definitions from the wiki article about
linear temporal logic.
Similar to propositional logic the sentences of this logic are based on propositional variables.
These are sentences that are true or false and cannot be split into smaller parts using predicates.

Lions and zebras never drink at the waterhole together.

We use sentences like "a lion drinks at the water hole" ($p$) and "a zebra drinks at the water hole" ($q$)
and do not split these sentences into smaller parts.
The sentence $p$ exactly means "at least one lion drinks at the water hole" and similar holds for $q$.
$$\lnot(p\land q)$$
means that not both a lion and a zebra are drinking at the water hole. This should be true for every point in time so we have
$$\operatorname{G}(\lnot(p\land q))$$

Even if you ask her infinitely often, she will never tell you.
The sentence "she will tell you" ($p$) is a proposition that can be true or false.

From my point of view the "even" part of the sentence does not contribute
anything to the logical content of the sentence and can be ignored.
The expression
$$\lnot p$$
means "she does not tell you" and similar to the previous example
we can use the operator $G$ to express that something never will happen:

$$ \operatorname{G}(\lnot p)$$

Once red, the light becomes green eventually after being yellow for a while.

We have the following elementary sentences: "The light is red"($r$), "the light is yellow" ($y$), "the light is green" ($g$).
Let us try to model this behavior with until-loops.

$$y\operatorname{U}g \tag 1$$
covers paths like

```<code>y,g,
</code>```

or

```<code>y, y, g
</code>```

and
$$r\operatorname{U}(y\operatorname{U}g) \tag 2 $$

covers paths like

```<code>r, y, g
</code>```

or

```<code>r, r, r, y, y, g,
</code>```

But $(1)$ also describes loops with $0$ iterations like

```<code>g
</code>```

and also $(2)$ allows

```<code>y, y, g
r,r,g
g
</code>```

To guarantee at least one iteration in $(1)$ one can use the next operator $X$

$$y \land \operatorname{N}(y \operatorname{U} g)$$

and instead of $(2)$ one uses

$$r \land \operatorname{N}(r \operatorname{U}(y \land \operatorname{N}(y \operatorname{U} g)))$$

But now the first $r$ is the first node of the path. The first $r$ should start anywhere so we use the $\operatorname{F}$ operator

$$\operatorname{F}(r \land \operatorname{N}(r \operatorname{U}(y \land \operatorname{N}(y \operatorname{U} g)))) \tag 3$$

Maybe we should also describe that the light can only have one color, so instead of  $(3)$ we get

$$\operatorname{F}((r\land \lnot y \land \lnot g) \land \operatorname{N}((r\land \lnot y \land \lnot g) \operatorname{U}((\lnot r\land y \land \lnot g) \land \operatorname{N}((\lnot r\land y \land \lnot g) \operatorname{U} (\lnot r\land \lnot y \land g)))))$$

Title: Sentences into LTL

Body:
I will use the definitions from the wiki article about
linear temporal logic.
Similar to propositional logic the sentences of this logic are based on propositional variables.
These are sentences that are true or false and cannot be split into smaller parts using predicates.

Lions and zebras never drink at the waterhole together.

We use sentences like "a lion drinks at the water hole" ($p$) and "a zebra drinks at the water hole" ($q$)
and do not split these sentences into smaller parts.
The sentence $p$ exactly means "at least one lion drinks at the water hole" and similar holds for $q$.
$$\lnot(p\land q)$$
means that not both a lion and a zebra are drinking at the water hole. This should be true for every point in time so we have
$$\operatorname{G}(\lnot(p\land q))$$

Even if you ask her infinitely often, she will never tell you.
The sentence "she will tell you" ($p$) is a proposition that can be true or false.

From my point of view the "even" part of the sentence does not contribute
anything to the logical content of the sentence and can be ignored.
The expression
$$\lnot p$$
means "she does not tell you" and similar to the previous example
we can use the operator $G$ to express that something never will happen:

$$ \operatorname{G}(\lnot p)$$

Once red, the light becomes green eventually after being yellow for a while.

We have the following elementary sentences: "The light is red"($r$), "the light is yellow" ($y$), "the light is green" ($g$).
Let us try to model this behavior with until-loops.

$$y\operatorname{U}g \tag 1$$
covers paths like

y,g,

or

y, y, g

and
$$r\operatorname{U}(y\operatorname{U}g) \tag 2 $$

covers paths like

r, y, g

or

r, r, r, y, y, g,

But $(1)$ also describes loops with $0$ iterations like

g

and also $(2)$ allows

y, y, g
r,r,g
g

To guarantee at least one iteration in $(1)$ one can use the next operator $X$

$$y \land \operatorname{N}(y \operatorname{U} g)$$

and instead of $(2)$ one uses

$$r \land \operatorname{N}(r \operatorname{U}(y \land \operatorname{N}(y \operatorname{U} g)))$$

But now the first $r$ is the first node of the path. The first $r$ should start anywhere so we use the $\operatorname{F}$ operator

$$\operatorname{F}(r \land \operatorname{N}(r \operatorname{U}(y \land \operatorname{N}(y \operatorname{U} g)))) \tag 3$$

Maybe we should also describe that the light can only have one color, so instead of  $(3)$ we get

$$\operatorname{F}((r\land \lnot y \land \lnot g) \land \operatorname{N}((r\land \lnot y \land \lnot g) \operatorname{U}((\lnot r\land y \land \lnot g) \land \operatorname{N}((\lnot r\land y \land \lnot g) \operatorname{U} (\lnot r\land \lnot y \land g)))))$$


问:
Interpretation and Usefulness of Direct Product
I'm currently reading A Brief on Tensor Analysis By J. Simmonds. It says,

Let the projection of a vector $\mathbf{v}$ on a vector $\mathbf{u}$ be denoted and defined as
$$\mathrm{Proj}_\mathbf{u}\mathbf{v}\equiv (\mathbf{v}\cdot \bar{\mathbf{u}})\bar{\mathbf{u}}.$$
(And I understand the Geometrical meaning of it as well as I know the implication.) Further, It says

We can generalize this. The direct product $\mathbf{u}\mathbf{v}$ (often denoted as $\mathbf{u}\otimes \mathbf{v}$) of two vectors $\mathbf{u}$ and $\mathbf{v}$ is a tensor that sends any vector $\mathbf{w}$ into a new vector according to the rule
$$\mathbf{u}\mathbf{v}(\mathbf{w})=\mathbf{u}(\mathbf{v}\cdot \mathbf{w})$$

But I really don't understand what's the geometrical meaning of it. For say $2D$ case, if $\mathbf{u}=2\hat{i}+3\hat{j}$ and $\mathbf{v}=-\hat{i}$ then what's this suppose to mean? Dot product supposed to give the length of the projection and then scale another vector by this amount.

答:
First of all, "direct product" is not a good name. That is usually used for other things. The symbol "$\otimes$" is usually called the "tensor product".

One way to think of it is that any linear map (i.e. a matrix) can be written as a sum of such tensor products.

For example, the matrix
$$ A = \begin{pmatrix} 1&2\\3&4 \end{pmatrix} $$
Any vector $(x,y)$ is sent to $(x+2y, \; 3x+4y)$. Call the rows of the matrix $a_1 = (1,2)$ and $a_2=(3,4)$. Also call the standard basis vectors $e_1=(1,0)$ and $e_2=(0,1)$. Then in your tensor notation, the linear map represented by $A$ is the same as
$$ e_1 \otimes a_1 + e_2 \otimes a_2 $$


问:
Hamiltonian cycle in an bipartite $(\frac{n}{2} + 1)$-regular graph with 2n vertices
Let G be a $(\frac{n}{2} + 1)$-regular bipartite graph with n vertices in each part. How to prove that this graph contains hamiltonian cycle?

I have already tried to use Ore's theorem or modify it to apply to bipartite graphs, but it didn't help. Also I've checked some papers (didn't find anything suitable in this situation).

Maybe you can give me some ideas how to solve this problem

答:
It’s possible to modify a proof of Ore’s theorem.

Let $V$ and $W$ be the parts of $G$. If $G$ has no Hamilton cycle, keep adding edges between $V$ and $W$ that do not create a Hamilton cycle until you cannot add any more without creating a Hamilton cycle; call the resulting bipartite graph $H$. There must be a $v\in V$ and a $w\in W$ that are not adjacent in $H$, so adding the edge $\{v,w\}$ would create a Hamilton cycle; let $P$ be the Hamilton path

$$v=u_1,u_2,\ldots,u_{2n}=w\,.$$

For $k=1,\ldots,n$ consider the vertices $u_{2k-1}\in V$ and $u_{2k}\in W$. If $H$ contained both of the edges $\{u_1,u_{2k}\}$ and $\{u_{2k-1},u_{2n}\}$, $H$ would contain the Hamilton cycle

$$u_1,u_2,\ldots,u_{2k-1},u_{2n},u_{2n-1},\ldots,u_{2k},u_1\,,$$

so at most one ot the edges $\{u_1,u_{2k}\}$ and $\{u_{2k-1},u_{2n}\}$ can be in $H$. But then

$$2\left(\frac{n}2+1\right)=n+2\le\deg v+\deg w\le n\,,$$

which is absurd, so $G$ must have a Hamilton cycle.


答:
Just solve them and plot the two:

The full forms have constants, that must be set for initial conditions.  For example $c_1 \cos t + c_2 \sin t$.  In the above, I've set all constants to $1$, just to reveal the periods... the focus of the question.

Title: Solutions that oscillate more rapidly

Body:
Which one of the following two equations has solutions that oscillate more rapidly?

$$x''+\sqrt{t^6+3t^5+1} x =0$$

$$ x''+2t^3 x =0$$

I was trying to use sturm separation theorem but not getting exact approach towards it!!

Is there any direct way to do it?

Edit: The book from where I have given this question has answer as first equation in the hints provided at the back of book.
Book name: "A Textbook  on Ordinary  Differential  Equations" by Shair Ahmad, Antonio Ambrosetti(Second Edition)-Chapter5-Exercise-D2


问:
why is the Ramsey number $R(3,4)$ not equal to $R(3,3)$?
I understand why R(3,3) is equal to 6 since we have a clique of 3 in a graph of 6 vertices. What I don't understand is that by definition of Ramsay number it says that for a R(r,s) we have to find a clique of either 3 or 4 in size. In this case, we can find a clique of 3 in a graph of 6 nodes unconditionally. Could someone explain to this hopeless student why R(3,4) = 9?</p>

答:
For $R(3,4)$ you must find either a clique of size $3$ or an independent set of size $4$. The graph

```<code>   *    *    *
   |    |    |
   *    *    *
</code>```

on $6$ vertices has neither a clique of size $3$ nor an independent set of size $4$. Every set of $3$ vertices contains two that are not adjacent, and every set of $4$ vertices contains two that are adjacent.

Body:
For $R(3,4)$ you must find either a clique of size $3$ or an independent set of size $4$. The graph

   *    *    *
   |    |    |
   *    *    *

on $6$ vertices has neither a clique of size $3$ nor an independent set of size $4$. Every set of $3$ vertices contains two that are not adjacent, and every set of $4$ vertices contains two that are adjacent.


问:
Calculating the number of integers divisible by 8
Given the set [10000,70000], in order to calculate the number of terms divisible by 8 I know of two ways to calculate the answer

1st is
$$\frac{70000-10000}{8}+1=7501$$

2nd is
$$⌊\frac{60001}{8}⌋=7500$$

the fact that the results differ is very peculiar to me, can anyone explain what is going on, and which one is more accurate?

答:
Your second method, generalized to the interval $[a,b],$ appears to be
$$\left\lfloor \frac{b - a + 1}8\right\rfloor.$$

The floor function gives you an integer, which you want, but it does this
simply by throwing out any fractional part.
When you use a formula like
$$\left\lfloor \frac x8\right\rfloor,$$
you get zero for the first seven positive values of $x$.
You don't get a positive result until $x \geq 8.$

Consider the simpler problem of counting multiples of $8$ in the set
$[10000, 10000].$ There is one element in that set and it is a multiple of $8$,
so the answer should be $1,$ but
$$\left\lfloor \frac{10000 - 10000 + 1}8\right\rfloor
= \left\lfloor \frac 18\right\rfloor = 0.$$
You'll get the same wrong answer for $[10000,10001],$ $[10000,10002],$ $[10000,10003],$ $[10000,10004],$ $[10000,10005],$ and  $[10000,10006].$

For $[10000,10007],$ you get
$$\left\lfloor \frac{10007- 10000 + 1}8\right\rfloor
= \left\lfloor \frac 88\right\rfloor = 1,$$
which is finally a correct answer.
But for $[10000,10008]$ you still get only $1$ from the formula while the correct answer is $2.$

The problem is that your formula only counts every eighth integer, but your interval has a multiple of $8$ at the first integer.
The only time the formula is correct is just before you reach the next multiple of $8,$ because that's the only time you can partition the set of integers into subsets with exactly $8$ integers in each subset and exactly one multiple of $8$ in each subset.

If your set started at $10001$ instead of $10000$ the formula would work fine, because the first seven integers starting at $10001$ are not divisible by $8$ and you only get a multiple of $8$ at the eighth integer added to the set.

The catch is that in general, in order to know how many multiples of some integer $n$ are in a set of consecutive integers it is not enough to know how many integers are in the set. The only time this information is sufficient is when the number of integers is an exact multiple of $n$. In any other case you need to know where the sequence starts (at least relative to the nearby multiples of $n$) in order to know whether there is an "extra" multiple of $n$ in the numbers that are left over after you have partitioned the rest of the numbers into subsequences of $n$ numbers each.


问:
Subsets - Ranking / Unranking
I am looking at the following codes:

It is lexicographic order related to ranking and unranking.

Here is also an example:

There is also the Gray code:

with the repective examples:

I haven't really understood the ranking and the unranking.

So we have a set and we want to find all the subsets  either with the lexicographic oder or with Gray code and then we search for a specific subset and the position is the rank?

$$$$

EDIT:

The subset $T \subseteq S$ can de described by the vector $x(T)$ of length $n$ where at each position there is $1$ if the respective element is contained, $0$ else.

As for Gray Code : It is defined recursively as follows. We start with $G^1=[0,1]$. in each of the next levels we take the previous one with a $0$ at the beginning and we add the previous one reversely with $1$ at the beginning, e.g. $G^2=[00,01,11,10]$, etc.

For the algorithms ranking and unranking we consider the relation between Gray Code and the binary representation.

The bits $b_i$ of Gray Code is $1$ if at the binary representation the bits $b_i$ and $b_{i+1}$ are different.

For the ranking we take the bits from the most important to the less important and where the respective element that we check contains in the subset we reverse the bit b (initially $0$). At each bit that is one we augment also the rank.

For the unranking we take the binary representation of rank and we add to the subset the elements for those that the consecutive bits were different.

答:
Your question example is part of these lecture notes.

SubsetLexRank computes an integer $0 \dots 2^n-1$ which indicates which of the $n$ elements is contained in the subset.

This method is used in many programming languages to express Boolean flags in one integer. The $n$ elements of the set are assigned to single-bit numbers which are integer powers of two:

```<code>S1 - 1 = 2^0      binary 000...0001
S2 - 2 = 2^1      binary 000...0010
S3 - 4 = 2^2      binary 000...0100
...
Sn -   = 2^(n-1)  binary 100...0000
</code>```

For $n$ elements of a set, there are $2^n$ ways to select a subset. Each selection can be expressed as $n$-bit binary number by adding up the numbers assigned to the selected elements.

If such $n$-bit binary number is given, method SubsetLexUnrank determines the subset which is represented by this number. The number is examined bit-by-bit. Whenever a bit is set, the respective set element is included in the subset.

The Lex methods simply increment the rank numbers in an arithmetic fashion. Subsequent ranks have difference of one.

The GrayCode methods order the rank numbers in such a way, that subsequent numbers are only different in one bit-position. Result is a different ordering of the subsets. Subsets which differ in one element are ordered in adjacent positions.

In GrayCodeRank, the loop variable $i$ is decremented from $n-1$ down to $0$. Starting from element $1$ to element $n$, the presence or selection state of a set element is tested. Whenever an element is found to be present in the subset, the Boolean variable $b$ is flipped between true and false. For true values of $b$, the respective bit is set in the rank variable $r$. Thus, the rank variable is a representation of selection changes. These changes can be re-played during unranking to retrieve the appropriate subset.

Gray code ranks for three elements:

subset
Gray code rank

{}
0

{3}
1

{2,3}
2

{2}
3

{1,2}
4

{1,2,3}
5

{1,3}
6

{1}
7

Note that subsets in adjacent rank positions $0 \dots 2^n-1$ differ in exactly one set element.

Summary:

The rank is a unique presentation of the set elements selected into the subset.

Unranking determines from a given rank, which elements are selected to be in the subset.

Title: Subsets - Ranking / Unranking

Body:
Your question example is part of these lecture notes.

SubsetLexRank computes an integer $0 \dots 2^n-1$ which indicates which of the $n$ elements is contained in the subset.

This method is used in many programming languages to express Boolean flags in one integer. The $n$ elements of the set are assigned to single-bit numbers which are integer powers of two:

S1 - 1 = 2^0      binary 000...0001
S2 - 2 = 2^1      binary 000...0010
S3 - 4 = 2^2      binary 000...0100
...
Sn -   = 2^(n-1)  binary 100...0000

For $n$ elements of a set, there are $2^n$ ways to select a subset. Each selection can be expressed as $n$-bit binary number by adding up the numbers assigned to the selected elements.

If such $n$-bit binary number is given, method SubsetLexUnrank determines the subset which is represented by this number. The number is examined bit-by-bit. Whenever a bit is set, the respective set element is included in the subset.

The Lex methods simply increment the rank numbers in an arithmetic fashion. Subsequent ranks have difference of one.

The GrayCode methods order the rank numbers in such a way, that subsequent numbers are only different in one bit-position. Result is a different ordering of the subsets. Subsets which differ in one element are ordered in adjacent positions.

In GrayCodeRank, the loop variable $i$ is decremented from $n-1$ down to $0$. Starting from element $1$ to element $n$, the presence or selection state of a set element is tested. Whenever an element is found to be present in the subset, the Boolean variable $b$ is flipped between true and false. For true values of $b$, the respective bit is set in the rank variable $r$. Thus, the rank variable is a representation of selection changes. These changes can be re-played during unranking to retrieve the appropriate subset.

Gray code ranks for three elements:

subset
Gray code rank

{}
0

{3}
1

{2,3}
2

{2}
3

{1,2}
4

{1,2,3}
5

{1,3}
6

{1}
7

Note that subsets in adjacent rank positions $0 \dots 2^n-1$ differ in exactly one set element.

Summary:

The rank is a unique presentation of the set elements selected into the subset.

Unranking determines from a given rank, which elements are selected to be in the subset.


问:
Trying to understand the idea behind path independence for line integrals
I’m trying to grasp the idea behind the line integral for conservative vector fields. If I have some vector valued function $F$ which can be represented as $\nabla f$ then I have a conservative field. Now what I’m not fully comprehending is that why the integral $$\int_C F \ dr$$ is path independent. I’ve looked several explanations for this and the analogy for FTC is quite clear so this makes sense in ”symbolic” form, but all the explanations I’ve seen usually explain this via some physical experiement such as traversing to the top of a mountain and since the gravitational field is conservative the work done to get to the top would be independent of the path I’m taking. I’m not a physicist and even this doesn’t quite make sense, I would assume that the work done from the gravitational field to me would be much greater if I start to go for example up sideways instead of directly walking up the mountain? If work is defined as $W=f \cdot d$ wouldn’t the $d$ be much greater if I’m taking a longer path and thus $W$ would get larger?

Is there some other kind of intuition for this that wouldn’t rely so much on physical concepts such as the gravitational, electric or magnetic fields?

Edit: Ah perhaps if I’m taking a path sideways the field vectors would be directed towards me at some angle and thus wouldn’t do as much work as if they would be directed right towards me?

答:
There are two main definitions for the definition of a conservative vector field $\mathbf{F}:\mathbb{R}^d\to\mathbb{R}^d$:

$\exists \Phi:\mathbb{R}^d\to\mathbb{R} \text{ such that }\mathbf{F}=\nabla\Phi$

In the special case that $d=2$ or $d=3$, we can also say $\nabla\times\mathbf{F}=\mathbf{0}.$

And of course often given as a definition is $\oint\mathbf{F}(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=0$, which is what you're asking about.

I'll show that the first and third definitions imply the second. Let's start with the first. Letting $O$ be a closed smooth curve in $\mathbb{R}^d$,

$$\oint_{O} \mathbf{F}\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}$$
Seeing as $O$ is smooth, we can parameterize it as $O=\{\mathbf{x}(t)~|~t\in(0,T)\}$, with of course $\mathbf{x}(0)=\mathbf{x}(T)$. With this in mind, we can represent the line element as
$$\mathrm{d}\mathbf{x}=\mathbf{x}'(t)\mathrm{d}t$$
So then
$$\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_0^T \nabla\Phi(\mathbf{x}(t))\boldsymbol{\cdotp}\mathbf{x}'(t)\mathrm{d}t$$
However, $\nabla\Phi(\mathbf{x}(t))=\left(\nabla\Phi\circ\mathbf{x}\right)(t)$, and so via the chain rule,
$$\mathrm{D}_t\left(\Phi\circ\mathbf{x}\right)(t)=\left(\nabla\Phi\circ\mathbf{x}\right)(t)\boldsymbol{\cdotp}\mathbf{x}'(t)=\nabla\Phi(\mathbf{x}(t))\boldsymbol{\cdotp}\mathbf{x}'(t)$$
Hence,
$$\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_0^T \mathrm{D}_t\left(\Phi\circ\mathbf{x}\right)(t)\mathrm{d}t$$
So by FTC,
$$=\left(\Phi\circ\mathbf{x}\right)(T)-\left(\Phi\circ\mathbf{x}\right)(0)$$
$$=\Phi(\mathbf{x}(T))-\Phi(\mathbf{x}(0))=0.$$
Which as per our parameterization, is precisely zero. One can take this idea further to say that for any smooth curve $C$, not necessarily closed, parameterized as $C=\{\mathbf{x}(t)~|~t\in[a,b]\}$ that
$$\int_C \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\Phi(\mathbf{x}(b))-\Phi(\mathbf{x}(a))$$

That is to say, it only depends on the endpoints.

In the special case of two or three dimensions, showing (3) implies (2) is practically instant. Let $\Omega$ be the region enclosed by $O$, that is to say, $\partial\Omega=O$. Using Stokes's theorem, one can say that
$$\oint_O \mathbf{F}(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_{\Omega}\left(\nabla\times\mathbf{F}\right)\boldsymbol{\cdotp}\mathrm{d}\mathbf{n}$$
But, the RHS is trivially zero since by assumption $\nabla\times\mathbf{F}=\mathbf{0}$. Therefore the LHS is zero as well.


问:
Solve $\sin(3x)\sin(4x)=\cos(4x)\cos(5x)$ (problem from a Swedish 12th grade ’Student Exam’ from 1931)
The equation
$$\sin(3x)\sin(4x)=\cos(4x)\cos(5x)\tag{1}$$
can be transformed by expanding in powers of $\cos(x)$ and letting $r=\cos(x)$;
$$128r^9-256r^7+160r^5-32r^3+r=0,$$
‘factor out’ the solution $r=0$;
$$128r^8-256r^6+160r^4-32r^2+1=0,$$
letting $s=2r$;
$$8s^4-32s^3+40s^2-16s+1=0,$$
letting $t=s-1$;
$$8t^4-8t^2+1=0,$$
letting $u=t^2$;
$$8u^2-8u+1=0,$$
which can be solved using the quadratic formula.
Going back through the steps gives the solutions to (1) as
$$x=\pm\arccos(...)+k\cdot2\pi$$
(18 solutions + period of $2\pi$).

Is there a simpler way to solve this equation?

答:
Hint

Use https://mathworld.wolfram.com/WernerFormulas.html

$$\cos x-\cos7x=\cos x+\cos9x$$

Use $$\cos9x=-\cos7x=\cos(\pi-7x)$$

Can you take it from here?

Or use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$0=\cos9x+\cos7x=2\cos x\cos8x$$


问:
Fallacy involving Euler's formula
If
$$
e^{ix} = \cos{x} + i \sin{x}
$$
which means
$$
e^{2\pi i} = \cos{2\pi} + i \sin{2\pi} = 1 + 0i = 1
$$
and
$$
a^{(bc)} = (a^b)^c
$$
then why is this wrong for some real number $x$?
$$
e^{ix} = e^{2\pi i(x/2\pi)} \\
= (e^{2\pi i})^{(x/2\pi)} \\
= 1^{(x/2\pi)} \\
= 1
$$
?

答:
The "rule" $a^{bc}=(a^b)^c$ simply does not hold for complex numbers. Indeed, it doesn't even hold for all real numbers (consider $a=-1$ and $b=2$ and $c=\frac12$).

This is a great example of how, as mathematicians, we need to remember more than just the equations we learn—we need to remember the precise conditions under which the equations hold.

Title: Fallacy involving Euler's formula


问:
Geodesics on a cylinder: straight line or helix?
It is known that geodesics on a cylinder are helical lines (helix). As a special case of a helix--- straight line. But one can take two points so cleverly that you can draw two helixes between them (see picture), one of which is a straight line. Both satisfy geodesics differential equation. The question: is, can they both be considered as geodesics or only the shortest one is "true" geodesic?

答:
Depending on the number of rotations executed between the end points  $(A,B)$ we have an infinite number of local length minimizing geodesics as helices shown on their cylindrical development.

There is one shortest among them running vertically down that you have sketched.

In development all helices are straight lines.

They are inclined to the generator line $L$ as

$$\tan^{-1}\dfrac{n\cdot 2 \pi R}{L}$$

whose periodicity can be visualized as below for $n=0,1,2,3$ in a development of multiple winding numbers associated with periodic wraps.


问:
How do you calculate how many decimal places there are before the repeating digits, given a fraction that expands to a repeating decimal?
If you have a fraction such as $$\frac{7}{26}=0.269230\overline{769230}$$ where there are a number of digits prior to the repeating section, how can you tell how many digits there will be given just the fraction?

I believe I could run through the standard long division algorithm until I come across the same remainder for the second time and then use the location of the first instance of this remainder to calculate the number of digits before the repeating section, but this feels very cumbersome.

After lots of reading online, I came across what looks like a formula for it from Wolfram MathWorld:

When a rational number $\frac{m}{n}$ with $(m,n)=1$ is expanded, the period begins after ${s}$ terms and has length ${t}$, where ${s}$ and ${t}$ are the smallest numbers satisfying $10^s\equiv10^{s+t}\pmod{n}$.

I know how to calculate the length of the period of a fraction, and so in the case of my original fraction we have $10^s\equiv10^{s+6}\pmod{26}$, but I don't know how to solve for ${s}$ in this equation!

So there are really two questions here - the one in the title, and a sneaky one about how to take logs in a modulo arithmetic equation.

答:
In the $2^{nd}$ section of this answer we get the answer using the high-school algorithm that the OP mentioned, but eschewed as being too cumbersome. But by studying the theory behind the algorithm and using elementary number theory, we will be able to understand, directly, the wolfram theory on decimal expansions at formulas/examples (7) - (9).

For the zero divisor $[10] \in {\textstyle \mathbb {Z} /26\mathbb {Z}}$, calculations show that

$\quad x \pmod{26} \text{ where } x \in [10^0, 10^1, 10^2, 10^3, 10^4, 10^5, 10^6, 10^7] = [1, 10, 22, 12, 16, 4, 14, 10]$

The numerator $7$ of the fraction $\large \frac{7}{26}$ carries over to a unit, $[7] \in {\textstyle \mathbb {Z} /26\mathbb {Z}}$.

So the unit $[7]$ 'goes for a ride' over the 'exponential graph' of zero-divisors generated by $[10]$,

$\quad [1\cdot7,10\cdot7,\, 22\cdot7,\, 12\cdot7,\, 16\cdot7,\, 4\cdot7,\, 14\cdot7]\quad \text{(not necessary to to calculate any of these residues)}$

Conclusion: In the decimal expansion of $\large \frac{7}{26}$, the repeating block of digits is of length $6$ and begins at the $2^{nd}$ fractional decimal digit ($10^7 \equiv 10^1 \pmod{26}$).

With this behind us, we can easily get the explicit answer - multiply the numerator by $10^7$ and keep the 7 quotient digits (padding $0s$ after the decimal point might be necessary) after dividing,

$\quad 7 \cdot 10^7 = 26\cdot2692307+18$

and (we've got the seven digits in the quotient),

$\quad \large \frac{7}{26} \approx 0.2\overline{692307}$

The same pattern (1 plus 6 block) occurs whenever the numerator of $\large \frac{n}{26}$ satisfies

$\quad 1 \le n \lt 26 \land n \text{ is odd } \land n \ne 13$

Analyzing the high-school algorithm (see some theory here), there are at most $26$ divisions (neophyte estimate) that have to be performed. So let us just jump into it!

Expand $\large \frac{7}{26}$:

Divide
Approximate (append $q$ digit)

$7\cdot 10 = 26 \cdot 2 + 18$
$\large \frac{7}{26} \approx 0.2$

$18\cdot 10 = 26 \cdot 6 + 24$
$\large \frac{7}{26} \approx 0.26$

$24\cdot 10 = 26 \cdot 9 + 6$
$\large \frac{7}{26} \approx 0.269$

$6\cdot 10 = 26 \cdot 2 + 8$
$\large \frac{7}{26} \approx 0.2692$

$8\cdot 10 = 26 \cdot 3 + 2$
$\large \frac{7}{26} \approx0.26923$

$2\cdot 10 = 26 \cdot 0 + 20$
$\large \frac{7}{26} \approx 0.269230$

$20\cdot 10 = 26 \cdot 7 + 18$
$\large \frac{7}{26} \approx 0.2692307$

Now the residue $18$ has already appeared and was used after calculating the first fractional decimal; the answer is summarized as follows:

$\quad \large \frac{7}{26} \approx 0.2\overline{692307}$


问:
What is the interval of convergence of the power series $\sum_{n=0}^\infty (\frac{\pi}{2} - \arctan(n))(x)^{n}$
What is the interval of convergence of the power series  $\sum_{n=0}^\infty (\frac{\pi}{2} - \arctan(n))(x)^{n}$

= $\sum_{n=0}^\infty \mathrm{arccot}(n)(x)^{n}$

Radius of convergence = $\lim_{k\rightarrow \infty}\biggl|\frac{a_{k}}{a_k+1}\biggr|$ = $\lim_{k\rightarrow \infty}\biggl|\frac{\mathrm{arccot}(k)}{\mathrm{arccot}(k+1)}\biggr|$ = ?

I'm stuck here. Any help is highly appreciated

答:
Let $ f $ defined at $ (0,+\infty) $, by

$$\forall x>0 \; f(x)=\arctan(x)+\arctan(\frac 1x)$$

It is easy to check that

$$\forall x>0 \; \;\;f'(x)=0$$
thus $ f $ is constant at $ (0,+\infty)$
and
$$\forall x>0\;\; f(x)=f(1)=\frac{\pi}{2}$$

So, $ \forall n\ge 1$,

$$\frac{\pi}{2}-\arctan(n)=\arctan(\frac 1n)$$
$$\sim \frac 1n \;\;(n\to+\infty)$$

Your series has the same radius of convergence than
$$\sum \frac{x^n}{n} \;which\; is\; R=1$$


问:
Is any compact metric totally disconnected space homeomorphic to a compact subspace of a Cantor space?
Every compact metric totally disconnected perfect space is homeomorphic to a Cantor space.

Is every compact metric totally disconnected space homeomorphic to a compact subspace of a Cantor space?

In other words, if you have a compact metric totally disconnected space can you embed it in a compact metric totally disconnected perfect space?

答: From general facts we know that a compact totally disconnected Hausdorff space is zero-dimensional (has a base of clopen subsets) and as we also have a metric space, we have a countable clopen base for our space. The characteristic functions of these form an embedding family into  $\{0,1\}^\omega$, which is (homeomorphic to) the Cantor set. In fact every compact Hausdorff totally disconnected space of weight $\kappa$ can be embedded into a Cantor cube of weight $\kappa$ in this way.</p>


问:
Counting The Number Of K-Bounded Good Lists Of Length N
Problem Statement:

A k-bounded list of length n is a sequence [x1, x2, . . . , xn] where each xj
is an
integer between 0 and k, inclusive. A good list is one that does not contain three
consecutive values 0, 1, 0. For example, for n = 4 and k = 2, [2, 0, 0, 1] and [1, 0, 0, 1]
are good lists while [0, 1, 0, 2] and [0, 0, 1, 0] are not.
Let good(n, k) be the number of good k-bounded lists of length n. For instance,
the good 1-bounded lists of length 2 are {[0, 0], [0, 1], [1, 0], [1, 1]}, so good(2, 1) = 4.
Similarly, you can check that good(4, 1) = 12 and good(3, 2) = 26.
Your task is to compute good(n, k) for the following values of n and k.

(a) good(7, 1)

(b) good(7, 3)

(c) good(20, 1)

Note: This is a question from a past contest and the official answers for the parts a, b and c are as follows:

(a) good(7, 1) = 65

(b) good(7, 3) = 15163

(c) good(20, 1) = 97229

My Approach: I came up with the following recurrence relations,

$$p(i,0) = p(i-1,0) + p(i-1,1) - p(i-2,0)$$ $$p(i,1) = [p(i-1,0) + p(i-1,1)]\times k$$
where $p(i,0)$ is the number of good lists of length $i$ ending with 0 and $p(i,1)$ is the number of good lists of length i not ending with 0. The idea behind this is that every good list of length $i$ is an extension of a good list of length $i-1$ but there can also be some bad extensions in the process, hence it is better to handle them separately. Therefore, in case a good list ends with 0,1 then we need to take care of the fact that while extending this list we don't count the lists where we place a 0 at the end, otherwise we will have 0,1,0 in the list. So what we do is that we just count all the lists of length $i$ ending with 0 and then subtract the lists of length $i-2$ that end with a 0. Whereas the case where we do not use a 0 at the end is a bit simpler.

I just filled the table for the above recurrence manually with the initial values as $p(0,0) = 0$, $p(0,1) = 0$, and $p(1,0) = 1$,  $p(1,1) = k$ (since it was a pen and paper exam) and could arrive at the answers which is the sum of $p(n,0)$ and $p(n,1)$ [C++ code for reference: https://ideone.com/wSVhVL]. But then I have been thinking whether this problem has a closed form solution, some function in n and k. I couldn't arrive at one until now. Can anyone provide some combinatorial insight? Is a closed form solution possible?

答:
I am going to use Chomsky-Schutzemberger method to find the generating function of this (you put DP as tag, so I guess you are interested in computational stuff). Consider the following automaton that accepts the good lists.

If you label $A,B,C,D$ the nodes of the automaton, then the following system of equations gives you the generating function of the number of strings of size $n$ that the automaton accepts.
$$A=1+kxA+xB, \, B=xB+xC+(k-1)xA+1,C=\, kxA+1.$$
We omit $D$ because is a state that dies. Doing the substitution, you get
$$(1-kx)A=1+xB=1+x\frac{x+kx^2A+(k-1)xA+1}{1-x},$$
So $$A=\frac{1+\frac{x^2+x}{1-x}}{1-kx-\frac{kx^3+(k-1)x^2}{1-x}},$$
hence $$A=\frac{1+x^2}{(1-x)(1-kx)-(kx^3+(k-1)x^2)}=\frac{1+x^2}{1-(k+1)x+x^2-kx^3}=\frac{1}{1-\left ( kx+\frac{x}{1+x^2}\right ) }.$$
This agrees with your findings.

One can try to factor the denominator to try to use partial fractions and get a formula. The problem is that this polynomial does not seem to have a nice factorization (for low $k'$s) that would be enough evidence to me that a "nice close formula" is not possible. OEIS just seems to have $k=1$.

Case k=1:
The OEIS gives that
$$a_n = \sum_{k=0}^{n+2} \binom{n+2-k}{2k}.$$
which says that somehow a tiling problem can be attach to this(reminds me of Fibonacci).

General case:
A very ugly but closed formula can be seen to be
$$a_n=\sum _{i=0}^n\sum _{\substack{a_0+2a_1+4a_2+\cdots +2\ell \cdot a_{\ell}+\cdots =n-i\\ a_0+a_1+\cdots =i}}\frac{i!}{\prod _{j=0}^{\infty} a_j!}(k+1)^{a_0}(-1)^{a_1+a_3+\cdots +a_{2\ell
 +1}+\cdots }$$


答:
$$\sin (3 x) \sin (4 x)=\frac{1}{2} (\cos (x)-\cos (7 x))$$
and
$$\cos (4 x) \cos (5 x)=\frac{1}{2} (\cos (x)+\cos (9 x))$$
$$\frac{1}{2} (\cos (x)-\cos (7 x))=\frac{1}{2} (\cos (x)+\cos (9 x))$$
$$\cos(7x)+\cos(9x)=0$$
In the following equations $k$ is any integer $k\in\mathbb{Z}$
$$2\cos(8x)\cos x=0$$
$$\cos(8x)=0\to 8x=\pm\frac{\pi}{2}+2k\pi\to x=\pm\frac{\pi}{16}+\frac{k\pi}{4}$$
$$\cos x=0\to x=\frac{\pi}{2}+k\pi$$

Formulae can be found here.

Title: Solve $\sin(3x)\sin(4x)=\cos(4x)\cos(5x)$ (problem from a Swedish 12th grade ’Student Exam’ from 1931)

Body:
The equation
$$\sin(3x)\sin(4x)=\cos(4x)\cos(5x)\tag{1}$$
can be transformed by expanding in powers of $\cos(x)$ and letting $r=\cos(x)$;
$$128r^9-256r^7+160r^5-32r^3+r=0,$$
‘factor out’ the solution $r=0$;
$$128r^8-256r^6+160r^4-32r^2+1=0,$$
letting $s=2r$;
$$8s^4-32s^3+40s^2-16s+1=0,$$
letting $t=s-1$;
$$8t^4-8t^2+1=0,$$
letting $u=t^2$;
$$8u^2-8u+1=0,$$
which can be solved using the quadratic formula.
Going back through the steps gives the solutions to (1) as
$$x=\pm\arccos(...)+k\cdot2\pi$$
(18 solutions + period of $2\pi$).

Is there a simpler way to solve this equation?


问:
Reference recommendations for Convex Geometry
So I'm an undergraduate with limited knowledge of analysis, and none of topology. I'm trying to study Convex Geometry, and I'm looking for recommendations for the same - in terms of books, video playlists, course webpages, etc.

It will be great if the book/resource is suited for self-study; but I don't mind if it's not!

One book I came across after a little search is Lectures on Convex Geometry by Daniel Hug and W. Weil - though I don't know much about it. Thank you!

答:
There are numerous references; here are my 3 favorites:

A superb introduction is
[1] Ball, Keith. "An elementary introduction to modern convex geometry." Flavors of geometry 31 (1997): 1-58.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.77.3836&rep=rep1&type=pdf

A wonderful but challenging  book is:
[2] Gilles Pisier, The volume of convex bodies and Banach space geometry,
Cambridge Tracts in Mathematics, vol. 94, Cambridge
University Press, Cambridge, 1989.

The combinatorial view:
[3] Danzer, L., B. Grünbaum, and V. Klee. "Helly’s Theorem and its Relatives (Proceedings of Symposia in Pure Mathematics VII)." American Mathematical Society (Providence, RI, 1963): 101-180.


问:
What is the relationship between $\mathbb{A}^n_k$ and $\operatorname{Spec} k[x_1, \dots, x_n]$ for an algebraically closed field $k$?
I have seen that due to Hilbert's Nullstellensatz that every maximal ideal of $k[x_1, \dots, x_n]$ is of the form $(x_1-a_1, \dots, x_n - a_n)$ and hence we obtain a bijection between the maximal ideals of $k[x_1, \dots, x_n]$ and $\mathbb{A}^n_k$.

Specifically the map $\operatorname{MaxSpec} k[x_1, \dots, x_n] \to \mathbb{A}^n_k$
given by $(x_1-a_1, \dots, x_n-a_n) \mapsto (a_1, \dots, a_n)$ is a bijection.

Now, I have seen in this question: What *is* affine space?, that apparently $\operatorname{Spec}k [x_1, \dots, x_n]$ is bijective to $\mathbb{A}^n_k$. But then doesn't this imply (taking the above into account) that $\operatorname{MaxSpec} k[x_1, \dots, x_n] = \operatorname{Spec} k[x_1, \dots, x_n]$? Then this would then imply that all prime ideals in $k[x_1, \dots, x_n]$ are maximal which is not the case, so this must be a contradiction, which leads me to believe that there is no bijection between $\operatorname{Spec}k [x_1, \dots, x_n]$ and $\mathbb{A}^n_k$ in general.

But I have seen, stated in examples in some Algebraic Geometry books, that for a field $k$, we have that $$\operatorname{Spec} k[x_1, x_2] = \mathbb{A}^2_k.$$

However, there is a question about this here: What do prime ideals in $k[x,y]$ look like? which seems to indicate otherwise (and which also makes the case that we don't have a bijection between $\operatorname{Spec} k[x_1, \dots, x_n]$ and $\mathbb{A}^n_k$ in general).

So my two questions are

What is the precise relationship between $\operatorname{Spec} k[x_1, \dots, x_n]$ and $\mathbb{A}^n_k$ for general $n$?

In the case $n=2$, do we know that $\operatorname{Spec} k[x_1, x_2] = \mathbb{A}^2_k$?

I think for question one, at the moment, all I can say is that there is an injection $\mathbb{A}^n_k \hookrightarrow \operatorname{Spec} k[x_1, \dots, x_n]$ since we have the bijection between $\operatorname{MaxSpec} k[x_1, \dots, x_n]$ and $\mathbb{A}^n_k$.

答:
You may think you are unambiguously talking about specific mathematical objects here, but "$\mathbb{A}^n_k$" has (at least) three different meanings and "$\text{Spec } k[x_1, \dots x_n]$" has (at least) two (maybe as many as four depending on how you count it). $\mathbb{A}^n_k$ can refer to

the set of points $k^n$, when $k$ is algebraically closed, regarded as a "concrete affine variety." This is, by the Nullstellensatz, the same as the maximal spectrum of $k[x_1, \dots x_n]$, but ignores the non-maximal prime ideals.

the topological space of prime ideals $\text{Spec } k[x_1, \dots x_n]$ with the Zariski topology. This is probably the least common meaning but it's worth pointing out. This is, of course, not the same as $k^n$.

the (affine) scheme $\text{Spec } k[x_1, \dots x_n]$ over $\text{Spec } k$, thought of either as a locally ringed space or via its functor of points, which is $R \mapsto R^n$ (for $R \in \text{CAlg}(k)$).

Different meanings of these symbols are appropriate to different levels of abstraction in your study of algebraic geometry. It's unfortunate that there isn't clearer terminology for distinguishing these but this is just something you have to deal with.

The unambiguous way to refer to $k^n$ is $\mathbb{A}^n(k)$.


问:
Field extension with infinite degree
I'm currently working on the proof of the following statement:

Let $L$ be a field and $K$ and $F$ be two subfields such that $F\subseteq K\subseteq L$. Then the following holds for the field extensions $L/F, L/K$ and $K/F$:

$$L/F \ \ \text{is infinite iff.} \ \ L/K \ \ \text{is infinite or} \ \ K/F \ \ \text{is infinite}$$

I've managed to prove the implication from left to right, and, from right to left in the case of $K/F$ is infinite.

How do I prove the other case, i.e. that if $L/K$ is infinite then so is $L/F$?

答:
The statement is equivalent to its contrapositive

if $L/K$ is finite and $K/F$ is finite, then $L/F$ is finite.

Let $\{x_1,\dots,x_m\}$ be a spanning set of $K$ as a vector space over $F$ and $\{y_1,\dots,y_n\}$ a spanning set of $L$ as a vector space over $L$.

Now consider $\{x_iy_j:1\le i\le m,1\le j\le n\}$ and conclude.

Hint: if $z\in L$, then
$$
z=\sum_{j=1}^n \alpha_jy_j
$$
with $\alpha_1,\dots,\alpha_n\in K$. Then…

For the converse, you want to prove that if $L/F$ is finite, then both $L/K$ and $K/F$ are finite. If you have a finite spanning set of $L$ as a vector space over $F$, then it is also a spanning set of $L$ as a vector space over $K$; moreover $K$ is a subspace of $L$ (both as vector spaces over $F$).

Title: Field extension with infinite degree


问:
How to simplify $(\lnot p \rightarrow \lnot q) \land p$ and match the right statement?
The problem is as follows:

"If Maria does not leave her house then she does not catch a cold, but Maria leaves her house" is equivalent to:

I. Maria does not leave her house.

II. It is not true that, Maria leaves her house and catches a cold.

III. Maria leaves her house.

Given these statements which of the alternatives match the right answer?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{Only III}\\
2.&\textrm{I and II}\\
3.&\textrm{Only II}\\
4.&\textrm{II and III}\\
\end{array}$

In my attempt to solve this problem I translated the statement in these terms of propositional logic.

If Maria does not leave her house then she does not catch a cold, but Maria leaves her house

Then

$\textrm{Maria leaves her house = p}$

$\textrm{Maria catches a cold = q}$

Then:

$(\lnot p \rightarrow \lnot q) \land p$

But how to simplify this expression?. Can someone help me?. Is this the right or the appropiate way to translate what it was mentioned?.

答:
Remember that $r\to s$ is equivalent to $\neg r\lor s$, so you have

$$\begin{align*}
(p\lor\neg q)\land p&\equiv(p\land p)\lor(\neg q\land p)\\
&\equiv p\lor(p\land\neg q)\,.
\end{align*}$$

Can you finish it from there?


问:
Prove that sequence of projections to nested subspaces is fundamental (= Cauchy sequence)
Let $H - $ Hilbert space and $Y_0 \supset Y_1 \supset Y_2 \supset \cdots - $ subspaces in $H$, $x_0 \in H$. Define a sequence $$x_{n+1} := P_{Y_n} x_n = P_{Y_n} P_{Y_{n-1}} \ldots P_{Y_0} x_0 = P_{Y_n} x_0.$$
Prove that this sequence $\{x_n\}$ is fundamental (=Cauchy) sequence, i.e $|x_n - x_m| \rightarrow 0$ when $n,m \rightarrow \infty$.

Body:
Put $$z_{n} = x_n - x_{n+1} = x_{n} - P_{Y_n} x_n$$
Hence $z_{n} $ and $Y_n $ are orthogonal. Moreover, $z_n =  P_{Y_{n-1}} x_{n-1} - P_{Y_n} x_n$ and thus $z_n \in Y_{n-1}$.

We have

$$x_n = (x_{n} - x_{n-1}) + (x_{n-1} - x_{n-2}) + \ldots + (x_{1} - x_0) + x_0 $$
$$= -z_{n-1} - z_{n-2} -\ldots - z_1 + x_0.$$

It follows that $z_{n-1} + z_{n-2} + \ldots + z_1 = x_0- x_n =  x_0-P_{Y_{n-1}} x_{0} $ and hence $z_{n-1} + z_{n-2} + \ldots + z_1$ and $Y_{n-1}$ are orthogonal.

We have $$x_0 = x_n + (z_{n-1} + z_{n-2} + \ldots + z_1 ),$$
where  $z_{n-1} + z_{n-2} + \ldots + z_1$ and $Y_{n-1}$ are orthogonal and $x_n  = P_{Y_{n-1}} x_{0} \in Y_{n-1}$. Thus
$$||x_0||^2 = ||x_n||^2 + ||z_{n-1} + z_{n-2} + \ldots + z_1 ||^2.$$

Similarly we get that all $z_i$ are orthogonal (e.g. $z_1$ and $Y_1$ are orthogonal and $z_2, \ldots, z_{n-1} \in Y_1$ and so on) and  $||z_{n-1} + z_{n-2} + \ldots + z_1 ||^2 = \sum_{i=1}^{n-1} ||z_i||^2$.

Hence $$||x_0||^2 = ||x_n||^2 + \sum_{i=1}^{n-1} ||z_i||^2$$
and it follows that $ \sum_{i=1}^{n-1} ||z_i||^2 \le ||x_0||^2 $ and thus
$$ \sum_{i=1}^{\infty} ||z_i||^2 \le ||x_0||^2 < \infty .$$

For $m \ge n$ we get $$||x_n - x_m||^2 = || (z_{n-1} + z_{n-2} + \ldots + z_1 + x_0) - (z_{m-1} + z_{m-2} + \ldots + z_1 + x_0)  ||^2$$
$$= ||z_{n+1} + z_{n+2} + \ldots + z_{m}||^2 = \sum_{i=n+1}^{m} ||z_i||^2 \to 0$$
as $n,m \to \infty$, q.e.d.


答:
$a\to b$ is shorthand for $\neg a\lor b$:
$$(\neg p\to\neg q)\land p\iff(p\lor\neg q)\land p\iff (p\land p)\lor(\neg q\land p)$$
The first term may be replaced with $p\land 1$:
$$\iff(1\land p)\lor(\neg q\land p)\iff(1\lor\neg q)\land p\iff p$$
Hence alternative 1 is the correct answer.

Title: How to simplify $(\lnot p \rightarrow \lnot q) \land p$ and match the right statement?

Body:
The problem is as follows:

"If Maria does not leave her house then she does not catch a cold, but Maria leaves her house" is equivalent to:

I. Maria does not leave her house.

II. It is not true that, Maria leaves her house and catches a cold.

III. Maria leaves her house.

Given these statements which of the alternatives match the right answer?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{Only III}\\
2.&\textrm{I and II}\\
3.&\textrm{Only II}\\
4.&\textrm{II and III}\\
\end{array}$

In my attempt to solve this problem I translated the statement in these terms of propositional logic.

If Maria does not leave her house then she does not catch a cold, but Maria leaves her house

Then

$\textrm{Maria leaves her house = p}$

$\textrm{Maria catches a cold = q}$

Then:

$(\lnot p \rightarrow \lnot q) \land p$

But how to simplify this expression?. Can someone help me?. Is this the right or the appropiate way to translate what it was mentioned?.


问:
What is the difference between dense and closed sets?
I am self studying topology and is a bit stuck on the difference between dense and closed sets. Intuitively, a dense set is a set where all elements are close to each other and a closed set is a set having all of its boundary points.

But to make this more concrete, can someone give me an example of a closed set that is not dense and a dense set that is not closed?

Thanks!

答:
No one talk about Zarisky topology so I wanted to write it. Zariski topology is taught at the beginning of algebraic geometry. Let $k$ be algebraically closed field and $\mathbb{A}^n_k$ is the set of all elements $(a_1,...,a_n)$ where $a_i's \in k$. In Zarisky topology, the closed sets are the zero set of ideals of $k[x_1,...,x_n]$.

In this topology, in $\mathbb{A}^n_k$, there is only one dense and closed set which is $\mathbb{A}^n_k$. The other closed sets are not dense. Besides, all nontrivial open sets are dense in this topology.


问:
Formula relating three coplanar vectors
Let $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ be Euclidean vectors in $\mathbb{R}^3$ such that:

All of the vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ are non-zero (i.e., $\mathbf{\vec{0}}\notin\{\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}\}$).

The vectors $\mathbf{\vec{u}}$ and $\mathbf{\vec{v}}$ are non-collinear (i.e., $\mathbf{\vec{u}}\wedge\mathbf{\vec{v}}\neq\mathbf{\vec{0}}$).

The vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ are coplanar (i.e., [$\mathbf{\vec{u}}$,$\mathbf{\vec{v}}$,$\mathbf{\vec{w}}]=0$).

Then, a fortiori, there exist real numbers $\lambda$ and $\mu$ such that:
$$\mathbf{\vec{w}}=\lambda\mathbf{\vec{u}}+\mu\mathbf{\vec{v}}$$

My question is the following:

"Is there a general formula for $\lambda$ and $\mu$ in terms of the vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ in $\mathbb{R}^3$ (or, if necessary, in $\mathbb{R}^2$)?"

EDIT:
I know of a more general formula relating four Euclidean vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$, $\mathbf{\vec{w}}$ and $\mathbf{\vec{x}}$ in $\mathbb{R}^3$, which I state as follows:
$$[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]\neq 0 \implies \mathbf{\vec{x}}=\displaystyle\frac{[\mathbf{\vec{x}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{u}}+\frac{[\mathbf{\vec{u}},\mathbf{\vec{x}},\mathbf{\vec{w}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{v}}+\frac{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{x}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{w}}$$
I ask my question above in search for an analogous formula which relates three coplanar Euclidean vectors, if there is one at all.

答:
The following is a generalization of Botnakov N.'s approach. Suppose we have $m$ linearly-independent nonzero (column) vectors
$v_1,v_2,\ldots,v_m \in\mathbb{R}^n$. If $u\in\mathbb{R}^n$ is in the span of
$\{v_k\}_{k=1}^n$, how do we express $u$ as a linear combination of these vectors?

To simplify notation let $V=(v_1,v_2,\ldots, v_m)\in\mathbb{R}^{n,m}$, i.e., $V$ is a matrix with the $v$'s as column vectors. We seek a vector $c\in\mathbb{R}^m$ such that
$u=V c = \sum_{k=1}^m v_k c_k.$ Since $V$ is not a square matrix in general, we can't simply invert it. But if we we act on the left with $V^\top$ we obtain $V^\top u = V^\top Vc$, and $V^\top V\in\mathbb{R}^{m,m}$ is invertible (its determinant is always non-negative). Hence we obtain $$c=(V^\top V)^{-1}V^\top u.$$

The matrix $V^+:=(V^\top V)^{-1}V^\top $ is known as the Moore-Penrose pseudoinverse of $V$. On the one hand, we have $V^+ V = I_m$ by construction. But in general $V V^+\neq I_n$: It is a left inverse but not a right inverse. (This owes to $V$ having linearly-independent columns.) But, as the above work shows, it does act as a right inverse on the span of the $v$'s—that is, as a right inverse on the column space of $V$.

This kind of computation is moreover typical of ordinary least-squares computations, with $V^\top u=V^\top Vc$ being the so-called Gauss normal equations. To illustrate the connection, suppose that $u$ does not lie in the column space of $V$. Then $u^+:=V^+ u =V (V^\top V)^{-1} V^\top u\neq u$: The vector $u$ isn't in the column space of $V$, so we cannot and do not obtain $u$ as a linear combination of the $v$'s. However, we do have $$V^\top u^+ = V^\top V^+u = V^\top V(V^\top V)^{-1}V^\top =V^\top u.$$
Therefore $V^\top (u-u^+)=0$, i.e., $u-u^+$ is perpendicular to the column vectors of $V$. That is, $u-u^+$ is the orthogonal projection of $u$ onto the column space of $V$. This implies that $u^+$ is the unique vector in the column space of $V$ which minimizes $\|u-u^+\|^2$—that is, it's the ordinary least-squares solution to the overdetermined system $u=Vc$.

Title: Formula relating three coplanar vectors


问:
Using Semi-circle find side of triangle
The figure below above shown a bicycle path. If semicircular portion $ABC$ is $100$ $\pi$ and $CD$ is $100$$ft$ then what is $AD$?

I have tried to find the diamenter of the circle and the Pythagorean, I am not able figure out my answer.

答:
The ratio of a circle's circumference to its diameter is $\pi$:
$$\frac{\text{circumference}}{\text{diameter}}=\pi.$$The length of the semicircle arc is half the circumference of what would be the full circle. Thus,
$$\frac{\text{semicircle}}{\text{diameter}}=\frac{\frac{1}{2}\cdot\text{circumference}}{\text{diameter}}=\frac{1}{2}\left(\frac{\text{circumference}}{\text{diameter}}\right)=\frac{\pi}{2}$$
Using that $\text{semicircle}=100\pi$, solve for the length of the diameter of the circle. Note that this is the same as the length of the line segment $AC$. Lastly, now apply the Pythagorean theorem.

Title: Using Semi-circle find side of triangle


问:
Absorbed Brownian motion CDF — is there a mistake in this derivation?
I am trying to understand the derivation of the CDF of a Brownian Motion absorbed at a value as shown in this link. The derivation in this link claims that for $a>0$, the Brownian Motion absorbed at a value $a$, defined as

$$B_a=\begin{cases}B_t&\text{for }t<T_a,\\
a&\text{for }t\geq T_a,\end{cases}$$

where $T_a=\inf\{t\geq0:B_t=a\}$ (so the event $\{t\geq T_a\}\equiv\{\max_{s\in[0,t]}B_t\geq a\}$), we obtain

$$\mathbb{P}(B_t\geq x)=\Phi\left(\frac{x}{\sqrt{t}}\right)+\Phi\left(\frac{2a-x}{\sqrt{t}}\right)-1.$$

I plotted the CDF, and it's not monotonically increasing. (I also worked out the PDF and it definitely goes negative at some point.) Is there a typo or mistake in the derivation outlined here? I've tried to nitpick at the use of the reflection principle but it seems logically sound to me.

答: The CDF is  $P(B_a(t) \le x)$. The formula for the CDF is correct, and monotone increasing, for $x<a$. The CDF has a jump discontinuity at $x=a$ and it equals 1 for all $x \ge a$.