问:
Calculating the limit of a geometrical construction
Suppose I have a rectangle ABCD, with AD larger than AB. I trace its diagonal AC, then the perpendicular to AC that goes through B. It intersects AC in $B_1$. I then trace the perpendicular to AD that goes through $B_1$, which intersects AD in $A_1$, and the perpendicular to CD that goes through $B_1$, which intersects CD in $C_1$.

We iterate the process to make $B_2$, $A_2$, $C_2$...

So on, so forth.

Clearly, $\lim_{n \to +\infty}$$C_n$ is D. But $\lim_{n \to +\infty}A_n$ is not D, and is instead somewhere on AD.

Question : Given the initial size AD and AB, with AD>AB, what's the length of $\lim_{x \to +\infty}A_n$D?

I think that if the initial lengths are $\{a,b\}$, the lengths on the next step are $$\left\{\frac{a^3}{a^2+b^2},\frac{b^3}{a^2+b^2}\right\},$$ but I have no idea where to go from there...

答:
The lengths of the next step you obtained are correct.

Now consider the ratio between the initial height and width, i.e. write $b = ka, k>1$.

The new lengths are given by:

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

$$\frac {b^3}{a^2+b^2} = \frac {k^3a^3}{a^2+b^2}$$

hence the new ratio of height and width is $k^3$. We can also write

$$\frac {b^3}{a^2 + b^2} = \frac {bk^2a^2}{a^2+k^2a^2} = \frac b{1+k^{-2}}$$

and so a recurrence relation becomes apparent. For the height, we are looking for:

$$\frac a{(1+k^2)(1+(k^3)^2)(1+((k^3)^3)^2)\dots} = \frac a{\prod_{n=0}^\infty (1+k^{2\cdot 3^n})}$$

Since $k > 1$, the infinite product diverges, so the height tends to zero as expected.

As for the width, the opposite occurs: we are looking for

$$\frac b{(1+k^{-2})(1+(k^3)^{-2})(1+((k^3)^3)^{-2})\dots} = \frac b{\prod_{n=0}^\infty (1+k^{-2\cdot 3^n})}$$

To show that the infinite product converges, we can check the equivalent series

$$\sum_{n=0}^\infty k^{-2\cdot 3^n} < \sum_{n=0}^\infty k^{-n} = \frac1{1-k^{-1}}$$

which converges as $0 < k^{-1} < 1$. It can also be expressed as the infinite sum:

$$\prod_{n=0}^\infty (1+k^{-2\cdot 3^n}) = 1 + k^{-2} + k^{-6} + k^{-8} + k^{-18} + \dots = \sum_{n \in S} k^{-n}$$

where $S$ is the set of all nonnegative integers that does not contain $1$ in their base $3$ representation (https://oeis.org/A005823).

A trivial upper bound is $\displaystyle \sum_{n = 0}^\infty k^{-2n} = \frac 1{1-k^{-2}} = \frac {k^2}{k^2-1}$, which shows that the limit of the width cannot be less than $\dfrac {b(k^2-1)}{k^2} = b\left(1-\dfrac1{k^2}\right)$. I am however unable to think of a closed form for the infinite product or the infinite sum.


问:
Help with continuity of a multivariable piecewise function

I need help finding if the limit of $g(x,y)$ at $(0,0)$ and at $(2,0)$ to see if it is continuous on these points.

I get confused, because normally these piecewise functions are defined such that $(x,y)$ different than some $(a,b)$ but here I have no idea how to do it because I can approach $(0,0)$ or $(2,0)$ from both pieces of the function...

And if I wanted to see the differentiability as well, what should I do?

I would really appreciate if you could explain this to me. Thank you !

答:
For $\vert y \vert \le 1$ and $y \neq 0$, you have

$$\vert g(x,y) \vert \le \vert x y \vert + \vert x \vert \vert y \vert^3 \le 2 \vert x \vert \vert y \vert.$$

As the right hand side of this inequality goes to zero as $y \to 0$ and $g(x,0)=0$ for all $x$, we get that $g$ is continuous everywhere.

ContentLicense: CC BY-SA 4.0


问:
Finding a limit using a definition
I want to prove, for $a,b$ positive numbers that $$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) = \cos a$$ by using the definition of a limit. In order to prove this, can I first prove by letting $a_n$ be the sequence which equals $1/n$, where $n = 1,2,3,\ldots$ then prove $a_n \to 0$ as $n$ approaches infinity, then apply this sequence $a_n$ to this equation like $$\lim_{n\to\infty} \cos \left( a + b\cdot a_n \right) = \cos \left( a + b \cdot \lim_{n\to\infty} a_n \right) = \cos(a)$$ as $n$ approaches infinity?

Title: Finding a limit using a definition

Body:
$$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) $$
$$\cos(a+\frac{b}{n} ) = \cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})$$
$$\lim_{n\to \infty} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{n\to \infty} (\dots) = \lim_{\frac{b}{n} \to 0} (\dots)$$
$$\lim_{\frac{b}{n} \to 0} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{x\to 0} \sin(x) = 0$$
$$\lim_{x\to 0} \cos(x) = 1$$
$$\lim_{\frac{b}{n} \to 0} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{\frac{b}{n} \to 0} (\cos(a)-0)$$
$$\cos(a)$$


问:
Problem 5.3 Character theory of finite groups (M. Isaacs)
Let G be a finite group. Let $H\subseteq G$ be a subgroup of G and suppose $\varphi$ is a class function of $H$ and $\psi$ is a class function of $G$. Show that $(\varphi\psi_H)^G=\varphi^G\psi$.

This problem will be used frequently throughout the book, so it would be very nice to figure it out, but I can't prove it.

I tried to prove in this way:

The character $(\varphi\psi_H)^G$ is afforded to $(V\otimes W_H)^G$ and the character $\varphi^G\psi$ to $(V^G\otimes W)$.
It remains to prove that $(V^G\otimes W)$ and $(V\otimes W_H)^G$ are isomorphic as G-modules, but I have no idea how to do it. In particular, it doesn't come in my mind which homomorphism I should use.

Thanks in advance.

答:
Recall that given a class function $f$ of $H$, the induced class function $f^G$ is defined as $g \mapsto \sum_{s\in S} f_0(s^{-1}gs)$, where $S$ is a set of representatives for the cosets in $G/H$, and $f_0(h)=f(h)$ if $h\in H$ and $0$ otherwise.

Now applying this to $\phi \psi_H$, we get $g \mapsto \sum_{s\in S} \varphi_0(s^{-1}gs)(\psi_H)_0(s^{-1}gs)$. Notice that
$\varphi_0(s^{-1}gs)(\psi_H)_0(s^{-1}gs)=\varphi_0(s^{-1}gs)\psi_H(s^{-1}gs)$ (just check the cases $s^{-1}gs\in H$ or $\not \in H$). In turn, $\psi_H(s^{-1}gs)=\psi(s^{-1}gs)$, and since $\psi$ is a class function for $G$, this equals $\psi(g)$. It follows that you can factor out $\psi(g)$ on the right, so you get $(\varphi \psi)^G=\varphi^G\psi$.


问:
Difference between $P(Y \ge y)$ and $P(Y > y)$.
In the book "Probability: Theory and Examples", there is this theorem:

Lemma 2.2.8 If $Y \ge 0$ and $p > 0$ then $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y > y) \,  \mathrm dy$.

The proof is by using Fubini's Theorem as follows

\begin{align*}
\int_0^\infty py^{p-1} \mathbb P (Y > y)  \mathrm dy
& = \int_0^\infty \int_\Omega py^{p-1} 1_{(Y > y)} \, \mathrm dP \, \mathrm dy \\
& = \int_\Omega \int_0^\infty py^{p-1} 1_{(Y > y)} \, \mathrm dy \, \mathrm dP \\
& = \int_\Omega \int_0^Y py^{p-1} \mathrm dy \mathrm dP  \\
& = \int_\Omega Y^p \mathrm dP = \mathbb E Y^p \\
\end{align*}

For me it seems totally fine to replace $Y > y$ with $Y \ge y$ in this proof, but the conclusion would be $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y \ge y) \,  \mathrm dy$. So if $Y$ can only take nonnegative integer values, we have $\mathbb E Y = \sum_{y \ge 0} P(Y \ge y)$, which is wrong.

So why we can not replace $Y > y$ with $Y \ge y$?

答: There is nothing wrong with: $\mathbb E Y = \sum_{y \ge 0} P(Y \ge y)$ as long as $Y$ takes non-negative integer values. See the Wikipedia section on Discrete distribution taking only non-negative integer values.</p>

ContentLicense: CC BY-SA 3.0


问:
Is there problems where the shell method cannot be applied?
The question is to find the volume of the solid formed by the revolution of the area $A$ around the $x$-axis. And the area $A$ is bounded by the curve $y=x^{\frac{1}{4}}$ and $y=x$ and $x=0$ and $x=1$.

I solved it by the washer method (I think I got an answer of $\frac{2\pi}{3}$).

But I tried to solve it using the shell method and I got lost here,

$V=\int_{0}^{1}y(x-x^{\frac{1}{4}})dy$

I know that I have to change the height of the cylinder to be in terms of $y$, but how do I do that? Which function do I use?

答:
Both methods can be used to integrate the volume.

$$\text{washer:}\>\>\>\>\>\int_0^1 \pi(x^{1/2}-x^2)dx = \frac\pi3$$

$$\text{shell:}\>\>\>\>\> \int_0^1 2\pi y(y-y^4)dy = \frac\pi3$$


答:
$g(x,y)$ is continuous at $(x_0,y_0)$ if $\forall \epsilon > 0$, $\exists \delta > 0$  $|| (x,y) - (x_0,y_0)|| < \delta \Rightarrow || g(x,y) - g(x_0,y_0)|| < \epsilon$.

Continuity at $(0,0)$:
Let $\epsilon > 0$.
Let $\delta = \min(\frac{\sqrt{\epsilon}}{2}, \frac{1}{2})$.
\begin{align*}
|| (x,y) - (0,0)|| < \delta &\Rightarrow |x| < \delta \land |y| < \delta \\
\Rightarrow || g(x,y) - g(0,0)|| &= |xy + xy^3 \sin(\frac{x}{y})| \\
&\leq |xy| +  |xy^3| < \delta^2 + \delta^4 \leq \epsilon \: (\text{check}!)
\end{align*}
Hence g is continuous at $(0,0)$. You could check for $(2,0)$.

$g(x,y)$ is differentiable at $(x_0,y_0)$ if $\exists (\alpha, \beta)$ s.t.
$\epsilon(h_1, h_2) := \frac{g(x_0 + h_1, x_0 + h_2) - g(x_0,y_0) - (\alpha, \beta) \odot (h_1, h_2)}{||(h_1,h_2)||} \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$. (where $\odot$ means dot product.)

You could check differentiability using this definition.

If you haven't seen these definitions before, then what you want to search for is 'continuity' and 'differentiability' of "multi-variable functions". Also there is another definition of continuity using sequences instead of $\epsilon-\delta$, you could again search for that (on google, or any calculus book on vector valued functions).

Edit (after seeing comment):

Intuitive understanding of continuity:

Continuity of $\mathbf{f: \mathbb{R} \rightarrow \mathbb{R}}$ at $\mathbf{x_0 \in \mathbb{R}}$.

Visualize $x_0$ on the real number line. The definition of continuity would mean "if you approach $x_0$ from any side, then it's corresponding value of $f(x)$ must approach $f(x_0)$. Note that since x is a real number, you can approach it from two sides - left and right leading to the definition of left hand limits and right hand limits etc.

Continuity of $\mathbf{f: \mathbb{R}^2 \rightarrow \mathbb{R}}$ at $\mathbf{(x_0,y_0) \in \mathbb{R}^2}$.

Visualize $(x_0, y_0)$ on the 2D plane (x-y plane). The definition of continuity would mean "if you approach $(x_0, y_0)$ from any side, then it's corresponding value of $f(x,y)$ must approach $f(x_0, y_0)$. Now however as $(x_0,y_0)$ is a point on the 2D plane, you can approach it from infinitely different directions. For example, you could approach this point via a line segment ending at $(x_0,y_0)$ with a slope of $k$ for any value $k$. (This actually leads to the definition of the $k$^th directional derivative). Alternatively you could approach it along any path which finally terminates at $(x_0,y_0)$ like a spiral etc.

If this is a bit hard to understand, let me know, I'll include some figures.

Title: Help with continuity of a multivariable piecewise function

Body:
I need help finding if the limit of $g(x,y)$ at $(0,0)$ and at $(2,0)$ to see if it is continuous on these points.

I get confused, because normally these piecewise functions are defined such that $(x,y)$ different than some $(a,b)$ but here I have no idea how to do it because I can approach $(0,0)$ or $(2,0)$ from both pieces of the function...

And if I wanted to see the differentiability as well, what should I do?

I would really appreciate if you could explain this to me. Thank you !


问:
The function $\arcsin(\sin(x))$
I have this question but I'm really not sure about my solution. Does this make sense?

Given the function $g(x) = \arcsin(\sin x))$, find its domain and where it is differentiable.

So I assume it's enough to say that $\sin$'s co-domain is $[-1,1]$ and so $\arcsin$'s domain, being its inverse function, is $[-1,1]$.

Now $\sin$ is differentiable on $\mathbb R$ and $\arcsin$ is differentiable on $(-1,1)$ (is it?) and so using the chain rule we can say that $g$ is differentiable at all $a \in \mathbb R$ for which $\sin(a) \neq 1$ or $\sin(a) \neq -1$. Which would be $\mathbb R $ \ {$-\pi/2, \pi/2$}.

Thanks!!

答:
I'm sorry if this sounds pedantic, but it seems that there are a few things about your post that are problematic:

You continually speak of the function $\arcsin(\sin(x))$, when in fact the function is $\arcsin \circ \sin$. Or you could speak of the function $g$ defined by $g(x)=\arcsin(\sin(x))$ for all $x$. Notice here how $x$ is a dummy variable that merely serves to illustrate what happens when you plug a number into the function. We could just as well speak of the function $g$ defined by $g(y)=\arcsin(\sin(y))$ for all $y$. It makes no difference. Saying that $\arcsin(\sin(x))$ is a function is an abuse of notation. If you already know and understand this, then feel free to move on to the next bullet point.

A function is only well-defined if you state what the domain is from the outset. So when you ask, 'what is the domain of $\arcsin(\sin(x))$', you should really be asking 'for what values of $x$ does $\arcsin(\sin(x))$ make sense?'

Now that we have got those technicalities out of the way, we can address the crux of your question. Consider how $\arcsin$ is a function that accepts inputs between $-1$ and $1$. The image of $\sin$ is $[-1,1]$, meaning that the outputs of the sine function always work as inputs to the arcsine function. And since the domain of $\sin$ is $\mathbb{R}$, the domain of $\arcsin \circ \sin$ is also $\mathbb{R}$.

As you have correctly pointed out, $\arcsin$ is differentiable on $(-1,1)$. At the points $(-1,-\pi/2)$ and $(1,\pi/2)$, the graph has a vertical tangent, and so the derivative doesn't exist. This means that $g'(x)$ only makes sense when $\sin x \in (-1,1)$. The $x$-values that we want to exclude are $-\pi/2$ and $\pi/2$, but, taking the periodicity of sine into account, we should exclude $\pi/2 \pm k\pi$ for any integer $k$. Hence, $\arcsin \circ \sin$ is differentiable on $\mathbb{R} \setminus \{x \mid x=\pi/2+k\pi, k \in \mathbb{Z}\}$.

Title: The function $\arcsin(\sin(x))$


问:
Is $(On,\in)$ + $(On, \in)$ a well-order?
the definition of a well-order we were given is that it is a left-narrow, well-founded linear order.
Therefore $(On,\in)$ is a well-order.
Furthermore we were said that the sum of two well-orders (which is basically glueing them together one after another) is also a well-order, but as I see it $(On,\in)$ + $(On, \in)$ is not left-narrow anymore.

Is there another definition or am I missing something else?
Thanks in advance

答:
In the $\mathsf{ZFC}$ context, unless otherwise specified all mathematical structures - topological spaces, linear orders, rings, groups, etc. - are sets. $(On,\in)$ is therefore not even a linear ordering since it's too big: it's a proper class. Meanwhile, you can prove that the sum of two well-orderings is again a well-ordering (a key point being that the union of two sets is always a set).

However, it sounds like your text is using a slightly odd approach here. "Left-narrow" appears to mean that every proper initial segment should be set-sized; under this definition, $(On,\in)$ is a well-ordering. This is not standard per the above, but if this is what your text is doing, then yes, $(On,\in)+(On,\in)$ is not a well-ordering anymore; indeed, $(On,\in)+{\bf 1}$ isn't a well-ordering! (Which is a good sign that we shouldn't include it.)

It's worth noting that we can talk - to a certain extent anyways - about "class-sized versions" of many kinds of mathematical structures. The issue (as long as we don't go to some class theory to simplify things) is that $\mathsf{ZFC}$ can't quantify over classes, so things like "Given any two class-well-orderings, one is isomorphic to an initial segment of the other" can't even be expressed in $\mathsf{ZFC}$, let alone proved.

Now the situation isn't entirely negative. We can talk about specific examples. Going back to the OP, the class $On\times\{0,1\}$ is definable by a formula $\psi$, and the relation $$(\alpha,i)\trianglelefteq (\beta,j)\quad\iff\quad i<j\mbox{ or }[i=j\mbox{ and }\alpha\le\beta]$$ on that class is definable by a different formula $\varphi$; intuitively, the pair $(\psi,\varphi)$ define "$On+On$." We can then prove, in $\mathsf{ZFC}$, that this class-sized "structure" has no infinite descending chains - that is, it's a class-well-ordering. With more work we can quantify over class-well-orderings whose defining formulas have bounded complexity in a precise sense.

But all of this is significantly messier than the "sets-only" situation, and most of the time is worth avoiding.


问:
Find the regression line of $X$ on $Y$

Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where  $0 ≤ x < ∞$, $0 ≤ y < ∞$

Now how to find the regression line.

My approach ,is to find marginal density functions $f_{X} (x)$ and $f_{Y} (y)$ then using that i will findout the standard deviation $\sigma_{x} \sigma_{y}$ and $\sigma_{xy}$

I got
$$f_Y (y)=\int_0^{\infty} xe^{-x(y+1)}dx$$ $$=-\frac x {y+1}e^{-x(y+1)}|_0^{\infty} +\frac 1 {y+1} \int_0^{\infty}e^{-x(y+1)}dx$$ $$=\frac 1{(y+1)^{2}}.$$
$f_X (x)=\int_0^{\infty} xe^{-x(y+1)}dy=e^{-x}$
Now the mean around origini is coming out to be $1$ but i am unable to find $\sigma^2_y$

By defination $\sigma^2_y=\int_{0}^{\infty}\frac{(y-1)^2}{(y+1)^2} dy$ which is divergent.

Please help me out ,i think i am making some calculation mistake.

答:
First, let's do a quick recap. As StubbornAtom commented, the regression line of $X$ on $Y$ is formally defined as the conditional expectation
$$\text{E}[X | Y = y] = \int x f_{X|Y}(x|y) dx$$
For $(X, Y)$ being bivariate normally distributed, one can further show that the regression line of $X$ on $Y$ is given by
$$\text{E}[X | Y = y] = \mu_x + \rho_{xy}(\sigma_x/\sigma_y)(y - \mu_y),$$
where $\rho_{xy} = \sigma_{xy}/(\sigma_x\sigma_y)$ is the correlation between $X$ and $Y$.

In this problem, we will directly calculate $\text{E}[X | Y = y]$ to obtain the regression line of $X$ on $Y$. As you have already (correctly) computed, the marginal density of $Y$ is
$$f_Y(y) = \int_0^{\infty} xe^{-x(y+1)}dx = \frac{1}{(y+1)^2},$$
such that the conditional density of $X$ given $Y$ is
$$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = x(y+1)^2 e^{−x(y+1)}.$$
The next step is to calculate
$$\text{E}[X | Y = y] = \int x f_{X|Y}(x|y) dx = \int_0^\infty x^2(y+1)^2 e^{−x(y+1)} dx.$$
This integral can be obtained by using Integration by parts twice. Here we borrow some useful facts from Exponential distribution to make the computation easier. Recall that if $W \sim \text{Exp}(\lambda)$ with $f_W(w) = \lambda e^{-\lambda w}$ for $w \geq 0$. Then we have $\text{E}[W] = 1/\lambda$ and $\text{Var}[W] = 1/\lambda^2$, which implies
$$\text{E}[W^2] = \int_0^\infty w^2 \lambda e^{-\lambda w} dw = (\text{E}[W])^2 + \text{Var}[W] = \left(\frac{1}{\lambda}\right)^2 + \frac{1}{\lambda^2} = \frac{2}{\lambda^2}.$$
Back to this problem, we then have
\begin{align*}
\text{E}[X | Y = y] &= \int_0^\infty x^2(y+1)^2 e^{−x(y+1)} dx\\
&= (y+1) \int_0^\infty x^2(y+1) e^{−x(y+1)} dx\\
&= (y+1) \cdot \frac{2}{(y+1)^2} \quad (\text{let } \lambda = y+1)\\
&= \frac{2}{y+1},
\end{align*}
which is the regression line of $X$ on $Y$.

Remark: When $(X, Y)$ is not bivariate normal (like in this problem), the regression line of $X$ on $Y$ depends on the joint distribution of $(X, Y)$, and may not necessarily be a straight line.

Title: Find the regression line of $X$ on $Y$


问:
Relation between Tensor Product of Modules and Multilinear Maps
In the case of a finite-dimensional vector space $V$ over some arbitrary field $\mathbb{F}$ it is well-known that there is the following canonical isomorphism:

$$\underbrace{V\otimes_{\mathbb{F}} \dots \otimes_{\mathbb{F}} V}_{r-\text{times}}\otimes_{\mathbb{F}} \underbrace{V^{\ast}\otimes_{\mathbb{F}}\dots\otimes_{\mathbb{F}} V^{\ast}}_{s-\text{times}}\overset{\text{can.}}{\cong} L^{r+s}(\underbrace{V^{\ast},\dots,V^{\ast}}_{r-\text{times}},\underbrace{V,\dots,V}_{s-\text{times}},\mathbb{F})$$

where $L^{r+s}(V^{\ast},\dots,V^{\ast},V,\dots,V,\mathbb{F})$ denotes the set of all multilinear maps from $(V^{\ast})^{r}\times V^{s}$ to $\mathbb{F}$.

Is the same also true for a module $M$ over some ring $R$? And if the answer is yes, does we need to assume that the ring is commutative, or does it hold in general?

(I ask because i would like to unterstand the equality between the set of all tensor fields and the set
$$\underbrace{\mathfrak{X}(\mathcal{M})\otimes_{C^{\infty}(\mathcal{M})} \dots \otimes_{C^{\infty}(\mathcal{M})} \mathfrak{X}(\mathcal{M})}_{r-\text{times}}\otimes \underbrace{\mathfrak{X}^{\ast}(\mathcal{M})\otimes_{C^{\infty}(\mathcal{M})}\dots\otimes_{C^{\infty}(\mathcal{M})} \mathfrak{X}^{\ast}(\mathcal{M})}_{s-\text{times}}$$
for some smooth manifold $\mathcal{M}$.)

答:
You need the ring to be commutative if you want to consider tensor products of modules, otherwise you can only define the tensor product of a right $R$-module by a left $R$-module and what results cannot generally be endowed with a structure of $R$-module.

So, let's assume $R$ is a commutative ring, so for any two modules $M,N$, $M\otimes_RN$ can be given a structure of $R$-module.

For a starter, you wish that $V\cong L(V^*,R)$, where $V^*=L(V,R)$. Unfortunately this holds quite rarely, even for finitely generated modules over a Noetherian ring. For instance, if $V$ is a finitely generated torsion group (a module over $\mathbb{Z}$), then $L(V,\mathbb{Z})=0$.

You have to restrict to finitely projective modules, in order that the thing works.


答:
As observed by $\textbf{Did}$, the lemma is equally valid when $P(Y > y)$ is replaced by $P(Y \ge y)$, since $P(Y > y)=P(Y \ge y)$ almost everywhere; this applies to every non-negative random variable $Y$.

In general, however, $P(Y \ge y)$ and $P(Y > y)$ are equivalent if and only if $P(Y = y)=0$, which is always true for continuous random variables, since a single point leaves the integral unaffected. For discrete random variables, this is not the case, since a point does affect the sum.

Finally, I am not sure where your formula for the expectation of a non-negative discrete random variable $Y$ comes from; as you noticed, it is easy to derive that

$$\operatorname{E}[Y] = \sum\limits_{y=1}^\infty P(Y \ge y).$$

Title: Difference between $P(Y \ge y)$ and $P(Y > y)$.

Body:
In the book "Probability: Theory and Examples", there is this theorem:

Lemma 2.2.8 If $Y \ge 0$ and $p > 0$ then $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y > y) \,  \mathrm dy$.

The proof is by using Fubini's Theorem as follows

\begin{align*}
\int_0^\infty py^{p-1} \mathbb P (Y > y)  \mathrm dy
& = \int_0^\infty \int_\Omega py^{p-1} 1_{(Y > y)} \, \mathrm dP \, \mathrm dy \\
& = \int_\Omega \int_0^\infty py^{p-1} 1_{(Y > y)} \, \mathrm dy \, \mathrm dP \\
& = \int_\Omega \int_0^Y py^{p-1} \mathrm dy \mathrm dP  \\
& = \int_\Omega Y^p \mathrm dP = \mathbb E Y^p \\
\end{align*}

For me it seems totally fine to replace $Y > y$ with $Y \ge y$ in this proof, but the conclusion would be $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y \ge y) \,  \mathrm dy$. So if $Y$ can only take nonnegative integer values, we have $\mathbb E Y = \sum_{y \ge 0} P(Y \ge y)$, which is wrong.

So why we can not replace $Y > y$ with $Y \ge y$?


问:
$T$ unbounded operator not closed $\implies$ resolvent of T is empty?
I am studying the subject of unbounded operators  and I'm wondering why if an operator is not closed than his resolvent is empty. Thanks !</p>

答:
Your definition of "resolvent" apparently says that $\lambda$ is in the resolvent if $\lambda I - T$ is injective and surjective, and has a bounded inverse.

We have the following lemma:

Let $A$ be an unbounded operator with domain $D(A)$ and image $R(A)$.  Suppose $A$ is injective, and consider $A^{-1}$ as an unbounded operator with domain $D(A^{-1}) = R(A)$.  Then $(A, D(A))$ is a closed operator iff $(A^{-1}, D(A^{-1}))$ is.

You can give a very quick proof by looking at the graphs of $A, A^{-1}$ as subsets of $H \oplus H$ and seeing how they relate to one another.

Now in our example, if you want $\lambda$ to be in the resolvent set of $T$ then we need to have $\lambda I - T$ injective and $R(\lambda I - T) = H$.  Moreover the inverse $(\lambda I - T)^{-1}$, whose domain is $H$, is a bounded and everywhere defined operator, so in particular it is a closed operator.  By the lemma, $\lambda I - T$ is closed as well, and this implies $T$ is closed.


问:
Are there counterexamples to: $\alpha^\gamma < \beta^\gamma \Rightarrow \alpha < \beta$?
We know that, for $\alpha$, $\beta$, $\gamma$ cardinals, if $\alpha \leq \beta$, then $\alpha^\gamma \leq \beta^\gamma$, and, in general, the usual operations defined on cardinals are compatible with the order relation.

I have been given the task to determine if the reversed implication holds, i.e. $\alpha^\gamma \leq \beta^\gamma \Rightarrow \alpha \leq \beta$.

I think the answer is no, given that $4^{\aleph_0}=2^{\aleph_0}=\mathfrak{c}$ and $4>2$. But what happens if we have a strict inequality, does this implication hold?

$$\alpha^\gamma < \beta^\gamma \Rightarrow \alpha < \beta.$$

I proved that the above implication is true for $\gamma=$ finite.

P.S. I am asking this question, because I realize an interesting thing: if we replace the exponentiation operator with multiplication or addition, the implication does hold, that is,
$$\alpha+\gamma < \beta+\gamma \Rightarrow \alpha < \beta.$$
$$\alpha\cdot\gamma < \beta\cdot\gamma \Rightarrow \alpha < \beta.$$

Also, there are counterexamples for division and subtraction, $\mathfrak{c}+\aleph_0=\mathfrak{c}\cdot\aleph_0=\mathfrak{c}+1=\mathfrak{c}\cdot1=\mathfrak{c}$, and $\aleph_0>1$.

答: Yes, it holds. Look at the contrapositive of your statement.</p>


问:
Use generating functions to determine how many four-element subsets of $S = \{1,2,3,4,\dots,15\}$ contain no consecutive integers?
Use generating functions to determine how many four-element subsets of $$S = \{1,2,3,4,\dots,15\}$$ contain no consecutive integers?

How do you approach this problem?

答:
so I was searching for this question and came up on an answer which is pretty neat and I found it to be a lot easier for me to understand.

Basically, you take any any 4 numbers within 1 and 15. and consider the differences between the numbers.

eg: say you took the set{2,4,6,11}

See that: 1 <= 2 < 4 < 6 < 11 < 15.

Now take the difference among the element to the previous one

like:  1(2-1),2(4-2),2(6-4),5(11-5),4(15-11)

Notice that their sums add up to 14(1 + 2 + 2 + 5 + 5). This condition is valid for all sets of numbers possible.

So now we can convert the question to a counting problem of the no. ways in which the 5 differences add up to 14:

i.e the solution for: $$ c1 + c2 + c3 + c4 + c5 = 14 $$

We get the conditions that c1 and c5  >= 0 since the first and 4th element could be 1 and 15 as well.

In order to ensure that consecutive numbers are not taken we can give the condition as c2,c3,c4 != 1
Also, c2,c3,c4 cannot be 0 as well since that would mean taking the same element twice. So c2,c3,c4 >= 2

so we can form our generating function as: $$ g(x) = (1 + x + x^2 +...)^2 (x^2 + x^3 + x^4 +...)^3$$
Now you can find the coefficient for $x^{14}$ from this generating function and you'll get the final answer as 495.

Check this image as well


问:
Making a logarithmic equation that starts at $(0,0)$ and passes through $(x, y)$?
I'm writing a computer program and for fading sound, it's best to do it in a logarithmic equation. What I need it to find a graph of the "volume" that starts at (0, 0) [x is the time, y is the volume] and goes to a certain volume after a certain time, increasing based on the logarithmic graph. What's a step by step way I can compute this? (So I can translate it into code)

Sorry, I just realized my question didn't make sense. By "ends at (x, y)" I mean passes through (x, y)

答: For clarity, I'm going to change notation and say it has to pass through $(x_0,y_0)$. Then we can start with $f(x) = A \, ln(x+1)$, which is logarithmic and passes through $(0,0)$. We need $y_0 = A \, ln (x_0+1)$, so we have $A = \frac{y_0}{ln(x_0+1)}$. So the answer is $f(x) = \frac{y_0}{ln(x_0+1)} \, ln(x+1)$.</p>


问:
How to find the inverse of this $3 \times 3$ block matrix?

Given $$E = \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}$$ define the following block matrix $$A = \begin{bmatrix}
I & 0 & 0 \\
0 & E & I \\
0 & I & -E
\end{bmatrix}$$ Determine $\left( 4 A^{-1} - A^3 \right)^n$ for all $n \in \mathbb{N}$.

I am stuck even at determining the inverse $A^{-1}$. What I normally do is use the Gauss-Jordan elimination or, for $2 \times 2$ matrices, I use the known identity. But I have absolutely no idea what to do with block matrices. I could write the entire block matrix explicitly and work with that, but that would be very painful to do. I am sure there is some elegant way to work with block matrices, which I hope someone would explain to me.

答:
$E^2 = I$

$A^{-1} = \begin{bmatrix} I\\&\frac 12 E & \frac 12 I\\&\frac 12 I &-\frac 12 E\end{bmatrix}$

However, it is not usually this easy.

As $(4A^{-1} - A^3) = A^{-1}(4I - A^4),$ I would be inclined to inspect $(4I - A^4)$

$4I - A^4$ gives a diagonal matrix with $3$ in the first $3$ entries and $0$ everywhere else.


问:
Calculation in geometrical interpretation of Lagrange's equation
I was reading the geometrical interpretation of Lagrange's equation and there it was written that if $f(x,y,z)=0$ be the solution of the Lagrange's equation $Pp+Qq=R$, then the direction cosines of the normal to the solution surface at any point are proportional to $p,q,-1$.
It comes like this:

$ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}$, i.e  $-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}},-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}},-1$, i.e $ \frac{\partial z}{\partial x},\frac{\partial z}{\partial y},-1$, i.e $ p,q,-1$

My question is how $-1$ comes in this calculation and how $-$ signs disappear in the $3$rd line from the first $2$ fractions?

There may be something silly that I am missing but I really can't figure it out.

答:
At first each fraction was divided by $-\dfrac{\partial f}{\partial z}$. So it becomes

$-\dfrac{\partial f/\partial x}{\partial f/\partial z},\quad-\dfrac{\partial f/\partial y}{\partial f/\partial z},\quad-1  \tag {1}$

Now here it was assumed that $z=\phi(x,y)$. So what we have is

\begin{align}
f(x,y,z)&=0 \implies \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial x}=0\\
\implies \dfrac{\partial z}{\partial x}&=-\dfrac{\partial f/\partial x}{\partial f/\partial z}
\end{align}

Similarly $\dfrac{\partial z}{\partial y}=-\dfrac{\partial f/\partial y}{\partial f/\partial z}$

Now plug these values in $(1)$ and we obtain

$\dfrac{\partial z}{\partial x},\dfrac{\partial z}{\partial y},-1\implies p,q,-1$


问:
Evaluate $\int_{-\infty}^{\infty}\frac{1-\cos(x)}{x^2}e^{ikx}dx$
Firstly $1-\cos(x)=2\big(\frac{1}{2}(1-\cos(x))\big)=2\sin^2(\frac{x}{2})$.

Also $$\sin^2(\frac{x}{2})=\Big(\frac{e^{\frac{1}{2}ix}-e^{-\frac{1}{2}ix}}{2i}\Big)^2=\frac{1}{4}(2-e^{ix}-e^{-ix})$$.

So the integral becomes: $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{(2-e^{ix}-e^{-ix})}{x^2}e^{ikx}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx$$.

We can now consider separate cases of $k$.

When $k=0$, we get $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{2-e^{ix}-e^{-ix}}{x^2}dx$$.

Now I am stuck. The problem now is that $\int_{-\infty}^{\infty}\frac{2}{x^2}dx$ is a divergent integral. And the same problem arises when $|k|=1$. Have I made an error somewhere?

Also the pole at $x=0$ is not simple - how could we use contour integration for a non-simple pole that lies on the real line (I've not seen this before)?

答:
You may use contour integration in the complex plane to evaluate integral.
You have written integral in the form
$$I(k)=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx$$
Now, you should consider 4 cases

a) $k>1$

b) $k<-1$

c) $k\in[0,1]$

d) $k\in[-1,0]$

We go to the complex plane and  use the contour (case a)

This includes $I(k)$, integral along a big half-circle of radius $R\to\infty$ counter clockwise and a small half-circle of radius $r\to 0$ around $x=0$ clockwise.

Integral around this contour $\oint{f}(x)dx=0$, because there are not poles inside the contour. On the other hand $\oint{f}(x)dx=I(k)+\int_R+\int_r=2\pi{i}Res{f}(x)\Rightarrow I(k)=-\int_R-\int_r$, if $Res{f}(x)=0$

Integral along a big half-circle $\to0$ at $R\to\infty$ due to Jordan lemma.
$I(k)=-\int_r =-\frac{1}{2}\int_r\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx\Rightarrow$ $${I(k)}=-\frac{1}{2}\lim_{r\to0}\int_{\pi}^0\frac{2 \exp(ikr e^{it}) -\exp(i(k+1)re^{it})-\exp(i(k-1)re^{it})}{r^2e^{2it}}ire^{it}dt$$

Expanding exponent into a series $\,\exp(ire^{it})=1+ire^{it}-\frac{1}{2}r^2e^{2it}-\frac{i}{3!}r^3e^{3it}+...$
$${I(k)}=\frac{1}{2}\lim_{r\to0}\int_0^{\pi}\frac{2 ikr e^{it} -i(k+1)re^{it}-i(k-1)re^{it}}{r^2e^{2it}}ire^{it}dt=\frac{\pi{i}}{2}\Bigl(2ik-i(k+1)-i(k-1)\Bigr)=0$$

Exactly in the same way $I(k)=0$ in the case b) (the only point - we have to close the contour in the lower half-plane and integrate around the point $x=0$ in the positive direction - counter clockwise)

In the case c) we go clockwise around $x=0$, but close the contour in the upper half-plane for first two terms and in the lower half-plane - for the third term of the integrand. In this case for the third term we will also get a residual at $x=0$ (the pole is inside the closed contour).

$${I(k)}=\frac{1}{2}\lim_{r\to0}\int_0^{\pi}\frac{2 ikr e^{it} -i(k+1)re^{it}-i(k-1)re^{it}}{r^2e^{2it}}ire^{it}dt-\frac{1}{2}2\pi{i} Res\Bigl(-\frac{e^{ix(k-1)}}{x^2}\Bigr)|_{x=0}=\frac{\pi{i}}{2}\Bigl(2ik-i(k+1)-i(k-1)\Bigr)+\pi{i}i(k-1)=\pi(1-k)$$

In the same way we evaluate the integral in the case d)

$I(k)=0$ for $k<-1$ and $k>1$

$I(k)=\pi(1-k)$ for $k\in[0,1]$

$I(k)=\pi(1+k)$ for $k\in[-1,0]$


问:
Is a class of $n< \infty$ infinite sets a finite class?
Question

Is a class of $n< \infty$ infinite sets a finite class?

My Thought

Suppose we have a class $S=\{N_1,N_2,...N_n\}$ where $N$ is a set with infinite members and $n<\infty$

Since $N_i \in S$, $i=\{1,2,...,n\}$ is itself a member of $S$, not its components. By definition, a class is finite if it has a finite $n$ members. Therefore, $S$ is a finite class.

What I was Wondering

Is my reasoning correct?

If not, what is the proper way to look at this?

答:
Yes, this is correct.

The correct way is to verify that the definition of "finite" holds for $S$. The slightly more "mathematical" definition of finite would be:

$S$ is finite if and only if there exists a bijection $f$ from $S$ onto a set of $\{0,\ldots,n-1\}$, where $n\in\Bbb N$.

We can assume that $N_i\neq N_j$, then the function $f(N_i)=i-1$ is a bijection, as wanted.


问:
Varying payment- with both increasing and decreasing annuity
I am trying to find the present value of an annuity-immediate such that the first payment
is $1000$, and each subsequent payment increases by $100$ until the
payments reach $2000$, but then decreases annually by $200$ until the final
payment of $200$. The effective annual rate of interest is $8$%.

Similar to another question I asked but with now a payment decreasing payment
(annuity- find the present value of varying payments)

so the payments would increases by $100$ until it reaches $2000$ but drops every $200$ until it reaches $200$ as the final:

$$1000, 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000$$
$$1800, 1600, 1400, 1200, 1000, 800, 600, 400, 200$$

I tried to understand it by using a table, similar to the response shown from my previous problem

$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Decreasing 2} & \text{Total} \\
\hline
1 & 1000 & 0 & 0 & 1000 \\
2 & 1000 & 100 & 0 & 1100 \\
3 & 1000 & 200 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
11 & 1000 & 1000 & 0 & 2000 \\
12 & 2000 & 0 & 200 & 1800 \\
13 & 2000 & 0 & 400 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 2000 & 0 &1600 & 200 \\
\end{array}$$

then I got

$$\require{enclose}
\begin{align}
PV
&= 1000(v + v^2 + \cdots + v^{11}) + 2000(v^{12} + v^{13} + \cdots + v^{20}) + (100v^2 + 200v^3 + \cdots + 1000v^{11}) \\ &\qquad + (200v^{12} + 400v^{13} + \cdots + 1600v^{20}) \\
&= 1000 a_{\enclose{actuarial}{11} i} + 2000 v^{11} a_{\enclose{actuarial}{9} i} + 100 v (Ia)_{\enclose{actuarial}{11}i} + 200 v^{11}(Da)_{\enclose{actuarial}{9}i}.
\end{align}$$

Did I do the level payment and equation correctly? When I went back to the previous problem I forgot to ask how the level payment was found for the second table (ie. where did 950 and 1450 come from?). How can I restructure this table to make the calculations less messy?

The first payment starts at 1000 then increases by 100, 10 times. Then it decreases by 200, 9 times so
$100(10) - 200(9) = -800 = 200 - 1000$

答:
so for the problem you asked, I noticed after $1950$ it jumped to $1800$. I think it was supposed to be 1900 etc. the first payment starts at 1000 and increases by 100 each time so the level of payment starting from day 1 to 9 is 1000-100 = 900. Then by the 10th payment, it starts to have a decreasing annuity of 50 so

$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing} & \text{Decreasing} & \text{Total} \\
\hline
1 & 900 & 100 & 0 & 1000 \\
2 & 900 & 200 & 0 & 1100 \\
3 & 900 & 300 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
9 & 900 & 1000 & 0 & 1900 \\
10 & 0 & 0 & 2000 & 2000 \\
11 & 0 & 0 & 1950 & 1950 \\
12 & 0 & 0 & 1900 & 1900 \\
13 & 0 & 0 & 1850 & 1850 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 0 & 0 & 1550 & 1550 \\
\end{array}$$

$$\require{enclose}
\begin{align}
PV
&
& 900 a_{\enclose{actuarial}{9} i} + 100 v (Ia)_{\enclose{actuarial}{9}i} + 50 v^{10}(Da)_{\enclose{actuarial}{10}i}.
\end{align}$$


问:
Given a point, a line and a circle; how can I find a point on the line having the following property?
First of all: I searched with different search engines and on StackExchange, but did not find a solution to my problem. This may be because its specific description is kind of long. For the same reason, I had to choose an unspecific title to this question due to the character limit.

Now, here is my problem:
Given

A point $D$

A line $l$

A circle $K$ (with center $C$ and radius $r$)

all in $\mathbb{R}^2$

I want to find a point $G$ such that:

$G$ is incident to $l$

Let the intersection of the line $GC$ with $K$ be $H$. It should hold that $G$ has the same distance to $D$ as to $H$, i.e. $\left\lVert\vec{GD}\right\rVert = \left\lVert\vec{GH}\right\rVert$

So far I tried to find some property that this point has that allows me to construct it, but haven't found one yet. I am also very inexperienced and have little to no knowledge in geometry, which might be why this is hard to me.

Edit (clarifications):

$D$ and $C$ may be assumed to lie on the same side of $l$ (I guess this can even be assumed w.l.o.g., as we could otherwise just mirror the point $D$).

$H$ is the intersection point that is further away from $G$ than $C$.

$D$ can be assumed to be exterior $K$.

$l$ can be assumed to intersect $K$ at two different points and to not pass through $C$.

I preferably look for geometric solutions or algebraic solutions in closed form.

答:
Reflect $D$ in the line $l$, denote the image $D^*$.

Draw the line $DD^*$.

Pick an arbitrary, but convenient, point $E$ on the line $l$ and draw the circle with center $E$ and radius $ED = ED^*$. It is very helpful if this circle $K(E, ED)$ intersects the the original one $K$ at two points.

Draw the radical axis of the two circles $K(E, ED)$ and $K$. If the two circle intersect, then the radical axis is the line that passes through both intersection points of the two circles.

Denote by $F$ the intersection point between the radical axis and the line $DD^*$

Construct the line through point $F$ and tangent to the circle $K$. There are two of them, so you have two solutions.

Denote by $T$ the point of tangency between the line from point 5 and the circle $K$.

The circumcircle of triangle $\Delta \, TDD^*$ is the circle you are looking for (there are two solutions).


问:
Expected value of a random variable in a piecewise function
I need to find the expected value of the following function

$$
F(x)=
\begin{cases}
 0 &\text{ if }\:x < 0 \\
 x^{3} &\text{ if }\: 0\leq x < 1\\
 1 &\text{ if }\: x\geq 1
\end{cases}
$$

I use the formula $E[x] = \int_{-\infty}^{\infty} x\cdot f(x)dx \rightarrow \int_{0}^{1}x\cdot x^3dx\, +\, \int_{1}^{\infty} x\cdot dx$

The second integral on the interval $[1, \infty]$ is infinity, which is definitely off.

But I am not sure about the right approach here.

Body:
$f(x)=F'(x)=3x^2$ for $0\le x\le 1$ and $f(x)=0$ otherwise.

$E(X)=\int_0^1 3x^3dx=\frac{3}{4}$.


问:
How to calculate different combinations of elements with restrictions
I have a set of $n$ different elements, and $m$, $m>n$, locations where I can put any element of that set.

The answer to the question “how many different combinations can I build with $m$ locations, each one taking anyone of the $n$ elements?” is just $n^m$.

However, imagine that now some elements $p_i$ of the original set must appear exactly $i$ times in the resulting combination of length $m$, while others don’t have any restriction on the number of times they appear.

How can I calculate the number of possible different combinations?

答:
I will assume that  the number of the special elements is $k$.

First choose $ki$ positions (from $m$) for the special elements, permute the elements arbitrarily over the positions and then fill the rest positions with all possible combinations of the rest elements ending with the expression
$$
\frac{m!}{(m-ki)!(i!)^k}(n-k)^{m-ki}.
$$


问:
Derivative Of Integral Vs Integral of Derivative.
let $f:\mathbb{R}×\mathbb{R} \rightarrow \mathbb{R}$

If $f_t(.,t) \in L^1(\mathbb{R})$ and $f(.,t) \in L^1(\mathbb{R})$, then, By Measure Theory:

$\frac{d}{dt}$$\int_{\mathbb{R}}$$f(x,t) dx$=$\int_{\mathbb{R}}$$f_t(x,t) dx$

But I can't understand why is this simply true when $f$ is a regular function? And in sense of what is this regularity?

Some help please?

答:
For any $h\gt 0$, $\frac{1}{h}(\int_Rf(x,t+h)dx-\int_Rf(x,t)dx)=\int_R\frac{f(x,t+h)-f(t,x)}{h}dx$  Take the limits as $h\to 0$ and get the desired equation.

$L_1$ is needed to insure limits exist.


问:
Ito representation of $W(T)$
I want to find Ito representation for variable $X = W(T)$, where $W$ is a Wiener process.

Let $W_{t}$ be a Wiener process on a standard filtered probability space $(\Omega ,{\mathcal {F}},{\mathcal {F}}_{t},P)$ and let ${\mathcal {G}}_{t}$ be the augmented filtration generated by $B$. If $X$ is a square integrable random variable measurable with respect to ${\mathcal {G}}_{\infty }$, then there exists a predictable process $\Theta$ which is adapted with respect to ${\displaystyle {\mathcal {G}}_{t}}$, such that

$$X = E[X] + \int_0^T\Theta_sdWs$$

I found somewhere on internet that solution to this problem is $W[T] = \int_0^TdWs$, however I'm not sure exactly what it means. Does it mean that this expression equals to $\Theta_s$ ? Could you please explain to me why this formula is proper for Ito representation ?

答:
This is simply the statement that when you compute the Ito integral $\int_0^T \Theta_s\,dW_s$ with $\Theta_s$ being the constant process $1$ (not depending on either $s$ or $\omega$), you get $\int_0^T 1\,dW_s = W_T$.  (This is an easy exercise that can be done directly from the definition of the Ito integral.)

Combining this with the fact that $E[W_T] = 0$ (from the definition of Brownian motion as a mean-zero Gaussian process), we see that
$$W_T = E[W_T] + \int_0^T 1\,dW_s$$
and so this is the Ito representation of $W_T$.  The notation $\int_0^T \,dW_s$ is just shorthand for $\int_0^T 1\,dW_s$.

Title: Ito representation of $W(T)$


问:
Integrate $\int \dfrac{\ln(x)}{(1-x^2)^{3/2}}\text{d}x$.

Integrate  $\int \dfrac{\ln(x)}{(1-x^2)^{3/2}}\text{d}x$

How can I solve that integral, only applying simple substitution, and by parts. It cannot be done any other way, I already did it by trigonometric and it comes out easy.

My try: Notice that,  $$\dfrac{\text{d}}{\text{d}x}\left ( \dfrac{x}{\sqrt{1-x^2}}\right )=\dfrac{1}{(1-x^2)^{3/2}}$$

Let $u=\ln x$, then $\text{d}u=\dfrac{1}{x}\text{d}x$. And, let $\text{d}v=\dfrac{1}{(1-x^2)^{3/2}}$. By (1) we have to  $v=\dfrac{x}{\sqrt{1-x^2}}$. Then,
\begin{align*}
    \int \dfrac{\ln(x)}{(1-x^2)^{3/2}}\text{d}x&=\dfrac{x\ln(x)}{\sqrt{1-x^2}}-\int \dfrac{x}{x\sqrt{1-x^2}}\text{d}x\\
    &=\dfrac{x\ln(x)}{\sqrt{1-x^2}}-\int \dfrac{1}{\sqrt{1-x^2}}\text{d}x\\
    &=\dfrac{x\ln(x)}{\sqrt{1-x^2}}-\arcsin(x)+C.
\end{align*}

答:
To begin with, if we make the substitution $x = \sin(u)$, we get:
\begin{align*}
\int\frac{\mathrm{d}x}{(1-x^{2})^{3/2}} = \int\frac{\mathrm{d}u}{\cos^{2}(u)} = \tan(u) = \frac{\sin(u)}{\cos(u)} = \frac{x}{\sqrt{1-x^{2}}}
\end{align*}

Hence we can apply integration by parts in order to obtain:
\begin{align*}
\int\frac{\ln(x)}{(1-x^{2})^{3/2}}\mathrm{d}x & = \frac{x\ln(x)}{\sqrt{1-x^{2}}} - \int\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}\\\\
& = \frac{x\ln(x)}{\sqrt{1-x^{2}}} - \arcsin(x) + c
\end{align*}
and we are done.

Hopefully this helps!


问:
Cauchy series "Product Like"
If I have:

$$\sum_{q=0}^{+\infty}\alpha_q \sum_{r=0}^q \beta_r\gamma_{q-r}$$

Is there a way to manipulate this expression in order to simplify it or at least rewrite into another form?

I've just noticed that it looks like a sort of Cauchy product

$$\sum_{q=0}^{+\infty}\beta_q \cdot\sum_{q=0}^{+\infty}\gamma_q=\sum_{q=0}^{+\infty}\sum_{r=0}^q\beta_r \gamma_{q-r}$$

but of course is missing the term $\alpha_q$ unfortunately. Do you have any idea?

答:
The common representations are
\begin{align*}
\sum_{q=0}^{\infty}\alpha_q\sum_{r=0}^q\beta_r\gamma_{q-r}
&=\sum_{\color{blue}{0\leq r\leq q<\infty}}\alpha_q\beta_r\gamma_{q-r}\\
&=\sum_{r=0}^\infty\beta_r\sum_{q=r}^\infty\alpha_q\gamma_{q-r}\\
&=\sum_{r=0}^\infty\beta_r\sum_{q=0}^\infty\alpha_{q+r}\gamma_{q}\\
\end{align*}

Title: Cauchy series "Product Like"


问:
Check a topology
Hello I am starting my topology studies, and I am having difficulties with this exercise

Let $\infty \in X$ be a given element in a set. Check that $\tau = \{ A \subset{X}|  \infty \not\in A  $ or $X-A$ is finite $\}$
defines a topology on $X$.

I have to check that:

$\varnothing$ and $X$ are open sets

The union of any number of open sets is an open set

The intersection of two open sets is an open set.

I don't know how to check it

Thanks

答:
We have a space $X$ with a distinguished point $\infty$.

Let $\tau$ be a topology (using the set of open sets definition) is defined as follows:

$$ \tau = \{ A \mathop| A \subset X \land (\infty \notin A \lor |X-A| \;\text{is finite}) \} $$

Let's check that condition 1 holds.

$\varnothing$ is an open set because $\infty$ is not in it.

$X$ is an open set because $|X-X|$ is empty and thus finite.

Let's check that condition 2 holds.

Let $F$ be a family of open sets. Suppose none of them contain $\infty$, then $\bigcup F$ also doesn't contain $\infty$ and is therefore open.

Suppose at least one element of $F$ contains $\infty$, let's pick one of them and call it $F_0$. Since $F_0$ is open and contains $\infty$ that means that $X-F_0$ is finite. $X-\bigcup F $ is necessarily a subset of $X-F_0$ and is therefore also finite. Therefore $\bigcup F$ is open.

Let's check that condition 3 holds.

Suppose we have two open sets $A$ and $B$.

If $A$ or $B$ or both fail to contain $\infty$, then $A \cap B$ also fails to contain $\infty$ and is therefore open.

If $|X-A|$ is finite and $|X-B|$ is finite then $|(X-A) \cup (X-B)|$ is finite and thus $|X - (A \cap B)|$ is finite. Therefore $A \cap B$ is open.

Title: Check a topology


问:
Does the empty model satisfy a contradiction?
Suppose we are working in a formulation of first-order logic that allows empty models. Then, does the empty model satisfy the formula $(x=x \land \neg x=x)$? And if so, why doesn't the empty model satisfy all formulas, since it satisfies a contradiction? Note, the formula I am referring to is an open formula, it doesn't have a quantifier. I would like some clarification of this point.

答:
The standard way to make sense of $\mathfrak{M}\models\varphi$ when $\varphi$ is a formula with free variables is to demand that $\mathfrak{M}$ satisfy every instantiation of $\varphi$; this is the same as $\mathfrak{M}$ satisfying the universal closure of $\varphi$. Under this definition, if we admit the empty structure we do indeed have $\emptyset\models(x=x)\wedge \neg(x=x)$.

However, it is no longer the case that a contradiction (in the sense of a formula gotten by substituting first-order formulas for propositional variables in an unsatisfiable propositional sentence) is explosive, as this example demonstrates. Variable-free contradictions are still explosive, of course. Basically, since we've changed our semantics we have to re-evaluate how our various notions relate to each other. And if we interpret "contradiction" as "unsatisfiable formula," then $(x=x)\wedge \neg(x=x)$ is no longer a contradiction ... as, again, this example demonstrates.

So there's no tension here at all, just a disconnect between two previously-equivalent properties arising because of a slightly broader semantics.

And if that's not how you're defining "$\mathfrak{M}\models\varphi$" when $\varphi$ has free variables, how are you defining it?


问:
Henselian ring which is not complete
Henselian ring is defined as a local ring which has the property that Hensel's lemma holds.

I understand that completeness is sufficient condition to be Henselian,
but is not necessary condition.

So, there should be example of a local ring which is Henselian, but not complete.

I would like to know such examples.

Thank you in advance.

答: The henselization of the $p$-adic integers is an example, and also $k[x^{1/2}, x^{1/4}, x^{1/8},\ldots]/(x)$.

Title: Henselian ring which is not complete


问:
Related to Taylor series of the natural log.
I'm trying to prove the Taylor series for the natural logarithm.

We know that $$\int_1^x \frac{dt}{t}=\ln x$$
Hence if I can know what is the Taylor Series for $1/t$ for $t>1$ I can solve the Taylor series for the natural log by integrating both sides, this is my Attempt:

Let $f(t)=1/t$ for $t>1$ and note that $f(1)=1$
$$\frac{1}{t}=f(1)+f^{(1)}(1)(t-1)+\frac{f^{(2)}(1)(t-1)^{2}}{2!}+\frac{f^{(3)}(1)(t-1)^{3}}{3!}+...$$
$$\frac{1}{t}=1-(t-1)+(t-1)^2-(t-1)^3+...$$
$$\int_{1}^x\frac{dt}{t}=t-\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3}-\frac{(t-1)^4}{4}+...\Bigr\rvert_{1}^x$$
$$\ln x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+..$$
I am sure that i didn't make any mistake, the problem is when i searched for the Taylor series of natural log on wikipedia i didn't found this formula, so Am i wrong?

Body:
The problem arose in your calculations
$$\int_{1}^{x} \frac{dt}{t}=\ln x$$

Let $f(t)=\frac{1}{t}$ for $t>1$
$$\frac{1}{t}=f(1)+f^{'}(1)(t-1)+\frac{f^{"}(1)(t-1)^{2}}{2!}+\frac{f^{'"}(1)(t-1)^{3}}{3!}+\dots \dots $$
$$\frac{1}{t}=1-1(t-1)+\frac{2(t-1)^2}{2}-\frac{6(t-1)^3}{6}+ \dots \dots$$

$$\int_{1}^{x} \frac{dt}{t}=1-1(t-1)+\frac{2(t-1)^2}{2}-\frac{6(t-1)^3}{6}+\frac{24(t-1)^4}{24}+\dots \dots \Bigr\rvert_{1}^{x}$$
$$\int_{1}^{x} \frac{dt}{t}=1-1(t-1)+(t-1)^2-(t-1)^3+(t-1)^4+\dots \dots \Bigr\rvert_{1}^{x}$$
$$In( x )= (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+\dots $$
$$In (1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6} \dots $$
you should also try this $\frac{1}{1-x}$ because you don't need Taylor series here, long division will do it
$$ \frac{1}{1-x} = 1+x+x^2+x^3+x^4+\dots \dots$$
You can integrate to see that,
$$-In(1-x) = x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots \dots $$

$$In (1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6} \dots $$
Combining this two gives the super $\log()$ series, because it has no restrictions and converges faster


问:
Prove that if $\vec v$ and $\vec w$ are orthogonal then $\|\vec v+\vec w\|=\|\vec v\|+\|\vec w\|$
So if $\vec V$ is a real vector space equipped with an inner product $\langle\cdot,\cdot\rangle$, then using the polar identity, it is not hard to show that the equality
$$
\|\vec v+\vec w\|=\|\vec v\|+\|\vec w\|
$$
holds if and only if
$$
\langle\vec v,\vec w\rangle=\|\vec v\|\cdot\|\vec w\|
$$
and by the Cauchy-Schwarz theorem the last equality holds if and only the vectors $\vec v$ and $\vec w$
are linearly dependent. So if $\vec v$ and $\vec w$ are orthogonal we conclude that the first equality does not holds, that is
$$
\|\vec v+\vec w\|<\|\vec v\|+\|\vec w\|
$$
by the triangular inequality. However this it seems to me strange because the Pythagoras' theorem must hold  for any $n$ orthogonal vectors: indeed if the vectors $\hat e_i$ for $i=1,...,n$ are the vectors of the canonical basis then
$$
\|\hat e_1+...+\hat e_n\|=\sqrt n=\underbrace{1+...+1}_{\text{n times}}=\|\hat e_1\|+...+\|\hat e_n\|
$$
and clearly this holds for any orthonormal basis. So why my first argument is incorrect? and how to prove the statement? Could someone help me, please?

答: It turns out that$$\sqrt n<\overbrace{1+1+\cdots+1}^{n\text{ times}}=n$$and that therefore$$\left\|\widehat{e_1}+\widehat{e_2}+\cdots+\widehat{e_n}\right\|<\left\|\widehat{e_1}\right\|+\left\|\widehat{e_2}\right\|+\cdots+\left\|\widehat{e_n}\right\|.$$


答:
There's something wrong with your equivalent statements. If it were true: say $v=1$ and $w=-1$ which are linearly dependent.

Equivalently (?), $-1=\langle v,w\rangle =||v||\cdot ||w||=1$?

Equivalently (?), $0=||v+w||=||v||+||w||=2$?

Title: Prove that if $\vec v$ and $\vec w$ are orthogonal then $\|\vec v+\vec w\|=\|\vec v\|+\|\vec w\|$

Body:
So if $\vec V$ is a real vector space equipped with an inner product $\langle\cdot,\cdot\rangle$, then using the polar identity, it is not hard to show that the equality
$$
\|\vec v+\vec w\|=\|\vec v\|+\|\vec w\|
$$
holds if and only if
$$
\langle\vec v,\vec w\rangle=\|\vec v\|\cdot\|\vec w\|
$$
and by the Cauchy-Schwarz theorem the last equality holds if and only the vectors $\vec v$ and $\vec w$
are linearly dependent. So if $\vec v$ and $\vec w$ are orthogonal we conclude that the first equality does not holds, that is
$$
\|\vec v+\vec w\|<\|\vec v\|+\|\vec w\|
$$
by the triangular inequality. However this it seems to me strange because the Pythagoras' theorem must hold  for any $n$ orthogonal vectors: indeed if the vectors $\hat e_i$ for $i=1,...,n$ are the vectors of the canonical basis then
$$
\|\hat e_1+...+\hat e_n\|=\sqrt n=\underbrace{1+...+1}_{\text{n times}}=\|\hat e_1\|+...+\|\hat e_n\|
$$
and clearly this holds for any orthonormal basis. So why my first argument is incorrect? and how to prove the statement? Could someone help me, please?


问:
Signed curvatures
Let $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ be two plane curves.

Show that, if $\textbf{$\tilde{\gamma}$}$ is obtained from $\gamma$ by applying an isometry $M$ of $\mathbb{R}^2$, the signed curvatures $κ_s$ and $\tilde{κ}_s$ of $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ are equal if $M$ is direct but that $\tilde{κ}_s = −κ_s$ if $M$ is opposite (in particular, $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ have the same curvature).

Show, conversely, that if $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ have the same nowhere-vanishing curvature, then $\textbf{$\tilde{\gamma}$}$ can be obtained from $\textbf{$\gamma$}$ by applying an isometry of $\mathbb{R}^2$.

$$$$

Could you give me some hints how we could show that? I don't really have an idea...

答:
$\large \textbf{First Part}$

The orthogonal $2\times 2$ matrices are of the two forms

$$
\begin{pmatrix}
\cos \theta & - \sin \theta\\
\sin \theta & \cos \theta
\end{pmatrix}, \qquad
\begin{pmatrix}
\cos \theta & \sin \theta\\
\sin \theta & -\cos \theta
\end{pmatrix}
$$

The first kind corresponds to $\textit{direct isometries: }$ a rotation (which is the composition of two reflections). The second kind corresponds to
$\textit{opposite isometries: }$ a reflection.

If $M$ is a reflection on a line $\ell$ then $M(v) = Pv + a$ where

$$P = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta\end{pmatrix}$$

for some angle $\theta$.

Now, $\tilde{\gamma} = P \gamma  + a$ and we can write $\dot{\gamma} = (\cos \varphi, \sin \varphi)$ and $\dot{\tilde{\gamma}} = (\cos \tilde{\varphi}, \sin \tilde{\varphi})$; furthermore we
$$
\dot{\tilde{\gamma}} = P \dot{\gamma} =
\begin{pmatrix}
\cos \theta & \sin \theta\\
\sin \theta & -\cos \theta
\end{pmatrix}
\begin{pmatrix}
\cos \varphi\\
\sin \varphi
\end{pmatrix} = (\cos (\theta - \varphi), \sin (\theta - \varphi)).
$$
Then we have the equality
$$(\cos \tilde{\varphi}, \sin \tilde{\varphi}) = (\cos (\theta - \phi), \sin (\theta - \phi))$$

which implies that
$$
\tilde{\varphi} = \theta - \varphi + 2\pi n, \quad \text{for some integer } n,
$$

and then

$$
\begin{align*}
\frac{d\tilde{\varphi}}{ds} &= - \frac{d\varphi}{ds}\\
\tilde{k}_{s} &= -k_{s}.
\end{align*}
$$

If $M$ is a rotation followed by a translation, then $M = P v + a$, where
$$
P =
\begin{pmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{pmatrix}
$$
and by the previous reasoning we have

$$
(\cos \tilde{\varphi}, \sin \tilde{\varphi}) = (\cos (\theta + \varphi), \sin(\theta + \varphi))
$$

which implies that
$$
\tilde{\varphi} = \theta + \varphi + 2\pi n, \text{ for some integer } n,
$$

hence, $\tilde{k}_{s} = k_{s}$.

$\large \textbf{Second Part}$

If $\gamma$ and $\tilde{\gamma}$ have the same non-zero curvature then their signed curvature are the same or differ by the sign. If the signed curvature are the same, theorem 2.2.6 from the book by Andrew Presly ensures that $\gamma$ and $\tilde{\gamma}$ differ by a direct isometry. If the signed curvatures differ in  sign, applying a reflection to one curve give us two curves with the same signed curvature, then these two last curves differ by a direct isometry and hence the original curves differ by an opposite isometry.

Title: Signed curvatures


问:
Surface Integral of a paraboloid below a plane
I'm trying to solve this problem:

Considering S the part of the paraboloid $z=x^{2}+y^{2}$ that is underneath the plane $z=3-2y$, calculate the integral:

$$\int\int_{S}\frac{1}{\sqrt{1+4z}}dS.$$

I have tried to use parametrization and the formula
$$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}$$

but I couldn't get anywhere. I believe I'm doing something wrong because I get that
$$ \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}=\iint 1 dA.$$ It would be of great appreciation if someone could help me out.

答:
$\displaystyle \iint_{S}\frac{1}{\sqrt{1+4z}} \ dS$

Surface $S$ is $z = x^2+y^2$, bound above by plane $z = 3 - 2y$

Now we know that,

$\displaystyle \iint_{S}\frac{1}{\sqrt{1+4z}} \ dS = \iint_{E}\frac{1}{\sqrt{1+4z}} \sqrt{1+ \bigg(\frac{\partial z}{\partial x}\bigg)^2 + \bigg(\frac{\partial z}{\partial y}\bigg)^2} \ \ dA$

where $E$ is the projection of the surface in $XY$ plane and $dA$ is the area element of the projection.

This indeed simplifies to $\displaystyle \iint_{E} 1 \cdot dA$

Now to find projection of the surface in $XY$ plane,

$z = x^2+y^2 = 3 - 2y \implies x^2+(y+1)^2 = 4$

We substitute $x = r \cos\theta, y = -1 + r\sin\theta$ and we have $0 \leq r \leq 2, 0 \leq \theta \leq 2\pi$

So the integral becomes $\displaystyle \int_0^{2\pi} \int_0^2 r \ dr \ d\theta = 4 \pi$


问:
What is larger -- the set of all positive even numbers, or the set of all positive integers?
We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out.  The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1).  This implies that N is twice as large as E

On second glance though, it seems less obvious.  Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then?  Or are they both equal in size?  Why?

(My background in Set theory is quite extremely scant)

答:
Your example is an example of a countable sets: http://en.wikipedia.org/wiki/Countable_set

As others have explained, if you find a bijection between these two sets then they are of the same cardinality or have the same number of elements.(I hope that statement is true :)) But in this case you have found one with N which is what makes it a countable set. f(x)=2x would be an example of a bijection between N and 2N. Since for every x in the set of natural numbers there is a corresponding number in the set of 2N (only one) then you can say that they have the same cardinality.

A cooler example would be the interval of the real numbers (0, 1) and R itself, those are not countable but their cardinality is the same.

Other notes, if you are wondering what aleph is: http://en.wikipedia.org/wiki/Aleph_number

ContentLicense: CC BY-SA 2.5


答: It's because the series for ln2 is conditionally convergent.  (see also Riemann's rearrangement theorem)</p>

Title: Explain why calculating this series could cause paradox?

Body:
$$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
      = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$
     $$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$

thanks.


问:
Converting word problem into algebra
A microprocessor plant that manufactures computer chips costs $\$400,000$ per day to operate. It costs $\$30$ to produce one chip which sell for $\$190$ dollars. How many chips does the plant have to make to break even on one day operation?

Let $X$ be the number of chips I need to make.

$400,000 = 190X-30X$ (but I'm not sure this is even right). I'll continue anyway, I'll check it later...

$400000 = 160X$

$X=2500\quad$ ?

$2500 * 190 = 475000$ (its wrong)

Can someone help me? I think that formula I have is very wrong and I can't figure out another one...

答: Your method of checking the answer is incorrect.  $2500\cdot 190=475000$ is the total gross income from producing 2500 chips, but the cost of producing those 2500 chips is not just 400000, but $400000+2500\cdot 30=475000$, since each of the 2500 chips costs \$30 to produce.</p>

Title: Converting word problem into algebra


问:
Calculating loan + interest recursively
I am trying to calculate a loan amount where the total principle depends on the payment amount.

Here's a simple example, leaving aside that a) no bank would actually do this, and b) my income is not specified:

I have $0 in the bank right now. I want to take out a loan for a $100 car, for which I will be making a regular annual payment, starting at the end of year 0 with an interest-only payment. My loan rate is 5%. In order to qualify for the loan, the bank wants to make sure that I have a balance equal to twice my annual payment amount throughout the life of the loan.

If I borrow exactly $100, my payment at the end of year 0 would be $5 ($100 x 5%). However, I need to borrow a little extra so that I have a $10 balance in the account (to satisfy the requirement that I maintain a balance of twice my annual payment). This means I actually need a loan of $110. However, a loan of $110 actually raises my payment amount to $5.50, which means I need to maintain a balance of $11 in my account.

This means that I actually need to borrow $111 to begin with, which raises my payment to $5.55, which means I actually need to borrow $111.10, etc.

I have worked it out to a summation that resembles the following, although I can't be certain this is accurate:

$$\sum_{n=1}^x (x-1-n)r^n$$

Where $r$ is the interest rate

I believe this would allow for me to iterate $x$ number of times over a formula that sums the additional principle needed for each incremental amount of interest. However, I'm not sure how to carry this out to infinity.

Ultimately, I need to plug this into Excel as part of a cash flow, but I wanted to think through the math first.

Any ideas?

答:
Rearrange summation into simpler terms:

$$\sum_{n=1}^x (x-1-n)r^n$$

$$\sum_{n=1}^x (x - 1)r^n - \sum_{n=1}^x nr^n$$

$$(x - 1)\sum_{n=1}^xr^n - \sum_{n=1}^x nr^n$$

Carry out to infinity, as you ask:

$$\lim_{x\to\infty}\left[(x - 1)\sum_{n=1}^xr^n - \sum_{n=1}^x nr^n\right]$$

Now we look here for the closed forms of these finite sum series.

$$\lim_{x\to\infty}\left[(x - 1)\left(\frac{1 - r^{x+1}}{1 - r} - 1\right) - r\frac{1-(x+1)r^x+xr^{x+1}}{(1-r)^2}\right]$$

But unfortunately, it looks like this won't converge to a limit. Since $0 \leq r < 1$, the $r^x$ and $r^{x+1}$ factors vanish and we are left with:

$$\lim_{x\to\infty}\left[(x - 1)\left(\frac{1}{1 - r} - 1\right) - r\frac{1}{(1-r)^2}\right]$$

This is just $Ax + B$ which has no limit as $x$ grows large.

But maybe the above ideas hint you in the direction of finding a closed form for your calculation so that you can avoid iterating or recursing.

Title: Calculating loan + interest recursively


问:
How does a 0-ball have volume 1?
I've been reading about n-balls and I found in the Wikipedia article that in dimension 0, the volume of a 0-ball is 1. I have searched for more information but I can't find any resource explaining this.

So, how does a 0-ball have volume 1?

答: $0$-dimensional space is just a single point and every ball of positive radius contains that point. Moreover, the measure in this space is just the counting measure. So the volume of the ball is $1$ because it contains one point.

Title: How does a 0-ball have volume 1?


答: They don't. There are no zero-dimensional balls. the Wikipedia article is not precise. The reason for the imprecision can be filed under "semantic sugar": it so happens that if you plug $n=0$ in the formula for the volume of an $n$-dimensional ball, where $n\geq 1$, you get $1$.

Title: How does a 0-ball have volume 1?

Body:
I've been reading about n-balls and I found in the Wikipedia article that in dimension 0, the volume of a 0-ball is 1. I have searched for more information but I can't find any resource explaining this.

So, how does a 0-ball have volume 1?


答:
You can also assume the form $f(x)=\sum\limits_{n=0}^\infty a_nx^n\ $ and try to calculate the coefficients of $f(x)=\ln(1+x)$ given that $f$ verifies the ODE: $$(x+1)f'(x)=1$$

$\begin{align}(x+1)\left(\sum\limits_{n=0}^\infty a_nx^n\right)'
&=(x+1)\sum\limits_{n=1}^\infty na_nx^{n-1}
=\sum\limits_{n=1}^\infty na_nx^{n}+\sum\limits_{n=1}^\infty na_nx^{n-1}\\
&=\sum\limits_{n=1}^\infty na_nx^{n}+\sum\limits_{n=0}^\infty (n+1)a_{n+1}x^{n}
=a_1+\sum\limits_{n=1}^\infty (na_n+(n+1)a_{n+1})x^{n}\end{align}$

Identifying with the constant series $1$ we get

$\begin{cases}a_1=1\\na_n+(n+1)a_{n+1}=0\end{cases}\iff \begin{cases}a_1=1\\a_{n+1}=-\dfrac {n}{n+1}a_n\end{cases}$

$\require{cancel}a_n=\left(-\frac{\cancel{n-1}}{n}\right)\left(-\frac{\cancel{n-2}}{\cancel{n-1}}\right)\left(-\frac{\cancel{n-3}}{\cancel{n-2}}\right)\cdots\left(-\frac{1}{\cancel{2}}\right)a_1=(-1)^{n-1}\times\dfrac 1n$

We have also $a_0=f(0)=\ln(1)=0$.

Title: Related to Taylor series of the natural log.

Body:
I'm trying to prove the Taylor series for the natural logarithm.

We know that $$\int_1^x \frac{dt}{t}=\ln x$$
Hence if I can know what is the Taylor Series for $1/t$ for $t>1$ I can solve the Taylor series for the natural log by integrating both sides, this is my Attempt:

Let $f(t)=1/t$ for $t>1$ and note that $f(1)=1$
$$\frac{1}{t}=f(1)+f^{(1)}(1)(t-1)+\frac{f^{(2)}(1)(t-1)^{2}}{2!}+\frac{f^{(3)}(1)(t-1)^{3}}{3!}+...$$
$$\frac{1}{t}=1-(t-1)+(t-1)^2-(t-1)^3+...$$
$$\int_{1}^x\frac{dt}{t}=t-\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3}-\frac{(t-1)^4}{4}+...\Bigr\rvert_{1}^x$$
$$\ln x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+..$$
I am sure that i didn't make any mistake, the problem is when i searched for the Taylor series of natural log on wikipedia i didn't found this formula, so Am i wrong?


问:
Probability of rolling a 6 immediately after a 1 is rolled
Question

Ann and Bob take turns to roll a fair six-sided die. The winner is the first person to roll a six immediately after the other person has rolled a one. Ann will go first. Find the probability that Ann will win.

Answer

$\mathbb{P} (\mathrm {Ann\ wins}) = \frac {36} {73}$

I have thought long and hard about this question but I am unable to even start. I have tried considering cases, but got stuck along the way. For example, it is trivial to calculate the probability that Ann wins if there are only three rolls (which is the minimum number of rolls needed for Ann to win). However, the problem easily becomes very complicated when we consider five rolls and more.

The suggested solution by my professor uses first-step decomposition, but it is a new concept to me and I am struggling to understand it.

If anyone can provide a detailed and intuitive explanation as to how this problem should be solved, that will be greatly appreciated!

答:
Let the probability of A winning from start be $s$, and from the brink of winning be $t$, then by symmetry, the corresponding state probabilities for $B$ will be $(1-s)\;$ and$\; (1-t)$, and a first step analysis gives

$\displaylines{s = \frac56(1-s) +\frac16(1-t)\\(1-s)=\frac56s +\frac16t\\(1-t) = \frac16\cdot0 +\frac46s +\frac16t}$

[Note: The zero multiplier in the last equation is because we want to find $\Bbb P$($A$ wins), so $B$ can't be allowed to win]

Wolfram gives the answer as $s= \frac{36}{73}, t = \frac{42}{73}$


问:
Prove that estimator isn't consistent.
Consider $X_1 \ldots X_n \sim N(\theta, 1)$ are i.r.v. Let's consider $\tau = \mathbb{P}(X_1 > 0)$. We want to find MLE estimator of $\tau$ and then do the following. Suppose that $X_1 \ldots X_n$ are not normally distributed, then we need to show that estimator isn't consistent.

My try:

I've proved that $\hat{\tau}_{MLE} = 1 - \Phi(-\bar{X})$. Now let's focus on the second part.
First of all, I suppose that we still consider the same estimator $\hat{\tau}_{MLE}$. Then if it's so why don't we have consistency? I mean, because of $-\bar{X} \xrightarrow{P} -\mathbb{E}(X)$ and because of $\Phi(x)$ is continuous function we should have $\hat{\tau}_{MLE} \xrightarrow{P} 1 - \Phi(-\mathbb{E}(X))$. Maybe I miss something?

答:
Define $f(x)=\frac{1}{C(x^4+1)}$ where $C=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{x^4+1}$. Take a SRS $X_1,\ldots,X_n\sim f$ and put $\bar{X}_n=\frac{X_1+\dots +X_n}{n}$.

By the weak law of a large numbers we have $\bar{X}_n\rightarrow \mathbb{E}(X_1)=0$ in probability. Therefore $1-\Phi(-\bar{X}_n)\rightarrow 1-\Phi(-0)=1/2$ in probability as well. But $1/2=P(X_1>0)$ so $1-\Phi(-\bar{X_n})$ is a consistent estimator for $P(X_1>0)$ even though $X_1$ isn't normal.

Title: Prove that estimator isn't consistent.


问:
How does one see whether to use a right- or left-side multiplication in a matrix equation?
I understand that the main question is too generic. (But after having typed out the full question/answer I think this 'question' might be more of a confirmation :). I will try my best to form a series of questions along with some (hopefully) clear examples. I'm working with matrices and can understand calculating with them normally. But I'm having trouble with the equations where the matrices themselves are not given.

Basic left- vs right-sided multiplication examples (where X must be determined) are:

AX = B, where we left-multiply with A-1 => A-1AX = A-1B. So we have X = A-1B

And

XA = B, where we right-multiply with A-1 => XAA-1 = BA-1. So we have X = BA-1

These are simple examples, but I fail to understand more difficult ones. I will give an example with its answer(s) and hope somebody can answer my questions about the answer.

Question:

Suppose A,B,X are square matrices.
A,X and A-AX are invertible.
(A - AX)-1 = X-1B. Determine X.

Solution:

(A - AX)-1 = X-1B

(A - AX)(A - AX)-1 = (A - AX) (X-1B).

Q1: Why does the solution use left-multiplication? Why not right: (A-AX)-1(A-AX)?

Q2: (has to do with Q1) If I'm not mistaken for all invertible A the following holds, right?
AA-1 = A-1A = I

Continuation of answer:

I = (A - AX) (X-1B)

I = AX-1B - AB

AX-1B = I + AB

(AX-1B)-1= (I + AB)-1

B-1XA-1= (I + AB)-1. Now B needs to be left-multiplied and A right-multiplied.

Q3: Is this because B is to the left of X and A to the right of X?

X= B(I + AB)-1A

答:
Q1: Why does the solution use left-multiplication? Why not right?

Because this way, that $X$ on the right will hit the $X^{-1}$ on the right. If you had used right-multiplication, these would not meet and you have a mess on the right.

Q2: (has to do with Q1) If I'm not mistaken for all invertible $A$ the following holds, right? $AA^{-1} = A^{-}1A = I$

Yes, that is true.

Q3: Is this because $B$ is to the left of $X$ and $A$ to the right of $X$?

Well, it's because $B^{-1}$ is to the left of $X$, and $A^{-1}$ is to the right. The left side of that equation transitions from $B^{-1}XA^{-1}$ to $BB^{-1}XA^{-1}A$, to just $X$.


问:
How do you work out $\sqrt{2} \pmod{31}$?
If $a^{n}\equiv 1 \pmod p$ with $n$ odd then $a^{n}=a^{1 + \frac{n-1}{2}\cdot 2}\equiv 1\pmod p$ so we have $a \equiv (a^{\frac{1-n}{2}})^{2} \pmod{p}$ and then we have $\sqrt{a}=a^{\frac{1-n}{2}} \pmod{p}$.

From this we have that since $2^{5}\equiv 1 \pmod {31}$ then $2 \equiv (2^{-2})^{2},$ so that $\sqrt{2}\equiv 2^{-2} \pmod {31}$.

I know that $2^{-2}\equiv 8 \pmod {31}$, (and this checks out since $8^2=64 \equiv2 \pmod{31}$) but I don't know how this was obtained and would like to know how do you see that $2^{-2} \equiv 8 \pmod{31}$ without guessing.

答:
$x^2 \equiv b^2 \pmod {31} \implies x^2 - b^2 = (x-b)(x+b) \equiv 0\pmod {31}$ but as $31$ is prime there are no zero divisors and $x\equiv \pm b\pmod{31}$ are the only two options.

Now we know by FLT then $2^{30}\equiv 1 \pmod {31}$ and so the order of $2$, call it $k$, will divide into $30$.  So as $2^{km} \equiv 1 \pmod {31}$ then $2^{km + 1}\equiv 2 \pmod {31}$.  So if we can find a even value of $km + 1$ then we have $(2^{\frac {km+1}2})^2 \equiv 2^{km+1} \equiv 2 \pmod {31}$.

If $k$ is odd then we can just use $\frac {k+1}2$.  But if $k$ is even we are s.o.l and there won't get us anywhere.

SO first what is $k$ so that $k|30$ and $2^k \equiv 1 \pmod {31}$?  It's not too hard to see that then the order of $2$ is $k = 5$ and $2^5 \equiv 1 \pmod {31}$.

And we are in luck $5$ is odd and so $(2^3)^2 \equiv 2^6 \equiv 2\pmod{31}$>

And so the only two solutions to $x^2 \equiv 2 \pmod {31}$ are $x \equiv \pm 2^3 \equiv \pm 8 \equiv 8, 23\pmod {31}$.

That's it.

Hmm I suppose I should explore if this wold be solvable for $\sqrt{m}$ where the order of $m$ is an even $k$....


问:
When are two localisations the same?
Consider $S$, the localisation of $\mathbb{Z}[x]$ at the multiplicative set $\{ 1, x^3, x^6, \ldots \}$. Now, $x^2/x^3$ is an element of $S$, and it seems to me that this behaves the same as $1/x$. Therefore, it seems we get the same result as if we had localised at $\{ 1,x,x^2,x^3,\ldots \}$ instead.

Am I correct that these two localisations are "the same"? If so, given a ring $R$, and multiplicative subsets $M, N \subseteq R$, when are $\mathrm{Loc}(R,M)$ and $\mathrm{Loc}(R,N)$ "the same" / canonically isomorphic?

答:
Yes, these two localizations are the same, and your logic is completely fine.

To answer your larger question, we define the saturation $\hat{S}$ of a multiplicatively closed subset $S\subset R$ to be the set of elements $r\in R$ so that there exists an $r'\in R$ so that $rr'\in S$.
It's not so difficult to see that $S^{-1}R$ is isomorphic to $\hat{S}^{-1}R$ as an $R$-algebra: clearly $R\to \hat{S}^{-1}R$ factors uniquely through $R\to S^{-1}R$ since every element in $S$ is invertible in $\hat{S}^{-1}R$; conversely, every element in $\hat{S}$ is invertible in $S^{-1}R$ as $r\cdot r'/(rr')=1$ so $R\to S^{-1}R$ factors uniquely through $R\to \hat{S}^{-1}R$.

This establishes that $S^{-1}R$ is always isomorphic as an $R$-algebra to $\hat{S}^{-1}R$, so if $S,T$ are two multiplicatively closed subsets having the same saturation, then $S^{-1}R\cong T^{-1}R$ as a $R$-algebras. Conversely, if $S$ and $T$ are two multiplicatively closed subsets with different saturations, then $S^{-1}R$ and $T^{-1}R$ are not isomorphic as $R$-algebras with their standard $R$-algebra structures: we might as well assume that $S$ and $T$ are saturated and that there's some element $s\in S$ not in $T$, and then $s$ has invertible image in $S^{-1}R$ but not $T^{-1}R$.

Title: When are two localisations the same?


问:
What's the $\int_0^T (T - s) dW_s$?
Let $W_{t}$ be a Wiener process on a standard filtered probability space $(\Omega ,{\mathcal {F}},{\mathcal {F}}_{t},P)$ and let ${\mathcal {G}}_{t}$ be the augmented filtration generated by $B$. If $X$ is a square integrable random variable measurable with respect to ${\mathcal {G}}_{\infty }$, then there exists a predictable process $\Theta$ which is adapted with respect to ${\displaystyle {\mathcal {G}}_{t}}$, such that

$$X = EX + \int_0^T\Theta_s dW_s$$

where $W$ is a Wiener process

I want to prove that Ito representation of $X = \int_0^T W(s) ds$ is in form for $\Theta_s = (T-s)$.

My work so far

Let's calculate the integral $$\int_0^T (T-s)dW_s = \int_0^T TdW_s - \int_0^TsdW_s = T\int_0^TdW_s - \int_0^TsdW_s=$$
$$= TW_T - \int_0^TsdW_s$$

And now I have two problems:

$(1)$ Why $EX = E[\int_0^T W(s)ds] = 0 $ ?

$(2)$ I'm not exactly sure how to calculate this integral $\int_0^TsdW_s$. I read that partial integration should be used but I don't see how.

Could you please give me a hand in justifying $(1)$ and $(2)$  ?

答:
Well for (1) it's only Fubini nothing more (as $E[W_s]=0$), for the (2) you have by stochastic integration by part (look for example Theorem 3) that :

$$TW_T-0.W_0=\int_0^TsdW_s+\int_0^TW_sds+\int_0^Td\langle s,W_s\rangle=\int_0^TsdW_s+\int_0^TW_sds+0$$

So $\int_0^T (T-s)dWs =\int_0^TW_sds$

QED

Title: What's the $\int_0^T (T - s) dW_s$?


问:
Putnam 1967 problem, integration
I have some problems understanding this solution of this problem:

Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that

$$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$

It says there that if we split this integral into $n$ parts to get

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$

we will have

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$

because for large $n$, $f$ is roughly constant over the range.

Could you tell me why $f$ is roughly constant over the range for $n$ large enough?

I would really appreciate all your insight.

答: It is saying $f$ is roughly constant over $[\frac{r}{n},\frac{r+1}{n}]$.</p>

Title: Putnam 1967 problem, integration


问:
Partial fraction decomposition proof
In the Wikipedia page for partial fraction decomposition there is a formula given (here):

I am trying to use this to find the partial fractions of $\frac{1}{Q(x)}$ however I don't know if there are limitations on what $Q(x)$ is other than it being a polynomial. I thought that question would be answered if I read the proof for the formula but I wasn't able to find it.

答:
For partial fractions, you need to fully factor the denominator polynomial $Q(x)$ into linear or quadratic factors, for real functions without getting into complex roots.  That is a requirement for the method to be helpful.

Once you have quadratic factors, you can complete the square on them, and then your original rational function will separate into terms that can be integrated using either $\ln$, (for linear terms), $u$ substitution (for $\frac{kx}{x^2 + c^2}$), or using $\tan^{-1}$.

Title: Partial fraction decomposition proof


问:
Closure of product equals product of closures: Application
One may prove that the Axiom of Choice is equivalent to the following statement $P$:

If $\{(X_i,\tau_i)\mid i\in I\}$ is a system of topological sets, and $\prod_{i\in I}X_i$ is equipped with the corresponding product topology, then
$$\overline{\prod_{i\in I}A_i}=\prod_{i\in I}\overline{A_i}$$
whenever $A_i\subseteq X_i$ for each $i\in I$.

I am looking for an interesting result whose proof relies on a direct application of $P$ (and which wouldn't be provable without assumption of $P$, or AC).

To give a different example, the Axiom of Choice is also equivalent to Tychonoff's Theorem. An application of Tychonoff's Theorem results in the De Bruijn–Erdős Theorem.

答:
If you assume that this theorem holds (in some model of set theory), it follows that AC holds too in that model. The proof is probably based on defining a product of topological spaces and subsets $A_i$ for a family of set that needs a choice function. I don't recall ever seeing this proof myself, it's probably not too hard;  I do show in this note how the Tychonoff theorem (a product of compact topological spaces is compact) also proves AC; I'd imagine the proof of "closures result implies AC" to be something like it.

So then you can use AC to prove all other results that depend on AC, like the De Bruijn–Erdős theorem that you mentioned. Any application of AC could be seen as an application of this result.

As AC can be proved from the closures result, it automatically follows that it cannot be shown without AC, as AC is independent from the other ZF axioms.


问:
Diagonalizable binary matrices over $\mathbb{Z}_4$
I'm trying to figure out the following question:

Are symmetric, binary $n\times n$ matrices with zeros on the diagonal, are diagonalizable over $\mathbb{Z}_4$?

I know that it isn't true that $\textit{any}$ binary matrix is invertible over $\mathbb{Z}_2$, but I'm not sure about what happens in $\mathbb{Z}_4$, in particular in the case where the diagonal is all-zero.

Thanks!

答:
First, notice that if $R$ is an invertible matrix with coefficients in $\mathbb{Z}_4$ then $\det(R)$ is an invertible element of $\mathbb{Z}_4$. So $\det(R)=1$  or $\det(R)=3$.

If $\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} \lambda  & 0\\ 0 & \mu \end{pmatrix}$ then  $\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a & b\\ \lambda a & \mu b\end{pmatrix}$

Hence,  $\det\begin{pmatrix} a & b\\ c & d\end{pmatrix}=ad-bc=(\mu -\lambda)ab$.

Next, notice that
$$-1.(ad-bc)=\det\left(\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\right)=\det\left(\begin{pmatrix} \lambda a & \mu b\\ \lambda c & \mu d\end{pmatrix}\right)=\lambda\mu (ad-bc).$$

If $ad-bc=1$ or $3$ the $\lambda\mu=-1$. Thus, $\mu-\lambda= 2$.

Therefore, $\det\begin{pmatrix} a & b\\ c & d\end{pmatrix}=2ab$.

This contradicts the assumption that $ad-bc=1$ or $3$.

So the matrix $\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$ can not be written as  $RDR^{-1}$, where $R$ is an invertible matrix with coefficients in $\mathbb{Z}_4$ and $D$ is a diagonal matrix.


问:
Derivative problem in R^n space
I have been trying to solve the following problem:

Let $f,g: \mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable functions and supose $f(x_0)=g(x_0)$, and $f(x)\leq g(x)$ in a certain region with center in $x_0$. Show that the Jacob Matrixes of $f$ and $g$ are equal at $x_0$.

Here's what I have so far:

$\cdot$First I started by defining the function $h(x)=f(x)-g(x)$. Therefore I have that $h(x)\leq 0$ and $h(x_0)=0$

$\cdot$ I write down the definition of the derivative at $x_0$
$$ \lim_{t\to\ 0} \frac{h(x_0 + tv)-h(x_0)}{t} $$
$$ \iff \lim_{t\to\ 0} \frac{h(x_0 + tv)-0}{t} $$

Suposing that $t$ approaches from the right (doesn't make diference), It's easy to conclude that if the limit is positive then $h(x_0+tv) > 0$, which is not possible base on the fact that $h(x)<0$.

What's left to conclude is that $h(x_0+tv)$ can't be negative, but I don't see where to go from here.

Please help me ending this proof or show me a better way to solve this problem.

答: Let $h=g-f$. Then $h \geq 0$ and $h(x_0)=0$ Hence, $h$ has  a local miminimum at $x_0$. This implies that all the partial derivatives of $h$ at $x_0$ are $0$. So $f$ and $g$ have the same partial derivatives at $x_0$.

Title: Derivative problem in R^n space


问:
Derivative Of Integral Vs Integral of Derivative.
let $f:\mathbb{R}×\mathbb{R} \rightarrow \mathbb{R}$

If $f_t(.,t) \in L^1(\mathbb{R})$ and $f(.,t) \in L^1(\mathbb{R})$, then, By Measure Theory:

$\frac{d}{dt}$$\int_{\mathbb{R}}$$f(x,t) dx$=$\int_{\mathbb{R}}$$f_t(x,t) dx$

But I can't understand why is this simply true when $f$ is a regular function? And in sense of what is this regularity?

Some help please?

答:
The OP clarified, in the comments below,  that by regular function it is meant smoth function that is $C^k$-functions ($k\geqslant 1$).

So, for all $t$, there is $\varepsilon>0$ and  there is a function $g$ such that, for all $u \in [t-\varepsilon, t+\varepsilon] $,
$$|f_t(\cdot,u)| \leqslant g(\cdot) \tag{1}$$
Let us assume that $g \in L^1(\mathbb{R})$.

Let $\{h_n\}_n$ be a sequence of positive real numbers such that $\lim_{n \to \infty}h_n = 0$.

\begin{align*}
\frac{d}{dt}\int_{\mathbb{R}}f(x,t) dx &= \lim_{n \to \infty}\frac{1}{h_n} \left ( \int_{\mathbb{R}}f(x,t+h_n) dx - \int_{\mathbb{R}}f(x,t) dx\right) = \\
& =  \lim_{n \to \infty}  \int_{\mathbb{R}} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx
\end{align*}

Let $g_n = \frac{f(\cdot,t+h_n)  - f(\cdot,t) }{h_n}$. Then $g_n \to f_t(.,t)$  pointwise.

On the other hand, from $(1)$, we have that for all $t$, there is $\varepsilon>0$ and  $g \in L^1(\mathbb{R})$ such that, for all $u \in (t-\varepsilon, t+\varepsilon) $,
$|f_t(\cdot,u)| \leqslant g(\cdot) $. Since $\lim_{n \to \infty}h_n = 0$, we have that there is $N$ such that, for all $n>N$, we have  $ [t, t+h_n] \subseteq (t-\varepsilon, t+\varepsilon)$ and so, using Mean Value Theorem, we have, for all $x$,

$$|g_n(x)|=\left |\frac{f(x,t+h_n)  - f(x,t) }{h_n} \right|\leqslant\sup_{u\in[t, t+h_n]}|f_t(x,u)|\leqslant g(x)$$
Now, we can apply the Lebesgue Dominated Convergence Theorem and we get

$$ \lim_{n \to \infty}  \int_{\mathbb{R}} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx =  \int_{\mathbb{R}}  \lim_{n \to \infty} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx = \int_{\mathbb{R}}f_t(x,t) dx$$

So we have:

\begin{align*}
\frac{d}{dt}\int_{\mathbb{R}}f(x,t) dx &= \lim_{n \to \infty}\frac{1}{h_n} \left ( \int_{\mathbb{R}}f(x,t+h_n) dx - \int_{\mathbb{R}}f(x,t) dx\right) = \\
& =  \lim_{n \to \infty}  \int_{\mathbb{R}} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx =\\
&  =   \int_{\mathbb{R}}  \lim_{n \to \infty} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx = \\
& = \int_{\mathbb{R}}f_t(x,t) dx  \tag{2}
\end{align*}

Important Remark: We could be tempted to answer the question by just directly writing $(2)$. The issue is that the step
$$\lim_{n \to \infty}  \int_{\mathbb{R}} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx=   \int_{\mathbb{R}}  \lim_{n \to \infty} \frac{f(x,t+h_n)  - f(x,t) }{h_n}dx $$
need not be true. In other words,  although $g_n(\cdot) \in L^1(X)$ , $f_t(\cdot,t) \in L^1(X)$ and $g_n(\cdot) \to f_t(\cdot,t)$ pointwise, we still have that
$$\lim_{n \to \infty}  \int_{\mathbb{R}}g_n(x)dx=   \int_{\mathbb{R}}  f_t(x,t)dx $$
need not be true.

To prove this step correctly, we need either Lebesgue Dominated Converge Theorem or Fatou's Lemma for $\limsup$. Both require that the sequence $\{|g_n|\}_n$ be dominated by a function in $L^1(\Bbb R)$. Here is exactly the point where we need $(1)$.

Remark 2 There is another way, using Fubini, to prove that, for smooth functions
$$\frac{d}{dt}\int_{\mathbb{R}}f(x,t) dx=\int_{\mathbb{R}}f_t(x,t) dx$$
But then we need to add the condition that
$$ \int_{[t,t+\varepsilon]} \int_{\Bbb R} |f_t(x,t)| \; dx \; dt < \infty $$


问:
Meagre subset of complete metric space has no dense subset?
I was assigned the exercise to show that in Complete metric space $(X,d)$ , if $G$ is a dense $G_{\delta}-$set, then $G$ is not meager. I proceeded as follows:

If $G \subseteq \bigcup \limits_{n=1}^{\infty} E_n$, where int$(\overline{E_n})=\emptyset$ for all $n$, then
$$X=\overline{G}\subseteq \overline{\bigcup \limits_{n=1}^{\infty} E_n} \subseteq \bigcup \limits_{n=1}^{\infty} \overline{E_n},$$
so $$X=\bigcup \limits_{n=1}^{\infty} \overline{E_n}.$$
But since $X$ is complete, one of the consequences of the Baire category theorem is that int$(\overline{E_N}) \ne \emptyset $ for some $N\in \mathbb{N}$, a contradiction. Thus $G$ is not meager.

But this argument doesn't seem to rely on the fact that $G$ is a $G_\delta-$set at all, only that $G$ is dense. Does this mean that a meager set cannot have a dense subset, or have a commited an error somewhere?

答:
So $G = \bigcap_n O_n$ where all $O_n$ are open. As $G$ is dense, all $O_n$ are open and dense and so $X\setminus O_n$ is closed and nowhere dense.

Note that $$X= G \cup (X\setminus G) = G \cup \bigcup_n (X\setminus O_n)$$

So if $G$ were meagre, so would $X$ be as a union of two meagre sets.
But this contradicts the Baire theorem as $X$ is complete (if $X=\bigcup_n E_n$ where all $E_n$ are closed and nowhere dense, none of them has any interior while the interior of $X$ is $X$..).


问:
How are these Fourier series coefficients obtained?
From

I wish to understand how equations $(4.8a)$ and $(4.8b)$ are obtained.

I have only seen the Fourier series be written as: $a_0 + \sum_{n=1}^{\infty} \left (a_n \cos(\frac{n\pi x}{l}) + b_n \sin(\frac{n\pi x}{l}) \right ) $ for a periodic piecewise smooth function $f(x)$ with period $2l$. Its Fourier coefficients being $a_0, a_n, b_n$  being written as $a_0 = \frac{1}{2l}\int_{-l}^{l}f(x)dx$, $a_n = \frac{1}{l}\int_{-l}^{l}f(x)\cos(\frac{n\pi x}{l})dx$, and $b_n = \frac{1}{l}\int_{-l}^{l}f(x)\sin(\frac{n\pi x }{l})dx$. These integrals, the $\textit{euler formulas}$, were obtained by saying the set $\{1, \cos(\frac{n\pi x}{l}),\sin(\frac{n\pi x }{l}) \}$ formed a $\textbf{basis}$ for the $2l$-periodic functions. The derivation process is taking the inner-product of test functions...these test functions being the elements of the basis.

Here, would the basis be $\{0, \cos(\frac{n\pi c t}{L}), \sin(\frac{n\pi c t}{L}), \sin(\frac{n \pi x}{L}))\}$?

答:
Trigonometric Polynomials

First, let us acknowledge the idea of trigonometric polynomials. Say, I have a periodic function $f$ i.e. $$f(t+T)=f(t),\;\forall t\in\mathbb{R}$$
Define $\displaystyle e_{k}(t):=e^{2jk\pi\frac{t}{T}}$ where $j=\sqrt{-1}$ and $k\in\mathbb{Z}$. So this can be interepreted as follow :
$$
''e_{k}(t)\;\text{has a period $T$ for all $k\in\mathbb{Z}$}''
$$
The same can be said for a polynomial $p$ of the form :
$$
p(t):=\sum_{k\in I}c_{k}e^{2jk\pi}
$$
where $I$ is any fixed, finite set of integers and the $c_{k}$ are arbitrary complex numbers.  By adding zero terms if necessary, we may assume that :
$$
p(t)=\sum_{k=-N}^{N}c_{k}e^{2jk\pi \frac{t}{T}}
\tag{1}$$

Before we dig in into the next concept of orthogonality we should note that $(1)$ can be represented in terms of $\text{sine}$ and $\text{cosine}$ as follow :
$$
p(t)=c_{0}+\sum_{k=1}^{N}\left(c_{k} e^{2k\pi\frac{t}{T}j}+c_{-k} e^{-2k\pi\frac{t}{T}j}\right)
$$
and by expanding the exponentials, this becomes :
$$
p(t)=\frac{T_{0}}{2}+\sum_{k=1}^{N}\left(\alpha_{k} \cos \left(2k\pi\frac{t}{T}\right)+\beta_{k} \sin \left(2k\pi\frac{t}{T}\right)\right)
$$
where for $k\geq0$ we have :
$$
\alpha_{k}:= c_{k}+c_{-k}\qquad\text{and}\qquad \beta_{k}:= j(c_{k}-c_{-k})
\tag{2}$$
The inverse formulas are :
$$
c_{k}=\frac{1}{2}(\alpha_{k}-j\beta_{k})\qquad\text{and}\qquad c_{-k}=\frac{1}{2}(\alpha_{k}+j\beta_{k})
\tag{3}$$
These relations are fundamental in determining the coefficients of the Fourier series.

Orthogonality

A simple computation shows that the following important relation holds
for the functions $e_{k}(t)$ :
$$
\int_{0}^{T}e_{k}(t)\overline{e}_{m}(t)\;\text{d}t=
\begin{cases}
T&\text{if $k=m$}\\
0&\text{if $k\neq m$}
\end{cases}
$$
We let $\mathcal{P}_{N}$ denote the set of all trigonometric polynomials $p$ of degree
less than or equal to $N$. $\mathcal{P}_{N}$ is obtained by letting the $c_{k}$ in formula $(1)$
vary over all possible values. If we endow this vector space, which has finite dimension equal to $\dim(\mathcal{P}_{N})\leq2N+1$
dimension with the scalar product

$$
\langle p,q \rangle := \int_{0}^{T}p(t)\overline{q}(t)\;\text{d}t
\tag{4}$$
the relation $(4)$ expresses the fact that the functions $e_{k}$ and $e_{m}$ are orthogonal :
$$
\langle e_{k},e_{m}\rangle =\begin{cases}0&\text{if $k\neq m$}\\
\|e_{k}\|_{2}=\sqrt{T}&\text{if $k=m$}
\end{cases}
$$
It follows that the vectors $e_{k}$ are independent and that
$\dim(\mathcal{P}_{N})=2N + 1$ If $p$ is of the form $(1)$, we have :
$$
\langle p,e_{k}\rangle=c_{k}\|e_{k}\|_{2}^{2}=Tc_{k}
$$
and that :
\begin{equation}
c_{k}:=\frac{1}{T}\int_{0}^{T}p(t)e^{-2k\pi \frac{t}{T}j}\;\text{d}t
\end{equation}
This is called Fourier's formula; it gives the coefficients $c_{k}$ explicitly in
terms of the function $p$. Now using $(2)$ and $(3)$ to obtain following formulas for the
coefficients $\alpha_{k}$ and $\beta_{k}$ for $k\geq0$ :
\begin{equation}
\alpha_{k}=\frac{2}{T}\int_{0}^{T}p(t)\cos\left(2k\pi\frac{t}{T}\right)\;\text{d}t\qquad\text{and}\qquad \beta_{k}=\frac{2}{T}\int_{0}^{T}p(t)\sin\left(2k\pi\frac{t}{T}\right)\;\text{d}t
\end{equation}


答: $2^{-2}$ is the multiplicative inverse of $2^2=4$, so you need to solve the congruence $4x\equiv1$ mod $p$ with $p=31$. Solving for multiplicative inverses in general can be done with the Euclidean algorithm, but in this case the solution $x=8$ may jump out at you. If it doesn't, note that $4$, in general, divides either $p+1$ or $p-1$. If it divides $p+1$, then $(p+1)/4$ is the mulitplicative inverse; if it divides $p-1$, then $-(p-1)/4$ is the multiplicative inverse. (Alternatively, $(p+1)/2$ is always the multiplicative inverse of $2$, so $((p+1)/2)^2$ is always the multiplicative inverse of $4$, so here we get $16^2=256=31*8+8\equiv8$ mod $31$.)

Title: How do you work out $\sqrt{2} \pmod{31}$?

Body:
If $a^{n}\equiv 1 \pmod p$ with $n$ odd then $a^{n}=a^{1 + \frac{n-1}{2}\cdot 2}\equiv 1\pmod p$ so we have $a \equiv (a^{\frac{1-n}{2}})^{2} \pmod{p}$ and then we have $\sqrt{a}=a^{\frac{1-n}{2}} \pmod{p}$.

From this we have that since $2^{5}\equiv 1 \pmod {31}$ then $2 \equiv (2^{-2})^{2},$ so that $\sqrt{2}\equiv 2^{-2} \pmod {31}$.

I know that $2^{-2}\equiv 8 \pmod {31}$, (and this checks out since $8^2=64 \equiv2 \pmod{31}$) but I don't know how this was obtained and would like to know how do you see that $2^{-2} \equiv 8 \pmod{31}$ without guessing.


问:
Continuity for nonequivalent norms in a linear space
If two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ defined on the same linear space $L$ are not equivalent. Is it possible to define a linear functional (which is not identically zero) $f:L\longrightarrow \mathbb{R}$ that is continuous for both norms?

答:
If I understand it correctly,  the questioner asks whether every  vector space equipped with two norms  admits a
nonzero linear functional  which is continuous relative to both norms.

I must say right away that I don't have an answer but, having spent some time puzzling about it, I found a very strange
equivalent condition that a counter-example  must satisfy and I'd like to share this here.

To fix notation let $X$ be our vector space and let $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$ be two norms on $X$.

Here is what I found:
a necessary and sufficient condition for there not to be any nonzero linear functional which is continuous relative to
both $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$ is that, for every $x$ in $X$, there are vectors $y$ and $z$ in $X$, such that

$x=y+z$,

$\Vert y\Vert _1<1$,  and

$\Vert z\Vert _2<1$.

Here is the proof:  suppose that the above condition is satisfied and, arguing by contradiction, that  there exists a nonzero
continuous linear functional relative to both norms.

By multiplying $f$ by a small enough number, we may assume that
the norm of $f$ is no bigger than 1 relative to both norms.  Since $f$ is nonzero, we may choose a vector $x$ such that
$f(x)\neq 0$, and again by scaling, we may assume that $f(x)=2$.   Write $x=y+z$, with  $\Vert y\Vert _1, \Vert z\Vert _2<1$,   and
observe that
$$
  2 = |f(x)| \leq    |f(y)| + |f(z)| \leq  \Vert y\Vert _1 + \Vert z\Vert _2  < 1+1,
  $$
a contradiction.

Conversely, suppose that there is no simultaneosly continuous linear functional and denote the open unit ball relative
to $\Vert \cdot \Vert _i$ by $B_i$.  Also let
$$
  B = B_1+B_2.
  $$
Our goal is thus to prove that $B=X$.   In any event, notice that $B$ is a convex, absorbing and balanced set, so we
may consider the so called Minkowsky functional, namely
$$
  p(x) = \inf\big\{|\lambda | : \lambda \in  {\mathbb R}\setminus \{0\},\ \lambda ^{-1} x\in  B\big\}.
  $$
Again arguing by contradiction, suppose that $B\neq X$, so we may pick some $x_0$ lying outside $B$, in which case clearly $p(x_0)\geq 1$.

Considering the linear functional defined on ${\mathbb R}x_0$ by $f(\lambda x_0)=\lambda $, we then see that
$$
  |f(\lambda x_0)| = |\lambda | \leq  |\lambda |p(x_0) = p(\lambda x_0),
  $$
so Hahn-Banach provides a linear functional $g$ on $X$,  extending $f$,  and satisfying $|g(x)|\leq p(x)$ for all $x$ in
$X$.

Observing that $p\leq 1$ on either $B_1$ or $B_2$, we deduce that
$$
  \sup_{x\in B_i}|g(x)| \leq 1,
  $$
which implies that $g$ is continuous for both norms, a contradiction.


答: To answer your question, this formula holds if and only if $Q$ has all distinct linear factors. Note that the Wikipedia article says that quite explicitly a few lines up.

Title: Partial fraction decomposition proof

Body:
In the Wikipedia page for partial fraction decomposition there is a formula given (here):

I am trying to use this to find the partial fractions of $\frac{1}{Q(x)}$ however I don't know if there are limitations on what $Q(x)$ is other than it being a polynomial. I thought that question would be answered if I read the proof for the formula but I wasn't able to find it.