问:
Linear equation | Finding the unknown values
I have a set of equations as follow,

$a_1\cdot x_1 + a_2\cdot x_2 + a_3\cdot x_3 + a_4\cdot x_4 + \dots + a_{11} \cdot x_{12} + a_{12}\cdot x_{12}  = 6000$

$x_1 + x_2 + x_3 + x_4 + \dots x_{11} + x_{12} = 100 $

$\{a_1,a_2,a_3, \dots , a_{11}, a_{12}\} \leq 60$

How can I find the values for all $a_i, \text{ and }  x_i$,not all values of $a_i$ can be equal to $60$ also if possible integer, else the general solution.

How the question get affected if I change the constrain on $a_i$ to some $X$, say $0 < a_i \leq X$,

Also if the constrain on $a_i$ is , say $a_i>0$.

If solvable in Matlab please show the code also .

If not solvable please mention why.

What I tried is
$$
\begin{bmatrix}
a_1&a_2 & a_3 &\dots & a_{11} & a_{12} \\
1 & 1 & 1& \dots & 1 & 1
\end{bmatrix}\begin{bmatrix}
x_1 \\ x_2 \\ x_3  \\ \vdots\\ x_{11} \\ x_{12}
\end{bmatrix} = \begin{bmatrix}
6000 \\ 100
\end{bmatrix}
$$

The rank will be almost $2$ so atmost I can get $10$ free variables. but since the values for $a_i$'s are also unknown I don't know how to proceed.

Body:
Assume that $\exists i\in \{1, 2, ..., 12\}: a_i < 60$, then
$$6000 = a_1 x_1 + ... + a_{12} x_{12} < 60x_1 + ...+ 60x_{12} = 60(x_1 + ...+x_{12}) = 6000 $$
is contradiction. So $\forall i \in \{1, 2, ..., 12\}: a_i \geq 60$, so $a_1 = a_2 = ... = a_{12} = 60$


问:
Evaluate $\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x$
I tried:
$$\left(\frac{x}{1+x}\right)^x=A$$
$$\ln(A)=x\ln\left(\frac{x}{1+x}\right)$$
$$\lim_{x\to 1^{+}}\frac{\ln x-\ln (x+1)}{\frac1x}=\lim_{x\to 1^{+}}\frac{\frac1x-\frac1{x+1}}{\frac{-1}{x^2}}=\frac{-1}{2}$$
I applied L'Hopital rule in last step.
I get $\lim(A)_{x\to 1^{+}}=e^{\frac{-1}{2}}$
But the final answer provided in the book is $e^{-1}$. where I made a mistake?

答:
If our function is wrote in this form, you can't use  l'Hopital.

Since, $e^x$ and $\ln$ are continuous functions we obtain:
$$\lim_{x\rightarrow1}\left(\frac{x}{1+x}\right)^x=\lim_{x\rightarrow1}e^{x\ln\frac{x}{1+x}}=e^{\lim\limits_{x\rightarrow1}x\ln\frac{x}{1+x}}=e^{1\cdot\ln\lim\limits_{x\rightarrow1}\frac{x}{1+x}}=e^{\ln\frac{1}{2}}=\frac{1}{2}.$$

ContentLicense: CC BY-SA 4.0


答:
As already noticed by continuity

$$\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x=\frac12$$

it seems that, fixed the typo, the original question is referring to

$$\lim_{x\to 0^{+}}\left(\frac{x}{1+x}\right)^x=\lim_{x\to 0^{+}}\frac{x^x}{(1+x)^x}=1$$

or to

$$\lim_{x\to \infty}\left(\frac{x}{1+x}\right)^x=\lim_{x\to \infty}\left(1-\frac{1}{1+x}\right)^x=\frac1e$$

note that in any case we don't need l'Hospital's rule to obtain the result.

Title: Evaluate $\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x$

Body:
I tried:
$$\left(\frac{x}{1+x}\right)^x=A$$
$$\ln(A)=x\ln\left(\frac{x}{1+x}\right)$$
$$\lim_{x\to 1^{+}}\frac{\ln x-\ln (x+1)}{\frac1x}=\lim_{x\to 1^{+}}\frac{\frac1x-\frac1{x+1}}{\frac{-1}{x^2}}=\frac{-1}{2}$$
I applied L'Hopital rule in last step.
I get $\lim(A)_{x\to 1^{+}}=e^{\frac{-1}{2}}$
But the final answer provided in the book is $e^{-1}$. where I made a mistake?


问:
Losing original values while solving inverse trig equations
While proving, say, that $3tan^{-1}\frac{1}{4} + tan^{-1}\frac{1}{20} = \frac{\pi}{4}- tan^{-1}\frac{1}{1985}.$

There is the possibility that one side would just be n$\pi$ added to the other. So just proving that the tan values of both sides are equal won't be enough, would it? Apparently this method is used a lot

Body:
$$(4+i)^3(20+i)(1985+i)$$

$$=(4^3-12+48i-i)(20+i)(1985+i)$$

$$=(52\cdot20-47+i(52+47\cdot20))(1985+i)$$

$$=(993+992i)(1985+i)$$

$$=(993\cdot1985-992+i(992\cdot1985+993))$$

Now $993\cdot1985-992=1985(1+992)-992=992\cdot1985+993$

Finally take argument in both sides


问:
Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$.
I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!

When $m$ is even, put $m=2k$.
$2^m-1=(2^k-1)(2^k+1)$
By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime.
So, these are integers to n-th power.
Put
$2^k-1=a^n,
2^k+1=b^n$ ($1≦a<b,a$ and $b$ are odd).
$2=(b-a)(b^{n-1}+...+a^{n-1})≧2n$
∴$n=1$

答:
If $n$ is odd, so $$p^n+1=(p+1)(p^{n-1}-p^{n-2}+...+1)=2^m,$$ which says $$p\equiv-1(\operatorname{mod}2^k)$$ and from here
$$p^{n-1}-p^{n-2}+...-p+1\equiv n(\operatorname{mod}2^k).$$
Can you end it now?


问:
Asymptotic behavior of a Fourier/Laplace transform
I see many results concerning the asymptotics of Fourier transforms. These link in particular the regularity/continuation properties of the function to the polynomial/exponential decay of its Fourier transform. However, these results often hold only in the real variable. I am interested in the Fourier transform "along the imaginary axis" instead.

Let us be more precise. I am interested by the digamma function $\psi = \frac{\Gamma'}{\Gamma}$, and in the function
$$h(\nu) = \exp\left(-\alpha \psi \left( \frac14 \pm \frac{i\nu}{2} \right)\right),$$

where $\alpha$ is a fixed parameter, say $\alpha > 1$. I am interested in the asymptotic behavior of the Fourier transform of $h$ at $+\infty$. More precisely,
$$\widehat{h}(x) = \int_{\mathbb{R}} h(\nu) e^{ix\nu} d\nu.$$

How to get asymptoptics when $x \to +\infty$ in this situation? I have no feeling about what determines it: size? variations? only asymptotics of $h$?

I had many trials, not convincing. Typically, just changing variables, I can get an expression of the shape
$$e^{-\frac{x}{2}} \int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{2xu} du$$

which looks more like a Laplace (?) transform than a Fourier transform. I was motivated by the fact that I am expecting for other reasons an exponential decay as above, so that I am hoping for a polynomial behavior in $x$ for the remaining integral. However, is the growth/decay estimate of this last integral easier to understand than the original one?

So my question could be synthzised into

Do we have $\int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{xu} du \ll x^A$ for a certain $A$?

答:
I'll provide a sketch answer, to illustrate a general process. It's all about poles! Admittedly I acted as though the function was the transform of something un smooth, like a counting function. Perhaps

$$h(z) := \psi\left(\frac 14 + \frac i2 z\right).$$

Contour integral approach

First consider the poles of $h(z)$, which can be deduced from those for the original Digamma function (in turn corresponding just to poles of $\Gamma$). These are simple poles at
$$z_n := (2n + 1/2)i,\qquad n \in \{0,1,2,3,\ldots\},$$

each with residue $2/i = -2i$ (since those for the digamma function each have residue $1$, which we have "scaled" by a factor of $i/2$).

Fix $S$ such that $ \mathrm{Im}(z_N) < S < \mathrm{Im}z_{N+1}$ for some $N$, and let $T>0$. Consider the following contour integral over the rectangle:

(My bad! The labels on the $x_n$ should start from $0$ in the picture.)

Fix $x>0$ for now.

For $f(z) := h(z) e^{ixz}$, Cauchy's Residue theorem gives that

$$\int_{A_T} f(z) = \int_{B_T} f(z) + \left(\int_{C_T} f(z) + \int_{D_T} f(z)\right) + 2\pi i\sum_{n=0}^N \mathrm{res}(f,z_n)$$

This simplifies in the limit with the following claim, which should follow from a naive uniform estimate on $|h|$ on $D_T$ and $C_T$. Since I haven't proven it, I'll label it as an assumption:

Assumption: As $T → ∞$,

$$ \int_{C_T} f(z)\ dz + \int_{D_T} f(z)\ dz \to 0.$$

Also noting that

$$\mathrm{res}(f,z_n) = -2i e^{ixz_n} = -2i e^{-\frac{4n+1}2x},$$

in the limit we would have an asymptotic expansion, with the resonances given precisely by the residues:

$$\hat h(x) = \lim_{T→ ∞}\int_{A_T} f(z)\ dz= 4\pi \sum_{n=0}^N e^{-x(4n+1)/2} + \int_{\mathbb R + iS} h(z) e^{ixz}\ dz.$$

To finish off, one would have to show that the last integral decays at a faster rate than the other terms (in terms of $x$), which I shall not do. Does Paley–Wiener still apply?

Full asymptotic expansion?

Perhaps simpler is to ignore the unrigourousness above and investigate the natural conjecture which results from taking $ N → ∞ $:

$$\hat h(x) = 4\pi \sum_{n=0}^∞ e^{-x(4n+1)/2} = 4\pi \frac{ e^{3 x/2}}{e^{2 x}-1}.$$

This conjecture would also come from some formal-series argument; it should be enough to show that the inverse transform of this gives the original $h$. I'm not convinced that it is true, but it's worth a pop.


问:
Showing that $\int_{-\infty}^{\infty} \exp(-x^2) \,\mathrm{d}x = \sqrt{\pi}$

Possible Duplicate:

  Proving $\\int_{0}^{+\\infty} e^{-x^2} dx = \\frac{\\sqrt \\pi}{2}$

The primitive of $f(x) = \exp(-x^2)$ has no analytical expression, even so, it is possible to evaluate $\int f(x)$ along the whole real line with a few tricks. How can one show that
$$
\int_{-\infty}^{\infty} \exp(-x^2) \,\mathrm{d}x = \sqrt{\pi} \space ?
$$

答:
Such an integral is called a Gaussian Integral

This link should help you out.

ContentLicense: CC BY-SA 3.0


问:
effective version of Mertens Theorem for the Euler product
I'm referring to the theorem given here, which is

$$\displaystyle\lim_{n\to \infty} \:\: \left(\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right)\right) \;\;\; = \;\;\; \exp(\hspace{.01 in}\gamma)$$

where $p$ ranges over the primes and $\gamma$ is the Euler-Mascheroni constant.

What I'm really interested in is a good (effective) lower bound for $\;\; \displaystyle\prod_{p\leq n} \: \left(1-\frac1p \right) \;\;$,

so if you have another way to get such a bound, that would work too.

Otherwise, is it explicitly known how large and $n$ is sufficiently large to get $$\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right) \; < \; c$$

where, for example, $\: c=2 \:$ or $\: c = \frac95 \;$?

答:
Here is some place to start

$$\prod_{p\leq x}(1-\frac{1}{p})=e^{\sum_{p\leq x}\ln(1-\frac{1}{p})}=e^{\sum_{p\leq x}-\frac{1}{p}-\frac{1}{2p^2}-\frac{1}{3p^3}...}=ce^{-\sum_{p\leq x}\frac{1}{p}}=c'e^{-\ln(\ln(x))}=\frac{c'}{\ln(x)}$$

Where the last estimate comes from $$\sum_{p\leq x} \frac{1}{p}=\ln(\ln(x))+O(1)$$

I think what you should do is bound the terms that aern't reciprocals of primes in the expansion of the logarithm, and then get a good estimate on the sum $\sum_{p\leq x}\frac{1}{p}$ by writing it possibly in terms of a steiljes integral over the prime counting function, and then using a chebyshev estimate for the convergent integral involving the prime counting function, to get an estimate on the constant $c'$.


问:
Fixed-point theorem for a component-wise convex function.
I'm looking for a proof or counterexample of the following statement:

Conjecture: Let $\phi :[0,1]^n \to [0,1]^n $ be smooth. Suppose that each component of $\phi$ is convex, nondecreasing (in the sense that $\phi_i (x) \leq \phi_i (y)$ whenever $x_k \leq y_k$ for all $1\leq k \leq n$). Let $\mathbf{1}_n$ be the vector with all entries equal to $1$ and suppose $\phi(\mathbf{1}_n) = \mathbf{1}_n$. Assume further that the Jacobian $\mathcal{J}_\phi (\mathbf{1}_n)$ at that point has an eigenvalue $\lambda$ strictly greater than $1$. Further, suppose that $\phi(e_i)$ is a strictly positive vector for every unit vector $e_i$. Then $\phi$ has a fixed point in its domain other than $\mathbf{1}_n$.

The $n=1$ case is rather trivial, and I'm looking to generalize it. If the above doesn't work, can we add more 'niceness' assumptions to $\phi$ in order to make it work (like being a diffeomorphism)?

Any help is greatly appreciated.

答:
Inspired by the previous answer, here's what I managed to come up with myself:

Let's assume, on top of what is stated in the question, that $\mathcal{J}_\phi(1_n)$ is a positive matrix and that the supposed eigenvalue $\lambda>1$ is also the dominant one. Then, we know that the corresponding eigenvector $v_\lambda$ has positive components, by the Perron-Frobenius theorem.

Since $\phi$ is smooth, we have that $$\phi(1_n+tv) = \phi(1_n)+\mathcal{J}_\phi(1_n)(tv)+o(t^2) = 1_n +\lambda t v + o(t^2).$$

Since $\lambda >1$ and $v$ is positive, we know that $\lambda tv < t v$ for $t < 0$. For $t$ negative but sufficiently close to zero, we hence obtain
$$\phi(1_n + tv)<1_n+tv. $$
Hence, by monotonicity of $\phi$, the box with opposite vertices $0_n$ and $1_n + tv$ gets mapped into itself. Then we can apply the Brouwer fixed point theorem to obtain what we wanted.

Remark: I really only need that $v$ is nonnegative (and nonzero), so assuming non-negative   $\mathcal{J}_\phi(1_n)$ and using the analog of the Perron-Frobenius theorem for non-negative matrices also works.

I also think that convexity of the components of $\phi$ can guarantee uniqueness of the fixed point.


问:
Prove that parallels of sides of a triangle drawn from the centroid, divides each sides in three parts.
I have been trying for hours, but I don't know if I should approach it through areas or Euclid's Fifth Postulate.

答:
This method does not use the centroid $2:1$ property I mentioned in the comments.

Construct the median $AE$.

Using the parallel lines, we have $\triangle AFD \sim \triangle ABE$ and $\triangle AGD \sim \triangle ACE$.

This gives the corresponding sides:

$$\frac {FD}{BE} = \frac {AD}{AE} = \frac {GD}{CE}$$

Since $AE$ is a median, $BE = EC$. Hence $FD = GD$.

By similar triangles or the intercept theorem (due to $AF // HD$), we have $AH = HG$.

Similarly, $HD = DI$, and by $DG // IC$ we have $HG = GC$.

Hence the side $AC$ is trisected, and using similar arguments we see that all sides are trisected.


问:
Example of an $R$-module and an automorphism of $R$ such that the derived module is not isomorphic to the initial one
I am looking for an example of a ring $R$, a left $R$-module ${}_{R}{M}$ and a ring automorphism $\psi:R \rightarrow R$ such that the derived module ${}_RM^\psi$ is not isomorphic to ${}_RM$. Note: I am only looking at morphisms for which $\psi(1)=1$.

${}_RM^\psi$ is defined as having the same abelian group and the exterior product $r*m:= \psi(r) \cdot m$, where $\cdot$ is the product in ${}_RM$.

The main difficulty I'm having is broader, namely that I barely know any "simple" rings which have nontrivial automorphism groups; there's $\mathbb{C}$, but its wild automorphisms are not given explicitly and seem therefore hard to work with, if not impossible. Perhaps polynomial rings might have concrete non-trivial automorphisms?

Thank you for your time and any suggestions.

答:
Let $R=\mathbb{F}_2^2$ and let $M=\mathbb{F}_2$. Then $M$ is an $R$-module via $(\bar{m},\bar{n})\cdot \bar{x}=\bar{m}\bar{x}$  for all $\bar{m},\bar{n},\bar{x}\in\mathbb{F}_2$.

Let $\psi$ be the ring automorphism of $R$ switching coordinates. Assume now that $u:M\to M^\psi$ is an isomorphism of $R$-modules. In particular, it is an automorphism of the additive group $\mathbb{F}_2$. Hence $u=\mathrm{Id}_{\mathbb{F}_2}$. Now for all $\bar{m},\bar{n},\bar{x}\in\mathbb{F}_2$, we should have $\psi(\bar{m},\bar{n})\cdot u(\bar{x})=\bar{n}\bar{x}=u((\bar{m},\bar{n})\cdot\bar{x})=\bar{m}\bar{x}$, which is not the case.

Consequently, $M$ and $M^\psi$ are not isomorphic as $R$-modules.


问:
Is the sequence $a_n=\cos(n)/\sqrt{n+1}$ in $L_2$?
I have been trying to show that the  sequence $a_{n}$ belongs to $L_2$,
$$ a_{n} =\frac{ \cos(n)} {\sqrt{n+1} } $$
But none of the test of convergence that i have tried, show that the following series converges.
$$\sum_{n=1} ^{\infty} |a_{n} |^{2}$$

I wanted to ask for a hint for proving that the sequence does belong to $L_2$ space. Or perhaps it does not?

答: Write $\cos^2n$ as $\cos(2n)/2+1/2$; the series $\sum_{n\geqslant 1}\cos(2n)/(n+1)$ is convergent (use Abel's transform) and the series  $\sum_{n\geqslant 1}1/(n+1)$ is divergent hence $(a_n)$ does not belong to $\ell^2$.


问:
Relationship between $f''$ and "$f$ is strongly convex".
Suppose $f$ is strongly convex and twice differentiable on some interval $I$:

(1) $\forall x,y \in I: \forall t \in (0,1) \implies f(tx + (1-t)y)<tf(x)+(1-t)f(y)$

(2) $\forall x \in I \implies f''(x) \in \mathbb{R}$

My question is what exactly can we say about $f''$ on $I$? For example obviously, $f'' \geq 0$, but there's more to it for the values of $x$ when $f''(x) = 0$. For example, $f(x) = x$ is definitely not strongly convex on any interval, but $f'' \geq 0: \forall x \in \mathbb{R}$.

I think a sufficient condition would be if $f''(z) = 0 \implies \forall \delta > 0: \exists y_1,y_2 \in (z-\delta,z+\delta):y_1<z<y_2$ and $f''(y_1) > 0 < f''(y_2)$. Or in other words, $f$ is not linear at $z$. This, combined with $f'' \geq 0$ on $I$ should ensure strong convexity.

I would like to clarify my understanding by knowing a near-miss example(s). For example, what if $f$ is a function such that is convex on $\mathbb{R}$ but, say, $f''(0) = 0$ and $\forall \delta > 0: \exists x_1, x_2 \in (-\delta, \delta): x_1 < 0 < x_2:f''(x_1)=0=f''(x_2)?$ In my head I see something that looks kind of like $f''(x) = \sin(\frac{1}{x}) + 1$, but I would like an example where I can express $f$ in elementary functions if possible?

答:
Let $I \subset \mathbb{R}$ be a non-empty interval.

Proposition. Let $f : I \to \mathbb{R}$ be a twice differentiable function. Then $f$ is strictly convex on $I$ if and only if $f'' > 0$ outside a set of empty interior.

Proof:
I do not assume that $f''$ is non-negative on $I$. But Darboux proved that derivative functions satisfy the conclusion of the intermediate value theorem. Since $f''$ is positive outside a set of empty interior, we deduce that $f''$ is non-negative on $I$.
Thus, the function $f$ is convex on $I$. So if $f$ is not strictly convex on $I$, there exists a non-empty interval $J \subset I$ such that $f_{\restriction J}$ is affine. Hence, the function $f''$ is identically zero on $J$, which contradicts our hypothesis. Reciprocally, if $f'' = 0$ on $J$ then $f_{\restriction J}$ is affine and so $f$ is not strictly convex on $I$.

Set $g(x) = x^4 \sin^2(\frac{1}{x})$. Remark that $g$ is non-negative and continuous on $\mathbb{R}$. Consider $f: \mathbb{R} \to \mathbb{R}$ the map defined by
$$\forall x \in \mathbb{R},\ f(x) = \int_0^x \Big(\int_0^y g(s)\, ds\Big) dy. $$
The function $f$ is twice differentiable and $f'' = g$. Thus, $f$ is strictly convex on $\mathbb{R}$ according to the above proposition.


答:
From $p^n + 1 \ge p+1 \ge 4$ we have $m \ge 2$.

Hence $p^n = 2^m - 1 \equiv -1\pmod 4$, and thus $p^n$ cannot be a square.

This forces $n$ to be odd.

From $p^n + 1 = (p+1)(p^{n-1} - p^{n-2} + \dots + 1) = 2^m$, we see that

$$(p^{n-1} - p^{n-2} + \dots + 1) \mid 2^m$$

Since $p$ is odd, this factor, which is a sum of $n$ odd numbers, is odd as well.

This forces $p^{n-1} - p^{n-2} + \dots + 1 = 1$, and thus $p^n+1 = (p+1)(1)$.

Title: Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$.

Body:
I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!

When $m$ is even, put $m=2k$.
$2^m-1=(2^k-1)(2^k+1)$
By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime.
So, these are integers to n-th power.
Put
$2^k-1=a^n,
2^k+1=b^n$ ($1≦a<b,a$ and $b$ are odd).
$2=(b-a)(b^{n-1}+...+a^{n-1})≧2n$
∴$n=1$


问:
Value of the variance
I'm trying to prove that this statement is true (maybe it's false), but I don't know how to do it:

If all the values of a statistic variable $X$ are in an interval of length 2, then the variance of $X$ cannot be grater than 4.

答: $\newcommand{var}{\operatorname{Var}}$$\var(X)$ is most generally defined by $E((X-\mu)^2)$ where $\mu$ is $X$'s expected value. If the support of $X$ is on an interval of length $2$, the mean must obviously fall within the support (property A), so any sample from $X$ can be at most $2$ away from the mean, i.e. $|X-\mu|\le2$ and thus $0\le(X-\mu)^2\le4$. Treating $(X-\mu)^2$ as a random variable, its expected value – $\var(X)$ – by property A must be in $[0,4]$, and the claim is proved.

Title: Value of the variance


问:
10 to the power of 3.5: $10^{3.5}$
So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!

答:
N.S. is exactly correct; $10^.5 = \sqrt{10}$.

For a small amount of reasoning behind it.

By exponent laws, we know $10^x \times 10^y = 10^{x+y}$. $10^.5 \times 10^.5 = 10^1 = 10$. Since we want to figure out what $10^.5$ is, let's sub in $x$ for it.

$x \times x = 10 \rightarrow \sqrt{x^2}=\sqrt{10} \to x = \sqrt{10}$

Title: 10 to the power of 3.5: $10^{3.5}$


问:
Evaluate: $\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n$
I tried this way : $$\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n=\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n$$

We know $\lim_{n\to \infty}\left(1+\frac1u\right)^u=e$. therefor in this case we have: $$\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^{\tfrac{b+n}{a-b}}=e$$
Here $n$ goes to infinity so we can ignore the number ($b$)added to it in numerator of the exponent. So $\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}$

Is my answer right? and is there any other approach to evaluate the limit?

答:
Your answer is right but, following your idea, we should proceed as follows

$$\left(1+\frac{a-b}{b+n}\right)^n=\left[\left(1+\frac{a-b}{b+n}\right)^\tfrac{b+n}{a-b}\right]^{\tfrac{n(a-b)}{b+n}}$$

For a rigorous justification of this step you can refer to the following

Why is $1^{\infty}$ considered to be an indeterminate form

About $\lim \left(1+\frac {x}{n}\right)^n$ (useful also for the alternative approach)

Proof of product rule for limits (relevant using that $f(x)^{g(x)}=e^{g(x)\log f(x)}$


问:
Proving collinear points - geometry
Straight line $p$ contains the orthocenter $H$ of an acute triangle $ABC$, where $p$ intersects sides $AB$ and $AC$ with $p$ intersecting $AC$ in the point $P$. Straight line $q$ also contains $H$, is perpendicular to $p$ and intersects $AB$ and $BC$ with $q$ intersecting BC in the point $Q$. Straight line containing $A$, parallel to $q$ and a straight line containing $B$, parallel to $p$ intersect in the point $R$. Prove that $P,Q$ and $R$ are collinear.

A question I came across in a practice book, I am not sure even where to start... I'm allowed to use trigonometry, similar triangles, point-line-plane.

答:
We draw the two altitudes $AX$ and $BY$ of the given scalene acute triangle $ABC$. For brevity, let $HP$ and $HQ$ equal $d$ and $e$ respectively. We also assume that $\measuredangle BCA=\omega$ and $\measuredangle HPA=\phi$.

We start our proof by doing some angle chasing in $\mathrm{Fig.\space 1}$. In the right angle triangle $AMP$,
$$\measuredangle PAM=90^o-\measuredangle MPA=90^o-\phi.$$

In a similar vein, by considering the right angle triangle $PYH$, we can state that,
$$\measuredangle YHP=90^o-\measuredangle HPY=90^o-\phi.$$

The line $HP$ is parallel to the line $BN$ and the altitude $BY$ is their transversal. Therefore, the two corresponding angles $\measuredangle YHP$ and $\measuredangle HBN$ are equal, i.e.
$$\measuredangle HBN=90^o-\phi.$$

As a consequence, in the right angle triangle $BNH$,
$$\measuredangle NHB=90^o-\measuredangle HBN=\phi. \tag{1}$$

Finally, consider the two right angle triangles $AXC$ and $CYB$, which share $\measuredangle BCA$. Since, we have assumed $\measuredangle BCA=\omega$,
$$\measuredangle CAX=\measuredangle YBC=90^o-\measuredangle BCA =90^o-\omega. \tag{2}$$

Now we have all the angles we need to proceed with the proof. First, consider the two triangles $AHP$ and $BQH$. They are similar triangles because of (1) and (2). Therefore, we shall write,
$$\frac{AP}{BH}=\frac{HP}{HQ}=\frac{d}{e}. \tag{3}$$

Let us pay our attention to the two right angle triangles $AMP$ and $BNH$. Because of (1), these two triangles are also similar. Hence, with the help of (3), we can state the following equation.
$$\frac{MP}{NH}=\frac{AP}{BH}=\frac{d}{e}. \tag{4}$$

Consider the right angle triangle $PMR$, where we denote $\measuredangle RPM$ as $\theta$, which then can be express as,
$$\theta = \tan^{-1}\left(\frac{MR}{PM}\right).$$

Since $HNRM$ is a rectangle, we have $MR=NH$. Therefore, using (4) we can write,
$$\theta = \tan^{-1}\left(\frac{NH}{PM}\right)= \tan^{-1}\left(\frac{e}{d}\right). \tag{5}$$

Now, visualize what happens if we join $P$ and $Q$. The triangle $PHQ$ becomes a right angle triangle. Therefore, using what we have assumed at the beginning of our proof $\left(\mathrm{i.e.}\space HP=d\space\space \mathrm{and}\space\space HQ=e \right)$, we can express $\measuredangle QPH$ as,
$$\measuredangle QPH = \tan^{-1}\left(\frac{HQ}{HP}\right)= \tan^{-1}\left(\frac{e}{d}\right) ,$$
which, according to (5), is equal to $\theta$. Therefore, $\measuredangle QPH = \measuredangle RPM$, which implies that the point $R$ lies on the line $PQ$.

Title: Proving collinear points - geometry


问:
Relation between $h:B\to \{0,1\}$ and prime filter $F \subseteq B$

$\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is a Boolean algebra. Prove that:

For any prime filter $F$, there is a homomorphism $h:B\to\{0,1\}$ such that $F = \{x\in B:h(x)=1\}$

For a homomorphism $h:B\to\{0,1\}$, the set $F = \{x\in B:h(x)=1\}$ is a prime filter in $\mathcal{B}$.

A filter $F$ is a prime filter if it is proper and if $a \lor b \in F$ implies either $a \in F$ or $b \in F$.

My work:
Isn't (i) obvious? I'm unsure how to prove (i) formally. It seems clear that we should be able to define a homomorphism $h$ such that if $f \in F$ then $h(f) = 1$, and $h(f) = 0$ if $f \notin F$. What am I missing?

If $h$ is a homomorphism, we know that $h(0) = 0$ so $0 \notin F$, i.e. $F$ is proper. If $a \lor b \in F$, then $h(a\lor b) = h(a)\lor h(b) = 1$. So either $h(a) = 1$ or $h(b) = 1$, hence $a \in F$ or $b \in F$. $F$ is a prime filter! Is this proof complete?

P.S. For reference, $F \subseteq B$ is a filter if:

$F \neq \phi$

If $x,y \in F$ then $x\land y\in F$

If $x\in F$ and $x\leq y$ then $y\in F$

答:
For (i), you still have to prove that $h$ thus defined is a homomorphism.

Now, if $h(a \vee b) = 0$, then $a \vee b \notin F$, whence $a,b \notin F$, yielding $h(a),h(b) = 0$ and $h(a) \vee h(b) = 0$.

If $h(a \vee b) = 1$, then $a \vee b \in F$, whence $a \in F$ or $b \in F$, yielding $h(a) = 1$ or $h(b) = 1$, and so $h(a) \vee h(b) = 1$.

So $h(a \vee b) = h(a) \vee h(b)$.

Likewise you can prove that $h(a \wedge b) = h(a) \wedge h(b)$.

It is quite clear that $h(0)=0$ (since $F$ is proper) and $h(1)=1$ (since $F$ is a filter).

Now, a lattice homomorphism between Boolean lattices which sends $0$ to $0$ and $1$ to $1$ is a BA homomorphism (i.e., it also preserves complements); this is very easy to verify.


答:
With $10^{0.5}$ you have to do a "half multiplication", not a multiplication by half. What this "half multiplication" is doesn't follow from any universal law, but only by extending in a coherent way the rules valid for integer exponents.

Since, for integer $m$ and $n$ you have
$$
a^{m+n}=a^m\cdot a^n
$$
you can derive also that
$$
(a^m)^n=a^{mn}
$$
(just repeat the multiplications and count the factors). So, what $a^{3.5}$ should mean? Well, a possible choice comes from doing
$$
a^7=a^{3.5\cdot 2}\overset{*}{=}(a^{3.5})^2
$$
where the equals sign marked with $*$ is where we apply an extension to the  rule above.

Thus one can try defining
$$
a^{3.5}=\sqrt{a^7}.
$$

This is how actually exponentiation to a rational is defined:
$$
a^{\frac{p}{q}}=\sqrt[q]{a^p}
$$
and it can be shown that the rules

$$
a^{x+y}=a^x\cdot a^y,\qquad
(a^x)^y=a^{xy}
$$
continue to hold for all rational numbers $x$ and $y$ (and positive $a$).

Title: 10 to the power of 3.5: $10^{3.5}$

Body:
So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!


问:
$h:\mathcal{B} \to \mathcal{B'}$ (onto) is an isomorphism iff $\forall x,y\in B$, $x\leq y$ iff $h(x)\leq h(y)$

Let $h: B\to B'$ be an onto map, where $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ and $\mathcal{B'}:=(B',\leq,\lor,\land,^c,0,1)$ are Boolean algebras. Show that $h$ is an isomorphism from $\mathcal{B}$ to $\mathcal{B'}$ iff for all $x,y\in B$, $x\leq y$ iff $h(x)\leq h(y)$.

To prove one side of the implication, let us start by assuming that $h$ is an isomorphism. So, $h$ is one-one and onto both. We can use the homomorphism axioms as stated here. Now, $x \leq y$ iff $x \land y = x$. So $h(x\land y) = h(x) \land h(y) = h(x)$. This implies $h(x)\leq h(y)$. All the arguments are reversible, and the double implication is proved.

I'm unable to show the other side of the implication, i.e. if for all $x,y\in B$, $x \leq y$ iff $h(x)\leq h(y)$ then $h$ is an isomorphism. We already know that $h$ is onto, so all that remains to be shown is that $h$ is one-one. Starting with $h(x) = h(y)$, it should suffice to prove $x=y$. How do I go about this? I have thought of writing $h(x)=h(y)$ as $h(x)\leq h(y)$ and $h(y)\leq h(x)$ (do we need to prove this as well?), but I didn't reach the conclusion after this.

Thanks!

答:
If $h(x) = h(y)$, then both
$$h(x) \leq h(y)$$
and
$$h(y) \leq h(x)$$
hold.

Now, from the equivalence it follows that
$$x \leq y$$
and
$$y \leq x,$$
and therefore, $x = y$, by anti-symmetry.


问:
raindrops - downward velocity antiderivative question

I am working through an antiderivative question but there is one part of the answer I don't understand.

Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10  and its downward acceleration is

$$
a=\left\{\begin{array}{ll}
9-0.9 t & \text { if } 0 \leqslant t \leqslant 10 \\
0 & \text { if } t>10
\end{array}\right.
$$

If the raindrop is initially 500  above the ground, how long does it take to fall?

So,
$$
v(t)=-9 t+.45 t^{2}-10
$$
$$
s(t)=-4.5 t^{2}+.15 t^{3}-10 t+500
$$

And then punch in t=10
$$
s(10)=-4.5(10)^{2}+.15(10)-10(10)+500=100
$$

It's the position result of a 100 I don't undertand. What is the 100 supposed to represent?

答:
$s(t)$ is a function which denotes the position of raindrops at time $t.$ When $t = 0,$ then $s(0) = 500.$ This tells us that the raindrops start falling at $500$ units. We find their position at $10$ seconds when $t = 10,$ which we see is $s(10) = 100.$ This tells us that at $10$ seconds, the raindrops are $100$ units above $0.$ However, the question of

If the raindrop is initially 500 above the ground, how long does it take to fall?

was not answered. If you would like to answer this question, then you need to set $s(t) = 0$ and solve for $t.$ This is a standard quadratic equation.


答: I think it is easier to write the expression as $$\frac{(1+\frac{a}{n})^{n}}{(1+\frac{b}{n})^{n}}$$ and take it from there.

Title: Evaluate: $\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n$

Body:
I tried this way : $$\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n=\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n$$

We know $\lim_{n\to \infty}\left(1+\frac1u\right)^u=e$. therefor in this case we have: $$\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^{\tfrac{b+n}{a-b}}=e$$
Here $n$ goes to infinity so we can ignore the number ($b$)added to it in numerator of the exponent. So $\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}$

Is my answer right? and is there any other approach to evaluate the limit?


问:
Accuracy of combining multiple predictive tests
Let’s say I have produced 100 neural network models that can predict the outcome of a binary event (for instance, whether it is going to rain tomorrow). The first model has a 51.25% accuracy, the second model has a 53.73% accuracy, the third model has a 52.68% accuracy, and so on......In total, the average accuracy across 100 models is 52%.

Let’s say the probability of whether it will rain tomorrow or not is 50%. I use all 100 models to make 100 individual predictions on whether it will rain tomorrow.

Since the average accuracy across 100 models is 52%, does that mean that 52% of the 100 predictions made will be correct, and 48% of the predictions will be wrong?

I’m only starting to learn probability and statistics, so forgive me if this answer is stupid, thanks!

答:
First of all, let's assume that each model makes predictions independently of each other. Secondly, let's simplify the problem to just 2 models.

We'll have model A is correct 51% of the time and model B correct 53% of the time - so an average of 52% accuracy.

Let A be the event that model A is correct, and B be the event that model B is correct.

$$P(A,B) = 0.51\times0.53 = 0.2703$$
$$P(A,B') = 0.51\times0.47 = 0.2397$$
$$P(A',B) = 0.49\times0.53 = 0.2597$$
$$P(A',B') = 0.49\times0.47 = 0.2303$$

So we have the probability of both models being correct as 0.2703, the probability of one model being correct is 0.4994 and the probability of both being wrong is 0.2303.

The same sort of logic would have to be applied with the exact numbers of your model accuracies, but you would end up with a small probability of exactly 52% of your models being correct, but definitely not 100%.

Let's suppose that each model was exactly 52% accurate. The probability of 52 out of 100 models being correct is $0.52^{52}0.48^{48}$ multiplied by how many ways that can happen which is $\binom{100}{52} = \frac{100!}{52!(100-52!)}$

$$\frac{100!}{52!(100-52!)} \times 0.52^{52} \times 0.48^{48} \approx 0.07965$$

We have about a 7.965% chance of exactly 52 being correct


问:
Proving that a square inscribed in a rhombus (with unequal diagonals) has sides parallel to the rhombus' diagonals

How would you prove that a square inscribed in a Rhombus (with unequal diagonals) has its sides parallel to the diagonals of the rhombus using elementary geometry?

Actually I have spent a LOT of time doing this, but couldn't prove it so please provide a mathematically rigorous proof.

答:
Take a look at triangles $AEI$ and $DGH$. Angles $\angle AIE$ and $\angle DGH$ are obviously equal (parallel sides). In the same way angles $\angle AEI$ and $\angle DHG$ are also equal. And EI=GH so by ASA triangles $AEI$ and $DHG$ are congruent. Therefore:

$$AE=DH$$

...and consequentially:

$$BE=CH$$

Denote with $F$ intersection of rhombus' diagonals. Now take a look at triangles $DHF$ and $AEF$: $\angle EAF=\angle HDF, AE=DH,AF=DF$, so by SAS triangles $DHF$ and $AEF$ are congruent and $\angle HFD=\angle EFA,FE=FH$. So points $E,F,H$ are colinear and line $EFH$ is a diagonal of the square. You can easily prove in the same way that $IFG$ is also a digalonal of the square.

Now draw perpendicular lines from $F$ to sides $AB$ and $CD$. Triangles $EFJ$ and $GFK$ are congruent by SSA ($FE=FG,FJ=FK,\angle EJF=\angle GKF$). Therefore:

$$EJ=KG$$

Triangles $BJF$ and $BKF$ are also congruent by ASA (all angles are equal and triangles share a common side $BF$). Consequentially:

$$JB=BK$$

It means that:

$$EJ+JB=BK+KG\iff BE=BG$$

Because of $FE=FG,BE=BG$ line $BF$ is the median of side $EG$. In rhombus $ABCD$, $BF$ is also median of diagonal $AD$. Because of that, lines $EG$ and $AD$ are parallel (perpendicular to the same median BF). It's trivial to show that all other square sides must be parallel to rombus' diagonals.

EDIT: You can use the same proof in case of inscribed rectangle, not just square. Reducing the problem to a square wasn't necessary.


问:
Random arrangements in a line
been having some trouble with this, not sure if I'm on the right track.

5 numbers are chosen uniformly at random from {1, 2, . . . , 10} and arranged uniformly at random in a line. What is the probability that the numbers appear in increasing order?

Am I correct in saying that the probability is  10C5 / (10C5 * 5!)  = 1 /5! ?

My thinking is that there are 10C5 possible combinations of numbers, and for each of those combinations there is only one arrangement in ascending order. Additionally there are 5! ways of arranging each possible combination, so the final probability is 1/5! ?

答: That is correct. Once you pick $5$ numbers, there is $1$ way to arrange them in increasing order, while there are $5!$ ways to order them.

Title: Random arrangements in a line


问:
Proving that rational numbers cannot have denominator zero (without referring to division by zero)
Given the set $S$ of all pairs $(a, b),$ with $a, b \in \mathbb{Z},$ the relation $Q$ on $S$ is defined by $(a,b)Q(c,d) \iff ad=bc$. How can I prove that $b$ cannot be equal to zero, without using the fact that you cannot divide by zero?

I'm guessing it's something to do with $Q$ being an equivalence relation, and perhaps using the property of transitivity?

Thanks!

答:
You know that $Q \neq \mathbb Z^2$, because, for example, $1 \neq 2$ and so $(1,1)$ and $(1,2)$ are not related by $Q$.

Now, if we allow $b=0$ (and likewise $d$), then from $0y = 0x$ we get $(0,0) Q (x,y)$, for all $x,y \in \mathbb Z$.

Therefore $Q$ would not be an equivalence relation.


问:
Proving origin to be removable singularity(Proof verification)
This particular question is from Ponnusamy Silvermann complex section 9.22 ( question no. 2) and I am unable to solve it.

Let f(z) be analytic in deleted neighborhood of origin and lim $z\to 0$ | zf(z) |=0 . Then show that origin is removable singularity of f(z).

I need to show that lim $z\to 0 f(z) $=M(finite) . But if it were infinite, then  $z \to z_0$ |zf(z) | would be indeterminate.

So, it's removable.

Is my proof correct?

答:
From $$\lim_{z\to 0}zf(z)=0,$$
we know than $z=0$ is  removable singularity of  $F(z):=zf(z)$.
Define $F(0)=0$, then $F$ is analytic at $z=0$, and its Taylor series of $F$ at $z=0$ is
$$F(z)=zf(z)=\sum_{n=1}^{\infty}a_nz^n,$$
($a_0=0$ is due to $F(0)=0$).
So the Taylor series of $f$ at $z=0$ is
$$f(z)=\sum_{n=1}^{\infty}a_nz^{n-1},$$
hence, $$\lim_{z\to 0}f(z)=a_1.$$
So the origin is removable singularity of $f(z)$.


问:
Show $\lvert x^{\beta}\rvert\le \lvert x \rvert^{\lvert\beta\rvert}$, for $\beta$ a multiindex
I am trying to show show that $\lvert x^{\beta}\rvert\le \lvert x \rvert^{\lvert\beta\rvert}$ where \beta is a multi-index. So I have to show $\lvert x_1^{\beta_1}\cdot\ldots\cdot x_n^{\beta_n}\rvert \le \sqrt{(x_1^2+\ldots+x_n^2)}^{\lvert \beta_1+\ldots+\beta_n\rvert}$. Should I show this by induction or is there a more straight-forward way?

I am stuck in the induction :
$\lvert x_1^{\beta_1}\cdot\ldots\cdot x_n^{\beta_n}\rvert^2\lvert x_{n+1}^{\beta_{n+1}}\rvert^2 \le(x_1^2+\ldots +x_n^2)x_{n+1}^{2\beta_{n+1}}$

答:
You can prove this directly, using the weighted AM-GM inequality and the
Cauchy-Schwarz inequality.

Set $w_i = \frac{\beta_i}{|\beta|}$. Then $\sum w_i =1$ and therefore
$$
 |x_1|^{w_1} \cdots |x_n|^{w_n} \le w_1 |x_1| + \ldots + w_n |x_n| \\
 \le \sqrt{w_1^2 + \ldots + w_n^2}  \sqrt{x_1^2 + \ldots + x_n^2}\\
\le  \sqrt{x_1^2 + \ldots + x_n^2}\, .
$$
The last inequality holds because
$$
\sqrt{w_1^2 + \ldots + w_n^2} \le \sqrt{w_1 + \ldots + w_n} = 1 \, .
$$


问:
$a\begin{pmatrix}1\\4\end{pmatrix}+b\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$ Find a and b

$$a\begin{pmatrix}1\\4\end{pmatrix}+b\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$$
Find a and b

I've found the answers to this to be $\frac{1}{13}$ and $\frac{4}{13}$ through google but just wondering, how would I go about working this through and finding the answer myself?

答:
You would set up two equations in the unknowns $a,b$, by using the fact that vectors are equal if and only if their components are equal. Thus, you get
$$a+3b=1,\quad\hbox{and}\quad 4a-b=0$$
which I am pretty sure you can solve.


问:
A limit of a fraction with sin in both the numerator and denominator
I need to find $$\underset{x\rightarrow\pi}{\lim}\frac{\sin\left(ax\right)}{\sin(bx)}$$

I'm thinking of transforming this fraction into something that I can then 'process' with $$\underset{y\rightarrow0}{\lim}\frac{\sin y}{y}=1$$

I just don't know how to do it. My idea was to introduce a new variable $t = \pi-x$, but that doesn't seem to bear any fruit: $$\underset{t\rightarrow0}{\lim}\frac{\sin\left(a\left(\pi-t\right)\right)}{\sin\left(b\left(\pi-t\right)\right)}$$ because the expression in the sin brackets in both cases still does not approach $0$.

How can I compute that limit?
I appreciate any help.

答:
If $a$ and $b$ are not integers with $\sin(b\pi)\neq 0$ then the limit is given by

$$\frac{\sin\left(a\pi\right)}{\sin(b\pi)}$$

for $a$ and $b$ integers we obtain an indeterminate form $\frac 0 0$ and we can proceed by l'Hospital's rule.

To proceed without l'Hospital's rule we can use that

$\sin ((2k)(\pi+x))=\sin (2kx)$

$\sin ((2k+1)(\pi+x))=-\sin ((2k+1)x)$

therefore by $t=\pi -x \to 0$ we reduce to

$$\frac{\sin\left(ax\right)}{\sin(bx)}=\frac{\sin\left(a(\pi +t)\right)}{\sin(b(\pi +t))}=\pm \frac{\sin\left(at\right)}{\sin(bt)}$$

with the sign depending upon the parity of $a$ and $b$.


问:
Solving complex equation $ z+3\bar{z}=(2+i\sqrt{3})|z|$
I struggle to solve this complex equation: $$ z+3\bar{z}=(2+i\sqrt{3})|z|$$
One way to resolve this is substitute |z| with $\sqrt{x^{2}+y^{2}}$ , $z=x+iy$, solve the cube etc,
but i did not reach the solution.

Any hint?. Thanks in advance.

答:
First note that $z=0$ is (trivially) a solution.

Now, for other possible solution(s), divide thoroughout by $|z|$ ($z \ne 0$) to give:

$e^{i\theta} + 3e^{-i\theta} = 2+i\sqrt 3$ where $\theta$ is the argument of $z$.

You can let $e^{i\theta} = w \implies e^{-i\theta}= \frac 1w$ and you get a quadratic you can solve.


问:
{ $e^{inx} : n \in Z$} forms an orthonormal basis of the hilbert space $L^2[0,2\pi]$
I am currently studying basics of Fourier analysis and I have read that { $e^{inx} : n \in Z$} forms an orthonormal basis of the hilbert space $L^2[0,2\pi]$. Now I am having a confusion regarding this. We know that a complete normed linear space cannot have a countable basis. Again $L^2[0,2\pi]$ is a complete normed linear space, the norm being induced form the inner product. Then how can it have a countable basis.

I am sure that I am missing on some details but I am not being able to figure it out at my own. It will be a great help if someone of you kindly help me out.

Thank you.

答:
This is the content of Riesz–Fischer theorem.

As @Sten noted, we are talking about Schauder basis here (you allow infinite linear combination, not just finite ones). The uncountable "Hamel basis" (each function is a finite combination...) is much larger and impossible to write down explicitly.


问:
Let $G$ be a group generated by two involutions $x$ and $y$. Show that there is some $N\unlhd G$ with $[G:N]=2$.
This is Exercise 3.35 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search, it is new to MSE.

The Details:

Presentations are not covered in the book so far, so, presumably, there is a way to answer it without thinking of $G$ given by (some quotient of) the free product

$$\Bbb Z_2\ast\Bbb Z_2\cong \langle x,y\mid x^2,y^2\rangle.$$

The combinatorial-group-theory tag does not apply.

Since I sometimes get involutions and idempotents mixed up, here is a

Definition: An element $f$ of a group $G$ with identity $e$ is an involution if $f^2=e$.

Since there is a plethora of definitions for a normal subgroup, the following is from the book cited above.

Definition 2: A subgroup $H$ of a group $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

The Question:

Let $G$ be a group generated by two involutions $x$ and $y$. Show that $G$ has a normal subgroup of index two.

Thoughts:

It won't help to consider, without loss of generality, that the normal subgroup - let's call it $N$ - is $\langle x\rangle$, say, since, although $x\langle x\rangle=\langle x\rangle x$, trivially, the same cannot be said in general for $y\langle x\rangle$ and $\langle x\rangle y$.

It does give me the idea, though, that, to test the normality of $N$ in $G$, we need only show that $xN=Nx$ and $yN=Ny$.

Exactly where the $[G:N]=2$ comes from is a mystery to me. My hope is that some pithy choice of generators for $N$ would work, but, well, that's using techniques not yet covered in the book.

I feel like this is a question I ought to be able to answer myself. I've given it a few days and this is all I have.

Please help :)

答:
Let $X=\mathbb{R}P^2\vee \mathbb{R}P^2$ be the wedge sum of two copies of $\mathbb{R}P^2$. Consider the continuous function $f:X\to \mathbb{R}P^2$ which is identity on one copy and is constant on the other copy. Then $$f_{*}:G=\pi_1(X)\to \pi_1(\mathbb{R}P^2)=\mathbb{Z}/2\mathbb{Z} $$ is a surjective homomorphism whose kernel is a subgroup of index $2$.

Remark: This homomorphism is actually the homomorphism $\phi$ which is presented in the comment to OP post.


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答: A simpler formula to problem 3 is to move row $r$, column $c$ to box $2^{r-1}(2c-1)$. For three dimensions you can move row $r$, column $c$, layer $l$ to box $2^{l-1}(2^{r}(2c-1)-1)$. Do you see why these work and how to generalize to $n$ dimensions?</p>

Title: A question with infinity


答:
Multiply the two matrices as shown and you will have following equations:

$$a+3b=1$$

$$4a-b=0$$
From (2), we have: $$b=4a$$
Putting in (1), we get: $$a+3(4a)=1$$
this implies, $$13a=1$$
So $$a=\frac{1}{13} $$ and $$b=\frac{4}{13}$$

Title: $a\begin{pmatrix}1\\4\end{pmatrix}+b\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$ Find a and b

Body:
$$a\begin{pmatrix}1\\4\end{pmatrix}+b\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$$
Find a and b

I've found the answers to this to be $\frac{1}{13}$ and $\frac{4}{13}$ through google but just wondering, how would I go about working this through and finding the answer myself?


问:
Basic questions on Ideal generated by a set.
Firstly Please correct me ,if i am assuming something wrong.

Let $(R,+,.)$ be a commutative ring ,not necessearily unital. Then
$I\subset R$ is Ideal if $I$ has following properties,
$(I,+) < (R,+)$ and
$\forall x\in I,r\in R \implies rx\in I$

My question is ,by defination ideals generated by say $1$ elements ,is given by

$$(a)=\{ar \mid \forall r\in R \}$$

How does $a\in (a)$ ,if we not assume unity ?
$a\in (a)\implies a=a.r$ for some $r\in R$

Similarly for $2$ elements say ,$$(a,b)=\{ar+bs \mid \forall r,s\in R \}$$

How does $a+b$ belongs to this ?
As this will lead to ,$a+b=ar+bs$ ! As ideals are also additive subgroup ,these all should belongs to $I$ ! Where am i making mistake ?

答:
The expresion of $(a)=\{ar\;|\;\forall r\in R\}$ is a consecuence of the definition of ideal generated by a set. The deffinition is:
\begin{equation}
(a)=\bigcap_{\substack{(I,+)<(R,+)\\
 a\in (I,+)}} I
\end{equation}
Or in general, for a set $A\subseteq R$
\begin{equation}
(A)=\bigcap_{\substack{(I,+)<(R,+)\\
 A\subseteq (I,+)}} I
\end{equation}
This clearly contains $a$. It is, in fact, the smallest ideal containing $a$. For Rings with unit, this is equivalent to the deffinition you have given.


答:
Let $z=x+iy$, then
$$4x-2iy=(2+i\sqrt{3})\sqrt{x^{2}+y^{2}}\iff 4x=2\sqrt{x^{2}+y^{2}}\  \text{and}\ -2y=\sqrt{3}\sqrt{x^{2}+y^{2}}.$$
By the  equations, we know $x\geq 0,y\leq 0$, and both the two equations implies
$$y=-\sqrt{3}x.$$
So $$z=x-i\sqrt{3}x,x\geq 0.$$

Title: Solving complex equation $ z+3\bar{z}=(2+i\sqrt{3})|z|$

Body:
I struggle to solve this complex equation: $$ z+3\bar{z}=(2+i\sqrt{3})|z|$$
One way to resolve this is substitute |z| with $\sqrt{x^{2}+y^{2}}$ , $z=x+iy$, solve the cube etc,
but i did not reach the solution.

Any hint?. Thanks in advance.


问:
$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$?
Let $(a_n)^i, i=1,2,3$, be complex sequences with $\limsup_{n\to \infty} \sqrt[n]{\lvert (a_n)^i\rvert}\leq 1, i=1,2,3$. I want to show that
$$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$$
Proof: Since $\limsup_{n\to \infty} \sqrt[n]{\lvert (a_n)^i\rvert}\leq 1, i=1,2,3$, for every $\epsilon>0$, there is a natural number $N_i$ such that, for every $n\geq N_i$, we have $\sqrt[n]{\lvert (a_n)^i\rvert}< 1+\epsilon \Leftrightarrow \lvert (a_n)^i\rvert < (1+\epsilon)^n$. Since
$$\sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq \sqrt[n]{\lvert (a_n)^1\rvert + \lvert (a_n)^2\rvert + \lvert (a_n)^3\rvert},$$
it follows for all $n\geq \max\{N_1,N_2,N_3\}$ that
$$\sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}<\sqrt[n]{3(1+\epsilon)^n}=\sqrt[n]{3} (1+\epsilon)$$
Since the limit superior is the largest accumulation point of a sequence, it immediatly follows that
$$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert}\leq 1$$
cause otherwise, we could make $\sqrt[n]{3}(1+\epsilon)$ as close to 1 as we want for all $n\geq N$ with $N$ sufficiently large which would contradict
$$\limsup_{n\to \infty} \sqrt[n]{\lvert \lvert ((a_n)^1, (a_n)^2, (a_n)^3 \rvert \rvert} > 1$$

答:
That is correct, but a little bit sloppy at the end. After
$$
\sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert}<\sqrt[n]{3} (1+\epsilon)
$$
for $n\geq \max\{N_1,N_2,N_3\}$ I would continue with
$$
 \limsup_{n \to \infty} \sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert}
\le \limsup_{n \to \infty} \sqrt[n]{3} (1+\epsilon) \\
= \lim_{n \to \infty} \sqrt[n]{3} (1+\epsilon) = 1 + \epsilon
$$
and since this holds for all $\epsilon > 0$ the desired conclusion
$$
 \limsup_{n \to \infty} \sqrt[n]{\Vert ((a_n)^1, (a_n)^2, (a_n)^3 )\Vert} \le 1
$$
follows.

(In other words, consider first what happens for $n\to \infty$, and then what happens for $\epsilon \to 0$.)


问:
Want to compute $E(\sum_{i = 1}^n X_i^2 | \sum_{i = 1}^n X_i = t)$ for a random sample
I have $X_1, X_2, \dots, X_n$ an iid sample. My idea was to use that $\sum_{i = 1}^n X_i^2 = (\sum_{i = 1}^n X_i)^2 - \sum_{i\neq j} X_iX_j$.

Thus $$E(\sum_{i = 1}^n X_i^2 | \sum_{i = 1}^n X_i = t) = t^2 - E( \sum_{i\neq j} X_iX_j| \sum_{i = 1}^n X_i = t) $$
Now since the $X_i$ are iid
\begin{align*}E(\sum_{i = 1}^n X_i^2 | \sum_{i = 1}^n X_i = t) &= t^2 - \sum_{i\neq j}(E(  X_1| \sum_{i = 1}^n X_i = t))^2\\
&= t^2 - (n-1)(E(  \sum_{i = 1}^n X_i| \sum_{i = 1}^n X_i = t))^2\\
&= t^2 - (n-1)t^2\\
&= (2-n)t^2
\end{align*}

Is this correct? I feel like I am going wrong somewhere.

答:
You are incorrect. A simple example: Let $X_1,X_2$ be two fair coin-tosses, i.e. $X_1,X_2$ are iid. $\text{Bern}(0.5)$-distributed. Their conditional expectation given the event $X_1+X_2=1$ is given by \begin{align*}\textbf{E}[X_1^2+X_2^2|X_1+X_2=1]&=\textbf{E}[(X_1+X_2)^2-2X_1X_2|X_1+X_2=1]\\&=\textbf{E}[(X_1+X_2)^2|X_1+X_2=1]+\underbrace{\textbf{E}[-2X_1X_2|X_1+X_2=1]}_{=0}=1.\end{align*} In your formula, $n=2$, so the result would be $0$.

Why are you incorrect? As has been pointed out, $\textbf{E}[X_1X_2|X_1+X_2=1]=\textbf{E}[X_1|X_1+X_2=1]\textbf{E}[X_2|X_1+X_2=1]$ does not hold in the example, even though $X_1$ and $X_2$ are independent.

I think what you have done is you have confused $\textbf{independence}$ with $\textbf{conditional independence}$ given an event $A$. Even though $X_1\text{ and }X_2$ are independent, they are not conditionally independent given $X_1+X_2=t\in\mathbb{R}$. They can't be (unless one of them is a constant): If you know one of the variables, you immediatly know the other (since the sum is given). The same holds true in the general case: If you know $n-1$ of the variables, you immediatly know the one that is left.

Is your calculation ever correct (in the non-trivial case)? Possibly. The calculation might work out "on accident", if the values just happen to overlap. Your justification however only really makes sense if $X_1,X_2,\dots,X_n$ are linearly uncorrelated given the event $A:=\{\sum X_i=t\}$. This is a very different (allthough $\textbf{not}$ stronger) condition than independence, which rarely holds. If someone is more interested on (conditional) independence I recommend the following lecture:

https://www.youtube.com/watch?v=JzDvVgNDxo8&list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo&index=5


问:
Finding convergence of Series using Comparison Test
Determine the Convergence of
$$ \sum_{n=1}^\infty n\sin \left (\frac{1}{n^3}\right) $$
Can i compare it to
$$  \left (\frac{1}{n^3}\right) \geq  n\sin \left (\frac{1}{n^3}\right) $$
and since $\left (\frac{1}{n^3}\right) $ it is a p-series with p>1 then it is convergent and the smaller series also converges
I'm not sure though how to prove that the original is smaller then $\left (\frac{1}{n^3}\right) $

答:
When $x\in(0,\pi/2)$,
$$\sin x<x.$$
So $\sin \frac{1}{n^3}<\frac{1}{n^3}$, and hence
$$0<n\sin \frac{1}{n^3}<\frac{1}{n^2}.$$
Take $p=2$ of $p-$series will work.

And also you can use  Limit Comparison Test:
$$\lim_{n\to\infty}\frac{n\sin \left (\frac{1}{n^3}\right)}{\frac{1}{n^2}}
=\lim_{n\to\infty}\frac{\sin \left (\frac{1}{n^3}\right)}{\frac{1}{n^3}}=1>0.$$


问:
Prove that $G$ is cyclic group of prime order
I'm at the beggining of a Group Theory course, and I'm trying to solve this problem:

Let $G = 1$ (being $1=\{e\}$, $e$ the neutral element in $G$) be a group containing no subgroup different from $1$ and $G$.
Prove that $G$ is cyclic of prime order.

I've consider using that, given $g\in G$, $n\in\mathbb{N}$, $r=|g|$, I can conclude that
$$g^n=e \Leftrightarrow r|n \text{ ($r$ divides $n$)},$$
but I'm not sure if it's useful without proving first that $G$ is cyclic, and I'm not sure how to do it. Any help will be appreciated.

答:
If $G$ is non-cyclic, then any non-identity element generates a non-trivial, proper (cyclic) subgroup

If $G$ is cyclic with $|G| = n$ non-prime, a generator of $G$ raised to the power of a non-trivial divisor of $n$ generates a proper, non-trivial (cyclic) subgroup


问:
Trace of product of two positive semi-definite matrix.
Let $\rho \in \mathfrak{L}(A)$ and $\sigma \in \mathfrak{L}(A)$. Now I want to know can we say that:

$\rho \geqslant 0 $ and $\sigma \geqslant 0 \quad \quad $ if and only if $\quad \quad Tr(\rho \sigma) \geqslant 0$

$\rho \geqslant 0 $ and $\sigma \geqslant 0 $ means that both of them are positive semi-definite operator and $Tr$ means trace. $A$ is a Hilbert space and $\mathfrak{L}(A)$ means all linear maps on $A$.

If the above theorem is true could you please help me how can I show it?

答:
Implication

$$\quad \quad Tr(\rho \sigma) \ge 0 \ \implies \ \rho \succcurlyeq 0 \ \text{and}
 \ \sigma \succcurlyeq 0$$

cannot hold; here is a counterexample:

$$\rho=I \ \text{and} \ \sigma=\begin{pmatrix}1&2\\2&0\end{pmatrix}.$$

We have $Tr(\rho \sigma)=1 > 0$, but $\sigma$ not positive definite.

For the other implication, see here.


问:
Multiplying a matrix equation by a matrix
$$A=\left[\begin{matrix}
1&2\\
-2&3\\
1&-1
\end{matrix}\right]{,}\ B=\left[\begin{matrix}
1&-2&-4\\
-1&4&9
\end{matrix}\right]{,}\ b=\left[\begin{matrix}
1\\
3\\
5
\end{matrix}\right]$$

$$ BA=I_2$$

Solving the equation  $$A x=b.$$
I start by multiplying with the matrix B.

$$ BAx=Bb \\
\left[\begin{matrix}
1&0\\
0&1
\end{matrix}\right]x=\left[\begin{matrix}
1&-2&-4\\
-1&4&9
\end{matrix}\right]\left[\begin{matrix}
1\\
3\\
5
\end{matrix}\right] =\left[\begin{matrix}
-25\\
56
\end{matrix}\right] $$
If $$x=\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right], $$ then
$$\left[\begin{matrix}
1&0\\
0&1
\end{matrix}\right]\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right]=\left[\begin{matrix}
-25\\
56
\end{matrix}\right] \\
\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right]=\left[\begin{matrix}
-25\\
56
\end{matrix}\right]. $$

Solution should be $$ x=\left[\begin{matrix}
-25\\
56
\end{matrix}\right], $$ but it's not, since
$$ Ax=\left[\begin{matrix}
1&2\\
-2&3\\
1&-1
\end{matrix}\right]\left[\begin{matrix}
-25\\
56
\end{matrix}\right]=\left[\begin{matrix}
87\\
218\\
-81
\end{matrix}\right]\ne b$$ and this equation doesn't have solutions at all.
So at what point does this go wrong? Is there some rule I don't know?

答:
I'd think about this in geometrical terms. A multiplication by a non-singular square matrix is just a change in coordinate system, and doesn't change the intrinsic nature of the equation. Therefore any solution to the equation after the multiplication is a solution to the original equation. Such a multiplication transforms the equation into an equivalent equation.

Compared to that a multiplication by a non-square matrix with fewer rows than columns is a projection onto a linear space with fewer dimensions. You lose information from the original equation. If the original equation does have a solution, then the projected version will have a projected instance of that solution, corresponding to the same variable vector $x$. But you might well get a solution as an artifact of the projection, which doesn't have a corresponding solution in the original.

To illustrate that analogy, think about the problem of intersecting two lines in 3D space. When they do intersect and you project them onto a plane, the projected lines will intersect in the projection of the original point of intersection. But you might have lines which are skew in 3d and still find an intersection in 2d.

A solution to the lower dimensional system is required for any solution of the higher dimensional system, but it is not sufficient.

You can make similar arguments for square but singular matrices, which you can think of as a projection onto a lower dimensional space followed by an embedding of that into a space of the original dimension.


问:
Any way of computing the number of co-prime numbers to $n$
For a certain problem, I need to compute
$$|\phi(n)|=|\{x\in\mathbb{N} : 0<x<n, gcd(x,n)=1\}|.$$
Is there any "formula" that works for any natural number $n$? Or is it just impossible. Any help will be appreciated.

答:
The Euler Totient function:

$$\phi(n) = n \prod_{p | n} \bigg( 1 - {1 \over p} \bigg)$$

counts the positive integers up to a given integer $n$ that are relatively prime to $n$


问:
How to prove that $n|(a-b)$, such that $a,b \in S$ and pick any $n + 1$ numbers from set $S$ using divisibility and Pigeonhole principle?
let $n$ be a positive interger , prove that if you pick any $n + 1$ numbers from any set $S$, there  are two distinct numbers $(a,b) \in S$ (in the set) such that $n|(a-b)$.

So, I tried to prove that  two of these $(a-b)$  have the same remainder , so their difference is divisible by n. I used divisibility and Pigeonhole principle but I m not really sure how to prove it.
Any suggestion is welcome.Thank you.

答: I am not sure what $S$ is. The integers$\mod n$ have $n$ distinct equivalence classes (represented by remainders). By the pigeonhole principle, if we take $n+1$ elements in $\mathbb{Z_{n}}$=$S?$, two of them must have the same remainder mod $n$. Then the difference of the two is a multiple of $n$.


问:
how to solve system of linear equations of XOR operation?
how can i solve this set of equations ? to get values of $x,y,z,w$ ?

$$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$
$$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$
$$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$
$$\begin{aligned}1=w \oplus y \oplus z \end{aligned}$$

this is not a real example, the variables don't have to make sense, i just want to know the method.

答: Fast Walsh–Hadamard Transform used to solve system of boolean linear equations with distorted right-hand sides. See implementation on https://fwht.codeplex.com/</p>


问:
Let $G$ be the group of rigid motions in $R_3$ of a tetrahedron. Show that $|G |= 12$.
Let $G$ be the group of rigid motions in $R_3$ of a tetrahedron. Show that $|G |= 12$.($|G|$ refers to the order of $G$)

Proof: We first note that the tetrahedron has $4$ vertices. In order for it to be a rigid motion, we want to preserve the distance between every pair of points. In other words, if one face is adjacent to another, we wish to ensure that after the transformation, the vertices are still adjacent. Take any vertex of the solid. We can "send" this vertex of the solid to any of the $4$ positions. This gives us $4$ options. Take any other vertex that is adjacent to it, we have $3$ other potential options (All vertices are adjacent in a tetrahedron, and one vertex is already "occupied"). Once the position of the $2$ vertices have already been specified, the action on the other vertices is completely determined. Thus we have $4 \times 3=12$ possible rigid motions.

Is this proof too informal? How can it be improved? Thanks for your help!

答: WLOG fix one side and rotate the $3$ distinct possible ways. This can be done in the same way for each side and this can be represented as a bijection to the set of all rigid motions and so $|G| = 12$. Is this idea not rigorous enough? Although I might be misunderstanding your definition of "rigid motion".


问:
A function having poles at a sequence that has a limit point
This question was asked in my complex analysis end term exam which I got from senior and I was unable to solve it.

If f(z) has poles at  sequence of points ${z_n}$ and ${z_n} \to {z_0}$ , then prove that f(z) doesn't have pole at $z =z_0$.

I don't have any intuition on which result should I use.

It's my humble request to you to kindly tell on how should I approach it's proof.

Thanks!!

答:
False statement in case $z_n=z_0$ for all $n$. True if this case is excluded.

A pole is an isolated singularity. This means if $f$ has pole at $z_0$ it must be analytic throughout some disk containing $z_0$ except for the point $z_0$. So the conclusion follows just by definition!.


问:
Compactness Of Sets And Metric Spaces
I was studying compactness property and got stuck on a point. I first saw the definition of compactness as : A subset $A$ of a metric space $(X, d)$ is said to be compact if for any collection of open subsets of $X$ that cover $A$, there is a finite subcollection that covers $A$. Now I saw problems in which there was compact metric space, but how can compactness be defined for metric spaces as metric spaces are themselves full spaces and hence how a collection of open sets will cover them and relative to which spaces those sets will be open??

答:
It might be better not to start with a definition of compact subsets of a topological space but with a definition of compact spaces.

This by:

If $(X,\tau)$ denotes a topological space then it is a compact topological space if any cover of $X$ by open sets contains a finite subcover.

Based on that you can secondly define the concept of compact subsets like this:

If $(X,\tau)$ denotes a topological space then a set $A\subseteq X$ is a compact subset of the space if the subspace $(A,\tau')$ is a compact subspace.

Here $\tau'$ denotes the subspace topology on $A$ inherited from topological space $(X,\tau)$ and defined as $\tau':=\{A\cap U\mid U\in\tau\}$.

If you like then you can easily translate this to metric spaces.

Title: Compactness Of Sets And Metric Spaces


问:
Show that $\int_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^n}$ is finite

Show that $\int_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^n}$ is finite.

Here $|x|^2 := x_1^2+\dots+x_n^2$ is the norm of $x$. The statement is clearly true for $n=1$. For $n=2$ we have $$ \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(1+x_1^2+x_2^2)^2}\le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(x_1^2+x_2^2)^2} \le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{x_1^2+x_2^2} \le\,\, ?$$ How do I proceed with this. I noticed that the integrandum is even, and I feel like I need to find an upper bound for this integral that is the $n$-th power of an integral that is finite and that can be easily computed (for example the $n=1$-integral); something like $$ \int_{\mathbb{R}} \frac{dx_1}{1+x_1^2}\int_{\mathbb{R}} \frac{dx_2}{1+x_2^2}\cdots\int_{\mathbb{R}} \frac{dx_n}{1+x_n^2}.$$ Thanks in advance.

答: $\frac 1 {1+|x|^{2}} = \frac 1 {1+x_1^{2}+x_2^{2}+...+x_n^{2}} \leq \frac  1 {1+x_i^{2}}$ for each $i$. Multiply these $n$ inequalities and integrate.


问:
Questions about the moment map
Let $(M,\omega)$ be a symplectic manifold endowed with a hamiltonian action of a torus $T$. Let $\mu : M \longrightarrow {Lie(T)}^*,$  be a moment map associated to this action. Let $S_M =\bigcap\limits_{m \in M} Stab(m)$, and $s_m$ be its lie algebra.

I have two questions about the moment map:

I) Why is the moment map  constant on each connected component of $M^T$ ?

II) I know that for each $m \in M $, the image of tangent map of $\mu $ at m is $ Im(T_m \mu)={(s_m)}^\bot = \lbrace \eta \in Lie(T) \mid
\langle \eta , X \rangle = 0 , \forall X \in s_m \rbrace $. How does this imply that the image of M by the moment map is an affine space directed by the linear space ${(s_m)}^\bot$ ?

答:
I think it is helpful always to build intuition for Hamiltonian circle actions and then bring it over to the general case.

For a Hamiltonian circle actions there is a vector field $X$ (i.e. the integral curves of the vector field are the orbits of $S^1$-action) and smooth function $H : M \rightarrow \mathbb{R}$, subject to the equation: $$dH = \omega(X,\cdot) .$$

Now note that by the above identity a point $p$ is fixed by $S^{1}$ $\iff$ $X|_p=0$ $\iff$ $dH|_p=0$.

Hence the derivative of the Hamiltonian is exactly zero on components of the fixed point set and hence $H$ has to be constant on them. For higher dimensional torus actions picking a basis for the Lie algebra and applying this argument to each of the components give the argument.

For $2$ I think you should clarify since the image of the moment map is not always an affine space although it is a convex polytope (By AGS convexity theorem).

Title: Questions about the moment map


问:
Why is the following method incorrect when deciding on the possible choices?
"Six men and three women are standing in a supermarket queue. Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?"

I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.

$$
 {3 \choose 1} \ \times \ {6 \choose 2} \ + \  {3 \choose 2} \ \times \ {6 \choose 1}  \ + \  {3 \choose 3}
$$

Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.

This approach is in fact correct. However, another approach came to my mind as well.

If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.

$$
{3 \choose 1} \ \times \ {8 \choose 2}
$$

This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.

答: Because you can choose Becky with ${3 \choose 1}$ and Kate with ${8 \choose 2}$, and in another way choose Kate with ${3 \choose 1}$ and Becky with ${8 \choose 2}$ all other choices being the same. You repeat cases depending on which order the women are chosen if there is more than $1$.


问:
Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

Let $X_i,\ i \geq 1$ be i.i.d. discrete random variables with mean $\mu$ and variance $\sigma^2.$ Let $k \gt 1.$ Define the sequence \begin{align*} Y_n : & = \dfrac {X_1 X_2 \cdots X_k + X_2 X_3 \cdots X_{k+1} + \cdots  + X_{n-k+1} X_{n-k+2} \cdots X_n} {n}. \end{align*} Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

How do I find that? Is there any easy way to proceed?

Thanks in advance.

EDIT $:$ What I get is that $$ \Bbb {E}\ \left [ {Y_n}^2 \right ]  = \dfrac {1} {n^2} \left [    \left (n - k + 1 \right )   \left ({\sigma}^2 + {\mu}^2 \right )^k + 2 \left [\sum\limits_{r = 0}^{k-2} \left (n-k-r \right ) \left (\mu^2 \right )^{r+1} \left ({\sigma}^2 + {\mu}^2 \right )^{k-r-1} + \dfrac {\left (n-2k+1 \right ) \left (n-2k+2 \right )} {2} \left ({\mu}^2 \right )^k \right ] \right ].$$ But then $$\begin{align*} \lim\limits_{n \to \infty} n^{\frac {3} {4}} \Bbb {Var}\ [Y_n] & = \lim\limits_{n \to \infty} \left [ \dfrac {\left (n-k+1 \right ) \left (n-k+2 \right)} {n^{\frac {5} {4}}} - \dfrac {\left (n-k+1\right)^2} {n^{\frac {5} {4}}} \right ]  {\mu}^{2k} \\ & = \lim\limits_{n \to \infty} \dfrac {\left (n - k + 1 \right )} {n^{\frac {5} {4}}} {\mu}^{2k} = 0.\end{align*}$$ Am I right? Can anybody please check my computation whether it is correct or not.

Thanks in advance.

Source $:$ This question appeared in ISI PhD entrance test in Mathematics held in $20$th September this year (TEST CODE : MTB) in the afternoon session (Question No. $9$).

答:
Let
$Z_i≔X_i⋯X_{i+k-1}$.

As upper bounds we use:

$$\mathbb{V}(Z_i)=\mathbb{V}\left(X_i\cdots X_{i+k-1}\right)=\mathbb{E}\left(X_i^2\cdots X_{i+k-1}^2\right)-\mathbb{E}\left(X_i\cdots X_{i+k-1}\right)^2\le2{\max{\left(\max_{i\in\left[n\right]}{\left(X_i^2\right)},\max_{i\in\left[n\right]}{\left(\left|X_i\right|\right)}\right)}}^k=:v
$$
and
$$
|\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)|
\le2{\max{\left(\max_{i\in\left[n\right]}{\left(X_i^2\right)},\max_{i\in\left[n\right]}{\left(\left|X_i\right|\right)}\right)}}^{2k}=:e
$$

Using these we can now estimate the variance of $Y_n$:

$$\begin{align}
\mathbb{V}\left(Y_n\right)&=\mathbb{V}\left(\frac{X_1X_2\cdots X_k+X_2X_3\cdots X_{k+1}+\cdots+X_{n-k+1}X_{n-k+2}\cdots X_n}{n}\right)\\&=\frac{1}{n^2}\mathbb{V}\left(Z_1+\ldots+Z_{n-k+1}\right)\\&=\frac{1}{n^2}\mathbb{V}\left(\sum_{i=1}^{n-k+1}Z_i\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\end{matrix}}^{n-k+1}Cov\left(Z_i,Z_j\right)\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\end{matrix}}^{n-k+1}{\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)}\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\left|i-j\right|\le k\\\end{matrix}}^{n-k+1}{\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)}\right)
\\&\le\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}{\ v}\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\left|i-j\right|\le k\\\end{matrix}}^{n-k+1} e\right)\\&=\frac{1}{n^2}\left(\left(n-k+1\right)v\right)+\frac{2}{n^2}\left(\left(n-k+1\right)ke\right)
\end{align}$$

Using this estimate, we can then conclude that the limit converges to $0$.


问:
Inverse of $g(x)=\frac{x^2+x}{2}$
While I am graphing a function $g$ in desmos, I want to make $g(y)$ be a function in $x$.

I doesn't mean $\frac{x^2+x}{2}$, but I want the function $f$ such that the graph $y=f(x)$ is same to the graph $x=f(y)$, i.e. the inverse of $g(x)=\frac{x^2+x}{2}$.

Can it be written out using simple funtion or is there a proof of negation?

Or is there any other way to graph the function (using desmos' simple technique) (I don't mean to write a program)?

What I have tried:

I think this is may be equation $$\frac{[f(x)]^2+[f(x)]}{2}=x$$, i.e. $$[f(x)]^2+[f(x)]=2x$$
If the R.H.S. is something like $x^2-\frac{1}{4}$ or $x-\frac{1}{4}$, it will be much simpler.

But I have no idea how to solve.

答:
Rewrite the equation as $x=f(a)$,
$$a^2+a-2x=0$$
$$a=\frac{-1\pm\sqrt{1+8x}}{2}$$

Title: Inverse of $g(x)=\frac{x^2+x}{2}$


问:
In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$

In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$. Also $EF$ and $AM$ intersect at $G$ with $GF = 36$ cm, $GE = x$ cm. Find $x$ .

What I Tried: Here is a picture :-

No idea for this. I am not getting any idea on how to use these facts together :-

$CM = MB$ and $AB:AC = 4:3$ and $AE:AF=2:1$ .

I don't see any similar triangles unfortunately. That's why I cannot proceed with this, angle-chasing is anyways not going to help. As some ideas I think I have to make an extra construction, but I don't know where. I also think areas can help, but did not find a way of it.

Can anyone help me? Thank You.

答:
Rename the points like on a picture: $AE:AD =2:1$

Draw a parallel through $B$ to $DE$ then if $AB = 4y$ then $AF = 2y$ and $AC = 3y$ so $AF:CF = 2:1$.

Draw a parallel through $F$ to $AM$ then $CI:IM = 1:2$, so if $MB = 3z$ then $MI = 2z$.

So $IM:MB = 2:3$ and thus $FH:HB = 2:3$. But $DG:GE = FH:HB$ so $x=54$.


问:
The rigorous definition of continuity in a topological space
Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.

Let $x \in S_1$.

In this page Proof Wiki, it is said that the mapping $f$ is continuous at (the point) $x$ (with respect to the topologies $\tau_1$ and $\tau_2$) if and only if: For every neighborhood $N$ of $f(x)$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f[M]\subseteq N$.

But in my knowledge, the condition should be $\forall y \in S_1 $, $y \in M \Rightarrow f(y) \in N$ instead of $f[M]\subseteq N$. The problem is that the two statements are not equivalent: the former statement is equivalent to $M \subseteq f^{-1}[N]$ because
$$
y\in M \iff \{y\} \subseteq M \implies f[y] \subseteq f[M] \iff f(y) \in f[M]
$$
As a result, we have a condition $f(y) \in f[M] \Rightarrow f(y) \in N$, which is equivalent to $M \subseteq f^{-1}[N]$.

Would you check this out?

答:
$$\forall y \in M: f(y) \in N$$ and $$f[M]\subseteq N$$

are exactly the same thing.


问:
Construct a DFA based on this language
Let $L \subseteq \Sigma^*$ be a language of any alphabet $\Sigma$. Let $move(L)$ describe the language which contains every string such that it takes the last letter in $x \in L$ and puts it to the front.

How can we construct a DFA $A$ for $move(L)$?

I started off by looking at what the DFA for L would be like:

Say $s = s_0 s_1 s_2 ... s_n$ where $s_i \in \Sigma$ is a string this DFA accepts

it will go by some path of states $q_0q_1...q_n$. I made the last state transition to the first in $A$ but something tells me this isn't right so I'm not sure how to move on.

答:
Let $\ A=\big(Q,\Sigma, \delta, q_0, F\big)\ $ be a DFA which accepts $\ L\ $. Let
\begin{align}
Q'&=\big(\Sigma\times Q\big)\,\cup \big\{q_0'\big\}\\
F'&=\big\{(s,q)\in \Sigma\times Q\,\big|\,\delta(q,s)\in F\big\}\\
\delta'(q',s)&=\cases{(s,q_0)&if $\ q'=q_0'$\\
\big(\sigma,\delta(q,s)\big)&if $\ q'=(\sigma,q)\in\Sigma\times Q\ $.}
\end{align}
Then $\ S'=\big(Q',\Sigma, \delta', q_0', F'\big)\ $ is a DFA which accepts $\ move(L)\ $.

If $\ \xi\in\Sigma^*\ $, and $\ q_n\ $ is the state of $ A\ $ after it has processed the string $\ \xi\ $, then $\ \big(s_0,q_n\big)\ $ will be the state of $\ A' $ after it has processed the string $\ s_0\xi\ $, and $\ \big(s_0,q_n\big)\in F'\ $ if and only if $\ \delta(q_n,s_0)\in F\ $—that is, $\ A'\ $ accepts the string $\  s_0\xi\ $ if and only if $\ A\ $ accepts the string $\ \xi s_0\ $.

Reply to query from OP in comments

In general, the minimal-state DFA for the language $\ L\ $ will have strictly fewer states than the minimal-state DFA for $\ move(L)\ $, so if you're given a DFA for $\ L\ $ it will not always possible to construct a DFA for $\ move(L)\ $ which has the same set of states.  If
\begin{align}
L=\ &\big\{a^{n_1}b^{n_2}c^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge1\,\big\}\\
\cup &\big\{a^{n_1}b^{n_2}d^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge1\,\big\}\\
\cup &\big\{a^{n_1}b^{n_2}e^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge1\,\big\}\ ,
\end{align}
for instance, it is easy to construct a $5$-state DFA which accepts $\ L\ $. However,
\begin{align}
move(L)=\ &\big\{ca^{n_1}b^{n_2}c^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge0\,\big\}\\
\cup &\big\{da^{n_1}b^{n_2}d^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge0\,\big\}\\
\cup &\big\{ea^{n_1}b^{n_2}e^{n_3}\,\big|\,n_1\ge0, n_2\ge0, n_3\ge0\,\big\}\ ,
\end{align}
and it is not difficult to show that any DFA which accepts $\ move(L)\ $ must have at least $7$ states (by using the Myhill-Nerode theorem, for instance).

What if $\ \epsilon\in L\ $?

Dromniscience's answer and LetmeKnow's comment below have alerted me to the fact that the above answer implicitly (and inadvertently on my part) makes an assumption which is not necessarily justified—namely that the move operation will eliminate the empty string $\ \epsilon\ $ if it happens to be in $\ L\ $.  Because $\ q_0'\not\in F'\ $ in the above definition of $\ A'\ $ the empty string $\ \epsilon\ $  can't be in the language accepted by $\ A'\ $.

However, since the OP doesn't specify how the move operation will deal with the empty string, it doesn't seem justified to me to assume that $\ \epsilon\not\in move(L)\ $ whenever $\ \epsilon\in L\ $. If, instead, $\ \epsilon\in L\ \implies\epsilon\in move(L)\ $ then the definition of $\ A'\ $ would have to be modified as follows:
\begin{align}
Q'&=\big(\Sigma\times Q\big)\,\cup \big\{q_0'\big\}\\
F'&=\cases{\big\{(s,q)\in \Sigma\times Q\,\big|\,\delta(q,s)\in F\big\}&if $\ q_0\not\in F$\\
\big\{(s,q)\in \Sigma\times Q\,\big|\,\delta(q,s)\in F\big\}\cup\big\{q_0'\big\}&if $\ q_0\in F$}\\
\delta'(q',s)&=\cases{(s,q_0)&if $\ q'=q_0'$\\
\big(\sigma,\delta(q,s)\big)&if $\ q'=(\sigma,q)\in\Sigma\times Q\ $.}
\end{align}

Title: Construct a DFA based on this language


问:
Find maximum for a function with two variables in the definition quantity
Hello fellow mathematicians,
I am trying to find maximum for the function:

```<code>f(x) =2xy^2 - 2x^2
</code>```

In the following definition quantity:

```<code>x^2+y^2<=1, x>=0
</code>```

It is obvious that the definition quantity is a half circle.
First I checked the stationary points, and I got the minimum (0,0)
Now I need to find maximum and I am sure it is on the edge, cuz I have not checked the edge yet. (not sure what you call this in English sorry, but on the edge of the definition quantity)

I think its obvious that the edge of the definition quantity is x^2+y^2 = 1. I've just removed the unequal sign and added in an equal sign.
But I don't know where to go from here. I want to find the maximum on the edge. Do I just put x=0 and x=1 and then solve for y? I'm not sure this is giving me the correct answer.

Kind regards,
Zebraboard

答:
Based on given constraints of $x \geq 0$ and $x^2 + y^2 \leq 1, \, 0 \leq x \leq 1$ and $-1 \leq y \leq 1$.

Please observe that at $y = 1, x = 0$, So $2xy^2 - 2x^2 = 0$. Similarly at $x = 1, y = 0$, so $2xy^2 - 2x^2 = -2$.

Also for any value of $x$ in the range, maximum value of our function will be when we maximize $y$. That reasoning leads us to finding maximum at the circumference.

Using the method of Lagrange multipliers,

$2xy^2 - 2x^2 = \lambda (x^2 + y^2 - 1)$

Differentiating wrt. $x$ and $y$, we get

$2y^2 - 4x = 2 \lambda x \implies \lambda = \frac{2y^2 - 4x}{2x} \, $ (for $x \ne 0$)

$4xy = 2 \lambda y \implies \lambda = 2x \,$ (for $y \ne 0)$

Equating the two we get, $y^2 = 2(x^2 + x)$

Substituting in our constraint $x^2 + y^2 = 1, x^2 + \frac{2}{3}x = \frac{1}{3}$

i.e. $(x+ \frac{1}{3})^2 = \frac{4}{9} \implies x = \frac{1}{3}$

That gives us $y^2 =  \frac{8}{9}$.

We get maximum value of $2xy^2 - 2x^2 = \frac{10}{27}$.


问:
Optimum way to find the determinant of the following matrix.
Given a matrix like the following, is there a way, only by using elementary transformations between rows and columns, to find its determinant? I know it would be easy to compute it or to try and find a pattern, but is there a faster way using that method?

The matrix goes like:

1, if $i = j$.

$\lambda$, if $i = j > 1$.

k, if $i \neq j$.

Where $i$ and $j$ indicate the row and the column, respectively.

答:
The direct way would be to compute its eigenvalues.

We want the eigenvalues of $A=
\begin{pmatrix}1&k&\ldots &k\\k&\lambda&\ldots&k\\\vdots&&\ddots&\vdots\\k&\ldots&k&\lambda\end{pmatrix}$
An $n \times n$ matrix.
Or $\dfrac{A-(\lambda+k)I}{k}=B$  is a matrix that has presumably only two non zero eigenvalues. The case $\lambda=1$ is easy so the eigenvalues of $A$ are $(kn+1-k,1-k,\ldots,1-k)$. Suppose $\lambda\neq 1$ we compute a non zero eigenvalue $r$ of $B$.  Set $a=\dfrac{1-\lambda+k}{k}-r$ and $b=1-r$ to annihilate the determinant of

$B-rI=\begin{pmatrix}a&1&\ldots &1\\1&b&\ldots&1\\\vdots&&\ddots&\vdots\\1&\ldots&1&b\end{pmatrix}$

To do this we solve a linear dependence system of rows so there is a non zero tuple $(x_1,\ldots,x_n)$
for  $2\le j\le n$:

$$\begin{cases}x_1a+\sum\limits_{i=2}^nx_i=0\\\sum\limits_{i\neq j}x_i+bx_j=0\end{cases}$$
doing some substracting with first equation  and each of the rest we get that all $x_j$ are equal for $j\ge 2$. Finally you solve $x_1+(n-2)x+bx=0$ with $x_1=\dfrac{-(n-1)x}{a}$.

This gives a second degree polynomial in $r$. Calculations are cumbersome but doable.


问:
Nondimensionalization of an ODE
I've been trying to reduce the number of parameters of the following ODE:
$$u'=ku+ruln(u)$$ where k, r are parameters.
When solving similar problems, we've used techniques like substitution. For example:
$$u'(x)=ku(x)$$ Let$$ y=kx$$ $$ w(y)=u(x)$$ Then $$w'(y)=(u(\frac{y}{k}))'$$ $$w'(y)=\frac{1}{k}ku((\frac{y}{k}))$$ $$w'=w$$
And we got rid of parameter k.
However, the problem with the first eq. is that we have log, which basically ''switches multiplication to addition''. To illustrate what I mean, here's my attempt:
$$\pi=kt$$ $$w(\pi)=u(t)=u(\frac{\pi}{k})$$ $$w'(\pi)=(u(\frac{\pi}{k}))'$$ $$w'(\pi)=\frac{1}{k}u'(\frac{\pi}{k})$$ $$w'(\pi)=\frac{1}{k}(ku(\frac{\pi}{k})+ru(\frac{\pi}{k})ln(u(\frac{\pi}{k})))$$ $$w'(\pi)=w(\pi)+\frac{r}{k}w(\pi)lnw(\pi)$$
So we kind of got rid of the parameter in front of the first term, but if we try to use substition as $z=Cw$ where C is some konstant, log will turn it to logC + logW which isn't very useful. I tried something like $z=e^\frac{r}{k}w$:
$$z'=e^\frac{r}{k}w'$$ $$z'=z+\frac{r}{k}zln(e^\frac{k}{r}z)$$ $$z'=z(1+1+\frac{r}{k}ln(z))$$
Also, I've forgot to mention: ODE with a given inital condition $u(0)=u_0$

答:
$$u'=ku+ru\ln(u)$$
$v(x)=\ln(u)$
$$e^v v'=ke^v+re^vv$$
$$v'=k+rv$$
This is a first order linear ODE that you can solve.

ADDITION in order to answer to your comment :

The aim of the change of function $v(x)=\ln(u)$ is to simplify the equation. This is the first step to remove the parameters $k$ and $r$ which is possible for the linear equation :
$$v'=k+rv$$
Let $v=k\,w$
$$\frac{dw}{dx}=1+r\,w$$
Let $x=\frac{t}{r}$
$$r\frac{dw}{dt}=1+r\,w$$
Let $z=r\,w$
$$\frac{dz}{dt}=1+z$$
This is the first order linear ODE without parameter in it.

Title: Nondimensionalization of an ODE


问:
Vanishing particular solution for $yy'-ay=-bx$
Consider a differential equation of the form:
$$y\frac{dy}{dx}-ay=-bx$$
where both both $a$ and $b$ are known real numbers
$$a,b\neq0.$$
When I attempt to solve this with a particular solution of:
$$y_p=Cx+D$$
Results to
$$CD-aD=0$$
$$C^2-aC=-b$$
By inspection, we can say that:
$$C=a$$
But this does not make sense since:
$$0=-b$$
Which is impossible!

I even tried using quadratic but same thing. This looks very simple, but I am now a bit lost here, what is the best way to solve this differential equation?

答:
$$yy'-ay=-bx$$
Try $y=kx$
$$k^2x-akx=-bx$$
$$k^2-ak+b=0$$
Solve for $k$. $k$ depends on both $a$ and $b$.
$$\Delta = a^2-4b \;\;,k=\dfrac {a \pm \sqrt {\Delta}}{2}$$


答:
You are counting each selection with two women twice and each selection with three women three times, once for each way you could have designated one of the women who is selected as the woman you have selected and the remaining woman or women as additional people.

Suppose you select two women, Abigail and Beatrice, and one man, Charles.  You count this selection twice:

\begin{array}{l l}
\text{woman} & \text{additional people}\\ \hline
\text{Abigail} & \text{Beatrice, Charles}\\
\text{Beatrice} & \text{Abigail, Charles}
\end{array}

Suppose you select three women, Abigail, Beatrice, and Charlotte.  You count this selection three times:

\begin{array}{l l}
\text{woman} & \text{additional people}\\ \hline
\text{Abigail} & \text{Beatrice, Charlotte}\\
\text{Beatrice} & \text{Abigail, Charlotte}\\
\text{Charlotte} & \text{Abigail, Beatrice}
\end{array}

Notice that
$$\binom{3}{1}\binom{6}{2} + \color{red}{\binom{2}{1}}\binom{3}{2}\binom{6}{1} + \color{red}{\binom{3}{1}}\binom{3}{3} = 84$$
where the factors in red are the number of ways you could have designated one of the women you have selected as the woman you have selected.

Title: Why is the following method incorrect when deciding on the possible choices?

Body:
"Six men and three women are standing in a supermarket queue. Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?"

I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.

$$
 {3 \choose 1} \ \times \ {6 \choose 2} \ + \  {3 \choose 2} \ \times \ {6 \choose 1}  \ + \  {3 \choose 3}
$$

Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.

This approach is in fact correct. However, another approach came to my mind as well.

If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.

$$
{3 \choose 1} \ \times \ {8 \choose 2}
$$

This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.


问:
To find the x and y-intercepts of the line $ax+by+c=0$
Please check if I've solved the problem in the correct way:

The problem is as follows:

Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.

My solution:

We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.

Now we solve the equation $ax+by+c=0$ for $y$:

$ax + by + c=0$

$ax + by = -c$

$by = -ax -c$

$y = \frac {-ax-c}{b}$

$\because x = 0$ at y-intercept,

$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$

$y = -\frac cb$.

The point at which the line crosses the y-axis is $(0,-\frac cb)$

Now we solve the equation $ax+by+c=0$ for $x$:

$ax+by+c=0$

$ax+by=-c$

$ax = -by-c$

$x = \frac {-by-c}{a}$

$\because y = 0$ at x-intercept

$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$

$x = -\frac ca$

The point at which the line crosses the x-axis is $(-\frac ca,0)$

答:
easier way is this equation of line intercept form
$$ \frac xa+\frac yb=1$$
so write the given eqn in this form
$$ax+by+c=0$$
$$ax+by=-c$$
$$\frac {ax}{-c}+\frac {by}{-c}=1$$
$$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$

so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)


问:
In how many ways can 3 dice add up to 16?
I am trying to use the stars and bars technique to calculate the number of ways 3 dice rolls can add up to 16. To my understanding, this can rephrased as:

$$x + y + z = 16 | x,y,z \in \mathbb{Z}^+ \land x,y,z \leq 6$$

I am not anywhere close to being a mathematician, so simple solutions are much appreciated. My main motivation to the question, is that I am trying to see if there is a cooler way to do it rather than count by hand, and stars and bars came to my mind as a potential solution. However, other simple solutions are also welcome.

答: The max sum of two dice is $12$ so the lowest die must be $\geq 4$. Then we have $4,6,6$. If the lowest die is $5$ we have $5,5,6$. If you count the dice as distinct then you can permute these possibilities.


问:
Is this a valid argument to show $|a|^q \in L^{(p/q)'}(\Omega)$?
Brezis' functional analysis book includes a solution to one of its exercises that is looking somewhat fishy to me: while attempting to formalize the last step in the solution, I have arrived at a viciously circular argument. Have I misinterpreted Brezis' argument? Here is the exercise, its solution and my (mis?)interpretation of such solution (I have transcribed the first two myself, so any typos are probably mine) (You can ignore case 2 ($p = \infty$); the important step is the last step in case 1 ($p < \infty$)):

Exercise 4.7: Let $1 \leq q \leq p \leq \infty$. Let $a(x)$ be a measurable function on $\Omega$. Assume that $au \in L^q(\Omega)$ for every function $u \in L^p(\Omega)$. Prove that $a\in L^r(\Omega)$ with $r = \frac{pq}{p-q}$ if $p < \infty$ and $r = q$ if $p = \infty$ [Hint: Use the closed graph theorem.]

Solution: Consider the operator $T: L^p(\Omega)\to L^q(\Omega)$ defined by $Tu = au$. We claim that the graph of $T$ is closed. Indeed, let $(u_n)$ be a sequence in $L^p(\Omega)$ such that $u_n \to u$ in $L^p(\Omega)$ and $au_n \to f$ in $L^q(\Omega)$. Passing to a subsequence we may assume that $u_n \to u$ a.e. and $au_n \to f$ a.e. Thus $f = au$ a.e. and so $f = Tu$. It follows from the closed graph theorem (Theorem 2.9) that $T$ is bounded and so there is a constant $C$ such that

$(S1)$ $ \| au \|_q \leq C \| u \|_p$ $\forall u \in L^p(\Omega)$

Case 1: $p < \infty$. It follows from $(S1)$ that

$\int |a|^q |v| \leq C^q \| v \|_{p/q}$ $\forall v \in L^{p/q}(\Omega)$

Therefore the map $v \mapsto \int |a|^q v$ is a continuous linear functional on $L^{p/q}(\Omega)$ and thus $|a|^q \in L^{ (p/q)' } $

Case 2: $p = \infty$. Choose $u = 1$ in $(S1)$.

How I interpreted the last step in the argument in case 1: from what I understand, when Brezis says "and thus $|a|^q \in L^{ (p/q)' } $", he is invoking Riesz's representation theorem: since the map $v \mapsto \int |a|^q v$ is a continuous linear functional on $L^{p/q}(\Omega)$, it follows that there exists a unique function $w \in L^{(p/q)'}(\Omega)$ such that $\int |a|^q v = \int wv$ $\forall v \in L^{p/q}$, from where, supposedly, it would follow that $w = |a|^q$, due to unicity.

The problem is that Riesz's theorem only guarantees the unicity in $L^{ (p/q)' }$; that is, it only tells us that there can't be two distinct functions both in $L^{ (p/q)' }$ representing the same continuous linear functional. Hence, we cannot directly apply (the unicity part of) such theorem to $|a|^q$ and $w$, since we have not yet proved $|a|^q \in L^{ (p/q)' }$ (in fact, that's what we would like to conclude!)

My question is, then: have I misinterpreted Brezis' argument or is it flawed? (Maybe he had something else in mind?)

答:
This first paragraph is not properly an answer but it may clarify things for future readers. The step from (S1) to
$$\int |a|^q |v| \leq C^q \| v \|_{p/q} \quad \forall v \in L^{p/q}(\Omega)$$
was to me a bit difficult to see but it is simply the fact that $v\in L^{p/q} \iff v^{1/q}\in L^{p}$. Raise (S1) to the power $q$ and apply it to $u = v^{1/q}$.

Let $\phi$ stand for $L^{p/q}(\Omega) \to \mathbb{K}, v \mapsto \int |a|^q v$. (Here $\mathbb{K}$ stands for $\mathbb{R}$ or $\mathbb{C}$.) This  $\phi$ is in $\left(L^{p/q}(\Omega)\right)^{*}$. Riesz Representation Theorem guarantees there is a unique $b$ in $L^{(p/q)^*}(\Omega)$ such that $\phi(v) = \int b v$ for all $v$ in $L^{p/q}(\Omega)$. Using this, the definition of $\phi$ and the linearity of integration, we get:
$$ \quad \int (|a|^q-b)v = 0 \quad \forall v \in L^{p/q}(\Omega).$$
So, as Luiz Cordeiro drew attention to, we want to show that this implies $(|a|^q-b) = 0$  almost everywhere.  It will follow from this:

$$
\left.
\begin{array}{l}
h: \Omega \to  \mathbb{K} \text{ is measurable}\\
\Omega \text{ is } \sigma\text{-finite}\\
\forall w\in L^r(\Omega) \;\; \int h w = 0
\end{array}
\right\} \implies h=0  \text{ a.e}.
$$

Proof. Since $\Omega$ is $\sigma$-finite, there are measurable subsets $X_n$ such that $|X_n|< \infty$ and $\bigcup_n X_n = \Omega$. Fix $n$ and let $E\subset X_n$ be measurable.
Since $|E| \leq |X_n|<\infty$, the indicator function $\mathbf{1}_{E}$ is in $L^{r}(\Omega)$ and so, by the property $h$ has, $\int h · \mathbf{1}_{E} = \int_{E} h  = 0$.
This is true for all measurable sets $E$ in $X_n$ and so $h = 0$ almost everywhere on $X_n$. Let us check this. Let $h_i = \mathrm{Im} h$ and $h_r = \mathrm{Re} h$ and write $h = i(h_i^+ -h_i^-)+ (h_r^+ -h_r^-)$ by decomposing $h_i$ and $h_r$ in negative and positive parts.  If $h$ were not $0$ almost everyhere on $X_n$, some $f$ in $\{h_i^+, h_i^-, h_r^+, h_r^-\}$ would have to be positive in a set of positive measure zero, and consequently the measure of $E = \{x\in X_n | f(x) >0\}$. If, say $f = h_i^+$, we would have $\mathrm{Im} \int_E h = \int_E h_i^+ > 0$ and then $\int_E h = i\mathrm{Im} \int_E h + \mathrm{Re} \int_E h$ would not be zero. The same would happen in the other cases. Therefore, $h$ is zero almost everywhere on $X_n$. In other words, $h=0$ in $X_n$ except perhaps on a null set $Z_n \subset X_n$. This can be said for all $n$ and so we conclude $h = 0$ almost everywhere on $\Omega$ by noticing that $h$ vanishes on $$\textstyle\bigcup_n (X_n \setminus Z_n) = \left(\bigcup_n X_n \right)\setminus \left(\bigcup_n Z_n\right) =  \Omega \setminus \left(\bigcup_n Z_n\right)$$
and that the set $\bigcup_n Z_n$ has measure zero. $\square$


答:
You don't have to actually move the last transition to the front. Instead, why don't you simply remember the first letter and check whether it will lead to a final state in the original DFA?

Please think it over before you continue reading.
If we denote the original DFA as $L = \langle Q, \Sigma, \delta, q_0, F\rangle$, we can construct a new DFA for $move(L) = \langle Q', \Sigma, \delta', q_{init}, F'\rangle$, where
\begin{align}
Q' &= \{q_{init}\} \cup \Sigma \times Q \\
F' &= \{(a, q)\mid \delta(a, q)\in F\} \\
\delta'((a, q), b) &= (a, \delta(q,b)) \\
\delta'(q_{init}, b) &= (b, q_0) \\
\end{align}
Since $\epsilon$ is never in $move(L)$, $F'$ suffices.

Title: Construct a DFA based on this language

Body:
Let $L \subseteq \Sigma^*$ be a language of any alphabet $\Sigma$. Let $move(L)$ describe the language which contains every string such that it takes the last letter in $x \in L$ and puts it to the front.

How can we construct a DFA $A$ for $move(L)$?

I started off by looking at what the DFA for L would be like:

Say $s = s_0 s_1 s_2 ... s_n$ where $s_i \in \Sigma$ is a string this DFA accepts

it will go by some path of states $q_0q_1...q_n$. I made the last state transition to the first in $A$ but something tells me this isn't right so I'm not sure how to move on.