答:
You can always add two inequalities in the the same direction: $P>Q$ and $R>S$
means $P+R >Q+S$. But you cannot subtract two inequalities in the same direction (sometime you may get right rsult and sometime wrong, so one avoids their subtraction). You are right in both the cases.

Body:
I know this is a very simple question, but I still have my doubts on this question:

Does $x<y$ and $a\le b\implies x+a<y+b$? Similarly, does $x>y$ and $a\ge b\implies x+a>y+b$? Why or why not?

Based on intuition, these seem to be true, but how would I "prove" this concretely? I'm not too sure if there are any rules that allow me to combine $<$ and $\le$(or $>$ and $\ge$) inequalities. It seems like there should be one, but I wasn't able to find any. Could someone help me on this?


问:
Find all the prime numbers $p$ which have a solution to $x^2\equiv14\bmod p$.

Find all the prime numbers $p$ which have a solution to $x^2\equiv14\bmod p$.

I found that for $p=2,5,7$ there is a solution and for $p=3$ there's no solution.

I tried using Legendre symbol because the given equation has a solution iff $(\frac{14}{p})=1 \iff (\frac{2}{p})(\frac{7}{p})=1$ but I get too many sub cases.

答:
The "second supplement to quadratic reciprocity" states that $\left(\frac2p\right)=1\iff p\equiv\pm1\bmod8$. It remains to work out $\left(\frac7p\right)$.
$$\left(\frac7p\right)\left(\frac p7\right)=(-1)^{3(p-1)/2}$$
Since $\left(\frac p7\right)=1\iff p\equiv 1,2,4\bmod7$ and the RHS is $1$ iff $p\equiv1\bmod4$, $\left(\frac7p\right)=1$ iff an even number of $p\equiv 1,2,4\bmod7$ and $p\equiv1\bmod4$ are true. Then $\left(\frac{14}p\right)=1$ iff an odd number of $p\equiv 1,2,4\bmod7$, $p\equiv1\bmod4$, $p\equiv\pm1\bmod8$ are true.

The last two conditions can be combined rather easily: an even number of them are true iff $p\equiv1,3\bmod8$. Thus $\left(\frac{14}p\right)=1$ iff an even number of $p\equiv1,2,4\bmod7$ and $p\equiv1,3\bmod8$ are true. By the CRT, this combines into $$p\equiv1,5,9,11,13,25,31,35,43,45,47,51\bmod56$$

All along we have ignored the exceptional cases, but the question already provides them. So we have the final result:
$$\left(\frac{14}p\right)=1\iff p\equiv1,2,5,7,9,11,13,25,31,35,43,45,47,51\bmod56$$

ContentLicense: CC BY-SA 4.0


问:
How many ways to arrange 10 people between a group of size 4 and a group of size 6?
I was trying to understand the answer given here which was very similar but I couldn't understand it well. Based on the answer in the link I'm thinking it would be

(10 nPr 4) * (6 nPr 6) = 3628800 different arrangements

I wouldn't be dividing between 2! since the groups are not of equal sizes and does not need to consider the reordering. Is this correct? I'm also not sure whether it would need to be solved with permutations or combinations.

答:
Since orders do not matter within each group, you should be using combinations rather than permutations

so you could have written $\,^{10}C_4 \,^6C_6$ or ${10 \choose 4}{6\choose 6}$, which as Brian M. Scott says in a comment is $210$

This differs from your answer by a factor of $4!\times 6!$.

To see this with a smaller example, if you were splitting three people into a group of two and a group of one, would you say $3$ possibilities or $6$?


问:
Euler's constant $e$ correctness of definition
Recently I started considering derivative (calculating it by definition) of an exponential function $a^x$. It requires couple of additional steps, one of them is to define constant $e$ as the only real number that satisfies:

$$ 1 = \lim_{h\to 0} \frac{a^h-1}{h} $$

And here I am stuck. Why such definition is correct? First we must prove that fot every real number $r$ exists such $a$ that

$$ r = \lim_{h\to 0} \frac{a^h-1}{h} $$

And I do not know how to proceed with it. Any suggestions? Concept of proof?

答: The online textbook CLP-1 Differential Calculus (by Feldman/Rechnitzer/Yeager) develops the derivative of exponential functions in exactly this way; they include a proof that there is a unique such number in Section 2.7 (especially Section 2.7.2).</p>


问:
What's the meaning of the unit bivector i?
I'm reading the Oersted Medal Lecture by David Hestenes to improve my understanding of Geometric Algebra and its applications in Physics. I understand he does not start from a mathematical "clean slate", but I don't care for that. I want to understand what he's saying and what I can do with this geometric algebra.

On page 10 he introduces the unit bivector i. I understand (I think) what unit vectors are: multiply by a scalar and get a scaled directional line. But a bivector is a(n oriented) parallellogram (plane). So if I multiply the unit bivector i with a scalar, I get a scaled parallellogram?

答:
The unit bivector represents the 2D subspaces.  In 2D Euclidean GA, there are 4 coordinates:

1 scalar coordinate

2 vector coordinates

1 bivector coordinate

A "vector" is then (s, e1, e2, e1^e2).

The unit bivector is frequently drawn as a parallelogram, but that's just a visualization aid.  It conceptually more closely resembles a directed area where the sign indicates direction.

OwnerDisplayName: rubenvb

ContentLicense: CC BY-SA 3.0


问:
function equality
The following really gets confusing:

As a simple example:

Is the function

$f(x) = 1$ the same function as $f(x) = \frac{x - 1}{x-1}$?

$f(x) = \frac{(x-1)(x+1)}{x-1}$ the same function as $f(x)= x + 1$?

Basically: does resolving and simplifying a function expression alter the function 'identity/essence'? I think it does because the domain changes ($1$ is put into the domain) and the only thing that matters are the exact values in the domain set and the range set, an algebraic expression is just a description.

So is it correct to say that one expression is an approximation of another function expressions? But that they are completely different functions, they just overlap?

Is there a bigger philosophy to be found somewhere?

答: By definition, it is necessary that for two functions to be the same, both functions must share the same domain as well as codomain. Your examples are not the same since $1$ is in the domain of one, but not the other. Although, you might be able to say they are "the same thing" in the sense that you are able take the restriction of $f$ to $\mathbb R - \{1\}$ so that $f|_{\mathbb R - \{1\}} = f_1$ where $f_{1}$ represents the function undefined at $x=1$.

Title: function equality


答:
I think that this depends on what you're working with. Sometimes if you get an algebraic expression and it makes sense to simplify it and then you have the function that you will continue working with. However, if we are being strictly formal, you are right. The functions

$$x \mapsto 1 \text{ and } x \mapsto \frac{x-1}{x-1}$$ are different because their domains are different. The first function has $\mathbb{R}$ as its domain, whereas the second one has $\mathbb{R} \setminus \{1 \}$. Formally a function is a triplet $(A, G, B)$, where $A$ and $B$ are the domain and codomain of the function and $G$ is what we call its graph, by which we do not refer to its graphical representation as a 2D plot in the $(x,y)$-plane, but rather its set of ordered pairs of arguments and function values. Formal function equality requires the domain, codomain and graphs to be equal for two functions to be equal.

In elementary calculus however, there are very few situations where this actually has a significance. When you study complex analysis, these kinds of questions become more cogent and the answer is also more interesting.

Title: function equality

Body:
The following really gets confusing:

As a simple example:

Is the function

$f(x) = 1$ the same function as $f(x) = \frac{x - 1}{x-1}$?

$f(x) = \frac{(x-1)(x+1)}{x-1}$ the same function as $f(x)= x + 1$?

Basically: does resolving and simplifying a function expression alter the function 'identity/essence'? I think it does because the domain changes ($1$ is put into the domain) and the only thing that matters are the exact values in the domain set and the range set, an algebraic expression is just a description.

So is it correct to say that one expression is an approximation of another function expressions? But that they are completely different functions, they just overlap?

Is there a bigger philosophy to be found somewhere?


问:
A(z1) B(z2) C(z3) D(z4) are four points on a circle |z| = 1 where z represents a complex number
A(z 1 )  B(z 2 )  C(z 3 )  D(z 4 ) are four points on a circle |z| = 1  where z represents a complex number . Given that line AB is parallel to line CD .  Prove that z 1  . z 2  =  z 3 .  z 4

MY ATTEMPT :
I tried using the condition of concyclic points that cross product of the numbers should be real , rotation taking center as origin and condition of parallel lines in complex form  but I didn't get any break through from that

Please help me

答:
My attempt would be as follows.
Given four complex numbers $z_i = e^{i \theta_i}$, $ i = 1,2,3,4$.

Further, we know the lines through $z_1$, $z_2$ and $z_3$, $z_4$ are parallel.
This translates to one of the two conditions

$$\frac{\theta_1 +\theta_2}{2} = \frac{\theta_3 +\theta_4}{2}$$ or $$\frac{\theta_1 +\theta_2}{2} = \frac{\theta_3 +\theta_4}{2} + \pi $$

They reflect the fact that for the lines to be parallel, their normals through the origin must coincide. Normals directions  are computed as the average of the arguments (angles $\theta_i$) of the complex numbers through which the lines are defined.

The claim that the products

$$ z_1 z_2 = e^{i (\theta_1 + \theta_2)}$$  and
$$ z_3 z_4 = e^{i (\theta_3 + \theta_4)}$$

follows immediately if the first conditions applies.
If the second condition above applies, one notes that points on the complex circle are defined modulo a $2 \pi$ rotation.


答:
$f(x) = 1$ and $g(x) = \frac{x - 1}{x - 1}$ are NOT same functions at all. They do not share the same domain set. The function $f$ can take values from all of $\mathbb{R}$ , whereas $g$ takes values only from $\mathbb{R} \setminus \{1\}$ .

When we we say that two functions are equal (or, identical), we mean that they take values from the same set, and they take two same value at each point in the domain set. So, in your question, if it's specified that $\text{domain}(f) = \mathbb{R}$ , then $f \not\equiv g$ .

However, in order to make $f \equiv g'$ forcefully, we may define $g' : \mathbb{R} \to \{1\}$ like : $$g'(x) = \begin{cases} g(x)&\quad\quad\text{if}~ x \neq 1\\
1 &\quad\quad\text{if}~ x = 1\\
\end{cases}$$
In that case, we can surely say that $f \equiv g'$ .

Title: function equality


问:
Why can't cosets in the polynomial ring $\Bbb Z[x]$ be written as a polynomial of lesser degree?
I am having a hard time understanding the reason that we cannot express cosets of the polynomial ring $\Bbb Z[x]$ with lower degree.

For example, I am given a function ${f(x) = 4x^3+2}$ in ${\mathbb Z[x]}$ and am asked why it is that I cannot express the cosets ${\mathbb Z[x]/(4x^3+2)}$    as    $${a + bx +cx^2 + (f(x))}$$

Usually when working with field coefficients of polynomials, such as ${\mathbb R}$ or ${\mathbb Q}$ I'd have the "condition" that ${x^3 = -1/2}$. But why can I not have the condition that ${4x^3 = -2}$ since those are still in ${\mathbb Z}$?

My professor has alluded to the fact that we cannot have cosets in ${\mathbb Z[x]}$ of this form, but he didn't elaborate as to why, and the provided text doesn't have anything about this topic that I can find.

答:
This is because long division of polynomials only works when the leading coefficient of $f$ is a unit.

Let me elaborate a bit. Consider any nonzero polynomial $f\in\mathbb{Z}[X]$. For two polynomials $P,R\in\mathbb{Z}[X]$, we have  $P+(f)=R+(f)\iff P=R+Qf$ for some $Q\in\mathbb{Z}[X]$.

Now assume further that given $P$, there exists an $R$ such that $\deg(R)<\deg(f)$ and $P+(f)=R+(f),$ so we would have $P=Qf+R$ with $\deg(R)<\deg(f)$. Viewing this equality in $\mathbb{Q}[X]$, by uniqueness of quotient and remainder in Euclidean division of polynomials, $Q$ and $R$ are necessarily respectively the quotient and remainder of the long division of $P$ by $f$.

However, if the leading coefficient of $f$ is not $\pm 1 $ (which are the only units of $\mathbb{Z}$), it may happen that these $Q$ and $R$ do not belong to $\mathbb{Z}[X]$.

In this case, $R$ will not exist.


问:
Solve the system of partial differential equations for 4 variable function
Let $G=G(x,y,u,v)$ and let consider the following system of equation
$$
\begin{cases}
\displaystyle \frac{\partial^2 G}{\partial x \partial y}=uv \,G,\\
\displaystyle x\frac{\partial G}{ \partial y}-y \frac{\partial G}{ \partial x}=v\frac{\partial G}{ \partial u}- u \frac{\partial G}{ \partial v}
\end{cases}
$$

By experiments I have found one solution $$G=e^{xu+yv}$$  but I think that must be one more solution.
Any ideas how to find all independent solutions?

PS. The first equation has solution

$$
G=C {{ e}^{{\frac {vyu+x \left( {\it f} \left( u,v \right)  \right) ^{
2}}{{\it f} \left( u,v \right) }}}}
$$
where  $f$ is arbitrary function.

答:
One can analytically solve the second PDE :
$$x\frac{\partial G}{ \partial y}-y \frac{\partial G}{ \partial x}=v\frac{\partial G}{ \partial u}- u \frac{\partial G}{ \partial v}$$
$$xG_y-yG_x-vG_u+uG_v=0$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dy}{x}=\frac{dx}{-y}=\frac{du}{-v}=\frac{dv}{u}=\frac{dG}{0}$$
A first characteristic equation comes from solving $\frac{dy}{x}=\frac{dx}{-y}$
$$x^2+y^2=c_1$$
A second characteristic equation comes from solving $\frac{du}{-v}=\frac{dv}{u}$
$$u^2+v^2=c_2$$
A third characteristic equation comes from solving $\frac{dx}{y}=\frac{du}{v}=\frac{dx}{\pm\sqrt{c_1-x^2}}=\frac{du}{\pm\sqrt{c_2-u^2}}$
$$\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{v}{u})=c_3$$
The general solution of the PDE $\quad G=\Phi(c_1,c_2,c_3)\quad$ is :
$$\boxed{G(x,y,u,v)=\Phi\left((x^2+y^2)\:,\:(u^2+v^2)\:,\:\left(\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{v}{u})\right) \right)}$$
$\Phi(X,Y,Z)$ is an arbitrary function of three variables :
$\quad\begin{cases}
X=x^2+y^2\\
Y=u^2+y^2\\
Z=\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{v}{u})
\end{cases}$

Alternatively this result can be obtained in polar coordinates instead of in Cartesian.
$$\begin{cases}
x=\rho\cos{\alpha}\\
y=\rho\sin{\alpha}
\end{cases}\quad
\begin{cases}
u=\mu\cos{\beta}\\
v=\mu\sin{\beta}
\end{cases}$$
After expressing $G_x$ and $G_y$ in terms of $G_\rho$ and $G_\alpha$ as well as $G_u$ and $G_v$ in terms of $G_\mu$ and $G_\beta$ the PDE simplifies a lot and becomes very simple :
$$\frac{\partial G}{\partial \alpha}+\frac{\partial G}{\partial \beta}=0$$
which general solution is :
$$\boxed{G(\rho,\mu,\alpha,\beta)=\Psi(\rho,\mu,\alpha-\beta)}$$
$\psi$ is an arbitrary function of three variables.

Back to Cartesian :
$$G(x,y,u,v)=\Psi\left(\sqrt{x^2+y^2}\:,\:\sqrt{u^2+v^2}\:,\:\tan^{-1}(\frac{y}{x})-\tan^{-1}(\frac{v}{u}) \right)$$
This is equivalent to the above result since $\Phi$ and $\Psi$ are arbitrary functions.

The above calculus leads to an infinity many solutions for the second PDE. Then one think to put the above general solution $G(x,y,u,v)$ into the first PDE
$\quad\frac{\partial^2 G}{\partial x \partial y}=uv \,G\quad$ in order to determine the function $\Phi$ and find the whole set of solutions of the coupled PDEs.

That is a different kettle of fish. So the problem is far to be fully solved. Hopping that the above calculus help.

NOTE :

The solution $G=c\:e^{xu+yv}$ corresponds to the particular case of
$$\Psi(\rho,\mu,\alpha-\beta)=c\:e^{\,\rho\,\mu\,\cos(\alpha-\beta)}.$$

$$G=c\:e^{-(xu+yv)}\quad\text{is another solution.}$$
As well as linear combinations :
$$G=c\:\cosh(xu+yv)$$
$$G=c\:\sinh(xu+yv)$$
$c$ : arbitrary constants.

More solutions can be found on the form :
$$G(x,y,u,v)=f(u^2+v^2)e^{xu+yv}+g(u^2+v^2)e^{-(xu+yv)}$$
where $f$ and $g$ are arbitrary functions.


答:
The segments $AB\parallel CD$ have a common median bisector which intersects the unit circle at two points; call one of them $z_*$. There are then two angles $\alpha$,  $\beta$ such that
$$z_1=e^{i\alpha}z_*\>,\quad z_2=e^{-i\alpha}z_*\>,\quad z_3=e^{i\beta}z_*\>,\quad z_4=e^{-i\beta}z_*\ .$$
It follows that $z_1z_2=z_*^2=z_3z_4$.

(By the way: It is not necessary to give each of your points two different names.)

Title: A(z1) B(z2) C(z3) D(z4) are four points on a circle |z| = 1 where z represents a complex number

Body:
A(z 1 )  B(z 2 )  C(z 3 )  D(z 4 ) are four points on a circle |z| = 1  where z represents a complex number . Given that line AB is parallel to line CD .  Prove that z 1  . z 2  =  z 3 .  z 4

MY ATTEMPT :
I tried using the condition of concyclic points that cross product of the numbers should be real , rotation taking center as origin and condition of parallel lines in complex form  but I didn't get any break through from that

Please help me


问:
Runge Kutta Fehlberg goes crazy when using adaptive h in python
So basically what i'm trying to do here is try to define a function that integrates for runge kutta. The most "physical" aspects of it i have already controlled for (it's not a problem of the values of the functions , or the physical parameters (initial conditions, etc), since i have controlled for those and they don't change the result) .
The problem i'm solvin is an ODE of second order (harmonic oscilator),

```<code>x''-μ(1-x^2 )x'+x=0 + CI x(0)=2, x'(0)=0, for t in [0,25], and μ=4
</code>```

which i have separated into two first-order ODES;

```<code>μ*(1-x**2)*v-x = v'
v = x'
</code>```

and we have inital conditions x0,v0,t0 and a interval on which to integrate t0,tf, and a step size h with which to integrate.
ε is my error tolerance.
The error I have is that when i activate the adaptivetoggle, so that i should use an adaptive h, x just goes to a value and v goes to 0, instead of oscilating like a harmonic oscilator should. I suspect the problem is only about that bit of code, because when i deactivate the adaptive toggle , everything runs just fine.
I am not sure of what is happening, my values should be oscilating, and small, and instead the error just goes to 0 ( it shouldn't, i think ) , v does too and x tends to something instead of oscilating.

The code i'm running is:

```<code>def rk452D(v, f, x0, v0, t0, tf, ε, h, adaptivetoggle):

x = x0; t = t0
valuesx = []
valuesv = []
while t<tf :
#We define the runge kutta functions on which to iterate while t is in [t0,tf], and they are of the kind (t,x,v)
    f1v = f(t,               x,                                                                                v0                                                                                )
    f1  = v(t,               x,                                                                                v0                                                                                )

    f2v = f(t + (1/4)*h,     x + (1/4)*f1,                                                                         v0 + (1/4)*f1v                                                                           )
    f2  = v(t + (1/4)*h,     x + (1/4)*f1,                                                                         v0 + (1/4)*f1v                                                                           )

    f3v = f(t + (3/8)*h,     x + (3/32)*f1      + (9/32)*f2  ,                                                     v0 + (3/32)*f1v      + (9/32)*f2v                                                        )
    f3  = v(t + (3/8)*h,     x + (3/32)*f1      + (9/32)*f2  ,                                                     v0 + (3/32)*f1v      + (9/32)*f2v                                                        )

    f4v = f(t + (12/13)*h,   x + (1932/2197)*f1 - (7200/2197)*f2 + (7296/2197)*f3,                                 v0 + (1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v                                 )
    f4  = v(t + (12/13)*h,   x + (1932/2197)*f1 - (7200/2197)*f2 + (7296/2197)*f3,                                 v0 + (1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v                                 )

    f5v = f(t + h,           x + (439/216)*f1   - 8*f2           + (3680/513)*f3  - (845/4104)*f4,                 v0 + (439/216)*f1v   - 8*f2v           + (3680/513)*f3v  - (845/4104)*f4v                )
    f5  = v(t + h,           x + (439/216)*f1   - 8*f2           + (3680/513)*f3  - (845/4104)*f4,                 v0 + (439/216)*f1v   - 8*f2v           + (3680/513)*f3v  - (845/4104)*f4v                )

    f6v = f(t + h/2,         x - (8/27)*f1      + 2*f2           - (3544/2565)*f3 + (1859/4104)*f4 - (11/40)*f5,   v0 - (8/27)*f1v      + 2*f2v           - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v )
    f6  = v(t + h/2,         x - (8/27)*f1      + 2*f2           - (3544/2565)*f3 + (1859/4104)*f4 - (11/40)*f5,   v0 - (8/27)*f1v      + 2*f2v           - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v )

   #Now we calculate the positions and velocities, for the fourth order runge kutta aproximation. Commented we have the fifth order approxiation results, which we use to estimate the error ( Error = abs(order5approximation-order4approximation)
    x4 = x + h*((25/216)*f1  + (1408/2565)*f3   + (2197/4104)*f4     -(1/5)*f5              )
    #x5 = x + h*((16/135)*f1  + (6656/12825)*f3  + (28561/56430)*f4   -(9/50)*f5  +(2/55)*f6 )

    v4 = v0 + h*((25/216)*f1v + (1408/2565)*f3v  + (2197/4104)*f4v    -(1/5)*f5v             )
    #v5 = v0 + h*((16/135)*f1v + (6656/12825)*f3v + (28561/56430)*f4v  -(9/50)*f5v +(2/55)*f6v)

  #If we want to use an adaptive h for our calculations,
    if adaptivetoggle == True :
       #We calculate error in x using fs, eror in v using fvs, and we take the smaller hnew of the two
       Errorx = abs((1/360)*f1  - (128/4275)*f3  - (2197/75240)*f4  + (1/50)*f5  + (2/55)*f6  )
       Errorv = abs((1/360)*f1v - (128/4275)*f3v - (2197/75240)*f4v + (1/50)*f5v + (2/55)*f6v )

       hnewx  = 0.9*h*((ε)/Errorx)**(0.25)
       hnewv  = 0.9*h*((ε)/Errorv)**(0.25)

       hnew = min(hnewx,hnewv)

       print(hnew)
       #if hnewx < hnewv:
        #   hnew = hnewx
       #if hnewx > hnewv:
         #  hnew = hnewv

       #After calculating hnew, we compare it to the integration step h we have used, to decide if we can keep our calculation or we have to repeat it using hnew.

       if hnew >= h :
          #we increment the loop,and take hnew for the next loop
          t += h
          h = hnew
          x = x4
          v0 = v4
          valuesx.append(x4)
          valuesv.append(v4)

       elif hnew < h:
          h  = hnew #we don't increment t , the loop variable, when we repeat the integration step using the new h integration step

    else :#if we don't want an adaptive h ( this works just fine)
       valuesx.append(x4)
       valuesv.append(v4)
       x = x4
       v0 = v4
       t+=h #increment the loop
return valuesx, valuesv
#Then we implement the function
#We define the two functions ( of the ODEs) to feed the RK solver
def f(t,x,v):
    return μ*(1-x**2)*v-x
def v(t,x,v):
    return v

#we feed it the parameters
μ  = 4; x0 = 2;  v0 = 0; t0 = 0; tf = 25; h  = 1; ε  = 10**-3
adaptivetoggle = True

solution = rk452D(v, f, x0, v0, t0, tf, ε, h, adaptivetoggle)
print(solution)
</code>```

答:
The error is so glaringly blatant, so blending and ubiquitous that it has become almost invisible. Straight-up not seeing the forest for the trees.

In the computation of the slopes f1, f1v, ... you do not use the step size h. You can see the effect of that if you enlarge the debug print command to also print t,h and Errorx, Errorv.

For better readability, single-point-of-failure and slightly better efficiency I'd suggest changing to the format

```<code>    ttmp = t + (12/13)*h
    xtmp = x  + h*((1932/2197)*f1  - (7200/2197)*f2  + (7296/2197)*f3 )
    vtmp = v0 + h*((1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v)
    f4v, f4 = f(ttmp, xtmp, vtmp ), v(ttmp, xtmp, vtmp )
</code>```

(or use a general vectorized version, without treating the components separately).

Then the initial step size h=1 is still too large as the controller regulates down to h=0.05, periodically raising up to h=0.6.

For faster testing, the modified slope computation is

```<code>        f1v = f(t,  x,  v0 )
        f1  = v(t,  x,  v0 )

        f2v = f(t + (1/4)*h,  x + h*(1/4)*f1,  v0 + h*(1/4)*f1v )
        f2  = v(t + (1/4)*h,  x + h*(1/4)*f1,  v0 + h*(1/4)*f1v )

        tstg = t + (3/8)*h   # stg like stage
        xstg = x  + h*((3/32)*f1  + (9/32)*f2 )
        vstg = v0 + h*((3/32)*f1v + (9/32)*f2v)
        f3v, f3 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + (12/13)*h
        xstg = x  + h*((1932/2197)*f1  - (7200/2197)*f2  + (7296/2197)*f3 )
        vstg = v0 + h*((1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v)
        f4v, f4 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + h
        xstg = x  + h*((439/216)*f1  - 8*f2  + (3680/513)*f3  - (845/4104)*f4 )
        vstg = v0 + h*((439/216)*f1v - 8*f2v + (3680/513)*f3v - (845/4104)*f4v)
        f5v, f5 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + h/2
        xstg = x  + h*(-(8/27)*f1  + 2*f2  - (3544/2565)*f3  + (1859/4104)*f4  - (11/40)*f5 )
        vstg = v0 + h*(-(8/27)*f1v + 2*f2v - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v)
        f6v, f6 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)
</code>```

One might change the acceptance condition to hnew >= 0.95*h or hnew >= 0.85*h to avoid too many step repetitions, with the factor 0.9 in the computation of hnew there is enough safety to allow that.

Body:
The error is so glaringly blatant, so blending and ubiquitous that it has become almost invisible. Straight-up not seeing the forest for the trees.

In the computation of the slopes f1, f1v, ... you do not use the step size h. You can see the effect of that if you enlarge the debug print command to also print t,h and Errorx, Errorv.

For better readability, single-point-of-failure and slightly better efficiency I'd suggest changing to the format

    ttmp = t + (12/13)*h
    xtmp = x  + h*((1932/2197)*f1  - (7200/2197)*f2  + (7296/2197)*f3 )
    vtmp = v0 + h*((1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v)
    f4v, f4 = f(ttmp, xtmp, vtmp ), v(ttmp, xtmp, vtmp )

(or use a general vectorized version, without treating the components separately).

Then the initial step size h=1 is still too large as the controller regulates down to h=0.05, periodically raising up to h=0.6.

For faster testing, the modified slope computation is

        f1v = f(t,  x,  v0 )
        f1  = v(t,  x,  v0 )

        f2v = f(t + (1/4)*h,  x + h*(1/4)*f1,  v0 + h*(1/4)*f1v )
        f2  = v(t + (1/4)*h,  x + h*(1/4)*f1,  v0 + h*(1/4)*f1v )

        tstg = t + (3/8)*h   # stg like stage
        xstg = x  + h*((3/32)*f1  + (9/32)*f2 )
        vstg = v0 + h*((3/32)*f1v + (9/32)*f2v)
        f3v, f3 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + (12/13)*h
        xstg = x  + h*((1932/2197)*f1  - (7200/2197)*f2  + (7296/2197)*f3 )
        vstg = v0 + h*((1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v)
        f4v, f4 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + h
        xstg = x  + h*((439/216)*f1  - 8*f2  + (3680/513)*f3  - (845/4104)*f4 )
        vstg = v0 + h*((439/216)*f1v - 8*f2v + (3680/513)*f3v - (845/4104)*f4v)
        f5v, f5 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

        tstg = t + h/2
        xstg = x  + h*(-(8/27)*f1  + 2*f2  - (3544/2565)*f3  + (1859/4104)*f4  - (11/40)*f5 )
        vstg = v0 + h*(-(8/27)*f1v + 2*f2v - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v)
        f6v, f6 = f(tstg, xstg,  vstg), v(tstg, xstg,  vstg)

One might change the acceptance condition to hnew >= 0.95*h or hnew >= 0.85*h to avoid too many step repetitions, with the factor 0.9 in the computation of hnew there is enough safety to allow that.


问:
Infinite fraction series
The infinite fraction $$f(n)=n+\frac{n}{n+\frac{n}{n+...}}$$ can be simplified to $$f(n)=\frac{n+\sqrt{n^2-4n}}{2}$$ However, I wanted to know if the fraction $$1+\frac{2}{3+\frac{4}{5+...}}$$ can be simplified further to get an answer using algebra or is the answer a transcendental number? On solving this in Desmos(till 25), I got 1.54149408254.
This is an image of the Desmos page.

答:
The Mathworld page on continued fraction constants gives the exact answer as
$$\frac1{\sqrt e-1}$$
which is indeed transcendental.

The value was known to Euler himself. See page 14 of this translation.

Title: Infinite fraction series


问:
Proof check for almost sure convergence
I encountered this problem from Durrett (Exercise 2.5.1):

Suppose $X_1, X_2, \dots$ are i.i.d. with $EX_i = 0$, $var(X_i) = C< \infty$. Use Kolmogorov's maximal inequality to prove that if $S_n = \sum_{i=1}^nX_i$ and $p > 1/2$, then $S_n/n^p \to 0$ a.s.

The problem says to prove it using Kolmogorov's maximal inequality, but I did not see how one would use it and I cannot really understand the solution proof to Durrett. So I came up with my own proof and I want to see if my proof is correct.

My Proof: Pick $\alpha$ such that $\alpha(2p-1) > 1$. Let the subsequence $n(m) = [m^\alpha]+1$ where $[m^\alpha]$ is the integer part of $m^\alpha$. Fix $\epsilon > 0$, then
$$
P(|S_{n(m)}| > \epsilon n(m)^p) \leq \frac{ES_{n(m)}^2}{\epsilon^2n(m)^{2p}} = \frac{C}{\epsilon^2n(m)^{2p-1}} \leq \frac{C}{\epsilon^2m^{\alpha(2p-1)}}.
$$
Since the RHS of the above is summable, by BC lemma, $S_{n(m)}/n(m)^p \to 0$ a.s.

Now since for any sequence $x_n$
$$
\frac{x_n}{n^p} \to 0 \Longleftrightarrow \frac{\max_{m\leq n}|x_m|}{n^p} \to 0,
$$
we have that
$$
\frac{\max_{k \leq n(m)}|S_{k}|}{n(m)^p} \to 0 \ a.s.
$$
Now since $\max_{k \leq n}|S_k|$ is increasing in $n$, then for $n(m) \leq n \leq n(m+1)$, we have
$$
\frac{\max_{k \leq n(m)}|S_k|}{n(m+1)^p} \leq \frac{\max_{k \leq n}|S_{k}|}{n^p} \leq \frac{\max_{k \leq n(m+1)}|S_{k}|}{n(m)^p}.
$$
Then we have
$$
\frac{\max_{k \leq n(m)}|S_k|}{n(m)^p}\frac{n(m)^p}{n(m+1)^p} \leq \frac{\max_{k \leq n}|S_{k}|}{n^p} \leq \frac{\max_{k \leq n(m+1)}|S_{k}|}{n(m+1)^p}\frac{n(m+1)^p}{n(m)^p}.
$$
Now since
$$
\frac{n(m)^p}{n(m+1)^p} = \left(\frac{[m^\alpha]+1}{[(m+1)^\alpha]+1}\right)^p \to 1
$$
and
$$
\frac{n(m+1)^p}{n(m)^p} = \left(\frac{[(m+1)^\alpha]+1}{[m^\alpha]+1}\right)^p\to 1,
$$
we can then conclude that
$$
\frac{\max_{k\leq n}|S_k|}{n^p} \to 0 \ a.s.
$$
and
$$
\frac{S_n}{n^p} \to 0 \ a.s.
$$

Is my proof correct? Thank you!

Proof Correction:
As per comment of Davide, the proof should correct the start of the argument and use Kolmogorov's maximal inequality:
$$
P(\max_{k\leq n(m)}|S_k| > \epsilon n(m)^p) \leq \frac{ES_{n(m)}^p}{\epsilon^2n(m)^{2p}} \leq \frac{C}{\epsilon^2m^{\alpha(2p-1)}}.
$$
Hence,
$$
\frac{\max_{k\leq n(m)}|S_k|}{n(m)^p} \to 0 \ a.s.
$$
The rest of the proof should follow in the same as my original proof.

答:
It seems that there is a gap in your proof: you proved that $S_{n(m)}/n(m)^p \to 0$ almost surely and the deterministic fact you mention, namely, $\frac{x_n}{n^p} \to 0 \Longleftrightarrow \frac{\max_{m\leq n}|x_m|}{n^p} \to 0$, would only allow to deduce that $\max_{1\leqslant k\leqslant m}\lvert S_{n(k)}\rvert/n(m)^p\to 0$ in probability. In other words, this gives no information about $S_j$ when $j$ is not of the form $n(k)$ for some integer $k$.

Nevertheless, the proof can be saved by using Kolmogorov's maximal inequality in order to bound $\mathbb P\left(\max_{1\leqslant k\leqslant m(n)}|S_{k}| > \epsilon n(m)^p\right)$ by a similar term as the one obtained in the opening post.

Title: Proof check for almost sure convergence


问:
Find $\lim_{x\to \infty}{(\frac{\sqrt{x^2-x}}{x} + \frac{1}{4}\sin{\frac{2}{x})^{x^2+\sin{3x}}}}$
I need to find
$$
\lim_{x\to \infty}\left(\frac{\sqrt{x^2-x}}{x} + \frac{1}{4}\sin\frac{2}{x}\right)^{x^2+\sin(3x)}
$$

So I do this:
$$
\lim_{x\to \infty}{{\left[\left(1+ \frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4x}\right)^{x}\right]}^{\frac{x^2+\sin{3x}}{x}}}=e^{{\lim_{x\to \infty}{{\frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4}}^{\frac{x^2+\sin{3x}}{x}}}}}=e^{*}
$$

$$
*={\lim_{x\to \infty}{{\frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4}}{\frac{x^2+\sin{3x}}{x}}}} = {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}{(x^2+\sin{3x})}}{4}}} \\
= {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}x^2+\sin{\frac{2}{x}}\sin{3x}}{4}}} = {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}x^2}{4}}} = \frac{1}{4} \frac{\sin{\frac{2}{x}}}{\frac{1}{x^2}}
$$

Applying L'Hopital I get that the the limit is $x$, which isn't true. Where am I wrong? Also, I'd like to avoid L'Hopital but I am unable to. Any suggestions would be appreciated.

答:
Take logarithm of the expression
$$
\lim_{x\to \infty}\left(\frac{\sqrt{x^2-x}}{x} + \frac{1}{4}\sin\frac{2}{x}\right)^{x^2+\sin(3x)}
$$
that gives
$$
(x^2+\sin 3x)\ln\left(\frac{\sqrt{x^2-x}}{x} + \frac{\sin(\frac{2}{x})}{4}\right)
= \frac{\ln\left(\frac{\sqrt{x^2-x}}{x} + \frac{\sin(\frac{2}{x})}{4}\right)}{\frac{1}{x^2+\sin 3x}}
$$
Now use L'Hospital, and remeber that what you get is the logarithm of the original limit


问:
Show |x| is not differentiable at x=0 using delta epsilon
I need to show that the absolute value of x is not differentiable at x=0 using epsilon - delta. Thank you.

My thoughts: The absolute value at 0 is where we have a sharp cusp, so it's obviously not differentiable there because that's where the sided limits do not match.  In terms of epsilon-delta, we need an epsilon that creates an instance where delta cannot be true to pose a contradiction.  I assume that an epsilon of 1/2 would work nicely for this cause, but am unsure where to get the contradiction.  If we plug in 1/2 for epsilon, 0 for x, and 1 for the other variable, c, then we end up with 1<delta and 1/2<Derivative.

答:
Note that, by definition of a derivative, for $\lvert x \rvert$ to be differentiable at $0$ the limit $$\lim_{x\to 0} \frac{\lvert x \rvert - \lvert 0 \rvert}{x - 0} = \lim_{x\to 0} \frac{\lvert x \rvert}{x}$$ must exist. Now you can apply the $\varepsilon$-$\delta$ definition to show this limit in fact doesn’t. Can you take it from here?

(If you need a hint at any point, see this answer for a full solution.)

P.S. The expression above may be written as $$\forall \varepsilon >0, \exists \delta > 0 \text{ s.t. } 0<\lvert x - 0 \rvert<\delta \implies \left\lvert \frac{\lvert x \rvert}{x} - L\right\rvert < \varepsilon $$ using $\varepsilon$-$\delta$ notation, where $L$ is the supposed limit. Please refer to my answer linked above for a detailed proof of why such $L$ could not exist.


问:
One-sample T-test p-value paradox?
Let $X$ be a random variable. According to CLT, the distribution of means of samples of $X$ converges to normal as the sample size grows.

Suppose we sample $X$ $n$ times and observe data $[x, x, x, ..., x, x+\epsilon]^T$ (that is, $n-1$ times $x$ and $x + \epsilon$ once, where $x \in \mathbb{R}$ and $\epsilon > 0$). If we compute the one sample $t$ statistic as

$t = \frac{\bar{X} - \mu}{s /\sqrt{n}}$

setting $\mu = x$ ($H_0$ is that $\mu = x$)  and compute the p-value, surprisingly enough, with increasing $n$, we observe decreasing $p$. As far as I know, the conclusion of this observation is that the larger the sample is (and therefore the more $x$'s we observe), the less likely $x$ is to be the true mean of $X$. But why? This feels counter-intuitive. Can anyone explain?

答:
Given the sample, it is easy to show that $$\bar x = x + \frac{\epsilon}{n}, \quad s^2 = \frac{\epsilon^2}{n}. $$  Then the test statistic for the hypothesis $H_0 : \mu = x$ is $$T \mid H_0 = \frac{\bar x - x}{s/\sqrt{n}} = \frac{\epsilon/n}{\epsilon/n} = 1.$$  This is independent of the sample size $n$.  Since $T$ is Student $t$-distributed with $n-1$ degrees of freedom, the limit as $n \to \infty$ is standard normal, thus $$\Pr[T > 1 \mid H_0] > \Pr[Z > 1 \mid H_0] \approx 0.158655.$$  This furnishes a strict lower bound on the $p$-value of this test for this kind of observed data, and you would never reject $H_0$ for any reasonable $\alpha$.

While it is true that $p$ decreases as an increasing function of $n$, despite the strength of evidence seemingly suggesting that the true mean is $x$, what the $p$-value is reflecting is the increased sensitivity of the test to deviation from the null mean as the sample size increases.  In other words, there are two competing effects at work here:

With more observations equaling $x$, the sample mean is tending toward the null mean.

With larger sample size, the test is more likely to reject $H_0$ for the same value of the test statistic.

Title: One-sample T-test p-value paradox?


问:
how to determine the domain of this function: $\sqrt{(\sin(x))^2 + \sin x}$?
let $x$ be in this interval: $[0, 2\pi]$. the exercise's request is to find the domain of this function:$$\sqrt{(\sin(x))^2 + \sin x}$$

Here's my attempt to solve it:

I found the domain of the square root: $$(sin (x))^2 + \sin x \geq 0$$
In this case, I've found that $$\sin x \ge 0 $$ and that: $$\sin x \ge -1 $$

in the trigonometric circle, I have found that the set of solution is only the 4th quadrant, since the sin of -1 is $3/2\pi$ (270 degree).

Is that correct?

答:
We must have

$$\sin x(\sin x+1)\ge0.$$

As $$\sin x+1\ge0$$ is always true, we seem to be left with

$$\sin x\ge 0.$$

But there is a trap*: when $$\sin x+1=0$$ the sign of $\sin x$ is irrelevant and finally $$x\in[0,\pi]\cup\left\{\frac{3\pi}2,2\pi\right\}.$$

*Credit to Buraian.

After the fact, a systematic solution is

$$\sin x>0\lor\sin x=0\lor\sin x=-1,$$

$$x\in(0,\pi)\cup\{0,\pi,2\pi\}\cup\left\{\frac\pi2,\frac{3\pi}2\right\}.$$


问:
What kind of math topics exist?

What kind of math topics exist?

The question says everything I want to know, but for more details: I enjoy studying mathematics but the problem is that I can't find any information with a summary of all math topics, collected together. I also googled this and took a look at other websites and searched this website but without success.

So if someone knows most of the topics, then please let me know them. The topics I am looking for are the ones from basics, to the university, and beyond university limit. Any comprehensive information you can provide would be useful.

Further thanks and have a nice day.

答:
You might find the "Princeton Companion to Mathematics" helpful.

http://www.amazon.com/Princeton-Companion-Mathematics-Timothy-Gowers/dp/0691118809

Title: What kind of math topics exist?


问:
What explains the bizarre behavior of the partial sums of this series?
In an attempt help answer the question of a fellow member, I recently tried to evaluate the integral

$$\int_{0}^{2}\frac{1-\cos(y^5)}{y^3}\text{ }dy$$

I figured I was unlikely to find a closed-form expression for the integral using only basic integration techniques, so I instead opted to find a series expression for it (maybe I can manipulate the series into a more familiar one?). Using the Maclaurin series of $\cos$ and some algebraic manipulation, I concluded that

\begin{align*}
\frac{1-\cos(y^5)}{y^3} &= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{y^{10n-3}}{(2n)!}\\
&= \frac{y^7}{2!}-\frac{y^{17}}{4!}+\frac{y^{27}}{6!}-\frac{y^{37}}{8!}+\cdots
\end{align*}

and thus that

\begin{align*}
\int_{0}^{2}\frac{1-\cos(y^5)}{y^3}\text{ }dy &= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2^{10n-2}}{(2n)!(10n-2)}\\
&= \frac{1}{8}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1024^n}{(2n)!(5n-1)}
\end{align*}

Not the prettiest series I've seen, but definitely not the ugliest!

Not seeing any tricks I could do to manipulate the series into one I'm familiar with, I figured the closed-form I was hoping for was out of the equation, so I settled with estimating the integral using the series I derived.

This is where things get interesting. Using Desmos for computation, I noticed that the partial sums $\{s_k\}$ quickly reached the billions in absolute value (according to the site, this first occurs at $k=11$), so I assumed I ran into a convergence issue. However, after raising the upper limit for $k$ to $50$ and varying $k$, I noticed that for $k\geq 19$, the partial sums quickly decreased in absolute value until $k=39$, where $s_k$ finally started approximating the integral rather well (the integral is approximately $0.47659$).

As you can probably guess, I'd like to know why this is happening. Most of the series I've worked with only need a few terms to approximate their sum with an error of at most $1$, even $0.1$ at times. This series isn't that much different than them, so I find it odd for it to behave so erratically and take so long to settle. Any assistance is appreciated.

答:
Here is a simpler example:

The exponential series $e^x=\sum_{k=0}^\infty{x^k\over k!}$ converges for all $x\in{\mathbb R}$. When you want to use this series to compute $e^x$ when $|x|\gg1$ you will see that the terms increase in absolute value up to a certain $k_*$ and only then start to decrease as desired. The value $k_*$ depends on the given $|x|$, and is easy to determine. In short words: We have to wait until $k\mapsto k!$ grows faster than $k\mapsto |x|^k$. This is the case when $k\gg |x|$.

In your example this effect is intensified by the ${}^5$ under the cosine.


问:
100 people choose any number of combinations from a pool of 90 options.
If 100 people can choose any number of combinations from a set of 90 options, where each option can only be selected once, but each person can select any number of options.

Say there is a product sortiment of 90 different products, [a, b, c...]

Each person can choose any combination of products, but can only choose each product once. So for example a person can choose [a] or [a, b, f, g] but not [a, a, c].

Now if 100 people has to make a selection and each selection doesn't affect other peoples selections. How many combinations does this equal?

EDIT: I forgot to add: [a, b, c] and [b, a, c] should count as the same combination.

答: Each person individually may or may not choose any option, but has to choose at least one option. Thus each person may choose in $2^{90}-1$ ways. Since the choices are independent between people, the answer is $(2^{90}-1)^{100}$.


答:
Look at the power set of the set of all options - that would have $2^{90} - 1$ elements (removing the empty set)

Now, each person can choose any one of these options, hence the total number of ways would be

$$(2^{90}-1)^{100}$$

Title: 100 people choose any number of combinations from a pool of 90 options.

Body:
If 100 people can choose any number of combinations from a set of 90 options, where each option can only be selected once, but each person can select any number of options.

Say there is a product sortiment of 90 different products, [a, b, c...]

Each person can choose any combination of products, but can only choose each product once. So for example a person can choose [a] or [a, b, f, g] but not [a, a, c].

Now if 100 people has to make a selection and each selection doesn't affect other peoples selections. How many combinations does this equal?

EDIT: I forgot to add: [a, b, c] and [b, a, c] should count as the same combination.


答:
$$(a^6+a^5+a^4+a^3+a^2+a+1)=0$$
$$\Rightarrow (a-1)(a^6+a^5+a^4+a^3+a^2+a+1) = a^7-1 = 0$$

So $a^7 = 1$ and $a^{2033} = a^{7 \cdot 290 + 3} = a^3$, and therefore $\frac{1}{a^{2033}} = \frac{1}{a^3}$.

Since $a^7 = 1$, $a = e^{2 \pi k i / 7}$, where $k =1, 2, \cdots, 6$ (note that we exclude $k = 0$ as $a \ne 1$).

Therefore, the expression equals $e^{6 \pi k i /7} + e^{-6 \pi k i / 7} = 2 \cdot \frac{1}{2} (e^{6 \pi k i /7} + e^{-6 \pi k i / 7}) = 2 \cos(6 \pi k /7)$ since $\cos x = \frac{1}{2} (e^{ix} + e^{-ix})$.

There are multiple values of this, but the principal value is when $k = 1$, which gives the value of the expression as $2 \cos(6\pi/7) = -2 \cos(\pi/7)$.

Title: How to find $a^{2033}+\frac{1}{a^{2033}}$ from the equation?

Body:
I was Junior High School Student, please explain the answer like you teach a Junior High School Student

Given :
$$P(x) = 325^6+325^5+325^4+325^3+325^2+325+2345$$

Given a number $a$, $P(a) = 2020$,
Find $$a^{2033} + \frac{1}{a^{2033}}$$

I have
$P(a) = 2020$

$325a^6+325a^5+325a^4+325a^3+325a^2+325a+2345$ = $2020$

$325(a^6+a^5+a^4+a^3+a^2+a+1)=0$

divided by $a^3$, I get :

$a^3+a^2+a+1+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=0$

$(a+\frac{1}{a})^3-3(a+\frac{1}{a})+(a+\frac{1}{a})^2-2+(a+\frac{1}{a})+1=0$

Suppose that $y=a+\frac{1}{a}$

$y^3+y^2-2y-1=0$, i'm stuck:)


问:
Linear Algebra - Another way of Proving a Basis?
If we have a Vector Space $V$, can we prove that a set $S$ of vectors $\in V$ is a basis of $V$ given that:

$S$ contains the same number of vectors as $\dim{(V)}$.

Every vector in a basis of $V$ can be written as a linear combination of the vectors in S

Example: Let $V$ be $\Bbb R_3$. The Standard Basis of $\Bbb R_3$ is $\{b_1,b_2,b_3\}=\{(1,0,0),(0,1,0),(0,0,1)\}$. Let $S$ be $\{v_1,v_2,v_3\}=\{(1,0,0),(1,1,0),(1,1,1)\}$. Then:

$$
\begin{align}
v_1 = b_1
\\
v_2 - v_1 = b_2
\\
v_3 - v_2 = b_3
\end{align}
$$
So:
$$
\begin{align}
c_1b_1+c_2b_2+c_3b_3 = (a,b,c)
\\
c_1(v_1)+c_2(v_2-v_1)+c_3(v_3-v_2) = (a,b,c)
\\
(c_1-c_2)v_1 + (c_2-c_3)v_2 + c_3v_3 = (a,b,c)
\end{align}
$$
therefore, since $\{b_1,b_2,b_3\}$ is independent (let $a = b = c = 0$) and spanning, $S$ is also independent and spanning so $S$ is a basis of $V$

If a set $S$ satisfies the before-mentioned conditions, is it a basis?

Edit: in response to Andres Caicedo, yes, $V$ is finite dimensional.

答:
Yes, every spanning set contains a basis: you just remove vectors that can be written as a linear combination of the others.  So we can remove vectors from $S$ to get a basis.  But the resulting basis must have $\dim V$ vectors and that's how many vectors $S$ has.  Therefore we removed $0$ vectors to get the basis.  The basis is $S$.

Similarly if $|S| = \dim V$ and $S$ is a linearly independent set then $S$ is a basis.


问:
Density of $X + Y$ when $X$ is discrete and $Y$ has a density

Let $X,Y$ be independent random variables where $X$ is discrete and $Y$ has a density $f$. Prove: $X + Y$ has a density.

I can not really deal with the "vague" specifications, e.g. what is the sample space? What is the measure function? What is the $\sigma$-algebra? I know, the reason is that the mathematic model does not matter in the end but for me, I have issues in understanding the assumptions.

Actually, I should real it like follows, shouldn't I?

Lez $(\Omega, \mathcal A, P), (\Omega', \mathcal A', P')$ be probability spaces and $X,Y: \Omega \to \Omega'$ be random variables with distributions

$$P^X: \mathcal A'\to[0,1], \quad A' \mapsto P(X^{-1}(A')), \qquad P^Y: \mathcal A'\to[0,1], \quad A'\mapsto P(Y^{-1}(A'))$$

where

$X$ is discrete, which means $X(\Omega)$ is finite,

$Y$ has density $f$, which means $P^Y(A') = \displaystyle\int_{A'} f(x) \, \mathrm dx$ (we only defined random variables with a density for distributions on $\mathbb R$ - so $(\Omega', \mathcal A') = (\mathbb R, \mathcal B(\mathbb R))$?)

So I have to prove that there is a function $g: \mathbb R \to \mathbb R$ with $\displaystyle P^{X+Y}(A') = \int_{A'} g(x) \, \mathrm dx$? Probably I should construct $g$, shouldn't I? But how?

I already got to $P^{X+Y}(A') = P((X+Y)^{-1}(A'))$ but at this point I'm stuck because $(X+Y)^{-1}(A') \neq X^{-1}(A') + Y^{-1}(A')$?

I'm really just very confused by all the terms. Any help would be highly appreciated.

答:
You have to take $\Omega'=\mathbb R$ and $\mathcal A'=\mathcal B$ for this to make sense.

If $x_1,x_2,..$ are the values of $X$ then $P(X+Y\leq z)=\sum_n P(Y \leq z-x_n )P(X=x_n)$. This can be writen as $\sum_n p_n\int_{-\infty} ^{z-x_n} f(t)dt $ where $p_n=P(X=x_n)$. By a change of variable this becomes $\sum_n p_n\int_{-\infty} ^{z} f(t-x_n)dt $. Thus $P(X+Y \leq z)=\int_{-\infty} ^{z} g(t)dt$ where $g(t)=\sum_n p_nf(t-x_n)$. Verify that $g$ is a non-negative measurable function whose integral equals $1$. This proves that $X+Y$ is absolutely continuous with density $g$.

I have repeatedly used the fact when the integrand is non-negative an infinite sum and an integral can be interchanged.


问:
Assymetric relation?
Hello I have a question about this task, if anyone had a similar problem it would help me.

The task is:
Is this relation asymmetric ?
$(x,y)\rho(a,b)\rightarrow x+y=a+b$

Thanks in advance !

答: Relation R is said to be asymmetric iff $~\forall p,q$ ($pRq \longrightarrow \neg (qRp))$. In our case if $(x,y)\rho(a,b)$ i.e. $x+y=a+b$, then $a+b=x+y,$ i.e. $(a,b)\rho(x,y)$. So $\rho$ is symmetric.

Title: Assymetric relation?


问:
Calculus Second Derivative with Inflection Point and Tangent Slope
Question: a function of the form $y=ax^3+bx^2+cx$ has an inflection point at $ A (1,5)$. If the slope of the tangent at $A$ is $4$, find the values of $a$,$b$, and $c$

Hi I don't quite get this question. Can someone please help? At most I have gotten done is $y''=6ax+2b$, and by plugging in the inflection point $(1,5)$ into that gives me $b=-3a$, but I think this might be wrong? Can someone help?

答:
What you've done so far is fine

Hint: At $A$, $y' = 4$

$$\implies 3a + 2b + c = 4$$

Now also, the point $(1,5)$ lies on the curve, so it must satisfy it's equation

$$5 = a + b + c$$

Along with the relation you already have, this makes 3 equations in 3 variables


问:
Connected $k-$regular graph does not have cut-edge for $k \ge 2$
This is a conjecture generalized from the result that "a connected k-regular bipartite graph does not have cut-edge."

Cut–edge is an edge whose deletion will result in a disconnected graph. I think we can prove this by arguing that every two vertex on this graph lies on a cycle. An intuition is that there is no tree like structure in such graph. If we consider a longest path that includes $v_i$ and $v_j$, $i\neq j$, since $\deg(v_i)>1$ for all vertex, the vertex at the end must connect to at least two vertices. This indicates this vertex must connect back to some vertex in this path or it is not the longest path.

How can we formalize this proof? If this conjecture is not true, can anybody give a counterexample?

答:
This is false. Consider a $3-$regular connected graph

which clearly has a cut-edge.


答: Here's a connected 3-regular graph with a cut-edge. Start with a path on vertices $1,2,3,\dots,14$. Add edges $1-3,1-7,2-5,4-6,8-14,9-11,10-13,12-14$. $7-8$ is a cut-edge.

Title: Connected $k-$regular graph does not have cut-edge for $k \ge 2$

Body:
This is a conjecture generalized from the result that "a connected k-regular bipartite graph does not have cut-edge."

Cut–edge is an edge whose deletion will result in a disconnected graph. I think we can prove this by arguing that every two vertex on this graph lies on a cycle. An intuition is that there is no tree like structure in such graph. If we consider a longest path that includes $v_i$ and $v_j$, $i\neq j$, since $\deg(v_i)>1$ for all vertex, the vertex at the end must connect to at least two vertices. This indicates this vertex must connect back to some vertex in this path or it is not the longest path.

How can we formalize this proof? If this conjecture is not true, can anybody give a counterexample?


问:
How does this hypotenuse/derivative/polynomial approximation work?
I am trying to understand an article regarding the forces of a vibrating string in terms of both longitudinal and transverse waves.

Here is the pertinent excerpt:

Equation 1 is obvious. It is just Pythagorean theory. Equation 2 is also obvious as it just converts this to derivatives. Equation 3 also makes sense - tension in a string is equal to the base tension plus the product of length deviation (extra stretch), Young's Modulus, and cross sectional area.

The part I don't get is equation 4. If you put equation 2 in equation 3 you definitely don't get equation 4.

I know equation 4 is some type of approximation, but I don't know how it works. I found another reference that seems to discuss this same approximation and they say of it: "It has been shown that by expanding both y and ξ as a series of polynomials and truncating at third order, the added force per unit length on the element in the longitudinal direction caused by the transverse displacement is given by [this equation]."

But I'm not a mathematician and I don't understand how this is represented by a series of polynomials or how they describe this approximation.

Does this make sense? Can anyone explain it in simple terms?

Thanks for any help. Very appreciated.

答:
For the sake of simplicity, let $$a = \frac{\partial \xi}{\partial x}, \quad b = \frac{\partial y}{\partial x}.$$  Then we can write $(3)$ as $$\frac{ds}{dx} = \sqrt{(a+1)^2 + b^2}.$$  The approximation in $(4)$ is then equivalent to the claim $$\sqrt{(a+1)^2 + b^2} - 1 \approx a + \frac{1}{2}b^2.$$  Is this reasonable?  Well, consider $$\left(a + \frac{1}{2} b^2 + 1\right)^2 = (a+1)^2 + (a+1)b + \frac{1}{4} b^4.$$  Although we do get the $(a+1)^2$ term, the rest doesn't seem to fit.

Is there another approach?  What about a series expansion about $b = 0$?  We get $$\sqrt{(a+1)^2 + b^2} = (a+1) + \frac{b^2}{2(a+1)} - \frac{b^4}{8(a+1)^3} + O(b^6).$$  So if $a \approx 0$, a low-order series expansion seems to be the motivation for the approximation.  But without the underlying physical context, I cannot say whether it is reasonable.


问:
Double root in Trigonometric equation?

Consider the function $ f( x) = \sin^2 x + \sin x$ in the domain $ [0,  2 \pi]$, find the subset of the domain where $f \geq 0 $

Factoring by treating as quadratic in $ \sin$:

$$ f = ( \sin x + 1 ) ( \sin x)$$

Hence, the roots are $ \sin x = \{ 0, -1 \}$, hence the roots in $x$ are $ \{ 0, \pi , 2 \pi, \frac{3 \pi}{2} \}$. Let us partition the interval using the roots :

$$ \left[ 0, \pi \right] \cup \left[ \pi, \frac{ 3 \pi}{2} \right] \cup \left[ \frac{ 3 \pi}{2} , 2 \pi \right]$$

From regular algebra, we know that a 'root' corresponds to a sign flip, hence since the function is negative in the second interval of a domain partition, it must be positive in the third ... but actually it's not! i.e: $f \leq 0 $ for $ x \in [ \frac{ 3 \pi}{2} , 2 \pi ]$ but there is a problem, there was only a single root at $ x= \frac{ 3 \pi}{2}$.

So, how did the sign not flip here even when it crossed a root? That is what property of the trigonometric function is which causes this?

答:
The property that causes this is the boundedness of the sine: there is only one solution per period of $\sin x=1$ and $\sin x=-1$, compared to two for $\sin x=a,a\in(-1,1)$.

If you plot the graphs $y=\sin x$ and $y=-1$ you will see that the latter line is tangent to the sine curve – i.e. $1+\sin x$ has only double root solutions.

Title: Double root in Trigonometric equation?


问:
Triplet of integers such that expression is prime for every prime number?
Using only elementary methods known to a middle schooler taking his first algebra course, is there a triplet of integers $(a, b, c)$ such that, for every prime number $p$, $(5a + 4)p^2 + (5b + 3)p + 2c$ is a prime number?

I know that if an even integer is a prime number, then it is $2$.

答:
No such $a,b,c$ exist.

Let $f(x)=(5a+4)x^2+(5b+3)x+2c$. Let $f(3)=q$. Then $f(kq+3)$ is a multiple of $q$ for all $k$. And by Dirichlet's Theorem $kq+3$ is prime for infinitely many $k$.

Now it's not possible to prove Dirichlet's Theorem by middle school methods, but at least the statement of Dirichlet's Theorem is quite simple. It says that if $\gcd(a,b)=1$ then there are infinitely many $n$ such that $an+b$ is prime.


问:
Isomorphism between $V$ and $U$
Let there be $V =$ $\mathbb{R}[x]$ space of all polynomials with 1 variable with
real coefficients

$U$ = { $p(x) \in\ V|p(0)=0$ }, In other words: $U$ is a subset in V of all of the polynomials:
 $a_nx^n + a_{n-1}x^{n-1} +......+ a_1x$

such that the free constant term equals to $0$

prove that $U$ and $V$ are isomorphic

MY APPROACH:
 I tried to build an injective ( one-to-one ) and
Surjective ( onto ) linear transformation / map. it is easy to see that there
is one, but it is not injective because to two polynomials, different
in the constant term, have the same output. I have tried to "fix" this
linear transformation but with no success at this moment

答: You want to define a map sending a polynomial in $V$, that is, something of the form $f(x)=c_0+c_1x+c_2x^2+\dots+c_mx^m$, to a polynomial in $U$, that is, a polynomial of the form $g(x)=a_1x+a_2x^2+a_3x^3+\dots+a_nx^n$ in a way that is injective. A way of thinking about what injectivity requires is that you don't want to lose any of the “distinctive” information that uniquely identifies the polynomial $f$. What does this “distinctive, identifying” information consist of? How may we define a polynomial which is in $U$ and which retains all of the distinctive information specific to $f$ (thus giving a $g$ which could not be the image of any other polynomial than $f$)?

Title: Isomorphism between $V$ and $U$


答: Try linear map $A:U\rightarrow V$ such that $A(a_nx^n+...+a_1x)=a_nx^{n-1}+...+a_1$.

Title: Isomorphism between $V$ and $U$

Body:
Let there be $V =$ $\mathbb{R}[x]$ space of all polynomials with 1 variable with
real coefficients

$U$ = { $p(x) \in\ V|p(0)=0$ }, In other words: $U$ is a subset in V of all of the polynomials:
 $a_nx^n + a_{n-1}x^{n-1} +......+ a_1x$

such that the free constant term equals to $0$

prove that $U$ and $V$ are isomorphic

MY APPROACH:
 I tried to build an injective ( one-to-one ) and
Surjective ( onto ) linear transformation / map. it is easy to see that there
is one, but it is not injective because to two polynomials, different
in the constant term, have the same output. I have tried to "fix" this
linear transformation but with no success at this moment


Title: Find $\lim_{x\to \infty}{(\frac{\sqrt{x^2-x}}{x} + \frac{1}{4}\sin{\frac{2}{x})^{x^2+\sin{3x}}}}$

Body:
I need to find
$$
\lim_{x\to \infty}\left(\frac{\sqrt{x^2-x}}{x} + \frac{1}{4}\sin\frac{2}{x}\right)^{x^2+\sin(3x)}
$$

So I do this:
$$
\lim_{x\to \infty}{{\left[\left(1+ \frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4x}\right)^{x}\right]}^{\frac{x^2+\sin{3x}}{x}}}=e^{{\lim_{x\to \infty}{{\frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4}}^{\frac{x^2+\sin{3x}}{x}}}}}=e^{*}
$$

$$
*={\lim_{x\to \infty}{{\frac{\sqrt{x^2-x} + x\sin{\frac{2}{x}}-4x}{4}}{\frac{x^2+\sin{3x}}{x}}}} = {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}{(x^2+\sin{3x})}}{4}}} \\
= {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}x^2+\sin{\frac{2}{x}}\sin{3x}}{4}}} = {\lim_{x\to \infty}{\frac{\sin{\frac{2}{x}}x^2}{4}}} = \frac{1}{4} \frac{\sin{\frac{2}{x}}}{\frac{1}{x^2}}
$$

Applying L'Hopital I get that the the limit is $x$, which isn't true. Where am I wrong? Also, I'd like to avoid L'Hopital but I am unable to. Any suggestions would be appreciated.

Body:
If
$$L=\lim_{x\to a} f(x)^{g(x)}\to 1^\infty,$$ then $$L=\exp[\lim_{x \to a}(g(x)(f(x)-1)]$$ $$= \exp[\lim_{x\to \infty} [(x^2+\sin 3x)((1-1/x)^{1/2}+\frac{1}{4}\sin (2/x)-1)]$$
Next use $(1+z)^k=1+kz+k(k-1)z^2/2+..., \sin z =z-z^3/6+...$ if $|z|<<1.$
$$L= \exp[\lim_{x\to \infty} [(x^2+\sin 3x) \left(1-\frac{1}{2x}-\frac{1}{8x^2}+...\frac{1}{4}(\frac{2}{x}-\frac{8}{6x^3}+..)-1\right)]$$
$$L=\exp[\lim_{x\to \infty} [(x^2+\sin 3x) \left(-\frac{1}{8x^2}+O(x^{-3}\right)]=\exp[\lim_{x\to \infty}-\frac{x^2+\sin 3x}{8x^2}]=e^{-1/8}.$$


问:
Why is the derivative of a quadratic invariant zero if the matrix is symmetric?
I was shown a proof in class, where it was said that all Runge-Kutta methods preserve the quadratic invariant $$I(y) = y^TCy \quad ,$$ where $C$ is a quadratic symmetric matrix, under the condition $$b_i a_{ij} + b_j a_{ji} - b_ib_j = 0 \quad .$$ In general I understood the form of the prove, however I am struggling with why $$\frac{d}{dt}I(y) = 2y^TCf(y) = 0 \quad, $$ holds and is zero as per my professor. If I try to perform the same step I arrive at the following form $$\frac{d}{dt}I(y) = (C+C^T)y  \quad, $$ where I do not see why it would be zero for a symmetric matrix C.

I would appreciate if someone could show me where I went wrong and especially why the statement of my professor is true. If any more of the underlying proof is necessary to show I can add it.

答:
The question seems not so much to be about Runge-Kutta methods but about the derivative of quadratic forms. There you get by the product rule
$$
\frac{d}{dt}I(y(t))=\frac{d}{dt}y(t)^TCy(t)=\dot y(t)^TCy(t)+y(t)^TC\dot y(t)
$$
Now the terms are scalars, thus invariant under transposition, apply that to the first term
$$
=y(t)^T(C^T+C)\dot y(t)
$$
and then apply the symmetry of $C$ to get $C^T+C=2C$.

That the derivative of a constant is zero is trivial, the last step then is to insert $\dot y(t)=f(y(t))$.


答:
If you have $a << 1$ , then

$$(1 + a)^n \approx 1 + na$$

This is because the terms after the linear term are so small that they can be ignored in comparison

$$\frac{ds}{dx} = \left[\left(1 + \frac{d \xi}{dx}\right)^2 + \left(\frac{\partial y}{\partial x}\right)^2\right]^{1/2}$$

$$\implies \frac{ds}{dx} = \left[1 + \left(\frac{d \xi}{dx} \right)^2 + 2 \frac{d \xi}{dx} + \left(\frac{\partial y}{\partial x}\right)^2\right]^{1/2}$$

Now I guess the assumptions taken are

$$\frac{d \xi}{dx}^2 \text{is ignored compared to other terms}$$

$$2 \frac{d\xi}{dx} + \frac{\partial y}{\partial x}^2 <<1$$

Hence you will get the terms you see. Whether these are reasonable I wouldn't know

Title: How does this hypotenuse/derivative/polynomial approximation work?

Body:
I am trying to understand an article regarding the forces of a vibrating string in terms of both longitudinal and transverse waves.

Here is the pertinent excerpt:

Equation 1 is obvious. It is just Pythagorean theory. Equation 2 is also obvious as it just converts this to derivatives. Equation 3 also makes sense - tension in a string is equal to the base tension plus the product of length deviation (extra stretch), Young's Modulus, and cross sectional area.

The part I don't get is equation 4. If you put equation 2 in equation 3 you definitely don't get equation 4.

I know equation 4 is some type of approximation, but I don't know how it works. I found another reference that seems to discuss this same approximation and they say of it: "It has been shown that by expanding both y and ξ as a series of polynomials and truncating at third order, the added force per unit length on the element in the longitudinal direction caused by the transverse displacement is given by [this equation]."

But I'm not a mathematician and I don't understand how this is represented by a series of polynomials or how they describe this approximation.

Does this make sense? Can anyone explain it in simple terms?

Thanks for any help. Very appreciated.


问:
Show that no linear map $F:X \rightarrow \mathbb {K}^{n}$ is injective
I have some troubles with this exercise.

Let X be an infinite dimensional normed vector space over $\mathbb {K}$ where $\mathbb {K}=\mathbb{R}$ or $\mathbb {K}=\mathbb{C}$. Let $n\geq 1$ be an integer. Show that no linear map $F:X\rightarrow \mathbb{K}^{n}$ is injective.

I am thinking that I have to show that the null-space of F is different from $\left\{0\right\}$, but I don't have any idea for how to show this.
Or else I am thinking that for $x_1 \neq x_2$ then $F(x_1)=F(x_2)$, where $x_1, x_2 \in X$, but I am not sure how to show this either.

Is there anyone who can help me with this exercise.

答: Suppose $f:X\rightarrow\mathbb{K}^n$ is injective. Let $x_1,...,x_{n+1}$ linear independent in $X$, $f(x_1),...,f(x_{n+1})$ is dependent. There exists $a_1,...,a_{n+1}$ not all equal to zero such that $a_1f(x_1)+...+a_{n+1}f(x_{n+1})=f(a_1x_1+..+a_{n+1}x_{n+1})=0$, we deduce that $a_1x_1+...+a_{n+1}x_{n+1}=0$ since $f$ is injective and $a_i=0$, since $x_1,..,x_{n+1}$ is independent.  Contradiction.


问:
Calculating the probability of someone having zero to n diseases given overlapping prevalence
I'm trying to calculate the proportion of people in a population who have a combination of disease $A$, $B$ and $C$. This can range from $0,0,0$ (the set of people with no disease) to $1,1,1$ (the set of people with all three diseases).

I would like to calculate the probability each of these 8 outcomes (from $0,0,0$ to $1,1,1$) given three inputs, which are the prevalence of each disease (e.g. 80% of people have disease $A$, 70% of people have disease $B$ and 50% of people have disease $C$). I would also like to generalise this to $n$ diseases in the future. It is important to be able to distinguish between each outcome, because they have different effects on people (e.g. $0,1,1$ has a different chance to kill someone than $1,0,1$).

Given that we know nothing about the relationship between these diseases and that the prevalences do not sum to 100%, I do not think this has a single solution. However, if we expect the diseases to be independent of each other, I assume this is solvable, but I do not know how to do it. If I reduce the diseases to two, this becomes easier. For example:

If the prevalence of Disease $A$ is $80\%$ and the prevalence of Disease $B$ is $70\%$:

$1,1$: The proportion of people with $A \& B$ = $0.8 * 0.7 = 0.56$

$1,0$: People with $A$ only = $0.8 - 0.56 = 0.24$

$0,1$: People with $B$ only = $0.7 - 0.56 = 0.14$

$0,0$: People with no disease = $1 - (0.56+0.24+0.14) = 0.06$

But in the case of three diseases, I get stuck. For example:

Prevalence of Disease $A$ is $80\%$, Prevalence of Disease $B$ is $70\%$, Prevalence of Disease $C$ is $50\%$

$1,1,1$ = $0.8 * 0.7 * 0.5 = 0.28$

From here, i'm not sure what to do. Perhaps I need to simultaneously solve the next set of equations e.g. $1,0,1$ and $1,1,0$, but i'm not sure.

Is this on the right track? Or does this problem become unsolvable with $>2$ diseases?

答:
+1 for the query, very good analysis so far.

To carry it into three people, using:

80% of people have disease A, 70% of people have disease B and 50% of people have disease C

and

However, if we expect the diseases to be independent of each other, I assume this is solvable, but I do not know how to do it.

Let $p_a = 0.8 =$ chance of person having disease A.

Let $p_b = 0.7 =$ chance of person having disease B.

Let $p_c = 0.5 =$ chance of person having disease C.

Let $q_a = 1 - p_a.$

Let $q_b = 1 - p_b.$

Let $q_c = 1 - p_c.$

You define 8 possibilities, where each person either does or doesn't have each of the 3 diseases.

Then, each of the 8 possibilities will compute to $s_a \times s_b \times s_c$, where if the possibility refers to:

the person having disease A, set $s_a = p_a,$ else $s_a = q_a.$

the person having disease B, set $s_b = p_b,$ else $s_b = q_b.$

the person having disease C, set $s_c = p_c,$ else $s_c = q_c.$


问:
$\|A\|,\|B\|\leq 1$. How do I prove $\left(\begin{array}{}1&A&AB\\A^*&1&B\\B^*A^*&B^*&1\end{array}\right)$ is positive.
$A,B$ are elements of a unital C*-algebra $\mathscr A$ with $\|A\|\leq 1$ and $\|B\|\leq 1$.

$X=\left(\begin{array}{}1&A&AB\\A^*&1&B\\B^*A^*&B^*&1\end{array}\right)\in M_3(\mathscr A)=\mathscr A\otimes M_3(\mathbb C)$ .

I tried to find a matrix such that $Y^*Y=X$ and failed.

Then I tried to show the invertibility of $X+\lambda$ for $\lambda>0$. This is equivalent to the invertibility of $X$ for $\|A\|,\|B\|<1$. And failed again.

Could you please give me some hint?

Body:
Hint
$$
\pmatrix{ 1&0&0\\-A^*&1&0\\0&-B^*&1}
\pmatrix{1&A&AB\\A^*&1&B\\B^*A^*&B^*&1}
\pmatrix{ 1&-A&0\\ 0&1&-B\\0&0&1}
=
\pmatrix{1&0&0 \\ 0& 1-A^*A &0 \\ 0&0&1-B^*B}
$$


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:
You can try plugging this in Desmos.com
$$f(x) = \frac{1}{1+e^{-cx}}$$
add a slider for $c$ and then vary it to see how to function changes. I would say that by setting $c>4$, you get closer to your desired red line.


问:
Find the vectors that span the subspace of $W$ in $R^3$
Question: Find the vectors that span the subspace of $W$ in $R^3$.
$$W = \{\begin{bmatrix}x\\y\\z\end{bmatrix}, 2x-y+3z=0 \}$$

I'm not very sure how to go about solving this problem. I know that the cross product of the vectors that span $W$ has to be a vector orthogonal to a normal vector of the plane $W$. Vector $<2, -1, 3>$ is the easiest normal vector to find. But then I'm unsure what to do next.

I'd much appreciate it if someone could show me how this is done. Thanks in advance!

Body: $2x-y+3z=0$ implies that $y=2x+3z$, if $(x,y,z)\in W$, $(x,y,z)=(x,2x+3z,z)=x(1,2,0)+z(0,3,1)$ implies that $(1,2,0)$ and $(0,3,1)$ span $W$.


答:
First find one non-zero vector perpendicular to the given normal $(2,-1,3)$; this is very easy (just substitute numbers in). I'll take $(-1,1,1)$ here. Then the second vector is the cross product of this with $(2,-1,3)$:
$$(-1,1,1)×(2,-1,3)=(4,5,-1)$$
The three vectors are mutually orthogonal, so any two vectors span a plane orthogonal to the third vector. Concretely, $(-1,1,1)$ and $(4,5,-1)$ span $W$.

Body:
Question: Find the vectors that span the subspace of $W$ in $R^3$.
$$W = \{\begin{bmatrix}x\\y\\z\end{bmatrix}, 2x-y+3z=0 \}$$

I'm not very sure how to go about solving this problem. I know that the cross product of the vectors that span $W$ has to be a vector orthogonal to a normal vector of the plane $W$. Vector $<2, -1, 3>$ is the easiest normal vector to find. But then I'm unsure what to do next.

I'd much appreciate it if someone could show me how this is done. Thanks in advance!


问:
Why Eulers' totient doesn't seem to work in following situation?
I have a question about Eulers' totient function related to RSA

(1) $N=PQ$

(2) $\varphi(N)=\varphi(P)\varphi(Q)=(P-1)(Q-1)$

Now, say $P = 3$ and $Q = 3$

From (1) it follows that $N=PQ=3\cdot3=9$

From (2) it follows that $\varphi(9)=\varphi(3)\varphi(3)=(3-1)(3-1)=2\cdot2=4$

But, actually $6$ numbers are coprime with $9$, being $1,2,4,5,7,8$, so actually $\varphi(9)=6$.

What am I missing here?

答:
The relation $\varphi(pq)=\varphi(p)\varphi(q)$ is only true if $\gcd(p,q)=1$. But here $\gcd(p,q)=\gcd(3,3)=3$.

Hence the two primes that multiply to $N$ in the RSA cryptosystem must be distinct.


问:
Real numbers such that $|x^2 + ax + b| \le {1\over2}$?
What are real numbers $a$, $b$ such that, for every $x$ satisfying $0 \le x \le 2$, it holds that $|x^2 + ax + b| \le {1\over2}$?

答:
From $x^2+ax+b\leq\frac{1}{2}$, we have $x_1\leq x\leq x_2$, where $$x_1=\frac{-a-\sqrt[]{a^2-4b+2}}{2}$$ and $$x_2=\frac{-a+\sqrt[]{a^2-4b+2}}{2}$$
From $x^2+ax+b\geq\frac{-1}{2}$, we have $x\leq x_3$ or $x\geq x_4$, where $$x_3=\frac{-a-\sqrt[]{a^2-4b-2}}{2}$$ and $$x_4=\frac{-a+\sqrt[]{a^2-4b-2}}{2}$$
It is easy to see that $x_1\leq x_3 \leq x_4 \leq x_2$.

Hence, the solution of the inequation is $x_1\leq x \leq x_3$ or $x_4\leq x \leq x_2$.

So, it must be: $x_1=0$ and $x_2=2$, from where $b=0.5$  and $a=-2$, and, in that case, $x_3=1$ and $x_4=1$ .


问:
prove by induction that $P\left(\bigcup\limits_{i=1}^{n} E_i\right) = 1-\prod\limits_{i=1}^{n}(1-P(E_i))$, $E_1,E_2,\ldots , E_i$ independent
Suppose $E_1,E_2,\ldots , E_i$ are independent events. prove by induction that $$P\left(\bigcup\limits_{i=1}^{n} E_i\right) = 1-\prod\limits_{i=1}^{n}(1-P(E_i))$$

The first step is easy. For $n=1$ we have
$$P\left(\bigcup\limits_{i=1}^{1} E_i\right) = P(E_1)= 1-(1-P(E_i))=1-\prod\limits_{i=1}^{1}(1-P(E_i))$$
I don't really see how to continue from here though. Any help would be appreciated!

答: As a set theoric proposition, it's not hard to prove that $\cup_{i=1}^n E_i=(\cap_{i=1}^n {E_i}^c)^c$. So $P(\cup_{i=1}^n E_i)=P((\cap_{i=1}^n {E_i}^c)^c)=1-P((\cap_{i=1}^n {E_i}^c)$. The $E_i$'s are independent, hence their complements are independent, too. So the latter term equals to $1-\Pi_{i=1}^n P({E_i}^c)=1-\Pi_{i=1}^n (1-P(E_i))$. QED.</p>


问:
How to declare formulas for banded pricing model?
I am making an application (using JavaScript) where users should be able to create pricing models with a table. A user will have an ability to choose a table model. The table models are internally being referred as tiered and banded models. Users will also be able to expand number of table rows and to manually enter some table values and a volume units based on which a monthly cost will be calculated.

I had to develop simple math formulas for every cell for both models and I managed to do it only for the tiered model. I'm posting an image here as a reference of what I did:

It was decided to use a really powerful JS library for math calculations called mathjs where I was able to produce values in certain cells based on the values from the other cells, much like Excel or Google Sheets do. From the image, it should be clear that users are able to enter numeric values in cells that are marked as null while the other values are the formulas (with the exception of A1 cell which is always set to 0). Also, a user should be able to enter value of the VOLUME variable somewhere outside of the table.

When this is translated to sheets, an example of a table with the real values looks like this:

Based on the values entered in columns B and C and in the VOLUME cell, it is possible to calculate values in other columns: A starting from A2 and entire D, E and F columns. Also, it's possible to calculate Monthly Cost which is a sum of values from the D column.

What I'm trying to do now is to create similar formulas for the banded model. I have to explain how tiered model works so that I can explain how the banded model should be constructed.

What's similar to both models is that they are charging based on the tier or band (basically a row) a volume units belong to. In the model from the second image (that is using tiered model), for the volume of e.g. 328, the tier is number 4. What's the difference between these two models is that tiered model is charging differently to the banded model for the volume units in previous tiers/bands.

Based on the image, the tiered model is charging the first 50 (B1 cell) volume units (out of these 328) by the cost/unit for that tier which is $0.50 (C1 cell). That is the maximum price shown in Tier Price column which for tier 1 is at cell E1: C1 * B1 which is $0.5 * 50 which is $25. So, the first 50 volume units are charged $25.

The next 50 (B2 - B1) volume units are being charged at cost of $0.45 (C2 cell). Maximum price for these 50 volume units is at cell E2: C2 * (B2 - B1) which is $0.45 * (100 - 50) which is $22.5.

The cost for these first 100 volume units is basically the sum of the tier costs of both tiers. That is at cell F2: $25 + $22.5 which is $47.5.

The next 150 (B3 - B2) volume units belong to tier 3 and are being charged at cost of $0.40 (C3 cell). Maximum price for these 150 volume units is at cell E3: C3 * (B3 - B2) which is $0.40 * (250 - 100) which is $0.40 * 150 which is $60.

The cost for the first 250 volume units is basically the sum of the tier costs of all three tiers. That is at cell F3: $60 + $47.5 which is $107.5.

Now, the next tier is tier 4 where 328 volume units belong. The first 250 volume units are being charged at price $107.5 while the remaining 78 (328 - 250) volume units are being charged as 78 * $0.35 which is $27.3. Now, the Monthly cost is the sum of $27.3 and $107.5 which is $134.8.

From what is now visible, what Actual price column represents is the cost of the volume units per that tier. If the volume units are larger than the maximum number of volume for that tier (column B), than the Actual price for that tier becomes the Tier price of that tier, but if it's the lower, then it's 0. That's why min and max functions are used.

The banded model is different to tiered model in a way that volume units are being charged at the price from a band that they belong to. I.e. if the table from the above example was using banded model, the monthly cost for 328 value units would be simple: take the cost/unit price of the band it belongs and multiply by the volume. In this case, the band is 4 so the cost/unit for band 4 is $0.35. So, the monthly cost of 328 volume units would be 328 * $0.35 which is $114,8.

What's noticeable is that the banded model is having slightly less monthly cost because all volume units are being charged at the same price, while this is not the case with the tiered model.

Now, what I need is the monthly cost formula for the banded model. For this to work, columns A, B and C which represent From, To and Cost/Unit values are required. The others ones can be made up.

If Actual price column is being used and if the Monthly cost would represent a sum of the Actual price values, then in this example of 328 volume units I need the following:

the values of Actual price cells that don't belong to tier 4 should be zeroes;

and the value of Actual price cell that belongs to tier 4 should be 328 * $0.35 which is $114,8.

Like this:

Basically, based on the VOLUME variable, I need to determine the band number and use the cost that belongs to that band to multiply it with the VOLUME. How can I achieve this only by using math functions that are supported by the mathjs library?

I'd love to use if statements much like I'm able to use it in Excel/Google Sheets, but the library is determining what I can use.

One note: using min and max functions, I think I managed to do something like this for the tiered model (per a single row/tier):

Now, I think I need this for the banded model (per a single row/band):

Is it possible to achieve this using mathjs functions? Maybe using mod could help here? Or maybe Monthly cost can be written somewhat different and not as a sum of values from the D column?

I am not certain how to achieve this at this moment and I need help. I am very sorry for the long post and thankful in advance for the answers.

答:
I wasn't really paying attention to the mathjs docs and now I have found a way to write conditional expressions which is all what I needed. When I wrote this in the question:

I'd love to use if statements much like I'm able to use it in Excel/Google Sheets, but the library is determining what I can use.

But I was wrong because mathjs actually supports if statements, it just have to be written as a ternary operator expression. So, my solution for Monthly cost would be:

```<code>VOLUME*(VOLUME<=B1?C1:VOLUME<=B2?C2:VOLUME<=B3?C3:VOLUME<=B4?C4:C5)
</code>```

With no need to have an Actual price column. Also, if a users adds another row, I'd have to add another ternary condition near the end of the expression - basically replace C5 with VOLUME<=B5?C5:C6.

It appears that this question should've been asked on StackOverflow forum because the solution is purely programming related. However, I'm open to another answers that don't include ternary condition, but use purely math functions, if this is possible.

Body:
I wasn't really paying attention to the mathjs docs and now I have found a way to write conditional expressions which is all what I needed. When I wrote this in the question:

I'd love to use if statements much like I'm able to use it in Excel/Google Sheets, but the library is determining what I can use.

But I was wrong because mathjs actually supports if statements, it just have to be written as a ternary operator expression. So, my solution for Monthly cost would be:

VOLUME*(VOLUME<=B1?C1:VOLUME<=B2?C2:VOLUME<=B3?C3:VOLUME<=B4?C4:C5)

With no need to have an Actual price column. Also, if a users adds another row, I'd have to add another ternary condition near the end of the expression - basically replace C5 with VOLUME<=B5?C5:C6.

It appears that this question should've been asked on StackOverflow forum because the solution is purely programming related. However, I'm open to another answers that don't include ternary condition, but use purely math functions, if this is possible.


问:
Find the limit of sequence without using derivatives
$$\lim_{n\to\infty}\large n(a^\frac{1}{n}-1) 0$$ $$a>0$$

is the problem. So far what I tried was transforming $$(a^\frac{1}{n}-1)$$

by using the formula $x^n-1 = (x-1)(x^{n-1} + x^{n-2} +...+1)$ but I fail to see how this would help me in this instance, or I just may be doing it wrong because it's the nth root that is giving me problems. Any help would be appreciated

Body:
$$L=\lim_{n\to \infty} n(a^1/n-1)$$
Let $t=1/n$, then
$$L=\lim_{t\to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{1+t\ln a+(t\ln a)^2/2+...-1}{t}=\ln a, a>0$$


问:
3-regular graph and two-way Euler circuit
A town-planner has built an isolated city whose road network consists of $2N$ roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his driving test, and is nervous about roundabouts. He starts driving from his house, and always takes the first edit at each roundabout he encounters. It turns out his journey incluldes every road in the town in both directions before he arrives back at the starting point in the starting direction. For what values of $N$ is this possible?

I have tried to turn this into an equivalent graph theory problem in which we can apply some results on Euler circuits or similar, but with no such rephrasals seem useful. Any help appreciated!

答:
Partial answer

Let me formalize. If you enter a roundabouts by road $i$ you leave it by road $(i \mod 3) +1$.

Let R.i be the road i of roundabout R.

$N=1$ is a solution of your problem.connect the two roundabout A and B as follow: for all i : A.i is connected to B.i.

We now show that for $N_1,N_2$ solution of your problem then $N_1+N_2+1$ is a solution as well.

Let $T_1,T_2$ be two towA_1s with respectively $2N_1,2N_2$ roundabout. Let $A_1,B_1$ be two roundabout connected in $T_1$ and $A_2,B_2$ connected in $T_2$. We construct a town $T_3$ as follow: we add two roundabout $C$ and $D$ and connect then as follow:

$A_1$ with $C.1$

$B_1$ with $C_2$

$C_3$ with $D_3$

$A_2$ with $D_1$

$B_2$ with $D_2$

$T_3$ is a solution of your problem with $2N_1+2N_2+2=2(N_1+N_2+1)$ roundabouts.

Thus every odd $N$ is a solution.

@Alex Ravsky comments tells us that $N=2$ is not a solution.
So may be Even numbers are impossible (i don't know yet). I'll try to think a a reduction with the same idea in order to prove this


问:
Finding the gradient of an expression
I'm trying to solve the following problem:

Let $\textbf{r} = (x,y) = x\textbf{i} + y\textbf{j}$ and $r = ||\textbf{r}|| = \sqrt{x^2+y^2}$.
Also $\textbf{r} \neq 0$.

Let $f(x,y) = r^m$. what is the correct expression for $\bigtriangledown$f?

I'm thinking $mr^{(m-2)}\textbf{r}$. what do you guys think?

答:
Worked explicitly:
$$\frac\partial{\partial x}(x^2+y^2)^{m/2}=2x(m/2)(x^2+y^2)^{(m-2)/2}=mxr^{m-2}$$
Similarly, the partial derivative on $y$ is $myr^{m-2}$. You are correct.

Title: Finding the gradient of an expression


问:
Example of isomorphic sheaves whose sections aren't isomorphic
Are there examples of sheaves $\mathcal{F}$ and $\mathcal{G}$ such that $\mathcal{F} \cong \mathcal{G}$ but it is not necessarily true that $\mathcal{F}(U) \cong \mathcal{G}(U)$ for every open set $U$? I was sure there are, since a surjective morphism of sheaves doesn't always induce a surjective morphism of sections for every open set. However, I'm having a hard time coming up with examples.

If the answer turns out that no such examples exist, my next question would be whether or not a natural isomorphism of functors $F: A \rightarrow B$ and $G: A \rightarrow B$ would imply that $F(X) \cong G(X)$ for every object $X$?

答:
There are two possible definitions for natural isomorphism. For a natural transformation $\eta: F \to G$, both of the following statements make sense.

Every arrow $\eta_X: F(X) \to G(X)$ is an isomorphism.

There is a natural transformation $\theta: G \to F$ such that $\eta \theta$ is $Id_G$, the identity natural transformation on $G$, and $\theta \eta = Id_F$. In other words: $\eta$ is an isomorphism in the functor category where $F$ and $G$ belong to.

These two are equivalent. The direction $2 \implies 1$ is easy, for every $X$ we directly see that the inverse of $\eta_X$ is $\theta_X$. For the converse, $1 \implies 2$, there is only one sensible candidate for $\theta$. Namely we take $\theta_X$ to be $\eta_X^{-1}$, the inverse of $\eta_X$. We only need to check that $\theta$ is then indeed a natural transformation. So let $f: X \to Y$ be any arrow. Using naturality of $\eta$ we get
$$
F(f) \theta_X = \theta_Y \eta_Y F(f) \theta_X = \theta_Y G(f) \eta_X \theta_X = \theta_Y G(f).
$$
It might help to draw the relevant picture if the above is confusing.

So in particular, whenever two sheaves $F$ and $G$ are isomorphic through a natural isomorphism we get that each $F(U)$ and $G(U)$ are isomorphic for every $U$. You mention that an epimorphism of sheaves is not always componentwise surjective. An isomorphism is a very special kind of epimorphism, namely a split epimorphism. A split epimorphism of sheaves is always surjective on every component (nice exercise).


答:
$$f(x,y) = r^m = \left(\sqrt{x^2 + y^2}\right)^{m}$$

$$\implies \nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} $$

$$\implies \nabla f = m\left(\sqrt{x^2 + y^2}\right)^{m-1} \cdot \frac{2x}{2\sqrt{x^2+y^2}}\hat{i} + m\left(\sqrt{x^2 + y^2}\right)^{m-1} \cdot \frac{2y}{2\sqrt{x^2+y^2}}\hat{j}$$

$$\implies \nabla f = mr^{m-2}\textbf{r}$$

Title: Finding the gradient of an expression

Body:
I'm trying to solve the following problem:

Let $\textbf{r} = (x,y) = x\textbf{i} + y\textbf{j}$ and $r = ||\textbf{r}|| = \sqrt{x^2+y^2}$.
Also $\textbf{r} \neq 0$.

Let $f(x,y) = r^m$. what is the correct expression for $\bigtriangledown$f?

I'm thinking $mr^{(m-2)}\textbf{r}$. what do you guys think?


问:
Determinants: A Special Condition
Under what conditions is

$$ \det(A_1 + \cdots + A_n) = \det(A_1)+\cdots+\det(A_n), $$

just curious.

答:
I'm not sure if you're looking for necessary and sufficient conditions, or just a list of sufficient ones.  Here is one sufficient condition: find any solution to $x_1+x_2+\ldots+x_n=0$, say $(a_1,a_2,\ldots,a_n)$, and let $A_i$ be a matrix full of $a_i$'s.  Then both sides are zero.

A bit more general: let $a_1+\ldots+a_{n-1}=0$, and define $A_i$ as above for $i=1,\ldots,n-1$.  Then for any $A_n$, both sides are equal to $\operatorname{det}(A_n).$

Title: Determinants: A Special Condition


问:
A problem involving complex numbers and polynomials to prove that $f(z) + f(−z) = 0$ for all $z ∈ C$.
Let $f : C → C$ be a function such that $f(z)f(iz) = z^2$ for all $z ∈ C$. Prove that $f(z) + f(−z) = 0$ for all $z ∈ C$. Also, find such a function.

What I did was take a $f(x)$ of the form or as a polynomial as $a_n x^n +  a_{n-1} x^{n-1} +\ldots+a_0$ and substituted $z$ and $iz$ and then some manipulations to do... But failed. I need help.

答:
Substituting $iz$ instead of $z$ in the functional equation

$$f(iz)f(-z) = -z^2$$

Adding the two, we get

$$f(iz)(f(z) + f(-z)) = 0$$


答:
Substituting $iz$ for $z$,
$$f(iz)f(-z)=(iz)^2=-z^2$$
Thus
$$f(iz)(f(z)+f(-z))=0$$
Either $f(iz)=0$ for all $z\in\mathbb C$, implying $f(z)=0$ and $f(z)+f(-z)=0$ over the same domain, or $f(z)+f(-z)=0$ outright.

An example $f$ is $f(z)=ze^{-i\pi/4}$.

Title: A problem involving complex numbers and polynomials to prove that $f(z) + f(−z) = 0$ for all $z ∈ C$.

Body:
Let $f : C → C$ be a function such that $f(z)f(iz) = z^2$ for all $z ∈ C$. Prove that $f(z) + f(−z) = 0$ for all $z ∈ C$. Also, find such a function.

What I did was take a $f(x)$ of the form or as a polynomial as $a_n x^n +  a_{n-1} x^{n-1} +\ldots+a_0$ and substituted $z$ and $iz$ and then some manipulations to do... But failed. I need help.


问:
Four-letter word contains no two consecutive equal letters.
This is taken from the book on Combinatorics, by Daniel Marcus.

Request vetting:

A16: Find the probability that a four-letter word that uses letters from $A,B,C,D,E$ contains (a) no repeated letters; (b) no two consecutive equal letters.

(a)
Total (Inclusion) ways to form four-letter word from five letters: $5^4$.

Ways to have repetitions of $A$:

Two $A$: Choose two positions for $A$ in $\binom{4}2= 6$ ways. Rest two positions can be filled in : $4\times 3= 12$ ways. So, by product rule: $72$ words. Also, the same for other $4$ letters, leading to $72\times 5= 360$ words.

Three $A$: Choose three positions for $A$ in $\binom{4}3= 4$ ways. Rest one position can be filled in : $4$ ways. So, by product rule: $16$ words. Also, the same for other $4$ letters, leading to $16\times 5= 80$ words.

Four $A$: Choose all four positions for $A$ in $\binom{4}4= 1$ ways. Also, the same for other $4$ letters, leading to $5$ words.

Sum of above three cases:
$360+80+5= 445.$

Left are: $5^4 - 445= 180.$

$p= \frac{180}{625}$

(b) no two consecutive same letters' case.

Approach#1:

To get, need subtract from $5^4$  the chances of having two consecutive same letters. This is divided into cases:

Case 1: Position #1,2 are the same, and position #4 can match too :  $5\times 1\times 4\times 5= 100$ cases.

Case 2: position #2,3 are the same: $5\times 4\times 1 \times 4= 80$ cases.

Case 3: Position #3,4 are the same, and position #1 can match too:
$5\times 4\times 1\times 5=100$

So, get : $625- 280= 345$ cases.

Approach#2:

Alternatively, can 'directly' calculate by having $5$ choices for the first position, then $4$ choices, then again need eliminate one choice (of the second element) to have $4$ choices for the third position, and same for the fourth position. This leads to: $5\times 4^3= 320$ choices possible.

The second approach seems correct, but not clear why first approach is faulty.

Seems there are too many factors to consider in the first approach & hence error-prone (to miss some case(s) as here). So, second approach seems better.

答:
You did (a) completely wrong. The number of four-letter words with no repeats is $5×4×3×2=120$, so the probability is $\frac{120}{625}=\frac{24}{125}$. The mistake is that words with two repeats of two letters have not been accounted for.

For (b), the problem with the first approach is that it fails to consider cases where both the first two and last two letters are the same (e.g. AABB). The second approach gives the right answer.


问:
Writing a word problem in an inequality
A phone card costs $5$ euros. This can be used to call someone for $60$ minutes within the zone. For a house phone one pays $14,25$ euros as a bundle and $0,15$ euros for every $12$ minutes. How long can one call per month so that the card is cheaper than with the house phone?

I already know it's at most 180 minutes with the card because $0,15 \cdot 5 = 0,75$ (you multiply by $5$ because $60 \div 12$ minutes $=5$. And $14,25 + 0,75 = 15$ so you can buy $3$ phone cards with it resulting in $180$ minutes. I don't know how to write it in inequality.

答:
Let's say that you speak $x$ minutes. Then, the total price for the house phone is
$$
h(x) = 14.25 + 0.15 \lceil \frac{x}{12} \rceil
$$
Phone cards, on the other hand, have a total cost of
$$
c(x) = 5 \lceil \frac{x}{60} \rceil
$$
Here $\lceil \ldots \rceil $ is the ceiling function. So for example $\lceil \frac{x}{12} \rceil$ is how many "units" of 12 minutes we need to buy.

Here are the graphs of the two functions: https://www.desmos.com/calculator/dh9gwbgokp

Can you continue from here?

Title: Writing a word problem in an inequality


问:
When will two cars pass each other
There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring ,
complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they
pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.

答: Suppose the slower car stands still for one hour.  How often will the faster car pass it?  Then stop the faster car and start the slower car for another hour.  How often will the slow car pass the stopped car?   Add.</p>

Title: When will two cars pass each other


问:
show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq  \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $

I've tried to use AM-GM directly on the LHS but that obviously failed.

Simplifying the question gives $a^2b + b^2c + c^2a \geq ab + bc + ac$

The fact that this is $a(ab) + b(bc) + c(ca) \geq ab + bc + ac$ makes me think that there's a way to proceed from here, but I'm not quite sure how.

Note : a,b,c are real positive numbers

答:
Case 1: Let $ab+bc+ca\ge a+b+c$, then the AM-HM of $a,b,c$ with weights as $1/b,1/c,1/a$ gives
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{1/a+1/b+1/c)^2}{1/(ab)+1/(bc)+1/ca)}=\frac{(ab+bc+ca)(1/a+1/b+1/c)}{a+b+c} \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$

Case 2: Let $(ab+bc+ca) \le a+b+c$, then AM-HM of $1/b,1/c,1/a$ with weights as $a,b,c$, we get
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{(a+b+c)^2}{ab+bc+ca} \ge ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$


问:
Isomorphism between $V$ and $U$
Let there be $V =$ $\mathbb{R}[x]$ space of all polynomials with 1 variable with
real coefficients

$U$ = { $p(x) \in\ V|p(0)=0$ }, In other words: $U$ is a subset in V of all of the polynomials:
 $a_nx^n + a_{n-1}x^{n-1} +......+ a_1x$

such that the free constant term equals to $0$

prove that $U$ and $V$ are isomorphic

MY APPROACH:
 I tried to build an injective ( one-to-one ) and
Surjective ( onto ) linear transformation / map. it is easy to see that there
is one, but it is not injective because to two polynomials, different
in the constant term, have the same output. I have tried to "fix" this
linear transformation but with no success at this moment

答:
Show that the map $\mathbb R[x]\to U$ given by $f\mapsto f\cdot x$ is linear, injective and surjective.

(It is the inverse to Timurs suggestion, but easier to handle IMHO.)

Title: Isomorphism between $V$ and $U$


问:
Visualizing $\textbf{Q}_p$ vs. $\textbf{F}_p((t))$?
I have a few questions.

How do I visualize the field $\textbf{Q}_p$ of $p$-adic numbers?

How do I visualize the field $\textbf{F}_p((t))$ of Laurent series of $\textbf{F}_p$?

How do I do 1 and 2 in a way as to make the similarities and differences between $\textbf{Q}_p$ and $\textbf{F}_q((t))$ as transparent as possible?

Hopefully this question can be answered in a way that does not go too much off the deep end. I just want to visualize things!

EDIT: I want pictures.

答:
Late answer, but I asked myself the same question and ended up trying to illustrate properties of the p-adic integers using the Chistyakov embedding.

The 3-adic world was the simplest one, where the following property : $$ord_p(a+b)≥\min(ord_p(a),ord_p(b))$$

can be seen as moving elements to the right. However, other properties (convergence, or visualisation for other values of $p$ were less successful).

More information and code can be found here: P-adic numbers visualization


问:
Book Suggestions for an Introduction to Measure Theory
Couldn't find this question asked anywhere on the site, so here it is! Do you guys have any recommendations for someone being introduced to measure theory and lebesgue integrals?

A mentor has suggested a book that's in french, but unfortunately I don't know french (heck I barely know english) - so english books only please!

Thanks in advanced!

EDIT: Did not realize this had been asked here but I'm going to leave this question open to see if there are any newer books (that question was originally asked in 2011). If I am doing something wrong here, just give me a gentle shoutout in the comments and I'll be understanding and close this

答:
I would say Fremlin: Measure theory is a good choice (i guess it's freely available on the web).
The best measure theory book i know is unfortunately in German (Elstrodt: Maß- und Integrationstheorie)

Hope your happy with Fremlin!


问:
Want to show $\sum_{n>N}n^{-1/2} \ll N^{-1/2}$
I feel like I can use the result $$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$ where $s=1/2$ to verify the $\ll$ approximation.

If I pick $A(n)=\sum_{x\leq t\leq n}\chi(t)$, then I know $|A(n)|\ll 1$. I think that will make the first term become $O(N^{-1/2})$, but am not sure how I should go about handling the integral term. Thanks!

Thanks for the help in advance!

答:
You have to edit the title as I think you actually meant $\sum_{n>N} \chi(n) n^{-1/2} \ll N^{-1/2} $. While this is valid for a Dirichlet character that is non-principal, it fails when $\chi$ is principal. Letting $$ S(y):= \sum_{1\leq n \leq y } \chi(n) $$ we get $$ \sum_{1\leq n \leq x}\chi(n) n^{-1/2} =\frac{S(x)}{x^{1/2}}+ \frac{1}{2} \int_1^x \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ Using this for $x=N$ and $x=M$ and subtracting we get $$ \sum_{N< n \leq M}\chi(n) n^{-1/2} =\frac{S(M)}{M^{1/2}}-\frac{S(N)}{N^{1/2}}+ \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ By the estimate $S(u ) =O(1)$ this  becomes $$\sum_{N< n \leq M}\chi(n) n^{-1/2} = O\left( \frac{1 }{M^{1/2}} \right)
-\frac{S(N)}{N^{1/2}}
+  \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u .$$ As $M\to \infty$  this converges because   the integral converges absolutely since it is $$
 O\left(  \int_N^M  \frac{ 1 }{u^{3/2 } } \mathrm d u \right)=
 O\left(  \frac{2}{N^{1/2}} -  \frac{2}{M^{1/2}} \right)=
  O(N^{-1/2})
  .$$ Hence, the limit satisfies $$\sum_{n=N+1 }^\infty
\chi(n) n^{-1/2} =
O\left( \frac{1 }{M^{1/2}} \right)
-\frac{S(N)}{N^{1/2}}
  O(N^{-1/2})
  ,$$ which, by $M >N$  and the estimate $|S(u)| =O(1)$ is $  O(N^{-1/2})
$.


问:
Is the orthogonal complement just the basis?
so I am trying to self study linear algebra and dear god that this has got to be the hardest branch of math I have encountered, and yes I love calculus. Right now, I am  taking up orthogonals and its related subtopics. One thing I can't verify  right now is if the orthogonal complement is just the basis of the subspace. For example if I was asked to find the orthogonal complement of the subspace $W$ on $\mathbb{R}^4$ where $W$:

$W= \{ [a-b-c, a, b+c]: a,b,c ∈ \mathbb{R} \}$

$= \{a[1,1,0]+b[-1,0,1]+c[-1,0,1]\}$

with this I could say that the orthogonal complement of $W$ is

$\{[1,1,0],[-1,0,1],[-1,0,1]\}$

Is my understanding correct or am I missing something?  Please do correct me if there are any inconsistencies. Thanks in Advance!

答: The orthogonal complement of $W$ is the set$$W^\perp=\{v\in\Bbb R^4\mid(\forall w\in W):v.w=0\}.$$It follows from this that $W^\perp$ is a subspace of $\Bbb R^4$, and therefore it cannot possibly consist only of those three vectors. And, actually, none of them belong to $W^\perp$.