问:
Product of numbers
Pair of numbers whose product is $+7$ and whose sum is $-8$.
Factorise $x^{2} - 8x + 7$.

I can factorise but it's just I can't find any products of $+7$ and that is a sum of
$-8$. Any idea? Thanks guys!

Thanks.

答:
your questions answer are already users have given so i'm not repeating them.
  sometimes factorization is not easy because number are too big to manually calculate or may be fractional so try this formula as alternative.
  roots for equation $(ax^2+bx-c)=0$ is,
  $x=\dfrac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ ,
  $\dfrac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$

Title: Product of numbers


问:
If $f:[a,b]\rightarrow \mathbb{R}$ is continuous, for any $x_{1},...,x_{n}\in [a,b]$ exist $c\in [a,b]$ s.t $f(c)=\frac{f(x_{1})+...+f(x_{n})}{n}$
My issue is that I don't understand this problem, so I don't know how to attack it. Could you give me some hints on how to approach this?</p>

答:
Let $(x_k)$ be a convergent sequence in $[a,b]$. Notice that the sequence $\big\{f(x_k)\big\}_{k=1}^{\infty}$ converges to $f(c)$ for some $c\in[a,b]$.

Because the sequence $$y_n=\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}$$ is the Cesaro Mean of the sequence $\big\{f(x_k)\big\}_{k=1}^{\infty}$, then $y_n$ converges to the same $f(c)$ as above.

ContentLicense: CC BY-SA 4.0


问:
If $x*y$ is the operation $x*y = xy/(x+y+1)$, then how does $x*y*z = xyz(x+y+1)/(x+y+z+xy+xz+1)$?
$$x*y = \frac{xy}{x+y+1}$$

What I got was,

$$x*(y*z) = x* \frac{yz}{y+z+1} =x\frac{\frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1}$$

The book's answer is

$$x*(y*z) = x*\frac{yz}{y+z+1} = \frac{xyz(y+z+1)}{x+y+z+xy+xz+1}$$

答:
$x*y=\frac{xy}{x+y+1}$

We put $  b=x*y=\frac{xy}{x+y+1}. $

$b*z=\frac{bz}{b+z+1}=\frac{\frac{xy}{x+y+1}z}{\frac{xy}{x+y+1}+z+1}$

$x*y*z=\frac{zxy}{xy+xz+yz+z+x+y+1}$

So: the book is wrong

ContentLicense: CC BY-SA 3.0


问:
Proving a generalization of Schwarz Lemma by calculating an integral
This problem comes from Gamelin, "Complex Analysis", Exercise IX.1.7. Basically it asks us to prove a generalization of the Schwarz Lemma without the assumption $f(0)=0$, by calculating a particular integral.

The statement of the problem is: let $f(z)=\sum_{k=0}^{\infty}a_k z^k$ be analytic in $B(0,1)$, with $|f(z)|\leq M$ for all $z\in B(0,1)$. By integrating $|f(z)|^2$ along $\partial B(0,r)$ for $r<1$, show that
$$\sum_{k=0}^{\infty} |a_k|^2\leq M^2.$$
(This will imply that $|f'(0)|=|a_1| \leq M$, which is what Schwarz Lemma states, but here it is not assumed that $f(0)=0$.)

Here is how I attempted to calculate the integral mentioned in the statement of the problem: using the change of variable $z=re^{i\theta}, dz=ire^{i\theta}d\theta$,
$$\begin{align}
\oint_{\partial B(0,r)} |f(z)|^2 dz & = \int_{0}^{2\pi} (\sum_{k=0}^{\infty}a_k r^k e^{i\theta k})(\sum_{k=0}^{\infty}\overline{{a_k} r^k e^{i\theta k}})ire^{i\theta}d\theta \\
& = \sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\int_{0}^{2\pi}a_{k_1}\overline{a_{k_2}}r^{k_1 + k_2}e^{i\theta (k_1 - k_2)}ire^{i\theta}d\theta \\
& = \sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}a_{k_1}\overline{a_{k_2}}ir^{k_1 + k_2 +1}\int_{0}^{2\pi}e^{i\theta (k_1 - k_2 +1)}d\theta,
\end{align}$$

and since
$$\int_{0}^{2\pi}e^{i\theta (k_1 - k_2 +1)}d\theta =
\begin{cases}
2\pi, & \text{if } k_2 = k_1 + 1 \\
0, & \text{otherwise}
\end{cases}$$

we have
$$\oint_{\partial B(0,r)} |f(z)|^2 dz = \sum_{k=0}^{\infty}a_k \overline{a_{k+1}}r^{2k+1}\cdot 2\pi i,$$

But I cannot see how to relate this to the result we wanted to prove. In particular I cannot make this into an expression that involves $\sum_{k=0}^{\infty} |a_k|^2$. However it's quite close to this, since $|a_k|^2 = a_k \overline{a_k}$ and we have terms of the form $a_k \overline{a_{k+1}}$ in the sum on the right hand side. This leads me to suspect that there is some mistake I've made in the computation of the integral; however I cannot see where it is.

If I am mistaken in any step above, can someone help me point it out? If not, can someone give me some hint at this problem? Any help is appreciated!

答:
Let $z=re^{i\theta}$,
\begin{align*}
\left|f(re^{i\theta})\right|^2
&=f(re^{i\theta})\cdot\overline{f(re^{i\theta})}\\[3pt]
&=\sum_{n=0}^{\infty}a_nr^ne^{i n\theta}\cdot
\sum_{n=0}^{\infty}\overline{a}_n r^ne^{-i n\theta}\\[3pt]
&=\sum_{m,n=0}^{\infty}a_m\overline{\alpha}_nr^{m+n}e^{-i (m-n)\theta}\\
&=\sum_{n=0}^{\infty}|a_n|^2r^{2n}
+\sum_{\substack{m,n=0\\ m\ne n}}^{\infty}a_m\overline{a}_nr^{m+n}e^{-i (m-n)\theta}.
\end{align*}
Note that
$$\frac{1}{2\pi}\int_{0}^{2\pi}e^{i n\theta}d\theta=0\ (n\ne 0),$$
we get
\begin{align*}
\frac{1}{2\pi}\int_{0}^{2\pi}\left|f(re^{i\theta})\right|^2d\theta
&=\frac{1}{2\pi}\int_{0}^{2\pi}\left(\sum_{n=0}^{\infty}|a_n|^2r^{2n}\right)d\theta\\[3pt]
&=\sum_{n=0}^{\infty}|a_n|^2r^{2n}.
\end{align*}
So
$$\sum_{n=0}^{\infty}|a_n|^2r^{2n}
=\frac{1}{2\pi}\int_{0}^{2\pi}\left|f(re^{i\theta})\right|^2d\theta
\leq M^2,\ 0<r<1.$$
Let $r\to 1^{-}$, we can get
$$\sum_{k=0}^{\infty} |a_k|^2\leq M^2.$$


问:
Orthogonality and Linear independence infinite dimension
I have a doubt, in a Hilbert space it is known that if I have $ \{x_k\}_{k=1}^n$ vectors in a finite space and these n elements are orthogonal then are linear independent. Is it true for a space of infinite dimension when if $(x_k, x)=0$ $\implies $ $x=0$?

答: It is apparent that the space is an inner product space, say $H$ and that the given vectors, say $\{e_k\}_{k\in I}$ for some indexing infinite set $I$, are mutually orthogonal. To that end, then yes, it is true that $H$ is infinite-dimensional ( what, however, is generally not true —in contrast to the $n$-dimensional case—is that $H=span\{\hat{e}_k\}_{k\in I}$, where $\hat{e}_k$ is the unit vector in the direction of $e_k$; even $span\{\hat{e}_k\}_{k\in I}$ may not be isomorphic to $H$.)


问:
Fubini's formula on fibered Riemannian manifolds.
Let $\pi:(M,g)\to(N,h)$ be a Riemannian submersion (that is, $\pi$ is a submersion satisfying $g|_{\ker(d\pi)^\perp}=\pi^*h|_{\ker(d\pi)^\perp}$). Each fiber of $\pi$ is a Riemannian submanifold of $M$, given by $(F_y,g_y):=\left(\pi^{-1}(y),g|_{\pi^{-1}(y)}\right)$. In the case that $\pi$ is a canonical projection on a Riemannian product manifold, and $f\in C^\infty_cM$ is a compactly supported smooth function, we can write Fubini's formula suggestively as
$$
\int_Mf(x)d\mu_g(x)=\int_N\left(\int_{F_y}f(x)d\mu_{g_y}(x)\right)d\mu_h(y)
$$
Where $\mu_g$ is the measure on $M$ induced by $g$, etc. Intuitively, it seems that this formula also should hold for a general Riemannian submersion $\pi$, but it is not so clear how to show this. $M$ need not look like a Riemannian product space, even locally, since $\ker(d\pi)^\perp$ need not be integrable. As a result, $g$ does not decompose nicely in adapted coordinates, making a direct computation difficult. I suspect there's a more straightforward geometrical argument, or a simple counterexample that I'm missing.

答:
It seems that the missing piece was was the use of an adapted orthonormal coframe (thanks to Ted Shifrin's suggestion) and a version of Fubini's theorem for differential forms, described in this answer. Let $\nu$ be an $n$-form on $N$, and $\mu$ be an $(m-n)$-form on $M$ (with both smooth and $\mu$ compactly supported). With suitable choices of orientation, we have
$$
\int_M\pi^*\nu\wedge\mu=\int_N\left(\int_{\pi^{-1}(y)}\mu\right)\nu(y)
$$
This can be shown straightforwardly by working on adapted local coordinates (and the hypotheses can be weakened significantly, but that isn't needed here).

We can w.l.o.g. consider the case where $\operatorname{supp}(f)$ is contained in the domain $U\subseteq M$ of a local adapted orthonormal frame $e^1,\cdots,e^m$, where $e^i=\pi^*(\widetilde{e}^i)$, and $\widetilde{e}^1,\cdots,\widetilde{e}^n$ are a local orthonormal frame on $\pi(U)\subseteq N$. Using the shorthand $e^{k\cdots l}:=e^k\wedge\cdots\wedge e^l$, and choosing appropriate orientations, we can write the integral in terms of these frames.
\begin{align*}
\int_Uf(x)d\mu_g(x)=&\int_Uf(x)e^{1\cdots m}(x) \\
=&\int_U\pi^*\left(\widetilde{e}^{1\cdots n}(x)\right)\wedge\left(f(x)e^{(n+1)\cdots m}(x)\right)
\end{align*}
This expression can be Fubinated, yielding the desired result.
\begin{align*}
=&\int_{\pi(U)}\left(\int_{U\cap F_y}f(x)e^{(n+1)\cdots n}(x)\right)\widetilde{e}^{1\cdots n}(y) \\
=&\int_{N}\left(\int_{F_y}f(x)d\mu_{g_y}(x)\right)d\mu_h(y) \\
\end{align*}


问:
Is this model a linear regression model?
We have,

$$Y_i = log_{10}(\beta_1X_{i1}) + \beta_2X_{i2} + \epsilon_i$$

So,

$$Y_i = log_{10}(\beta_1) + log_{10}(X_{i1}) + \beta_2X_{i2} + \epsilon_i$$

I don't think this is a general linear regression model because of the term $log_{10}(\beta_1)$ as I believe this makes the model non-linear in the parameters. However, I also think that since $\beta_1$ is a constant, $log_{10}(\beta_1)$ is also a constant, and thus the model should still be linear in the parameters. In addition, I am not sure how to think about the term $log_{10}(X_{i1})$ since it does not have a coefficient. What is its role in the model?

答:
$$Y_i = \log_{10}(\beta_1) + \log_{10}(X_{i1}) + \beta_2X_{i2} + \epsilon_i$$ is not linear since
$$\frac{\partial Y_i}{\partial \beta_1} \neq 0$$ But
$$Y_i = \beta_3+ \log_{10}(X_{i1}) + \beta_2X_{i2} + \epsilon_i$$ is linear.

When done, make $\beta_1=10^{\beta_3}$.


问:
How can $\lambda$ in Poisson distribution be constant if derived from binomial distribution for $\lambda=np$ while $n$ goes to infinity?

My Problem. Above is the derivation for Poisson from binomial which I quoted from somewhere. I understand this derivation mostly, but my struggle is that I do not understand why we can say $\lambda=np$, and $\lambda$ is constant in Poisson distribution.

My trial. My thought is that in a binomial distribution the probability of a success (i.e. for a certain event to happen) for each trial is constant, so $p$ is constant. Then deriving Poisson from binomial, we let $n$ approach infinity, and should not $\lambda$ go to infinity as a result as $\lambda=np$?

My comment. Any kind of idea or help would be appreciated. Thank you!

答: Implicitly we assume at the very beginning that $\lambda$ is a constant. $p$ decreases as $n\to\infty$ so as to keep the product $\lambda=np$ constant (otherwise the expression for $P(X=k)$ would not remain a valid probability).


问:
Is this a valid linear transformation?
(1) Which of the following is a linear transformation?

$ \quad $ (a) $ T ( \begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  0 \\ x \\y \end{pmatrix}$, $ \quad $ (b) $ T ( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ) = \begin{pmatrix}  y \\ x \end{pmatrix}$ $ \quad $ (c) $ T (\begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  x^2 \\ y^2 \\ 0 \end{pmatrix}$

$ \quad $ (d) $ T (\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix}  1 \\ x \\y \end{pmatrix}$ $ \quad $ (e) $ T ( \begin{pmatrix}\ x \\ y  \\ z \end{pmatrix} ) = x \begin{pmatrix}  1 \\ 1 \\ z \end{pmatrix}$

I am new to this topic, am I correct in saying that only (a) is a linear transformation?

(2) $ T (\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}) = \begin{pmatrix}  2 \\ 2 \\2 \end{pmatrix}$ , $ \quad $ $ T (\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}) = \begin{pmatrix}  1 \\ -1 \\ -1 \end{pmatrix}$, $ \quad $ $ T (\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}) = \begin{pmatrix}  0 \\ 0 \\ 0 \end{pmatrix}$ $ \quad $

$ \quad $$ M = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & -2  \end{pmatrix}. $

$ \quad $ Then,  $ T (\begin{pmatrix} x \\ y \\ z \end{pmatrix}) = \begin{pmatrix}  2 &  1 & 0 \\ 2 & -1 & 0 \\ 2 & -1 & 0 \end{pmatrix} M^{-1}  \begin{pmatrix} x \\ y \\ z \end{pmatrix} $  ?

答: $(a)$ and $(b)$ are.  The others do not possess the properties that $T(cv)=cT(v)$ and $T(v+w)=T(v)+T(w)$.

Title: Is this a valid linear transformation?


答:
I think you need to understand the definition of linear transformation. Note that:

Let $V$ and $W$ vector spaces over field $F$. Let $T: V\to W$ a function, so $T$ is said linear transformation if, and only if,

$1.$ $\forall v_{1},v_{2}\in V: T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$.

$2.$ $\forall \alpha \in F, \forall v \in V: T(\alpha v)=\alpha T(v)$.

Now, you can prove in this sense that
\begin{eqnarray}
T \text{is a linear transformation}  \iff \forall \alpha, \beta\in F, \forall v_{1},v_{2} \in V: T(\alpha v_{1}+\beta v_{2})=\alpha T(v_{1})+\beta T(v_{2})
\end{eqnarray}

Now, for your question you need to see if $T$ is a linear transformation, so:

b) $T$ Is a linear transformation, because
\begin{eqnarray}
T\left(\alpha \begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta \begin{pmatrix} a \\ b \\ c \end{pmatrix} \right)
&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \\ \alpha z +\beta c \end{pmatrix}\\
&=&\begin{pmatrix} \alpha y +\beta b \\ \alpha x +\beta a \end{pmatrix}\\
&=&\alpha \begin{pmatrix} y \\ x \end{pmatrix} +\beta \begin{pmatrix} b \\ a \end{pmatrix}\\
&=&\alpha T\begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \\ c \end{pmatrix}
\end{eqnarray}

a) Also $T$ is linear transformation (I think you can do this part).

c) $T4 is not linear transformation (I think you can do this part).

d) $T$ is not linear transformation, because
\begin{eqnarray}
T\left( \alpha \begin{pmatrix} x \\ y \end{pmatrix} +\beta\begin{pmatrix} a \\ b \end{pmatrix} \right)&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&=&\begin{pmatrix} 1 \\ \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&\not=& \alpha \begin{pmatrix} 1 \\ x \\ y \end{pmatrix}+\beta \begin{pmatrix} 1 \\ a \\ b \end{pmatrix}\\
&=& \alpha T \begin{pmatrix} x\\ y \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \end{pmatrix}
\end{eqnarray}

e) $T$ is not linear transformation ( I think you can do this part).

Title: Is this a valid linear transformation?

Body:
(1) Which of the following is a linear transformation?

$ \quad $ (a) $ T ( \begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  0 \\ x \\y \end{pmatrix}$, $ \quad $ (b) $ T ( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ) = \begin{pmatrix}  y \\ x \end{pmatrix}$ $ \quad $ (c) $ T (\begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix}  x^2 \\ y^2 \\ 0 \end{pmatrix}$

$ \quad $ (d) $ T (\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix}  1 \\ x \\y \end{pmatrix}$ $ \quad $ (e) $ T ( \begin{pmatrix}\ x \\ y  \\ z \end{pmatrix} ) = x \begin{pmatrix}  1 \\ 1 \\ z \end{pmatrix}$

I am new to this topic, am I correct in saying that only (a) is a linear transformation?

(2) $ T (\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}) = \begin{pmatrix}  2 \\ 2 \\2 \end{pmatrix}$ , $ \quad $ $ T (\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}) = \begin{pmatrix}  1 \\ -1 \\ -1 \end{pmatrix}$, $ \quad $ $ T (\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}) = \begin{pmatrix}  0 \\ 0 \\ 0 \end{pmatrix}$ $ \quad $

$ \quad $$ M = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & -2  \end{pmatrix}. $

$ \quad $ Then,  $ T (\begin{pmatrix} x \\ y \\ z \end{pmatrix}) = \begin{pmatrix}  2 &  1 & 0 \\ 2 & -1 & 0 \\ 2 & -1 & 0 \end{pmatrix} M^{-1}  \begin{pmatrix} x \\ y \\ z \end{pmatrix} $  ?


问:
Show that $(0, 1)$ fails Heine-Borel.
So I think I understand the Heine-Borel property, but I want to make sure I've wrapped my head around it. With the help of @C Squared I was able to come up with the following, but I want to make sure it's okay. Thanks!

Let $I_k = (\frac{1}{k}, 2)$. Observe that $\bigcup_{k=1}^\infty I_k$ covers $(0, 1)$. Let $x\in (0, 1)$. It follows by the achimedean principle that there exists some natural number $k$ such that $\frac{1}{k} < x$. We may now conclude,
\begin{equation*}
0 < \frac{1}{k} < x < 1 < 2
\end{equation*}
\begin{equation*}
x \in (\frac{1}{k}, 2), k \in \mathbb{N}^+.
\end{equation*}
so,
\begin{equation*}
x\in\bigcup_{k=1}^\infty (\frac{1}{k}, 2).
\end{equation*}
therefore,
\begin{equation*}
(0, 1) \subset (1, 2) \cup (\frac{1}{2}, 2) \cup  (\frac{1}{3}, 2) \cup \dots = \bigcup_{k=1}^\infty I_k
\end{equation*}
Now, Heine-Borel insists that there exists a finite subcover for every cover of $(0, 1)$ that also covers $(0, 1)$. So to show that $(0,1)$ is not compact we must show that the finite subcovers of $\bigcup_{k=1}^\infty I_k$ do not cover $(0,1)$.

Suppose a finite subcover ($\bigcup_{k=1}^n I_k$) of $\bigcup_{k=1}^\infty I_k$. Then we have,
\begin{equation*}
(0, 1) \not\subset \bigcup_{k=1}^n (\frac{1}{k}, 2),
\end{equation*}
because $\frac{1}{n+1} \in (0, 1)$ but $\frac{1}{n+1} \not\in (1, 2) \cup \dots \cup (\frac{1}{n}, 2)$ (by the definition of this set).

There exists a finite subcover that does not cover $(0, 1)$, so the set fails Heine-Borel and is not compact.

答:
Using the Heine-Borel theorem, you need only show that $(0,1)$ is not closed. Since $0,1$ are limit points of $(0,1)$, and $0,1\not\in[0,1]$, then $(0,1)$ is not closed, therefore it cannot be compact.

Alternatively, if you suppose that $(0,1)$ is compact, then consider the cover $$\mathcal{U}=\bigcup_{k=1}^{\infty}(\frac{1}{k},2)$$ By assumption, $\mathcal{U}$ admits a finite sub-cover, $$\mathcal{U}'=\bigcup_{k=1}^n(\frac{1}{k},2)$$ for some $n\in\mathbb{N}$. Now, notice that $(\frac{1}{n},2)=\mathcal{U}'$.

We claim that $(0,1)$ is not a subset of $(\frac{1}{n},2)$.

Fix $n\in\mathbb{N}$. Since $\frac{1}{n+1}\in (0,1)$ and $\frac{1}{n+1}\not\in(\frac{1}{n},2)$, then $(0,1)$ is not a subset of $(\frac{1}{n},2)$, so $\mathcal{U}'$ is not a cover of $(0,1)$. Since $n$ was arbitrary, then no finite sub-cover exists.

Because we have found a cover of $(0,1)$, namely $\mathcal{U}$, that does not admit a finite sub-cover, then $(0,1)$ cannot be compact.

Title: Show that $(0, 1)$ fails Heine-Borel.


问:
Prove that if $n$ and $m$ are coprime then $2^n-1$ and $2^m-1$ respectively are coprime too
How to prove that if $n$ and $m$ are coprime then  $2^n-1$ and $2^m-1$  are coprime too?
I tried with Bezout's identity without results.

Thanks

Body: Show that $\gcd(2^n-1, 2^m-1) = 2^{\gcd(m, n)} - 1 = 2^1-1= 1$.


问:
Asymptotic estimate as $N \rightarrow \infty$ of $\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$
Looking for the exact if possible, otherwise the asymptotic expansion and best estimate of the error terms as $N \rightarrow \infty$ of the two fractional sums $$\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$$  My literature search has not found any examples similar to this.  I have some material for the divisors terms such as $\left\lfloor{N/a}\right\rfloor^k$ and $\left\{{N/a}\right\}^k$.  This is part of the calculation of the summations over the floor function of this argument.

From Benoit Cloitre. On the circle and divisor problem. November 2012 we have
$$\lim_{x \rightarrow \infty} \sum_{n = 1}^{x} \left\lfloor{\frac{x}{{n}^{2}}}\right\rfloor \sim
\zeta \left({2}\right) x
+ \zeta \left({\frac{1}{2}}\right) \sqrt{x}
+ O \left({{x}^{\theta}}\right)$$

Where $\theta = 1/4 + \epsilon$ is the estimated best error.

答:
Asymptotic is $(1 - \gamma) N$, where $\gamma$ is Euler–Mascheroni constant.

Proof

For any $x, y$:
$$
\begin{array}\\
\{x \pm y\} &= x \pm y - [x \pm y] \\
&= x \pm y - [[x] + \{x\} \pm [y] \pm \{y\}] \\
&= x - [x] \pm \{y\} - [\{x\} \pm \{y\}].
\end{array}
$$

Now set $x = \frac Nn, y = \frac N{n^2}$ to break the sum apart:

$$
\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}
= \underbrace{\sum\limits_{n = 1}^{N} \frac Nn}_{(1)}
- \underbrace{\sum\limits_{n = 1}^{N} \left[ \frac Nn \right]}_{(2)}
\pm \underbrace{\sum\limits_{n = 1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3)}
- \underbrace{\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right]}_{(4)}.
$$

$(1)$ is Harmonic series, $(1) = N \ln N + \gamma N + \frac 12 + o(1)$.

$(2)$ is divisor summatory function, $(2) = N \ln N + N(2\gamma - 1) + O(\sqrt N)$.

$(3) = \underbrace{ \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} \left\{ \frac{N}{n^2} \right\} }_{(3.1)}
+ \underbrace{ \sum\limits_{n = \left[ \sqrt N \right]+1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3.2)}.
$

$
(3.1) \leq  \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} 1 \leq \sqrt N.
$

$(3.2) = \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{N}{n^2}
\leq N \cdot \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n (n-1)}
= N \cdot \left( \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n-1} - \frac{1}{n} \right)
= N \left( \frac{1}{\left[ \sqrt N \right]} - \frac{1}{N} \right)
\leq \frac {N}{\sqrt{N} + 1} - 1.
$

$(4) = O(\sqrt N)$. Proof is very technical and is written below.

Putting $(1)$, $(2)$, $(3)$, $(4)$ together, and leaving only leading asymptotic terms we have

$$
\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}
= (1 - \gamma) N + O(\sqrt N).
$$

Proving $(4) = O(\sqrt N)$

We want to show that $\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right] = O(\sqrt N)$.

$$
\sum\limits_{1}^{N} [...] = \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]} }[...] + \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N } [...], \\
$$

We split the sum in such a way, so that

First part doesn't have too many summands.

In the second sum we have $n > \frac{N}{\left[\sqrt N \right]} \geq \sqrt N$, which means we can "drop" braces: $\left\{ \frac{N}{n^2} \right\} = \frac{N}{n^2}$.

It will be convenient to work with the second sum later.

First sum is $O(\sqrt N)$ because the $[...]$ part equals either $-1$, $0$ or $1$:
$$
\left| \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} [...] \right| \leq  \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} 1 = O(\sqrt{N}) .
$$

We'll split second sum even further, so that we can also "drop" braces for $\left\{ \frac Nn \right\}$:

$$
\sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...]
= \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} \sum\limits_{\frac{N}{k + 1} < n \leq \frac Nk} [...].
$$

Note, that $\frac{N}{k + 1} < n \leq \frac Nk \implies k \leq \frac Nn < k + 1 \implies \left\{ \frac Nn \right\} = \frac Nn - k$.

$$
[...]
= \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right]
= \left[ \frac Nn - k \pm \frac {N}{n^2} \right]
= \left[ N \frac{n \pm 1}{n^2}  \right] - k.
$$

When "$\pm$" is "$+$", the $[...]$ is either $0$ or $1$. We want to find for how many $n$ it is $1$.

$$
\left[ N \frac{n + 1}{n^2}  \right] - k = 1 \iff  N \frac{n + 1}{n^2} \geq k + 1 \iff \frac{k+1}{N}n^2 - n - 1 \leq 0, \\
\text{where} \; n \in \left( \frac{N}{k+1}; \frac Nk \right].
$$

Solving quadratic inequality gives $n \in \left( \frac{N}{k+1}; \frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} \right] $. Length of this semi-interval is

$$
\frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} - \frac{N}{k+1}
= \frac{N}{k+1} \frac{-1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2}
= \frac{N}{k+1} \frac{-1 + 1 + 4 \frac{k+1}{N}}{2 \left(1 + \sqrt{1 + 4 \frac{k+1}{N}} \right) }
= \frac{2}{1 + \sqrt{1 + 4 \frac{k+1}{N}}} < 1.
$$

This means that at most $1$ integer $n$ can be inside that semi-interval.

When "$\pm$" is "$-$", the logic is similar, in that case there can be at most $2$ integer $n$ for which $[...] \neq 0$.

Finally, for the second sum we have $$
\left| \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...] \right|
\leq \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} 2
= O(\sqrt N).
$$


问:
Integral of a two dimensional function
A function is defined by
$$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt$$
We are to decide the partial derivatives of the function.

I am quite unsure whether this is a trick question because we could do an approximation of the integrand with the Taylor series, but I can't see how that would simplify the work.

Could I just get a direction to as how I should approach this problem? I would very much not like to just get the solution.

Note that this is one of the easiest questions on the paper.

答:
You have
$$f(x,y)=\int_{a(x,y)}^{b(x,y)} g(t) \, dt$$ Using the fundamental theorem of calculus
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial b(x,y)}{\partial x}\, g(b(x,y))-\frac{\partial a(x,y)}{\partial x}\, g(a(x,y))$$
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial b(x,y)}{\partial y}\, g(b(x,y))-\frac{\partial a(x,y)}{\partial y}\, g(a(x,y))$$

Just apply.

Title: Integral of a two dimensional function


问:
Relationship between $_nC_2$ and $\sum_{k=1}^{n-1} k$
$$_nC_2 = \frac{n(n-1)}{2}$$

and so is the following sum: $$\sum_{k=1}^{n-1} k = \frac{n(n-1)}{2}$$

I'm curious if there's any ingenious alternative interpretation of the combination $_nC_2$ in terms of the sum of numbers.

答:
Think of a square consisting of $n \times n$ unit squares. A diagonal of the square contains $n$ unit squares and it splits the rest of the square into $2$ equally sized "triangles" of $x$ unit squares each. We then have
$$n+2x=n^2 \Rightarrow x=\binom{n}{2}$$


问:
A map on the $2n$-sphere and degree zero
Let $f : \mathbb{S}^{2n} \to \mathbb{S}^{2n}$ is continue e $f(x) \neq -f(-x)  \quad\forall x$. Prove that $f$ has degree zero.

Thanks!!

答: Since $f(x) \ne -f(-x)$ for all $x$, the function $\varphi(x, t) = ((1-t)f(x) + (t)f(-x))/|(1-t)f(x) + (t)f(-x)|$ is well-defined.  In fact $\varphi$ defines a homotopy from $f$ to $g$, where $g$ is the map $g(x) = f(-x)$.  Since $g$ is just the composition of the antipodal map $-\mathbf{1}$ followed by $f$, the degree of $g$ is $\deg(g) = (-1)^{2n+1} \deg(f) = -\deg(f)$.  But if $f$ is homotopic to $g$, then ...</p>


问:
$f \in D\left[x\right]$ is invertible iff $f \in D$
If $D$ is a domain, I must show that $f \in D\left[x\right]$ is invertible iff $f \in D$.

Is it important that $D$ is a domain? It would happen if $D$ is a ring?

答:
The element zero is never a unit of a ring, so let's only consider nonzero elements as potential units.

First note that if $R$ is a ring and $h\in R[x]$ then

$h\in R{\,\setminus}\{0\}$ implies $\deg(h)=0$.$\\[4pt]$

$h\in R[x]{\,\setminus}R$ implies $\deg(h) > 0$.

Next show that if $D$ is a domain and $f,g$ are nonzero elements of $D[x]$ then
$$\deg(fg)=\deg(f)+\deg(g)$$
Regarding your initial statement, your claim (corrected to exclude zero) was that if $D$ is a domain then

$f\in D[x]$ is a unit iff $f\in D{\,\setminus}\{0\}$.

First consider the "only if" direction.

Thus let $D$ be a domain and suppose $f\in D[x]$ is a unit.

Then $fg=1$ for some $g\in D[x]$, hence
$$0=\deg(1)=\deg(fg)=\deg(f)+\deg(g)$$
so we can't have $\deg(f) > 0$, thus we must have $f\in D{\,\setminus}\{0\}$.

Hence if $D$ is a domain, $f\in D[x]$ is a unit only if $f\in D{\,\setminus}\{0\}$.

To show that the "if" direction can fail, suppose $D$ is a domain but not a field. Then there must exist a nonzero element $f\in D$ which is not a unit in $D$. Let $f\in D$ be a nonzero nonunit of $D$ and suppose $f$ is a unit of $D[x]$.

Then $fg=1$ for some $g\in D[x]$ with $g\not\in D$, but then
$$0=\deg(1)=\deg(fg)=\deg(f)+\deg(g)=0+\deg(g)=\deg(g) > 0$$
contradiction.

So for the case where $D$ is a domain, you have "only if" but not "iff".

For the case where the ring of coefficients is not a domain, the "only if" direction can also fail.

For example, let $R=\mathbb{Z}_4$ and let $f$ in $R[x]$ be given by $f=2x+1$.$\;$Then
$$
f^2=(2x+1)^2=4x^2+4x+1=0+0+1=1
$$
so $f$ is a unit in $R[x]$, but $f\not\in R$.


Title: Integral of a two dimensional function

Body:
A function is defined by
$$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt$$
We are to decide the partial derivatives of the function.

I am quite unsure whether this is a trick question because we could do an approximation of the integrand with the Taylor series, but I can't see how that would simplify the work.

Could I just get a direction to as how I should approach this problem? I would very much not like to just get the solution.

Note that this is one of the easiest questions on the paper.

Body:
\begin{align}
& f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt = \int_0^{xy} e^{t^2} \, dt - \int_0^{x^2} e^{t^2} \, dt. \\[10pt]
& \frac d {du} \int_0^u e^{t^2} \, dt = e^{u^2}. \\[10pt]
& \frac d {dx} \int_0^u e^{t^2}\,dt = \frac d {du} \int_0^u e^{t^2} \, dt \times \frac{du}{dx}
\end{align}


问:
Bounding $E[X]E[e^{-X}]$ for a non-negative random variable with finite moments
Let $X$ be a random variable such that $X \geq 0$ and $\exists \varepsilon > 0: E[e^{X \varepsilon}] < \infty$. Is it true that $\exists$ a real number $B < \infty$ such that $E[X]E[e^{-X}] \leq B$ for all such $X$? If so, please provide a constructive proof that derives a real value for $B$.

The question is motivated by the observation that as $X$ increases point-wise, $e^{-X}$ decreases point-wise. If $X$ is non-random then $Xe^{-X} \leq \frac{1}{e}$. The question asks whether the fact that $Xe^{-X}$ is a bounded function over $X \geq 0$ may be generalized to $X$ being a non-negative random variable with new goal being to bound $E[X]E[e^{-X}]$.

I provide no detail regarding the specific distribution of $X$ because I am asking whether a hard real-valued bound exists for all random variables satisfying the given criteria. Otherwise, $B = E[X]$ suffices for a particular random variable.

答: If $X$ is $0$ or $N$ with equal probability, then $\mathbb E[X] = \frac N2$ and $\mathbb E[e^{-X}] = \frac{1 + e^{-N}}{2} > \frac12$, so $\mathbb E[X] \mathbb E[e^{-X}] > \frac N4$ and no upper bound can exist.


答:
There is no such $B$, this is just wrong.

For example taking $X_n$ such that :$\mathcal{L}(X_n)= \frac{1}{2} \delta_0 +\frac{1}{2}\delta_n$

Title: Bounding $E[X]E[e^{-X}]$ for a non-negative random variable with finite moments

Body:
Let $X$ be a random variable such that $X \geq 0$ and $\exists \varepsilon > 0: E[e^{X \varepsilon}] < \infty$. Is it true that $\exists$ a real number $B < \infty$ such that $E[X]E[e^{-X}] \leq B$ for all such $X$? If so, please provide a constructive proof that derives a real value for $B$.

The question is motivated by the observation that as $X$ increases point-wise, $e^{-X}$ decreases point-wise. If $X$ is non-random then $Xe^{-X} \leq \frac{1}{e}$. The question asks whether the fact that $Xe^{-X}$ is a bounded function over $X \geq 0$ may be generalized to $X$ being a non-negative random variable with new goal being to bound $E[X]E[e^{-X}]$.

I provide no detail regarding the specific distribution of $X$ because I am asking whether a hard real-valued bound exists for all random variables satisfying the given criteria. Otherwise, $B = E[X]$ suffices for a particular random variable.


答:
You are getting confused because the proof is missing a few prior details. Also, it abuses notation by making you think only a single binomial distribution is involved above. This is not the case. Here is a more accurate rendition of what the proof is showing:

Fix $\lambda > 0$. And suppose $X_n \sim Binomial(n, p_n)$ is a sequence of binomial variables such that:

each has a different success probability $p_n$, and

they satisfy the very special condition $np_n = \lambda$ for all large enough $n$.

Then, the proof shows $\lim\limits_{n \to \infty} P(X_n = k) = P(X = k)$ where $X \sim Poisson(\lambda)$.

So the proof isn't really deriving the Poisson distribution from the Binomial distribution. That is, the proof isn't showing Poisson is some special case of the Binomial.

Rather, it is saying that a very special collection of binomial variables, satisfying a very special condition $np_n = \lambda$, starts behaving more and more like the Poisson distribution for large $n$.

Title: How can $\lambda$ in Poisson distribution be constant if derived from binomial distribution for $\lambda=np$ while $n$ goes to infinity?

Body:
My Problem. Above is the derivation for Poisson from binomial which I quoted from somewhere. I understand this derivation mostly, but my struggle is that I do not understand why we can say $\lambda=np$, and $\lambda$ is constant in Poisson distribution.

My trial. My thought is that in a binomial distribution the probability of a success (i.e. for a certain event to happen) for each trial is constant, so $p$ is constant. Then deriving Poisson from binomial, we let $n$ approach infinity, and should not $\lambda$ go to infinity as a result as $\lambda=np$?

My comment. Any kind of idea or help would be appreciated. Thank you!


问:
Show that $[0, \infty)$ fails the Heine-Borel property.
I would like to thank @C Squared for helping me out with open covers. I have updated this page with a correct answer.

I want to show $[0, \infty)$ fails the Heine-Borel property. With some help, I was able to come up with a method of proof so I want to know if I got it right on this one. Any corrections would be appreciated, thank you!

Observe that $\bigcup_{k=1}^\infty (-k, k)$ covers $[0, \infty)$ by letting $x\in [0, \infty)$. It follows by definition of the set that,

\begin{equation*}
0 \leq x,
\end{equation*}
By the archimedean property, it follows that there exists $k\in\mathbb{N}$ such that,

\begin{equation*}
-k < 0 \leq x < k
\end{equation*}
\begin{equation*}
x \in\bigcup_{k=1}^\infty (-k, k)
\end{equation*}
so,
\begin{equation*}
[0, \infty) \subset\bigcup_{k=1}^\infty (-k, k).
\end{equation*}

Now observe that $\bigcup_{k=1}^n (-k, k)$ does not cover $[0, \infty)$ because $k+1\in[0,\infty)$ but $k+1\not\in\bigcup_{k=1}^\infty (-k, k)$-by the definition of the set.

\begin{equation*}
[0, \infty) \not\subset \bigcup_{k=1}^n (-k, k)
\end{equation*}

Therefore $[0, \infty)$ fails the Heine-Borel property and is not compact.

答: Suppose for contradiction, that $[0,\infty)$ was compact. Then by the Heine-Borel theorem, $[0,\infty)$ is closed and bounded. However, we know that $[0,\infty)$ is not bounded, therefore we have reached a contradiction. Our assumption that $[0,\infty)$ was compact must be false.


答:
When $\lambda$ is called a "constant" in this context, that means that among the probabilities $\dfrac{\lambda^k e^{-\lambda}}{k!},$ the number $\lambda$ remains the same as $k$ varies within the set $\{0,1,2,3,\ldots\}.$

But also, within this proof, $\lambda=np$ does not change as $n$ and $p$ change. As $n$ gets bigger, $p = \lambda/n$ gets smaller and $\lambda$ remains the same.


问:
How do I evaluate a definite integral involving trigonometric functions?
Evaluate $$\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds$$

I know that the antiderivative of $\dfrac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}x + C$ but I am a little unsure how I would change the integrand to get hat form. Could someone give me a hint or explanation on how to evaluate this integral?

答:
The easiest way is to do the integral yourself. Let $s = 16 \sin(t)$. We then have $ds = 16 \cos(t) dt$. Hence, we get
$$\int_0^{8 \sqrt{2}} \dfrac{ds}{\sqrt{256-s^2}} = \int_0^{\pi/4} \dfrac{16 \cos(t) dt}{\sqrt{256-256 \sin^2(t)}} = \int_0^{\pi/4} \dfrac{16 \cos(t) dt}{16 \cos(t)}= \int_0^{\pi/4}dt = \dfrac{\pi}4$$</p>


问:
proof using Brownian motion scaling property
I am trying to prove $A_t^+$ has the same distribution as $tA_1^+$ using the scaling property, where
$$
A_t^+ = \int_0^t\mathcal 1_{[0,\infty)}(W_s)\,ds
$$
I tried to scale the Brownian motion but it cannot lead to the right conclusion. I wonder which step is wrong. Here is what I attempted.
$$
\begin{aligned}
tA_1^+ & = t\int_0^1\mathcal 1_{[0,\infty)}(W_s)\,ds \\
& = \sqrt{t}\int_0^1\mathcal 1_{[0,\infty)}(\sqrt{t} W_{s})\,ds \\
& = \frac{1}{\sqrt{t}}\int_0^1\mathcal 1_{[0,\infty)}(W_{ts}) \,d(ts) \\
& = \frac{1}{\sqrt{t}}\int_0^t\mathcal 1_{[0,\infty)}(W_u) \,du \\
& = \frac{1}{\sqrt{t}}A_t^+
\end{aligned}
$$

答:
You see, you let $\sqrt{t}$ enter the parenthesis of the indicator, which does not make sense. And you don't need that to have the indicator $1_{ \ge 0}(\sqrt{t}W_s)$ because it already is:

$$
1_{ \ge 0}(W_s) = 1_{ \ge 0}(\sqrt{t}W_s)
$$


问:
Algebra Confusion
For the problem below how did they get the final expression?

It looks like they subtracted a $7$ from $64$, but in all honesty I am lost on how they ended up with the final expression.

$$
\begin{aligned}
7^{(k+1)+2}+8^{2(k+1)+1} &=7^{k+3}+8^{2 k+3} \\
&=7 \cdot 7^{k+2}+8^{2} \cdot 8^{2 k+1} \\
&=7 \cdot 7^{k+2}+64 \cdot 8^{2 k+1} \\
&=7\left(7^{k+2}+8^{2 k+1}\right)+57 \cdot 8^{2 k+1}
\end{aligned}
$$

Original image

Title: Algebra Confusion

Body:
It's because $64\cdot8^{2k+1}=7\cdot8^{2k+1}+57\cdot8^{2k+1}$:

$7\cdot7^{k+2}+\color{blue}{64\cdot8^{2k+1}}=7\cdot7^{k+2}+\color{blue}{7\cdot8^{2k+1}+57\cdot8^{2k+1}}=7(7^{k+2}+8^{2k+1})+57\cdot8^{2k+1}$


问:
For $X ∼ Pois(λ)$, find $E(2^X)$, if it is finite.
For $X \sim Pois(λ)$, find $\Bbb E(2^X)$, if it is finite.

Hint: Use these facts, $$\begin{align}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}&=e^\lambda\\\sum_{k=0}^{\infty}\frac{(k+1)\lambda^k}{k!}&=e^\lambda+\lambda e^\lambda\end{align}$$

The lecture ended before we had time to cover this section and I have no notes to work from. I am having trouble getting to the answer because I am getting confused with my work. Any help would be much appreciated.

I have this so far but I am not sure if it is correct,
$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$ for $x \in\{ 0, 1, 2, \ldots\}$

$$\begin{align}\mathbb E(2^X) &= \sum_{x=0}^{\infty} 2^x \frac{e^{-\lambda}\lambda^x}{x!}\\&= e^{-\lambda} \mathbb\sum_{x=0}^{\infty} \frac{(2\lambda)^x}{x!}\\&= e^{-\lambda} e^{2\lambda}\end{align}$$
So then $\Bbb E (2^X) = e^\lambda$

答:
\begin{align}
\operatorname E(2^X) = {} & \sum_{x=0}^\infty 2^x \Pr(X=x) = \sum_{x=0}^\infty 2^x \cdot\frac{\lambda^x e^{-\lambda}}{x!} \\[10pt]
= {} & e^{-\lambda} \sum_{x=0}^\infty \frac{(2\lambda)^x}{x!} \\
& \text{The factor $e^{-\lambda}$ can be pulled out of} \\
& \text{the sum because it does not change} \\
& \text{as $x$ goes from $0$ to $\infty$.} \\[10pt]
= {} & e^{-\lambda} \cdot e^{2\lambda} = e^{\lambda}.
\end{align}


问:
A circle of radius $r$ is inscribed into a triangle.
A circle of radius $r$ is inscribed into a triangle $ABC$. Tangent lines to this circle parallel to the sides of the triangle cut out three smaller triangles, $\triangle A_cB_cC$, $\triangle A_bBC_b$, $\triangle AB_aC_a$. The radii of the circles inscribed in these smaller triangles are equal to $1$, $2$ and $3$, respectively. Find $r$.

I have no idea how to start...

答:
Let $|CE|=H_c$ and $|CD|=h_c$ be the altitudes
of similar triangles $\triangle ABC$ and $\triangle A_cB_cC$.
Then
\begin{align}
\frac{|CD|}{r_c}
&=
\frac{|CE|}{r}
\tag{1}\label{1}
,\\
\frac{H_c-2r}{r_c}
&=
\frac{H_c}{r}
\tag{2}\label{2}
,\\
H_c &= \frac{2r^2}{r-r_c}
\tag{3}\label{3}
.
\end{align}

Similarly, two other altitudes of  $\triangle ABC$
in terms of $r,r_a,r_b$ are

\begin{align}
H_a &= \frac{2r^2}{r-r_a}
\tag{4}\label{4}
,\\
H_b &= \frac{2r^2}{r-r_b}
\tag{5}\label{5}
,
\end{align}

and we can apply a well-known relation

\begin{align}
\frac1r&=
\frac1{H_a}+\frac1{H_b}+\frac1{H_c}
\tag{6}\label{6}
\end{align}

to find out that $r$ in terms of $r_a,r_b,r_c$ is just
\begin{align}
r&=r_a+r_b+r_c
\tag{7}\label{7}
.
\end{align}

The original question would be solved by now,
but we can do more than that: we can completely solve the $\triangle ABC$.

Using known  Heron-like formula for the area, we have

\begin{align}
S&=
\frac1{\sqrt{
{(\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(-\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}-\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}+\tfrac1{H_b}-\tfrac1{H_c})}
}}
\\
&=\frac{r^{7/2}}{\sqrt{r_a r_b r_c}}
\tag{8}\label{8}
.
\end{align}

Next, we can find the semiperimeter $\rho$
and circumradius $R$ of $\triangle ABC$:

\begin{align}
\rho&=\frac Sr
=\frac{r^{5/2}}{\sqrt{r_a r_b r_c}}
\tag{9}\label{9}
,\\
R&=
\frac{2\,S^2}{H_a H_b H_c}
=\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c}
\tag{10}\label{10}
.
\end{align}

Now we are ready to find the three side lengths of $\triangle ABC$
as the roots of cubic equation in terms of $\rho,r,R$:

\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,\rho\,r\,R&=0
\tag{11}\label{11}
.
\end{align}

In particular, for $r_a=1,\ r_b=2,\ r_c=3$ we have

\begin{align}
r&=6
,\quad
S=216
,\quad
\rho=36
,\quad
R=15
\tag{12}\label{12}
,
\end{align}

\eqref{11} becomes

\begin{align}
x^3-72\,x^2+1692\,x-12960&=0
\tag{13}\label{13}
\end{align}

with three roots $\{18,\, 24,\, 30\}$,
that is, the sought triangle is
the famous $3-4-5$ right-angled triangle,
scaled by $6$.

Note that the side lengths
are inversely proportional to corresponding radii of incircles.

For another example, the picture illustrates
a solution for $r_a=7,\ r_b=5,\ r_c=3$.
In this case we have $r=15$ and the side lengths are

\begin{align}
a&=\tfrac{120\sqrt7}7
,\quad
b=\tfrac{150\sqrt7}7
,\quad
c=\tfrac{180\sqrt7}7
\tag{14}\label{14}
.
\end{align}

Edit

In fact, solution of the cubic equation \eqref{11}
is unnecessary: since the area and the altitudes are known,
the side lengths can be found explicitly as

\begin{align}
a&=r\,(r-r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{15}\label{15}
,\\
b&=r\,(r-r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{16}\label{16}
,\\
c&=r\,(r-r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{17}\label{17}
.
\end{align}


问:
Finding the MLE based on a set of derived random variables
I've encountered a sample problem in my statistics notes which I don't understand.

Suppose $X_1,...,X_{100}$ are r.v.'s which are Poisson-distributed with parameter $\frac{\mu}{100}$. We're looking for the MLE for $\mu$. However, we can only observe $Y_1,...,Y_{100}$ which are Bernoulli distributed with value $0$ if $X_i=0$ and $1$ otherwise (this is due to technical constraints). We find that the MLE for $p=e^{{-\frac{\mu}{100}}}$ is the sample proportion $\frac{1}{100} \sum_{i=1}^{100} Y_i$. Based on this, the notes say we can derive that

$-100\log(\frac{1}{100} \sum_{i=1}^{100} Y_i)$

is the MLE for $\mu$

Does anyone have any idea how we arrive at this result? It's been driving me crazy.

答:
In order to get the same result of your solution, we must amend a typo in the text

which are Bernoulli distributed with value $0$ if $X_i=0$ and 1 otherwise

It must be the converse:

They are Bernoulli distributed with value $1$ if $X_i=0$ and $0$ otherwise

Now it works because now your rv's become $B\Big(e^{-\frac{\mu}{100}}\Big)$

The MLE of $g(\mu)=e^{-\frac{\mu}{100}}$ is obviously

$$\widehat{g(\mu)}=\overline{Y}_{100}$$

For invariance property we can write

$$g(\hat{\mu})=\overline{Y}_{100}$$

$$e^{-\frac{\hat{\mu}}{100}}=\overline{Y}_{100}$$

now solve in $\hat{\mu}$ and get your result

If you use exatly the text you posted, the result is not very different...try, FYK


问:
Delta of options : A mathematical explanation
Consider the Black-Scholes equation $$\begin{equation}\label{eq3}
\frac{\partial{V}}{\partial{t}}+\frac{1}{2}\sigma^2S^2\frac{\partial^2{V}}{\partial{S}^2}+(r-D)S \frac{\partial{V}}{\partial{S}}-rV=0,~~~~S\in (0,\infty),~~~t\in(0,T)
\end{equation}$$
where $D$ is the dividend yield, $\sigma$ is the market volatility, $r$ is the interest rate.

Doubt : The delta $\triangle =\frac{\partial{V}}{\partial{S}}$ of an option mathematically equivalent to the rate of change of option price with respect to the change in asset price. How this is connected with the amount of shares need to be short selled to hedge the position changes?

答: The basic idea is to create a portfolio that (to first order) does not change in value when the share price changes.  If you have a portfolio of put options and shares it has a certain value. If the stock price goes up, the shares are more valuable and the put options are less valuable.  Let us say the put options decline in value by $0.50$ when the the share price increases by $1$, which is the delta you define.  If you own twice as many puts as shares, the first derivative of your portfolio value with respect to share price is $0$ so you don't care whether the stock goes up or down.


问:
Finding an order of $n$ such that a product is exponentially small in $n$
I am looking for a function $f$ s.t. if $t=O(f(n))$, where $O$ stands for Big-O notation, then:

$$\frac{t}{n}\times\frac{t-1}{n}\times\cdots \times \frac{2}{n}\sim e^{-an} \tag{1}$$

i.e. when the LHS of $(1)$ is exponentially small in $n$?

Say $t=n/2$, so $t=O(n)$, then the LHS of $(1)\leq \left(\frac{1}{2}\right)^n$ so it is exponentially small in $n$. But I was wondering if I can do better than that like what if $t=O(\sqrt{n})$ or something like that in which case I don't know if the LHS of $(1)$ would still be exponentially small?

答:
Starting from @Carl Schildkraut'answer, you want to solve for $f(n)$ the equation

$$\frac{[f(n)]!}{n^{f(n)-1}}= e^{-an}$$ To simplify notations, let $f(n)=y$ and $k=\frac{1}{n}e^{-a n}$ to make
$$y!=k\, n^y\tag 1$$

Looking at this question of mine, you will find a superb approximation proposed by @robjohn who is a very eminent user in this forum.

Adapted to $(1)$, it will give
$$y=e\,n\, \exp\Big[W\left(\frac{1}{2 e n}\log \left(\frac{k^2}{2 \pi  n}\right)\right) \Big]-\frac 12$$ where $W(.)$ is Lambert function.

Back to the original notations, this will give
$$f(n)=e\, n\,\exp\Big[W\left(-\frac{a}{e}-\frac{3 \log (n)+\log (2 \pi )}{2 e n}\right) \Big]-\frac 12$$


答:
Assume we are long a call option, $C$. If the price of the asset, $S$, underlying it declines, the value of $C$ decreases and the long position loses money. To counter the decline in price of the underlying we want to short $\Delta$ units of the underlying asset. What this looks like is if $\Pi$ is the value of the portfolio of long one call option and short $\Delta$ units we have

$$\Pi = C - \Delta S$$

We want $\Pi$ to be insensitive to small changes in the price of the underlying, $S$. That is we want

$$\frac{\partial \Pi}{\partial S} = \frac{\partial C}{\partial S} - \Delta\frac{\partial S}{\partial S} = 0.$$

Solving for $\Delta$ gives

$$\Delta = \frac{\partial C}{\partial S}$$

We can also see this from a Taylor series approximation. If we let $V(S, t)$ be the value at time $t$ of a European option on an underlying with spot price $S$, and no dividend. $V(S, t)$ is infinity many times differentiable in both $S$ and $t$. We now expand $V$ around $(S, t)$

$$V(S + dS, t + dt) = V(S, t) + dS\frac{\partial V}{\partial S} + dt\frac{\partial V}{\partial t} + \frac{(dS)^2}{2}\frac{\partial^2 V}{\partial S^2} + \frac{(dt)^2}{2}\frac{\partial^2 V}{\partial t^2} + dSdt\frac{\partial^2 V}{\partial S \partial t} $$

We approximate $(dS)^2 \approx \sigma^2S^2dt$ and say that $dV = V(S + ds, t + dt) - V(S, t)$ we have

$$dV \approx \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt $$

We can recognize the terms on the right as the options Delta, Theta, and Gamma. Now using this in a portfolio $\Pi$ we have

$$\Pi = V - \Delta S$$
$$d\Pi = dV - \Delta dS$$

Which is approximatly

$$d\Pi \approx \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt - \Delta dS $$

So to remove the sensitivity to price we have
$$\Delta = \frac{\partial V}{\partial S}$$

Leaving only Theta and Gamma sensitivity

$$d\Pi \approx \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt$$

Title: Delta of options : A mathematical explanation

Body:
Consider the Black-Scholes equation $$\begin{equation}\label{eq3}
\frac{\partial{V}}{\partial{t}}+\frac{1}{2}\sigma^2S^2\frac{\partial^2{V}}{\partial{S}^2}+(r-D)S \frac{\partial{V}}{\partial{S}}-rV=0,~~~~S\in (0,\infty),~~~t\in(0,T)
\end{equation}$$
where $D$ is the dividend yield, $\sigma$ is the market volatility, $r$ is the interest rate.

Doubt : The delta $\triangle =\frac{\partial{V}}{\partial{S}}$ of an option mathematically equivalent to the rate of change of option price with respect to the change in asset price. How this is connected with the amount of shares need to be short selled to hedge the position changes?


问:
How I can solve this limit only with algebra
I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.

$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$

first all I use the logarithms properties and rewrite like this:

$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$

and tried to make to this expression goes to  $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2

I can't reach to this because the limit goes to 0 instead of ∞

答:
Your first step is half right.  Leave the $\frac 1x$ as a factor for a bit.  We have
$$\log{\sqrt\frac{1 + x}{1 - x}}=\frac 12((\log 1+x)-\log(1-x))$$
Now expand the logs as a Taylor series, which works when $x$ is small.  As we are taking the limit as $x \to 0$, we are given (or can require) that $x$ be small.  We have
$$\log(1+x) \approx x-\frac {x^2}2+\frac {x^3}3\\
\log(1-x) \approx -x-\frac {x^2}2-\frac {x^3}3\\
\log{\sqrt\frac{1 + x}{1 - x}}=\frac 12((\log 1+x)-\log(1-x))\approx x+\frac {x^3}3$$
and plugging that in the limit is $1$.


答:
Hint to proceed with your idea: $$\frac{1}{\sqrt{256-s^{2}}}=\frac{1}{\sqrt{256\left( 1-\frac{s^{2}}{256}\right) }}=\frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}.$$

I haven't learned u substitution yet

Added. An antiderivative is

$$\int \frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}ds=\arcsin \frac{s}{16},$$

because

$$\frac{d}{ds}\arcsin \frac{s}{16}=\frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}=\frac{1}{\sqrt{256-s^{2}}}.$$

Title: How do I evaluate a definite integral involving trigonometric functions?

Body:
Evaluate $$\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds$$

I know that the antiderivative of $\dfrac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}x + C$ but I am a little unsure how I would change the integrand to get hat form. Could someone give me a hint or explanation on how to evaluate this integral?


答:
You are misunderstanding the question.  The Heine-Borel property doesn't refer to $[0,\infty)$ being compact.  The Heine-Borel property refers to considering $[0, \infty)$ as a metric space and seeing if the Heine-Borel property is true in the space.

And $[0,\infty)$ has the Heine-Borel property because the Heine-Borel theorem is true on $[0,\infty)$.  A subset $X \subset [0,\infty)$ is compact if and only if $X$ is closed and bounded.

But $(0,\infty)$ fails the Heine-Borel property because the Heine-Borel theorem is not true on $(0,\infty)$.  The subset $(0,1]$ is not compact ($\{(\frac 1n, 2)\}$ is an open cover with not finite subcover) but $(0,1]$ is closed and bounded.  That is, it is closed in $(0, \infty)$ (but not in $\mathbb R$ or even in $[0,\infty)$.

It is closed because it contains all its limit points because $0$ does not exist in the space $(0, \infty)$ and is not a limit point.

Title: Show that $[0, \infty)$ fails the Heine-Borel property.

Body:
I would like to thank @C Squared for helping me out with open covers. I have updated this page with a correct answer.

I want to show $[0, \infty)$ fails the Heine-Borel property. With some help, I was able to come up with a method of proof so I want to know if I got it right on this one. Any corrections would be appreciated, thank you!

Observe that $\bigcup_{k=1}^\infty (-k, k)$ covers $[0, \infty)$ by letting $x\in [0, \infty)$. It follows by definition of the set that,

\begin{equation*}
0 \leq x,
\end{equation*}
By the archimedean property, it follows that there exists $k\in\mathbb{N}$ such that,

\begin{equation*}
-k < 0 \leq x < k
\end{equation*}
\begin{equation*}
x \in\bigcup_{k=1}^\infty (-k, k)
\end{equation*}
so,
\begin{equation*}
[0, \infty) \subset\bigcup_{k=1}^\infty (-k, k).
\end{equation*}

Now observe that $\bigcup_{k=1}^n (-k, k)$ does not cover $[0, \infty)$ because $k+1\in[0,\infty)$ but $k+1\not\in\bigcup_{k=1}^\infty (-k, k)$-by the definition of the set.

\begin{equation*}
[0, \infty) \not\subset \bigcup_{k=1}^n (-k, k)
\end{equation*}

Therefore $[0, \infty)$ fails the Heine-Borel property and is not compact.


问:
Find ranges and kernels of a linear transformation
$ T \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix}  x-z \\ x-y \\ 0 \end{pmatrix}$

Basis for Range = { $ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},  \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}    $ }

Basis for Kernel = { $ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $ }

I am not sure if the bases for the subspaces: range and kernel, are correct?

答: Your kernel seems fine. For your range you should have two vectors since it is two dimensional. For the range your vectors are not linearly independent. Using the standard $e_1$ and $e_2$ vectors would suffice to be a basis of range, since they both lie outside of the kernel and are linearly independent. (Here $e_j$ is the vector with all zeros and 1 in the $j$th position)


问:
$\mathcal{I}\vDash \alpha$ if and only if $\mathcal{I}\vDash\forall x\alpha$ in FOL

For any interpretation $\mathcal{I}$ and wff $\alpha$, prove that $\mathcal{I}\vDash \alpha$ if and only if $\mathcal{I}\vDash\forall x\alpha$. Moreover if $\alpha^*$ denotes the closure of $\alpha(x_1,...,x_n)$ i.e. $\alpha^* = \forall x_1...\forall x_n$, we have $\mathcal{I}\vDash \alpha$ if and only if $\mathcal{I}\vDash \alpha^*$

Here's my work:

One direction seems simple, i.e. $\mathcal{I}\vDash \alpha$ implies $\mathcal{I}\vDash \forall x\alpha$. By definition, $\mathcal{I}\vDash \alpha$ means $\mathcal{I},v\vDash \alpha$ ($v$ satisfies $\alpha$ in $\mathcal{I}$) for all valuations $v$ in $\mathcal{I}$. For an arbitrary valuation $v_0$, $\mathcal{I},v_0\vDash\forall x\alpha$ iff for each valuation $v_0'$ that is $x$-equivalent to $v_0$, $\mathcal{I},v_0\vDash \alpha$. This is true since $\mathcal{I},v\vDash \alpha$ for all $v$. Hence, $\mathcal{I}\vDash\forall x\alpha$.

I started with $\mathcal{I}\vDash\forall x\alpha$, which by definition means that $\mathcal{I},v\vDash\forall x\alpha$ for all valuations $v$ in $\mathcal{I}$. For a particular $v$, we know that $\mathcal{I},v\vDash\forall x\alpha$ iff for all $v'$ $x$-equivalent to $v$, we have $\mathcal{I},v'\vDash \alpha$ (this holds for all $v$). How do I take it from here?

I think once "$\mathcal{I}\vDash \alpha$ if and only if $\mathcal{I}\vDash\forall x\alpha$" is proved, the statement concerning $\alpha$ and $\alpha^*$ is obvious! I'd appreciate any help, thanks!

For reference (some definitions):

Two valuations $v,v'$ are $x$-equivalent if they match on all variables but $x$, i.e. $v(y)=v'(y)$ for all $y\neq x$.

$\mathcal{I}\vDash \alpha$ means $\mathcal{I},v\vDash \alpha$ ($v$ satisfies $\alpha$ in $\mathcal{I}$) for all $v$ in $\mathcal{I}$.

$\mathcal{I},v\vDash\forall x\alpha$ iff $\mathcal{I},v'\vDash\alpha$ for all $v'$ $x$-equivalent to $v$.

答: Just figured it out! Since $v$ is $x$-equivalent to itself, putting $v' = v$ in (2) does the trick.


Title: How I can solve this limit only with algebra

Body:
I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.

$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$

first all I use the logarithms properties and rewrite like this:

$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$

and tried to make to this expression goes to  $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2

I can't reach to this because the limit goes to 0 instead of ∞

Body:
Let $a = \lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$. Then:

$$\exp(a) = \lim_{x\to 0} \exp \left(\log({\frac{1 + x}{1 - x}})^{\frac{1}{2x}} \right) = \lim_{x\to 0} \left(\frac{1 + x}{1 - x} \right)^{\frac{1}{2x}} = \lim_{x\to 0} \left(\frac{1/x + 1}{1/x - 1} \right)^{\frac{1}{2x}}$$

Now let $u = \frac{1}{x}$. Then we have:
$$\lim_{u \to \infty} \left(\frac{u + 1}{u - 1} \right)^{\frac{u}{2}} =\lim_{u \to \infty} \left(\frac{u-1}{u-1} + \frac{2}{u - 1} \right)^{\frac{u}{2}} = \lim_{u \to \infty} \left(1 + \frac{2}{u - 1} \right)^{\frac{u-1}{2}} \left(1 + \frac{2}{u - 1} \right)^{\frac{1}{2}}$$
$$= e \cdot 1 = e$$

Therefore $\exp(a) = e \Rightarrow \boxed{a = 1}.$


答:
Again it is important that you remember the definition of $\color{blue}{image}$ (what you call range) and the definition of $\color{blue}{kernel}$ of a linear transformation.

Below, I will give a detailed answer to the problem and leave the calculation details to you.

First, let's start by remembering the Image definition of a linear transformation and then the definition of kernel by a linear transformation.

Image of a linear transformation:

Let $V$ and $W$ vector spaces and let $T:V \to W$ a linear transformation, so  the image of a linear transformation is defined by $$\mathbf{im}(T)=\left\{w\in W, \exists v \in V: T(v)=w\right\}$$

Kernel of a linear transformation:

Let $V$ and $W$ vector spaces and let $T:V \to W$ a linear transformation, so  the kernel of a linear transformation is defined by $$\mathbf{ker}(T)=\left\{v\in V: T(v)=0_{W}\right\}$$

Now, in your problem you have a linear transformation defined by $T: \mathbb{R}^{3}\to \mathbb{R}^{3}$ such that $$T\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} x-z \\ x-y \\ 0 \end{pmatrix}$$

So, using the definition of $\color{blue}{image}$ we have, $$\mathbf{im}(T)=\left\{\begin{pmatrix} a \\ b \\ c \end{pmatrix}\in \mathbb{R}^{3}, \exists \begin{pmatrix} x \\ y \\ z \end{pmatrix}\in \mathbb{R}^{3}:T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} a \\ b \\ c \end{pmatrix} \right\}$$
Now, you need to solve
$$\mathbf{im}(T)=\left\{\begin{pmatrix} a \\ b \\ c \end{pmatrix}\in \mathbb{R}^{3}, \exists \begin{pmatrix} x \\ y \\ z \end{pmatrix}\in \mathbb{R}^{3}:\underbrace{T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} a \\ b \\ c \end{pmatrix}}_{\color{blue}{solve}} \right\}$$

Now, using the definition of $\color{blue}{kernel}$ so, you have
$$\mathbf{ker}(T)=\left\{\begin{pmatrix} x\\ y \\ z \end{pmatrix} \in \mathbb{R}^{3}: T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \right\}$$
Now, you need to solve,
$$\mathbf{ker}(T)=\left\{\begin{pmatrix} x\\ y \\ z \end{pmatrix} \in \mathbb{R}^{3}: \underbrace{T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}}_{\color{blue}{solve}}\right\}$$

Note that:

If you solve (that is, find the conditions that fall on the associated systems of linear equations) and always look for the consistency of the system of linear equations (in the case of the image) and the solutions (in the case of the kernel), then you have solved the requested problem.

Also, notice that in the problem you are not asked to find bases for the vector subspaces, they ask you to find the vector subspaces themselves.

However if you want to find a basis for the kernel and a basis for the image, you start with $\mathbf{im}(T)$ and $\mathbf{ker}(T)$ as vector subspaces with their respective conditions, then you can easily find a basis for the vector subspaces.

Title: Find ranges and kernels of a linear transformation

Body:
$ T \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix}  x-z \\ x-y \\ 0 \end{pmatrix}$

Basis for Range = { $ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},  \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}    $ }

Basis for Kernel = { $ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $ }

I am not sure if the bases for the subspaces: range and kernel, are correct?


答:
Hint: Use the substituition $s=16\cdot\sin x$.
\begin{align}
\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds
=
&
\int_0^{8\sqrt{2}} \dfrac{1}{16\sqrt{1-\left(\frac{s}{16}\right)^2}}ds
\\
=
&
\int_{16\sin0}^{16\sin(\frac{\pi}{4})} \dfrac{1}{16\sqrt{1-\left(\frac{s}{16}\right)^2}}ds
\\
=
&
\int_{0}^{\frac{\pi}{4}} \dfrac{\cos(x)}{\sqrt{1-\left(\sin(x)\right)^2}}dx
\end{align}


问:
$\exists c>0$ so that $f$ can be chosen to satisfy $||f|| \le c||g||$
Let $K$ be compact and $X$ be Banach subspace of $C(K)$. Let $E \subseteq _{closed} K$ so that $\forall g \in C(E), \exists f \in X$ with $f|_{E} =g$. To show: $\exists c>0$ such that $f$ can be chosen to satisfy $||f|| \le c||g||$.

I think we have to apply bounded inverse theorem somewhere but I'm not sure how to approach.

答: This is part of the proof of Open Mapping Theorem. Apply the theorem to the bounded linear operator $T: C(K) \to C(E)$ defined by $T(f)=f|_E$ (the restriction of $f$ to $E$).


问:
ODE without solutions
Is it possible to find functions $a,b,c$ defined on $\mathbb{R}$, such that
\begin{equation*}
ay'+by=c
\end{equation*}
has no global solution and $a(x)$ and $b(x)$ are never both zero ?

If $c=0$ there is always $y=0$ but I don't know about the general case. I think there is always a solution but I don't know how to prove it and it doesn't fall into the Cauchy-Lipschitz realm.

My question comes from the theory of pasting solutions: when $a(x)$ is zero the affine subset of solutions can be of dimension 0,1,2 according as to how solutions of homogenous equations paste around the singularity. I wonder if it can be empty.

The condition on $a$ and $b$ is to avoid trickery counter-examples like $x^3y'+3x^2y=2x$ whose solutions $y=\frac{1}{x}+\frac{\lambda}{x^3}$ are not global.

答:
If $a(x)\ne 0$, then this is not posible. The general solution is
$$
y= \int\frac{c(x)}{a(x)} \exp\left(\int\dfrac{b(x)}{a(x)}\, dx\right)\, dx\;\cdot\;
\exp\left(-\int\dfrac{b(x)}{a(x)}\, dx\right),
$$
or, if we denote as $G(x)$ some antiderivative
of $\frac{b(x)}{a(x)}$ and as $F(x)$ some antiderivative of $\frac{c(x)}{a(x)} \exp\left(G(x)\right)$,
$$\tag{1}
y=(F(x)+C)\cdot \exp(-G(x))=C \exp(-G(x))+F(x)\exp(-G(x)).
$$
Indeed,
$$
y'= C \exp(-G(x))(-G'(x))+F'(x)\exp(-G(x))+
F(x)\exp(-G(x))(-G'(x))
$$
$$
=-C \frac{b(x)}{a(x)}\exp(-G(x))+\frac{c(x)}{a(x)} \exp\left(G(x)\right)\exp(-G(x))-\frac{b(x)}{a(x)}
F(x)\exp(-G(x));
$$
substituting this into the original equation, one obtains
$$
-C b(x)\exp(-G(x))+c(x) -b(x)
F(x)\exp(-G(x))
+C b(x) \exp(-G(x))+b(x)F(x)\exp(-G(x))=c(x).
$$
Consider the initial value problem
$$
a(x)y'+b(x)y=c(x),\quad y(x_0)=y_0.
$$
For any the initial point $(x_0,y_0)$ from (1) we have
$$\tag{2}
y_0=C \exp(-G(x_0))+F(x_0)\exp(-G(x_0))
$$
For any $(x_0,y_0)$ the corresponding constant $C$ can be found:
$$
C=\exp(G(x_0))\left( y_0- F(x_0)\exp(-G(x_0)) \right),
$$
thus, (1) gives a solution for any initial point. Hence, it is a general solution.

Finally, note that (1) gives us a continuous function because the antiderivative of a continuous function is continuous. Hence, the solution is globally defined.

Title: ODE without solutions


问:
Show that $(0, 1)$ fails Heine-Borel.
So I think I understand the Heine-Borel property, but I want to make sure I've wrapped my head around it. With the help of @C Squared I was able to come up with the following, but I want to make sure it's okay. Thanks!

Let $I_k = (\frac{1}{k}, 2)$. Observe that $\bigcup_{k=1}^\infty I_k$ covers $(0, 1)$. Let $x\in (0, 1)$. It follows by the achimedean principle that there exists some natural number $k$ such that $\frac{1}{k} < x$. We may now conclude,
\begin{equation*}
0 < \frac{1}{k} < x < 1 < 2
\end{equation*}
\begin{equation*}
x \in (\frac{1}{k}, 2), k \in \mathbb{N}^+.
\end{equation*}
so,
\begin{equation*}
x\in\bigcup_{k=1}^\infty (\frac{1}{k}, 2).
\end{equation*}
therefore,
\begin{equation*}
(0, 1) \subset (1, 2) \cup (\frac{1}{2}, 2) \cup  (\frac{1}{3}, 2) \cup \dots = \bigcup_{k=1}^\infty I_k
\end{equation*}
Now, Heine-Borel insists that there exists a finite subcover for every cover of $(0, 1)$ that also covers $(0, 1)$. So to show that $(0,1)$ is not compact we must show that the finite subcovers of $\bigcup_{k=1}^\infty I_k$ do not cover $(0,1)$.

Suppose a finite subcover ($\bigcup_{k=1}^n I_k$) of $\bigcup_{k=1}^\infty I_k$. Then we have,
\begin{equation*}
(0, 1) \not\subset \bigcup_{k=1}^n (\frac{1}{k}, 2),
\end{equation*}
because $\frac{1}{n+1} \in (0, 1)$ but $\frac{1}{n+1} \not\in (1, 2) \cup \dots \cup (\frac{1}{n}, 2)$ (by the definition of this set).

There exists a finite subcover that does not cover $(0, 1)$, so the set fails Heine-Borel and is not compact.

答:
As a subset of $\mathbb R$, $(0,1)$ is not compact.  That is because $\mathbb R$ has the Heine-Borel property and $(0,1)$ is not closed and bounded.  And you successfully gave an example of an open cover with no finite subcover.

But that is not what the question is asking.  The question is asking to think of $(0,1)$ as a space in and of itself, does not have the Heine-Borel property (which means that in $(0,1)$ as a space that the Heine-Borel Theorem is FALSE).

And to do that we consider the the set $(\frac 0, \frac 12]$.  This set is CLOSED in $(0,1)$ because it contains all its limit points (note $0$ does not exist in $(0,1)$ so $0$ is not a limit point of $(\frac 0, \frac 12]$).  And the set is bounded.

But it is not compact. Let $I_n = (\frac 1{n+2}, 1 - \frac 1{n+2})$.  $\{I_n\}$ is an open cover of $(\frac 0, \frac 12]$.  (For any $x \in (0, \frac 12]$ there is a $k\ge 3$ so that $\frac 1k < x$ so $x \in I_{k-2}$).  But it has no finite subcover as for any finite $\{I_n\}$ there is a $m= \min n$ and so if $0< x < \frac 1{m+2}$ we have $x$ not being covered.

So $(0,1)$ doesn't have the Heine-Borel Property because in $(0,1)$ the Heine Borel theorem is FALSE.

Title: Show that $(0, 1)$ fails Heine-Borel.


问:
How can I use the first derivative test in an optimization problem?
This is a short and simple one, but I've solved the following optimization problem:

Find two numbers whose sum is $60$ and whose product is a maximum.

where $x$ and $y$ are variables.

Here's my work:

Sum = $60$, Product = $xy$

Solving for y:

$$60 - x = y$$
$$P = x(60 - x) = 60x - x^2$$
$$P' = 60 - 2x \Rightarrow  60 - 2x = 0 \Rightarrow x = 30$$
$$60 - x = y, y = 30$$

I'm not sure how exactly I can use the first derivative test to determine/prove the max/min. Max would be for this specific problem though.

How can I determine the maximum value by using the first derivative test when having only one critical point?

答:
Here is one relevant statement of the First Derivative Test.

In your case, you have found that $P'(30)=0$. You can also observe that $P'(x) > 0$ when $x\in(0,30)$, and that $P'(x) < 0$ when $x\in(30,60)$. The First Derivative Test then tells you that $x=30$ is a local maximum.


问:
Evaluating $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}}\, dx \,dy \,dz$ by converting to spherical coordinates

I would like to know how to evaluate the following triple integral with the help of spherical coordinates

$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}} \,dx \,dy\, dz$$

The relations between Cartesian coordinates and spherical ones are given by

${\displaystyle {\begin{aligned}x&=r\,\sin \theta \,\cos \varphi \\y&=r\,\sin \theta \,\sin \varphi \\z&=r\,\cos \theta \end{aligned}}}$

I know that a function is generally integrated over $\mathbb{R}^3$ by the following triple integral

$$ \ \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty }f(r,\theta ,\varphi )r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi .$$

I found a numerical solution with Wolfram Alpha (0.960592), I tried to change the bounds of integration from Cartesian to spherical but I got a different numerical value.

Could someone please give a detailed solution showing how to change the limits of integration?

Thanks.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \root{{x^{2} + y^{2} + z^{2}}}\dd x\,\dd y\,\dd z} =
\iiint_{\pars{0,1}^{\,\,3}}r\,\,\dd^{3}\vec{r}
\end{align}
Note that

$\ds{\nabla\cdot\pars{r\vec{r}} =
\pars{1 \times{\vec{r} \over r}}\cdot\vec{r} + r\nabla\cdot\vec{r} = 4r \implies r = {1 \over 4}
\nabla\cdot\pars{r\vec{r}}}$ such that
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \root{{x^{2} + y^{2} + z^{2}}}\dd x\,\dd y\,\dd z}
\\[5mm] = &\
{1 \over 4}
\iiint_{\pars{0,1}^{\,\,3}}\nabla\cdot\pars{r\vec{r}}\,\,\dd^{3}\vec{r} =
{1 \over 4}\iint_{\partial\pars{0,1}^{\,\,3}}\,\,
r\,\vec{r}\cdot\dd\vec{S}
\end{align}
where I used Gauss-Ostrogradsky Divergence Theorem.

The contribution to the integral from the faces which intersect the origen of coordinates vanishes out because $\ds{\vec{r} \perp \dd\vec{S}}$.

Each contribution from the remaining three faces are equal between them. So, I evaluate just one integral in one of the faces ( for instance, the one which is perpendicular to $\ds{\hat{z}}$ and multiply the integral result  by $\ds{3}$ ). Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \root{{x^{2} + y^{2} + z^{2}}}\dd x\,\dd y\,\dd z}
\\[5mm] = &\
3\pars{{1 \over 4}\int_{0}^{1}\int_{0}^{1}\root{x^{2} + y^{2} + 1^{2}}\dd x\,\dd y}
\\[5mm] = &\
{3 \over 4}\int_{0}^{\pi/2}\int_{0}^{\infty}
\root{\rho^{2} + 1}\ \times
\\[2mm] &\
\bracks{\rho\cos\pars{\phi} < 1}
\bracks{\rho\sin\pars{\phi} < 1}\rho\,\dd\rho\,\dd\phi
\\[5mm] = &\
{3 \over 8}\int_{0}^{\pi/2}\int_{0}^{\infty}
\root{\rho + 1}\ \times
\\[2mm] &\ \phantom{{3 \over 8}}
\bracks{\rho <
\min\braces{{1 \over \cos^{2}\pars{\phi}},
{1 \over \sin^{2}\pars{\phi}}}}
\dd\rho\,\dd\phi
\\[5mm] = &\
{3 \over 8}\int_{0}^{\pi/4}
\int_{0}^{1/\cos^{2}\pars{\phi}}
\root{\rho + 1}\dd\rho\,\dd\phi
\\[2mm] + &\
{3 \over 8}\int_{\pi/4}^{\pi/2}
\int_{0}^{1/\sin^{2}\pars{\phi}}
\root{\rho + 1}\dd\rho\,\dd\phi
\\[5mm] = &\
{1 \over 4}\int_{0}^{\pi/4}
\braces{\bracks{{1 \over \cos^{2}\pars{\phi}} + 1}^{3/2} - 1}\dd\phi
\\[2mm] + &\
{1 \over 4}\int_{\pi/4}^{\pi/2}
\braces{\bracks{{1 \over \sin^{2}\pars{\phi}} + 1}^{3/2} - 1}\dd\phi
\\[5mm] = &\
{1 \over 2}\
\underbrace{\int_{0}^{\pi/4}
\bracks{\sec^{2}\pars{\phi} + 1}^{3/2}\,\dd\phi}
_{\ds{{\root{3} \over 2} + {\pi \over 6} + 2\on{arccoth}\pars{\root{3}}}}\ -\
{\pi \over 8}
\\[5mm] = &\
\bbx{{\root{3} \over 4} - {\pi \over 24} + \on{arccoth}\pars{\root{3}}} \approx 0.9606 \\ &
\end{align}


问:
Need help with calculating this sum: $\sum_{n=0}^\infty\frac{1}{2^n}\tan\frac{1}{2^n}$
I need help with calculating this sum:
$$\sum_{n=0}^\infty\frac{1}{2^n}\tan\frac{1}{2^n}$$</p>

答: Hint: use the fact that $\cot{x}-2\cot{2 x}=\tan{x}$</p>


答:
Your problem is as follows:

\begin{equation}
\begin{aligned}
\max_{x,y} xy\\
\textrm{s.t.} \quad & x+y = 60\\
\end{aligned}
\end{equation}

By a simple substitution, you let $x = 60 - y$. Your problem now becomes an unconstrained optimization problem:

\begin{equation}
\begin{aligned}
\max_{y} \quad 60y-y^2\\
\end{aligned}
\end{equation}

When is this maximum? Take your first derivative test. You also have a sufficiency condition using your second derivative test.

Also, take for instance, the parabola function, you only have one critical point and it is indeed a maximum/or minimum.

Title: How can I use the first derivative test in an optimization problem?

Body:
This is a short and simple one, but I've solved the following optimization problem:

Find two numbers whose sum is $60$ and whose product is a maximum.

where $x$ and $y$ are variables.

Here's my work:

Sum = $60$, Product = $xy$

Solving for y:

$$60 - x = y$$
$$P = x(60 - x) = 60x - x^2$$
$$P' = 60 - 2x \Rightarrow  60 - 2x = 0 \Rightarrow x = 30$$
$$60 - x = y, y = 30$$

I'm not sure how exactly I can use the first derivative test to determine/prove the max/min. Max would be for this specific problem though.

How can I determine the maximum value by using the first derivative test when having only one critical point?


问:
A poset which has no upper bound for some chain and no maximal element
Zorn's lemma states that if every chain of a partially ordered set $A$ has an upper bound in $A$, then $A$ contains at least one maximal element.

The question is: does there exist a partially ordered set $B$ such that

1)Some number of chains of $B$ have no upper bound in $B$ (while some chains of $B$ do)

2)$B$ contains no maximal elements?

答: The set of natural numbers $\mathbb{N}$ ordered by the usual $<$ has this property: infinite chains have no upper bound, finite chains do, and there's clearly no maximal element.


答:
You have $f'(x) = 60 - 2x$. Using a sign diagram:

$$\begin{array}{c|c|c} x & 0 & 30 & 60 \\ \hline
f'(x)& + & 0 & - \end{array}$$

and since the derivative goes from positive to negative (sketch this out), you have a local maximum.

Alternatively, using the second derivative test $f''(x) = -2$, and when $f''(x)$ is negative, you have a local maximum (think of a sad face).

It's worth noting that you do not need calculus: find the axis of symmetry of the parabola, then substitute that value in. Quadratics with a negative $x^2$ term are always concave up.


问:
Calculate the number of possible passwords
I would like to know how many possible 8 length passwords can be generated with 72 distinct characters: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()

No, other limitations, the characters can definitely repeat though.

Thanks.

答:
So we have $8$ characters, and $72$ options for each character, so we have
$$ 72\cdot 72 \cdot 72 \cdot 72 \cdot \ldots = \boldsymbol{72^8}$$

passwords.


答:
The number of possibilities for each password character is $72$. There are $8$ such characters in the password. Hence total number of possibilities is given by
$$N = 72\times 72\times 72\times 72\times 72\times 72\times 72=72^8$$

Title: Calculate the number of possible passwords

Body:
I would like to know how many possible 8 length passwords can be generated with 72 distinct characters: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()

No, other limitations, the characters can definitely repeat though.

Thanks.


问:
$f \in D\left[x\right]$ is invertible iff $f \in D$
If $D$ is a domain, I must show that $f \in D\left[x\right]$ is invertible iff $f \in D$.

Is it important that $D$ is a domain? It would happen if $D$ is a ring?

答:
As I said in a comment, the precise statement should be that for a domain $D$ one has $D[X]^\times=D^\times$, where $R^\times$ denotes the multiplicative group (or set of invertible elements) of a ring $R$.

All you need for this is the relation $\deg(PQ)=\deg(P)+\deg(Q)$ for $P,Q\in D[X]$, where $\deg(P)\in\{-\infty\}\cup\Bbb N$ is the degree, taken to be $-\infty$ (only) for $P=0$, and addition produces $-\infty$ whenever one of its arguments is so. This relation is easily established under the hypothesis that $D$ is a domain, since in the computation of $PQ$ there can be no additive cancellation in degree $\deg(P)+\deg(Q)$, the only contribution being the product of the leading terms of $P$ and $Q$, and the product of their leading coefficients cannot be $0$ because it is computed in the domain$~D$.

So here is the argument for $D[X]^\times=D^\times$. Since the only addition of two elements of $\{-\infty\}\cup\Bbb N$ producing the result $0$ is $0+0=0$, one has $\deg(PQ)=0\iff\deg(P)=0\land\deg(Q)=0$. So $PQ=1$ implies that $P$ and $Q$ are (nonzero) constant polynomials whose constant terms are mutually inverse elements of$~D$, in other words $D[X]^\times\subseteq D^\times$. But conversely it is obvious that mutually inverse elements of$~D$ become mutually inverse elements of $D[X]$ when viewed as constant polynomials: $D^\times\subseteq D[X]^\times$.


问:
Determine convergence and limit
I want to determine whether each of the sequences converges, and if so, to what limit.

$$\begin{align} 1. \quad \quad & a_n=\frac{n^{2}}{n+1}-\frac{n^{2}+1}{n}\\
2. \quad \quad & a_n=\sqrt{n+1}-\sqrt{n}
\end{align}$$

For the first one, I tried to use ratio test, but it's not working. For the second, I tried to multiply it by $\sqrt{n+1}+\sqrt{n}$ and then devided it. But I'm stuck again and don't know which test should I use to determine its convergence.

答:
The first one is equivalent to

$$\frac{-n^2-n-1}{n(n+1)}$$

for the second one your way is right, when multiply by

$$\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$$

you should obtain $1$ for the numerator.

Title: Determine convergence and limit


Body:
$f(x) :=\log (\dfrac{1+x}{1-x})^{1/2}=$

$(1/2)\log (1+x)-(1/2)\log (1-x);$

$f'(0)=$

$\lim_{x \rightarrow 0}\dfrac{\log (\dfrac {x+1} {x-1})^{1/2}-\log 1}{x}=$

$(1/2)(1)-(1/2)(-1)=1.$

Note:

$f'(x)=$

$(1/2)\dfrac{1}{1+x}-(1/2)\dfrac{1}{1-x}(-1).$


问:
A curious corollary: there always exists some $c$ such that $2g'(c)h'(c)+g(c)h''(c)=0$
In a recent question, it was proved that if $f,g:[a,b]\to\mathbb R$ are two twice differentiable functions such that $f(a)=f(b)=g(a)=g(b)=0$ and both $g$ and $g''$ are non-vanishing on $(a,b)$, then $\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$ for some $c\in(a,b)$.

Now, given any $g$ that satisfies the above premises, if we take $f=gh$, where $h$ is some twice differentiable function, then $f$ is also twice differentiable and the condition that $f(a)=f(b)=0$ is automatically satisfied. Therefore, by the above result,
\begin{aligned}
\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}
&\ \Rightarrow\ h(c)=\frac{g''(c)h(c)+2g'(c)h'(c)+g(c)h''(c)}{g''(c)}\\
&\ \Rightarrow\ g''(c)h(c)=g''(c)h(c)+2g'(c)h'(c)+g(c)h''(c)\\
&\ \Rightarrow\ 2g'(c)h'(c)+g(c)h''(c)=0.\\
\end{aligned}
Hence we obtain the following corollary:

Let $g:[a,b]\to\mathbb R$ be a twice differentiable function such that $g(a)=g(b)=0$ and both $g$ and $g''$ are non-vanishing on $(a,b)$. For every twice differentiable function $h:[a,b]\to\mathbb R$, there exists a point $c\in(a,b)$ such that
$$ \bbox[5px,border:2px solid red]
{
2g'(c)h'(c)+g(c)h''(c)=0.
}
$$

I find this corollary more intriguing than the original problem, because $h$ is basically unrelated to $g$. Are there any direct or intuitive proofs that do not refer to the original problem? Can the assumptions that $g$ and $g''$ are non-vanishing be weakened or even removed from the corollary? (In particular, since the boxed statement does not involve $g''$, I wonder if the corollary is still true when $g$ is only differentiable or $C^1$.)

答:
That corollary can be proved directly by applying Rolle's theorem to $g^2h'$. That is not surprising because the statement about $\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$ is proved by applying Rolle's theorem to $\phi = f'g - fg'$. Substituting  $f=gh$ gives $\phi =g^2h'$.

The non-vanishing of $g''$  on $(a, b)$ in the other question guarantees that one can divide by $g''(c)$, so that condition is not needed here.

Since the corollary only involves $h'$ and the second derivative of $g$ is not needed, it would be more naturally stated as

Let $g, h:[a,b]\to\mathbb R$ be differentiable functions such that $g(a)=g(b)=0$ and $g$ is non-vanishing on $(a,b)$. Then there exists a point $c\in(a,b)$ such that
$$
2g'(c)h(c)+g(c)h'(c)=0.
$$

By applying Rolle's theorem to $g^\alpha h$ with $\alpha > 0$ it can be generalized as follows:

Let $g, h:[a,b]\to\mathbb R$ be differentiable functions such that $g(a)=g(b)=0$ and $g$ is non-vanishing on $(a,b)$. For every $\alpha > 0$ there exists a point $c\in(a,b)$ such that
$$
\alpha g'(c)h(c)+g(c)h'(c)=0.
$$


问:
irreducible elements, primitive elements, and UFD.
I am trying to understand irreducible elements, primitive elements, and UFD from Milne's algebraic geometry book. There is something which is confusing me.

In his book which is freely available. He mentions that irreducible elements of A[x] are irreducible elements of A. But that doesn't account for all irreducible elements right ? How does that follow from Gauss lemma? I also don't see why primitive elements are irreducible?

https://www.jmilne.org/math/CourseNotes/AG.pdf

答:
It seems like you have parsed the content of the corollary incorrectly. It says that an polynomial $f\in A[x]$ is irreducible over $A[x]$ if and only if one of these two things hold:

$f$ is a constant polynomial $f=a$, where $a$ is an irreducible element of $A$.

$f$ is a non-constant primitive polynomial that is irreducible over $F[x].$

($\leftarrow$) If $f=a$ where $a$ is irreducible, and $a=bc$ then one of $b$ and $c$ is a unit in $A,$ and hence a unit in $A[x].$ Let $f$ be a nonconstant primitive polynomial
and $f(x)=g(x)h(x).$ If $g$ and $h$ are both non-constant, then $f$ is reducible over $F[x]$. If $g$ is constant, then since $f$ is primitive, $g$ must be a unit in $A$, hence in $A[x].$

($\rightarrow$) If $f$ is a constant polynomial that is reducible in $A$, then this nontrivial factorization is a nontrivial factorization in $A[x].$ If $f$ is a nonconstant polynomial that is not primitive, then it has a factorization $f(x) = ag(x)$ where $a$ is not a unit in $A$, and this is a nontrivial factorization over $A[x]$. Lastly, if $f(x)$ is primitive but reducible in $F[x],$ then $f(x) = g(x)h(x)$ where $g$ and $h$ are in $F[x]$ and of smaller degree than $f,$ since all nonzero constants are units in $F[x].$ Then, we can clear denominators and factor out gcds from the right hand side and get $$df(x) = c g'(x)h'(x)$$ where $g'(x)$ and $h'(x)$ are primitive polynomials in $A[x]$ ($d$ is the product of the cleared denominators and $c$ is the product of the gcds of the coefficients factored out of the resulting polynomials in $A[x]$). The product $l(x)=g'(x)h'(x)$ is primitive by Gauss's lemma, so $df(x) = c l(x).$ And since $f$ and $l$ are both primitive, $c=ud$ where $u$ is a unit in $A,$ so, cancelling $d$ from both sides of the equations gives $f(x) = ul(x)=u g'(x)h'(x).$ And since the degree of $g'$ and $h'$ are less than the degree of $f,$ this is a nontrivial factorization in $A[x].$


问:
Divide polynomials and conclude:
I have to divide $x^3-a^3$ by $x-a$ and conclude that $x^3 - a^3 = (x-a)*(x^2+ax+a^2)$, but, I'm not sure about how to divide this polynomials, basically doing what I think I should do I will get just "$x^2+a^2$", from where in the division came the "$ax$"?</p>

答:
Division of polynomials works a lot like the division of integers (more formally, like the integers, we can divide using a Euclidean algorithm).

Hence, to divide $x^{3}-a^{3}$ by $x-a$, we find the largest multiplicative factor that we can multiply the remainder by, and subtract it as follows until we get a remainder of $0$, or a term having degree smaller than the original divisor. The sum of the multiplicative factors will be the quotient:

$$x^{3} - a^{3} = x^{2}(x-a) + r_{1}(x)$$
$$\implies r_{1}(x) = ax^{2} - a^{3}$$
$$ax^{2}-a^{3} = ax(x-a) + r_{2}(x)$$
$$\implies r_{2}(x) = a^{2}x - a^{3}$$
$$a^{2}x - a^{3} = a^{2}(x-a) + r_{3}(x)$$
$$\implies r_{3}(x) = 0$$

Since our remainder is now 0, we have completely factored the expression. Thus, we get $x^{3} - a^{3} = (x^{2} + ax + a^{2})(x-a)$.

Title: Divide polynomials and conclude:


问:
Regular expression for all strings with even length and not containing “ab” and “ba”
I have a problem with this question:

we have a language with alphabet {a, b, c}, all strings in this language have even length and does not contain any substring "ab" and "ba" for example these strings acceptable: "accb", "aa", "bb", "bcac", and these strings not acceptable "ccab", "abca", "cabc" and so on.. actually I think this is not regular language but I can't prove it, anyone can help me by give a regular expression for this language or prove that this language is not regular? or any help that I can figure out how I must think about it.

i tryed (aa)(cc)(bb)+(bb)(cc)(aa)+(ac+ca)+(bc+cb) and ((a+c)(a+c))+((b+c)(b+c))+(ac+bc)(ca+cb)+(ca+cb)(ac+bc) but this not work

答:
Your language is regular. All you need to know is that regular languages are closed under Boolean operations (union, intersection, complement, set difference, ...)

Let $A = \{a,b,c\}$ be the alphabet. The set $E$ of words of even length is $(AA)^*$ and hence it is regular. The set $F$ of words containing either $ab$ or $ba$ as a factor is $A^*(ab + ba)A^*$ and it is also regular. Your language is $E \setminus F$ and thus it is regular.


问:
How to solve $\lim_{x \to 0}\frac{\arctan x -\sqrt{x +1} -1}{x}$?
$$\lim\limits_{x \to 0}\dfrac{\arctan x -\sqrt{x +1} -1}{x}$$

Is this $\dfrac00$ form?

Can I use L'Hôpital's rule like this:

$$\lim\limits_{x \to 0}\dfrac{\arctan x -\sqrt{x +1} -1}{x}$$

$$=\lim\limits_{x \to 0}\dfrac{\dfrac{1}{x^2 +1} -\dfrac{1}{2\sqrt{x +1}}}{1}=\dfrac12$$

Or this limit does not exist!??

Only has $\mathop{\lim}\limits_{x \to 0^+}\dfrac{\arctan x -\sqrt{x +1} -1}{x}$ and $\mathop{\lim}\limits_{x \to 0^-}\dfrac{\arctan x -\sqrt{x +1} -1}{x}$!??

答:
HINT

For the correct expression, we can avoid l'Hospital by

$$\dfrac{\arctan x +\sqrt{x +1} -1}{x}=\dfrac{\arctan x }{x}+\dfrac{\sqrt{x +1} -1}{x}$$

and refer to standard limit for the first one (i.e. by $y=\tan x \to 0$) and the rationalization trick for the second one.


问:
Chebyshev's inequality question on theorem

Chebyshev's inequality: suppose $\varphi: R \rightarrow R$ has $\varphi \ge 0$. Let $A \in \mathcal{R}$ and $ i_A=\{ \textbf{inf} \varphi(y);y \in A \}$. Then $$i_AP(X \in A) \le E(\varphi(x):X \in A) \le E \varphi(X)$$.

I have hard time understanding what does $i_AP(X \in A)$ indicate (the infimum of a function that has probability of all elements in a Borel set?) and why does the inequality $$i_AP(X \in A) \le E(\varphi(x):X \in A) \le E \varphi(X)$$  make sense? Thanks in advance.

答:
The symbol $i_A$ is supposed to be a number, so $i_A P[X \in A]$ is the multiplication of the number $i_A$ with the number $P[X \in A]$.

You can prove it by the following inequality of random variables: Assuming $\phi$ is a nonnegative function, we have
$$ \left(\inf_{y \in A} \phi(y)\right) 1_{X \in A} \leq \phi(X) 1_{X \in A} \leq \phi(X)$$
where $1_{X \in A}$ is an indicator function that is 1 if $X \in A$ (and 0 else). Taking expectations gives
$$\underbrace{\left(\inf_{y \in A} \phi(y)\right)}_{i_A} P[X \in A]  \leq \underbrace{E[\phi(X) 1_{X \in A}]}_{E[\phi(X): X \in A]} \leq E[\phi(X)]$$


问:
A case in which the lower Riemann integral is additive
Let $A$ and $B$ be two open subsets of $[a,b]$.

I claim that

$$\underline {\int_a^b} \chi_A + \chi_B = \underline {\int_a^b} \chi_A + \underline {\int_a^b} \chi_B$$ where $\chi_A$ is the characteristic function of $A$.

Since the lower Riemann integral is superadditive, I should prove that $$\underline {\int_a^b} \chi_A + \chi_B \le \underline {\int_a^b} \chi_A + \underline {\int_a^b} \chi_B$$

Do I have to use the Heine-Borel theorem in some way ?

Many thanks in advance.

答:
Suppose $\underline{\int_{[a, b]}}(\chi_A+\chi_B)>\underline{\int_{[a, b]}}\chi_A+\underline{\int_{[a, b]}}\chi_B$, choose a partition $\mathcal{P}_1: a=t_0<t_1<\dots<t_n=b$ whose corresponding lower sum $L_1$(of $\chi_A+\chi_B$) lies between them.

Call an interval $[r, s]$ problematic if
$$\min_{[r, s]}(\chi_A+\chi_B)>\min_{[r, s]}\chi_A+\min_{[r, s]}\chi_B.$$

Observe that if $[r, s]$ is problematic, then every point of $[r, s]$ must be in one of $A, B$ and at the same time it contains points from $A\setminus B, B\setminus A$. In particular, the left side above should be $1$.

In the partition above, some $[t_i, t_{i+1}]$ should be problematic, else the lower sum of $\chi_A+\chi_B$ will be equal to the sum of lower sums of $\chi_A, \chi_B$. Let $P_1$ be the collection of problematic intervals in $\mathcal{P}_1$.

Subdivide elements of $P_1$ in half to arrive at a new partition $\mathcal{P}_2$ with lower sum $L_2$ of $\chi_A+\chi_B$. Since $L_1\leq L_2, \mathcal{P}_2$ must also contain some problematic intervals, and every problematic interval of $\mathcal{P}_2$ must be contained in a problematic interval of $\mathcal{P}_1$. Let $P_2$ be the collection of problematic intervals in $\mathcal{P}_2$.

Inductively define $\mathcal{P}_n, L_n, P_n, n\geq 1$. The collection $\cup_{n\geq 1}P_n$ is partially ordered by inclusion and contains chains of any finite length (simply take an element of $P_n$ and go backwards to obtain a chain of length $n$) and finitely many chains of a given length (because each $P_n$ is finite).

Suppose $C_1\supset C_2\supset\dots$ is an infinite chain in $\cup_{n\geq 1}P_n$. Note that each $C_i$ is a problematic interval and since we are halving the lengths, the lengths of these intervals decreases to $0$. Therefore, $\cap_{n\geq 1} C_n$ is a singleton $\{z\}$ (we need an infinite chain for this to work). Since each $C_i$ is problematic, $z\in A\cup B$ and every $C_i$ has a point from $A\setminus B$ and one from $B\setminus A$.

Thus, we have a sequence in $A\setminus B$ converging to $z$ and a sequence in $B\setminus A$ also converging to $z$. Since $z\in A\cup B$, we have a contradiction as $A, B$ are open.

Now, why does an infinite chain exist? See Given an infinite poset, show that it contains either a infinite chain or an infinite totally unordered set. or Prove that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.

Therefore the lower integrals must be equal.


问:
Partial derivatives and constraints on variables
I think I have a problem with a basic understanding of partial derivatives when changing variables.

I do understand that if we just change variables, e.g. $(x_1, x_2, x_3) \rightarrow (y_1,y_2,y_3)$, the partial derivatives change as $\frac{\partial}{\partial x_i} \rightarrow \sum\limits_{j=1}^{3}\frac{\partial y_j}{\partial x_i}\frac{\partial }{\partial y_j}$. However, I do not understand what happens if we reduce the number of variables by introducing constraints. For example, suppose that equation $g(x_1,x_2,x_3)=0$ is fulfilled. Now we have 2 degrees of freedom left, so how does one calculate the partial derivatives with respect to the new variable. For concreteness, let's assume that the new variables are $(x_1, x_2)$ and $x_3 = f(x_1, x_2)$, so what are $\frac{\partial}{\partial x_1}$ and $\frac{\partial}{\partial x_2}$?

答:
If you have a function $h(x,y,z)$, and you define $z=f(x,y)$, then the chain rule says

$$ \frac{\partial}{\partial x} h(x,y,f(x,y)) = \frac{\partial h}{\partial x} + \frac{\partial h}{\partial z} \frac{\partial f}{\partial x} $$

and similarly for $y$. In this formula, the terms $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial z}$ on the right-hand side should be understood to be evaluated at $z=f(x,y)$.


答: Alternative approach.  Since $x$ and $y$ sum to $60$, the problem can be re-interpreted as assuming that $x = 30 + k$, $y = 30 - k$, and you want to maximize $xy = (30 + k)(30 - k) = 900 - k^2.$


问:
Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)?
Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is
$$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$
I am wondering why this is non-negative?

答:
Using taylor series:
$$\frac {\ln(1-x)}{x}=-\frac {x+\frac {x^2}2+...}x=-
(1+\frac x2+...)$$
Which is decreasing, similarly:
$$\frac {\ln(x)}{1-x}=-\frac{\ln(1-(1-x))}{1-x}=-(1+\frac{1-x}2+...)$$
Which is increasing, then its reciprocal $\frac {1-x}{\ln(x)}$ is decreasing.

Since these two are negative, then they multiplied together is increasing.


问:
Finding $\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$$

Okay this seems weird to me-- couldn't i just write down the answer 0 because $\int_{0}^{0} f(x)dx=0$? (The answer is $-1$ btw)
There was even a hint for the question: $\displaystyle \lim_{x\to0}\frac{e^{-x^2}-1}{x^2}$ may be useful.

Am I supposed to go  $$\displaystyle \int_{\varepsilon}^{0}\frac{e^{x^2}}{x^2}dx+\int_{0}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{x^2}}{x^2}dx\,\,\,?$$ And I couldn't even go L'Hospital with that.....
Any help would be appreciated ! Thanks!

Of course, it would be nice if the answer was simply 0!

答:
The value of this limit is $-1$. $$\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac {e^{-x^2}} {x^{2}}dx =\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac {e^{-x^2}-1} {x^{2}}dx +\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac 1 {x^{2}} dx$$. You can easily evaluate the limit of  second term by evaluating the integral explicitly. For the first term observe that $\frac {e^{-x^2}-1} {x^{2}}$ is bounded function. For any bounded function $f$ we have $$\left|\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} f(x)dx\right| \leq M\left(\frac {\epsilon} {1+\epsilon}-\epsilon\right) \to 0$$ where $M$ is a bound for $|f|$.

For boundedness of $\frac {e^{-x^2}-1} {x^{2}}$ note that it is continuous on $(0,1]$ and it has  a finite limit, namely $0$, at $0$. (Use series expansion for the limit. You can also use L'Hopital's Rule if you prefer).


问:
Prove exists $c\in(0,1)$ such that $f''(c)=0$.
$y=f(x)$ has one intersection with the straight line that passes thorugh $(0,f(0))$ and $(1,f(1))$ at $(x_0,f(x_0))$, where $0<x_0<1$,
Where $f$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$，
Prove exists $c\in(0,1)$ such that $f''(c)=0$.

since it involved the line $y=x$, I thought of let $g(x)=f(x)-x$, but this gets me nowhere-- can it be proved by contradiction? suppose there does not exist $f''(c)=0$ then by Darboux $f'$ is strictly monotonic-- I don't know how to continue next! Any help would be appreciated! Thanks!

答:
As $(x_0,f(x_0))$ belongs to the line through $(0,f(0)),(1,f(1))$; we have that the slope of this line
$$= \frac{f(x_0)-f(0)}{x_0-0} = \frac{f(1)-f(x_0)}{1-x_0} = k \text{ (say)}$$
So, by Lagrange's MVT, $\exists\; c_1\in (0,x_0), c_2\in (x_0,1) \text{ s.t. } f'(c_1)=f'(c_2)=k$

$\therefore$ By Lagrange's MVT again, $\exists\; c\in (c_1,c_2)\subseteq (0,1) \text{ s.t. } f''(c) = 0$


问:
Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.

I found these pairs work by mod 3 and mode 4:

Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$

Case 2, $2^m \le 3^n$, $(0,0), (1,1), (1,3), (3, 2), (5, 4)$

How to prove there's no others?

I also found this post: Pairs forming perfect square. There's good discussions. Do we have any complete solution?

答:
This is a partial solution concerning $m\ne 1$.

First we consider the cases $m\ge 2$.

For case 1 ($2^m \ge 3^n$), write $2^m = 3^n + k^2$.

Taking modulo $3$, we have $(-1)^m \equiv k^2 \pmod 3$. This gives $m$ is even.

Write $m = 2s$, then $3^n = 2^m-k^2 = (2^s-k)(2^s+k)$, and both $2^s \pm k$ are powers of $3$.

Summing them up, however, gives $2^{s+1}$ which is not a multiple of $3$. Hence $2^s-k = 1$.

$2^s+k = 2^{s+1}-1$ is a power of $3$. [This gives $s=1, m=2, n=1$.]

For case 2 ($2^m \le 3^n$), write $3^n = 2^m + k^2$.

Taking modulo $4$, we have $(-1)^n\equiv k^2 \pmod 4$. This gives $n$ is even.

Write $n=2t$, then $2^m = 3^n-k^2 = (3^t-k)(3^t+k)$, and both $3^t \pm k$ are powers of $2$.

If we write $3^t-k = 2^x, 3^t+k =2^y$, we have $2^x+2^y = 2^x(1+2^{y-x})$.

Summing them up, however, gives $2 \times 3^t$. Since $3^t$ is odd, we must have $2^x=2$ and $1+2^{y-1} = 3^t$.

[This equation has $2$ solutions: $y=2, t=1$ and $y=4, t=2$.]

Both cases above end with solving an equation of the form $1+a^b= c^d$, which by Catalan's Conjecture/Mihăilescu's theorem only has the solution $a=2,b=3,c=3,d=2$, for integers $a,b,c,d >1$. However there are elementary proofs for this special case $a,c \in \{2,3\}$, which I choose to omit (unless requested)

For the case $m=0$, write $3^n = 1+k^2$. By modulo $3$, no solutions exist for $n \ge 1$. This forces $n=0$.

For the case $m=1$, write $3^n = 2+k^2$. By modulo $3$ we have $k^2 \equiv 1\pmod 3$, and by modulo $4$ we have $(-1)^n \equiv 2+k^2$, forcing $n$ to be odd and $k^2\equiv 1 \pmod 4$. $k$ must be an odd number not divisible by $3$, but as of right now I am not sure how to proceed.


问:
Interpretation of dP in Radon-Nikodym Theorem
[Radon-Nikodym Theorem]

Let $(\Omega, \Sigma, P)$  be a probability space. Suppose that $(\Omega, \Sigma, \mu)$ be a measure space with $\mu(A)=0$ implies $P(A)=0$, then there exist a function $f:X \rightarrow \mathbb{R}$ such that

\begin{equation}
P(A)=\int_A f d\mu.
\end{equation}

Further, function $f$ is called Radon-Nikodym derivative of $P$ with respect to $\mu$.
(So far I know, why this definition makes sense, because it's similar with derivative in calculus, that is, $g'$ is derivative of $g$ because $g(x)=\int g'(x) dx$).

In calculus, we also know differential, that is, if $y=f(x)$ be a differentiable function, then differential is defined by $dy=f'(x)dx$. Could we interpret $dP$ as a differential like in calculus? How could we interpret $dP$ in Radon-Nikodym context?

答:
The Radon-Nikodym theorem is very general. To tone down the generality lets consider $f$ a continuous function and $g$ a differentiable function. This will allow us to understand the problem better and make interpretations from there.

In this nice Riemann-Stieltjes context we arrive at the following calculation:

$F(x) = \int_a^x f\ dg = \int_a^x f\cdot g'\ dt$
(Page 255 - Proposition 2.10 - Haaser & Sullivan - Real Analysis)

We see that $F'(x)=f(x)\cdot g'(x)$. $F$ could be called the antiderivative of $f$ with respect to $g$.

In the case that $g=x$ then this new definition reduces to the old definition of antidifferentiation.

In some sense antidifferentiating $f$ with respect to $g$, means to find a function $h$ such that $f=h'\circ g$. Then $h\circ g$ will be the antiderivative of $f$ with respect to $g$ as in the following calculation.

$(h\circ g)'=(h'\circ g) \cdot g' = f\cdot g'$.

So, $P$ can be interpreted as the antiderivative of $f$ with respect to $\mu$, meaning that there is some function $h$ such that $P=h\circ \mu$, $f=h'\circ \mu$ and $P'=f\cdot \mu'$. Of course, the symbols I just wrote may not make much sense in this different context.


问:
1-1 and onto functions between $A=\{z \in \mathbb{C} | |z| >1\}$ and $B=\{z \in \mathbb{C} | z\neq 0\}$
I am solving question of a institute in which I don't study and I was unable to solve this question.

Let $A=\{z \in  \mathbb{C} \mid |z| >1\}$ and $B=\{z \in \mathbb{C} \mid z\neq 0\}$. Then which of the follwoing holds:

A There exists a continuous onto function $f: A \to B$

B There exists a continuous 1-1 function $f: B \to A$

C There exists a non- constant analytic function $f: A \to B$

D There exists a non- constant analytic function $f: B \to A$

I have done C, D but no progress can be made on A , B . I have studied complex analysis from Bak , Newman in case its helpful. I am unable to use any concrete reasoning for A , B.

Please tell me how to solve it.

答: Verify that $e^{z}$ is a continuous function from $A$ onto $B$ and $\frac {1+|z|} z$ is a continuous one-to-one function from $B$ into $A$.