问:
Confusion about what are dilatations (my understanding is different from the text)
The confusion comes from a geometry text in which the concept of dilatation is given as follows (I use image for the purpose of a faithful reproduction of the problem):

This proposition and definition appears in a chapter on affine spaces and morphisms. The space is denoted as $(X,\vec X, \Phi)$ where $X$ is the affine space, $\vec X$ the underlying vector space and $\Phi$ the group action. $T(X)$ is the set of all translations $\{\Phi(\vec\xi)=t_{\vec\xi}|\vec\xi\in\vec X\}$. ${\rm{GL}}(\vec X)$ is the linear group of $\vec X$. $K$ is the scalar field and $K^*=K-\{0\}$. ${\rm{GA}}(X)$ is the affine group of $X$. $L$ is the homomorphism ${\rm{GA}}(X)\ni f\to\vec f\in{\rm{GL}}(\vec X)$. $H_{a,\lambda}$ denotes homothety of center $a$ and ratio $\lambda$: $x\to a+\lambda\vec{ax}$. Some facts follows: 1) $\ker L=T(X)$. 2) Let $f\in{\rm{GA}}(X)$ be such that $\vec f=\lambda{\rm{Id}}_{\vec X}$ where $\lambda\in K,\lambda\ne0,1$, it can be shown that there is a unique $a$ such that $f=H_{a,\lambda}$. I write the above here to avoid inconsistency on terminology. Now comes my question.

In my understanding of group homomorphism, the inverse image of a point in range is a coset with regard to the kernel. Since the operation in ${\rm{GA}}(X)$ is composition and the kernel is $T(X)$, the elements in coset has the form $H_{a,\lambda}\circ t_{\vec\xi}$ and I have shown for verification that $\overrightarrow{H_{a,\lambda}\circ t_{\vec\xi}}$ is indeed $\lambda{\rm{Id}}_{\vec X}$. So the inverse image ${\rm{Dil}}(X)$ of set $K^*{\rm{Id}}_{\vec X}$ should be the union of all aforementioned cosets, i.e., the "product" (actually the semidirect product) of all $H_{a,\lambda}$ for $a\in X$ and $\lambda\in K^*$(including $1$) with $T(X)$, instead of the "union" of them as I underlined in the text in red. Each coset contains a unique homothety $H_{a,\lambda}$ while all other elements in the coset (that is, dilatations) are a composition of the homothety with a translation. Now my question is simple: can you please confirm my understanding (and hence that the text is incorrect in using the word "union")?

Perhaps another lemma in the text related to dilatations may be helpful in clarifying what the elements of ${\rm{Dil}}(X)$ is:

In the above lemma, the author first says $f$ is a dilatation. Then he goes on to assume $f\in T(X)$ in the first case, a translation, and then $f$ is a homothety of center $0$ in the second case. This is another place where I got confused. From common writing rules, $f\in T(X)$ and $f$ is a homothety should be two concrete examples of $f$ being a dilatation. Am I right? If I'm correct, it can be inferred that dilatation can take a form of either translation or homothety. If the author means translations and homotheties are the only dilatations, this is consistent with his definition by means of "union". But are there any other form of dilatations, those in my point of view in the coset, that are not translations or homotheties and therefore do not fall into any case in the lemma but that have its own result (e.g., $f(b)$ is on line $D^\prime$). I would greatly appreciate it if anyone having background of college geometry could help me clarify what are the dilatations. Please let me know if you have any difficulty in understanding my question (say, what $\vec f$ is). Thank you.

答:
For reference purposes, the book in question is M. Berger's Geometry I; 2.3.3.12 is on p. 40.

You seem to have answered your own question: Indeed, any dilatation is either a homothety (with ratio not equal to $1$) xor it is a translation.

Let me recap aspects of the theory in my own slightly different notation (the details can be found in Berger's book). For $\mathbb{K}$ a field, an affine space over $\mathbb{K}$ is a triple $(X,V,a_\bullet)$, where $X$ is a set, $V$ is a vector space over $\mathbb{K}$ and $a_\bullet: (V,+)\to \operatorname{Bij}(X)$ is a faithful and transitive (hence also free) group action of $V$ considered as an abelian group. By the freeness of the action we have a structure or transition map or cocycle $\Theta:X\times X\to V$ that takes a pair $(x_1,x_2)$ and sends it to the unique solution $v\in V$ to the equation $a_v(x_1)=x_2$. We can thus define the difference of two points of $X$:

$$-x_1+x_2 := \Theta(x_1,x_2) =\mbox{ the unique vector that takes $x_1$ to $x_2$}.$$

(I will avoid using this abbreviation, though it's very useful as an intuition pump.)

Using the transition cocycle we have isomorphisms of $(V,+)$-sets:

$$\forall x_0\in X: \Theta_{x_0}:=\Theta(x_0,\bullet): X\xrightarrow{\cong} V, x\mapsto \Theta(x_0,x),$$

and if we fix any $x_0$ in $X$ we can use $\Theta_{x_0}$ to endow $X$ with a $\mathbb{K}$-vector space structure. If we do so this vector space is denoted by $X_{x_0}$. This means explicitly that in $X_{x_0}$ the vector space operations work like so:

$$cx_1+x_2 = \Theta_{x_0}^{-1} (c\;\Theta_{x_0}(x_1)+\Theta_{x_0}(x_2)), \quad\quad (\dagger)$$

and that the zero vector of $X_{x_0}$ is $x_0$. Observe that this is the first instance when scaling comes into the picture.

If $(X,V,a_\bullet)$ and $(Y,W,b_\bullet)$ are two $\mathbb{K}$-affine spaces, a morphism of $\mathbb{K}$-affine spaces $f:(X,V,a_\bullet)\to (Y,W,b_\bullet)$ from the former to the latter is by definition a function $f:X\to Y$ such that

$$\exists x_0\in X: f_{x_0}:=H_{f(x_0)}\circ f \circ \Theta_{x_0}^{-1} : V\to W \mbox{ is $\mathbb{K}-$linear}. \quad\quad(\ast)$$

($H$ is supposed to be capital eta, which I used for the transition cocycle of the latter affine space.)

Changing the basepoint has the following effect for any function $f$ from $X$ to $Y$ whatever:

$$\forall x_1,x_2\in X, \forall v\in V: f_{x_2}(v) = f_{x_1}(\Theta(x_1,x_2)+v)-f_{x_1}(\Theta(x_1,x_2)).$$

This means in particular that the test ($\ast$) for a function $f:X\to Y$ to be affine is independent of the basepoint $x_0$ chosen; which in turn means that if for some $x_0\in X$: $f_{x_0}$ is linear, then $f_x$ is linear for any $x\in X$, since they are all the same linear map, which is denoted by $L(f)$.

As expected the set of all affine automorphisms of $X$ form a group $GA(X)$ and $L: GA(X)\to GL(V)$ hence defines a group homomorphism, which we may call the linear part homomorphism. Recall that the kernel $\ker(L)$ of $L$ in this situation is the group of all affine automorphisms of $X$ whose linear part is $\operatorname{id}_V$. Using the pointwise realization ($\ast$) at an arbitrary point $x_0\in X$ we see that

$$\forall f\in \ker(L): f=\Theta_{f(x_0)}^{-1}\circ \operatorname{id}_V \circ \Theta_{x_0}=\Theta_{f(x_0)}^{-1}\circ \Theta_{x_0} = a_{\Theta(x_0,f(x_0))}.$$

In other words the kernel of $L$ is precisely the image of $a_\bullet$ in $\operatorname{Bij}(X)$:

$$\operatorname{im}(a_\bullet) = \{a_v| v\in V\} = \ker(L)=: T(X).$$

Another important definition is that of a homothety. For $x_0\in X$ and $\lambda \in \mathbb{K}^\times $ (which is the set of all units in $\mathbb{K}$), the homothety $h_{x_0,\lambda}$ with center $x_0$ and ratio $\lambda$ is by definition the function

$$h_{x_0,\lambda}:X\to X, x\mapsto a_{\lambda\Theta(x_0,x)}(x_0).$$

Three observations are in order:

Obs. 1: $\forall x_0\in X:h_{x_0,1}=\operatorname{id}_X$.

Obs. 2: $\forall x_0\in X,\forall \lambda\in\mathbb{K}^\times: L(h_{x_0,\lambda})=\lambda \operatorname{id}_V$.

Obs. 3: $\forall x_0,x_1 \in X, \forall \lambda\in\mathbb{K}^\times:$ if $x_1$ is a fixed point of $h_{x_0,\lambda}$, then either $\lambda=1$ (in which case Obs. 1 kicks in), or $x_1=x_0$. In words, nontrivial homotheties have unique fixed points, which are their centers.

By Obs. 3, if we have a fixed ratio $\lambda$ for a homothety, a point in $X$ completely determines the homothety (as the one that fixes said point). Let us denote by $\operatorname{Htt}(X)$ the set of all homotheties of $X$ and by $\operatorname{Htt}_\lambda(X)$ ($\lambda\in \mathbb{K}^\times$) the set of all homotheties of $X$ with ratio $\lambda$. Thus

\begin{align*}
\operatorname{Htt}(X)
&= \biguplus_{\lambda\in\mathbb{K}^\times} \operatorname{Htt}_\lambda(X)
= \left(\biguplus_{\lambda\in\mathbb{K}^\times\setminus1} \operatorname{Htt}_\lambda(X)\right) \uplus \operatorname{Htt}_1(X)\\
&= \left(\biguplus_{\lambda\in\mathbb{K}^\times\setminus1} \operatorname{Htt}_\lambda(X)\right) \uplus \{\operatorname{id}_X\}.
\end{align*}

Here $\uplus$ stands for (internal) disjoint union and the last equality is because of Obs. 1.

By Obs. 2, for any $\lambda\in\mathbb{K}^\times$, $\operatorname{Htt}_\lambda(X)\subseteq L^{-1}(\lambda \operatorname{id}_V)$. Now the crucial lemma is:

Lemma: Let $\lambda\in\mathbb{K}^\times\setminus 1$. Then  the inclusion map $\operatorname{Htt}_\lambda(X)\to L^{-1}(\lambda \operatorname{id}_V)$ is surjective whose inverse is given by:

$$L^{-1}(\lambda \operatorname{id}_V) \to \operatorname{Htt}_\lambda(X), f\mapsto h_{x_f,\lambda},$$

where

$$x_f:=\Theta_{x_1}^{-1}\left(\dfrac{1}{1-\lambda}\Theta_{x_1}(f(x_1))\right)$$

is the unique fixed point of $f$ that is independent of the basepoint $x_1\in X$.

(This is Prop.2.3.3.10 on p. 40 of Berger's book; there it looks much more pleasant because the transition cocycle is suppressed. Observe that the whole point of introducing $X_{x_1}$ actually is executing this suppression systematically; see ($\dagger$).)

By definition a dilatation of $X$ is an affine automorphism whose linear part is of the form $\lambda \operatorname{id}_V$ for some $\lambda\in\mathbb{K}^\times$ and the set of all dilatations of $X$ is denoted by $\operatorname{Dil}(X)$. Putting everything together we finally have:

\begin{align*}
\operatorname{Dil}(X)
&= L^{-1}(\mathbb{K}^\times \operatorname{id}_V)
= \biguplus_{\lambda\in\mathbb{K}^\times} L^{-1}(\lambda \operatorname{id}_V)
= \left(\biguplus_{\lambda\in\mathbb{K}^\times\setminus1} L^{-1}(\lambda \operatorname{id}_V)\right) \uplus L^{-1}(1\operatorname{id}_V)\\
&= \left(\biguplus_{\lambda\in\mathbb{K}^\times\setminus1} \operatorname{Htt}_\lambda(X) \right)\uplus \ker(L)
= \left(\operatorname{Htt}(X)\setminus\{\operatorname{id}_V\} \right)\uplus T(X).
\end{align*}

Thus we see that indeed a dilatation is either a homothety or it's a translation. If it's both; it it's trivial.

There is a heuristic picture that I think is helpful: Take $\mathbb{K}=\mathbb{R}$. We have the real line without its origin, parametrizing the center of the general linear group. Above it floats the set of all dilatations of $X$. $L$ projects down. Above $1\in\mathbb{R}^\times$ we see $T(X)$, which is nothing but $V$. Above any other point $\lambda\in\mathbb{R}^\times$ we see $\operatorname{Htt}_\lambda(X)$, which are parametrized by their fixed points, so we see a copy of $X$. Of course by the definition of an affine space one can also consider $X$ to be a copy of $V$.

On the other hand you are right that the dilatations can be considered as a semidirect product by way of considering homotheties. Indeed we have a split short exact sequence

$$1\to T(X)\to \operatorname{Dil}(X)\xrightarrow{L} \mathbb{K}^\times \to 1,$$

with $h_{x_0,\bullet}: \mathbb{K}^\times  \to \operatorname{Dil}(X), \lambda\mapsto h_{x_0,\lambda}$ providing the splitting ($x_0$ an arbitrary basepoint) (See 2.3.3.11 on p. 40 of Berger's book). Thus $\operatorname{Dil}(X)\cong T(X)\rtimes \mathbb{K}^\times$, just as $GA(X)\cong T(X)\rtimes GL(V)$.


问:
Are homotopy-preserving maps continuous?
I was thinking some about weak homotopy equivalences and what happens when you drop the continuity assumption.
For example, it's not too hard to come up with non-continuous functions that induce isomorphisms on all homotopy groups (and a bijection on connected components): let $X$ be any contractible space, then the identity map $\operatorname{codisc}X\to X$ from the codiscrete topology on $X$ is not continuous in general, but it trivially induces isomorphisms of all $\pi_n$ as both spaces are contractible.

My question is if continuity remains unnecessary if I strengthen the assumptions a bit from just inducing homomorphisms of homotopy groups to actually just preserving homotopies:

Question: Does there exist a non-continuous map $f:X\to Y$ such that $\def\bS{\mathbb{S}}fg:\bS^n\to Y$ is continuous for all continuous $g:\bS^n\to X$ and $n\geq0$, where $\bS^n$ is the $n$-sphere?

I want to believe that the answer is yes, but I am drawing blanks for an explicit example.
I thought to use length constraints on homotopies like what happens with the long line (each individual homotopy is "too small" to see everything), but I couldn't put it into practice.

If the answer to the question is indeed yes, then how nice can $X$ and $Y$ be with $f$ still existing? For instance, can $X$ and $Y$ be compactly generated and (weakly) Hausdorff?

答:
If you do not impose some hypothesis related to path-connectedness on $X$, then there are essentially arbitrarily nice counterexamples.  For example, let $X$ any totally path-disconnected space (e.g., $\mathbb{N}\cup\{\infty\}$ which is additionally compact Hausdorff) and $Y$ be any space.  Then every map $f:X\to Y$ has continuous composition with every map from a sphere, so such a counterexample exists unless every map from $X$ to $Y$ is continuous.

Note that spheres are a bit of a red herring here--it's equivalent to ask for the same condition with $\mathbb{R}$ (or $[0,1]$) in place of spheres.  One direction is easy (any interval in $\mathbb{R}$ can be written as a quotient of a sphere so if you have continuity after composing with maps from spheres you get continuity after composing with maps from $\mathbb{R}$).  The other direction is nontrivial (see below).

More generally, say a space $X$ is $\mathbb{R}$-generated if for every space $Y$ and every function $f:X\to Y$, $f$ is continuous iff its composition with each continuous map $\mathbb{R}\to X$ is continuous.  I discuss such spaces in detail in this answer.  In particular, every locally path-connected first-countable space is $\mathbb{R}$-generated, as is every CW-complex, and every $\mathbb{R}$-generated space is locally path-connected.  If $Z$ is $\mathbb{R}$-generated and $f:X\to Y$ has the property that its composition with every continuous map $\mathbb{R}\to X$ is continuous, then $f$ also has the same property with respect to maps $Z\to X$.  In particular, taking $Z$ to be the spheres, this proves the nontrivial direction mentioned in the previous paragraph.  It follows that the $\mathbb{R}$-generated spaces are exactly the spaces $X$ for which no counterexample $f:X\to Y$ exists to your question.

ContentLicense: CC BY-SA 4.0


问:
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$

If

$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:

$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.

答:
We have$$x^2+y^2+z^2+t^2=x(y+z+t)\iff x^2-\frac34x^2+(y-\frac x2)^2+(z-\frac x2)^2+(t-\frac x2)^2=0$$ i.e.$$\left(\frac x2\right)^2+\left(y-\frac x2\right)^2+\left(z-\frac x2\right)^2+\left(t-\frac x2\right)^2=0$$
We are done


问:
Projective plane and concurrency of three lines formula
I was reading AMD's answer here , and he considers three plane which represent three lines, now for three planes to intersect other than origin it must be that the all the normals together are not perpendicular, using this we can get the condition on the scalar triple product of coefficents i.e: if our lines are given as:

$$ a_1 x + b_1  y + c_1 =0 $$
$$ a_2 x + b_2 y + c_2 = 0$$
$$ a_3 x + b_3 y +c_3 = 0$$

Then, if:

$$ v= <a_1 , b_1 , c_1>$$
$$ u = <a_2,b_2,c_2>$$

And,

$$ q=<a_3,b_3,c_3>$$

Must have scalar triple product as zero:

$$ q \cdot ( v \times u ) = 0$$

Is a 'required' condition for the lines to intersect, but is it sufficient for it to always gaurentee that it would? As in how do we know that when we put $z=1$ , that is where the planes intersect?

答:
Yes it is a necessary and sufficient condition if you include as "point of intersection" the point at infinity in the case of parallel lines.

If the two lines $l_1: \; a_1x+b_1y+c_1=0,\quad l_2:\; a_2x+b_2y+c_2=0$ intersect in a point than $l_3=\lambda l_1 +\mu l_2$ also passes though that point. And the their $3 \times 3$ determinant is null.

Viceversa, if the determinant is null, then the rows are dependent and one can be expressed as a linear combination of the others.

-- reply to your comment --

In  analytic or affine geometry, the introduction of the homogeneous coordinates render the panorama more ..homogeneous, in fact.

In particular we have the interesting duality points/lines in 2D, and points /planes in 3D lines.

in 2D, two lines always have a common point same as two points have a common line.

Translated to algebra, given
$$l_1: \; a_1x_1+b_1x_2+c_1x_0=0,\quad l_2:\; a_2x_1+b_2x_2+c_2x_0=0$$
then
$$l_1 \cap l_2$$ always has one non-trivial solution, either in terms of a common point $(x_1,x_2,x_0)$ in the above, and in terms of a line common to two points in
$$a x_{1,1}1+bx_{1,2}+cx_{1,0}=0 \; \cap \; a x_{2,1}1+bx_{2,2}+cx_{2,0}=0$$

Then you know that in a homogeneous system prescribing the common point among three lines, a necessary and sufficient condition is that the determinant of the coefficients ($3 \times 3$) be null, because otherwise (rank of the matrix $3$) you have the only solution $(0,0,0)$ which does not correspond to a defined point.

Then if the rank of the matrix is $2$, you have a homogeneous solution $\lambda (\xi, \eta, \zeta)$ which is one point,

and if the rank is $1$ you get a homogeneous solution depending on two parameters, which is a line and which is the same as any of the three in the system.

The same, reverted, for three points sharing a common line.


问:
Line graph $L(G)$ is bipartite if and only if the graph is isomorphic to path or even cycle.
Let $G$ be a connected graph.

Line Graph $L(G)$ is bipartite if and only if the graph is isomorphic to path or even cycle.

I have shown that if the graph is a path or even cycle then the line graph is bipartite.

I am having difficulty in proving the other part.

答:
The "only if" direction: If $G$ has a vertex of degree $k=3$ or higher then $L(G)$ on the edges incident to $v$ is isomorphic to $K_k$ and so is not bipartite.

If $G$ has an odd cycle $C$ of length $k$; $k$ an odd integer; then writing the edges of $C$ in consecutive order as $e_0e_2\ldots e_{k-1}$, then $e_0$ is adjacent to $e_1$ in $L(G)$, and $e_2$ is adjacent to $e_3$ in $L(G)$ and so on. So $L(G)$ has a cycle of the same [odd] length $k$ as well, so $L(G)$ is not bipartite if $G$ is not bipartite.

The "if" direction: If $C$ is a cycle, then $L(C)$ is also a cycle of the same length as $C$. If $C$ is an even cycle then, then so is $L(C)$.

If $P$ is a path of length $k$ i.e., $k$ vertices and $k-1$ edges, then $L(P)$ is also a path, but with length $k-1$ i.e., $k-1$ vertices and $k-2$ edges.

Can you put this all together.


问:
How to express by a single equation a region delimited by confocal ellipse and hyperbola
I am trying to find the equation of a shape formed by a conic (ellipse) that excludes its intersection with another conic (hyperbola). The hyperbola and ellipse are assumed to share common foci. But they are general conics  in 2D which means they can be rotated about the z-axis and translated along x and y axes. As it can be seen in the figure, the conics are rotated by an angle theta and not centered at the origin.

In the figure, I want the equation of resulting shape shown in grey shade (formed by the two conics). This new shape can be shown as follows:

Please let me know if it is possible to get an equality representing this new shape in the quadratic form as follows:

$ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

答:
Such a curve having the shape of a double-sided axe (https://en.wikipedia.org/wiki/Labrys) with mixed elliptical/hyperbolic boundaries can be given by the general equation

$$x^2+k|y^2-1|=k^2 \tag{1}$$

with a "behavorial change" when variable $y$ crosses values $y=1$ or $y=-1$ as can be seen on the following graphics for different values of $k$:

Fig. 1: Curves with equation (1) for different values of $k$, with $k=0.25$ for the most internal one (in two parts) and $k=2$ for the most external, with steps $0.25$. For our question, only values $k>1$ (blue curves) make sense.

It remains now to tilt the "axe" using the following coordinates change in (1):

$$\begin{cases}x&=& \ \ \ (\cos \theta) X + (\sin \theta) Y \\ y&=&-(\sin \theta) X + (\cos \theta) Y \end{cases}$$

(It looks like a pun: a change of axes inducing a different axe...)

Explanations:

I have obtained (1) by imposing common foci to the ellipse and the hyperbola. How can it be done ? If $f$ is the common distance from the origin to the foci, it is known that $f^2=a^2-b^2$ for an ellipse and $f^2=a^2+b^2$ for a hyperbola with canonical equations $\dfrac{x^2}{a^2}\pm\dfrac{y^2}{b^2}=1$ ; one obtains in particular $f=\sqrt{k^2-1}$.

As a consequence, the elliptical arcs are orthogonal to the hyperbolic arcs. This wasn't necessary (see the solution by @David K).

Case $k=1$ is particular as can be seen on Fig. 1 : the elliptical arcs become circular arcs.

Edit: Here is the Matlab program I have written for the generation of the figure:

```<code>   clear all;close all;hold on;axis equal
   for k=0.25:0.25:2;
      c='r';
      if k>1;c='b';end;
      e=ezplot([num2str(k),'*abs(y^2-1)+x^2-',num2str(k^2)]);
      get(e);set(e,'linecolor',c);
   end;
   u=2.5;plot([-u,u],[0,0],'k','linewidth',1.5);
   v=2;plot([0,0],[-v,v],'k','linewidth',1.5);
</code>```

Body:
Such a curve having the shape of a double-sided axe (https://en.wikipedia.org/wiki/Labrys) with mixed elliptical/hyperbolic boundaries can be given by the general equation

$$x^2+k|y^2-1|=k^2 \tag{1}$$

with a "behavorial change" when variable $y$ crosses values $y=1$ or $y=-1$ as can be seen on the following graphics for different values of $k$:

Fig. 1: Curves with equation (1) for different values of $k$, with $k=0.25$ for the most internal one (in two parts) and $k=2$ for the most external, with steps $0.25$. For our question, only values $k>1$ (blue curves) make sense.

It remains now to tilt the "axe" using the following coordinates change in (1):

$$\begin{cases}x&=& \ \ \ (\cos \theta) X + (\sin \theta) Y \\ y&=&-(\sin \theta) X + (\cos \theta) Y \end{cases}$$

(It looks like a pun: a change of axes inducing a different axe...)

Explanations:

I have obtained (1) by imposing common foci to the ellipse and the hyperbola. How can it be done ? If $f$ is the common distance from the origin to the foci, it is known that $f^2=a^2-b^2$ for an ellipse and $f^2=a^2+b^2$ for a hyperbola with canonical equations $\dfrac{x^2}{a^2}\pm\dfrac{y^2}{b^2}=1$ ; one obtains in particular $f=\sqrt{k^2-1}$.

As a consequence, the elliptical arcs are orthogonal to the hyperbolic arcs. This wasn't necessary (see the solution by @David K).

Case $k=1$ is particular as can be seen on Fig. 1 : the elliptical arcs become circular arcs.

Edit: Here is the Matlab program I have written for the generation of the figure:

   clear all;close all;hold on;axis equal
   for k=0.25:0.25:2;
      c='r';
      if k>1;c='b';end;
      e=ezplot([num2str(k),'*abs(y^2-1)+x^2-',num2str(k^2)]);
      get(e);set(e,'linecolor',c);
   end;
   u=2.5;plot([-u,u],[0,0],'k','linewidth',1.5);
   v=2;plot([0,0],[-v,v],'k','linewidth',1.5);


问:
proof) if set A,B has no common elements, and n(A)=m, n(B)=n $A \cup B$ has m+n elements.
Related Question
Proof that cardinality of the union of disjoints sets is the sum of cardinality of each set

If $A$ and $B$ have $n$ and $m$ elements respectively with $A\cap B=\emptyset$. Prove that $A\cup B$ has $m+n$ elements.

theorem 1.3.4 Introduce to Real Analysis, Bartle 4th

I'm stuck on prove $h$ is bijective

my proof is down below:

Let $f$ is a bijection $\mathbb{N}_m$ onto A, $g$ is bijection $\mathbb{N}_n onto B

For $A=\{a_1,a_2, \cdots ,a_m \}$, Let $\mathbb{N}_m=\{1,2,\cdots,m\}$

$f(a_1)=1, f(a_2)=2, \cdots, f(a_m)=m$. and

For $B=\{b_1,b_2, \cdots ,b_n \}$, Let $\mathbb{N}_n=\{m+1,m+2,\cdots,m+n\}$

$g(b_1)=m+1, g(b_2)=m+2, \cdots ,g(b_n)=m+n$

then defined function $h$ is bijective $\mathbb{N}_{m+n}$ onto $A\cup B$

How to proof Surjectivity, Injectivity with section

What did I forgot this question...?

P.S) I can't understand after seeing solutions

PP.S) I think the hint is Definintion 1.3.1 (b): if $ n \in \mathbb{N}$ a set S is said to have $n$ elements

答:
Let me write some stuff to establish some common terms that I think will clarify things a little bit. Given a set $A$, the symbols $`` n(A)=m"$ mean that $m$ is natural number (here, the natural numbers set is $\mathbb{N}=\{0,1,2,3,4,...\})$ and there is a biyective function $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$. Let me give you some examples.

(a) Suppose $n(A)=0$. Then, by definition, we know there is a biyection between $A$ and the set $\{k\in\mathbb{N} : 0<k\leq 0\}$. This means that there is a biyection between $A$ and the empty set (there is no natural number $k$ that satisfies the relations $0<k\leq 0$). In colloquial terms, we say that $A$ has no elements. Since the empty set is the only set that satisfies this property, we deduce that $A=\emptyset$.

(b) Now let's assume that $n(A)=m$, with $m>0$. What does this mean?
If we take the biyection $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$ that exists by definition, then, if we look a little closer, we can see that the set $\{k\in\mathbb{N} : 0<k\leq m\}$ is simply the set $\{1,...,m\}$, that you denote by $\mathbb{N}_{m}$. In this case, we say that $A$ has $m$ elements.

(c) Take $a,b$ and $c$, three distinct real numbers. I claim that, if we let $A=\{a,b,c\}$, then $n(A)=3$. In order to do this, the only thing we have to show is that there is a biyective function between the sets $A$ and $\{1,2,3\}$. For example, if we send $a\to 1$, $b\to 2$, and $c\to 3$, then this produces a biyective map between $A$ and  $\{1,2,3\}$; hence, we can conclude that $n(A)=3$. Here, we say that $A$ has $3$ elements.

Now, we can try to proof the result you mention. Let $A$ and $B$ be a pair of disjoint sets (this is, $A\cap B=\emptyset$), such that $n(A)=m$ and $n(B)=n$. Fix a pair of biyective functions $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$ and $g:B\to \{k\in\mathbb{N} : 0<k\leq n\}$. Our goal is to show that $n(A\cup B)=m+n$; in other words, we must find a biyection between $A\cup B$ and $\{k\in\mathbb{N} : 0<k\leq m+n\}$ (in colloquial terms, this means that $A \cup B$ has $m+n$ elements).

First, if $n(A)=m=0$, then by the example explained in (a), we know that $A=\emptyset$. So, $A\cup B=B$, $m+n=n$, and $\{k\in\mathbb{N} : 0<k\leq m+n\}=\{k\in\mathbb{N} : 0<k\leq n\}$ . In this case, if we let $h=g$, then $h$ is the desired biyection. Similarly, if we have $n(B)=n=0$, then we can argue that $h=f$ is the biyection we are looking for.

Now, asumme that $n(A)=m>0$ and $n(B)=n>0$. Then, from (b), we know that $f$ is a biyection between $A$ and $\{1,...,m\}$, $g$ is a biyection between $B$ and $\{1,...,n\}$, and we are looking for a biyection $h$ between $A\cup B$ and $\{1,...,m,m+1,...,m+n\}$. Our aim is to construct $h$ in some way related to $f$ and $g$.

Consider $h: A\cup B \to \{1,...,m,m+1,...,m+n\}$ given by

$$ h(x) = \begin{cases} f(x), & \text{if} \hspace{2mm} x \in A, \\
m+g(x), & \text{if} \hspace{2mm} x\in B.
  \end{cases} $$

Since $A\cap B=\emptyset$, we know that $h$ is a well defined function. I claim that $h$ is biyective.

To see that $h$ is surjective, take $k\in \{1,...,m,m+1,...,m+n\}$. Then, we have two cases. If $1\leq k \leq m$, then use that $f$ is surjective to find $x\in A$ such that $f(x)=k$. It is clear that $h(x)=k$. Now, if $k=m+\ell$ form some $1\leq \ell \leq n$, employ that $g$ is surjective to find $x\in B$ such that $g(x)=\ell$. Then, again, we have that $h(x)=k$. This shows that $h$ is surjective.

In order to prove that $h$ is injective, first observe that $h[A]\subseteq \{1,...,m\}$ and $h[B]\subseteq \{m+1,...,,m+n\}$. Now, suppose that $x,y\in A\cup B$ satisfy the equality $h(x)=h(y)$. From the observation we get two cases. If $h(x),h(y)\in \{1,...,m\}$ then, from the definition of $h$, the equality $h(x)=h(y)$ becomes $f(x)=f(y)$ and, since $f$ is injective, we get that $x=y$. On the other hand, if $h(x),h(y)\in \{m+1,...,m+n\}$ then again, from the definition of $h$, the equality $h(x)=h(y)$ becomes $m+g(x)=m+g(y)$. So, $g(x)=g(y)$ and, since $g$ is injective, we get that $x=y$. We conclude that $h$ is injective.

From the two previous arguments, we get that $h$ a biyection between $A\cup B$ and the set $ \{1,...,m,m+1,...,m+n\}$, as desired.

Hope this helps.


问:
Question about the factors of a number (apart from 2 consecutive numbers)

Pr. 13. In a class, 25 students were lined up. The teacher wrote a number on the board.

• The first student said the number was divisible by 1.

• Student number 2 said it was divisible by 2.

• Student 3 said it was divisible by 3.

...

• This went on until the 25th student.

The teacher said everyone was correct except two boys, named Jack and Jill, who were wrong. They were standing one immediately after the other in the line.

What are their positions in the line?

I had no idea how to start, so I listed the factors and multiples under 25 of the numbers under 25:

```<code>1: 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25
2: 1,4,6,8,10,12,14,16,18,20,22,24
3: 1,6,9,12,15,18,21,24
4: 1,2,8,12,16,20,24
5: 1,10,15,20,25
6: 1,2,3,12,18,24
7: 1,14,21
8: 1,2,4,16,24,
9: 1,3,18
10: 1,2,5,20
11: 1,22
12: 1,2,3,4,6,24
13: 1
14: 1,2,7
15: 1,3,5
16: 1,2,4,8
17: 1
18: 1,2,9
19: 1
20: 1,2,4,5
21: 1,3,7
22: 1,2,11
23: 1
24: 1,2,3,4,6,8,12
25: 1,5
</code>```

I thought I could choose 17 and 19, but they aren't consecutive!

How shall I attempt and solve this question?

答:
If all of the responses had been correct, then the number would have to be a multiple of

$$2^4 \times 3^2 \times 5^2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23.$$

The obvious try, which works, is that neither $(16)$ nor $(17)$ divides the number.


问:
Is the centralizer of a subfield $Z$ of a central simple algebra $A$ over $F$ a central simple algebra over $Z$?
Notation and Terminology:

An algebra over a field is always associative and unitary, and subalgebras always possess the unity.

A subfield of an $F$-algebra $A$ is a commutative $F$-subalgebra $B$ of $A$ such that for every $b\in B$ there is $c\in B$ such that $bc=cb=1$.

An algebra $A$ over a field $F$ is said to be a central simple algebra of degree $n$ over $F$ if there is an extension field $K$ of finite dimension over $F$ such that the scalar extension $A_K:=K\otimes_FA$ is isomorphic as $K$-algebra to $M_n(K)$, the set of all square matrices of size $n$ over $K$, and in this case we say that $K$ is a splitting field of $A$ over $F$.

Question:

If $A$ is a central simple algebra of degree $n$ over $F$ and $Z$ is a subfield of $A$ of dimension $m$ over $F$, is the centralizer $C_A(Z)$ a central simple algebra of degree $\frac{n}{m}$ over $Z$?

My Attempt:

I know the following facts:

Fact 1: Let $A\cong M_n(F)$ be an $F$-algebra and $B\cong M_m(F)$ be an $F$-subalgebra of $A$. Then the centralizer $C_A(B)\cong M_{\frac{n}{m}}(F)$ and $A\cong B\otimes_FC_A(B)$ as $F$-algebras.

Fact 2: Let $A$ be a finite dimensional $F$-algebra and $B\cong M_m(F)$ be an $F$-subalgebra of $A$. Then $A\cong B\otimes_FC_A(B)$.

Fact 3: Let $A$ be a finite dimensional $F$-algebra and $B$ be a central simple $F$-subalgebra of $B$ with splitting field $K$, then $A\cong B\otimes_FC_A(B)$ and $C_{A_K}(B_K)=(C_A(B))_K$.

Fact 4: Let $A\cong M_n(F)$ be an $F$-algebra and $B$ be a central simple $F$-subalgebra with splitting field $K$, then $C_A(B)$ is a central simple $F$-subalgebra of $A$ with splitting field $K$.

Fact 5: If $A\cong M_n(F)$ is an $F$-algebra and $Z$ is a subfield of dimension $m$ over $F$, then the centralizer $C_A(Z)\cong M_\frac{n}{m}(Z)$ as an $Z$-algebra.

If $D$ is a central simple algebra of degree $n$ over $F$ and $Z$ is a subfield of $D$ of dimension $m$ over $F$, then, by the regular representation of $D$, $D$ is an $F$-subalgebra of an $F$-algebra $M\cong M_{n^2}(F)$ and, if $D_0=C_M(D)$, then $D_0$ is central simple over $F$ and $M\cong D\otimes_F D_0$, so $D\subseteq C_M(D_0)$, $C_M(D_0)$ is central simple over $F$ and $M\cong C_M(D_0)\otimes_F D_0$, so $C_M(D_0)=D$. Also, if $A=C_M(Z)$, then $A\cong M_{\frac{n^2}{m}}(Z)$ as an $Z$-algebra, $Z\subseteq A$ and $D_0\subseteq A$. Moreover, if $Q=A\cap D$, then $Q=C_D(Z)=C_A(D_0)$, so $A\cong Q\otimes_FD_0$ as $F$-algebra.

I do not know if $D_0$ is a central simple algebra over $Z$, but if is so, then I can conclude that $C_D(Z)=Q$ is central simple over $Z$.

答: Indeed, $D$ is a subalgebra of an algebra $M\cong M_{n^2}(F)$, so if $D_0=C_M(D)$, then $D_0$ is central simple of degree $n$ over $F$ and also $M=D\otimes_FD_0$, so $(D_0)_Z=Z\otimes_F D_0\subseteq D\otimes_F D_0=M$, so if $A=C_M(Z)$ then $(D_0)_Z=Z\otimes_FD_0\subseteq A$ because $Z\subseteq D$ and $Z$ is commutative, and $A\cong M_{\frac{n^2}{m}}(Z)$, hence, as $D_0$ is central simple of degree $n$ over $F$, then $(D_0)_Z$ is central simple of degree $n$ over $Z$, so if $E=C_M(D_0)$, as $D_0$ is central simple over $F$, then $M=E\otimes_FD_0$, so $\mathrm{dim}(E)=\mathrm{dim}(D)$, but $D\subseteq E$, so $D=E$. Let $Q=C_A(D_0)$, then $Q=A\cap D=C_D(Z)$ and $Q=C_A((D_0)_Z)$ is central simple of degree $\frac{n}{m}$ over $Z$ and $A=Q\otimes_Z(D_0)_Z$, so that $m\mid n$, so $C_D(Z)$ is central simple of degree $\frac{n}{m}$ over $Z$.


问:
Given vectors v = [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

Given vectors v= [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

The solution is [2, 3], but I need to know how.

v in the same basis as b1 and b2 is [5,-1]

but if v has to be defined in terms of b1 and b2, then

vb = 2b1, 3b2 = [2, 3] because this vector v is now being shown with b1
and b2 as new basis vectors.

I understand b1, but get confused how projection b2 gets calculated the way it does.

The solution provided completed projection b2 as follows:

Proj. b2 = $\frac{v.b2}{|b2|^2}$ = $\frac{(5x1)+(-1x-1)}{-1^2 + (-1^2)}$ = $\frac{6}{2}$ = 3

How does b2 [1, -1] become [-1, -1] in the calculation?

And even still, shouldn't that equal -2, not 2? Giving answer -3, not 3?

I seem to calculate  1^2, + (-1^2) = 0, and arrive at 6/0, which obviously isn't right. Where am I going wrong?

答:
For the denominator, the -1 is a mistake. There should not be a negative sign in front of the 1. In fact the writer acts as if there is no negative sign there.

You're going wrong because you're doing $1^2+-1^2=0$. You should instead be doing $1^2+(-1)^2=2$

For the numerator, $b_2$ has not become (-1,-1). It simply dots $b_2$ with v.

$$\begin{split}v\cdot b_2&=\left(\begin{matrix}5\\-1\end{matrix}\right)\cdot \left(\begin{matrix}1\\-1\end{matrix}\right)\\
&=5*1+-1*-1\\&=6\end{split}$$

The correct calculation without all these errors looks like

$$\begin{split}\frac{5*1+-1*-1}{1^2+(-1)^2}
&=\frac 6 2\\&=3\end{split}$$


问:
How to find the roots of a cubic polynomial?
All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what if you can't do that?

For example, how would you find the roots of the following equation:

$x^3+7x^2+16x+12=0$

I know the roots are $-2$, $-2$ and $-3$ but don't know how to get them.
If I somehow guess them, is there a way to tell which one is the double root?

答: Start out by checking the positive and negative factors of 12. Once you find one factor that makes the polynomial equal to zero, say $x = -2$, divide the polynomial by the corresponding factor $x+2$. You can use synthetic division or long division. Once you do that you get the quadratic factor $x^2+5x+6$. Factor the quadratic to get $(x+2)(x+3)$, therefore:$$x^3+7x^2+16x+12=(x+2)(x+2)(x+3)=0$$ This means $x = -2$, $-2$, and $-3$ as you pointed out but now you can see clearly that one factor repeats.


问:
Combinatorial proof of a Stirling number identity.
Consider the identity
$$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$
where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity
$$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$
which essentially states that the number of even subsets of a set is equal to the number of odd subsets.

Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?

As an added note, there is a similar identity
$$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$
for $n \geq 2$.
A combinatorial interpretation of this one would also be appreciated.

答:
For the sake of completeness I include a treatment using generating functions.
The exponential generating function of the Stirling numbers of the second kind is
$$ G(z, u) = \exp(u(\exp(z)-1))$$
so that
$$ {n \brace k} = n! [z^n] \frac{(\exp(z) - 1)^k}{k!}.$$
It follows that
$$\sum_{k=0}^n (-1)^k k! {n \brace k} = n! [z^n] \sum_{k=0}^n (1-\exp(z))^k
= n! [z^n]  \sum_{k=0}^\infty (1-\exp(z))^k,$$
where the last equality occurs because the series for $(1-\exp(z))^k$ starts at degree $k.$
But this is just
$$ n! [z^n] \frac{1}{1-(1-\exp(z))} = n! [z^n] \exp(-z) = (-1)^n,$$
showing the result.</p>

ContentLicense: CC BY-SA 3.0


Title: How to find the roots of a cubic polynomial?

Body:
All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what if you can't do that?

For example, how would you find the roots of the following equation:

$x^3+7x^2+16x+12=0$

I know the roots are $-2$, $-2$ and $-3$ but don't know how to get them.
If I somehow guess them, is there a way to tell which one is the double root?

Body:
Factorize as follows
$$x^3+7x^2+16x+12\\
=(x^3+7x^2+12x)+(4x+12)\\
=x(x+4)(x+3)+4(x+3)\\
=(x+3)(x^2+4x+4)
$$


问:
Linear algebra cylinder problem
We have a plane. On that plane there is a line $p$:
$$p:x-1=2y-2=z+1$$
This plane also touches a cylinder with axis $x=y=z$. Find the equation of the plane and the diameter of the cylinder.

What I tried:
I solved for parameter $t$ from the line:
$$t = x-1 \implies x = t+1$$

$$t = 2y-2 \implies y = \frac{t+2}{2}$$
$$t = z+1\implies z = t-1$$

We also know that the point $T(1,1,-1)$ is on the line thus it also needs to be on the plane.
Then Ive put the parametrically defined coordinates of the line into the equation of the plane.
$$ax+by+cz = a(t+1)+b(\frac{t+2}{2})+z(t-1)$$

I do not know how to continue.

The solution is $x-z = 2$ which is the equation of the plane and the radius of the cylinder is $\sqrt{2}$

答: The normal $(a,b,c)$ to the plane must   be orthogonal to the axis of the cylinder and to the given line $p$, so $a+b+c=0$ and $a+b/2+c=0$. This forces $b=0$ and $a=-c$ so $(1,0,-1)$ is a normal. Since $T=(1,1,-1)$ is on the plane, you deduce the equation of the plane is indeed $x-z=2$. The origin is on the axis of the cylinder, and the normal direction from the origin reaches the plane at $(1,0,-1)$, so the radius of the cylinder is $2^{1/2}$.

Title: Linear algebra cylinder problem


问:
Let $(x_n)$ be a sequence s.t l. $|x_m - x_n|\le\frac1{1+|m-n|}$ for all $m$ and $n$. Find (with proof) $\lim x_n$.
Suppose $(x_n)$ is a sequence that satisfies

$$ |x_m - x_n| \leq \dfrac{1}{1+|m-n|} $$

for all $m$ and $n$. Find (with proof) $\lim x_n$

ATTEMPT:

If we put $m = n+1$, then

$$ |x_{n+1} - x_n | \leq \dfrac{1}{1+n} $$

From this condition we see that $\lim x_n = L$ exists. On the other Hand, if we choose $n=m-1$, we obtain

$$ |x_{m} - x_{m-1} | \leq \dfrac{1}{2} $$

And, as $m$ can by any number, then we may not conclude that the sequence converges as it may oscillate. Can I get some hints on how to solve this? Is my reasoning reasonable? or am I overthinking and it is a simple exercise.

答:
HINT: Fix $m\in\Bbb N$. Then for each $n>m$ we have

$$|x_n-x_m|\le\frac1{1+(n-m)}=\frac1{n-(m-1)}\,,$$

which tends to $0$ as $n$ increases without bound. Thus, $\lim\limits_{n\to\infty}x_n=x_m$. This is true for each $m\in\Bbb N$, so ... ?


问:
The area between coordinate axes and tangent to $f(x)$ is $2x f(x)$. Is this correct?
I was checking the answers of the coordinate axes has minimum area question.

In short, the mentioned question asks the minimum area formed by a tangent to $y=4-x^2$ and coordinate axes. I noticed one of the answers to the question mentioned an interesting property:

Using similar triangles, it can be proven that the area of the triangle is twice the product $xy$ where $y=f(x)$.

This property seems valid for the mentioned question, as well as $y=a/x$ and several other forms that I tried. However, I cannot prove or disprove.

Can we prove or disprove the property? In what conditions is the property valid?

答:
In another answer from your link it is proven that:
$$
A=\frac{(4+x^2)^2}{4x}\ne2x(4-x^2).
$$


问:
How to solve this second-order ODE?
what technique should I use to solve this $v'' e^{v'} -2v=0$ analytically.

Here is what I did:

$v''e^{v'}=2v$ integrate both sides wrt x to get $e^{v'}=v^2 +c$
then I took the natural logarithm to both sides, this gives:

$v'=\ln(v^2 +c)$ integrate again by parts this gives: $v=x \ln(v^2+c) - \int \frac{2v}{v^2 +c} dx = x \ln(v^2+c) - \ln(v^2 +c) + c_1$

答:
Using shorthand you have
\begin{equation}
v''e^{v'} = 2 v.
\end{equation}
Multiply through by $v'$ to get
\begin{equation}
v''e^{v'}v' = 2 vv'.
\end{equation}
Integrate both sides:
\begin{align*}
\int v'' e^{v'} v' dt &= \int 2 v v' dt \\
\int e^{v'} v' dv' &= \int 2 v dv.
\end{align*}
Use integration by parts to compute the first integral on the left; you can do the final one directly. That’ll get you to a first order equation.

Title: How to solve this second-order ODE?


问:
Difficulty with question about countable product of copies of $\mathbb{N}$
I had a lot of trouble trying to figure this problem out. Can someone explain how to properly do this?

Problem:Assume that $\mathbb{N}$ has the usual order. Let $\mathbb{N}^{\omega}$ denote the Cartesian product of a countable number of copies of the space $\mathbb{N}$. It can be endowed with the dictionary order in a natural way. Show that $\mathbb{N}^{\omega}$ with the dictionary order topology is uncountable, is not well ordered, and any set that does not have a least element does have a limit point.

Edit:I realized I showed a misunderstanding for the first part of the problem since for some reason I assumed each $s_{i,j} \in \{1,...,9\}$ instead of $\mathbb{N}$.

Attmempt:Suppose that $\{S_1,S_2,...\}$ is an enumeration of the elements of $\mathbb{N}^{\omega}$.Where $f:\mathbb{N} \rightarrow \mathbb{N}^{\omega}$ given by $f(n)=S_n$, is bijective.We will produce an element in $\mathbb{N}^{\omega}$ that is not on this list. Then there is no surjection from $\mathbb{N}$ onto $\mathbb{N}^{\omega}$, and our assumption that $f$ is bijective is false.
A list of elements in the enumeration is given by:
$$S_1=(s_{1,1},s_{1,2},...)$$
$$S_2=(s_{2,1},s_{2,2},...)$$
$$\vdots$$
$$S_n=(s_{n,1},s_{n,2},...)$$
$$\vdots$$
Where $s_{i,j} \in \mathbb{N}$ for all $i,j \in \mathbb{N}$.
Consider the element $A=(a_1,a_2,...) \in \mathbb{N}^{\omega}$ such that $a_k=7$ if $s_{k,k} \in \{1,2,3,4,5,6\}$, and $a_k=3$ if $s_{k,k} \in \{7,8,9\}$. $f$ is bijective so $A \in \mathbb{N}^{\omega}$ must appear on the list. However, the $k$'th component of $a_k$ of $A$ and $s_{k,k}$ of $S_k$ differ by between $4$ and $7$, so $A$ does not occur on this list and $\mathbb{N}^{\omega}$ is uncountable.
We will construct a set of elements that has no least element.  Let $S=\{S_1,S_2,...\}$ be the infinite set with elements given by:
$$S_1=(1,2,3,4,5,6,7,...)$$
$$S_2=(1,1,2,3,4,5,6,...)$$
$$S_3=(1,1,1,2,3,4,5,...)$$
$$S_4=(1,1,1,1,2,3,4,...)$$
$$S_5=(1,1,1,1,1,2,3,...)$$
$$\vdots$$
Then the succesor of every element in the set is less than it's predecessor in the lexicographical ordering. Since there is no smallest element in this nonempty subset of $\mathbb{N}^{\omega}$,$\mathbb{N}^{\omega}$ is not well ordered. Let $S$ be any subset of elements in $\mathbb{N}^{\omega}$ that does not have a limit point.

Note for the part below $(\downarrow$).I ran out of time, and had to turn in but knew it was wrong. I felt like it was wrong, because for a decreasing sequence of elements (in the dictionary order) must have reach a point where all the components are identical, up to a point where, the the components become eventually constant, to a point. However I had difficulty expressing this mathematically.

Any subset $S$ of $\mathbb{N}^{\omega}$ without a minimum contains a sequence of elements of the form $(a_1,a_2,a_3,...)$, where $a_i=a_j$ for $i \neq j$ holds for all terms after some index $t$ for all indices $n$ with $1\leq t \leq n<k$, and $a_k>a_n$. Consider the element in $\mathbb{N}^{\omega}$ having each entry the element $a_n \in \mathbb{N}$ after $a_t$ . Then any open set in $\mathbb{N}^{\omega}$ containing this element, contains elements in $S$. Since $S$ has no minimum there is an element $(a_1,a_2,...) \in S$ where $a_{t+1}=\dots=a_n$ for each natural number $n$, so that any open interval containing the element with all entries $a_n$ must contain an element of this form.

答:
Suppose the sequence $S$ below of elements in $\mathbb{N}^{\omega}$ (dictionary order) has no least element. We’ll show that $S$ has a limit point.

$\begin{gather}
S_1=(s_{11},s_{12},\dots)\\
S_2=(s_{21},s_{22},\dots)\\
\vdots\\
S_n=(s_{n1},s_{n2},\dots)\\
\vdots\\
\end{gather}
$

Then it contains a strictly decreasing subsequence $T = S_{i_1},S_{i_2},\dots$, which can be constructed as follows.

Choose $i_1=1$. Now $S_{i_1}$ is not the least element of $S$, so there is a smallest index $i_2>i_1$ for which $S_{i_2}<S_{i_1}$. $\color{red}{\text{Note}}$ that by choosing $i_2$ as the smallest such index, $S_{i_2}$ is then smaller than every term preceding it in all of $S$.

But while $S_{i_2}$ is the least element of $S_1,S_2,\dots,S_{i_2}$, it is not the least element of $S$, because $S$ has no least element. Choose the smallest $i_3>i_2$ for which $S_{i_3}<S_{i_2}$. As $\color{red}{\text{noted}}$, the terms of $S$ smaller than $S_{i_2}$ do not precede $S_{i_2}$, so there is such an $i_3>i_2$.

Proceeding inductively, we have the following decreasing sequence with no least element, which is a subsequence of $S$.

$\begin{gather}
T_1=(t_{11},t_{12},\dots)\\
T_2=(t_{21},t_{22},\dots)\\
\vdots\\
T_n=(t_{n1},t_{n2},\dots)\\
\vdots\\
\end{gather}
$

It suffices to show that $T$ has a limit point in $\mathbb{N}^{\omega}$, because any limit point of $T$ is a limit point of its supersequence $S$.

The sequence $T$ is decreasing in dictionary order, so for each $i$, the sequence of integers in the $j$th positions, $t_{1j},t_{2j},\dots$, is decreasing for any $j$. The natural numbers are well-ordered, so $t_{1j},t_{2j},\dots$ is eventually constant and has a limit $t_j=\lim_{i\to\infty}t_{ij}$.

Then $\lim_{k\to\infty}T_k= (t_1,t_2,\dots)$.


问:
Factor $y^3-32y+64$
I only remember learning how to factor 2nd degree polynomials, but I don't have a straight method into factoring something like $y^3-32y+64$ into $(y-4)(y^2+4y-16)$.

Would appreciate some help into figuring out a way to do so for general problems like these, thanks in advance!

答:
HINT

The possible rational roots of such polynomial belong to the set $\{\pm1,\pm2,\pm4,\pm8,\pm16,\pm32\pm64\}$.

After some few attempts, we find out that $y = 4$ is a root.

Based on such result, we can manipulate such expression as follows
\begin{align*}
y^{3} - 32y + 64 & = (y^{3} - 16y) - (16y - 64)\\\\
& = y(y^{2} - 16) - 16(y-4)\\\\
& = y(y+4)(y-4) - 16(y-4)\\\\
& = [y(y+4) - 16](y-4)\\\\
& = (y^{2} + 4y - 16)(y-4)
\end{align*}

Can you take it from here?

Title: Factor $y^3-32y+64$


问:
What is the higher dimensional analogy of $x^3, x^4$, etc.?
We know that a paraboloid, $x_1^2 + x_2^2 = x^T x$ is the generalization of a quadratic $x^2$.

Is there some function similar to $x^Tx$ that can be used to represent cubic, quartic, quintic...polynomials?

I was thinking of $x x^Tx$ for $x^3$ but obviously that doesn't work. Or perhaps $x_1^3 + x_2^3$ is the natural generalization that you simply cannot be put into a dot product between two vectors.

答:
What you want is a special case of a homogenous function (that's functions satisfying $f(\lambda x)=\lambda^k f(x)$ for some $\lambda$, just like $x^k$ in 1d). In particular, you want those homogenous functions which are generated by a symmetric multilinear map: Let $M_k:V^k\longrightarrow W$ be a multilinear map, where $V,W$ are vector spaces. Then you want those functions $f:V\longrightarrow W$ of the form $f(x)=M_k(\underbrace{x,\dots,x}_{k~\textrm{times}})$. Such functions will always be homogenous, like powers are in 1d. Such functions come up in the generalized Taylor's theorem, for instance, similar to what Hyperplane mentions in their answer. These kinds of homogenous functions are also special because they allow us to express polynomials in multiple variables: For every polynomial $P\in\mathbb R[X_1,\dots,X_n]$ of degree $m$, there are unique symmetric multilinear maps $M_k:\underbrace{\mathbb R^n\times\dots\times\mathbb R^n}_{k~\textrm{times}}\longrightarrow\mathbb R$  for $k=0,\dots,m$ such that
$$P(X)=M_0+M_1(X)+\dots+M_m(\underbrace{X,\dots,X}_{m~\textrm{times}}),$$
where $X:=(X_1,\dots,X_n)$. And that's what we're really doing with integer powers of $x$ most of the time, isn't it?

If you want to have it more abstractly, you can use the symmetric $k$-th power $S^k(V)$ of $V$, which is similar to the tensor product (a quotient space of the tensor product, to be precise), with a universal property uniquely suited to describe the above maps: The symmetric power $S^k(V)$ is the unique (up to unique isomorphism) vector space for which there exists a symmetric multilinear map $\vee :V^k\longrightarrow S^k(V),~(v_1,\dots,v_k)\mapsto v_1\vee\dots\vee v_k$, called the symmetric product, with the property that every other symmetric multilinear map $m:V^k\longrightarrow W$ can be written uniquely as $m(v_1,\dots,v_k)=\tilde m(v_1\vee\dots\vee v_k)$, where $\tilde m:S^k(V)\longrightarrow W$ is linear. So you can abstractly describe "integer powers of $x$" in multiple dimensions as linear maps whose domain is the symmetric power $S^k(V)$.

To get slightly more down-to-earth, a description of how the symmetric power looks: It's essentially made up of linear combinations of elements of the form $v_1\vee\dots\vee v_k$, where
$$v_1\vee\dots\vee v_i\vee\dots\vee v_j\vee\dots\vee v_k=v_1\vee\dots\vee v_j\vee\dots\vee v_i\vee\dots\vee v_k$$
and
$$v_1\vee\dots (v_i+\lambda w_i)\vee\dots\vee v_k=v_1\vee\dots v_i\vee\dots\vee v_k+\lambda(v_1\vee\dots w_i\vee\dots\vee v_k)$$
meaning that the $\vee$ is symmetric and linear. Essentially like the tensor product, which is made up of linear combinations of elements of the form $v_1\otimes\dots\otimes v_k$, but $\otimes$ is only linear, not symmetric.


问:
Is this equivalent to NP?
If we had a question Q(n) for which there is an alogorithm A(n) that can solve a problem in polynomial time P(n)  and can be checked in logarithmic time Log(n).

Then we could define a new set of questions $$Q'(n)=Q(2^n)$$ which then has an algorithm that can solve it in non-polynomial time but can be checked in polynomial time.

So $Q(n)$ is in P and $Q'(n)$ is in NP. If this was the case then $P\neq NP$.

Therefor is it correct to say a proof of $P\neq NP$ would be solved if there was a set of questions which could be solved in polynomial time but checked in logarithmic time?

答:
Even throwing on the implicit sharpness conditions (you presumablywant $Q$ to be solvable in polynomial time but not in logarithmic time), this isn't enough.

The issue is that $Q'$ only "samples" a small fraction of $Q$: maybe, even though general inputs to $Q$ take polynomially-long to solve, whenever I input a power of $2$ the answer happens to be very easy to figure out! (E.g. maybe the answer to $Q(n)$ is always "no" if $n$ is a power of $2$ - then $Q'$ is really quite simple, regardless of how complicated $Q$ itself is.)

To make this work, you would need to define $Q'$ so that it grabs "appropriately complicated" $Q$-instances. Moreover, the way we pick these $Q$-instances would itself have to be simple, or we wouldn't be able to conclude $Q'\in\mathsf{NP}$. It's not at all clear how to do any of that.

Title: Is this equivalent to NP?


答: From your parametrization, (x,y,z) = (1,1,-1) + t(1, 1/2, 1).  Meanwhile the axis of the cylinder is (0,0,0) + s(1,1,1).  You want to find the point of closest approach between those two lines.  One way is to take the cross product of the direction vectors of the lines to get (-1/2, 0, 1/2).  The plane has to be perpendicular to that, so -1/2x + 1/2z = d, and using (1,1,-1) we find d = -1.  The formula for the distance from a point to a plane is $D = \frac {|ax_1 + by_1 + cz_1 + d|}{{(a^2 + b^2 + c^2)}^{1/2}} $ and you can use any point in the other plane, such as 0,0,0 because the cylinder axis goes through the origin.

Title: Linear algebra cylinder problem

Body:
We have a plane. On that plane there is a line $p$:
$$p:x-1=2y-2=z+1$$
This plane also touches a cylinder with axis $x=y=z$. Find the equation of the plane and the diameter of the cylinder.

What I tried:
I solved for parameter $t$ from the line:
$$t = x-1 \implies x = t+1$$

$$t = 2y-2 \implies y = \frac{t+2}{2}$$
$$t = z+1\implies z = t-1$$

We also know that the point $T(1,1,-1)$ is on the line thus it also needs to be on the plane.
Then Ive put the parametrically defined coordinates of the line into the equation of the plane.
$$ax+by+cz = a(t+1)+b(\frac{t+2}{2})+z(t-1)$$

I do not know how to continue.

The solution is $x-z = 2$ which is the equation of the plane and the radius of the cylinder is $\sqrt{2}$


答: Say you have the cubic polynomial $ax^3+bx^2+cx+d$, start by checking all the integer factors of $d$ divided by the integer factors of $a$. In your case, $a=1$ and $d=64$.. When you do so, $y=4$ will satisfy the polynomial. Once you have one factor, $y-4$, do $(y^3-32y+64)$ divided by $(y-4)$ by long division to get the quadratic factor $y^2+4y-16$. Factor the quadratic to get the other factors, if there are any more rational ones.

Title: Factor $y^3-32y+64$

Body:
I only remember learning how to factor 2nd degree polynomials, but I don't have a straight method into factoring something like $y^3-32y+64$ into $(y-4)(y^2+4y-16)$.

Would appreciate some help into figuring out a way to do so for general problems like these, thanks in advance!


答:
I disagree with the approaches stated in the other responses which I have just skimmed.  There is no stipulation that the roots are rational or that the 3rd degree polynomial will bend to normal attempts at factoring.

In my opinion, for a new student, there is only one question:  What is the tried and true method for attacking problems of this type?

Answer: The Cubic Equation.

Addendum

Per trailing comment/request:

Use the Wikipedia article to attack the equation

$y^3 - 32y + 64 = 0.$

The first thing to notice is that the polynomial constitutes a depressed cubic, so the place to start is the "Cardano's formula" section, where

$p = -32$ and $q = 64$,

Next, following the algorithm,

$4p^3 + 27q^2 = -114688 < 0$

which indicates that there are three real roots.

At this point, the easiest approach is to then turn to the
"Trigonometric and hyperbolic solutions" section.

Within this section, the first thing to do is to label the  angle $\theta$ as the angle whose cosine is

$\frac{3q}{2p}\sqrt{\frac{-3}{p}} =
\frac{192}{-64}\sqrt{\frac{-3}{-32}} = \frac{-3}{4}\sqrt{\frac{3}{2}}.$

This gives $\theta \approx 156.716283^\circ.$

Taking $\alpha = \theta/3 - (k \times 120^\circ) ~: ~k \in \{0,1,2\}$

yields the three intermediate values of

$\cos(52.23875609^\circ) = 0.612372435 = r_1$

$\cos(-67.7612439^\circ) = 0.378466979 = r_2$

$\cos(-187.761243^\circ) = -0.99083941 = r_3$.

The next step is to double check for math errors

noting that the formula is

$t = 2\sqrt{\frac{-p}{3}} \times r = 8\sqrt{\frac{2}{3}} \times r$.

Thus, $t_1 = 8\sqrt{\frac{2}{3}} \times 0.612372435 = 3.999999995.$

Actually plugging in $t_1 = 4$ confirms that $4$ is a root of the equation, so the rest of the math can be dispensed with, since the equation now routinely factors into a quadratic equation, for the remaining two roots.

At this point, someone fairly new to cubic equations show draw the following conclusions:

1.

While it is unclear, in the depressed cubic, what will happen when $4p^3 + 27q^2 \geq 0$, one of the lessons to be learned when $4p^3 + 27q^2 < 0$ is that apparently, there is no easy way to calculate $\cos(\theta/3)$ when the computation of $\cos(\theta)$ is known. It is probably not a good idea to chase your tail via the formula

$\cos(3x) = 4\cos^3(x) - 3\cos(x)$.

2.

You can work with angles in degrees or radians.  Either way, carry all of the decimal places afforded by your calculator.

3.

In general, you probably won't be asked to solve a cubic equation that is not routinely factorable by methods indicated by the other responses.  However, the wikipedia article is the industrial strength approach, when you are having trouble with other approaches.

I emphasize that the OP could routinely have used a calculator, walked down the same path that I did, and then solved the cubic using nothing more than wits.

In conclusion, it is not a bad idea to learn other methods.  Where I disagree with the other approaches is in priorities.  I would require that the algebra student be able to wade through the wikipedia article, following instructions, without drowning.  I see no reason for an algebra student to regard the article as onerous.

It is merely time consuming.

Title: Factor $y^3-32y+64$


问:
Where to start when learning math (again)?
I have a few questions I hope you can help me answer.

First, I'll introduce myself. I'm a finance undergraduate student in Australia, but I'm originally from Norway. Throughout school I always loved math, but I ended up studying finance. The last year or so I have started to realise that I should have done Computer Science or Engineering in stead, as I would like to see myself in a quant role when I finish studying. Even so, I've decided to finish this finance degree.

The last month or so I've started programming in C++, and refreshing up on my maths knowledge (Khan Academy, algebra, precalculus). My problem is I don't know where to go from here, and what order I should be learning the different branches of math.

I've picked up a book on linear algebra today; "Linear algebra and its applications" by Gilbert Strang. Do you think this is an alright place to start? Or do I start with basic calculus? And where can I go next? Differential equations?

I hope some of you can help me in the right direction. Thanks.

答:
It is simply because there is not enough space for me to write in comment. So here we go.

As a finance major, there are many potential options for you. What math you should or ought learn really depends on what direction you would want to go in the future.
In financial industry(so to speak, altough I don't really feel it is an industry), there are many job scopes heavily rely on mathematics.
To name a few,

quantitative analysts, a.k.a. quants that many mathematicians end up
with.

risk analyst, as far as I know, many of my mates doing this kind of work now.

programmer, every industry needs programmers, so does finance.

However the math background for those jobs differs from each other. For the first two, you need more 'theoretical knowledge' of math with the specific emphasis on analysis (real analysis, stochastic analysis, modern probability, etc.), but in contrast, the math for programmers leans to numerical analysis and (theoretical) computer science.

To sum up, I want to address here is that to study math in a 'pragmatic' way, it's better to know what direction to go in the first place. However, on the other hand, you would be only benifited from understanding math in the finance field, so why not learn as much as you can.

And in the end I humbly suggest two subjects worth studying thoroughly: analysis and numerics. the fellows on math.SE will offer you many recommendations on study materials and the study procedure. My opinion would be nothing more than redundancy, at best.
Unfortunately, I failed to train myself in undergraduate, and now it takes me enormous time to catch up, only referred to analysis.

I wish you good luck with the study of mathematics with notice that finance industry is open to people with different background (engineering, physics, mathematics, ecnomics...).


问:
Non sober spaces, products and Alexandroff topology
According to the Scholium 1.1.4 in Part C of Peter's Johnstone ,,Sketeches of the Elephant'' the free frame $F(A)$ over a set $A$ can be realized as $\mathcal{O}(X)$ for suitable topological space $X$. The space $X$ can be constructed as $KA^{op}$ (the family of finite subsets of $A$ with the inverse order) with the Alexandroff topology (where a set is open iff it is lower set) but also $X$ can be realized as $S^A$ where $S$ is the Sierpiński space. It is known that two topological spaces $X,Y$ are homeomorphic iff the frames $\mathcal{O}(X)$ and $\mathcal{O}(Y)$ are isomorphic provided these two spaces are sober. Since the cardinality of $KA$ is less than that of $S^A$ these two spaces clearly are not homeomorphic. Therefore one of them can not be sober. Somewhere I have read that the product of sober spaces is again sober.

Could someone point me some reference for this result?

If it is so, then $KA^{op}$ cannot be sober:

However I would also like to see a direct proof of the fact that $KA^{op}$ is not sober.

答: To prove that $KA^{op}$ is not sober (for $A$ infinite), just observe that $KA^{op}$ is irreducible: any two basic open sets have nonempty intersection, since you can take the union of the corresponding finite subsets of $A$.  But $KA^{op}$ does not have a generic point, since there is no largest finite subset of $A$.


问:
Suppose $g(u) = \int_{2}^{u} \sqrt{1 + t^3} \ dt$. Find the value of $(g^{-1})'(0)$ if it exists.
I want some verification to see if I am doing this question right. Here is the question, and my attempt:

Suppose $g(u) = \int_{2}^{u} \sqrt{1 + t^3} \ dt$. Find the value of $(g^{-1})'(0)$ if it exists.

So first I found the derivative of $g(u)$, which was $$g'(u) = \sqrt{1 + u^3}$$

Then I let $v = g'(u)$. The domain of $g'$ is $u\in[-1, \infty)$ and the range of $g'$ is $v\in[0,\infty)$. Now solve for $u$, so
\begin{align*}
v &= \sqrt{1 + u^3} \\
v^2 &= 1 + u^3 \\
u^3 &= v^2 - 1 \\
u &= \sqrt[3]{v^2 - 1}
\end{align*}

So the inverse is $u = \sqrt[3]{v^2 - 1}$, or $(g^{-1})'(v) = \sqrt[3]{v^2 - 1}$. The domain of $(g^{-1})'$ is $v\in[0,\infty)$, and the range for $(g^{-1})'$ is $u\in[-1,\infty)$.

Now find $(g^{-1})'(0)$, so sub $v = 0$.
\begin{align*}
(g^{-1})'(v) &= \sqrt[3]{v^2 - 1} \\
(g^{-1})'(0) &= \sqrt[3]{0^2 - 1} \\
(g^{-1})'(0) &= \sqrt[3]{-1} \\
(g^{-1})'(0) &= -1 \\
\end{align*}

Since the value $-1$ is within the range, this value exists. Therefore, the value of $(g^-1)'(0)$ is $-1$.

This is my attempt, not sure if it is correct...Would like some suggestions or advice, if it is off somewhere.

答:
You're not doing it correctly, you're calculating $(g')^{-1}$ instead of $(g^{-1})'$. To do it correctly, you need the formula
$$ (g^{-1})'(x) = \frac{1}{g'(g^{-1}(x))}$$
Specifically
$$ (g^{-1})'(0) = \frac{1}{g'(g^{-1}(0))}$$
You properly calculated $$g'(u) = \sqrt{1+u^3} $$
so you just need to find $g^{-1}(0)$ which is such a value $u$ for which $g(u) = 0$, that is
$$ 0 = \int_{2}^u\sqrt{1+t^3}dt$$
even without solving the integral it can be noticed that this equation is satisfied for $u=2$. We have then

$$ (g^{-1})'(0) = \frac{1}{g'(g^{-1}(0))} = \frac{1}{g'(2)} = \frac{1}{\sqrt{1+2^3}} = \frac13$$


答:
That's much too "sophisticated" for me!

To write v = [5, −1] in terms of b1 = [1, 1] and b2 = [1, −1] means to write [5, -1]= a[1, 1]+ b[1, -1].  We can write this as
[5, -1]= [a, a]+ [b, -b]= [a+ b, a- b].

So we want a+ b= 5 and a- b= -1.
Adding the two equation eliminates b: 2a= 4 so a= 2.
Then 2+ b= 5 so b= 3.

[5, -1]= 2[1, 1]+ 3[1, -1].

Title: Given vectors v = [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

Body:
Given vectors v= [5, −1], b1 = [1, 1], and b2 = [1, −1] all written in the standard basis, what is v in the basis defined by b1 and b2?

The solution is [2, 3], but I need to know how.

v in the same basis as b1 and b2 is [5,-1]

but if v has to be defined in terms of b1 and b2, then

vb = 2b1, 3b2 = [2, 3] because this vector v is now being shown with b1
and b2 as new basis vectors.

I understand b1, but get confused how projection b2 gets calculated the way it does.

The solution provided completed projection b2 as follows:

Proj. b2 = $\frac{v.b2}{|b2|^2}$ = $\frac{(5x1)+(-1x-1)}{-1^2 + (-1^2)}$ = $\frac{6}{2}$ = 3

How does b2 [1, -1] become [-1, -1] in the calculation?

And even still, shouldn't that equal -2, not 2? Giving answer -3, not 3?

I seem to calculate  1^2, + (-1^2) = 0, and arrive at 6/0, which obviously isn't right. Where am I going wrong?


问:
Show that for any real constants a and b, where b>0, (n + a)^b = θ(n^b)
This problem is from "Introduction to Algorithms" Cormen 3rd edition. It's problem 3.1-2 from the book. I have ready many many websites, including this one, on how to solve this problem. My motivation is that I'm a grad student and taking "Introduction to Algorithms". I don't know how to solve this problem in a way that makes sense to me. I could loosely copy some elses work and hopefully not get docked on grading, but I want to get a doctorate in computer science after my master's degree. I'm not cutting corners, taking less classes to get straight A's and thoroughly understand everything in all my classes.

Here is the problem: Show that for any real constants a and b, where b>0, (n + a)^b = θ(n^b)

The way to set it up is using the definition of theta asymptotic notation, which is: θ(g(n)) = {f(n): there exists positive constants c1, c2 and n0 such that
0 <= c1g(n) <= f(n) <= c2g(n) for all n >= n0}.

So initially setting up the inequality is:
0 <= c1*(n^b) <= (n + a)^b <= c2*(n^b)

I don't know how to get from this point to the answer, which is c1=(1/2) and c2=2. I already have seen the answers to the first six pages that come up on Google by typing in the problem question. One of them was from this website. I thoroughly read all the solutions and some are done a lot different (or seem that way on the surface).

I need a detailed explanation. Here is an example. Changing (n+a) to (n + |a|). Why use the absolute value? Because a negative value is possible for a when n > 0? Why? Show examples. Explain each step in depth. I don't remember everything from every math class I took. I did read the appendices for the Algorithms book and read chapter 3 all the way through. So, yes I did study how to manipulate equations recently. I did my part and still don't know how to solve this problem all the way through making sense the entire way and can defend each step cogently.

Can anyone help?

答:
\begin{align*}
(n + a)^b ≤ (n + |a|)^b & \text{ where } n > 0\\
          ≤ (n + n)^b & \text{ for } n ≥ |a|\\
          = (2n)^b\\
          = c1 · n^b & \text{ where } c1 = 2^b
\end{align*}

Thus
\begin{align*}
(n + a)^b = Ω(n^b)\\
(n + a)^b ≥ (n − |a|) b& \text{ where } n > 0\\
          ≥ (c′_2n)^b & \text{ for } c′_2 = 1/2 \text{ where }n ≥ 2|a|\\
\end{align*}
as  $n/2 ≤ n − |a|$, for $n ≥ 2|a|$.

Thus
$$(n + a)^b = \Omega(n^b)$$

The result follows from 1 and 2 with $c_1 = 2^b$, $c_2 = 2^{−b}$, and $n_0 ≥ 2|a|$.


问:
Proving that two curves are orthogonal
Let $α(s)=(x(s)),y(s))$ be a regular plane curve that is parameterized by arc length, and let $n(s)$ be its normal vector. Consider the family of curves:

$β(s,r)=α(s)+rn(s),−ϵ≤r≤ϵ$

I need to prove that if $s_0$ and $r_0$ are constants, then $β(s,r_0)$ and $β(s_0,r)$ are orthogonal.

So far, I've tried to show the dot product of $β(s,r_0)$ and $β(s_0,r)$ is equal to $0$, but  I don't see where to go from here. Any hints?

答:
For constant $r=r_0$, the tangent vector to $\beta(s,r_0)$ is
$$
T_1(s)=\frac{\partial\beta}{\partial s}=\alpha'(s)+r_0n'(s).
$$
For fixed $s=s_0$ the tangent vector to $\beta(s_0,r)$ is$$
T_2(r)=\frac{\partial\beta}{\partial r}=n(s).
$$
Their dot product when $r=r_0$ and $s=s_0$ is
$$
T_1\cdot T_2=\alpha'(s_0)\cdot n(s_0)+r_0n(s_0)\cdot n'(s_0)
$$
The first product above is zero because $\alpha'(s)$ is tangent to the curve, and the second is zero because $n(s)$ is a unit vector.

Title: Proving that two curves are orthogonal


问:
Addition of subspaces
Prove or provide a counterexample

Let $V_1, V_2, W$ be subspaces of the vector space $V$, such that $V_1+W=V_2+W$

Then can we conclude that $V_1=V_2$?

答:
What have you tried? Think about the spaces $V_1$ spanned by $(1,0,0)$ and $(0,1,0)$ and $V_2$ spanned by $(1,0,0)$ and $(0,0,1)$. Consider $W$ spanned by $(0,1,1)$.

Then $V_1 + W = V_2 + W = \mathbb{R}^3$. Are $V_1$ and $V_2$ equal as vector subspaces?

Title: Addition of subspaces


问:
$A \in \mathbb{R}^{2 \times 2}$, $A(I_2-A)=I_2 \Rightarrow A^2-I_2$ is invertible
I've been asked to either prove or disprove that for a 2 by 2 matrix $A$ and the 2 by 2 identity matrix $I$, if $A(I-A) = I$ then $A^2-I$ is invertible.

To detail what I've attempted so far, if necessary:

I started by trying to find a counterexample to no avail, so I tried to prove it; but that also didn't quite work out. The things I did show (assuming no silly mistakes were made) are:

$A$ is invertible, and $A^{-1}=I-A$, which results directly from $A(I-A)=I$ by the definition of matrix invertibility (so $I-A$ is also invertible).

$A^2 = A-I = -A^{-1}$ through left distributivity.

$A^3 = -I$, stemming directly from #2 by right-multiplying both sides by $A$.

$A^2-I = A-2I = (A+I)(A-I)$, using #2 (left side to middle) and left & right distributivity (right side to left side).

As for #4, which feels significant because it involves the expression I'm investigating, I already know that $(A-I) = A^2$ which is invertible (because $A$ is, and also because of #2 above), so I could "reduce" the problem to proving that $A+I$ is invertible because then $(A^2-I)\cdot((-A)\cdot(A+I)^{-1})=I$ and we're done; but I couldn't find a way to do that, either.

Would greatly appreciate some assistance or a nice hint :)

答: If $(A+I)x=0$ then $A^{2}x=A(-x)=-Ax=x$ and  $A(I-A)x=Ax-A^{2}x=-x-x=-2x$. By hypothesis this implies that $x=0$ so the kernel of $A+I$ is $\{0\}$. Hence $A+I$ is invertible. Also $det (A(I-A))=1$ so $det (A) \neq 0$ and $det (I-A) \neq 0$. It follows now that $A-I$ and $A+I$ are both invertible. Hence $A^{2}-I=(A-I)(A+I)$ is also invertible.


问:
A question regarding the Riemann sphere
According to Wikipedia, in the complex plane, an open set, say $X$, is simply connected if both $X$ and its complement are connected on the Riemann sphere.

Consider the complement of the closed unit disk, this set is open, and both the complement and the set are connected (at least that's what I would assume); however, this set isn't simply connected.

My question, is it possible to construct the complement of the closed unit disk using two disjoint open sets, i.e., is $\{\infty\}$ considered an open set?

Help me to understand, using this definition, why the set is not simply connected, and perhaps the significance of using the Riemann sphere in defining simple connectedness.

答: The set is simply connected in the Riemann sphere. Any contour around the hole can be deformed continuously into a contour around the point at infinity and you can collapse the latter into a point (the point at infinity, say).</p>

Title: A question regarding the Riemann sphere


问:
challenging digit sum problem
I have an interesting problem having to do with digit sum! but the problem is challenging for a novice like me.  I have copied the interesting problem in the lines under.

Consider a number $N$ which is created by concatenating the numbers $1$ through $\underbrace{99\ldots99}_{k\ 9\text{'s}}$ inclusive. For example, if $k=2$, then $N=1234567891011\dots9899$. Find a formula for the digit sum of $N$ in terms of $k\ge1$.

this is very sophisticated problem for a novice like me but i like the challenge! i will be detailing the strategy i am thinking of in the following lines!

so it is going from $1$ to $99...999$ so the number of 1 and 9 will be the same! same with any pair that can be taken! so hear me out here.... we can count the number of 1's in N! now i will call this x! the answer will be $(1+2+3+4+5+6+7+8+9)x=45x$
now the problem is to find x! this is a complicated issue that i cannot seem to get anywhere with! i am asking for everyone's help on this problem! please enter in your thoughts on how to continue.

答:
The sum of the numbers $0-9$ is $45$. So if $k=1$, the digit sum is $45$.

Now say $k=2$. The numbers $0-9$ will cycle ten times as ones digits. So that sum is $450$. As tens digits, the tens digit $0$ will be repeated ten times (although this will not show in the actual number) then $1$ is repeated ten times, and so on until the numbers $0-9$ are repeated ten times each, the same as the ones digits. Total: $900$.

We notice a pattern: The ones digits will cycle as many times as $10^{k-1}$. The tens digit will do the same, in a different order. This pattern continues. But each time, the sum is $45$.

So the sum will be $45k\cdot{10^{k-1}}$.

Title: challenging digit sum problem


问:
Proof that a Markov matrix, $M$, has$ |\det(M)|=1$ iff $M$ is a permutation matrix
I've read in a number of places (eg here) that if a matrix $M$ is Markov (aka stochastic), then $|\det(M)|\leq 1$, with equality $|\det(M)|=1\iff$ $M$ is a permutation matrix. It is fairly easy to show the first part of this result, based on the eigenvalues of $M$ being of magnitude 1 or less, but I'm struggling with the second part.

If anyone could provide a proof (or reference) of this result, I would be much obliged.

答:
If we denote entries in the matrix $M$ in $k$ row by $a_{k,1},a_{k,2},\ldots,a_{k,n}$ then $a_{k,i} \ge 0$ and $\sum_{i=1}^{i=n}a_{k,i} = 1$ so $a_{k,i} \le1$.

By Hadamard's inequality $$\det(M) \le \prod_{i=1}^{i=n}\sum_{k=1}^{k=n} a_{k,i}^{2}$$
This can be $1$ only if $a_{k,i} = 1$ for any $k$. In this case $M$ is a permutation matrix.


问:
$\pm 1$ combination of binomial coefficients that gives $0$
Let $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$ be the binomial coefficient where $n\ge 1$, $0\le k\le n$. I am wondering if the following is true

Let $n\ge 1$ be integers and for any $0\le k\le n$, $\lambda_k\in\mathbb{N}$. Then
$$\sum_{k=0}^n \binom{n}{k}(-1)^{\lambda_k}=0$$ if and only if $\lambda_k\equiv k\mod 2$ or $\lambda_k\equiv k+1\mod 2$.

In non-mathematical terms, the only way you can make a $\pm 1$-combination of binomial coefficients to be $0$ is that you arrange the $+1$ and $-1$ alternatively.

The "if" part follows directly from binomial expansion of $$0=(1-1)^n=\sum_{k=0}^n\binom{n}{k}(-1)^k.$$

Body:
Here is a somewhat less trivial counterexample than the ones I put in the comments: Let $n=14$, let the exponents be $1,-1,1,-1,-1,1,-1,-1,1,1,1,-1,1,-1,1$.

In numbers, this is $$1-14+91-364-1001+2002-3003-3432+3003+2002+1001-364+91-14+1=0$$


问:
Sigmoid function with a longer, straighter middle
Can I modify the following function $y = 1 / (1 + e^{-x})$ such that I can make the curve straighter in the middle?

I would like the function to resemble a straight line until it approaches $0$ or $1$.

Here is an example diagram, the red line is desired:

答:
Alternate Answer: Change the Question

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\eqa}{\approx}$
$\newcommand{\abs}[1]{{\left\lvert #1 \right\rvert}}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\brlr}[1]{\left.#1\right|}$
$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\intcc} [2]  {{\left[#1:#2\right]}}$
$\newcommand{\intoo} [2]  {{\left(#1:#2\right)}}$
$\newcommand{\intoc} [2]  {{\left(#1:#2\right]}}$
$\newcommand{\intco} [2]  {{\left[#1:#2\right)}}$
$\newcommand{\ff}{\mathrm{f}}$
$\newcommand{\fg}{\mathrm{g}}$
$\newcommand{\fphi}{\mathrm{\phi}}$
$\newcommand{\dx}{\mathrm{dx}}$
$\newcommand{\du}{\mathrm{du}}$
$\newcommand{\dv}{\mathrm{dv}}$

The original question was in essence how to modify the function $y\eqd\frac{1}{1+e^{-4mx}}$ such that it's slope $m$ about $x=0$ was approximately maintained for a greater interval $x\in\intcc{-a}{a}$. Previously it was shown that this game is not winnable under the rules of the game specified by the equation. So, given that there is no solution, of course one solution is to change the question.

To find a function $\ff(x)$ that more or less satisfies the original requirements, we could start by listing what the requirements are and/or think they should be and/or what would help make life interesting:
$$\begin{align*}
    \lim_{x\to-\infty}\ff(x) &= 0
       && \text{(constraint 1)}
  \\\lim_{x\to+\infty}\ff(x) &= 1
       && \text{(constraint 2)}
  \\\ff(0) &= \frac{1}{2}
       && \text{(constraint 3)}
   \\\forall x\in\R\,,\ff(x) &> 0
       &&  \text{(constraint 4)}
       && \text{(positive)}
   \\\forall x\in\R\,,\ff(x) &< 1
       &&  \text{(constraint 5)}
       && \text{(bounded)}
   \\x_1<x_2\implies \ff(x_1)&<\ff(x_2)
       && \text{(constraint 6)}
       && \text{(strictly monotonically increasing)}
   \\\forall \abs{x}<0.45\,,\deriv{}{x}\ff(x) &\eqa 1
       &&  \text{(constraint 7)}
   \\\forall |x|>0.55\,,\deriv{}{x}\ff(x) &\text{ is "very small"}
       &&  \text{(constraint 8)}
   \\\ff(x)&\in C^{\infty}
       &&  \text{(constraint 9)}
       && \text{(Smooth---continuous in all derivatives)}
\end{align*}$$
Since so much of what we want in $\ff(x)$ is constrained by the derivative
$\fphi(x)\eqd\deriv{}{x}\ff(x)$, we could start by defining $\fphi(x)$ and then integrating $\fphi(x)$ to recover $\ff(x)$,
as in $\ff(x)=\int_{-\infty}^x\fphi(u)\du$.
To do this, we can look at the constraints and see what that implies ($\implies$) about $\fphi(x)$:
\begin{align*}
    \text{(constraint 7)} &\implies &\fphi(x) &= 1    && \text{for $\abs{x}<0.45$}
  \\\text{(constraint 8)} &\implies &\fphi(x) &\eqa 0 && \text{for $\abs{x}>0.55$}
  \\\text{(constraints 2, 5)} &\implies &\int_{-\infty}^{\infty}\fphi(x)\dx &=1
  \\\text{(constraints 4, 6)} &\implies &\fphi(x) &> 0 && \forall x\in\R
  \\\text{(constraint 1)} &\implies &\lim_{x\to-\infty}\fphi(x) &=0
  \\\text{(constraint 3)} &\implies &\fphi(-x)&=\fphi(+x) && \text{(symmetric about $x=0$)}
\end{align*}

The (constraint 9) should fall out naturally from the method, using integration,
which tends to smooth out discontinuous functions into smooth ones
(as reflected in the legacy of the Fourier Expansion
in some ways trumping the Taylor Expansion).

Finding such a $\fphi(x)$ is a challenge in and of itself.
But in looking for a sigmoid function $\ff(x)$, we may want to start with a $\fphi(x)$ that is itself a function of a sigmoid,
but with perhaps better promise of finding its anti-derivative in the literature (with, say, a $\tanh$ function or something).
One sigmoid function that is similar to $\sigma(x)=\frac{1}{1+e^{-4mx}}=\frac{e^{4mx}}{1+e^{4mx}}$
is
\begin{align*}
  \fg(x) &= \frac{1}{2} \brs{ 1+\tanh(bx) }
          = \frac{1}{2} \brs{ 1+\frac{e^{bx}-e^{-bx}}{e^{bx}+e^{-bx}} }
          = \frac{1}{2} \brs{ 1+\frac{1-e^{-2bx}}{1+e^{-2bx}} }
          = \frac{1}{2} \brs{ 1+\frac{2-1-e^{-2bx}}{1+e^{-2bx}} }
       \\&= \frac{1}{2} \brs{ 1+2\frac{1}{1+e^{-2bx}} - \frac{1+e^{-2bx}}{1+e^{-2bx}} }
          = \frac{1}{2} \brs{ 1+2\sigma(2bx) - 1 }
          = \sigma(2bx)
\end{align*}

Well, that certainly doesn't look like the $\phi$ we are desperately looking for; but,
we could use the old trick of forming a polynomial, not of $x$, but of the function $\fg(x)$.
Say, let $y\eqd\fg(x)$ and let $\phi(x)\eqd y^3$ for example.
This is an ``old" trick because polynomials of cosines for approximation was used by
Chebyshev (Chebyshev Polynomials
where the harmonic form can be converted to polynomial form)
and in the computation of Daubechies-$p$ scaling functions (in wavelet theory).

You can experiment around with the function $\fphi(x)\eqd 1-\tanh^p(bx)$ to find values of $b$ and $p$ that give you a $\phi(x)$
that is close to the requirements listed above (welcome to use the R code below).
If you happen to pick $b=10$ and $p=6180$, then you just so happen to have picked the same one I did (amazing!):

To get $\ff(x)$ from this $\phi(x)$, simply integrate from $-\infty$ to $x$:
$$\ff(x)\eqd \int_{-\infty}^x \phi(u)\du = \int_{-\infty}^x \brs{1 - (\tanh(bu))^p}\du$$
I say "simply" in the sense that a mathematician might say it---which means it's not really that simple.
But with a little help from I.S. Gradshteyn and I.M. Ryzhik,
the integral can be solved (but maybe not how one might like it to be solved):
\begin{align}
  \boxed{\ff(x)}
    &\eqd \int_{u=-\infty}^{u=x} \phi(u)\du
  \\&\eqd \int_{u=-\infty}^{u=x} \brs{1 - \tanh^p(bu)}\du
  \\&= \int_{u=-\infty}^{u=x}1\du - \int_{u=-\infty}^{u=x} \tanh^p(bu)\du
  \\&= \int_{u=-\infty}^{u=x}1\du - \int_{u=-\infty}^{v/b=x} \tanh^p(v)\frac{1}{b}\dv
    && \text{where $v\eqd bu$ $\implies$ $\du = \frac{1}{b}\dv$}
  \\&= \int_{u=-\infty}^{u=x}1\du - \frac{1}{b}\int_{u=-\infty}^{v=bx} \tanh^{2n}(v) \dv
    && \text{where $n\eqd p/2$}
  \\&= \brlr{u}_{u=-\infty}^{u=x}
     - \frac{1}{b}\brlr{v}_{v=-\infty}^{v=bx}
     + \brlr{\frac{1}{b}\sum_{k=1}^n \frac{\tanh^{2n-2k+1}(v)}{2n-2k+1}}_{v=-\infty}^{v=bx}
    && \text{by Gradshteyn and Ryzhik page 119}
  \\&= \boxed{\frac{1}{b}\sum_{k=1}^n \frac{\tanh^{2n-2k+1}(bx)}{2n-2k+1}
     - \frac{1}{b}\sum_{k=1}^n \frac{(-1)}{2n-2k+1}}
\end{align}
And so now we should have something close to what was originally asked for:

```<code>#============================================================================
# R script file
#============================================================================
#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
#install.packages("R.utils");
#install.packages("rootSolve");
 require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 x = seq( from=-2.5, to=2.5, length=1000 )

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
linear = function(x,slope,yintercept)
{
   result = slope * x + yintercept
}

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
sigmoid_logistic = function(x,slope)
{
   b = 4 * slope
   result = 1 / (1 + exp(-b*x))
}

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
sigmoid_tanh = function(x,slope)
{
   b = 4 * slope
   result = (exp(x) - exp(-x)) / (exp(x) + exp(-x))
}

#---------------------------------------
# Logistic sigmoid to power p
#---------------------------------------
sigmoid_logp = function(x,slope,p)
{
   a = 2^(1/p)-1
   b = slope * (1 + a)^(2 + p -1) / (a * p )
   result = (1 / (1 + a*exp(-b*x)))^p
}

#---------------------------------------
# tanh^p(bx)
#---------------------------------------
#tanhp = function(x,b,p)
#{
#   result = 0.5 * (((exp(b*x) - exp(-b*x)) / (exp(b*x) + exp(-b*x)))^p + 1)
#}

#---------------------------------------
# Logistic sigmoid function
# https://books.google.com/books?id=OeUKAAAAYAAJ&pg=PA15
# https://archive.org/details/integralstable00peirrich/page/59/
# https://archive.org/details/integralstable00peirrich/page/81/
#---------------------------------------
tanhp = function(x,b,p)
{
  #result = 1 - ((exp(b*x) - exp(-b*x)) / (exp(b*x) + exp(-b*x)))^p
   result = 1 - (tanh(b*x))^p
}

#---------------------------------------
# Logistic sigmoid function
# https://ia800806.us.archive.org/7/items/GradshteinI.S.RyzhikI.M.TablesOfIntegralsSeriesAndProducts/Gradshtein_I.S.%2C_Ryzhik_I.M.-Tables_of_integrals%2C_series_and_products.pdf
# 2.424 (3) page 119
#---------------------------------------
inttanhp = function(x,b,p)
{
   n = p/2
   inta = 0
   for( k in c(1:n) )
   {
     inta = inta + (-1) / (2*n-2*k+1)
   }
   inta = inta / b
   intx = 0
   for( k in c(1:n) )
   {
     intx = intx + (tanh(b*x))^(2*n-2*k+1) / (2*n-2*k+1)
   }
   intx = intx / b
   result = intx - inta
}

#---------------------------------------
# Display
#---------------------------------------
 colors = c( "blue", "red", "orange", "black", "purple","green");
#plot ( x, tanhp(x,2,1) , col=colors[1], lwd=3, type='l', xlab="x", ylab="y" )
#plot ( x, tanhp(x,1/(0.55-0.45),6180) , col=colors[1], lwd=3, type='l', xlab="x", ylab="y", ylim=c(-0.1,1.1), xlim=c(-1.5,1.5))
 plot ( x, inttanhp(x,1/(0.55-0.45),6180) , col=colors[2], lwd=3, type='l', xlab="x", ylab="y" )
#lines( x, linear(x,1,1/2)       , col=colors[6], lwd=2, type='l' )
#lines( x, linear(x,0,1/2)       , col=colors[6], lwd=2, type='l' )
 legend("topleft", legend="f(x); b=10; p=6180", col="red", lwd=3, lty=1:1)
 grid()
</code>```

Body:
Alternate Answer: Change the Question

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\eqa}{\approx}$
$\newcommand{\abs}[1]{{\left\lvert #1 \right\rvert}}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\brlr}[1]{\left.#1\right|}$
$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\intcc} [2]  {{\left[#1:#2\right]}}$
$\newcommand{\intoo} [2]  {{\left(#1:#2\right)}}$
$\newcommand{\intoc} [2]  {{\left(#1:#2\right]}}$
$\newcommand{\intco} [2]  {{\left[#1:#2\right)}}$
$\newcommand{\ff}{\mathrm{f}}$
$\newcommand{\fg}{\mathrm{g}}$
$\newcommand{\fphi}{\mathrm{\phi}}$
$\newcommand{\dx}{\mathrm{dx}}$
$\newcommand{\du}{\mathrm{du}}$
$\newcommand{\dv}{\mathrm{dv}}$

The original question was in essence how to modify the function $y\eqd\frac{1}{1+e^{-4mx}}$ such that it's slope $m$ about $x=0$ was approximately maintained for a greater interval $x\in\intcc{-a}{a}$. Previously it was shown that this game is not winnable under the rules of the game specified by the equation. So, given that there is no solution, of course one solution is to change the question.

To find a function $\ff(x)$ that more or less satisfies the original requirements, we could start by listing what the requirements are and/or think they should be and/or what would help make life interesting:
$$\begin{align*}
    \lim_{x\to-\infty}\ff(x) &= 0
       && \text{(constraint 1)}
  \\\lim_{x\to+\infty}\ff(x) &= 1
       && \text{(constraint 2)}
  \\\ff(0) &= \frac{1}{2}
       && \text{(constraint 3)}
   \\\forall x\in\R\,,\ff(x) &> 0
       &&  \text{(constraint 4)}
       && \text{(positive)}
   \\\forall x\in\R\,,\ff(x) &< 1
       &&  \text{(constraint 5)}
       && \text{(bounded)}
   \\x_1<x_2\implies \ff(x_1)&<\ff(x_2)
       && \text{(constraint 6)}
       && \text{(strictly monotonically increasing)}
   \\\forall \abs{x}<0.45\,,\deriv{}{x}\ff(x) &\eqa 1
       &&  \text{(constraint 7)}
   \\\forall |x|>0.55\,,\deriv{}{x}\ff(x) &\text{ is "very small"}
       &&  \text{(constraint 8)}
   \\\ff(x)&\in C^{\infty}
       &&  \text{(constraint 9)}
       && \text{(Smooth---continuous in all derivatives)}
\end{align*}$$
Since so much of what we want in $\ff(x)$ is constrained by the derivative
$\fphi(x)\eqd\deriv{}{x}\ff(x)$, we could start by defining $\fphi(x)$ and then integrating $\fphi(x)$ to recover $\ff(x)$,
as in $\ff(x)=\int_{-\infty}^x\fphi(u)\du$.
To do this, we can look at the constraints and see what that implies ($\implies$) about $\fphi(x)$:
\begin{align*}
    \text{(constraint 7)} &\implies &\fphi(x) &= 1    && \text{for $\abs{x}<0.45$}
  \\\text{(constraint 8)} &\implies &\fphi(x) &\eqa 0 && \text{for $\abs{x}>0.55$}
  \\\text{(constraints 2, 5)} &\implies &\int_{-\infty}^{\infty}\fphi(x)\dx &=1
  \\\text{(constraints 4, 6)} &\implies &\fphi(x) &> 0 && \forall x\in\R
  \\\text{(constraint 1)} &\implies &\lim_{x\to-\infty}\fphi(x) &=0
  \\\text{(constraint 3)} &\implies &\fphi(-x)&=\fphi(+x) && \text{(symmetric about $x=0$)}
\end{align*}

The (constraint 9) should fall out naturally from the method, using integration,
which tends to smooth out discontinuous functions into smooth ones
(as reflected in the legacy of the Fourier Expansion
in some ways trumping the Taylor Expansion).

Finding such a $\fphi(x)$ is a challenge in and of itself.
But in looking for a sigmoid function $\ff(x)$, we may want to start with a $\fphi(x)$ that is itself a function of a sigmoid,
but with perhaps better promise of finding its anti-derivative in the literature (with, say, a $\tanh$ function or something).
One sigmoid function that is similar to $\sigma(x)=\frac{1}{1+e^{-4mx}}=\frac{e^{4mx}}{1+e^{4mx}}$
is
\begin{align*}
  \fg(x) &= \frac{1}{2} \brs{ 1+\tanh(bx) }
          = \frac{1}{2} \brs{ 1+\frac{e^{bx}-e^{-bx}}{e^{bx}+e^{-bx}} }
          = \frac{1}{2} \brs{ 1+\frac{1-e^{-2bx}}{1+e^{-2bx}} }
          = \frac{1}{2} \brs{ 1+\frac{2-1-e^{-2bx}}{1+e^{-2bx}} }
       \\&= \frac{1}{2} \brs{ 1+2\frac{1}{1+e^{-2bx}} - \frac{1+e^{-2bx}}{1+e^{-2bx}} }
          = \frac{1}{2} \brs{ 1+2\sigma(2bx) - 1 }
          = \sigma(2bx)
\end{align*}

Well, that certainly doesn't look like the $\phi$ we are desperately looking for; but,
we could use the old trick of forming a polynomial, not of $x$, but of the function $\fg(x)$.
Say, let $y\eqd\fg(x)$ and let $\phi(x)\eqd y^3$ for example.
This is an ``old" trick because polynomials of cosines for approximation was used by
Chebyshev (Chebyshev Polynomials
where the harmonic form can be converted to polynomial form)
and in the computation of Daubechies-$p$ scaling functions (in wavelet theory).

You can experiment around with the function $\fphi(x)\eqd 1-\tanh^p(bx)$ to find values of $b$ and $p$ that give you a $\phi(x)$
that is close to the requirements listed above (welcome to use the R code below).
If you happen to pick $b=10$ and $p=6180$, then you just so happen to have picked the same one I did (amazing!):

To get $\ff(x)$ from this $\phi(x)$, simply integrate from $-\infty$ to $x$:
$$\ff(x)\eqd \int_{-\infty}^x \phi(u)\du = \int_{-\infty}^x \brs{1 - (\tanh(bu))^p}\du$$
I say "simply" in the sense that a mathematician might say it---which means it's not really that simple.
But with a little help from I.S. Gradshteyn and I.M. Ryzhik,
the integral can be solved (but maybe not how one might like it to be solved):
\begin{align}
  \boxed{\ff(x)}
    &\eqd \int_{u=-\infty}^{u=x} \phi(u)\du
  \\&\eqd \int_{u=-\infty}^{u=x} \brs{1 - \tanh^p(bu)}\du
  \\&= \int_{u=-\infty}^{u=x}1\du - \int_{u=-\infty}^{u=x} \tanh^p(bu)\du
  \\&= \int_{u=-\infty}^{u=x}1\du - \int_{u=-\infty}^{v/b=x} \tanh^p(v)\frac{1}{b}\dv
    && \text{where $v\eqd bu$ $\implies$ $\du = \frac{1}{b}\dv$}
  \\&= \int_{u=-\infty}^{u=x}1\du - \frac{1}{b}\int_{u=-\infty}^{v=bx} \tanh^{2n}(v) \dv
    && \text{where $n\eqd p/2$}
  \\&= \brlr{u}_{u=-\infty}^{u=x}
     - \frac{1}{b}\brlr{v}_{v=-\infty}^{v=bx}
     + \brlr{\frac{1}{b}\sum_{k=1}^n \frac{\tanh^{2n-2k+1}(v)}{2n-2k+1}}_{v=-\infty}^{v=bx}
    && \text{by Gradshteyn and Ryzhik page 119}
  \\&= \boxed{\frac{1}{b}\sum_{k=1}^n \frac{\tanh^{2n-2k+1}(bx)}{2n-2k+1}
     - \frac{1}{b}\sum_{k=1}^n \frac{(-1)}{2n-2k+1}}
\end{align}
And so now we should have something close to what was originally asked for:

#============================================================================
# R script file
#============================================================================
#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
#install.packages("R.utils");
#install.packages("rootSolve");
 require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 x = seq( from=-2.5, to=2.5, length=1000 )

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
linear = function(x,slope,yintercept)
{
   result = slope * x + yintercept
}

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
sigmoid_logistic = function(x,slope)
{
   b = 4 * slope
   result = 1 / (1 + exp(-b*x))
}

#---------------------------------------
# Logistic sigmoid function
#---------------------------------------
sigmoid_tanh = function(x,slope)
{
   b = 4 * slope
   result = (exp(x) - exp(-x)) / (exp(x) + exp(-x))
}

#---------------------------------------
# Logistic sigmoid to power p
#---------------------------------------
sigmoid_logp = function(x,slope,p)
{
   a = 2^(1/p)-1
   b = slope * (1 + a)^(2 + p -1) / (a * p )
   result = (1 / (1 + a*exp(-b*x)))^p
}

#---------------------------------------
# tanh^p(bx)
#---------------------------------------
#tanhp = function(x,b,p)
#{
#   result = 0.5 * (((exp(b*x) - exp(-b*x)) / (exp(b*x) + exp(-b*x)))^p + 1)
#}

#---------------------------------------
# Logistic sigmoid function
# https://books.google.com/books?id=OeUKAAAAYAAJ&pg=PA15
# https://archive.org/details/integralstable00peirrich/page/59/
# https://archive.org/details/integralstable00peirrich/page/81/
#---------------------------------------
tanhp = function(x,b,p)
{
  #result = 1 - ((exp(b*x) - exp(-b*x)) / (exp(b*x) + exp(-b*x)))^p
   result = 1 - (tanh(b*x))^p
}

#---------------------------------------
# Logistic sigmoid function
# https://ia800806.us.archive.org/7/items/GradshteinI.S.RyzhikI.M.TablesOfIntegralsSeriesAndProducts/Gradshtein_I.S.%2C_Ryzhik_I.M.-Tables_of_integrals%2C_series_and_products.pdf
# 2.424 (3) page 119
#---------------------------------------
inttanhp = function(x,b,p)
{
   n = p/2
   inta = 0
   for( k in c(1:n) )
   {
     inta = inta + (-1) / (2*n-2*k+1)
   }
   inta = inta / b
   intx = 0
   for( k in c(1:n) )
   {
     intx = intx + (tanh(b*x))^(2*n-2*k+1) / (2*n-2*k+1)
   }
   intx = intx / b
   result = intx - inta
}

#---------------------------------------
# Display
#---------------------------------------
 colors = c( "blue", "red", "orange", "black", "purple","green");
#plot ( x, tanhp(x,2,1) , col=colors[1], lwd=3, type='l', xlab="x", ylab="y" )
#plot ( x, tanhp(x,1/(0.55-0.45),6180) , col=colors[1], lwd=3, type='l', xlab="x", ylab="y", ylim=c(-0.1,1.1), xlim=c(-1.5,1.5))
 plot ( x, inttanhp(x,1/(0.55-0.45),6180) , col=colors[2], lwd=3, type='l', xlab="x", ylab="y" )
#lines( x, linear(x,1,1/2)       , col=colors[6], lwd=2, type='l' )
#lines( x, linear(x,0,1/2)       , col=colors[6], lwd=2, type='l' )
 legend("topleft", legend="f(x); b=10; p=6180", col="red", lwd=3, lty=1:1)
 grid()


问:
Help integrating the contour integral $\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz$ around the unit circle
Consider the integral $$\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz$$

where C is the anti-clockwise oriented unit circle and q is a complex constant.

I have no clue how to integrate this. I tried using the residue theorem but it just doesn't come out correctly. Any help would be appreciated. The reason I am stuck is because it seems like there are an infinite amount of poles at z =0, so am very confused

答:
First get rid of the $e^{1/z}$ term, make the change of variable $w = 1/z$. Then
the integral can be written
\begin{equation}
-\int_{C'}\, \frac{e^w}{w - q}\, dw
\end{equation}
and we're still on the same circle, but now going clockwise. The only pole is at $w = q$, so if $q$ is outside the unit disk the integral is zero. If it's inside, the residue is $-e^q$, so, accounting for the direction of the integration, the integral is $2\pi i e^q$.