问:
Understanding the differential $dx$ when doing $u$-substitution
I just finished taking my first year of calculus in college and I passed with an A.  I don't think, however, that I ever really understood the entire $\frac{dy}{dx}$ notation (so I just focused on using $u'$), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important.

Take this problem for an example:

$$ \int 2x (x^2+4)^{100}dx $$

So solving this...

$$u = x^2 + 4 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}$$

And so now I'd have: $\displaystyle \int (u)^{100}du$ which is $\displaystyle \frac{u^{101}}{101} + C$ and then I'd just substitute back in my $u$.

I'm so confused by all of this.  I know how to do it from practice, but I don't understand what is really happening here.  What happens to the $du$ between rewriting the integral and taking the anti-derivative?  Why is writing the $dx$ so important?  How should I be viewing this in when seeing an integral?

答:
Okay, so this is a slightly tricksy problem, because at first people think "Oh, it's fine, I just multiply through by this thing and it's fine" but then they think "Hang on, we've never actually defined what we meant by this... it's just some shorthand trickery" - but then if they eventually do differential geometry you think "Aha! So it did make sense all along!"

The main message I want you to take away is that things like $\mathrm d x$ are actually well defined things called differential forms which you don't really need to understand in any detail at all to get how they work.

The way they end up working in integration and changes of variable is roughly the following: they come together to make a volume form which just tells you how much volume a small range of your parameters corresponds to. (I say "come together" because if  you are doing many integrations like in $\int \int f(x,y) \;\mathrm d x \mathrm d y$ then you get a bunch of the $\mathrm d [...]$ things all together.) More precisely, remember how you can roughly define integration as a limit of a sum like
$$\int_a^b f(x) \mathrm d x \equiv \lim_{N\to\infty}\sum_{n=1}^N f(x_n) \left(x_n-x_{n-1}\right)$$
where $x_0=a,x_N=b$ and the other points $x_i$ are chosen in between, such that all gaps $\delta_n = x_n-x_{n-1} \to 0$ (say, uniformly) as $N\to \infty$.
Here, $\delta_n = x_n-x_{n-1}$ is providing some measure of how important the bit of space between $x_{n-1},x_n$ is in computing the integral. The $\mathrm d x$ is what keeps track of that information.

Suppose you then try $u=x^2$ or $x=\sqrt u$. Then in general
$$\int_{a^2}^{b^2} f(\sqrt u) \mathrm d u \equiv \lim_{N\to\infty}\sum_N f(\sqrt{u_n}) \left(u_n-u_{n-1}\right) = \lim_{N\to\infty}\sum_N f(x_n) \left(x_n^2-x_{n-1}^2\right)\neq \int f \mathrm d x$$
because the weight is different!

But notice that $x_n^2-x_{n-1}^2 = (x_n-x_{n-1})(x_n+x_{n-1}) \approx (x_n-x_{n-1})(2x_n)$ in the limit of fine spacing, so
$$\int_{a^2}^{b^2} f(\sqrt u) \frac{\mathrm d u}{2x} = \int_a^b f(x) \; \mathrm d x$$

We're really analyzing the difference between the volumes of the little patches of space when we play with the differentials. The trick is to realize that in general, just like here, $\mathrm d u = u'(x) \mathrm dx$. In higher dimensional integrals, you will discover that the generalization to e.g.
$$\int f(x,y) \;\mathrm d x\mathrm d y = \int f(u,v) \; J \; \mathrm d u\mathrm d v$$
where $u=u(x,y),v=v(x,y)$ involves a quantity $J$ called the Jacobian (determinant) which uses all the possible derivatives of $u,v$ with respect to $x,y$ in a particular way.

The notation
$$\frac{\mathrm d u}{\mathrm d x} = \lim_\text{fine spacing}\frac{\delta u}{\delta x} = \lim_{x_n-x_{n-1}\to 0}\frac{u_n-u_{n-1}}{x_n-x_{n-1}} = u'(x)$$
is now seen to be just a suggestive notation which works for the case of only one variable changing. It's used because it makes it clear how the volume form should be replaced.

When there are many variables, this notation breaks down because the factors are all mixed up together and people write partial derivatives, which you'll see soon if you haven't already, instead. It turns out that it makes sense to use a generalization of the
$$\mathrm d u = u'(x) \mathrm d x$$
law called the chain rule in which, for $u=u(x,y)$ for example
$$\mathrm d u = u_x \mathrm d x + u_y \mathrm d y$$
where $u_x(x,y)$ is the derivative of $u$ with respect to $x$ when we just think of $y$ as a constant.

You'll have to wait until differential geometry courses to see how to use this to get the Jacobian factor; it turns out that rather than just writing the forms together, you should technically define something called a wedge product such that $a\wedge b = -b\wedge a$ for one-forms like $\mathrm d x$; then you get
$$\mathrm d u \wedge \mathrm d v = (u_x \mathrm d x + u_y \mathrm d y)\wedge(v_x \mathrm d x + v_y \mathrm d y) = (u_x v_y-u_y v_x) \mathrm d x \wedge \mathrm d y$$
so that the Jacobian is (one over) $(u_x v_y-u_y v_x)=\det \pmatrix{u_x & u_y \\ v_x & v_y}$.

You can get this result directly from thinking about little patches of volume, however, so you'll see this far earlier than any differential form stuff. I just thought that, since you were curious, you should have had the full story mentioned to you along the way.


问:
Have trouble understanding Matlab code of stochastic integrals
Hell, I have trouble understanding the matlab code of approximating stochastic integrals.

```<code>rng(100)
T = 1; N = 500; dt = T/N;

dW = sqrt(dt) * randn(1,N);
W = cumsum(dW);

ito = sum([0, W(1:end-1)].*dW)
strat = sum((0.5*([0, W(1:end-1)] + W) + 0.5*sqrt(dt)*randn(1,N)).*dW)
</code>```

The ito refers Ito integrals, whose formula is $\sum_{j=-0}^{N-1} W(t_j)(W(t_{j+1})-W(t_j))$

But I just cannot understanding it from the coding perspective. I think W refers to $W(t_{j+1})+W(t_j)$ which is not part of the formula.

Similarly, strat refers to Stratonovich integrals, whose formula is $ \sum_{j=-0}^{N-1} W(t_j)(\frac{t_j + t_{j+1}}{2})(W(t_j) - W(t_j))$. But I cannot find the $W(t_j) - W(t_j)$ part in the coding.

Please help me by point out which part of the coding refers to which part of the formula.

Thank you so much

答:
dW are increments, which are iid normals with mean $0$ and variance $\Delta t$. In math notation we write the increments as $W(t_{j+1})-W(t_j)$ but numerically the information flows the other way: you generate increments and then reconstruct the values of W from the increments.

You get W at a particular time by summing up the increments up to that time, since the process starts up at zero. This is what "cumsum" is doing.

Note that in the Ito form of the integral, you multiply the next increment by the present value of $W$. The vector W in the code only contains $W(t_j)$ for $j \geq 1$, so the factor that should hit the first increment is actually $W(t_0)=0$. This is why there is that zero in the first entry. (I actually would say this is bad notation at least for the Ito case and that it would be better to have this $0$ inside the $W$ vector i.e. W=[0 cumsum(dW(1:end-1))] but it's not a big deal.)

That strat line actually involves much more math to see why it is correct. The idea is that if you know $W(s)=a$ and $W(t)=b$ with $s<t$ then $W(s+u(t-s))$ for $u \in [0,1]$ is a Brownian bridge. A single time sample from a Brownian bridge is normally distributed with mean $a+u(b-a)$ and variance $u(1-u)(t-s)$, which is ultimately what they implemented in that line, though it is somewhat obfuscated that this is what is happening.

I think a more obviously correct implementation of what they achieved in the code snippet above for the Stratonovich integral would be:

```<code>N=1000; dt=T/N;
dW_fine=sqrt(dt)*randn(1,N);
dW_coarse=dW_fine(1:2:end)+dW_fine(2:2:end); %N must be even
W=cumsum(dW_fine);
strat=dW_coarse' * W(1:2:end)
</code>```

This costs more memory but doesn't really cost significantly more CPU time.

ContentLicense: CC BY-SA 4.0

Body:
dW are increments, which are iid normals with mean $0$ and variance $\Delta t$. In math notation we write the increments as $W(t_{j+1})-W(t_j)$ but numerically the information flows the other way: you generate increments and then reconstruct the values of W from the increments.

You get W at a particular time by summing up the increments up to that time, since the process starts up at zero. This is what "cumsum" is doing.

Note that in the Ito form of the integral, you multiply the next increment by the present value of $W$. The vector W in the code only contains $W(t_j)$ for $j \geq 1$, so the factor that should hit the first increment is actually $W(t_0)=0$. This is why there is that zero in the first entry. (I actually would say this is bad notation at least for the Ito case and that it would be better to have this $0$ inside the $W$ vector i.e. W=[0 cumsum(dW(1:end-1))] but it's not a big deal.)

That strat line actually involves much more math to see why it is correct. The idea is that if you know $W(s)=a$ and $W(t)=b$ with $s<t$ then $W(s+u(t-s))$ for $u \in [0,1]$ is a Brownian bridge. A single time sample from a Brownian bridge is normally distributed with mean $a+u(b-a)$ and variance $u(1-u)(t-s)$, which is ultimately what they implemented in that line, though it is somewhat obfuscated that this is what is happening.

I think a more obviously correct implementation of what they achieved in the code snippet above for the Stratonovich integral would be:

N=1000; dt=T/N;
dW_fine=sqrt(dt)*randn(1,N);
dW_coarse=dW_fine(1:2:end)+dW_fine(2:2:end); %N must be even
W=cumsum(dW_fine);
strat=dW_coarse' * W(1:2:end)

This costs more memory but doesn't really cost significantly more CPU time.


问:
What is the name of an algebraic structure which is both a tensor space and a module.
I am studying a set of high-dimensional arrays of complex numbers, which is both a tensor space (defined with the notion of addition and complex scalar product) and a module (defined with the same notion of addition and the notion of the product with a ring element). What is the formal name of this kind of structure?  Is there some literature that discusses this structure in more detail?

I think I can even embed a subset of this structure in it. The subset has a topology and is locally homeomorphic to a small open set of the unnamed structure, and has properties analogous to a canonical manifold. Since the unnamed structure is a module, can I call the constrained set a generalized manifold? Since the unnamed structure is also a vector space (i.e., a tensor space), the subset is also a canonical manifold.

What is the name of this topological subset?  The topological subset is established over a new commutative algebra called T-Algebra. The commutative t-algebra is both a finite-dimensional commutative ring and a vector space. I leave the URL of the t-algebra as follows if somebody wants to know more about the algebra.

https://github.com/liaoliang2020/talgebra

Thanks for any help.

答:
At this point, I'd just like to address the first question along with a comment you added later:

What is the formal name of this kind of structure?

The copy of $\mathbb{C} $ using the $C$'s action is the SAME to the $\mathbb{C}$ on $G$.

That being said, I do not see any reason to call this anything other than a $C$ module over the $\mathbb C$ algebra $C$.  Depending on how you use it, there may be justification for looking upon it as a $C, \mathbb C$ bimdoule, but I have not read your whole exposition.

I skimmed the linked arxiv papers briefly, and would offer the following feedback. I realize your articles aren't really directed at theoretical mathematicians, but passages like this are apt to cause reflexive eyerolls for algebraists:

With appropriate operations,
the t-scalars are trinitarian in the following sense. (1) T-scalars are generalized complex
numbers. (2) T-scalars are elements of an algebraic ring. (3) T-scalars are elements of
a linear space.

After parsing, an algebraist will simply see this:

It's a $\mathbb C$ algebra.

Firstly, I'd recommend not using the phrase "algebraic ring" to mean "ring (in the sense of abstract algebra)" My impression is that the former is a hallmark of authors who are only tenuously familiar with rings.  There are, actually, specialized uses of the term.

Secondly, given your comments I've echoed at the top of this post, no algebraist would make such a big deal about the ring and linear space being separate. They'd just call it a module over a $\mathbb C$ algebra.

The first item on the list is also ostensibly subsumed under the observation you are working with a $\mathbb C$ algebra. And the whole trinitarian thing is a bit weird.

If this passage is any sign, there is a lot of reduction that could be made in order to eliminate unnecessarily fancy appellations in your work.  Simplifying terms will boost readability and transmissibility, and it will avert a lot of face-palming by some of your readers.

Perhaps even the word "t-scalar" or "t-algebra" should probably be replaced by something more informative.  While "t-scalar" might be original, it looks like "T-algebra" is already something established.

"Inventing and naming a new algebra that has uses" is looked upon as a 19th century pursuit.  It seems like the venture is alive and well though. Nowadays the focus is on defining useful classes of algebras and what their properties are.  I'm not saying a paper on a particular algebra is automatically undesirable, but it will be important to know what classes of algebras are relevant to it so that you can leverage existing theory about those classes and focus on your applications.

I'd like to help with the second half of the post, but there are still some things to be clarified before that is possible.


问:
What is $I \otimes I$?
Supposing that $I$ is the identity on some Hilbert space $H$, what does the transformation $I \otimes I$ correspond to on $H \otimes H$? Is it the identity on $H \otimes H$? It seems like it should be but I'm new to this so I wanted to make sure.

答:
As OP suggests in the comments:

Let's have a more general look at operators of the form $u\otimes v$. Recall that the Hilbert space tensor product of two Hilbert spaces $H,K$ is the Hilbert space completion of tha algebraic tensor product $H\odot K$ under the norm induced from the (unique) inner product on $H\odot K$ satisfying $\langle x\odot y, x'\odot y'\rangle=\langle x,x'\rangle_H\cdot\langle y,y'\rangle_K$ for all elementary tensors.

If $u\in B(H)$ and $v\in B(K)$ the operator $u\otimes v$ is defined as the unique operator that is defined on $H\otimes K$ that satisfies
$$u\otimes v(x\odot y)=u(x)\odot v(y)$$
for all elementary tensors. It is proven that this is well-defined and actually $\|u\otimes v\|=\|u\|\cdot\|v\|$. You can check out Murphy's book for more on that and for detailed proofs.

Now how does one show that $I\otimes I$ is the identity map on $H\otimes  H$? Well, since $H\odot H$ is dense in $H\otimes H$ it suffices to show that $I\otimes I$ is the identity on $H\odot H$. But $H\odot H$ is linearly spanned by the elementary tensors, so it is enough to show that $I\otimes I$ is the identity on elementary tensors. But this is obvious by definition:
$$I\otimes I(x\odot y)=I(x)\odot I(y)=x\odot y.$$

This is in general how one works with operators of the form $u\otimes v$. Not only that, but one should also keep in mind the following "operation" rules:

$$(u_1\otimes v_1)(u_2\otimes v_2)=(u_1u_2)\otimes (v_1v_2)$$
and
$$(u\otimes v)^*=u^*\otimes v^*$$
where $-^*$ denotes the adjoint.

Title: What is $I \otimes I$?


问:
Can someone explain how the square root of two negative numbers multiplied together doesn't equal a positive/negative number.
On a test I got a question asking to simplify $\sqrt{-25}\times \sqrt{-3}$

I answered $5\sqrt 3$ but apparently it's only $-5\sqrt 3$ because of some rule of imaginary numbers, could someone please explain to me why $\sqrt{-25}\times \sqrt{-3} \ne \pm 5\sqrt 3$

答:
You presumably tried to use the "rule" $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. However, this only holds when $a$ and $b$ are positive! It is not applicable here.

Instead, we need to compute this product directly. $\sqrt{-25} = 5i$ and $\sqrt{-3} = i\sqrt{3}$, so $$\sqrt{-25} \sqrt{-3} = (5i)(i \sqrt{3}) = -5\sqrt{3}.$$


问:
Expectation for the sum of E[X^2]
Is this correct?

expected value for E[X^2]

答:
Yes, the proof in the image is correct. Given $E[X_i^2] = \sigma^2+\mu^2$, we have:

$$ E\left[\sum_{i=1}^n {X_i^2}\right] = \sum_{i=1}^n {E[X_i^2]} $$

This first step follows from the linearity of the expectation operator.

Then, we can substitute in $E[X_i^2] = \sigma^2+\mu^2$:

$$ E\left[\sum_{i=1}^n {X_i^2}\right] = \sum_{i=1}^n {(\sigma^2+\mu^2)} $$

Since $\sigma^2+\mu^2$ is a constant, we can factor it out of the summation:

$$ E\left[\sum_{i=1}^n {X_i^2}\right] = (\sigma^2+\mu^2) \sum_{i=1}^n {1} $$

Then, the sum of $1$ iterated $n$ times is simply $n$:

$$ E\left[\sum_{i=1}^n {X_i^2}\right] = n(\sigma^2+\mu^2) .$$

Title: Expectation for the sum of E[X^2]


问:
Weighted sum of arrangements
I was pondering this problem (not a homework):

What are the coefficients {a_i} that cancel all powers except the biggest one ?

The values for n=1,2,3 are easy to find, Wolfram-alpha helped find the next ones.

Is there a formula to generate them?

答:
We have that

$$\sum_{k=1}^n {n\brace k} x^{\underline{k}} =
\sum_{k=1}^n {n\brace k} \frac{x!}{(x-k)!} = x^n.$$

with the Stirling numbers of the second kind.

To see this combinatorially write it as

$$\sum_{k=1}^n {x\choose k} {n\brace k} k! = x^n.$$

Let $x$ be the number of elements of some set. Then $x^n$ counts the
number of $n$-tuples with components chosen from this set. This is the
RHS. The LHS  classifies the tuples according to the number of different
values that  appear, say there are $k$ of these. Then we must first
choose these values which  gives ${x\choose k}.$ Next we partition the
$n$ slots into $k$  non-empty disjoint subsets where the slots in a
subset all hold the same value  from the $k$ that we have chosen. We can
assign the latter to the subsets  in $k!$ different ways. This gives the
factor ${n\brace k}\times k!$ and  hence the LHS. (We have shown that it
holds  for positive integers $x.$ With both sides being polynomials in
$x$ it  holds for all $x$ where we define ${x\choose k} =
x^{\underline{k}}/k!.$)

Title: Weighted sum of arrangements


问:
Prove that $r_{k+1}(3) \leq (k+1)( r_k(3) - 1) + 2$
I'm tasked to prove that $r_{k+1}(3) \leq (k+1) (r_k(3) - 1) + 2$ and then use the result to come up with an upper bound for $r_n(3)$.

I'd appreciate a bit of a walkthrough, as I've had a hard time with this.

答:
HINT: Suppose that we have an edge coloring of $K_n$ with $k+1$ colors that has no monochromatic triangle. Let $v_0$ be a vertex of the graph, and for $i=1,\ldots,k+1$ let $N_i$ be the set of vertices connected to $v_0$ by an edge of color $i$.

Show that for each $i$, no two vertices in $N_i$ can be connected by an edge of color $i$. Conclude that the complete subgraph induced by $N_i$ is edge-colored with $k$ colors and has no monochromatic triangle.

What is the maximum possible number of vertices in $N_i$?

What is the maximum possible number of neighbors of $v_0$?

What is the maximum possible value of $n$?


答:
The use of infinitesimals can only be formalized an analysed with care when working with non-standard analysis. When people first study calculus and the most elementary books define $dx$ as $\Delta x$ approaching zero, usually people get confused think "but this should be zero", and in standard analysis it'll really be.

Things are much easier than this, however, we just have to throw away those $dx$ and $dy$. Why? Simply because modern mathematics adhered to methods that are more sophisticated, simpler, and because if you are going to proceed in mathematics you'll really need those modern methods when studying analysis or differential geometry (where $dx$ receives a true definition and gains a fundamental role).

Now you might ask the exact same question I've asked when my I first encountered the rigorous framework: "this guy is mad! My book talks about infinitesimals, my teacher told me it's all right, he must be mad", but it's not like that. I'll show you two examples: the first one is meant to show you that when working with integrals those things appear just as mnemonic rules that allows you to remember the true formula easier when you're starting. The second is to show you when things become confusing when using infinitesimals without care.

First, consider your function:

$$\int2x(x^2+4)^{100}dx$$

The idea is that we see that this can be rewritten in some convenient form. Note that if we set $f(x) = x^2+4$, then $f'(x)=2x$. Then we are integrating:

$$\int f'(x) (f(x))^{100}dx$$

Now, if we set $g(x)= x^{100}$ note that by composition we are integrating:

$$\int g(f(x))f'(x)dx$$

Now, if $G(x)$ is a primitive of $g$ recall the chain rule, the integrand is just $(G(f(x)))'=G'(f(x))f'(x)$, so since the indefinite integral is a primitive, and the integrand is a derivative the result is simply:

$$\int (G(f(x)))'dx = G(f(x))$$

Ande since $g(x) = x^{100}$ the obvious primitive is $G(x) = x^{101}/101$ and hence:

$$\int 2x(x^2+4)^{100}dx = \frac{(x^2+4)^{101}}{101}$$

We can "remember" that by saying that we set $u = x^2+4$ and $du=2xdx$, so is just a rule to remember how to find the formula that's just an application of the chain rule.

The second example is the chain rule itself. Usually people write:

$$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}$$

But look, on the lhs you are differentiating not $f$, but rather $f\circ u$. So, on the left $f$ means one thing, on the right it means another thing! So using this language of infinitesimals the wrong way arround may lead to confusions and may hide from you the true nature of what you are studying. The book I've used when I moved from the infinitesimals treatment to the rigorous one was Spivak's Calculus. Try it! He develop everything formally, without appeal to those "undefined" creatures and shows you where they appear just as ways to remember formulas.

I hope this helps you. Good luck!

Title: Understanding the differential $dx$ when doing $u$-substitution

Body:
I just finished taking my first year of calculus in college and I passed with an A.  I don't think, however, that I ever really understood the entire $\frac{dy}{dx}$ notation (so I just focused on using $u'$), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important.

Take this problem for an example:

$$ \int 2x (x^2+4)^{100}dx $$

So solving this...

$$u = x^2 + 4 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}$$

And so now I'd have: $\displaystyle \int (u)^{100}du$ which is $\displaystyle \frac{u^{101}}{101} + C$ and then I'd just substitute back in my $u$.

I'm so confused by all of this.  I know how to do it from practice, but I don't understand what is really happening here.  What happens to the $du$ between rewriting the integral and taking the anti-derivative?  Why is writing the $dx$ so important?  How should I be viewing this in when seeing an integral?

ContentLicense: CC BY-SA 3.0


问:
Expressing a indefinite integral using a definite integral with the same function
I was wondering how we can express a indefinite integral $\int f(x) \, dx$ with the function $F(x) = \int_a^x f(t)\,dt$.

I was experimenting with some functions, for example, if $f(x) = 3x^2$, then

\begin{equation}
\int f(x) \, dx = \int 3x^2 \, dx = x^3 + C.
\end{equation}

On the other hand,

\begin{equation}
F(x) = \int_a^x f(t) \, dt = \int_a^x 3t^2 \, dt = [t^3]^x_a = x^3 - a^3.
\end{equation}

That means what $F(x)$ and $\int f(x) \,dx$ differs is only by a constant.

I was wondering if there is a proof for $F(x) + C = \int f(t)\,dt$ and the proof works for integrals that do not cannot be expressed as an elementary function (e.g. the error function (maybe))

P.S. I was thinking if this has to do with the derivatives of $F$ and integral of $f$, but if there is no upper and lower limit in the indefinite integral, we cannot use FTC. That's why I do not know how to proceed.

Thanks you!

Edit: Yes. I have forgotten to mention that $f$ should be a continuous function, in order to make things more neat and to apply FTC. However, I very much thank you all for the responses, especially those who talked about discontinuous functions as well. Cheers.

答:
Here is the full Monty, i.e., the overly pedantic answer to the question extracted from the discussion with @ParamanandSingh whose comments point us in the right direction.

The trouble with calculus courses and textbooks (as opposed to real analyis courses and textbooks) is that the goal is to get the students able to work computationally with derivatives and integrals without necessarily a thorough understanding of what they are doing or what the rigorous foundations really are.

Here are some details that should be known relative to this question.

If  $f:(a,b)\to \mathbb R$ is a function defined on an open interval  $(a,b)$  then the statement
$$\int  f(x)\,dx = F(x) + C $$
means that   $F'(x)=f(x) $  for all  $x $  in the open interval  $(a,b)$  and that $C$ represents an arbitrary constant.

If this is true then  $f$ is said to be a derivative on $ (a,b) $  and  $F$ is said to be its primitive (or its antiderivative or an indefinite integral).

It is essential that the interval in question is mentioned.  Calculus textbooks seldom do however.  Calculus questions and answers seldom do.

The Fundamental Theorem of the Calculus (FTC)  as it is usually taught
asserts that if   $f:(a,b)\to \mathbb R$ is continuous at every point of the open interval $ (a,b) $ then

(i)    $f$  is indeed a derivative on  $(a,b)$.
(ii)   $f$ is Riemann integrable on any interval  $[s,t]\subset (a,b)$.
(ii)    Pick any  $a<c<b$  and then the function
$$      F(x) = \int_c^x f(t)\,dt   \tag{1}$$
is a primitive for   $f$  on the interval $ (a,b)$.  Here the integral is interpreted in the Riemann sense.

This formula (1)  always gives a primitive for a continuous  $f$ but in special cases, as every calculus student discovers, there are techniques (integration by parts, by substitution, etc.)  that allow you to write an explicit formula for   $F(x)$.   Those techniques can fail even for very simple functions $f$.   That doesn't mean there is no primitive or indefinite integral, it just means that (1) or some series or limit etc. will be your only choice for a formula.

This formula (1)  offers a constructive method (the method of Riemann integration) for determing a primitive from a continuous function.   In theory then we always can work with primitives of continuous functions, even if no series or formula etc. is availble.

Now let's drop continuity.   A function  $f:(a,b)\to \mathbb R$ is given.  How will we know it is a derivative, i.e., how will we know if it has an indefinite integral on this interval?

Maybe, with luck you can construct a function   $F(x) $  and check at each point that   $F'(x)=f(x)$.

If not then maybe look for some other properties that   $f$ has that will imply it is a derivative.    The problem in this generality has no current satisfactory solution.   There are special cases, but even these are not elementary.

So leaving the world of continuous functions the general problem of determining conditions under which a function is or is not a derivative is not elementary and not particularly accessible.

Let us suppose, however, that we do discover that   $f$ is a derivative on  $(a,b)$,  but we do not have a procedure yet to find its primitive.  We know a primitive exists here but how to   express it?

Can we still write
$$     \int f(x)\,dx  = \int_c^x f(t)\,dt  + C ?   \tag{2}$$

(i)  If   $f $  is known to be a derivative on $(a,b)$ and if  $f $ is Riemann integrable
on each interval   $[s,t]\subset (a,b)$  then  (2)  is valid.  (Easy calculus proof.)

(ii)  If   $f$   is known to be a derivative on $(a,b)$ and if  $f $ is  Lebesgue integrable
on each interval  $ [s,t]\subset (a,b) $ then  (2)  is valid.  (Real analysis level proof.)

(iii)   If   $f$   is known to be a derivative on $(a,b)$  then $ f$  must be   Denjoy-Perron-Henstock-Kurzweil integrable
on each interval   $[s,t]\subset (a,b)$  and (2)  is valid.  (Real analysis level proof.)

In case (i) there is a simple constructive process (due to Cauchy and Riemann)  for finding  a primitive   from $f$.

In case (ii) there is the constructive method of Lebesgue for   finding a primitive from $f$.

In case (iii)  the methods of Perron and Henstock-Kurzweil are not constructive.   The method of Denjoy is constructive but it is way too sophisticated to even sketch here.

Short answer:  For continuous functions FTC does all the work and the story is very nice and straightforward.  For discontinuous functions---a real can of worms.


问:
Negation of an almost surely statement
The paper am reading proves a statement of the following form:

For all $\delta>0$, $$  X<\delta \hspace{1cm} \text{almost surely}$$

where $X$ is real-valued function on some probability space $(\Omega,\mathcal{F},P)$. The proof is by contradiction. It begins by saying:

Suppose there exist a set $A\in \mathcal{F}$  with $P(A) > 0$, and a constant $\delta>0$, such that
$$ X(ω) ≥ δ $$
for any $\omega \in A$.

The proof then proceeds by showing that this assumption leads to a contradiction. But I don't see why the second statement is the negation of the original one.

Any ideas?

答:
Suppose that the set $\{X<\delta\}$ is measurable. Then $X<\delta$ almost surely means that $\mathbb P(\{X<\delta\})=1$, or equivalently $\mathbb P(\{X\ge\delta\})=0$. Its negation is therefore $\mathbb P(\{X\ge\delta\})>0$, or equivalently there exists a measurable set $A\subset\{X\ge\delta\}$ such that $\mathbb P(A)>0$.

Isn't $X$ a random variable? In that case, $\{X<\delta\}$ is measurable. If $\{X<\delta\}$ is not measurable, then your proof is not correct.

Take for example $(\Omega,\mathcal F)=([0,1],\{\emptyset,[0,1]\})$ and $\mathbb P(\emptyset)=0$, $\mathbb P([0,1])=1$. Let $X:x\in[0,1]\mapsto x$ and $\delta=\frac12$. Then $\{X<\delta\}=[0,\frac12)$, so we do not have that $X<\delta$ almost surely. We do not have either the existence of a measurable set $A\subset\{X\ge\delta\}=[\frac12,1]$ such that $\mathbb P(A)>0$.

Title: Negation of an almost surely statement


问:
Degree of a smooth proper function, smooth homotopy and their pullbacks
Let $M$ and $N$ be compact manifolds, and $F, G: M \rightarrow N$ two smooth maps such that $F \sim G$, where $\sim$ means smooth homotopic. Then a well-known result is that the degree of $F$ is the same as the degree of $G$.

A corollary is the following: let $F :M \rightarrow M$, then if $F \sim id_M$, where $id_M$ is the identity function on $M$, then the degree of $F$ is 1. My question is if the converse hold as well. Is it true that if we have a map from a compact manifold to itself with degree 1, then this must be homotopic to the identity?

My attempt: if we know that the degree of $F$ is 1, then we have, by definition of degree $$\int_M F^{\star}\omega = \int_M \omega$$
Hence we must have that $F^{\star}$, the pullback of $F$, is equal to the pullback of the identity. Does the fact that two functions have the same pullback imply that they are smooth homotopic? Thanks!

答:
Consider the map $F:S^1\times S^1\rightarrow S^1\times S^1$ which is the product of the degree $-1$ maps $f:S^1\rightarrow S^1$. The fundamental class of $S^1\times S^1$ is the product $\Omega=[S^1]\otimes[S^1]\in H_2(S^1\times S^1)\cong H_1(S^1)\otimes H_1(S^1)\cong\mathbb{Z}$ and we have
$$F_*\Omega=f_*[S^1]\otimes f_*[S^1]=(-[S^1])\otimes(-[S^1])=\Omega.$$
Still, $F$ is not homotopic to the identity, since it does not induce the identity on $H_1(S^1\times S^1)\cong\mathbb{Z}\oplus\mathbb{Z}$.

I'll leave it up to you to rewrite the above in terms of forms if that is how you would prefer to see it.


问:
The square is not a submanifold of $\mathbb{R}^2$
Leb $X$ be the square in $\mathbb{R}^2$

$
X = \{(x,y) \in \mathbb{R}^2 : |x| + |y| = 1\}
$

It's so easy to show that $X$ is a differentiable manifold of dimension one. But, it's not possible that $X$ be a regular submanifold of $\mathbb{R}^2$ and I dont know prove it.

答:
Expanding on C. Falcon's answer:
For anyone wondering why the derivative at the vertex must be zero; let $c:I \to X$ be a curve such that $c(0)=v$ and $c'(0) \not = 0$. Let $f:I \to \mathbb{R}^2$ be such that $f(t) = \frac{c'(t)}{|c'(t)|}$. By hypothesis, $f$ is well defined and obviously continuous. Now, since $f(t)$ is the (unit) direction of the tangent vector to the square at the point $c(t)$ then, $f(t)$ is equal to the (unit) slope of the side of the square in which $c(t)$ lives. In particular,
\begin{equation}
\lim_{t \to 0^+} f(t) \not = \lim_{t \to 0^-} f(t)
\end{equation}
A contradiction since $f$ is continuous.

P.D. I'm still a newb so any corrections or comments will be greatly appreciated.


答:
To prove the given inequality, you should recall the proof that $r(3,3)\le 6$, and try a similar strategy. For example, to prove that $r(3,3,3)\le 3\cdot (r(3,3)-1)+2=17$, consider a particular vertex, $v$ in a graph whose edges are colored red, blue and green. Since $v$ is the endpoint of $16$ edges, there must be some color for which $v$ is the endpoint of $\lceil 16/3\rceil =6$ edges of that color. WLOG the color is red, and let $S$ bet the set of vertices joined to $v$ by a red edge. There are then two cases:

If any two vertices $w$ and $u$ in $S$ are connected by a red edge, then $\{u,v,w\}$ is a red triangle.

Otherwise, $S$ is only joined together by blue and green edges. Since $6\ge r(3,3)$, there must be either a blue or red triangle.

You need to generalize that argument to all $k$. Instead of using $\lceil 16/3\rceil =6$, you will use
$$
\left\lceil (k+1)(r_k(3)-1)+1 \over k+1 \right\rceil=r_k(3)
$$

For an upper bound on $r_k(3)$, I will use a weaker form of the inequality you were tasked to prove:
$$
r_{k+1}(3)\le (k+1)r_k(3)$$
You can apply this repeatedly to get a non-recursive bound on $r_k(3)$. For example, shortening $r_k(3)$ to $r_k$:
\begin{align}
r_5
  &\le 5\cdot r_4
\\&\le 5\cdot 4\cdot r_3
\\&\le 5\cdot 4\cdot 3\cdot r_2
\\&\le 5\cdot 4\cdot 3\cdot 2\cdot r_1
\\&=   5\cdot 4\cdot 3\cdot 2\cdot 3
\\&=   5!\cdot 3
\end{align}
Generalizing this pattern, you can prove $r_k(3)\le 3\cdot k!$. You can get a better upper bound by using the full inequality instead of the simplified, but it will be messier.

Title: Prove that $r_{k+1}(3) \leq (k+1)( r_k(3) - 1) + 2$

Body:
I'm tasked to prove that $r_{k+1}(3) \leq (k+1) (r_k(3) - 1) + 2$ and then use the result to come up with an upper bound for $r_n(3)$.

I'd appreciate a bit of a walkthrough, as I've had a hard time with this.


问:
Construct the vertex of an angle given three points on the plane.
Suppose you are given three points A,B,C on the plane. A is on the bisector of angle XOY, B is on the side OX and C is on the side OY. How can you find the locus of the vertices(find all possile points O)?

答:
Taking $P=(x,y)$, the locus of $P$ is given by
$$\cos\angle APB=\cos\angle APC\quad\to\quad\frac{\overrightarrow{PA}\cdot\overrightarrow{PB}}{|\overrightarrow{PA}||\overrightarrow{PB}|} = \frac{\overrightarrow{PA}\cdot\overrightarrow{PC}}{|\overrightarrow{PA}||\overrightarrow{PC}|}$$ For the specific case of $A=(a_x,a_y)$, $B=(-d,0)$, $C=(d,0)$ the equation becomes

$$\left(\,x (x - a_x) + y (y - a_y)\,\right) \left(\,a_y (x - a_x) - a_x (y - a_y)\,\right) + d^2 (x - a_x) (y - a_y) = 0\tag{$\star$}$$

representing a cubic curve passing through $A$ (twice), $B$, and $C$:

When $A$ lies on the perpendicular bisector of $\overline{AB}$ (ie, when $a_x=0$ in $(\star)$), the equation factors

$$x \left(\;a_y \left(x^2 + y^2 - d^2\right) - y \left(a_y^2 - d^2\right)\;\right) = 0 \tag{$\star'$}$$

so that the locus is the union of that perpendicular bisector with the circumcircle of $\triangle ABC$:


问:
First group cohomology and composition factors
Let $G$ be a finite group. Let $k$ be a field ($\text{char}(k)=p>0$).
Let $P(k)$ be the projective cover of $k$. Assume that for any nontrivial simple $kG$-module $M$ we have $H^1(G,M)=0$. Does it mean that all composition factors of $P(k)$ are trivial?</p>

答:
Yes. $H^1(G,M) = \operatorname{Ext}_{kG}^1(k,M)$ describes how many ways $M$ can be a submodule of $N$ with $N/M \cong k$. $P(k)$ can only have composition factors that are “connected” to $k$ by paths of nonzero $\operatorname{Ext}_{kG}^1$. We prove this by induction on the Loewy length, that is, by considering the semisimple modules $J^n P(k) / J^{n+1} P(k)$ and their extensions $J^{n-1} P(k) / J^{n+1} P(k)$ where $J$ is the Jacobson radical of $kG$.

Lemma: $P(k) / J^2 P(k)$ only has trivial composition factors.

Proof: Let $Q$ be a maximal submodule of $J P(k)$ setting $M=J P(k)/Q$ to be the simple quotient module. Consider the short exact sequence $0 \to M \to P(k)/Q \to k \to 0$. Assume by way of contradiction that $M$ is not the trivial (central/principal) module $k$. Since $\operatorname{Ext}_{kG}^1(k,M) = 0$, the sequence splits. However, $J P(k)/Q$ is superfluous in $P(k)/Q$, so never has a supplement, much less a complement. This contradiction shows $M=k$. $\square$

Proposition: $P(k)$ only has trivial composition factors.

Proof: Suppose $P(k)/J^n P(k)$ only has composition factors isomorphic to $k$, and let $R$ be a maximal submodule of $J^n P(k)$. Set $M=J^n P(k)/R$ to be the simple quotient, and assume by way of contradiction that $M$ is not the trivial (central/principal) module $k$. Consider $$0 \to M \to J^{n-1} P(k)/R \to J^{n-1} P(k)/ J^n P(k) \to 0$$ and notice $J^{n-1} P(k)/ J^n P(k) \cong k^m$ for some positive integer $m$. Hence $$\operatorname{Ext}_{kG}^1(J^{n-1} P(k)/ J^n P(k),M) = \operatorname{Ext}_{kG}^1(k^m, M) = \operatorname{Ext}_{kG}^1(k,M)^m = 0^m = 0$$ so the sequence splits. However, $M = J^n P(k) / R = J ( J^{n-1} P(k)/R )$ is superfluous, so not a direct summand. This contradiction shows that in fact $M$ must be isomorphic to $k$. Since $J^n P(k) / J^{n+1} P(k)$ is isomorphic to the direct sum of such $M$, the induction proceeds to show that $J^n P(k) / J^{n+1} P(k)$ and hence $P(k) / J^{n+1} P(k)$ only has trivial composition factors. Since $P(k)$ has finite Loewy length, $P(k)$ itself only has trivial composition factors. $\square$

The more intuitive method view $P(k)$ made up of layers. The top layer is a single copy of $k$, and the second layer is proven in the lemma to only have copies of $k$ again. What can happen in the third layer? Well the third layer of $P(k)$ is actually the second layer of $J (k)$ and since $J P(k)$ is a homomorphic image of $P(k)^m$, its second layer also only has copies of $k$. Thinking of this in terms of the original copy of $P(k)$, we see the third layer of $P(k)$ also only has trivial composition factors. In other words the $n$th layer of $P(k)$ is the second layer of $J^{n-2} P(k)$, and if $J^{n-2} P(k)$ only has trivial top composition factors, then it is a homomorphic image of $P(k)^m$, etc.

This kind of thing is uncommon in finite groups. I believe this is only the case when the principal block contains only one simple module. I think this is the case when $G/O_{p'}(G)$ is a $p$-group, but I don't have my references around to double check.


问:
Arithmetic Mean/Geometric Mean Inequality
We have to use the arithmetic mean/geometric mean inequality.
Let $a$ and $b$ be fixed positive numbers. Find the value of $x>0$ that minimizes the given expression and determine the minimum value of the expression,
$ax+\frac{b}{x}$.

I don’t know how to get started by transforming the expression to be able to use the mean inequality. We were told not to solve using calculus.

答:
The $AM-GM$ inequality tells us that for non-negative $a$ and $b$ we have

$$\frac{a+b}{2}\geq \sqrt{ab}$$

Since $x>0$ then $ax>0$ and by $AM-GM$

$$\frac{ax+\frac{b}{x}}{2}\geq\sqrt{ax\frac{b}{x}}$$

Or $$ax+\frac{b}{x}\geq2\sqrt{ab}$$

with equality when $ax+\frac{b}{x}=2\sqrt{ab}$. Can you end it now?


问:
Prove that $(z^a)^b = z^{ab}e^{2kb\pi i}$
I'm trying to prove that
$$
(z^a)^b = z^{ab}e^{2kb\pi i},
$$
where $a,b,z\in\mathbb{C}$ and $k\in\mathbb{Z}$.

So far, I have that
\begin{align*}
(z^a)^b & = (\exp(a\log z))^b \\
        & = \exp(b\log(\exp(a\log z))),
\end{align*}
and I know that the complex logarithm can be defined such that $\log z=\log|z|+i\theta+2k\pi i$, where $\theta\in(-\pi,\pi]$ is the principal argument of $z$.

I feel like I've got all the components I need, but I'm not quite sure how to piece it together to get the desired result.

答: Instead use $(z^a)^b=(e^{a\ln z+2k\pi i})^b=e^{ab\ln z+2kb\pi i}=e^{ab\ln z}e^{2kb\pi i}=z^{ab}e^{2kb\pi i}$.


问:
How to find the length of the projection of $v$ onto $u$ without knowing the dot product formula?
Let's say I have two $2$-D vectors, $v$ and $u$. I want to find the length of the projection,$P$, of $v$ onto $u$ as depicted by the red arrow in the figure below:

Now I know this problem becomes trivial if I use the dot product. However, I'm still sort of struggling to understand the dot product so I want to see if I can arrive to the answer without using it (and maybe the dot product will pop out along the way).

I've been trying to use the Law of Cosines as my starting idea:
$$\vert\vert{u-v}\vert\vert^2 = \vert\vert{u}\vert\vert^2 + \vert\vert{v}\vert\vert^2 -2\vert\vert{u}\vert\vert*\vert\vert{v}\vert\vert\cos{\theta}$$

I also know that:

$$cos{\theta} = \frac{P}{\vert\vert{v}\vert\vert}$$

Therefore:

$$P = \frac{\vert\vert{u}\vert\vert^2 + \vert\vert{v}\vert\vert^2 - \vert\vert{u-v}\vert\vert^2}{2\vert\vert{u}\vert\vert}$$

答:
Let the coordinates for $\mathbf{u}$ and $\mathbf{v}$ be $(u_x, u_y)$ and $(v_x, v_y)$ respectively. We can derive the expression for $\operatorname{cos}\theta$ using the coordinates. We need some trigonometric identity along the way.

As we can see in the illustration, the angle between $\mathbf{u}$ and the x-axis is $\alpha$, and the angle between $\mathbf{v}$ and $\mathbf{u}$ is $\theta$. We have

$$
\operatorname{tan}\alpha = \frac{u_y}{u_x},\;
\operatorname{tan}(\alpha + \theta) = \frac{v_y}{v_x}.
$$

We need to the trig identity to solve for $\operatorname{tan}\theta$

$$
\operatorname{tan}(a - b) = \frac{\operatorname{tan}(a) - \operatorname{tan}(b)}{1+\operatorname{tan}(a) \operatorname{tan}(b)}  .
$$

To solve for $\operatorname{tan}\theta$
\begin{align}
\operatorname{tan}\theta = \operatorname{tan}(\alpha + \theta - \alpha)
&= \frac{\frac{v_y}{v_x} - \frac{u_y}{u_x}}{1 + \frac{u_yv_y}{u_xv_x}}\\
&= \frac{u_xv_y - u_yv_x}{u_xv_x + u_yv_y}.
\end{align}

From $\operatorname{tan}\theta$, we can easily derive $\operatorname{cos}\theta$,

$$ \operatorname{cos}\theta = \frac{u_xv_x + u_yv_y}{u_x^2v_x^2 + u_x^2v_y^2 + u_y^2v_x^2 + u_y^2v_y^2}.$$

With the formula for the projection

$$ p = ||\mathbf{v}|| \operatorname{cos} \theta,$$

we have

$$
p = \frac{u_xv_x + u_yv_y}{u_x^2v_x^2 + u_x^2v_y^2 + u_y^2v_x^2 + u_y^2v_y^2} (v_x^2 + v_y^2)
=  \frac{u_xv_x + u_yv_y}{u_x^2 + u_y^2}.
$$

So without using the dot-product, we can derive the expression of the projection using the vector coordinates. However, we can easily see that the numerator is the dot product and the denominator the norm of $\mathbf{u}$, i.e.,

$$
p = \frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{u}||}
$$

So indirectly, we end up with a form which shows that the dot-product indeed works. :)


问:
Intuitive way to memorize certain points about the jocobian transformation of a random variable?
Suppose we have $X$ and $Y$ two random variables and $U$=$f_1(X,Y)$ and $V$=$f_2(X,Y)$.

Let's suppose we have the two or one of these densities $f_{(X,Y)}$ and $f_{(U,V)}$. What I don't get is when trying one in function of the other is :

I sometimes have trouble understanding where to multiply the determinant but what I got is that when we are calculating the jacobian determinant's elements if I derive with respect to $x$ and $y$ then the equality is $f_{(X,Y)}$=$|Je|$$f_{(U,V)}$ and if I derive with respect to $u$ and $v$ then the equality $f_{(U,V)}$=$|Je|$$f_{(X,Y)}$ ???

Suppose that the two functions $f_{(X,Y)}$ and $f_{(U,V)}$  don't map to the same space. When writing the double integrals when do I change the interval I'm integrating in?

I just want an intuitive way to understand these things so I could memorize them and never forget them.

答:
Jacobian determinants admit the intuitive notation $$\frac{\partial u\partial v}{\partial x\partial y}:=\frac{dudv}{dxdy}=\left|\begin{array}{cc}
\tfrac{\partial u}{\partial x} & \tfrac{\partial v}{\partial x}\\
\tfrac{\partial u}{\partial y} & \tfrac{\partial v}{\partial y}
\end{array}\right|.$$Then $dudv=\frac{\partial u\partial v}{\partial x\partial y}dxdy$ is a multivariant extension of the univariate chain rule $du=\frac{\partial u}{\partial x}dx$.


问:
A question about Vectors
Consider a vector $a = (x,y)$.

I know that this notation $|a|$ means the magnitude of the vector which is calculated by $\sqrt{x^2 + y^2}$.

But what does this notation $\|a\|$ mean?

Thank you.

答:
Like much of notation, it could depend on your instructor's personal style.

But, in general, the mapping $v\mapsto||v||$ refers to an arbitrary norm on the space.  The go-to example for norms is magnitude or absolute value, so usually we refer to it as $|v|$.  But if the problem you're talking about had a different norm function under consideration, that could explain why there was both.

Or it was a typo.

Title: A question about Vectors


问:
Proof that $\binom{2N-1}{N} = \sum_{j=0}^{N-1} \binom{N+j-1}{j}$
This is part of a stochastic excercise. The sample solution states $$\binom{2N-1}{N}$$, but my solution is $$\sum_{j=0}^{N-1} \binom{N+j-1}{j}$$.
I calculated both for some N, like N=2, N=5 and both solutions provided the same result. Acutally I am pretty sure that my solution is equal to the one given, but I am unable to proof it. The main struggle is that $j$ is part of both entries of the binomial coefficient, what makes it hard to use the defintion of the binomial coefficient with the factorial.
Like I said, it's not like I know that this is true, but I am pretty sure. But if someone got an counterexample or can proof that this is not the same I would be thankful too.

答:
Let $ n\in\mathbb{N}^{*} $, note that, $ \left(\forall j\in\mathbb{N}\right) $, we have : \begin{aligned}\binom{n-1+j}{j}&=\binom{n-1+j}{n-1}\\ &=\binom{n+j}{n}-\binom{n-1-j}{n}\end{aligned}

Thus : \begin{aligned} \sum_{j=0}^{n-1}{\binom{n-1+j}{j}}&=\sum_{j=0}^{n-1}{\left[\binom{n+j}{n}-\binom{n-1-j}{n}\right]}\\ &=\binom{2n-1}{n} \end{aligned}


问:
Dihedral group $D_4$ (symmetries of a square) is isomorphic to a subgroup of $S_4$ (permutation group)
I have this problem that I have been stuck on for a while. I know that I can write the elements of $D_4$ as products of the identity, rotation by $90$ degrees and a vertical reflection and I have already written the $8$ elements of $D_4$ as their cycle notation is $S_4$ but have no idea what to do from there and would really appreciate the help!

答: Hint: Label the corners of a square. Permute them.


问:
If $v_1, v_2, ..., v_n$ are linearly independent and so are $w_1, w_2, ..., w_n$, what can we say about $v_1+w_1, v_2+w_2, ..., v_n+w_n$?
As the title says, what can we say about $v_1+w_1, v_2+w_2, ..., v_n+w_n$ if $v_1, v_2, ..., v_n$ and $w_1, w_2, ..., w_n$ are linearly independent vectors of $R^n$?

I've tried approaching this problem by observing that $dimV=dimW=n$, where $V=[v_1, v_2, ..., v_n]$ and $W=[w_1, w_2, ..., w_n]$. Then $v_1+w_1, v_2+w_2, ..., v_n+w_n$ are linearly independent if and only if $dim(V+W)=n$. It follows that $dim(V \cap W)=n$, but then I get stuck.

答:
Hint.

If $v_k$ are LI then they form a basis for $\mathbb{R}^n$ and then

$$
w_k = \sum_j a_j^k w_j
$$

or

$$
W = A V
$$

Now $V+W = (A+I_n)V$ so if $A+I_n$ is rank $n$ then $V+W$ are a LI set. Note that it is well possible that  $\text{rank}(A+I_n)< n$ As an example take $A = -I_n$


问:
Does $\sum_{n=1}^{\infty}(-1)^{n}\cos(\frac{\pi}{n})$ converges?
My attempt:

I used the fact that if $\lim_{n \rightarrow \infty} a_{n} \neq 0 \Rightarrow \sum_{n=1}^{\infty} a_{n}$ doesn´t converges. So, I checked that

\begin{align*}
\lim_{n \rightarrow \infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right |=1
\end{align*}
i.e.,
\begin{align*}
\lim_{n \rightarrow \infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right |\neq0 \Longrightarrow \sum_{n=1}^{\infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right | \text{does not converges}
\end{align*}

My doubt:

Is this implication correct?
\begin{align}
\sum_{n=1}^{\infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right | \text{does not converges} \Longrightarrow \sum_{n=1}^{\infty} (-1)^{n}\cos(\frac{\pi}{n})  \text{ does not converges}
\end{align}

答: Since $ \lim\limits_{n\to +\infty}{\left(-1\right)^{n}\cos{\left(\frac{\pi}{n}\right)}} $ is not $ 0 $, then your series diverges.


问:
Solutions to a linear equation over a subfield of a finite field
I know that if we have a line defined over a finite field $\mathbb{F}_p=K$ then this line has p+1 solutions in $P^2(K)$, $p^2+1$ in $P^2(\mathbb{F}_{p^2})$ and so on. But what about the opposite? If the line is defined over $\mathbb{F}_{p^2}$ how many solutions does it have in $P^2(K)$?

答: A line in $P^2$ is defined by an equation $ax+by+cz=0$ where $a,b,c$ are not all $0$.  The set of solutions $(x,y,z)\in K^3$ to such an equation is just a vector subspace of $K^3$ whose dimension is $3-d$ where $d$ is the dimension of the span of $\{a,b,c\}$ as a vector space over $K$.  The solutions in $P^2(K)$ are the just the projectivization of this vector subspace, which has $\frac{p^{3-d}-1}{p-1}$ elements.  In the case where $a,b,c\in\mathbb{F}_{p^2}$, this $d$ can only be $1$ or $2$.  In the case $d=1$, you get $p+1$ elements in $P^2(K)$ and the line is actually defined over $K$ (any two of $a,b,c$ are linearly dependent over $K$, so you can rescale to make one of them $1$ and then the rest are also in $K$).  In the case $d=2$, you get just $1$ element on $P^2(K)$.


问:
Finding hidden relations between objects with minimum effort
A room has 100 light bulbs and there is a switch panel outside the room with 100 switches each one controlling one light bulb. The switches are not labelled so it’s not clear which switch controls which bulb.

Assuming that that the status of the light bulbs can only be verified by entering the room (one trip to check the status of all 100 light bulbs), what’s the minimum number of trips into the room needed to label each switch with the corresponding light bulb?

I’ve tried setting half of the set in one state and the other half in the opposite state, such as in a binary system. Then you enter the room and check the status. You then divide the two groups into two further subgroups with opposite states, enter the room again and so forth. This gives 7 total trips. But I am not sure if it’s the minimum.

How do one go about proving the minimum number?
Also, what’s the minimum number of trips for n light bulbs?

答: Your binary strategy works, and is optimal. Suppose you entered the room only six times. For each switch, there is a vector $(s_1,\dots,s_6)$, where for each $i\in\{1,\dots,6\}$, $s_i$ was the state of that switch when you entered the room for the $i^{th}$ time. Since there are $100$ light bulbs and $64$ possible state vectors, there would have to exist two switches with the same state vector. But that means you always did the same thing to those switches, so you cannot distinguish them.


问:
Generating valid inequalities to describe the convex hull of a set
How to generate the valid inequalities needed to describe the below sets' convex hull?

```<code>S := {(x, y) : x + y ≥ b, x ≥ 0 integer, y ≥ 0}

S := {(x, y) : x + y ≥ b, x ≥ d integer}
</code>```

What are the general guidelines to follow in such cases?

答:
The general guideline for two variables is to plot the region.

Because $b$ is not integer, the convex hull of the first one is slightly more complicated than in the comment from @Kavi; there is one more inequality related to $(x,y)=(\lfloor b \rfloor, b - \lfloor b \rfloor)$ and $(x,y)=(\lceil b \rceil, 0)$.

The second one has a couple of cases depending on $b \le d$ or $b > d$.


问:
Prove that the sequence defined as $x_{n+1}=\frac{2x_n+5}{6}, x_0=\frac{1}{2}$ converges
Prove that the sequence $(x_n)$ defined as $x_{n+1}=\frac{2x_n+5}{6}, x_0=\frac{1}{2}$ converges.
I would like to have a feedback if my proof is correct, please)

To show that $(x_n)$ converges, we have to show that it is bounded and monotonic.

First we show by induction that the sequence is increasing. Suppose $x_{n+1}>x_n$. The previous inequality is true for $n=0$, because $1/2<1$. So we can suppose that the inequality holds $\forall n\ge0$. Now we have to show that it works for $n+1$. So, we would like to show that $x_{n+2}>x_{n+1} \ \forall n\ge 0$. But by our induction hypothesis we have that:

$x_{n+1}>x_n \iff 2x_{n+1}>2x_n \iff \dfrac{2x_{n+1}+5}{6}>\dfrac{2x_n+5}{6} \iff x_{n+2}>x_{n+1}$

So we conclued that $x_{n+2}>x_{n+1}$ is true $\forall n\ge0$

We show right now that $(x_n)$ is bounded above. Suppose that $x_n<2$. The inequality holds for $n=0$, so we can suppose that $x_n<2$ is true $\forall n\ge0$. We want to show that it is true for $n+1$, i.e $x_{n+1}<2$. By induction hypothesis we have:

$x_n<2 \iff 2x_n<4 \iff 2x_n+5<9 \iff \frac{2x_n+5}{6}<\frac{9}{6}<2 \iff x_{n+1}<2 \ \forall n\ge0$

as wanted.

So we showed that the sequence $(x_n)$ is convergent.

答:
If $ \{x_n\}_{n\in\mathbb{N}} $ converges to $ L $, then

$$L=\frac{2L+5}{6} \iff L=\frac 54$$

$$x_{n+1}-\frac 54=\frac{2x_n+5}{6}-\frac 54$$
$$=\frac 13\left(x_n-\frac 54\right)$$
$$=\left(\frac 13\right)^2\left(x_{n-1}-\frac 54\right)$$
$$=\left(\frac 13\right)^{n+1}\left(x_0-\frac 54\right)$$
$$=-\frac 14\left(\frac 13\right)^n$$

the limit is $L=\frac 54$.


问:
How to minimize exposed surface for half a pie?
I bought a large pie (of radius $R$). I cut off a half and gave it to my friend. This exposed an area or $2Rh$ --  where $h$ is the pie's thickness -- to air.  I watched one Numberphile video too many, and now I want to rearrange my half of the pie so that the exposed area is minimal. I can get $\sqrt{2}Rh$ like this:

Main question: Can I do better?

Alternative question:  What if I had this in mind already when I was cutting off my half?

That is, can I divide a pie into two equivalent parts such that one (and thus also the other) of the remaining parts can be rearranged to have even smaller exposed area?
(Here "equivalent" means "such that they can be in turn subdivided  into finite number of subparts which can be matched into equal pairs", equal, in turn,  being "movable to each other, with the motion matching up the pieces inherited from the original boundary of the pie".)

My Osgood-Denjoy–Riesz pie-cutting knife is being resharpened, so piecewise-smooth cuts only, please.

答:
Great question!

I will answer the two-dimensional case, where your cake is simply a disk of unit radius and you wish to minimize the perimeter not originally part of the disk's border. This encompasses all dissections performed with vertical cuts and rearrangements of the pieces on a flat surface.  (I'm normalizing to radius $1$, since the problem is obviously invariant up to scaling.)

By the isoperimetric inequality, any shape of area $\frac{\pi}2$ must have a perimeter of at least $\pi\cdot\sqrt{2}$, which is less than the length-$\pi$ amount of side frosting you have available. So the best we can hope for is $(\sqrt{2}-1)\pi\approx 1.301$ unfrosted perimeter; your $\sqrt{2}$ solution is already pretty close. (Note that this argument doesn't rely on anything other than having $\pi$ perimeter available to us, so making weirder cuts won't improve on this bound.)

However, it turns out that we can get as close to this limit as we would like. The idea is to make (something very close to) a circle of radius $1/\sqrt{2}$, and ensure that we use the frosted border to cover as much of this as we can. I will handwave some of these arguments, because I think excessive rigor would just muddy the intuitive idea, but you could formalize this with a lot of epsilon-delta hand-wringing and gross trigonometry if you really wanted to.

Our strategy is as follows:

Pick some very large $N$, and place $N$ points equally spaced on the semicircular frosted boundary. Cut off the frosted bits by cutting along these $N$ points, as shown below for $N=7$ (removed regions and frosting in red):

Build a convex polygon with $N-1$ sides that match the length of the cuts in the previous step, and some finite number of other sides ($O(N)$ more certainly suffice), arranged so that the result looks very close to a circle of radius $\pi/\sqrt{2}$ (with correspondingly similar perimeter) and has the same area as the grey region above. (You have more than enough degrees of freedom to make this happen easily - let me know if I should elaborate more on this point.)

Apply the Wallace–Bolyai–Gerwien theorem to turn the grey region into our new circular-ish polygon using finitely many pieces. (Note that because the theorem only uses translation and rotation of the pieces, we don't have to worry about turning a piece upside down and smearing frosting on the floor.)

Glue the red bits back onto our $N-1$ free sides, covering up (in the limit) $\pi$ of our perimeter and leaving (in the limit) only $(\sqrt{2}-1)\pi$ uncovered.

So if we restrict ourselves to making vertical cuts in the cake, the minimum unfrosted area is $(\sqrt{2}-1)\pi Rh$.

In the 3-dimensional case, things will depend on the height of the cake. If $h$ is very small, you can do some silly things: for instance, cut little beveled squares out of the top frosted part of the cake, then put $6$ such frosted squares together in a cube and put the leftovers inside the cube; they will at least not see the "outside air", though perhaps this is considered cheating. (I suspect one can use similar constructions to still have arbitrarily little surface area anywhere, though.)

Using the isoperimetric inequality in 3 dimensions for surface area, we can work out that with a radius of $1$ and a height of $h$, we will need an excess area of at least $\pi(3h)^{2/3}-\pi(h+1)$, which is positive between $x\approx 0.81521$ and $x\approx 5.4115$ (this is assuming the bottom of the cake is frosted; if not, the range of such $x$ would be a bit larger). So perfection cannot be achieved at least for medium-tall cakes.

In general, trying to prove any more complex lower bounds on $3D$ solutions seems pretty complicated.


问:
Is there a way to better write chains of multivariate functions?
In a single variable, if $f, g, h$ are functions with compatible domains and codomains, we use the notation $f \circ g \circ h$ to have a more readable version of $f(g(h(x)))$.

Now consider for example the function $S(X,Y,Z)$ where $X(u,v), Y(u,v), Z(u,v)$ are functions of 2 variables and $u,v$ are functions of the variable $t$. Writing the full composition of $S$ in terms of $t$ would involve a lot of parenthesis and is kinda hard to read.

Is there a similar notation convention to chain multivariate functions as there is for univariate ones?

答:
A shorter notation I have seen is $S(X(\vec x),Y(\vec x),Z(\vec x))$. If you don't want to name the variables, there's also $S(X(-),Y(-),Z(-))$ (or replace $-$ with $\cdot$), even though I don't know if that would be good enough.

It feels (to me) that $S\circ(X,Y,Z)$ should be valid, but I have not seen many such usages.


问:
The columns of AB are linearly dependent. Are the columns of B linearly dependent? show why or why not this would be true.
The columns of AB are linearly dependent. Are the columns of B linearly dependent? show why or why not this would be true or provide an example.</p>

答: Assuming $A$ and $B$ are matrices, no because if $A$ is the zero matrix, and $B$ is any invertible matrix, $AB=0$ and therefore has linearly dependent columns but $B$ has linearly independent columns by Invertible Matrix Theorem.


问:
Prove: Every permutation matrix is the product of elementary matrices of interchanges (no replacement or scaling)
I know that a permutation matrix is made up of standard basis vectors, but I'm not exactly sure how to put this in words.</p>

答:
Hint

The symmetric group is generated by the transpositions.  Then use the homomorphism between the permutation matrices and the symmetric group.


问:
Direct Sum Question
Let $U$ and $W$ be 3-dimensional subspaces of a 4-dimensional vector space $V$, and suppose that $U\neq W$ .  Is $V$ the direct sum of $U$ and $W$?

答:
No, that's not possible.

Note that for a finite number of factors, the direct sum is essentially the same as the product.

(Here, the direct sum $U\oplus W$ is not isomorphic to a subspace of $V$. The normal sum $U+W$ would be equal to $V$, but since $U\cap V \neq 0$  the sum is not a direct sum.)

Title: Direct Sum Question


答: As you have $\dim U \cap W =2$, $U+W$ can’t be a direct sum.

Title: Direct Sum Question

Body: Let $U$ and $W$ be 3-dimensional subspaces of a 4-dimensional vector space $V$, and suppose that $U\neq W$ .  Is $V$ the direct sum of $U$ and $W$?


问:
Solve the inequality $\frac{1}{2}(e^x+e^{-x}) \leq e^{cx^2}$ with series
The inequality I have to solve is: find $c\in\mathbb{R}$ such that the inequality $\frac{1}{2}(e^x+e^{-x}) \leq e^{cx^2}$ is verified for all $x\in\mathbb{R}$.

My exercises book suggests to use the series expansion, this is my attempt: I've written $\frac{1}{2}(e^x+e^{-x})=\cosh x$ and I've expanded both sides:
$$\cosh x \leq e^{cx^2} \iff \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} \leq \sum_{k=0}^\infty \frac{(cx^2)^k}{k!} \iff \sum_{k=0}^\infty \frac{c^k x^{2k}}{k!}-\sum_{k=0}^\infty \frac{x^{2k}}{2^k k!} \geq 0$$
$$\iff \sum_{k=0}^\infty \left(c^k-\frac{1}{2^k}\right) \frac{x^{2k}}{k!}\geq0$$
Since $\frac{x^{2k}}{k!} \geq 0$ for all $x\in\mathbb{R}$, the only possibility for the series to be positive is if it is
$$c^k-\frac{1}{2^k} \geq 0 \iff c^k \geq \frac{1}{2^k} \iff |c| \geq \frac{1}{2} \iff c \geq \frac{1}{2} \lor c \leq -\frac{1}{2}$$
However the answer in the book considers only $c \geq \frac{1}{2}$; I've considered $|c|$ for even $k$, where am I wrong? Is it somehow related to the fact that for $c \leq -\frac{1}{2}$ I am summing negative numbers and so there is some problems in the inequality resolution and so I must impose $c \geq 0$ as well? Thanks to anyone who will answer.

答:
your problem arises from the conclusion that $c^k\geq\frac{1}{2^k}$ imples that $|c|\geq\frac{1}{2}$ for take $c=2$ then using your conclusion we get $2\leq-\frac{1}{2}$ which is wrong.
on the other hand, notice that the right hand side of the inequality is positive, hence the left hand side being greater than the right, it is always positive and therefore $c$ cannot be negative, so that $c\geq\frac{1}{2}$ $$.
$$
in general $c^k\leq\frac{1}{2^k}$ imples that $|c|\leq\frac{1}{2}$, not the other way around.


问:
Why continuum hypothesis implies the unique hyperreal system, ${}^{\ast}{\Bbb R}$?
On page 33, Robert Goldblatt, Lectures on Hyperreals(1998):

Now it has been shown under certain set-theoretic assumption called continuum hypothesis the choice of $\mathcal F$ is irrelevant: All quotients of $\Bbb {R}^{\Bbb N}$ with respect to nonprincipal ultrafilters on $\Bbb N$ are isomorphic as ordered fields.

I don't know why this is true. It seems to me if we want to prove  continuum hypothesis implies the unique hyperreal system, we have to show that, for any two free ultrafilters $\mathcal {F}_{1}$ and $\mathcal {F}_{2}$ on $\Bbb N$, there is an bijection $f: \Bbb N \to \Bbb N$ such that there it's also a bijection from $\mathcal {F}_{1}$ to $\mathcal {F}_{2}$. How to show this holds with $\bf CH$ and fails without it.

答:
This is not an answer; it’s just showing that your approach can’t work.

The Rudin-Keisler order on $\beta\omega\setminus\omega$ is defined as follows. For $p,q\in\beta\omega\setminus\omega$ write $p\le_{RK}q$ iff there is an $h:\omega\to\omega$ such that $A\in p$ iff $h^{-1}[A]\in q$ for each $A\subseteq\omega$, i.e., $q=\{h^{-1}[A]:A\in p\}$. It turns out that $p\le_{RK}q\le_{RK}p$, written $p\equiv_{RK}q$, iff $h$ can be taken to be a bijection. Your conjecture, then, is that $\mathsf{CH}$ implies that $p\equiv_{RK}q$ for all $p,q\in\beta\omega\setminus\omega$. However, this is false. There are only $2^\omega$ bijections from $\omega$ to $\omega$, but $|\beta\omega\setminus\omega|=2^{2^\omega}>2^\omega$: with or without $\mathsf{CH}$ there aren’t enough bijections in ${}^\omega\omega$ for all free ultrafilters on $\omega$ to be Rudin-Keisler equivalent. (In fact under $\mathsf{CH}$ it’s extremely easy to construct non-equivalent ultrafilters.) Thus, the uniqueness of the ordered field $\Bbb R^\omega/p$ under $\mathsf{CH}$ must come from some other source.

If Goldblatt doesn’t cite a reference for the result, you might look for one in H.J. Keisler’s survey The hyperreal line in P. Ehrlich (ed.) Real numbers, generalizations of reals, and theories of continua, Kluwer Academic Publishers, $1994$, pp. $207$-$237$; I’ve not seen it, but I know that it at least mentions some closely related results.


问:
Why is the period of $\sec$ and $\csc 2\pi$ instead of $\pi$?
Hi and thanks in advance.

When $\tan x = \frac{(2x + 1)\pi}{2}$, the function is undefined, resulting in a period of simply $\pi$.
However, $\csc$ and $\sec$ also include undefined points. Taking this discontinuity into account, why are their periods $2\pi$ instead of $\pi$?

答:
The period of $\csc x$ is $2\pi$ because the period of $\sin x$ is $2\pi$, and $\csc x = \frac{1}{\sin x}$. It doesn't have to do with the discontinuity. The same thing holds for $\sec x$.

The period of $\tan x$ is the outlier here. Note that $\sin (x + \pi) = -\sin x$, and similarly, $\cos (x + \pi) = -\cos x $. So, after a rotation of $\pi$, neither $\sin x$ nor $\cos x$ has returned to its starting point. But, here's what happens to tangent:

$$ \tan (x+\pi) = \frac{\sin (x + \pi)}{\cos (x + \pi)} = \frac{-\sin x}{-\cos x} = \frac{\sin x}{\cos x} = \tan x . $$

The same thing happens with $\cot x$. So, while most trig functions have a period of $2\pi$, $\tan x$ and $\cot x$ have a period of $\pi$.


问:
Proving $\cos(a+b)=\cos a\cos b-\sin a\sin b$ starting from $\sin(a+b)$
The question is to prove the compound angle identity $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ starting from the $\sin$ compound angle identity. Before this, the task wants me to show that $\sin(\frac \pi 2 - x) = \cos(x)$ and I did not have any problems there. I guess I have to use this fact somehow so thats what I've tried:

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$
$$\sin\left(\frac \pi 2 - (a+b)\right)=\sin\left(\frac \pi 2\right)\cos(a+b)-\cos\left(\frac \pi 2\right)\sin(a+b)$$
$$\cos(a+b)=1\cos(a+b)-0\sin(a+b)$$
$$\cos(a+b)=\cos(a+b)$$
And that brings me nowhere. How can I prove this identity in the required way?

答:
Solution to the problem:
$$\sin((\frac \pi 2-a)-b)=\sin(\frac \pi 2 -a)\cos(b)-\cos(\frac \pi 2-a)\sin(b)$$
$$\sin((\frac \pi 2-a)-b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$
$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$


问:
Is this math game always winnable?
Kent Haines describes the game of Integer Solitaire, which I find to be excellent for young kids learning arithmetic. I'm sure they will be motivated by this game to get a lot of practice.

Kent asks a question about his game, which I find very interesting, and so I am asking here, in the hopes that Math.SE might be able to answer.

The child draws 18 cards from an ordinary deck of cards, and then regards the cards to have values Ace = 1, 2, 3, ..., Jack = 11, Queen = 12, King = 13, except that Black means a positive value and Red means a negative value.

Using 14 of the 18, the child seeks to find solutions of four equations:

For example, a successful solution would look like:

Question. Does every set of 18 cards admit a solution?

Kent Haines says, "I have no idea whether all combinations of 18 cards are solvable in this game. But I have played this game for five years with dozens of students, and I have yet to see a combination of 18 cards that is unsolvable."

Follow up Question. In the event that the answer is negative, what is the probability of having a winning set?

For the follow up question, it may be that an exact answer is out of reach, but bounds on the probability would be welcome.

答:
Unsatisfyingly, a counterexample is (all black):

$$(5,5,6,6,7,7,8,8,9,9,10,10,J,J,Q,Q,K,K)$$

which does not satisfy the last two equations, since

$$\_+\_+\_ \ge 5+5+6 =16>13 = K$$

Extending this result, we need at least $22$ cards to guarantee a solvable $14$-tuple since we have the $21$-card counterexample

$$(3,4,4,5,5, \dots , K, K)$$

where $3+4+4+5+5+6 = 27 > 26 = 2K$, so the last two equations cannot both be satisfied. I do not know whether a counterexample to $22$ cards exists at this moment.

Title: Is this math game always winnable?


问:
Set of functions that form a basis for a Topology
I'm trying to prove the following statement: Define $X=\lbrace f| f: [0,1]\rightarrow[0,1]\text{ is a function}\rbrace$. For each subset $A$ of $[0,1]$, define
\begin{equation}
B_A=\lbrace f\in X|f(x)=0, \forall x \in A \rbrace.
\end{equation}
Prove that the set $B=\lbrace B_A|A \subseteq [0,1]\rbrace$ is a basis for a topology on $X$.

I tried doing the following: let $f \in X$ with $f \in B_{A_1} \cap B_{A_2}$. Then I need to prove that there is a basis element $B_A$ in $B$ such that $f \in B_A \subseteq B_{A_1} \cap B_{A_2}$. I got stucked in here: if $f \in B_{A_1} \cap B_{A_2}$ then
\begin{equation}
f(x)=0, \forall x\in B_{A_1} \cap B_{A_2}.
\end{equation}
How can I proceed from here? Only hints, not answers, please.

答:
If a map $f$ vanishes on $A_1$ and $A_2$, on which minimal subset (for the inclusion order) does $f$ vanishes?

Note: you also have to prove that the base elements cover $X$.


问:
Show that exists an epimorphism $\phi_i: T \rightarrow G_i$ and find $\ker(\phi_i)$
Let $G_1,...,G_n$ be groups and $T:=G_1× \cdots × G_n$.

Show that $ \forall i \in \{1, \cdots , n \} $ exists an epimorphism $\phi_i: T \rightarrow G_i$ and  find
$\ker(\phi_i)$.

I'm not sure if what I have done is correct:

Let $\phi_i (g_1, \cdots , g_n) = g_i $ for $i\in \{ 1, \cdots , n\}$

We want to see first if $\phi$ is a homomorphism, so let's consider:

$(g_1, \cdots , g_n),(h_1, \cdots , h_n) \in T$

then

$\phi_i ((g_1, \cdots , g_n) \cdot (h_1, \cdots , h_n)) = \phi_i ((g_1h_1, \cdots , g_nh_n)) = g_ih_i$

$\Rightarrow \phi_i (g_1, \cdots , g_n) \phi_i(h_1, \cdots , h_n) = g_ih_i $

And now we want to see $\phi_i$ is surjective. Consider $g_i \in G_i$

$\Rightarrow \phi_i (e, \cdots , g_i , \cdots, e) = g_i$

And since then, $\phi_i$ is surjective. On the other hand

$\ker(\phi_i)=\{ (g_1, \cdots , g_{i-1}, e , g_{i+1}, \cdots , g_n) | g_i \in G_i \}$

答:
Yes, your solution is fine.

These homomorphisms are natural consequences of the universal property of the direct product.


问:
Proving a function from a set with a well-founded relation into the ordinals satisfies some property.
In Jech Set Theory, while proving that for every well-founded relation $E$ on a set $P$ there exists a unique function $\rho$ from $P$ into the ordinals satisfying $\rho(x) = \sup\{\rho(y)+1 : y E x \}$, he defines the following sets: $$P_0=\emptyset, \hspace{1mm} P_{\alpha+1}=\{ x\in P : \forall y \hspace{1mm} (yEx \rightarrow y\in P_{\alpha} \}, \hspace{1mm} P_{\alpha}=\bigcup_{\xi<\alpha}{P_{\xi}}$$ if $\alpha$ is limit. Then $P_0 \subset P_1 \subset \ldots \subset P_\theta = P$ for some $\theta$ and $\rho(x)$ is defined to be the least $\alpha$ such that $x \in P_{\alpha+1}$. Since $x E y$ implies $\rho(x) < \rho(y)$, he claims that it is easy to verify that this function $\rho$ satisfies $\rho(x) = \sup\{\rho(y)+1 : y E x \}$. To verify this, I have done the following:

Clearly, $\rho(y)+1 \leq \rho(x)$ for all $y E x$. If $\alpha = \rho(x)$ and $\alpha = \beta +1$, then, $x \in P_{\alpha+1}$, but $x \not \in P_\alpha=P_{\beta+1}$. So, there is some $y E x$ such that $y \not \in P_\beta$. For that $y$, $\rho(y)=\beta$ and so $\rho(y)+1=\alpha$. So, $\rho(x) = \alpha = \sup\{\rho(y)+1 : y E x \}$. If $\alpha$ is limit, then $\forall \xi < \alpha$, $x \not \in P_\xi$ and so all $P_\xi$'s miss some $y E x$, so these $\xi$'s cannot be the supremum. So, $\rho(x) = \alpha = \sup\{\rho(y)+1 : y E x \}$ and we are done.

Is this reasoning correct? I think I have overcomplicated it a lot so I would be happy if you can show some easier ways, thank you.

答:
It appears to be correct, though your argument for limit $\alpha$ could be expressed more clearly. However, the division into two cases isn’t actually necessary.

If $\rho(x)=\alpha$, then $x\in P_{\alpha+1}\setminus P_\alpha$. Let $\beta=\sup\{\rho(y)+1:yEx\}$. If $yEx$, then $y\in P_\alpha$, so $\rho(y)<\alpha$, and hence $\rho(y)+1\le\alpha$. Thus, $\beta\le\alpha$. On the other hand, $y\in P_{\rho(y)+1}\subseteq P_\beta$ for each $yEx$, so $x\in P_{\beta+1}$, and therefore $\beta\ge\alpha$. Thus,

$$\rho(x)=\alpha=\beta=\sup\{\rho(y)+1:yEx\}\,.$$


问:
Solving an exponential equation with different bases
Solve the equation $2^x + 5^x = 3^x + 4^x$. I can figure out two special solutions $x=0$ and $x=1$, and I try to prove that they are the only two solutions. However, I find it hard to do so because I can't prove the monotony given there are also exponential in the derivative. Any hints to that?</p>

答:
Since $f(a)=a^x$ is concave for $x\in[0,1]$ and convex for $x\not\in(0,1)$, the definitions of concavity and convexity say
$$
\begin{align}
\color{#C00}{3^x}+\color{#090}{4^x}
&=\color{#C00}{\left(\frac23\cdot2+\frac13\cdot5\right)^x}+\color{#090}{\left(\frac13\cdot2+\frac23\cdot5\right)^x}\tag1\\[3pt]
&{\ge\atop\le}\color{#C00}{\left(\frac23\cdot2^x+\frac13\cdot5^x\right)}+\color{#090}{\left(\frac13\cdot2^x+\frac23\cdot5^x\right)}{{\quad\text{if }x\in[0,1]}\atop{\quad\text{if }x\not\in(0,1)}}\tag2\\[6pt]
&=2^x+5^x\tag3
\end{align}
$$
Explanation:

$(1)$: write $3$ and $4$ as convex combinations of $2$ and $5$

$(2)$: definition of $\text{concavity}\atop\text{convexity}$

$(3)$: combine like terms

Furthermore, equality holds only if $f$ is linear on $\{2,3,4,5\}$, and that happens iff $x\in\{0,1\}$.


问:
Arithmetic Mean/Geometric Mean Inequality
We have to use the arithmetic mean/geometric mean inequality.
Let $a$ and $b$ be fixed positive numbers. Find the value of $x>0$ that minimizes the given expression and determine the minimum value of the expression,
$ax+\frac{b}{x}$.

I don’t know how to get started by transforming the expression to be able to use the mean inequality. We were told not to solve using calculus.

答:
Ares has provided an excellent algebraic solution that certainly answers your question. Based on how this site is supposed to work, that answer is better than mine. But I would like to also provide an answer for you that does not jump straight into an algebraic solution; but takes a moment to practice some important mathematical skills for problem solving which come in handy when we are stumped! It is a useful skill in math to learn how to explore a problem if we don't know where to begin on a solution.

What will the function $f(x) = ax + b/x$ look like? Are we convinced it has a minimum?

When $x$ is small $ax$ is going to be small and not contribute much to the function and $b/x$ will be large and dominate the function. For small $x$ we can expect the function to approximately look like $b/x$, which for any positive value of $b$ will look approximately like the function $1/x$ which is a decreasing function.

As $x$ grows, $b/x$ will start to get small and contribute less to the function and $ax$ will grow and begin to dominate the function. As $x$ gets very large the function will get closer and closer to the function $ax$, which for all positive values of $a$ is an increasing linear function.

From our work above we expect that for all positive values $a,b$ our function will switch from decreasing to increasing which guarantees a minimum.

The function out to look something like this:

If you really want to play around with graphing this function for intuition, you could make a tool like this on desmos which allows you to toggle the values of $a,b$ (though with more practice I don't think this is necessary)

Where might the minimum occur?

We know the minimum should be wherever the function "switches" from $b/x$ to $ax$, but this will depend on the chosen values of $a,b$.

If $a$ is large, the function will switch to $ax$ sooner because $ax$ will grow faster. This makes the minimizer smaller.

If $b$ is large, the function will switch to $ax$ longer because it will take a larger $x$ to make $b$ small. This will make the minimizer bigger.

Moving to the algebra

At this point I am convinced of the existence of the minimum, and I am convinced it will be some function of the numbers $a,b$. I feel more grounded in the problem. I know I need to use the AM-GM inequality which in its simple case states that for any nonnegative $a,b \in \mathbb{R}$:

$$\frac{a+b}{2} \geq \sqrt{ab}.$$

Since $x > 0, $a > 0$, $b > 0$ then $ax$ and $b/x$ are both positive real numbers. At this point, the only logical thing to try is top plug the numbers into the inequality to yield

$$\frac{ax + \frac{b}{x}}{2} \geq \sqrt{ax\frac{b}{x}}, \hspace{2mm} \text{ for all} x \in \mathbb{R}$$,

The part about "for all $x \in \mathbb{R}$" is very important!

At this point you can pick up where Ares solution starts. Good luck!


问:
Prove that $\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$ is greater than $\frac{2}{3}$.
I have to prove that the product
$$\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$$
is greater than $\displaystyle\frac{2}{3}$.

I've tried to write $k^{3}-1$ as $(k-1)(k^{2}+k+1)$ or another ways but I couldn't finish it.

Body:
$$\begin{align} \prod_{k=2}^{99} \frac{k^3-1}{k^3+1}
&= \prod_{k=2}^{99} \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\
&= \Biggl(\prod_{k=2}^{99} \frac{k^2+k+1}{k+1}\Biggl) \Biggl(\prod_{k=2}^{99} \frac{k-1}{k^2-k+1} \Biggl) \space \space \text{(Separability of Multiple Products)} \\
&= \frac{\prod_{k=2}^{99} \frac{k^2+k+1}{k+1}}{\prod_{k=2}^{99} \frac{k^2-k+1}{k-1}} \\
&= \frac{\prod_{k=2}^{99} \frac{(k+1)^2-(k+1)+1}{k+1}}{\prod_{k=2}^{99} \frac{(k-1)^2+(k-1)+1}{k-1}} \\
&= \frac{\prod_{k=2}^{99} (k+1)-1+\frac{1}{k+1}}{\prod_{k=2}^{99} (k-1)+1+\frac{1}{k-1}} \\
&= \frac{\prod_{n=3}^{100} n-1+\frac{1}{n}}{\prod_{n=1}^{98} n+1+\frac{1}{n}} \\
&= \Biggl(\prod_{k=3}^{98} \frac{(n+\frac{1}{n})-1}{(n+\frac{1}{n})+1}\Biggl) \Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl) \\
&= \Biggl(\prod_{k=3}^{98} 1-\frac{2}{n+1+\frac{1}{n}}\Biggl) \Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl) \\ \end{align}$$
At this stage, we can make a numeric calculation:

$$\Biggl(\prod_{k=3}^{98} 1-\frac{2}{n+1+\frac{1}{n}}\Biggl)=\frac{7}{13} \cdot \frac{13}{21} \cdot \frac{21}{31} \cdots \frac{98 \cdot 99+1-2\cdot 98}{98 \cdot 99+1}=\frac{7}{98 \cdot 99 +1}$$
$$\Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl)=\frac{2}{3} \cdot \frac{(98 \cdot 99+1)\cdot(99 \cdot 100+1)}{7 \cdot 99 \cdot 100}$$

The resultant product is: $$\frac{2}{3} \cdot \frac{(98\cdot 99+1) \cdot (99 \cdot 100+1)}{7 \cdot 99 \cdot 100} \cdot \frac{7}{98 \cdot 99+1}=\frac{2}{3} \cdot \frac{9901}{9900} \gt \frac{2}{3}$$


问:
The equation $x^3 + y^3 = z^3$ has no integer solutions - A short proof
Can someone provide the proof of the special case of Fermat's Last Theorem for $n=3$, i.e., that
$$
x^3 + y^3 = z^3,
$$
has no positive integer solutions, as briefly as possible?

I have seen some good proofs, but they are quite long (longer than a page) or use many variables. However, I would rather have an elementary long proof with many variables than a complex short proof.

Edit. Even if the bounty expires I will award one to someone if they have a satisfying answer.

答: There’s a wonderful elementary (and fairly short) proof in this paper by S.Dolan.</p>


问:
Binary floating point precision question
How do we know what is the degree of precision of an 8 bit floating point number and what fraction of the possible value of the number might it vary by and still
be represented by the same bits in my floating point 8-bit memory?

What we written in class and I can't seem to understand is:

$$1.00111011 - 1.00111001 \implies 2\times 2^{-7} \text{precision}$$

Can somebody explain what we just did, and what is the general procedure for such tasks.

答:
Run the following C program:

```<code>double x = 1.00111011;
double y = 1.00111001;

printf("\n0x%llx", *((long long int*) & x));
printf("\n0x%llx", *((long long int*) & y));
</code>```

It will show that both 8-byte floating-point numbers in fact do not have the same byte representation:

```<code>0x3ff0048c08e251c4
0x3ff0048bee0a5f2a
</code>```

You can get a visual explanation by using an IEEE 754 calculator.

If you run the same experiment with single-precision 4-byte float variables, you get:

```<code>0x3f802460
0x3f80245f
</code>```

In this case, the difference is just one bit.
The difference of the two numbers is the smallest one you can express with this data type. So, you probably tried to reduce the second number as far as possible without reaching zero difference.

You can find a float calculator here.

Body:
Run the following C program:

double x = 1.00111011;
double y = 1.00111001;

printf("\n0x%llx", *((long long int*) & x));
printf("\n0x%llx", *((long long int*) & y));

It will show that both 8-byte floating-point numbers in fact do not have the same byte representation:

0x3ff0048c08e251c4
0x3ff0048bee0a5f2a

You can get a visual explanation by using an IEEE 754 calculator.

If you run the same experiment with single-precision 4-byte float variables, you get:

0x3f802460
0x3f80245f

In this case, the difference is just one bit.
The difference of the two numbers is the smallest one you can express with this data type. So, you probably tried to reduce the second number as far as possible without reaching zero difference.

You can find a float calculator here.


问:
Help me verify this simple claim about continuous function on the real line

Proposition: Let $y$ be a fixed number in $\mathbb{R}$, given $f$ continuous and suppose that $f(x) \geq 0$ for all
$x,y \in\mathbb{R}$ and $f(x) = 0$ only when $x = y$, then there
exists some $k > 0$ such that $f(x) \geq k$ for all $x \neq y$.

I think this is a trivial consequence of continuous function on the real line. Maybe directly from the epsilon-delta definition.  But I cannot remember exactly what the sequence of arguments are, or perhaps a theorem that can show this. I'm also concerned about $\geq$ vs $>$ in the latter half of the statement, but it seems to be intuitive also.  Can someone verify the proposition and provide a line of reasoning.

答: This is wrong. $f(x)=x^2$ only vanishes at $0$ but can be arbitrarily small for $x \neq 0$.


问:
Proof of measurable function
If X is no. Of times the coin is tossed until head appears how do we show that X: S -> R is F-B(R) measurable?</p>

答:
We want to show that $X^{-1}(-\infty,x]\in F$ for all $x\in\mathbb{R}$, i.e., $\{\omega:X(\omega)\leq x\}\in F$. I suppose that $S=\{(H), \ (TH), \ (TTH), \ (TTTH),\dots\},$ where $H$ stands for heads and $T$ for tails. Note that $X$ takes countably infinite number of values. Hence, we can write $X^{-1}(-\infty,x]$ for each $x\in\mathbb{R}$ as follows:
$$
X^{-1}(-\infty,x] = \left\{
        \begin{array}{ll}
            \emptyset & \quad x < 1 \\
             \{(H)\}& \quad 1\le x  <2\\
\{(TH)\}& \quad 2\le x  <3 \\
\{(TTH)\}& \quad 3\le x  <4 \\
\ \ \vdots
        \end{array}
    \right.
$$
Therefore, $X^{-1}(-\infty,x]\in F$ for each $x\in\mathbb{R}$ and the claim follows.

Title: Proof of measurable function


问:
Absolute convergence of a real series
I need to show that the following series:

$\sum_{i=1}^n (-1)^n\dfrac{x^2+n}{n^2} $

Is uniformly convergent on any bounded interval, but not absolutely convergent for any real $x$. My first thought was to use the Weierstrass M-Test, however this is pointless as if the above series "passed" the test, it would have to be absolutely convergent.

In assessing uniform convergence, I considered partial sums. Is it possible to show that the partial sums can get arbitrarily close together, (in a Cauchy sense) which would then imply the series converges uniformly?

答: Hint: For uniform convergence, look at the natural estimate of the remainder for series satisfying Leibniz' test (a.k.a the alternating series test).</p>

Title: Absolute convergence of a real series


答:
I dont think this is true.

Consider

$$f:\mathbb{R} \rightarrow \mathbb{R}, x \mapsto |x|$$

Let $y = 0$. Then $f(x) \geq 0$ and $f(x) = 0 \iff x = 0$. But there can't exist $k>0$ such that $f(x) \geq k$ for all $x$, since $f(\frac{k}{2}) = \frac{k}{2} < k$ for any $k>0$.

Title: Help me verify this simple claim about continuous function on the real line

Body:
Proposition: Let $y$ be a fixed number in $\mathbb{R}$, given $f$ continuous and suppose that $f(x) \geq 0$ for all
$x,y \in\mathbb{R}$ and $f(x) = 0$ only when $x = y$, then there
exists some $k > 0$ such that $f(x) \geq k$ for all $x \neq y$.

I think this is a trivial consequence of continuous function on the real line. Maybe directly from the epsilon-delta definition.  But I cannot remember exactly what the sequence of arguments are, or perhaps a theorem that can show this. I'm also concerned about $\geq$ vs $>$ in the latter half of the statement, but it seems to be intuitive also.  Can someone verify the proposition and provide a line of reasoning.


问:
Absolute continuity of "approximately uniform" measures
Let $\mu$ be a Borel probability measure on the torus $\mathbb{T} = \mathbb{R}/\mathbb{Z}$.  Suppose $\mu$ is "approximately uniform" in the sense that there exists a constant $C > 0$ such that
$$
C^{-1} \ \leq \ \frac{\mu(B_r(x))}{\mu(B_r(y))} \ \leq \ C
$$
for all $x,y \in \mathbb{T}$ and all $r>0$ (here $B_r(x)$ is the ball of radius $r$ centered at $x$).  Does this imply that $\mu$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{T}$?

答:
Yes.

Claim: For $n \in \mathbb{N}$, $\mu(B_{1 / 2n}(0)) \lesssim \mathcal{L}^{1}(B_{1 / 2n}(0))$, with the implicit constant independent of $n$.

Before we prove this claim, let's see why it helps. By approximate uniformity, the claim actually gives us that $\mu(B_{1 / 2n}(x)) \lesssim \mathcal{L}^{1}(B_{1 / 2n}(x))$ for all $x \in \mathbb{T}$, where the implicit constant is the same for all $x$ and $n$. Moreover, one can either be more careful in the proof or use the claim and some reasoning to see that $\mu(B_{r}(x)) \lesssim \mathcal{L}^{1}(B_{r}(x))$ for all $r \leq 1$, with the implicit constant independent of $x$ and $r$. Having the constant independent of both $x$ and $r$ gives absolute continuity.

Proof of claim: The balls $B_{1 / 2n}\big(\frac{k}{n}\big)$ are disjoint for $0 \leq k < n$, so
$$1 \geq \sum_{k = 0}^{n - 1} \mu\Big(B_{1 / 2n}\Big(\frac{k}{n}\Big)\Big) \geq (1 + (n - 1) C^{-1}) \mu(B_{1 / 2n}(0)),$$
which means that
$$\mu(B_{1 / 2n}(0)) \leq \frac{1}{1 + (n - 1) C^{-1}}.$$
The right-hand side is comparable to $\mathcal{L}^{1}(B_{1 / 2n}(0))$ with the implicit constant independent of $n$, which finishes the proof.


答:
The basic trigonometric functions are $\sin$ and $\cos$. $\cos\theta$ and $\sin\theta$ are, respectively, the $x$- and $y$-coordinates that you arrive at after tracing an arc of $\theta$ units counterclockwise around the unit circle, starting from the point $(1,0)$.

From this, the other four functions can be defined in the following way:
\begin{align}
\sec\theta &= \frac{1}{\cos\theta} & \csc\theta &= \frac{1}{\sin\theta} \\[6pt]
\tan\theta &= \frac{\sin\theta}{\cos\theta} & \cot\theta &= \frac{\cos\theta}{\sin\theta}
\end{align}
Because $\tan\theta$ is defined as the ratio of $\sin\theta$ to $\cos\theta$, it has a nice geometric interpretation as the slope of the line segment joining the origin with the point $(x,y)=(\cos\theta,\sin\theta)$. From this it follows that
$$
\tan(\theta + \pi) = \tan\theta \, ,
$$
since a rotation of $\pi$ units simply maps the point $(x,y)$ to $(-x,-y)$. By contrast, $\sec$ and $\csc$ are simply the reciprocal functions of $\cos$ and $\sin$, meaning that their periodicity must be $2\pi$. If you want more of a justification, then consider that $\cos(\theta + \pi) = -\cos\theta$, and so $$\sec(\theta + \pi) = \frac{1}{\cos(\theta + \pi)} = \frac{1}{-\cos\theta} = -\sec\theta \, .$$
A similar argument can be made for $\csc$. Hence, as Amaan has already mentioned, $\tan$ is the outlier.

Title: Why is the period of $\sec$ and $\csc 2\pi$ instead of $\pi$?

Body:
Hi and thanks in advance.

When $\tan x = \frac{(2x + 1)\pi}{2}$, the function is undefined, resulting in a period of simply $\pi$.
However, $\csc$ and $\sec$ also include undefined points. Taking this discontinuity into account, why are their periods $2\pi$ instead of $\pi$?