答: Consider $f(x,y,z) = x^2 + y^2$. The solutions of $f(x,y,z)=0$ are the $z$ axis. Then just compose with an affine transformation that sends your line to the $z$ axis and you're done.

Title: Does there exist a differentiable function from $\mathbb R^3$ to $\mathbb R$ whose zeros are a line?

Body:
Suppose we are given a line in $\mathbb R^3$.  Is there a differentiable function, $f$, from $\mathbb R^3$ to $\mathbb R$ such that the solutions to $f(\vec{x})=0$ are precisely the points on the line?

I would like to be able to describe a line using the zeros of a single function rather than using a parametrization.

EDIT: I deleted a false claim about not being able to use a polynomial.


问:
Poisson distribution with $E(X^2) = 12.$
I tried to resolve this issue, but I don’t know if it’s correct, could anyone appreciate the resolution and help me?

The emission of a radioactive source is such that the number of particles emitted in a
time period, given in seconds, $X$, has Poisson distribution with $E[X^2] = 12.$
observed the issue during that period, what is the probability to be issued $4$ or
more particles?

The key is that is that the mean and variance of a Poisson distribution are both equal to λ. In particular
$\lambda=\text{Var}(X)=E[X^2]-(EX)^2=12-\lambda^2\iff\lambda^2+\lambda-12=0 \quad (\lambda>0)$

We have $\lambda = 3$
\begin{align*}
  P(X=x)=\frac{e^{-\lambda}.\lambda^{x}}{x!} \Rightarrow P(X\geq 4)&= 1-P(X<4)\\
  &=1-\left[\frac{e^{-3}\cdot 3^{0}}{0!}+\frac{e^{-3}\cdot 3^{1}}{1!}+\frac{e^{-3}\cdot 3^{2}}{2!}+\frac{e^{-3}\cdot 3^{3}}{3!}\right]\\
  &= 1-[e^{-3}\cdot13]\\
  &=1-0,6472\\
  &=0,3528.
\end{align*}

Thanks!

答:
Since $Var(X)=\lambda=E[X^2]-(E[X])^2=E[X^2]-\lambda^2$ you get the quadratic equation $\lambda^2+\lambda-12=0$ hence $\lambda=-4$ or $\lambda=3$, therefore $\lambda=3$. Then
$$
P(X\ge 4)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3),
$$
which you easily compute with the pmf.

ContentLicense: CC BY-SA 4.0


问:
Lagrange Multiplier Problem - Distance from point With a Circle Constraint
Given a point $P:(x_0,y_0)$ in $\mathbb{R}^2$, and a constraint function
$$x^2+y^2=R^2$$where $R$ is the radius of the circle. The distance from $P$ to any point on the circle is to be minimized using the method of Lagrange Multiplier. The distance $d$ is given by
$$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}.$$
Then the Lagrange function becomes
$$\mathcal{L}(x,y,\lambda)=d(x,y)+\lambda\ (x^2+y^2-R^2)$$
with the optimality conditions $\partial_x\mathcal{L}=0$, $\partial_y\mathcal{L}=0$ and $\partial_\lambda\mathcal{L}=0$. These conditions yield:
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(x-x_0)+2\ \lambda\ x=0$$
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(y-y_0)+2\ \lambda\ y=0$$
$$x^2+y^2-R^2=0$$
How should I continue from this point?</p>

ContentLicense: CC BY-SA 3.0

Body:
$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(x-x_0)=-2\ \lambda\ x$

$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(y-y_0)=-2\ \lambda\ y$

$\dfrac{x}{x-x_0}=\dfrac{y}{y-y_0} \ = \ -\dfrac{1}{2\lambda}\left[(x-x_0)^2+(y-y_0)^2\right] \ \to \ x(y-y_0)=y(x-x_0) $

$ \to \ xy_0=yx_0 \  \to \ x^2=\dfrac{(x_0R)^2}{x_0^2+y_0^2} \ , \  y^2=\dfrac{(y_0R)^2}{x_0^2+y_0^2} $


问:
Prove that the infinite sum of the difference of even and odd values of the Riemann zeta function is 1/2
I am interested in finding closed form solutions for the positive odd integers of the Riemann zeta function, of which only 1 is known. Please forgive me if this is already proven or well-known, but in running some calculations it appears that the resulting sum when the positive odd integer values of the Riemann zeta function are subtracted from the positive even integer values thereof, the result is ½; that is to say:
$$\sum_{n=1}^{\infty} \biggl(\zeta(2n)-\zeta(2n-1)\biggr) = {1\over 2}$$
such that:
$$=\biggl(\zeta(2)-\zeta(1)\biggr)+\biggl(\zeta(4)-\zeta(3)\biggr)+\biggl(\zeta(6)-\zeta(5)\biggr)+...+\biggl(\zeta(2n)-\zeta(2n-1)\biggr)$$
The positive even integer values of $\zeta(n)$ are well known, but no closed form expression is currently known for the positive odd integers of $n>1$, so I've included the decimal equivalents here:
$$=\biggl(\frac{\pi^2}{6}-1\biggr)+\biggl(\frac{\pi^4}{90}-1.20205690315959428539\biggr)+\biggl(\frac{\pi^6}{945}-1.03692775514336992633\biggr)+\biggl(\frac{\pi^8}{9450}-1.00834927738192282683\biggr)$$
Performing this calculation, as $n$ increases, the sum approaches $\infty$.

If this is currently unproven and is true, it perhaps may lead to further revelations regarding the Riemann zeta function and/or finding closed form expressions for the positive odd values of $n$.

答: A possible corrected interpretation (to get rid of $\color{red}{\zeta(1)}=\infty$) is to consider the sum of $$\zeta(2n)-\zeta(2n+1)=\sum_{k=\color{red}{2}}^\infty(k^{-2n}-k^{-2n-1})=\sum_{k=2}^\infty(k-1)k^{-2n-1}$$ over $n>0$. After interchange of the summations, we get a telescoping series: $$\sum_{n=1}^\infty\big(\zeta(2n)-\zeta(2n+1)\big)=\sum_{k=2}^\infty\frac{k-1}{k^3}\left(1-\frac{1}{k^2}\right)^{-1}=\sum_{k=2}^\infty\frac{1}{k(k+1)}=\frac12.$$


问:
Find the closed formula for the following sequence $3, 3, 4, 8, 26\cdots$
Find the closed formula for the following sequence$:\quad 3, 3, 4, 8, 26\cdots$

I have been struggling with this one for a while, and am failing to see what I'm missing. It looks like it might be related to Fibonacci's number, but I'm not sure how.

Body:
You can find this sequence starting at $n=20$ in the [OEIS database][1] (sequence A061602).
Basically it is the sum of the factorials of the digits of $n$, so for the case of $n_1=20, n_2=21$, etc., you have:
$$(2!+0!),(2!+1!),(2!+2!),(2!+3!),(2!+4!),...$$
$$=(2+1),(2+1),(2+2),(2+6),(2+24),...$$
$$=3,3,4,8,26$$
[1]: https://oeis.org/A061602


问:
Integer partitioning under constraints on number of parts and maximum elements
I have an use case which requires storing and then looking up the integer partitions of numbers. So before I go and actually store all the integer partitions, I am actually trying to check if they are feasible.I have 3 questions associated with it.

Question 1

I am not actually storing all the partitions of the number rather I am storing only those which have 16 parts and maximum element used is 128 . Hence 2048 has 1 partition and everything greater than that has 0 partitions satisfying the constraints. So is there an upper bound or an approximation to find number of such partitions i.e p(n,k,m) which is number of partitions of n into k partitions with maximum partition element used being m?

Question 2

I found out that

p(16,16,128) = 1 which is 1+1+...+1(16 times)

p(2048,16,128) = 1 which is 128+128+...+128(16 times)

p(17,16,128) = 1 which is 1+1+...+2

p(2047,16,128) = 1 which is 128+128+...+127

And I checked for few more and they were similar. As in p(n,k,m) = p(2048-(n-16),k,m) and we could derive the other partition with the one I had. Is this actually true?

Question 3

What is the best algorithm that can be used to generate such integer partitions under constraints? I had seen some code on the internet which does fairly well but it doesn't eliminate the possibilities fast enough i.e for p(2048.16,128) it takes a lot of time even though there is only one such partition.

答:
You may be interested in the following (infinite) (binary) tree:

It contains all partitions of any integer $n$ at its $n$-th level, and also as a sub-tree with same root. In the picture, for instance, the partitions of $7$ are in the dotted areas.

Notice that partitions of given length are arranged in specific sub-trees, as well as partition with given maximum part.

This make this tree very convenient for fast listing of (specific kinds of) integer partitions (Question 3); it is also helpful to derive expressions to count them (Question 1).

More details are available in: The Lattice of integer partitions and its infinite extension. See in particular Algorithm 1 and Corollary 2.


问:
Cone Area From Rotated Triangle
Given a right triangle $ABC$ rotated $2\pi$, one full circle, around line $BC$. Find area of the formed cone.

Everyone knows that the formula for a cone includes a $\frac13$ in it due to some integral calculus.

Why can the area of a cone not be visualized as a triangle rotated about an axis? Intuitively this makes sense. The area of a triangle is $\frac12bh$. The distance it is rotated is $2\pi$. By that this should give us a formula $bh\pi$. This seems to work for a  cone with $r = 3$ and height $4$. (Aka a $3,4,5$ Right Triangle Rotated).

Is there a non-calc based approach for understanding why this does not work?

答: The area is the area of the base ($\pi r^2$ where $r = \overline{AB}$) and the area of the curved portion is $\pi R^2 \theta$ where $R = \overline{AC}$ and $\theta$ is the angle subtended by the surface if flattened onto a plane.  The length of the perimeter is $\pi r^2$ and that is the proportion of the full $2 \pi$ central angle by $\frac{2 \pi r}{2 \pi R}$.

Title: Cone Area From Rotated Triangle


问:
is there any Mathematical notation to show must happen or may happen events?
There are two types of events

1- it will must happen

2- it may or may not happen depending on previous circumstances.

```<code>Must happening event: NT ∪ {NTn} and ETH ∪ { ETHn }  (given: n=1)
May happen event: NA ∪ {NAn} and NH ∪ {NHn}   ( given: n=1 ∨ n>=1)
</code>```

Must happening event is we will add 1 new element in set.

May happen event is we may add new elements in set if it is not already there.

Do we have any mathamtical notaions for that? I am not in math major so I am not sure, what i am looking is really math or something else.

答:
You put the probability tag on your question, but there is nothing about probability in the question.   So I will answer both for probability theory, and for other ideas that fit better with the way that you put things.

You have at least two ideas here.  One concerns whether something must happen or might not happen.  The other is the idea is that the "must" and "may" are conditional on other things happening.  A third idea concerns adding things to a set.

Without probability:

Note that if something must happen, then it also may happen.  If you mean to use "may" to imply not must, then what you mean is possibly, and possibly not.  We also say "necessarily" instead of "must" in mathematical logic.

The the standard notation for necessarily is "$\Box$".  The standard notion for possibly is "$\diamond$".

So event $A$ must happens, is $\Box A$.  To say that $A$ might happen, but it might not, you can write $\diamond A \;\&\; \neg\! \diamond\! A$, where $\neg$ means not.

For the conditional, it looks like you might mean if/then: If $n=1$ then $A$.  If you want a symbol for if/then, people use various kinds of right arrows, as in $(n=1) \rightarrow A$.  In mathematical logic some people use "$\supset$" instead of an arrow; this symbol has a different meaning in set theory.

Using probability:

In probability theory, to say that an event $A$ must happen, you say that its probability is 1: $\mathsf{P}(A)=1$.  (Actually, this is not precisely correct, but at this stage of learning, it will be close enough.)

If you want to say that event $A$ has some chance of occurring, but is not guaranteed to occur, you can say that it has a probability is less than 1: $\mathsf{P}(A) < 1$.

If you want to say that $A$ cannot happen, that's $\mathsf{P}(A) = 0$.  (Again, this is not strictly correct, but it's a good starting point for now.)

If you want to think in terms of probability, and that's why you're using "given", then if/then is not the right relationship.  Then I think you want to use conditional probability.  The conditional probability that $A$ will occur, given that $B$ occurs (or is true), is written $\mathsf{P}(A|B)$.  For most purposes this can be defined as:

$$\mathsf{P}(A|B) = \frac{\mathsf{P}(A \& B)}{\mathsf{P}(B)}$$

Finally, I think you know, basically, how to notate unions between sets.  Just remember that if $a$ is the sort of thing that can go in set $S$, since the union operator connects two sets, write $\{a\} \cup S$, where $\{a\}$ is the set containing only $a$.


问:
Aproximating function by polynomials of degree at least $2$.
I was working in the next exercise but I don't know what to do. The exercise is the next:

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suposse that for some $\alpha>1$ the inequality $$\left|\dfrac{f(x)}{x^{\alpha}} \right|\leq 1$$holds for all $x\in (0,1]$. Prove that $f$ can be aproximated uniformly by polynomials of the form $a_2x^{2}+\cdots+a_nx^{n}$ with $n\geq 2$.

I really don't know how to proced. My first idea was to think in Stone-Weierstrass theorem because then I can find a sequence of polynomials that converges uniformly to $f$ but the problem is the degree. How can I "delete" the first terms of the polynomials and don't lose the convergence? Is it possible? Anothe idea was to think in Taylor series but I'm not sure that $f$ is an infinitely many differentiable function. Any hint? I really aprecciate any help you can provide me. Thanks.

答:
The set of all polynomials $p$ with $p'(0)=0$ is a subalgebra of $C(0,1]$ which contains constants and separates points. Hence, given any continuous function $f$ and $\epsilon >0$ we can find a polynomial $p$ such that $p'(0)=0$ and $|f(x)-p(x)| <\epsilon $ for all $x$. Note that $|p(0)| <\epsilon$ if $f(0)=0$. Let $q(x)=p(x)-p(0)$. Then $q(0)=q'(0)=0$ and $|f(x)-q(x)| <2\epsilon$ for all $x$.

I did not use the full force of the hypothesis. I only need $f(0)=0$.

A much simpler proof: There is a polynomial $p$ such that $f(\sqrt x)-p(x)| <\epsilon $ for all $x$. If $q(x)=p(x)-p(0)$ then  $|f(\sqrt x)-q(x)| <2\epsilon $ for all $x$. Just replace $x$ by $x^{2}$ to finish the proof.


问:
Kernel of a surjective homomorphism of free modules over a Noetherian ring
$R$ is Noetherian, $f:R^m \rightarrow R^n$ is surjective. Is $\ker(f)$ free?</p>

答:
Let $K = \operatorname{ker} f$.  We have a short exact sequence

$0 \rightarrow K \rightarrow R^m \rightarrow R^n \rightarrow 0$.

Since $R^n$ is free, it is projective, so the sequence splits: $R^m \cong K \oplus R^n$.  Thus $K$ is finitely generated projective, which over certain rings already implies it must be free: e.g. this is true if $R$ is a PID.

Going deeper, the condition $R^m \cong K \oplus R^n$ says precisely that $K$ is stably free.  Such modules need not be free in general, but to give non-free examples is a subtle matter.  (For the cognoscenti: stably free modules are precisely the modules which algebraic K-theory cannot distinguish from free modules.)

A discussion of stably free modules -- including some examples of non-free ones -- can be found in $\S$ 6.5 of these notes.


问:
Partial Derivatives - Find the tangent line
So I'm trying to find the tangent line of the form $(ax + by = c)$ to the level curve at point $(1,1)$ of $$f(x,y) = 3x^2y^2+2x^2-3x+2y^2$$

I'm not quite sure how to go about this.
If I take both derivatives ($f_x$ and $f_y$) and plug in the points individually it doesn't really make much sense, I was hinted towards trying implicit differentiation but that also seems a bit weird. Would appreciate some help, thanks in advance!

答:
Hint:

A level curve of a surface is a curve where $f(x,y)$ is constant. Since we want the level curve that contains $(1,1)$, we plug in this point to get $f(1,1)=4$. So we want to find the line tangent to $$4=3x^2y^2+2x^2-3x+2y^2$$ through the point $(1,1)$. Now, you should use implicit differentiation to find $\frac{dy}{dx}$.

If you are looking to use the partial derivatives instead of the implicit differentiation, for a level curve $F(x,y)=k$, that $$\frac{dy}{dx}=\frac{-F_x}{F_y}.$$

Either way, once you have $\frac{dy}{dx}$, use point-slope form $y-y_0=m(x-x_0)$ to get the equation of the tangent line.


答:
First note that the triangle is not rotated through "a distance of $2\pi$" but through an angle of $2\pi$, which is not the same thing.

Also, you seem to be assuming that the volume will be the area multiplied by the angle of rotation, which is not true.

A similar result which is true is Pappus' theorem, which states that the volume swept out by a plane figure when rotated is equal to its area, multiplied by the distance travelled by its centroid.

Title: Cone Area From Rotated Triangle

Body:
Given a right triangle $ABC$ rotated $2\pi$, one full circle, around line $BC$. Find area of the formed cone.

Everyone knows that the formula for a cone includes a $\frac13$ in it due to some integral calculus.

Why can the area of a cone not be visualized as a triangle rotated about an axis? Intuitively this makes sense. The area of a triangle is $\frac12bh$. The distance it is rotated is $2\pi$. By that this should give us a formula $bh\pi$. This seems to work for a  cone with $r = 3$ and height $4$. (Aka a $3,4,5$ Right Triangle Rotated).

Is there a non-calc based approach for understanding why this does not work?


答:
It appears that the equation of the curve is
$$3x^2y^2+2x^2-3x+2y^2=4$$ as $(1,1)$ falls on it.
D. w.r.t $x$:
$$6xy^2+6x^2yy'+4x-3+4yy'=0 \implies y'=\frac{3-4x-6xy^2}{6x^2y+4y}$$
$m=y'(1)=\frac{-7}{10}$
So the equation of the tangent on the curve at point $(1,1)$ is
$$y-1=\frac{-7}{10}(x-1)\implies 7x+10y=17.$$

Title: Partial Derivatives - Find the tangent line

Body:
So I'm trying to find the tangent line of the form $(ax + by = c)$ to the level curve at point $(1,1)$ of $$f(x,y) = 3x^2y^2+2x^2-3x+2y^2$$

I'm not quite sure how to go about this.
If I take both derivatives ($f_x$ and $f_y$) and plug in the points individually it doesn't really make much sense, I was hinted towards trying implicit differentiation but that also seems a bit weird. Would appreciate some help, thanks in advance!


问:
Prove $f$ is not of bounded variation function.
Let $f(x)=\dfrac{1}{x}$, for $0<x\leq 1$ and $f(x)=0$, for $x=0$. Prove $f$ is not of bounded variation function.

This is my attemp.

\begin{align}
V(f,[0,1])&=\sum\limits_{k=1}^{n} \left\vert f{(x_k)}-f{(x_{k-1})}\right\vert\\
&=\left\vert f{(x_1)}-f{(x_{0})}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(x_n)}-f{(x_{n-1})}\right\vert
\\
&=\left\vert f{(x_1)}-f{(0)}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(1)}-f{(x_{n-1})}\right\vert
\\
&= \left\vert\dfrac{1}{x_1}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\\
&= 1-0\\
&=1
\end{align}
Now I confused to prove $f$ is not of bounded variation function, because $V(f,[0,1])=1<M$, for some $M>0$.
So, anyone can correct my answer if my answer is incorrect?

答:
Bounded variation requires the bound to hold for every finite set $x_k$, not just for one set.

To find a set where the total variation is large, you need some $x_i$ to be positive but small.  In fact, you can do it with just one intermediate point. Let  $M$ be given (with $M>1$) and let $x_1=1/M$

$$|f(0)-f(x_1)|+|f(x_1)-f(1))| = M + (M-1)=2M-1> M$$


问:
Proof that inequality involving logarithms and exponentials holds.
Let $a\in(0,1)$ and $x\geq 1$. I want to show that

$$g(a,x)=-\ln(a)\cdot a^x\cdot x -1+a^x\leq 0.$$

My attempt: I can define $G(x)=\frac{1}{x}(1-a^x)$ and then write

$$\int_a^1z^{x-1}\,dz=G(x).$$

Then, as $0\leq z^{x-1}\leq 1$ for every $x\geq 1$ and $z\in(0,1)$, the dominated convergence theorem allows me to exchange integral and derivative to get

$$G'(x)=\int_a^1\log(z)z^{x-1}dz\leq 0,$$

Finally, direct computation yields $G'(x)=\frac{g(a,x)}{x^2}$, so we conclude that $g(a,x)\leq 0$.

My question is if anyone have any ideas to show that $g(a,x)\leq 0$ directly. Thanks!

答:
Here is another direct way just using that $e^t > 1+t$ for $t>0$.

Set $a =\frac 1b$ with $b>1$. Your inequality becomes

$$\ln b\cdot b^{-x}\cdot x - 1 +b^{-x}\leq 0$$

Now, multiply by $b^x$ and rearrange

$$\ln b \stackrel{x>0}{\leq} \frac{b^x-1}{x}$$

which is true because of

$$\frac{b^x-1}{x} = \frac{e^{x\ln b}-1}{x}> \frac{1 + x\ln b - 1}{x} =\ln b$$

Done.


问:
Eigenvalue of f and df
Given 1 is not an eigenvalue of $df$ at $x_0$, take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood, think of $f$ as a map from open ball in $\mathbb{R}^n$ (say $B$), to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$  such that $f(0)=0.$ Then $det(f - id)(0)\neq 0$.

But I thought it should be det$(df-id)(0) \neq 0$? Did we assume f is linear map so f = df here?

答:
This question is answered by Ted Shifrin.

Yes this should be det$(df - id)_0 \neq 0$. I can't assume that $f$ is linear.

Title: Eigenvalue of f and df


问:
Solving $\frac{\cos(2x)^2}{\sin(2x)^2} = 1 +\frac{3}{\cos(2x)}$ for $x$
Professor said it should be easy....

How can one find $x$ here?

$$\frac{\cos(2x)^2}{\sin(2x)^2} = 1 +\frac{3}{\cos(2x)}$$

I've already tried my textbooks and Mathematica. Could it just be undefined?

答:
$$\frac{\cos(2x)^2}{\sin(2x)^2} = 1 +\frac{3}{\cos(2x)}$$ Let $c=\cos(2x)$ to make
$$\frac{c^2}{1-c^2}=1+\frac 3c\implies 2 c^3+3 c^2-c-3=0$$ Using the method for cubic equations, the first thing to notice is that $\Delta=-307$; so there is only one real root.

Using the hyperbolic method for this case, the solution is given by
$$c=-\frac{1}{2}+\sqrt{\frac{5}{3}} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{12
   }{5}\sqrt{\frac{3}{5}}\right)\right)$$ and then
$$x=\pm \frac 12 \cos^{-1} (c)\sim 0.224024$$ which is not the simplest result.

However, you can approximate the result considering that you loof for the zero of function
$$f(x)=\frac{\cos(2x)^2}{\sin(2x)^2} - 1 -\frac{3}{\cos(2x)}$$ which not very convenient because of the vertical asymptotes. Since the solutions cannot be $0$,$\frac \pi 4$, $\frac \pi 4$, consider instead
$$g(x)=\cos(2x)\sin^2(x) f(x)=\cos ^3(2 x)-\sin ^2(2 x) \cos (2 x)-3 \sin ^2(2 x)$$ Since the function is even, we just need to focus on the positive root.

By inspection or graphing, we can notice that the root is close to $0$. So, trying with Taylor series, we have
$$g(x)=1-22 x^2+\frac{130 x^4}{3}+O\left(x^6\right)$$ Ignoring the higher order terms, we face a quadratic in $x^2$ and the smallest root is then given by
$$x=\sqrt{\frac{1}{130} \left(33-\sqrt{699}\right) }\sim 0.224660$$ which is not bad as an approximation.

Addendum (for you personal curiosity)

We could write $g(x)$ as an infinite series

$$g(x)=1+\sum_{n=1}^\infty (-1)^n \frac{2^{2 n-1} \left(3\ 2^{2 n}+3^{2 n}+1\right) }{(2 n)!}\,x^{2 n}$$ If we truncate it to any order to make it  a truncated Taylor series, we can inverse it and have for example
$$x=t+\frac{65 }{66}t^3+\frac{11923 }{4840}t^5+\frac{108895
   }{13552}t^7+\frac{45684073 }{1533312}t^9+O\left(t^{11}\right)$$ where $t=\sqrt{\frac {1-g(x)}{22}}$. Making $g(x)=0$ as desired, the above would give
$$x=\frac{13209412121291}{12571563755520 \sqrt{22}}\sim 0.224018$$


答:
$$\pi=\{a=0\leq x_1<x_2<...<x_n\leq b=1\}$$take $x_1=\epsilon$ now
$$ \left\vert\dfrac{1}{x_1}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert=\\
 \left\vert\dfrac{1}{\epsilon}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{\epsilon}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \\
\left\vert\dfrac{1}{\epsilon}-0\right\vert
+\left\vert\dfrac{1}{1}-\dfrac{1}{\epsilon}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \frac2{\epsilon}-1+|\frac1{x_2}-\frac1{x_3}|+...\\
\geq \frac2{\epsilon}-1\\
\to \infty$$

Body:
Let $f(x)=\dfrac{1}{x}$, for $0<x\leq 1$ and $f(x)=0$, for $x=0$. Prove $f$ is not of bounded variation function.

This is my attemp.

\begin{align}
V(f,[0,1])&=\sum\limits_{k=1}^{n} \left\vert f{(x_k)}-f{(x_{k-1})}\right\vert\\
&=\left\vert f{(x_1)}-f{(x_{0})}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(x_n)}-f{(x_{n-1})}\right\vert
\\
&=\left\vert f{(x_1)}-f{(0)}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(1)}-f{(x_{n-1})}\right\vert
\\
&= \left\vert\dfrac{1}{x_1}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\\
&= 1-0\\
&=1
\end{align}
Now I confused to prove $f$ is not of bounded variation function, because $V(f,[0,1])=1<M$, for some $M>0$.
So, anyone can correct my answer if my answer is incorrect?


问:
ambiguity concerning the fundamental group of the circle
Let $\alpha(s)$ be the loop in $S^1$ starts from $1$ and goes to $-1$ and back along the upper hemisphere and let $\beta(s)$ be the loop that goes from $1$ to $-1$ and back along the lower hemisphere. Since both loops are not surjective they are null-homotopic so that their product is null-homotopic as well.

However, we can interpret the product $\alpha(s)\cdot \beta(s)$ in the following way:
We let $h(s)$ be the path from $1$ to $-1$ along the upper hemisphere and then we go around the a full circle starting and ending at $-1$ and denote this by $\gamma(s)$ (note that $\gamma(s)$ is not null-homotopic). Then we go back from $-1$ to $1$ along the lower hemisphere and denote this $g(s)$.

But this defines and isomorphism of loops between  the fundamental groups based at $1$ and $-1$ i.e. $H(\gamma)= [h\cdot \gamma \cdot h^{-1}]=  [h\cdot] [\gamma][ \cdot h^{-1}]$. But since $g(s)$ homotopic to $h^{-1}(s)$ (since they are both null-homotopic) it follows that $H(\gamma)\neq 0$.

So, we conclude that $\alpha \cdot \beta$ is null homotopic and not null-homotopic and this where I am confused.

答:
There are at least two problems with this argument:

$h$ and $g$ are not loops, but paths, so one cannot do loop-equivalence-classes calculations with them;

$h$ and $g^{-1}$ are not the same function.

(If one allows homotopies of paths that don't fix the far endpoint, then every path from a given basepoint is homotopic to every other.)


Title: Prove $f$ is not of bounded variation function.

Body: $|\frac{1}{x_1}-0|+|\frac{1}{x_2}-\frac{1}{x_1}|+... +|\frac{1}{x_n}-\frac{1}{x_{n-1}}|$ is not equal to $1$.  It is $[\frac{1}{x_1}-0]+[\frac{1}{x_1}-\frac{1}{x_2}]+... +[\frac{1}{x_{n-1}}-\frac{1}{x_n}]$. This works out to $\frac 2 {x_1} -1$ which is not bounded.


答:
Since my comments were all over the place and wildly incorrect, I will give a better answer here.

In the context of the isomorphism $H$, $[h^{-1}]$ denotes the PATH-homotopy class of $h^{-1}$. While it is true that $h^{-1}$ and $g$ are homotopic, they are not PATH-homotopic. This because the homotopy from $h^{-1}$ to $g$ does not fix the endpoints throughout the entire deformation. This is obvious because the homotopy from $h^{-1}$ to some constant map cannot possible be a path-homotopy since no constant path shares endpoints with $h^{-1}$. Thus, $[h^{-1}] \neq [g]$.

Title: ambiguity concerning the fundamental group of the circle

Body:
Let $\alpha(s)$ be the loop in $S^1$ starts from $1$ and goes to $-1$ and back along the upper hemisphere and let $\beta(s)$ be the loop that goes from $1$ to $-1$ and back along the lower hemisphere. Since both loops are not surjective they are null-homotopic so that their product is null-homotopic as well.

However, we can interpret the product $\alpha(s)\cdot \beta(s)$ in the following way:
We let $h(s)$ be the path from $1$ to $-1$ along the upper hemisphere and then we go around the a full circle starting and ending at $-1$ and denote this by $\gamma(s)$ (note that $\gamma(s)$ is not null-homotopic). Then we go back from $-1$ to $1$ along the lower hemisphere and denote this $g(s)$.

But this defines and isomorphism of loops between  the fundamental groups based at $1$ and $-1$ i.e. $H(\gamma)= [h\cdot \gamma \cdot h^{-1}]=  [h\cdot] [\gamma][ \cdot h^{-1}]$. But since $g(s)$ homotopic to $h^{-1}(s)$ (since they are both null-homotopic) it follows that $H(\gamma)\neq 0$.

So, we conclude that $\alpha \cdot \beta$ is null homotopic and not null-homotopic and this where I am confused.


问:
dimension of graph
I have a really rudimentary question hope to clear out:

If $f: V \rightarrow V$, where $f$ is a linear map and $V$ is a vector space. Then what is the dimension of $\operatorname{graph}(f)$? To my understanding, $\operatorname{graph}(f)$ is $V \times V$, but with $\dim V$. Is this correct?

答:
As I said this question suppose to be a basic one, so user15464 answered it. Though other comments provide valuable resource and insight.

graph($f$) is the subset of $V \times V$ given by {$(v,f(v)):v\in V$}. It is homeomorphic to $V$ via projection onto the first coordinate.

For dimensions of manifolds, the dimension of the graph is the same as the dimension of $V$, since graph($f$) and $V$ are diffeomorphic.

Title: dimension of graph


问:
Which of the following admit a unique solution?
Which of the following two-point boundary value problems admit a
unique solution?

$1$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u(0) = u(1) = 0$

$2$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u(0) = u'(1) = 0$

$3$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u'(0) = u'(1) = 0$

$4.$ None of these

My attempt :$ u''=-2x \implies  r^2=-2$

so  $r=  \pm \sqrt2i$

Hence  $u(x)= c_1 \cos (\sqrt2x) + c_2 \sin (\sqrt2x)\tag1$

if $u(0)=0$

then $u(0)=c_1=0 $

If $u(1) =0$ ,then put $c_1 =0 $ in $(1)$, then we get $c_2=0$

Here we can not uniquely determine $c_2$ , so  there are infinitely many solution

Similarly we can use the same method for option  $2)$ and $3)$

So  i think that answer will be none of these i,e option $4$ will be correct

答:
$u''=-2x.$ Integrate w.r. x to get
$u'=-x^2+A$ again integrate to get
$$u(x)=-x^3/3+Ax+B~~~(1)$$ $u(0)=0$ gives $B=0$, then $u(1)=0$ gives $0=-1/3+A \implies A=1/3$. So we get the unique solution as $u(x)=-x^3/3+1/3$

Next. from (1) $u(0)=0 \implies B=0$ and $u'(1)=0$ gives u'(x)=-x^2+A \implies $A=2$, so the solution is  $u(x)=2x-x^2$

$u'(0)=0$ gives $A=0$ then $u'(x)=-x^2$, where $u'(1) \ne 0$. So no solution.


Title: Which of the following admit a unique solution?

Body:
Which of the following two-point boundary value problems admit a
unique solution?

$1$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u(0) = u(1) = 0$

$2$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u(0) = u'(1) = 0$

$3$$. −u''(x) = 2x$ in $0 < x < 1 $ and $u'(0) = u'(1) = 0$

$4.$ None of these

My attempt :$ u''=-2x \implies  r^2=-2$

so  $r=  \pm \sqrt2i$

Hence  $u(x)= c_1 \cos (\sqrt2x) + c_2 \sin (\sqrt2x)\tag1$

if $u(0)=0$

then $u(0)=c_1=0 $

If $u(1) =0$ ,then put $c_1 =0 $ in $(1)$, then we get $c_2=0$

Here we can not uniquely determine $c_2$ , so  there are infinitely many solution

Similarly we can use the same method for option  $2)$ and $3)$

So  i think that answer will be none of these i,e option $4$ will be correct

Body:
Both 1 and 2 are the right answers:

U"=-2x,  U'=$-x^2+C$,  U=-$\frac{x^3}{3}+Cx+D$

1: U(0)=U(1)=0, D=0, C=$\frac{1}{3}$, $\to$ U= -$\frac{x^3}{3}$+$\frac{1}{3}$x

U(0)=U'(1)= 0,  D=0,C=1  $\to$ U= -$\frac{x^3}{3}$+x


问:
Intuition/references for why Möbius transformations are uniquely defined by three points?
Assume that I have a Möbius transformation

$$M(z)=\frac{az+b}{cz+d}$$

with real coefficients $a$, $b$, $c$, and $d$, and complex argument $z$. I have been interested in gaining intuition on why this transformation is uniquely defined by three points $(z_1,z_2,z_3)$ and their images $(w_1,w_2,w_3)$. I combed the StackExchange and found this answer by user Lehs, which suggests the reason is that the equation above can be rewritten as:

$$M(z)=\frac{a}{c}\frac{z+\frac{b}{a}}{z+\frac{d}{c}}=A\frac{z+B}{z+C}$$

where

$$A=\frac{a}{c}$$
$$B=\frac{b}{a}$$
$$C=\frac{d}{c}$$

This sort of explains how the four coefficients $a$, $b$, $c$, and $d$ can be reduced to three coefficients $A$, $B$, and $C$, but leaves me with two new questions:

Why is Möbius transformation usually expressed in terms of $a$, $b$, $c$, and $d$ if there is a seemingly more succinct way of representing this transformation?

An earlier comment in the same question also mentions 'symmetry', and suggests that while the transformation is uniquely determined, the coefficients $a$, $b$, $c$, and $d$ are not. I can see that multiplying the four coefficients with the same non-zero factor would yield the same $A$, $B$, and $C$. Does this mean that $A$, $B$, and $C$ uniquely define a transformation?

If you have any other comments which may aid my intuition, I would greatly appreciate it. Also, if you have any references you could suggest with regards to intuition about the Möbius transformation, or which detail any of the results I collected here, that would be amazing.

答:
I have been interested in gaining intuition on why this transformation is uniquely defined by three points and their images.

A little bit of intuition comes from counting dimensions. While there are four variables written for a transformation, there are really only three degrees of freedom since they can all be multiplied by a scalar without changing the transformation. Similarly, there are three degrees of freedom in specifying three points on the complex projective line. (I will make the executive decision to assume we are talking about complex coefficients, with real coefficients as a possible special case.) This intuition is useful, however not enough to turn into a proof - in principle, it only guarantees there are "discretely" many orbits (acting on $3$-tuples of distinct points) and discrete stabilizers, since discrete sets are $0$-dimensional.

One way to show $\mathrm{GL}_2\mathbb{C}$ (or, WLOG, $\mathrm{PSL}_2\mathbb{C}$) acts sharply $3$-transitively is that only the identity fixes all three points $0,\infty,1$ and we can arrange for any three points $a,b,c$ to become $0,\infty,1$ via

$$ M(z)=\frac{z-a}{z-b}\cdot\frac{c-b}{c-a}, $$

called a cross-ratio. The factor $z-a$ ensures $a\mapsto 0$, the factor $z-b$ ensures $b\mapsto\infty$, and the second fraction is a normalizaton factor that ensures $c\mapsto1$. The ratios must be suitably interpreted using the rules $\infty/\infty=1$ and anything$/0=\infty$, which means $a,b,c$ (distinct) can also be e.g. $\infty$.

Why is Möbius transformation usually expressed in terms of a, b, c, and d if there is a seemingly more succinct way of representing this transformation?

Not every transformation can be written that way, actually. But even ignoring that, the form with $a,b,c,d$ highlights the fact that Mobius transformations originate from matrix transformations. It's no coincidence that composition of Mobius transformations is represented by matrix multiplication! One starts with $2\times2$ matrices acting on $\mathbb{C}^2$ (or $\mathbb{R}^2$, if you wish), and then forms the projective line $\mathbb{CP}^1$ (or Riemann sphere $\mathbb{C}\cup\{\infty\}$) by simply turning a pair of coordinates into a fraction, $(w,z)\mapsto w/z$.

Does this mean that A, B, and C uniquely define a transformation?

Technically, no. If $c=0$ then $A=C=\infty$ but then there is no way to figure out what coefficient should be out in front of $(z+B)$. But if $c\ne0$ then $A,B,C$ do uniquely determine $M$.


问:
Slowpoke and Doubles probability -- does it converge as the numbers of laps increases?
I posted this on puzzling stack exchange three months ago and it was immediately closed as off-topic, for being a "fairly straightforward probability calculation".

I have the solution to the problem, which I do not give here in case you might want to solve it yourself. My two questions for this forum are: 1) is this indeed an interesting problem to solve and not "fairly straightforward", and 2) how can I prove that the answer for the probability of N laps converges over time (i.e. as N increases)? (This second part I have been unable to solve.)

=== Here is the original post: ===

I created this dice and paper game for my kids years ago, and at the same time gave myself an interesting puzzle.

This game is a horse race between two horses: Slowpoke and Doubles.

With a sheet of lined writing paper, draw vertical lines to make six columns down the paper. The first five columns belong to Slowpoke, and the sixth column belongs to Doubles.

Each row across the paper represents one lap around a racetrack. Before the race starts, decide on the number of laps that will be run. You can then have fun predicting which horse will win the race.

To begin the race, roll a pair of dice. If you roll a double -- (1,1), (2,2), (3,3), (4,4), (5,5), or (6,6) -- then draw an X in Doubles' column in the first lap. Doubles has just completed the first lap!

If you roll anything other than doubles, then draw an X in Slowpoke's first column in the first lap. Slowpoke has made it one fifth of the way around the track.

Keep rolling the dice. Each time a doubles comes up, Doubles completes another lap.

Each time anything else comes up, Slowpoke advances across the current lap, left to write. Draw an X in the next free column. It will take five non-doubles rolls for Slowpoke to complete the lap.

In this way, roll after roll, one or the other horse advances with another X, until finally, one of the two horses completes their last lap and wins!

My puzzle for you is:

Is one of the two horses favored to win the race?

If so, what is the probability that a given horse wins a race of N laps?

Enjoy!

=== UPDATE:

@user2661923 I found solving the probability for Doubles is easier than for Slowpoke. It follows a simple pattern. (I'm using spoiler blocks in case others find it interesting to solve on their own.)

For one lap:

 $$ \frac {6^4 + (6^3\cdot 5) + (6^2\cdot 5^2) + (6\cdot 5^3) + 5^4} {6^5} $$

For two laps:

 $$ \frac {6^9 + (6^8\cdot 5) + (6^7\cdot 5^2) + (6^6\cdot 5^3) + (6^5\cdot 5^4) + (6^4\cdot 5^5) + (6^3\cdot 5^6) + (6^2\cdot 5^7) + (6\cdot 5^8) + 5^9} {6^{11}} $$

For n laps:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k\cdot 6^{(5n - 1) - k}} {6^{6n - 1}} $$

Which simplifies to:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k} {6^{k + n}} $$

This last equation I would like to prove continually approaches, but never reaches, 1/2.

答:
Partial Answer Only

Let $p = (1/6), q = (5/6)$

If I understand the problem correctly, after exactly $(6N - 1)$ rolls there are
two mutually exclusive possibilities.  Either there have been at least $(N)$ doubles
or there have been fewer than $(N)$ doubles.

If there have been at least $N$ doubles, then have been no more than $(5N-1)$ non-doubles.
In this scenario, Doubles must have crossed the finish line, and Slowpoke could not
have crossed the finish line.  If there were fewer than $N$ doubles, then the
situation is reversed: Slowpoke had to have crossed the finish line, and Doubles
could not have crossed the finish line.

In R Bernoulli trial's of an independent event, with probability of success
$= p$, and $q = (1-p)$ the probability of exactly
$k$ successes, for $k \in \{0,1,2, \cdots, R\}$ is $\binom{R}{k} p^kq^{(R-k)}$.

Here, this means that the probability of at most $S$ successes is

$\sum_{k=0}^S \binom{R}{k} p^kq^{(R-k)}$,

while the probability of at least $(S+1)$ successes is

$\sum_{k=(S+1)}^R \binom{R}{k} p^kq^{(R-k)}$.

Therefore, you can simply plug in the numbers of

$R = (6N-1)$, $S = (N-1)$, and $p = (1/6)$

to solve the general case of $N$ laps.

Examining the case of $N = 1$ gives Slowpoke's chances as

$$\left(\frac{5}{6}\right)^5 \approx 0.40,$$

while $N = 2$ gives Slowpoke's chances as

$$\left(\frac{5}{6}\right)^{11} + (11) \times (1/6) \times \left(\frac{5}{6}\right)^{10}
~\approx 0.43.$$

This suggests that as $N \to \infty$, Slowpokes chances are strictly increasing, and
approach $(1/2)$ from below.  This takes some proving however.  Shown below, is
analysis that goes as far as I can take it.

If you form the sequence of Slowpokes chances, based on the number of laps, as
$\{t_1, t_2, \cdots \}$, you can then simplify the analysis by expressing

$$t_n = \frac{a_n}{b_n} ~\text{where} ~b_n = 6^{(6n-1)}.$$

Then, the numerator, $a_n$ will equal

$$a_n = \binom{6n-1}{0} 5^{(6n-1)} + \binom{6n-1}{1} 5^{(6n-2)} + \cdots + \binom{6n-1}{n-1} 5^{(5n)}.$$

Somebody with a better knowledge of statistics than me, is going to have to establish that

as $n \to \infty$, $\frac{a_n}{b_n}$ is strictly increasing and approaches $(1/2)$ from below.


问:
Cosine Series of $-\frac{1}{5}\sin x$ on $[0,\pi]$
I am really confused as to why $\sin x$ has a Fourier cosine series. I thought that since $\sin x$ is an odd function, then $a_0$ and $a_n$ both equal $0$, and we find the coefficient for $b_n$. However, in the examples I have looked over, they are calculating $a_0$ and $a_n$ and subjecting $b_n$ to be equal to zero. Why is this? Maybe I am misunderstanding the how the interval ties into this. This problem is on the interval $[0,\pi]$. Please help me!

Also- I thought that Cosine series and Sine series are merely methods of shortcuts. If we know that $f(x)$ is an even function, then we only need to solve for $a_0$ and $a_n$. If we know that $f(x)$ is an odd function, then we only need to solve for $b_n$. Is this an incorrect way of thinking?

Thank you in advance!

答:
On the figure below

The arc of sine function in $0<x<\pi$ is drawn in red.

The function drawn in black is the Fourier cosine series.

This makes you understand the meaning of the problem ?


问:
If you have a set of eigenvectors with full multiplicity, is it always possible to orthogonalize them?
a)  construct a (diagonalizable) 2×2 complex symmetric matrix not admitting an orthogonal basis of eigenvectors

b)  construct a 2×2 complex symmetric matrix which cannot be diagonalized

The point of this exercise is to show that theorems that works for self adjoint or symmetric real matrices do not work for complex symmetric matrices. I had no problem with part b but I don't understand part a. If we have a diagonalizable complex symmetric matrix, that means we have a full set of eigenvalues with full geometric multiplicity otherwise, we wouldn't be able to diagonalize it. If that is the case, then we also have a full set of eigenvectors. We can just apply gram-schmidt and get a set of orthonormal vectors. The only way part a can be true is if Gram-schmidt doesn't work. Why would that be the case?

答:
It is often easier to try rank-one matrices first.

Part (b) is easy. Just pick a nonzero vector $u$ such that $u^Tu$ (note: not $u^\ast u$) is equal to zero. Then $uu^T$ is complex symmetric (this should be obvious) but not diagonalisable (do you see why?).

For part (a), find a pair of vectors $u$ and $v$ such that both $u^Tu$ and $u^\ast v$ are nonzero but $u^Tv=0$. Then the matrix $A=uu^T$ will do the trick because $u$ and $v$ are the eigenvectors of $A$ corresponding to two different eigenvalues but they are not orthogonal to each other.

Now, to answer your question:

If we have a diagonalizable complex symmetric matrix, that means we have a full set of eigenvalues with full geometric multiplicity otherwise, we wouldn't be able to diagonalize it. If that is the case, then we also have a full set of eigenvectors.

You are correct at this point.

We can just apply gram-schmidt and get a set of orthonormal vectors. The only way part a can be true is if Gram-schmidt doesn't work. Why would that be the case?

You always get an orthogonal basis of $\mathbb C^n$ by performing Gram-Schmidt orthogonalisation on an eigenbasis. However, the outcome is not necessarily an eigenbasis, and we haven't any reason to expect that it is.


问:
What is the relation?
The function $x^2 = y\quad$ limits two areas $A$ and $B$:

$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.

$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.

What are the relations between $a$ and $b$, when $Vb = Va$?

..................

I have come to the solution that:

$Vb = (\pi b^2 )/ 2$

$Va = (\pi a^5) / 5$

so the relation between them is:

$2.5b^2 = a^5$

Is that the final solution or is it more?

答: If your calculations are correct this is what you should have found.</p>

Title: What is the relation?


问:
How to think about finding the Nash equilibrium of $f(x,y) = xy$
Let $x, y \in (-\infty, \infty)$

I know that  $f(x,y) = xy$ has a equilibrium at $(0,0)$, but I do not know how to logically arrive at this solution.

I know that a Nash equilibrium is when $\min_x\max_y xy = \max_y \min_x xy$

Start from the LHS, we have,
$$\min_x\max_y xy$$
The inner player goes first, which implies $y$ player will choose $\infty$, and $x$ will choose $-\infty$, yielding a value of $-\infty$.

For, $$\max_y \min_x xy$$

The inner player goes first and chooses $x = -\infty$, and outer player y will choose $-\infty$, yielding $\infty$.

It does not seem to me that the players can arrive at $(0,0)$. Because whoever goes second sets the value of the game.

What is missing?

答:
If $x > 0$, then $xy$ can be made as large and positive as we want, if we choose $y$ to be large and positive. So, for $x > 0$, $\sup_y xy = \infty$.

If $x < 0$, then $xy$ can still be made as large and positive as we want, this time choosing $y$ to be large and negative. So, for $x < 0$, $\sup_y xy = \infty$.

If $x = 0$, then no matter what $y$ chooses, the result of $xy$ is $0$. Thus, if $x = 0$, $\sup_y xy = 0$. Therefore,
$$\sup_y xy = \begin{cases} 0 & \text{if } x = 0 \\ \infty & \text{otherwise.}\end{cases}$$

Now, if we are taking the infimum (or indeed minimum) of this extended real function over all possible choices for $x$, we clearly need to choose $x = 0$, for a minimum of $0$.

Similar logic shows that
$$\inf_x xy = \begin{cases} 0 & \text{if } y = 0 \\ -\infty & \text{otherwise.}\end{cases}$$
Thus, taking the maximum over $y$, we get $0$, which occurs when $x = 0$. Therefore, $(0, 0)$ is the Nash equilibrium.


问:
UMVU estimator for iid observations from uniform distribution
Let $X_1, \cdots, X_m$ and $Y_1,\cdots,Y_n$ be independent i.i.d observations from the uniform distributions $U(0,\theta_1)$ and $U(0,\theta_2)$ , respectively. Find the UMVU estimator of $\frac{\theta_1}{\theta_2}$ .

I know how to find UMVU estimators for $\theta$ or $\theta_2$ by themselves but I am not sure what would be procedure to find UMVU of $\frac{\theta_1}{\theta_2}$.

答:
Start in the following way

$$\mathbb{E}\left[\frac{X_{(m)}}{Y_{(n)}}\right]=\mathbb{E}\left[X_{(m)}\cdot\frac{1}{Y_{(n)}}\right]=\mathbb{E}[X_{(m)}]\cdot\mathbb{E}\left[\frac{1}{Y_{(n)}}\right]$$

You know the distribution of Max thus you can go on in your calculations...


问:
How to minimize a 2D function given a constraint?
How can I minimize
$$
2a^2 + b^2
$$
given the constraint that
$$
a+b = 1
$$

I am able to obtain the answer using Wolfram Alpha, but I'm not sure how to go through the math.

答:
Here is a way using the well known Cauchy-Schwarz inequality (C.S.):

$$1^2 = \left((\sqrt{2}\cdot a)\frac 1{\sqrt 2} + b\cdot 1\right)^2\stackrel{C.S.}{\leq}(2a^2+b^2)\left(\frac 12 + 1\right)$$
Hence,

$$2a^2+b^2 \geq \frac 23$$

Equality is reached if $(\sqrt{2}\cdot a, b)$ is parallel to $\left(\frac 1{\sqrt 2},1\right)$, which happens for $a=\frac 13$ and $b= \frac 23$.


问:
Dimension of Example Quotient Space
Let $V$ be the vector space across the set of all integers, with subspace $W$ as the vector space across all even numbers, both with the field as the integers. The quotient space $V/W=\{v+W|v\in{V}\}=\{W,X\}$ where $X$ is the set of all odd numbers, right? Since if $v$ is even, $v+W$ is all the evens, and if $v$ is odd, $v+W$ is all the odds. So, if $\dim{V/W}=\dim{V}-\dim{W}$, $\{W,X\}$ should have dimension 0, so why? Why does $\{W,X\}$ have dimension 0? If it did have a basis, wouldn't that basis just be $W$? Or is there no way to find the dimension of this set for some reason, defaulting it to 0?

答:
It appears that you are confused between different algebraic subjects. The theory of vector spaces (called linear algebra) starts by assuming the choice of a field of numbers (called scalars in the context of linear algebra). That is a structure in which the four arithmetic operations between numbers are always defined, with the exception only of division by zero, and return a number. The most popular choice seems to be the field $\Bbb R$ of real numbers (probably because geometric intuition works best for that choice), but one can perfectly well choose to work with the rational numbers $\Bbb Q$ as scalars, or on the contrary allow all complex numbers $\Bbb C$ as scalars, and many more sophisticated choices are possible. But one cannot choose the integers $\Bbb Z$ as scalars, because many divisions of integers like $24\div 15$ cannot be performed with an integer as result.
There is an important reason to insist that division by any nonzero scalar is always possible, though this reason is not immediately clear when one starts to set up linear algebra.

The definition of vector spaces require that the set of vectors be closed under addition, subtraction, and multiplication by scalars. If you apply that definition to the set of integers, you will find that (whatever field you chose to use) there are scalars that you cannot always multiply by; for the most modest choice of $\Bbb Q$ as scalars, you cannot always multiply by$~\frac23$, or more generally by any fraction that is not an integer (and if you choose $\Bbb R$ or $\Bbb C$ as field, there are even more problematic scalars). So $\Bbb Z$ is not a vector space, over any field.

It is possible to set up a theory that is not linear algebra, in which one does not insist that division by nonzero scalars is always possible. One then has a ring of scalars, and the notion of vector space is replaced with that of a so-called module over the ring. But dropping the division requirement does not simplify things; on the contrary. Rings that are not fields can be complicated structures that are hard to work with, and often one needs additional assumptions on the ring used to ensure that some complications will not happen; this leads to a great variety of kinds if rings with additional requirements weaker than that of being a field (like principal ideal domains, which are fairly nice rings and of which $\Bbb Z$ is an example). The theory of modules over rings is harder to develop than that of vector spaces, and one of the reasons for this is that there does not exists a neat notion like dimension. Therefore what you want to argue cannot be done in (the $\Bbb Z$-module of) the integers; the notion of dimension simply does not apply there. For what its worth, you can form the submodule $W$ of the even integers in the module $V$ of all integers, and form the quotient module $V/W$, which has two elements as you correctly argue. But that's all one can say; $V/W$ has no dimension (nor do $V$ and $W$).

One might wonder where in linear algebra theory the requirement of dividing by scalars really plays a role. The most basic place where this is crucially the case is the statement: if one has a linear combination of distinct vectors with nonzero scalar coefficients that gives the zero vector, then and one of those vectors can be expressed as linear combination of the others. For instance if $\alpha u+\beta v+\gamma w=\vec0$ (with the Greek letters scalars and the Romans vectors), then $u=-\frac\beta\alpha v-\frac\gamma\alpha w$. Working in a module over a ring that is not a field, one can easily have such linear relations for which it is not possible to express any of the elements $u,v,w$ in the relation in terms of the others. The possibility to use such expressions is vital to obtain the theoretical result that makes the notion of dimension possible (namely that all choices of a basis for a given vector space involve the same number of vectors).

Title: Dimension of Example Quotient Space


问:
Existence of a non-zero T- invariant subspace on $\mathbb{Q}^{4}$
For any linear transformation $T:\mathbb{Q}^{4} \to \mathbb{Q}^{4} $, does there always exist a non-zero T- invariant subspace?

As we know

For any linear transformation $T:\mathbb{R}^{4} \to \mathbb{R}^{4}$, there exists a non-zero T- invariant subspace.

Does completeness property cause some differences for T- invariant subspaces?

答: Take any $4 \times 4$ matrix of rational numbers whose characteristic polynomial is irreducible over $\Bbb{Q}$, and form a linear operator $T$ on $\Bbb{Q}$ based on this matrix. If $V \le \Bbb{Q}^4$ is some proper non-trivial invariant subspace of $T$, then $T|_V$ is an operator on $V$, whose characteristic polynomial has degree $0 < \dim(V) < 4$, and which divides the characteristic polynomial of $T$. But, this can't happen, as the characteristic polynomial of $T$ is irreducible.


问:
Finding limit of sequence $a_n=\frac{\alpha(n)}{n}$
Given sequence $a_n=\frac{\alpha(n)}{n}$, where $\alpha(n)$=total no. of primes by which $n$ is divisible, find $\lim_{n\to \infty}a_n$.

Example: $\alpha (8)=1$, as $2$ is the only prime by which $8$ is divisible.

Observing that $\alpha(n)\lt n$ for all $n\in \mathbb N$, the given sequence is bounded. Then I tried using the fact that every $n$ has prime factorisation.

But I don't know how to proceed further. Any hints?

答: Hint: You can have a stricter bound than just $n$ for $\alpha(n)$. If every prime factor were a $2$ and repeated prime factors were allowed, then $\alpha(n) = \log_2(n)$, which is an upper bound because other prime factors have to be larger than $2$. Then, an upper bound on the limit is $$\lim_{n \to \infty}\frac{\log_2(n)}{n}$$


问:
finer topology and topology induced by norms
I have two questions. First: Let $d_1,d_2$ be metrics in $M$ and consider the open balls $$B_r^1(x)=\{y \in M; d_1(x,y)<r\}$$ $$B_r^2(x)=\{y \in M; d_2(x,y)<r\}.$$

a) Proof that, if for all $x \in M$ and $r>0$ there is $\delta>0$
satisfying $B_\delta^2(x) \subset B_r^1(x)$, then the topology $\tau_2$ generated by $d_2$ is finer then $\tau_1$ generated by $d_1$.

b)Consider two norms $\|\cdot\|_1,\|\cdot\|_2$ in a vector space. If we have $$\alpha\|x\|_1\leq\|x\|_2\leq \beta \|x\|_1$$ then $\|\cdot\|_1,\|\cdot\|_2$ generate the same topology.

For the first question, i just need to prove that every open ball in $\tau_1$ is in $\tau_2$? Let $a \in B_r^1$, then $d_1(x,a)<r$ and proof that $d_2(x,a)<\delta$? I'm just confused with this.

For the second question, i saw this question: Equivalent metrics generate the same topology. We have the same steps? Sounds logical to me, but my textbook does not have nothing on topology induced by norms.

答:
As to a): Let $O$ be open for $d_1$. Let $x \in O$. Then there is some $r>0$ such that $B_r^1(x) \subseteq O$ as $x$ is a $\tau_1$-interior point of $O$.

Then the condition ensures that there some $\delta>0$ such that $B_\delta^2(x) \subseteq B_r^1(x)$, so that $x$ is a $\tau_2$ (or $d_2-$) interior  point of $O$. As $x \in O$ was a arbitrary, $O \in \tau_2$. So $\tau_1 \subseteq \tau_2$ and the claim has been shown.

For the rest, see my answer at the quoted question here. We really apply $(a)$ twice one you realise how the condition is fulfilled from the inequalities. The inequalities for the norm immediately imply those for the corresponding metrics $d(x,y)=\|x-y\|$ with the same constants, of course. So the fact that we have a norm is not really relevant.


问:
prove $\prod_{i=0}^{p-1}(y+ix) \equiv y^p - x^{p-1}y \mod p$ where $p$ is prime
The coefficient of $y^p$ is obviously 1 and coefficient of $x^{p-1}y$ is $(p-1)!$ which by wilson's theorem is $-1 \pmod{p}$

Now further thinking leads me to all the coefficients as being stirling numbers of first kind. And all the stirling numbers ${p \brack k}$ for $1 < k < p$ are divisible by $p$ if $p$ is prime.

But how to prove this? If each stirling number of first kind can be written as a binomial coefficient or product of binomial coefficients of $p$, then we are all set.

Thank you

答:
Hint: Consider the polynomials (over the field $\Bbb F_p$)
\begin{align}
f(x)&=\prod_{i=0}^{p-1}(x-i)&
g(x)&=x^p-x
\end{align}
They have the same roots, leading coefficient and same degree, hence $f(x)=g(x)$.


问:
When does $\lim_{x \to \infty} \frac{f(x)}{g(x)}=\infty \Rightarrow \lim_{x \to \infty} f(x)-g(x) = \infty$?
Suppose $f,g$ are real continous function and $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to \infty} g(x) = \infty$. When does $\lim_{x \to \infty} \frac{f(x)}{g(x)}=\infty \Rightarrow \lim_{x \to \infty} f(x)-g(x) = \infty$?

So in other words I am looking for a result to determine that $f$ "grows faster" than $g$. Say for instance I am looking at $h(x) = x^7-100x-\log(1+x^{50})$ with $f(x) = x^7$ and $g(x) = 100x+\log(1+x^{50})$. I would like to compare them as a fraction as then we have L'Hopital's rule or statements about exponential growth vs polynomial growth etc.
Any help is appreciated!

答: You cannot prove it, since it is not true. Just take, for instance $f(x)=e^x+x$ and $g(x)=e^x$.


答:
You could have done it a bit faster and get more than the limit using generalized harmonic numbers.

$$S_n=\sum_{k=1}^{3n^2}\frac{1}{\left(8 n^3+2k\right)^{3/2}}=\frac 1{2\sqrt 2}\Bigg[H_{4 n^3+3 n^2}^{\left(\frac{3}{2}\right)}-H_{4
   n^3}^{\left(\frac{3}{2}\right)} \Bigg]$$

Now, using the asymptotics
$$H_{p}^{\left(\frac{3}{2}\right)}=\zeta \left(\frac{3}{2}\right)-\frac 2{p^{1/2}}+\frac  1{2p^{3/2}}-\frac  1{8p^{5/2}}+O\left(\frac{1}{p^{9/2}}\right)$$ apply it twice and continue with Taylor series to get
$$S_n=\frac{3}{16 \sqrt{2}\, n^{5/2}}\Bigg[1-\frac{9}{16 n}+\frac{45}{128 n^2}-\frac{1713}{4096
   n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$

Title: Calculate the limit $\lim\limits_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$

Body:
I have to calculate the limit $$\lim_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$$

I tried to use Sandwich Theorem like this $$\frac{3n^{2}}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}} \leq \Bigg(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\Bigg) \leq \frac{3n^{2}}{\sqrt[3]{(8n^{3}+2)}^{2}}$$

And for result I got that limit is $\frac{3}{4}$

Is this correct?


问:
Strange AP Calculus BC question help?
The question is $\int_0^3\frac{1}{(1-x)^2}$. I got an answer (from u-substitution) however the solution manual says that the integral does not converge. Someone told me that the integral is undefined at $x = 1$, but if we look at $\frac {1}{x}$ that function is undefined at $x=0$ yet we can still integrate it. </p>

<p>Can someone explain what is going on?! </p>

答:
You have an improper integral due to your function being unbounded on the interval. Anyways, you can interpret it in terms of limits of Riemann integrals as follows:$$\int_0^3\frac1{(1-x)^2}\,\mathrm{d}x=\lim_{a\to1^-}\int_0^a\frac1{(1-x)^2}+\lim_{b\to1^+}\int_b^3\frac1{(1-x)^2}\,\mathrm{d}x$$
You can determine the indefinite integral of $1/(1-x)^2$ with ease:$$\int\frac{\mathrm{d}x}{(\underbrace{1-x}_{\text{u}})^2}=-\int\frac{\mathrm{d}u}{u^2}=\frac1{1-x}$$
Using the fundamental theorem of calculus we may then rewrite our definite integrals as follows:$$\int_0^a\frac{\mathrm{d}x}{(1-x)^2}=\frac1{1-a}-1\\\int_b^3\frac{\mathrm{d}x}{(1-x)^2}=-\frac12-\frac1{1-b}$$Recognize that in the limit as $a\to1^+$ and $b\to1^-$ both of these are unbounded, leaving us with a divergent integral.

Alternatively, recognize $\dfrac1{1-x}=1+x+x^2+\dots$; it is clear this series diverges for $x>1$. You could use this to reason this reason about the behavior of our integral and understand why it diverges (though it is not generally enough to show that it does indeed diverge).

Title: Strange AP Calculus BC question help?


问:
Prove $\mathbb{E}(X|Y,Z) = E(X|Y)$ a.s. given that $Z$ is independent of $Y$ and of $X$.
I am stuck trying to prove the following statement involving conditional expectation of random variables:

Suppose $Z$ is independent of $Y$ and of $X$, and $E|X|<\infty$, then  $\mathbb{E}(X|Y,Z) = E(X|Y)$ a.s..

Here's my attempt: by definition of conditional expectation, it suffices to show the following two conditions:

$\int_A \mathbb{E}(X|Y,Z) dP=\int_A XdP $ for all $A\in \sigma(Y)$.

$\mathbb{E}(X|Y,Z)$ is $\sigma(Y)$ measurable.

The first condition is easy, as for every $A\in \sigma(Y)$, $A\in \sigma(Y,Z)$. Thus by definition it automatically follows that $\int_A \mathbb{E}(X|Y,Z) dP=\int_A XdP $.

However, for the second condition, I have little clue as to what to do. In particular, I haven't figure out a way to utilize the independence assumption to prove measurability. I would appreciate thoughts on whether this is the right approach.

答: This is false. Let $A,B,C$ be pair-wise independent but not jointly independent. Take $X=I_A,Y=I_B$ and$Z=I_C$. Then the the integral of LHS over $Y^{-1}(\{1\}) \cap Z^{-1}(\{1\})$ is $P(A\cap B \cap C)$  whereas RHS is $P(A)$ so the integral of RHS over $Y^{-1}(\{1\}) \cap Z^{-1}(\{1\})$ is $P(A)P(B \cap C)=P(A)P(B)P(C)$.


答:
This is true as stated now. Set $\alpha(x) = f(x) / g(x) $. Then

$$\lim_{x \to \infty} f(x) -g(x) = \lim_{x \to \infty} \alpha(x) g(x) - g(x) =\lim_{x \to \infty} (\alpha(x) -1) g(x) = \infty $$

Because both terms of the product have limit equal to infinity.

Body:
Suppose $f,g$ are real continous function and $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to \infty} g(x) = \infty$. When does $\lim_{x \to \infty} \frac{f(x)}{g(x)}=\infty \Rightarrow \lim_{x \to \infty} f(x)-g(x) = \infty$?

So in other words I am looking for a result to determine that $f$ "grows faster" than $g$. Say for instance I am looking at $h(x) = x^7-100x-\log(1+x^{50})$ with $f(x) = x^7$ and $g(x) = 100x+\log(1+x^{50})$. I would like to compare them as a fraction as then we have L'Hopital's rule or statements about exponential growth vs polynomial growth etc.
Any help is appreciated!


答:
$g(x) = 100x+\log(1+x^{50}) < 100x + \log(2x^{50}) = 100x + \log 2 + 50 \log x < 100x + 50x+\log2$ as $\log_{10} x < \ln x < x$ for all $x > 1$.

Thus you are comparing $x^7$ to $150x$, which means that the first grows faster.

Title: When does $\lim_{x \to \infty} \frac{f(x)}{g(x)}=\infty \Rightarrow \lim_{x \to \infty} f(x)-g(x) = \infty$?


问:
Understanding of Tangent Plane in R3
Given a sphere $(x-2)^2+(y-1)^2+(z+1)^2=9$, steps to find the equation of the tangent plane to this sphere at point (1,-1,1) are as follows:

$$(x-2)^2+(y-1)^2+(z+1)^2=9$$
$$(x-2)^2+(y-1)^2+(z+1)^2-9=0$$
$$(x-2)^2+(y-1)^2+(z+1)^2-9=f(x,y,z)$$

$$\nabla f(1,-1,1)=(-2,-4,4) $$
$$f_x(1,-1,1)(x-1)+f_y(1,-1,1)(y+1)+f_z(1,-1,1)(z-1)=0$$

When setting the equation of the sphere to 0, then setting the equation equal to a fourth variable, $f$ is not a sphere anymore.

If the sphere equation is altered in this way, how does the partial derivatives of $f$ associate back to the sphere in that they are apart of the tangent plane equation for the sphere???

答:
The equation of the tangent plane is:
$$-2(x-1)-4(y+1)+4(z+1)=0 \implies x+2y-2z=1.$$

Title: Understanding of Tangent Plane in R3


问:
Calculating weighted geometric returns
Consider the quarterly sales figures and returns in the table below:

Qtr
Sales ($m)
Return (%)

Q4 19
8

Q1 20
10
25

Q2 20
13
30

Q3 20
15.6
20

Q4 20
23.4
50

The year on year return for Q4 20 is:
$$
Ret = \frac{R_4}{R_1}-1 = \frac{23.4}{8} - 1 = 1.925 = 192.5\%
$$
Which can also be found geometrically as:
$$
Ret = \left(\prod(1+R_n)\right)-1
$$
$$
Ret = (1.25 * 1.3 * 1.2 * 1.5) - 1 = 1.925 = 192.5\%
$$
Suppose you want to find year on year performance where each of the quarters had a different weighting. E.g. the Q1-3 at 20% and Q4 at 40%. This should give returns from Q4 more importance.

What is the mathematical method for applying weights to geometric returns?

答:
In this case, you would just have a year-on-year return of $1.2^3 \cdot 1.4 - 1 \approx +142 \%$ return, where I assume you mean that quarters 1, 2 and 3 all have a $20 \%$ quarterly return. Although $1.2 < 1.4$, $1.2^3 > 1.4$ so the $1.4$ does not contribute as much to the total product.

What you probably want is a weighted average where for example, the rate of return for Q4 is twice as important as the other quarters. We can assign quarters $1, 2, 3$ a weight of $1$, and quarter $4$ a weight of $2$. However, since the sum of the weights is now $5$ where it was previously $4$, we need to take the $4/5$th root so that the total sum of weights is the same, which gives:

$$\left(1.2^1 \cdot 1.2^1 \cdot 1.2^1 \cdot 1.4^2 \right)^{4/5} - 1 \approx 165 \%$$

In general, with weightings $w_1, w_2, \cdots w_n$ and growth rates $r_1, r_2, \cdots r_n$, the weighted percentage growth rate will be:

$$({r_1}^{w_1} \cdot {r_2}^{w_2} \cdots {r_n}^{w_n})^{n/(w_1+w_2+\cdots+w_n)} - 1$$

Title: Calculating weighted geometric returns


答:
As was pointed out to you, the integrand fails to exist at $x=1$, so the integral must be rewritten as the sum of two improper integrals, one from $0$ to $1$ and the other from $1$ to $3$. When you try to evaluate the latter, you get this:

$$\begin{align*}\int_1^3\frac{dx}{(1-x)^2}&=\lim_{a\to 1^+}\int_a^3\frac{dx}{(1-x)^2}\\\\
&=\lim_{a\to 1^+}\left[\frac1{1-x}\right]_a^3\\\\
&=-\frac12-\lim_{a\to 1^+}\frac1{1-a}\;,
\end{align*}$$

and the limit in the last line clearly does not exist. Thus, the improper integral diverges, as does your original integral.

Note that you cannot integrate the function $f(x)=\frac1x$ over any interval that contains $0$, and the same goes for the function $f(x)=\frac1{x^2}$: in both cases you will get a divergent improper integral.

Title: Strange AP Calculus BC question help?

Body:
The question is $\int_0^3\frac{1}{(1-x)^2}$. I got an answer (from u-substitution) however the solution manual says that the integral does not converge. Someone told me that the integral is undefined at $x = 1$, but if we look at $\frac {1}{x}$ that function is undefined at $x=0$ yet we can still integrate it. </p>

<p>Can someone explain what is going on?! </p>


问:
Statistics question in cosmology
The following are the approximate parameters for the Benchmark model in cosmology:
$H_0 = 73.8 km s^{-1} Mpc^{-1})$, $\Omega_{m,0} = 0.266$, and $\Omega_{\Lambda,0} = 0.734$.

Given that we know $H_0$ very precisely to this value in a flat universe (meaning neglecting the negligible contribution from radiation, $\Omega_{m,0} + \Omega_{\Lambda,0} = 1$) with $\Omega_{\Lambda,0}$ being an unknown value very close to the given number, if we want to make a measurement of $\Omega_{\Lambda,0}$ with an accuracy of 5%, what is the maximum percent error we can tolerate on our measurement of luminosity distance at z = 0.5, 1.0, and 1.5?

Extra note: luminosity distance can be approximated as
$$d_L \approx \frac{c}{H_0} z (1 + \frac{1 - q_0}{2} z)$$

Where
$$q_0 = \frac{1}{2} \Omega_{m,0} - \Omega_{\Lambda,0}$$

Due to covid, I've been unable to take the statistical physics course that was part of my plan, so it has been since the mid 2000's since I've had a stats course. There are a bunch of follow-up questions, but I'm hoping once I grasp the approach to this one I will be able to do them myself.

答:
Write your equation as $d_L(z) = A(z)+B(z) \Omega_{\Lambda,0}$

Then,

$ \Omega_{\Lambda,0} =  {d_L - A \over B}$

and an error $\epsilon $ in $d_L$ translates into an error $\delta$ in $\Omega$:

$ \Omega_{\Lambda,0}+\delta =  {d_L - A \over B}+{\epsilon \over B} $.

The relative error in $\Omega$ is ${\delta \over \Omega} = {\epsilon \over d_L -A} = {{\epsilon \over d_L} \over 1-A/d_L}$ compared to the relative error in the luminosity $\epsilon \over d_L$.

You are left with the task of calculating  $A,B$.

PS: A course in statistical physics does not deal with these issues. I believe it is discussed in lab courses. See experimental error.

Title: Statistics question in cosmology


问:
Partial integration problem in calculus
This is a problem that I have in my calculus class. However I don't know how to solve it. I was wondering if anyone has some tips on solving the problem.

Given this information:
$f\left(0\right)=b,\:f\left(2\right)=0,\:f'\left(x\right)=\cos \left(x^5\right),$

Solve: $\int x^3\cdot f\left(x\right)dx$

I'm supposed to use partial integration here, but I am just stuck. Any tips or hints would be awesome! Thanks in advance :-)

Title: Partial integration problem in calculus

Body:
$$\begin{aligned}&\int x^3 f(x)\;\text{d}x \;\;=\;\; \frac{x^4}{4} \cdot f(x) \;-\; \int \frac{x^4}{4} \cdot \cos(x^5)\;\text{d}x\\[5mm] =\;\;& \frac{x^4}{4} \cdot f(x) \;-\; \frac{1}{5 \cdot 4}\int5x^4 \cdot \cos(x^5)\;\text{d}x \\[5mm] =\;\;& \frac{x^4}{4} \cdot f(x) \;-\; \frac{1}{20}\int^{x^5}\cos(x)\;\text{d}x \\[5mm] =\;\;& \frac{x^4}{4} \cdot f(x) \;-\; \frac{1}{20}\cdot\sin(x^5)\end{aligned} $$

Integrating $\;f'(x)\;$ to get $\;f(x)\;$ is beyond me though; I hope this helps.


问:
Dividing by dominant term - can we always apply the Combination Rules for convergent sequences?
In a book I'm reading they introduced the usual "Combination Rules for convergent sequences" like so:

If $\lim_{n \to \infty} a_n = l$ and $\lim_{n \to \infty} b_n = m$, then:
$$\lim_{n \to \infty} (a_n + b_n) = l + m$$
$$\lim_{n \to \infty} k a_n = kl$$
... etc.

Next, they introduced a way how to enable applying these rules to more complex algebraic fractions in order to find its limit.

The idea is, for a given fraction, divide the numerator and denominator by the dominant term.
Then, we can find the limit of the simpler result.

Example:

Given is $a_n = \frac{(2n+1)(n+2)}{3n^2 + 3n}$ for $n=1, 2, ...$. The dominant term is $n^2$, then divide in the numerator and denominator by $n^2$ and find:

$$a_n = \frac{(2n+1)(n+2)}{3n^2 + 3n} = \frac{(2+1/n)(1+2/n)}{3+3/n}$$

Then we can apply the combination rules to find

$\lim_{n \to \infty} a_n = \frac{(2+0)(1+0)}{(3+0)} = \frac{2}{3}$.

What I don't fully understand:

In the definition of these combination rules, we assume that the limits of the sequences we combine exist, since we say: "If $\lim_{n \to \infty} a_n = l$ and $\lim_{n \to \infty} b_n = m$, then: ..."

To my understanding, for a given algebraic fraction the limit might or might not exist.

Therefore:

As described in the strategy above, we divide through by the dominant term - but how does this ensure that we fulfill the prerequisites that we need in order to apply the combination rules (limits exist)?

Couldn't there be an algebraic fraction which still doesn't have a limit even after dividing by the dominant term?
In such case, we still couldn't apply these combination rules in my view.

I think my question boils down to: Can I just always find the dominant term, divide the algebraic fraction by it and expect to find a useful limit?

答:
Yes, the product, quotient, sum, scalar, $\dots$ properties only apply to convergent sequences, and the statement $\lim_{n\to\infty}a_n=l$ says both that $a_n$ is a convergent sequence and that $a_n$ converges to $l$.

If $a_n=1/n$ and $b_n=n$, then $\lim_{n\to\infty} a_n \cdot b_n =1$ but the limit of $b_n$ does not exist.

If $a_n=n$ and $b_n=-n$, then $\lim_{n\to\infty}a_n +b_n=0$ but the limit of neither $a_n$ nor $b_n$ exist.

By using the phrase dominant term, I am assuming you mean rational functions for algebraic functions. The dominant term is the highest degree of a monomial in either polynomial of the quotient and this aligns with dividing by the largest power of $n$ in your example.

A rational function has a horizontal asymptote, or limit at infinity, if and only if the degree of the polynomial in the numerator, dividend, is at most the degree of the polynomial in the denominator, divisor.  If the dividend has a larger degree then you can examine the quotient and remainder to show divergence.

Dividing by the dominant term is equivalent to multiplying by the convergent sequence $d_n=(1/n^k)/(1/n^k)=1$. Since multiplying by 1 gives the same sequence, the existence of a limit is invariant whether you divide by the dominant term.

As we saw earlier, the converse of the properties are false. Meaning, if we break your example into a product of three sequences, $$(2n+1) \cdot (n+2) \cdot \frac{1}{3n^2+3n}$$ we see that, individually, two of these factor sequences are divergent, but this does not mean that the product diverges. If  we multiply by $1=(1/n^2)/(1/n^2)$, out of cleverness with rational functions not because this is a technique that works in general, we get $$(2+1/n) \cdot (1+2/n) \cdot \frac{1}{3+3/n}$$ a product of three sequences that are convergent individually. In this form, the properties tell us that the limit is 2/3.


问:
Making Sense of the weighted Least Squares method from ordinary Least Squares

It is stated that the equation for the best $\hat{x}$ uses the covariance matrix $V$ such that
$$
A^TV^{-1}A\hat{x}=A^TV^{-1}b
$$
and the error vector is $\vec{e}=(b-Ax)^TV^{-1}(b-Ax)$ and we need to minimize $e=\sum_{i=1}^m\dfrac{(b-Ax)_i^2}{\sigma_i^2}$

How do I make sense of this idea of weighted least squares in comparison with ordinary least square method ?

And how does the inverse of variances become the respective weights ?

My Understanding

Least square

We have a system $Ax=b$ and $b$ need not have to be in the column space of $A$.

We need the vector $A\hat{x}$ in $C(A)$ thats best fit to $b$, ie., closest to $b$.

So we need to minimize the error vector $\vec{e}=b-Ax$, ie., we need to minimize $e=e_1^2+\cdots+e_m^2=(b-Ax)^T(b-Ax)=||b-Ax||^2=\sum_{i=1}^m(b-Ax)_i^2$

Weighted Least square

I think here we are assigning weights for each errors such that each error is not equally likely, say $w_i$

Then we need to minimize the error $e=\sum_{i=1}^m w_ie_i=(b-Ax)^TW(b-Ax)=\sum_{i=1}^m w_i(b-Ax)_i^2$.

But how does the matrix for the weights $W$ is the inverse of the covariance matrix $V^{-1}$ ?

Reference: Page 557, Introduction to Linear Algebra, Gilbert Strang

Note: I am actually looking for an exaplanation based on linear algebra, by not going deep into statistics

答:
I don't think we can understand where $V^{-1}$ comes from without a bit of statistics, so here's a statistics viewpoint (but using notation similar to Strang's notation). Suppose there is an $m \times n$ matrix $A$ and a vector $\bar x \in \mathbb R^n$ such that
$$
Y = A \bar x + \epsilon.
$$
Here $A$ and $\bar x$ are non-random but $\epsilon$ is a normally distributed random vector with mean $0$ and covariance matrix $V$. So $Y$ is a normally distributed random vector with mean $A \bar x$ and covariance matrix $V$. You can think of $Y$ as being a noisy measurement of the value of $A \bar x$.

Let $b \in \mathbb R^m$ be the observed value of $Y$. Our goal is to estimate $\bar x$, given $A$ and $b$. Let $f_Y$ be the probability density function for a normally distributed random vector with mean $Ax$ and covariance matrix $V$. Notice that $f_Y$ depends on $x$. A natural estimate of $\bar x$ is the maximum likelihood estimate
\begin{align}
\hat x &= \arg \max_x f_Y(b) \\
&= \arg \max_x \frac{1}{\sqrt{(2 \pi)^m |V|}}e^{-\frac12 (b - Ax)^T V^{-1}(b - Ax)}
\end{align}
Maximizing $f_Y(b)$ is equivalent to minimizing $-\ln(f_Y(b))$, which is equivalent to minimizing
$$
E(x) = (b - Ax)^T V^{-1} (b - Ax).
$$
That is how we arrive at our weighted least squares problem. Setting the gradient of $E$ equal to $0$ yields Strang's equation (10).

If the components of the random variable $\epsilon$ are independent, then $V$ is diagonal, so the formula for $E(x)$ simplifies to Strang's equation (11).


问:
Expression for the Lee form
On a Hermitian manifold $(M,J,g)$ with associated fundamental form $\omega(X,Y)=g(JX,Y)$ we have an $L^2$ inner product $\langle \cdot,\cdot \rangle$ on $p$-forms $\alpha,\beta$ given by integrating $\alpha \wedge * \beta$ where $*$ denotes the Hodge star operator and we define $d^*$ as the adjoint of the differential $d$ with respect to $\langle \cdot,\cdot \rangle$. I have read many places now two equivalent definitions of this so-called Lee-form $\theta$:

$\theta$ is the unique 1-form such that $d\omega^{n-1}=\theta \wedge \omega^{n-1}$ where $\omega^{n-1}$ denotes the $(n-1)$-fold wedge product of $\omega$ with itself, $2n=\text{dim}(M)$.

$\theta=J(d^*\omega)$ where $J$ acts on 1-forms $\beta$ as $(J\beta)(X)=-\beta(JX)$.

My question is: Why do these two definitions of $\theta$ define the same object, or at least why does the expression for $\theta$ in 2. satisfy the relation in 1.? I've found and proved concrete expressions for $d$ and $d^*$ in terms of the interior product operator and Levi-Cevita connection, so I thought that I should be able to simply compute the relevant quantities, in say an orthonormal frame $e_1,...,e_{2n}$ for which $\omega=e^1 \wedge e^{2n}+\cdots+e^n \wedge e^{n+1}$ where $e^i$ denotes the dual of $e_i$. Considering that both forms in 1. are $2n-1$ forms I thought it should not be that difficult to actually verify 1. given 2., but I don't seem to gain anything usefull from my computations, especially considering that I don't have a nice expression for $\omega^{n-1}$. Is there a better way to go about this than simply direct computation, maybe using some properties of the operators involved instead? Any hints or suggestions as to how I could proceed will be very much appreciated as I seem to be stuck at the moment. Thank you for your time!

答:
Okay I think I got it actually, so I'll post an answer for future reference or in case it might be usefull for other people at some point:
The co-differential $d^*$ on a Hermitian manifold satisfies $d^*=-*d*$ and $*$ is an isometry, so in particular injective. It can also be checked that $\omega^{n}=n!\text{Vol}$ where $\text{Vol}$ is the Riemanian volume form. Hence identity 1. is equivalent to $\frac{1}{n!}d^*\omega=-*(\theta \wedge \omega^{n-1})$. Also, we can pick a basis $e^1,...,e^{2n}$ for which $\theta=ae^1$ and $\omega=e^1 \wedge e^2+\cdots+e^{2n-1}\wedge e^{2n}$ (so we take $e^2=Je^1,e^4=Je^3,...$). Using the fact that 2-forms commute one can be convinced that $*(\theta \wedge \omega^{n-1})=\frac{1}{a(n-1)!}e^2=\frac{1}{a^2(n-1)!}J\theta$ and so with an appropiate choice of $a$ we see that 1. and 2. are in fact equivalent.

Title: Expression for the Lee form


问:
is it possible to find a monic quadratic polynomial that factors in Z but not in Z3?
Only using integer coefficients, is it possible to find a monic quadratic polynomial that factors normally but doesn’t factor in field/mod 3?</p>

答:
No.  If we had $$x^2 + Ax + B = (x+a)(x+b)$$ over $\mathbb{Z}$, then we'd have $$x^2+[A]x+[B] = (x+[a])(x+[b])$$ over $\mathbb{Z}/(3\mathbb{Z})$, where $[x] = x+3\mathbb{Z}$ is the quotient map.

This is true just because $x \mapsto x+3\mathbb{Z} : \mathbb{Z} \to \mathbb{Z}/(3\mathbb{Z})$ is a ring homomorphism, and as such, it can generalize to any ring homomorphism (at least for monic polynomials, to avoid one factor becoming a unit).


问:
Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$
I need the closed-form for the $n$-th derivative ($n\geq0 $):

$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$

Thanks!

By following the suggestion of Hermite polynomials:

$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^n}{\partial x^n}e^{-x^2}$$

and doing the variable change $x=\pi a y^{-\frac{1}{2}}$, I obtain:

$$\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n\frac{\partial^n}{\partial y^n}$$

and therefore

$$H_n(\pi a y^{-\frac{1}{2}})=(-1)^{n+1}e^{\frac{\pi^2a^2}{y}}2\left(\frac{\pi a}{y^{\frac{3}{2}}}\right)^n\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}$$

Finally

$$\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n$$

Is this correct?

答:
Related problem: (I). Here is a formula for the $n$th derivative of integer order of $e^{\frac{c}{x}}$

Formula 1: This formula is valid only for $n \in \mathbb{N} \cup \left\{ 0\right\}$.

$$\left( \rm e^{\frac{c}{x}}\right)^{(n)}  = {{\rm e}^{{\frac{c}{x}}}}\sum _{s=0}^{n}
  \sum _{k=0}^{n} \left( -1 \right)^{-k-s}\left[\matrix{n\\k+s}\right] \left\{\matrix{k+s\\s}\right\}{c}^{s}{x}^{-s-n},\quad n\in \mathbb{N} \cup \left\{ 0\right\} $$

where $\left[\matrix{n\\k+s}\right]$ and $\left\{\matrix{k+s\\s}\right\}$ are the Stirling numbers of the first kind and the second kind respectively.

Formula 2: Here is a unified formula which gives a complete solution to the problem of differentiation and integration of real order of the function in terms of the MeijerG function

$$\left(\rm e^{\frac{c}{x}}\right)^{(n)} = \left( -1 \right)^{n+1}{a}^{-n} \left( -1 \right) ^{n}G^{1, 1}_{1, 2}\left(-{\frac {a}{x}}\, \Big\vert\,^{1}_{1+n, n}\right)
,\quad n\in \mathbb{R}.$$

The last formula gives

1) derivatives of real order if $n>0$,

2) anti-derivatives of real order if $n<0$.

Note: The $n$th derivative of integer order of the function $e^{\frac{1}{x}}$ has to do with Lah nembers.


问:
Dummit and Foote 14.2.4: Let $p$ be a prime. Determine the elements of the Galois group of $x^p -2$
Dummit and Foote 14.2.4: Let $p$ be a prime. Determine the elements of the Galois group of $x^p -2$.

My instinct tells me this has to be related to the dihedral group with order relating to $p$, but I'm not too sure where to start. I'm not wanting a whole solution, but rather a hint as to where to start.  The splitting field is $\mathbb{Q} (2^{\frac{1}{p}}\zeta ^i _p)$, but how should this relate? Can you give me some guidance?

答:
Prove that the splitting field is $\Bbb Q(2^{1/p}, \zeta_p)$ with degree $p(p-1)$. (Hint: $p$ and $p-1$ are coprime.)

Look at what an automorphism can do to those two generators. (Hint: any element must be sent to itself or one of its conjugates.)

Define an injection from the Galois group to some concrete group. (Hint: It won't be the dihedral group because the field has degree $p(p-1)$.)

Use cardinalities to show that it is a bijection.


问:
Finding an angle in a circle with triangles
good day! Excuse me, I am new to finding angles using a circle, and I don't know how to find X using trigonometric functions and everything else about triangle angles. What is the 100° talking about? Or, how can one proof that BE is parallel to AC? (Maybe it isn't!) What I know is the angle of ACD is going to be 78. Any help, guidance, approach would be warmly welcomed.

答:
The $100º$ refers to $\angle BOA$. Whenever you see an angle on the arc of a circle, this means that the two points are separated by that angle: in this case, $A$ and $B$ are separated by an angle of $100º$.

If $\angle BFC + \angle ACF = 180º$, then $BF$ is parallel to $AC$ (these would be co-interior angles, if you're curious). Therefore, we have to find $\angle BFC$ and $\angle ACF$ somehow.

It is often helpful to draw an extra line or two: for this problem, connect $B$ and $C$ together. Now, can you find:

$\angle DCA$?

$\angle BCA$ using the inscribed angle theorem (with angles $2 \theta$ and $\theta$)?

$\angle BCF$ using the alternate angle theorem for circles?

$\angle CEB$ using the sum of opposite angles in a cyclic quadrilateral?

and these facts will lead you to find the two angles mentioned above, and hence whether $BF$ is parallel to $AC$.


问:
The Cross Ratio of a Mobius Transformation - Definition Clarification
so the Mobius transformation $f(z)$ of an extended complex plane preserves the following cross ratio
$$\frac{(w_1-w_4)(w_3-w_2)}{(w_1-w_2)(w_3-w_4)}=\frac{(z_1-z_4)(z_3-z_2)}{(z_1-z_2)(z_3-z_4)}$$
where $w_i$ and $z_i$ belongs to the extended complex plane. I have a silly question about the numeration. Let's say I want to recover that cross ratio in a couple of years, and I will know that I need to put $$(a-b)(c-d)$$ on the top. But, what goes on the bottom $$(a-c)(b-d)\text{ or }(a-d)(b-c)\text{ and two more options. }$$
So, to answer that question, I consider all 4 cases and simply did the renumerations. So, I got that the choice does not matter as I always will obtain the desired
$$\frac{(z_1-z_4)(z_3-z_2)}{(z_1-z_2)(z_3-z_4)}.$$
Is there a shorter way to argue that? Or, is there another intuition that stands behind that definition?

答:
The cross-ratio – as I learned it many years ago – is defined as
$$
 (z_1, z_2, z_3, z_4) = T(z_1)
$$
where $T$ is the (unique) Möbius transformation which maps $(z_2, z_3, z_4)$ to $(1, 0, \infty)$, in that order. This definition holds for all $z_k$ in the extended complex plane, and it easy to remember.

For finite values $z_k$ that is
$$
(z_1, z_2, z_3, z_4) = \frac{(z_1 −z_3)(z_2 −z_4)}{(z_1 − z_4)(z_2 − z_3)}
$$
which coincides with the definition given in Wikipedia.

One could also map $(z_2, z_3, z_4)$ to any other permutation of $(1, 0, \infty)$, that gives $6$ possible definitions of the cross-ratio. I assume that the actual definition is just by convention. The important properties (it is invariant under Möbius transformations, and it is a real number exactly if the four points line on a circle or a line) are not affected by that choice.


问:
Whether the integral is converging or diverging. $\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1}$
Find whether the integral is converging or diverging.
$\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1}$

My solution:
$\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1} = \int_{-\infty} ^{ 0} \frac{x \ dx}{x^2+1} + \int_{0} ^{ \infty} \frac{x \ dx}{x^2+1}$

Now if in the first part if I substitute $t=-x $  and $ dt=-dx$ and then I swap the limits the equation becomes $ - \int_{0} ^{ \infty} \frac{t \ dt}{t^2+1} + \int_{0} ^{ \infty} \frac{x \ dx}{x^2+1}$. Now I can just replace $t$ with $ x$ and the whole thing becomes $0$. Therefore it is converging.

The solution:
The solution says diverging. The inegrals were found with respect to a placeholder $eg. R $ and then the limit of $R$ to $\infty$ was put. And hence we get $log\ \infty$ which is $ \infty$ which is divergent.

My question is:  Why is there a difference in solutions between both the approaches ?  Can I not add integrals like I did in the final step when infinity is involved ? And in the log method, I shall get one integral as $+ \infty$ and the other as $- \infty$ so wouldn't the whole thing still be zero ?

答:
By definition, this is an improper Riemann integral. Your integral exists if BOTH of the following limits exist

$$\lim_{a\rightarrow \infty}\int_{0}^{a}\frac{x}{1+x^2}dx,\qquad
\lim_{b\rightarrow -\infty}\int_{b}^{0}\frac{x}{1+x^2}dx$$

Neither of these limits exist (the non-existence of one limit is enough to
conclude the non-existence of the integral). The first limit
is
$$\lim_{a\rightarrow \infty}\int_{0}^{a}\frac{x}{1+x^2}dx=\frac{1}{2}
\lim_{a\rightarrow \infty} \log(1+a^2)=\infty.$$
Similarly
$$\lim_{b\rightarrow -\infty}\int_{b}^{0}\frac{x}{1+x^2}dx=-\frac{1}{2}
\lim_{b\rightarrow \infty} \log(1+b^2)=-\infty.$$

What you are trying to compute is the principal value of the integral which is indeed zero.

$$P.V.\int_{-\infty}^{\infty}\frac{x}{1+x^2}dx= \lim_{a\rightarrow \infty}\int_{-a}^{a}\frac{x}{1+x^2}dx=
\lim_{a\rightarrow \infty}(\log(1+a^2)-\log(1+a^2))=0
$$
where $P.V.$ stands for the principal value of the integral.
Another typical example is the integral $\int_{-1}^{1}\frac{dx}{x}$
which does not exist but has principal value, also zero.


问:
Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.

Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.

  Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:

   $$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$

  i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.

I don't get the reasoning completely.

答:
$c$ can take only $1, 3, 7, 9$ . [$abc$ is prime]

$b$ can take only $1, 2, 3, 4, 5, 6, 7, 8, 9$. [$b\neq 0$ as $-4ac$ can't be a square of a number]

$a$ can take $1, 2, 3, 4, 5, 6, 7, 8, 9$.

$b^2-k^2=4ac$.

Thus $b$ and $k$ both should be both odd or both even and $b>k$ as $c$ can't be $<0$.

$4ac=\text{Multiple of 4,12,28 or 36}$ as $c$ can take only $1, 3, 7, 9$.
$$(b,k)=(2,0),(4,0),(6,0),(8,0),(4,2),(6,2),(8,2),(6,4),(8,4),(8,6)(3,1),(5,1),(7,1),(9,1),(5,3),(7,3),(9,3),(7,5),(9,5),(9,7).$$
For all this values of $b$ and $k$, $a$ and $c$ can't assume such values such that $abc$ becomes a prime number. Thus, $b^2-4ac$ is never a perfect square when $abc$ is prime.


问:
How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.

答:
The key here is to think in terms of radial co-ordinates rather than cartesian.  Whenever you see a complex exponential it's best to think of the geometric interpretation first, if possible.

The radial representation of a complex number $z = a + bi$ is $Re^{i\theta}$

where $R = |z|$ and $\theta=arctan(b/a)$.

So for the number i we ask ourselves what is R, and what is $\theta?$

We look where the point i is on the complex plane - it's on the y axis 1 unit away from the origin. So R is 1 and $\theta$ is $\frac{\pi}{2}$.  (Note that you can add any multiple of $2\pi$ to $\theta$ and things stay the same.)

So plug these R, $\theta$ values into the radial form of z to get the radial representation for i which is  $1.e^{i\frac{\pi}{2}} = e^{i\frac{\pi}{2}}$

So $e^{ix} = i = e^{i\frac{\pi}{2}}$ and a naive answer is $x = \frac{\pi}{2}$.
But remember we could add any multiple of $2{\pi}$ to our angle for i so we need to add $2n{\pi}$ for the general answer.

Hence $x = \frac{\pi}{2}+2n\pi \quad n\in \mathbb{Z}$

Title: How to solve $e^{ix} = i$


答:
To highlight the main problem, let me introduce a simpler problem. Let $(a_n)$ be the sequence
$$
a_n =
\begin{cases}
1 & \text{for } n = 1,2,3,\ldots, \\
-1 & \text{for } n = -1,-2,-3,\ldots, \\
0 & \text{for } n = 0.
\end{cases}
$$
The infinite sum $\sum_{n \in \mathbb{Z}} a_n$ is divergent. However, you could argue  that the positive and negative parts cancel each other - this is not rigorous, but can be summed up as the following observation:
$$
\sum_{n=-N}^{N} a_n \longrightarrow 0 \quad \text{as } N \to \infty.
$$

Your problem is similar. There is no rule saying that the difference of two infinite quantities is zero, therefore the first method (as it's written) is wrong. However, it is true that
$$
\int_{-R} ^{R} \frac{x \ dx}{x^2+1} \longrightarrow 0 \quad \text{as } R \to \infty,
$$
and the effect of this procedure is sometimes called the principal value of the integral. Note that this strongly depends on choosing the same $R$ on both sides, for example $\int_{-R} ^{2R} \frac{x \ dx}{x^2+1}$ would give you a different answer.

Title: Whether the integral is converging or diverging. $\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1}$

Body:
Find whether the integral is converging or diverging.
$\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1}$

My solution:
$\int_{-\infty} ^{ \infty} \frac{x \ dx}{x^2+1} = \int_{-\infty} ^{ 0} \frac{x \ dx}{x^2+1} + \int_{0} ^{ \infty} \frac{x \ dx}{x^2+1}$

Now if in the first part if I substitute $t=-x $  and $ dt=-dx$ and then I swap the limits the equation becomes $ - \int_{0} ^{ \infty} \frac{t \ dt}{t^2+1} + \int_{0} ^{ \infty} \frac{x \ dx}{x^2+1}$. Now I can just replace $t$ with $ x$ and the whole thing becomes $0$. Therefore it is converging.

The solution:
The solution says diverging. The inegrals were found with respect to a placeholder $eg. R $ and then the limit of $R$ to $\infty$ was put. And hence we get $log\ \infty$ which is $ \infty$ which is divergent.

My question is:  Why is there a difference in solutions between both the approaches ?  Can I not add integrals like I did in the final step when infinity is involved ? And in the log method, I shall get one integral as $+ \infty$ and the other as $- \infty$ so wouldn't the whole thing still be zero ?


问:
calculating the area of a circle by integrating over straight lines

The Problem

Imagine having a circle of radius $R$. We all know the area of this circle is $A=\pi R^2$ and the circumference is $2\pi R$. I would like to know why this approach for calculating the area is wrong, and why it gives me the circumference instead:

If I have a line of length $R$, I can rotate it around its edge a full rotation ($2\pi$ rad) and it would cover the whole circle, meaning it should give me the area, so we have the area be equal to $A=2\pi R$ or alternatively $A=\int\limits _{0}^{2\pi}Rd\theta =2\pi R$.

Note: By rotating the line I mean taking the original line (purple in the image) and then placing its end (which isn't in the center of the circle) on some other point on the circumference of the circle, as seen in the picture (the red lines), doing so for every possible point on the circle would look like the entire area of the circle is colored red, since its all covered by the rotations of this one purple line.

Note: I tried to look at other similar SE questions such as this one, but they didn't answer my question well.

Context

The reason I thought of this is because a few days ago I tried to calculate the electric field of a ring centered at $\left(0,0,0\right)$ with some inner radius $R_1$  and outer radius $R_2$ and charge density $\sigma = \frac{\lambda}{r^{3}}$, at the point $\left(0,0,z\right)$ for some $z$. since the ring is symmetrical in the $x,y$ directions the field has affect in the $z$ direction at that point. I thought about finding the field one "line" in the ring causes, and multiply that by $2 \pi$, but I found this approach to be incorrect.

答:
Using that reasoning, the length of a line segment is always $0$. Because it is made of individual points, which have length $0$, so that

$$\int_0^L 0\,dx=0.$$

For the circle, the correct approach is to decompose the disk in tiny triangles of long sides $r$ and short side $\delta r$. Every triangle has the area

$$\frac{\delta }2\sqrt{r^2-\dfrac{(\delta r)^2}4}$$

Doing this approximation, you drop the areas of the circular segments along the short sides.

One can show that as $\delta r$ tends to $0$,

$\delta r$ represents both the short side and the corresponding circular arc,

the area of the triangles is well approximated by $\dfrac{r\delta r}2$,

the total area of the segments is neglectable.

Finally, summing over a full turn,

$$A=\frac r22\pi r.$$


问:
Can you use an equals sign if the answer is rounded to two decimal places?
For my latest maths PSMT, the teacher has specified that equals signs should only be used if the values are actually equal. Does that mean I should use an approximation sign if I'm rounding the values to two decimal places?

This is what I've been doing so far...

\begin{align}
&x = -5.057588.... \\
&x = -5.06\;\text{(2 d. p.)} \\
\end{align}

Should I be using an approximation sign instead? Should I remove the $\text{(2 d. p.)}$ at the end?

答: I would rather use $$x=−5.057588\ldots \\ x\approx -5.06$$ since they are not exactly equal. I Agree with your teacher.


问:
Expectation value of number of rounds played in a card game
Suppose we have a game where each round, a random winner is chosen among $m$ players. The game ends when a person has won $k$ rounds. What's the expectation value of total rounds played in the game?

This is a question I came up with when playing a card game and has been bothering me ever since.

答:
For convenience of the expressions below I will redefine $k\mapsto k+1$, so that $k+1$ wins of a player are required to end the game.

In terms of generating functions the probability that the game will end in
$n+1$ rounds is:

$$
P_n(m,k)=\frac{n!C_{n}(m,k)}{m^{n}}\tag1
$$
where
$$
C_n(m,k)=[x^n]\frac{x^k}{k!}\left(\sum_{i=0}^k\frac{x^i}{i!}\right)^{m-1},\tag2
$$
with $[x^n]$ being the coefficient extractor.

The expected number of rounds can then be computed as:
$$
R(m,k)=\sum_{n\ge0}(n+1)P_n(m,k)=\sum_{n=k}^{km}\frac{(n+1)!C_{n}(m,k)}{m^{n}}.\tag3
$$

For the case $m=2$ it is not hard to find the closed form of the above expressions:
$$\begin{align}
P_n(2,k)&=\binom{n}{k}\frac1{2^n},\quad k\le n\le 2k,\\
R(2,k)&=(2k+1)\left[1-\binom{2k}k\frac1{4^k}\right]+1.
\end{align}$$


问:
For $\mu(X)<\infty$ and $f_n<\infty$ a.e. s.t. $\forall M$ $\exists n$ s.t. $\mu(\{f_n>M\})\geq\delta$, then $f_n$ diverges on set of positive measure

Let $f_n$ be a sequence of measurable finite a.e. functions on a space
  $X$ of finite measure.  Given $\delta >0$, suppose that for all $M>0$
  there exists an $n$ such that $\mu(\{f_n>M\})\geq\delta$.  Prove that
  $f_n$ is unbounded on a set of positive measure.

I'm looking for a clean way to do this that makes it appear as obviously true as possible.  The best I could come up with was to set $M=\left \lceil \frac{\mu(X)}{\delta} \right \rceil$, to figure out how many disjoint sets of measure $\delta$ it would take to sum up to the measure of $X$.  And then look at the collection

$$\{f_{n_1}>1\},\{f_{n_2}>2\},...,\{f_{n_M}>M\},\{f_{n_{M+1}}>M+1\}$$

And conclude that these sets cannot be pairwise disjoint modulo a set of measure zero.  And then keep repeating the process again and again to find a subsequence  $f_{n_k}$ and a descending sequence $E_k$ of sets of positive measure on which the subsequence $f_{n_k}$ increases without bound.

Assuming this is even correct, is there a simpler way to solve this?

答:
For an integer $M$, let $A_M = \bigcup_{n=1}^\infty \{f_n > M\}$.  Since  $\mu(\{f_n > M\}) \ge \delta$ for at least one $n$, we have $\mu(A_M) \ge \delta$.  Moreover, it should be clear that $A_1 \supset A_2 \supset \dots$.  Set $A = \bigcap_M A_M$; by continuity from above (which holds because our measure is finite), we have $\mu(A) = \lim_{M \to \infty} \mu(A_M) \ge \delta$.  But for $x \in A$, for every $M$ there is an $n$ such that $f_n(x) > M$, which is to say that $\{f_n(x)\}$ is unbounded.

As an alternative formulation: suppose $f$ is a measurable function with $f < \infty$ almost everywhere.  Then, again using continuity from above, $0 = \mu(f = \infty) = \lim_{M \to \infty} \mu(\{f > M\})$.  So for any $\delta > 0$, there exists $M$ such that $\mu(\{f > M\}) < \delta$.  Take $f = \sup f_n$ and you will see this is precisely the contrapositive of the desired statement.


问:
$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

key points: This is supposed to be solved with knowledge of just L'hospital's rule.
*The value of limit i.e, k is finite and non zero

I have tried manipulation using logarithm which didn't turn out to be of good use. I think multiplying by conjugate should work but that just  gives complex derivatives  and doesn't seem to lead to any answer.
Am I missing some basic point or is there a specific procedure for this problem.

答:
Hints: If $c <\frac 1  4$ then the limit on the left is $-\infty$.

If $c>\frac 1 4 $ write LHS as $\lim x^{4c} [(1+\frac 5  x+\frac 3 {x^{4}})^{c} -x^{1-4c}]$ to see that the limit is $+\infty$.

Finally, let $c=\frac 1  4$. Then LHS is $\lim x[(1+5x^{-1}+3x^{-4})^{1/4}-1]$. This limit is $\frac 5  4$ so we must have $c=\frac 1 4$ and $k=\frac  5 4 $.

For the case $c=\frac 1  4$ bring $x$ to he denominator and apply L'Hopital's Rule.