问:
Does the order of the Fibonacci sequence's initial values matter?
I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example:

The Fibonacci sequence is defined by the initial conditions $a_0=0$,
$a_{-1}=1$, and the recursion relation $a_n=a_{n-1}+a_{n-2}$ for
$n>0$.

Using this definition, I am able to work through the author's example to get the correct generating function:
$$G(x)=\frac{1}{1-x-x^2}$$

However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result:

$$G(x)=\frac{1+x}{1-x-x^2}$$

This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first $(1,2,3,5,8,13,...)$. Where did I go wrong? I would have thought that the sequence should be identical in both cases.

Many thanks for any help!

My calculations:

The author starts with a general expression for a generating function for a sequence $\{a_n\}$:
$$G(x)=\sum_{n=0}^{\infty}a_nx^n$$

He then looks at the special case where $\{a_n\}$ is a linear recurrence of range $r$:
$$a_n=\sum_{i=1}^rc_ia_{n-i}$$

and proves that this leads to:
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$

So, for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$. So we get:

$$g(x)=a_{-1}+a_{-2}+a_{-1}x$$

Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts ($a_{-i}$ for $i=1,2,\dots,r$). But just for the Fibonacci example, we have initial conditions $a_0=0$ and $a_{-1}=1$. So I assume these are just off by 1. Therefore, substituting $a_{-1}=0$ and $a_{-2}=1$ gives the author's result above and using $a_{-2}=0$ and $a_{-1}=1$ gives my result above.

答:
There is already a nice description of generating functions in Somos answer. I will focus here on the question you had about the author's general formula.

$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$

You are right in saying, 'for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$.' You are also correct in observing that you weren't given $a_{-2}$ and the formula requires it.

Also, technically, since the recursion given in the exercise only holds for $n > 0$, $a_{-2}$  is indeterminate in the exercise. So this is 'bad' notation on the author's part if you are expected to use this general theorem, in my opinion.

So, to alleviate this I propose you consider an extension to the recursion, where the recursion holds for $n\geq 0$, (as it must whenever the theorem you have stated and it's associated formula should apply). In such a case, although you aren't given $a_{-2}$ you can find its value from the recursion itself since:

$$a_{0} = a_{-1} + a_{-2}$$

So $a_{-2} = a_{0} - a_{-1} = -1$. Plugging it all in:

$$ G(x) = \frac{x(x^{-1}) + x^{2}(-x^{-2}+ x^{-1})}{1-x-x^{2}} = \frac{x}{1-x-x^{2}} $$

So, it yields the expected generating function for the case where $a_{0}=0$ and $a_{1} = 1$.

So, I understand the confusion you had. However to answer the question as to what happened, when you shift the indices for the initial condition, it will result in shifting the indices throughout the sequence which emerges from the recursion and this will change the generating function.


问:
Is $ \mathbb{Q} \subset L=\mathbb{Q}[X]/(X^4+1)$ a Galois extension?
Let $L=\mathbb{Q}[X]/(X^4+1)$. Is $L/\mathbb{Q}$ a Galois extension?

$[L:\mathbb{Q}] = 4$. How do I show that there there are 4 elements in $\mathrm{Aut}(L/\mathbb{Q})$?

答:
Seeing this as a subfield of $\mathbb C$ by sending $X$ to $e^{i\pi/4}$, we can use the polar decomposition and write
$$
e^{i\pi/4} = \cos(\pi/4) + i \sin(\pi/4) = \frac{\sqrt 2 + \sqrt 2 i}2.
$$
So $L = \mathbb Q(\sqrt 2, \sqrt{-2})$, thus our only options are $[L:\mathbb Q] \in \{2,4\}$. If you are only interested in your answer, you can see that $\mathbb Q(\sqrt 2)$ is a real subfield of $\mathbb C$, and therefore could not contain $\sqrt{-2}$. Therefore $[L: \mathbb Q] > 2$, so it must equal $4$. Showing $L/\mathbb Q$ is Galois is now easy since the other roots of $X^4+1$ are $\zeta = e^{i\pi/4}$, $\zeta^3, \zeta^5$ and $\zeta^7$, all elements of $L$, so $L$ is a splitting field for $X^4+1$. The other powers of $\zeta$ are $\pm 1$ and $\pm i$.

For a technique that's a bit more abstract, you can try factoring $X^4 + 1$:
$$
X^4 + 1 = (X^2)^2 + 1 = (X^2 - i)(X^2 + i)
$$
(where $i$ is some square root of $-1$ here). Since we are trying to find the square root of $i$ and $-i$ which are primitive fourth roots of unity, let us consider a primitive eighth root of unity, say $\zeta$. We get
$$
X^4+1 = (X-\zeta)(X+\zeta)(X-i\zeta)(X+i \zeta) = (X-\zeta)(X-\zeta^7)(X-\zeta^3)(X-\zeta^5).
$$
The field generated by the roots of $X^4+1$ is the same as that generated by the roots of $X^8 - 1$ because $X^8-1 = (X^4-1)(X^4+1)$ and roots of $(X^4+1)$ squared are roots of $(X^4-1)$ since $\zeta^4 = -1$ means $(\zeta^2)^4 = \zeta^8 = 1$. If $\zeta$ is a primitive root of unity, then the roots of $X^8-1$ are just $\zeta^i$ for $0 \le i \le 7$. So $L$ is spanned as a vector space by $1,\zeta,\zeta^2,\zeta^3$, and the other powers are just these times a minus sign, i.e. $\zeta^4 = -1$, $\zeta^5 = -\zeta$, and so on. In particular, $[L: \mathbb Q]$ has degree at most $4$.

But it has degree at least $4$ since the irreducible polynomial $X^4 + 1$ has a root in it. It's a bit hard to show it's irreducible over $\mathbb Q$ without looking directly at its roots over $\mathbb C$; one way is to apply the $\mathbb Q$-algebra automorphism of $\mathbb Q[X]$ sending $X$ to $X+1$ which maps $X^4+1$ to $(X+1)^4+1 = X^4 + 4X^3 + 6X^2 + 4X + 2$ and notice that this is Eisenstein with the prime $p=2$. Automorphisms preserve factorizations in UFDs, and therefore also preserve irreducible polynomials in free polynomial algebras like $\mathbb Q[X]$.

Showing that $L/\mathbb Q$ is Galois is actually the easy part once you know that $X$ in $\mathbb Q[X]/(X^4+1)$ is a primitive $8^{\text{th}}$ root of unity (which is easily seen since $X^4 = -1$ and $X^8 = 1$; the roots of unity form a group, and Lagrange's theorem says that the order of an element of a group divides the order of that group, so this root has order $8$, and therefore is primitive). So you can see $L$ as the splitting field of $x^8-1$ or $x^4+1$, your choice; in any case, it is a Galois extension, and therefore has $[L:\mathbb Q] = 4$ automorphisms.

Hope that helps,

ContentLicense: CC BY-SA 4.0


问:
"Infinity" in mathematics and an elementary question on dimension.
Infinity and an elementary question on continuous manifolds, surjective maps and a dimension formula. The well known "Peano curve" is a continuous surjective map

$$ f(t): I \rightarrow X$$

where $I:=[0,1]$ and  $X:=I \times I$ is the unit square in $\mathbb{R}^2$.

There is an equivalence relation on $I$: Two points $s,t$ are equivalent ($s\cong t$) iff $f(s)=f(t)$, and we may pass to the quotient space $I/\cong$. There is the product map

$$\tilde{f}: I \times I \rightarrow X \times X $$

defined by $\tilde{f}(s,t):=(f(s),f(t))$ and the inverse image $R:=\tilde{f}^{-1}(\Delta) \subseteq I\times I$ equals the equivalence relation $\cong$. The diagonal $\Delta \subseteq X \times X$ is a smooth submanifold and the equivalence relation $R$ is "continuous" in the sense that it is the inverse image of a smooth equivalence relation on $X$ under a continuous map $\tilde{f}$. The equivalence relation $R$ gives rise to a continuous bijection

$$b: I/R \cong X$$

of sets. By definition $dim(I)=1, dim(X)=2$. The set $I/R$ is a quotient of $I$ by a continuous equvalence relation on $R$, hence $I/R$ is a "smaller set" than $I$ since we identify points in $I$ when we pass to the equivalence relation. Do you agree this is a "strange" phenomenon? What "happens" to the dimension? Intuitively it should drop. For fractals there is a notion of dimension - the Hausdorff dimension. Is it possible to define the dimension $dim_q(I/R)$ of the quotient $I/R$? If this is possible and $dim_q(I)=1$, we should "intuitively" have the formula $dim_q(I/R) \leq 1$.

Example 1. In general given a surjective map

$$f: M \rightarrow N$$

where $M,N$ are $C^k$-manifolds and $f$ is a surjective map of $C^k$-manifolds - the dimension should drop: Hence $\dim(N) \leq \dim(M)$. This property holds for $C^k$-manifolds when $k$ is "large enough". A continuous manifold is a fundamental object in mathematics and we want the formula to hold in this case as well.

Note. The dimension of a continuous manifold $M$ is a way to measure the "size" of the manifold, and when we take the quotient $M/R$ of $M$ by a "continuous equivalence relation" $R \subseteq M \times M$, we "identify points" and hence the dimension of the quotient should drop. Hence the quotient should not have a continuous bijection with a manifold of higher dimension. The underlying sets of the continuous manifolds $X:=\mathbb{R}$ and $Y:=\mathbb{R}^2$ have the same cardinality, hence there is a set-theoretic bijection $b: X \cong Y$. For the dimension $dim(X),dim(Y)$ to be "well defined", there should be no continuous surjection from $X$ to $Y$.

Question. Is there a way to "change" the definition of the "category of sets" such that the dimension formula holds for the category of continuous manifolds (with corners) and surjective continuous maps? We want to define the category $C:=Cont(\mathbb{R})$ whose objects $Ob(C)$ are finite dimensional continuous manifolds and whose morphisms $Mor(C)$ are continuous maps of continuous manifolds. In this category the following Lemma should hold:

Lemma. If $f: X \rightarrow Y$ is a surjective map in $C$ then $dim(Y) \leq dim(X)$.

Why should this formula hold? If $f$ is surjective then $Y$ is "smaller" than $X$ and hence $dim(Y) \leq dim(X)$ should hold.

This is an "elementary question" on set theory in a sense, and should be interesting for many of the readers of this forum. If you find a solution, please post it here.

A similar question has been asked here

https://mathoverflow.net/questions/383367/infinity-in-mathematics-the-peano-curve-and-other-paradoxes

Example 2. A set is "Dedekind infinite" iff (this is a wikipedia citation):

"In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is equinumerous to A. Explicitly, this means that there exists a bijective function from A onto some proper subset B of A. A set is Dedekind-finite if it is not Dedekind-infinite (i.e., no such bijection exists). Proposed by Dedekind in 1888, Dedekind-infiniteness was the first definition of "infinite" that did not rely on the definition of the natural numbers."

So in some cases being a strict subset of itself is used to define a "Dedekind infinite set". Has the theory of finite/infinite sets been developed without this definition? I ask for references.

https://en.wikipedia.org/wiki/Dedekind-infinite_set

答:
Here is an answer, of sorts, to your questions:

(i) Do you agree this is a "strange" phenomenon? (ii) What "happens" to the dimension? (iii) Intuitively it should drop.

Answer: (i) No, I do not find it strange at all. I also find it completely normal that there are continuous nowhere differentiable functions and that there is a continuous function mapping the Cantor set onto the unit interval (Cantor function). Such functions even appear "naturally" in real analysis and if one teaches RA one usually gives these as examples of functions which are continuous but not absolutely continuous (a monotonic nonconstant continuous function whose derivative vanishes almost everywhere).

(ii) Dimension increased, that's all.

(iii) The intuition that dimension should drop is based on a "naive" understanding of continuous functions which suffices for undergraduate calculus but which one typically looses after taking a real analysis class. Incidentally, there are some interesting inequalities regarding the behavior of dimension under continuous maps; the earliest ones are called Hurewicz formulae. Here is one of these inequalities.

Definition. Let $f: X\to Y$ be a map of topological spaces. Then $\dim(f)$ is defined as the supremum of dimensions of the subsets $f^{-1}(y), y\in Y$. In particular, if $f$ is a finite-to-one map, $\dim(f)=0$.

Theorem. Suppose that $X, Y$ are separable metrizable spaces (say, $R^n$ and $R^m$ respectively) and $f: X\to Y$ is a closed map (i.e. a map sending closed subsets to closed subsets). Then
$$
\dim(X)\le \dim(Y) + \dim(f).
$$
For instance, if $f$ is finite-to-one (as in the case of the classical Peano curve or the Cantor function mapping the Cantor set onto $[0,1]$), then $\dim(X)\le \dim(Y)$, hence, one gets the opposite of the inequality that you expected.

One can find a proof of this result for instance in

Hurewicz, Witold; Wallman, Henry, Dimension theory, Princeton: Princeton University Press. 165 p. (1948). ZBL0036.12501.

Not a question but a sentiment expressed in the Note: "For the dimension (),() to be "well defined", there should be no continuous surjection from  to ."

No, (as it was pointed out in comments) for a topological dimension to be well-defined it should be invariant under homeomorphisms and it is. There are other notions of dimension in topology, ones which make sense for spaces more general than manifolds. The earlier one is the Lebesgue covering dimension. It is defined in such a way that it is manifestly invariant under homeomorphisms; then one proves that it equals $n$ for (nonempty) $n$-dimensional manifolds. Ditto for the cohomological dimension, etc.

Regarding the Question after the Note: As it was pointed out in comments, you should not mess with the category of "sets" but rather modify the category of "spaces" to exclude examples of morphisms which increase the dimension (whatever the definition of the dimension is). Such a request came earlier from Alexander Grothendieck who asked for something similar in his Esquisse d'un Programme; he was also troubled by Peano curves. See for instance, the discussion here.

(a) One solution comes from logic, the o-minimal theory. Somebody more knowledgeable in logic should provide the details. Otherwise, just read one of the references given in the Wikipedia article.

(b) As a geometer, I would argue for the "Lipschitz category." You can find many more details in my answer here, especially on how to define this category without metrics and how to relate it to Banach star-algebras. The thing is that Lipschitz maps of (compact, or, more generally, complete 2nd countable, locally compact) metric spaces cannot increase the Hausdorff dimension because if the $\alpha$-Hausdorff measure of a metric space $X$ is zero then for every Lipachitz map $f: X\to Y$, the $\alpha$-Hausdorff measure of $f(X)$ is also zero.

(c) As a topologist, I also like the categories of PL and DIFF (piecewise-linear and smooth manifolds respectively), where morphisms are, respectively, PL and smooth. Again, morphisms cannot increase the dimension. PL category naturally extends to the one of simplicial complexes which still have the same property (plus fibers of simplicial maps are simplicial subcomplexes, which is something else that Grothendieck did not like about the category of manifolds). Still, if one is interested in topological manifolds, the Lipschitz viewpoint is the best because of the following theorem due to Dennis Sullivan:

Theorem. Every topological manifold of dimension $n\ne 4$ admits a Lipchitz structure, i.e. an atlas with Lipschitz transition maps. Moreover, this structure is unique, i.e. every homeomorphism is isotopic to a bilipschitz homeomorphism.

See:

Sullivan, Dennis, Hyperbolic geometry and homeomorphisms, Geometric topology, Proc. Conf., Athens/Ga. 1977, 543-555 (1979). ZBL0478.57007.

In other words, in order to study topological manifolds (of dimension $\ne 4$) one looses nothing by working in the Lipschitz category. It was later proven by Donaldson and Sullivan that dimension 4 is a genuine exception, for instance, the E8-manifold does not admit a Lipschitz structure:

Donaldson, S. K.; Sullivan, D. P., Quasiconformal 4-manifolds, Acta Math. 163, No. 3-4, 181-252 (1989). ZBL0704.57008.

Which (if any) of these categories you should use, is up to you, and depends on what your further objective is.


问:
Sigma of 50 element sample vs sigma of the average of 5 samples of 10 elements
I would like to understand a statement I found on a book concerning the explanation of confidence interval which is not clear to me.

We have two different situations and I would like to understand if the standard deviation is different and if it is bigger in case a or b.

Case a) We have 50 elements and I calculate the average ($\bar x$) and standard deviation (σ).

Case b) We take 5 samples of 10 elements each of the same population. We calculate $\bar x$ and σ and then from the averages and standard deviations we calculate the global $\bar x$ and σ.

The average should be the same but what about the standard deviation of the Case b in comparison with Case a? And why?

Thank you in advance for any eventual reply!

Giacomo

答:
(a) Samples of size $n = 50$ from $\mathsf{Norm}(\mu = 60, \sigma = 7):$ We estimate $\mu$ by $A_{50} =\bar X$ and
$\sigma$ by $S_{50}^2.$

Also, $49S_{50}^2/\sigma^2 \sim \mathsf{Chisq}(\nu=49).$ Then a 95% CI for $\sigma^2$ is of the form $\left(\frac{49S_{50}^2}{U},\, \frac{49S_{50}^2}{L}\right),$ where $L=63.88$ and $U=70.22$ cut probabilities $0.025$ from the lower and upper tails, respectively, of
$\mathsf{Chisq}(\nu = 49).$ Take square roots of endpoints
to get a 95% CI for $\sigma.$

```<code>qchisq(c(.925,.975), 49)
[1] 63.88477 70.22241
</code>```

(b) If you take five samples of size $n=10$ from this normal distribution you will have five variance estimates
$S_1^2, S_2^2, \dots, S_5^2.$ Then you can "pool" these
five variance estimates to get an estimate $S_p^2$ as
below:

$$S_p^2 = \frac{9S_1^2 + 9S_2^2 + \cdots + 9S_5^2}{5(9)}.$$
Then $45S_p^2/\sigma^2\sim\mathsf{Chisq}(\nu=45).$
Because the sub-sample sizes $(10)$ are all equal, $S_p^2$ is just the average of the five $S_j^2, j=1,2,3,4,5,$

Then a 95% CI for $\sigma^2$ is of the form
$\left(\frac{45Sp^2}{L},\,\frac{45Sp^2}{U}\right),$ where $L$ and $U$ are from $\mathsf{Chisq}(\nu=45).$
Again here, you can take square roots of endpoints to
get a CI for $\sigma.$

```<code>qchisq(c(.925,.975), 45)
[1] 59.28717 65.41016
</code>```

In a one-factor ANOVA with five levels of the factor
and ten replications at each level, this is the
standard method of estimating the (assumed identical) $\sigma^2$ within each level. In an ANOVA, the means
of the five levels have other uses, so this method is
reasonable.

However, if you are just estimating $\sigma^2$ (or $\sigma),$ then pooling is not an efficient procedure. You have "lost" four degrees of freedom by making five
estimates of $\mu$ (from the five small-sample means)---instead of only one (from the one large sample). With fewer degrees of freedom, the CIs will tend to be longer if
you use $S_p^2$ than if you use $S_{50}^2.$

Because $S_{50}^2$ and $S_p^2$ are both unbiased estimators
of $\sigma^2 = 7^2 = 49,$ the two estimates will tend
to be about the same size. However, $S_p^2$ (based on fewer degrees of freedom) will tend to be more variable.

This is illustrated by the two simulations in R below:

One large sample:

```<code>set.seed(221)
m = 10^5;  v.50 = numeric(m)
for(i in 1:m) {
 v.50[i] = var(rnorm(50, 60, 7))
}
mean(v.50)
[1] 49.00551
var(v.50)
[1] 96.81574
sd(v.50)
[1] 9.839499
</code>```

Five small samples:

```<code>set.seed(221)
m = 10^5;  v.p = numeric(m)
for(i in 1:m) {
 v.p[i] = mean(replicate(5, var(rnorm(10,60, 7))))
}
mean(v.p)
[1] 49.00486
var(v.p)
[1] 105.2702
sd(v.p)
[1] 10.26013
</code>```

Finally, the title of your question may contemplate
looking at the variance of the five means of 10 observations. If the purpose is to estimate the population variance, this might be a catastrophically bad ides: These means have variance $\sigma^2/10,$ so you'd have to multiply by $10$
to estimate the population variance. Such an
estimate would be extremely variable (based on
only 4 degrees of freedom).

```<code>set.seed(221)
m = 10^5;  v.a = numeric(m)
for(i in 1:m) {
 v.a[i] = var(replicate(5, mean(rnorm(10,60, 7))))
}
mean(10*v.a)
[1] 49.01282
var(10*v.a)
[1] 1208.783
sd(10*v.a)
[1] 34.76755
</code>```

Body:
(a) Samples of size $n = 50$ from $\mathsf{Norm}(\mu = 60, \sigma = 7):$ We estimate $\mu$ by $A_{50} =\bar X$ and
$\sigma$ by $S_{50}^2.$

Also, $49S_{50}^2/\sigma^2 \sim \mathsf{Chisq}(\nu=49).$ Then a 95% CI for $\sigma^2$ is of the form $\left(\frac{49S_{50}^2}{U},\, \frac{49S_{50}^2}{L}\right),$ where $L=63.88$ and $U=70.22$ cut probabilities $0.025$ from the lower and upper tails, respectively, of
$\mathsf{Chisq}(\nu = 49).$ Take square roots of endpoints
to get a 95% CI for $\sigma.$

qchisq(c(.925,.975), 49)
[1] 63.88477 70.22241

(b) If you take five samples of size $n=10$ from this normal distribution you will have five variance estimates
$S_1^2, S_2^2, \dots, S_5^2.$ Then you can "pool" these
five variance estimates to get an estimate $S_p^2$ as
below:

$$S_p^2 = \frac{9S_1^2 + 9S_2^2 + \cdots + 9S_5^2}{5(9)}.$$
Then $45S_p^2/\sigma^2\sim\mathsf{Chisq}(\nu=45).$
Because the sub-sample sizes $(10)$ are all equal, $S_p^2$ is just the average of the five $S_j^2, j=1,2,3,4,5,$

Then a 95% CI for $\sigma^2$ is of the form
$\left(\frac{45Sp^2}{L},\,\frac{45Sp^2}{U}\right),$ where $L$ and $U$ are from $\mathsf{Chisq}(\nu=45).$
Again here, you can take square roots of endpoints to
get a CI for $\sigma.$

qchisq(c(.925,.975), 45)
[1] 59.28717 65.41016

In a one-factor ANOVA with five levels of the factor
and ten replications at each level, this is the
standard method of estimating the (assumed identical) $\sigma^2$ within each level. In an ANOVA, the means
of the five levels have other uses, so this method is
reasonable.

However, if you are just estimating $\sigma^2$ (or $\sigma),$ then pooling is not an efficient procedure. You have "lost" four degrees of freedom by making five
estimates of $\mu$ (from the five small-sample means)---instead of only one (from the one large sample). With fewer degrees of freedom, the CIs will tend to be longer if
you use $S_p^2$ than if you use $S_{50}^2.$

Because $S_{50}^2$ and $S_p^2$ are both unbiased estimators
of $\sigma^2 = 7^2 = 49,$ the two estimates will tend
to be about the same size. However, $S_p^2$ (based on fewer degrees of freedom) will tend to be more variable.

This is illustrated by the two simulations in R below:

One large sample:

set.seed(221)
m = 10^5;  v.50 = numeric(m)
for(i in 1:m) {
 v.50[i] = var(rnorm(50, 60, 7))
}
mean(v.50)
[1] 49.00551
var(v.50)
[1] 96.81574
sd(v.50)
[1] 9.839499

Five small samples:

set.seed(221)
m = 10^5;  v.p = numeric(m)
for(i in 1:m) {
 v.p[i] = mean(replicate(5, var(rnorm(10,60, 7))))
}
mean(v.p)
[1] 49.00486
var(v.p)
[1] 105.2702
sd(v.p)
[1] 10.26013

Finally, the title of your question may contemplate
looking at the variance of the five means of 10 observations. If the purpose is to estimate the population variance, this might be a catastrophically bad ides: These means have variance $\sigma^2/10,$ so you'd have to multiply by $10$
to estimate the population variance. Such an
estimate would be extremely variable (based on
only 4 degrees of freedom).

set.seed(221)
m = 10^5;  v.a = numeric(m)
for(i in 1:m) {
 v.a[i] = var(replicate(5, mean(rnorm(10,60, 7))))
}
mean(10*v.a)
[1] 49.01282
var(10*v.a)
[1] 1208.783
sd(10*v.a)
[1] 34.76755


问:
Does this notion of project in Banach spaces exist?
Let $X$ be a Banach space and let $W$ be a linear subspace. Consider some $v$ in $X$ and let the projection of $v$ onto $W$ be defined as the $u$ in $W$ st. $\forall w\in W,\|v-u-w\|=\|v-u+w\|$.
If there is a unique $u$ for which this holds, call this $u$ the projection of $v$ onto $W$.

I would like to know if this notion of projection is well defined, that is, does $u$ exist and is $u$ unique.

答:
For general Hilbert spaces $H$ and the usual definition of the orthogonal projection $P:H \rightarrow H$ onto a closed subspace $W \subset H$, we have $P(v) \in W$ satisfying
$$
||v - P(v) - w|| = ||v - P(v)|| + ||w|| = ||v - P(v) + w||, \forall w \in W
$$
where an orthogonal decomposition was used (Parseval's identity for Hilbert spaces).
Actually, using a parallelogram rule like expansion we can see
$$
||(v - u) - w||^2  - ||(v-u) + w||^2 = 0 \quad \iff \quad \langle (v-u),w \rangle = 0, \forall w \in W.
$$
From this it follows that $(v-u) \in W^\perp$, and that your definition is equivalent to the the usual definition of the orthogonal projection on Hilbert spaces, unique since $v$ has a unique expansion with respect to $W \oplus W^\perp$.

This also means that for a general non-closed subspace of a Hilbert space, the usual definition of orthogonal complement is not well-defined (i.e. does not exist, I do not think uniqueness is ever a problem due to strict convexity of the norm, but you can ignore this comment), so your definition isn't as well.

For general Banach spaces I am not sure to be honest, maybe someone else can shed some light on this.

The usual orthogonal projection for Hilbert spaces can be generalised to Banach spaces in the following way:
As you probably know, another possible way (as opposed to the inner product way) is to project to the unique $u \in W$ such that $||v - u||$ is minimized.
Of course only a norm is used here, so it can be generalised to Banach spaces.
However, it seems that for general subsaces it is not certain that this minimum is attained, as can be read on p. 133 of Conway, an introduction to functional analysis (http://users.math.uoc.gr/~frantzikinakis/FunctionalGrad2015/Conway.pdf).
The generalisation goes well at least for reflexive Banach spaces (again p.133):
There Conway states:

"It is not generally true that the distance from a point to a linear subspace is attained. If $M \subset X$ [$M$ is a vector subspace of a Banach space $X$] call $M$ proximinal if for every $x \in X$ there is a $y \in M$ such that $|| x-y || = dist(x, M)$. So if $X$ is reflexive, Corollary 4.6 implies that every closed linear subspace of $X$ is proximinal".
(Compare this, if you have seem some non-linear functional analysis/calculus of variations, to the fact that the above norm, for a reflexive separable Banach space $X$, has a unique minimum, since the norm is weakly sequentially lower semicontinuous and weakly sequentially precompact and strictly convex.)

Again, I cannot see immediately how your definition relates to the one of proximinal subspaces and the generalisation of the orthogonal projection in Banach spaces.
The only thing I can say, together with the Calculus of Variations perspective taken above is that, if the function $f:\mathbb{R} \rightarrow \mathbb{R}$ (or $\mathbb{C}$), $f_w(\lambda) = ||v -u - \lambda w|| - ||v - u + \lambda w||$ is differentiable at zero, for all $w \in W$, then I do believe find that the two notions/generalisations are equavalent.


问:
A path between two points
Say we have a set $S\subset\Bbb{R^n}$, and $p\in\text{int}(S), q\in\text{int}(\Bbb{R^n}\setminus S)$. I have to prove that every path between $p,q$ goes through $\partial S$ (the boundary of S).
So I let $\gamma:[0,1]\rightarrow \Bbb{R^n}$ be a continuous path s.t $\gamma(0)=p, \gamma(1)=q$, and I want to prove that there exists $c\in [0,1]$ s.t $\gamma(c)\in\partial S$.

My attempt was assuming by contradiction that such $c$ doesn't exist and reach a contradiction, but I don't know how to even start. Intuitively, this statement makes sense but I don't know how to prove it.

I did notice that $\gamma ([0,1])$ is compact, though I don't know if it helps. I want to show that if there doesn't exists such $c$, then $\gamma([0,1])\subseteq\text{int}(S)$ or $\gamma([0,1])\subseteq\text{int}(\Bbb{R^n}\setminus S)$.

答:
For every set $S$ we have that $$\Bbb R^n = \rm{int}(S) \cup \partial S \cup \rm{int}(\Bbb R^n \setminus S)$$

where the union is (pairwise) disjoint. $p$ is in the leftmost set, $q$ in the rightmost. So if $\gamma[[0,1]] \cap \partial S = \emptyset$, we could write

$$\gamma[[0,1]]  = \left((\gamma[[0,1]] ) \cap \rm{int}(S)\right) \cup \left((\gamma[[0,1]] ) \cap \rm{int}(\Bbb R^n \setminus S)\right)$$

which makes the connected set $\gamma[[0,1]]$ a disjoint union of two non-empty ($p$ and $q$ are witness to that) and disjoint relatively open sets, contradiction.

So the intersection with $\partial S$ is not empty.

Title: A path between two points


问:
Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively.
As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$  cannot be a square when $x\neq3$  and $x\neq2$  respectively, where $x$ is a natural number.

In order to do this, I have to use the fact that if $x$ is any natural number >1, any natural number $n\geq x$ can be represented uniquely in the form

$n=c_0+c_1x+c_2x^2+...+c_mx^m$,

where $0\leq c_i \leq x-1$ for $i=0,1,2,...,m-1$  and
$0< c_m \leq x-1$.

答: Hint: IF either of these numbers were squares, then you could express them as $N^2$, where $N$ is some number. By what you know, you can write $N=c_0+c_1x+c_2x^2+\ldots+c_mx^m$ uniquely. Notice that $N$ uses at most three powers of $x$: $N=c_0+c_1x+c_2x^2$, since when you square it you need an $x^4$ to appear as the highest power. Now square $N$ out and show you can't have nonnegative $c_i$ which give you the coefficients of your numbers. For example, for $x^4+x^3+x^2+x+1$, you immediately get $c^2_0\equiv 1\pmod{x}$. </p>

ContentLicense: CC BY-SA 3.0


问:
Matrices and images
Let T: $\mathbb{R}^4 \to \mathbb{R}^5$ defined by

$T (x_1, x_2, x_3, x_4) = (x_4, x_1, x_3, x_2, x_1 - x_3)$

A) Find the standard matrix for the linear transformation $T$

B) Find the image of $x = (1, 2, 1, 2)$ by $T$

Can anyone help me out with this question. I have been through my notes but can not understand how to do it. Hopefully, a detailed solution would be the best so I can get a better understanding.

答: Each component of the right hand side is a dot product of $(x_1,x_2,x_3,x_4)$ with the first row of the matrix.  So the first row $a$ is the solution to $(a_1,a_2,a_3,a_4) \cdot (x_1,x_2,x_3,x_4) = x_4$.

Title: Matrices and images


问:
How to show that this $2n \times n^2$ matrix has rank $2n-1$?
The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is

\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1
\end{bmatrix}

So, basically, it splits into two parts:

the upper part (from row $1$ to row $n$) has $n$ identity matrices $I_n$.

the lower part is $O_n^1, \ldots, O_n^n$, where each $O_n^i$ is an $n \times n$ matrix whose $i$-th row is all one whereas the other entries are zero.

By some examples and intuition, I am aware that the rank of such a matrix is $2n-1$, but how should I rigorously prove it?

答:
Things will perhaps be clearer if we write these $2n$ rows from $\Bbb{R}^{n^2}$ as
$n \times n$ matrices.

For $n=3$ we are looking at
$$\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & 1\\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}  0 & 0 & 0\\ 0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}.$$

Call them $A_1, \ldots, A_{2n}$. We will prove that $\{A_1, \ldots, A_{2n-1}\}$ is linearly inependent and $A_{2n}$ is in its span.

Assume $$0 = \sum_{i=1}^{2n-1}\alpha_iA_i = \begin{bmatrix} \alpha_{1}+\alpha_{n+1} & \alpha_{2}+\alpha_{n+1} & \cdots & \alpha_{n}+\alpha_{n+1} \\
\alpha_{1}+\alpha_{n+2} & \alpha_{2}+\alpha_{n+2} & \cdots & \alpha_{n}+\alpha_{n+2}\\
\vdots & \vdots  & \ddots & \vdots \\
\alpha_{1}+\alpha_{2n-1} & \alpha_{2}+\alpha_{2n-1} & \cdots & \alpha_{n}+\alpha_{2n-1}\\
\alpha_{1} & \alpha_{2}& \cdots & \alpha_{n}\end{bmatrix}$$
so first we get $\alpha_1=\cdots=\alpha_n = 0$ and then $\alpha_{n+1}=\cdots=\alpha_{2n-1}=0$.

On the other hand, we have
$$A_{2n}= \sum_{i=1}^n A_i - \sum_{i=n+1}^{2n-1}A_i.$$

Therefore $\dim\operatorname{span}\{A_1, \ldots, A_{2n}\} = 2n-1$.


问:
Identify your two nonzero solutions to this homogenous system
Question: I am given a homogenous equation $A\vec{x}=\vec{0}$ where $\vec{x}$ is a vector and $\vec{0}$ is the $0$ vector. I have solved it and in parametric vector form it is:

$\begin{pmatrix}
x_1\\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{pmatrix}=x_2\begin{pmatrix}
1.5\\
1 \\
0 \\
4 \\
0
\end{pmatrix}+x_4\begin{pmatrix}
4\\
0 \\
-6\\
1\\
0
\end{pmatrix}$

The equations are $\:\: x_1 = 1.5 x_2 + 4 x_4;\:\: x_2=x_2 (\text{free});\:\: x_3=-6x_4;\:\: x_4=x_4 (\text{free});\:\: x_5=0$

The question I'm asked, and I can find no reference point in our study materials is to identify my two non-zero solutions to the system. Call them vector $\vec{u}$ and vector $\vec{v}$.

I'm also to determine whether $3 \cdot \vec{u}$ is a solution and whether the sum $\vec{u}+\vec{v}$ of $\vec{u}$ and $\vec{v}$ is a solution. I can't figure out the two solutions part. I'm sure I'll get everything else once I just understand what is being asked.

答:
You can let $(x_2 = 0$ and $x_4 = 1)$ and $(x_2 = 1$ and $x_4 = 0)$ give you two solutions $\vec{u}$ and $\vec{v}$

Since we know $A\vec{u} = \vec{0}$ and $A\vec{v} = \vec{0}$

then by the rules of matrix multiplication:

$A(3\vec{u}) =_\text{(product with a scalar)} 3(A\vec{u}) =3\cdot \vec{0} =\vec{0}$ thus $3\vec{u}$ is also a solution

and

$A(\vec{u} +\vec{v}) =_{\text{(distributivity)}} A\vec{u} +A\vec{v} = \vec{0}+\vec{0} = \vec{0}$

and $\vec{u}+\vec{v}$ is another solution as well.

When you solved your system all solutions are of the form $x_2 \cdot \vec{u} + x_4 \cdot \vec{v}$ for arbitrary $x_2, x_4 \in \mathbb{R}$. When they ask for a nonzero solution, that is because the zero solution $\vec{x} = \vec{0}$ is always a solution to any homogeneous system $A\vec{x} =\vec{0}$, namely just plug it in $A\vec{0} = \vec{0} \:\: \text{(true)}$ so it's more interesting when there exists a solution other than $\vec{0}$. You could plug in any $x_2,x_4$ as long as the resulting vector is not $\vec{0}$, and since they ask for two, you would plug in different pairs of numbers for $x_2,x_4$.

Nonzero means a solution $\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} \not= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \vec{0}$ of $A\vec{x} = \vec{0}$ thus $x_1,x_2,x_3,x_4,x_5$ are not all equal to zero.

This will all be obvious once you get to basis of the nullspace.


问:
what is the meaning of this symbolics rel ${I^.}?$
I have some confusion in symbolics

My confusion is given below and marked in red circle

What is the meaning of this symbolics $\operatorname{rel}\dot{I}$?

My attempt: I know that  $I$ represent closed interval $[0,1]$   but my confusion is  about a point  which is lie above on  $I$ and  word rel

答: I would be surprised if there wouldn't be a definition in your textbook. Anyway, "rel $\dot I$" means relative to the boundary of $I$, i.e. relative to $\{0,1\}$. And this means that the homotopy $H : u * u^{-1} \simeq c \text{ rel } \dot I$ is stationary on $\{0,1\}$, i.e, that $H(i,s) = u(0) = c(0)$ for all $s \in I$ and $i = 0, 1$.


答:
The recursion $a_n=a_{n-1}+a_{n-2}$ is "encoded" in the denominator of the generating function: $1-x-x^2$. The numerator is determined by the values of $a_0$ and $a_1$. In particular,
$$
\overbrace{\left(1-x-x^2\right)\vphantom{\sum^\infty}}^\text{denominator}\overbrace{\sum_{n=0}^\infty a_kx^k}^{\substack{\text{generating}\\\text{function}}}=\overbrace{a_0+(a_1-a_0)\,x\vphantom{\sum^\infty}}^\text{numerator}+\sum_{n=2}^\infty\overbrace{(a_n-a_{n-1}-a_{n-2})\vphantom{\sum^\infty}}^0\,x^n
$$
Thus, the generating function is
$$
\frac{a_0+(a_1-a_0)x}{1-x-x^2}
$$

I prefer the definition of the Fibonacci Numbers that starts with $F_0=0$ and $F_1=1$, whose generating function is
$$
\frac{x}{1-x-x^2}
$$
because, among other things, we have
$$
n\mid m\implies F_n\mid F_m
$$
Furthermore,
$$
F_{-n}=(-1)^{n-1}F_n
$$
Lucas Numbers also obey the same recursion as Fibonacci Numbers, and their generating function is
$$
\frac{2-x}{1-x-x^2}
$$

Title: Does the order of the Fibonacci sequence's initial values matter?

Body:
I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example:

The Fibonacci sequence is defined by the initial conditions $a_0=0$,
$a_{-1}=1$, and the recursion relation $a_n=a_{n-1}+a_{n-2}$ for
$n>0$.

Using this definition, I am able to work through the author's example to get the correct generating function:
$$G(x)=\frac{1}{1-x-x^2}$$

However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result:

$$G(x)=\frac{1+x}{1-x-x^2}$$

This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first $(1,2,3,5,8,13,...)$. Where did I go wrong? I would have thought that the sequence should be identical in both cases.

Many thanks for any help!

My calculations:

The author starts with a general expression for a generating function for a sequence $\{a_n\}$:
$$G(x)=\sum_{n=0}^{\infty}a_nx^n$$

He then looks at the special case where $\{a_n\}$ is a linear recurrence of range $r$:
$$a_n=\sum_{i=1}^rc_ia_{n-i}$$

and proves that this leads to:
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$

So, for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$. So we get:

$$g(x)=a_{-1}+a_{-2}+a_{-1}x$$

Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts ($a_{-i}$ for $i=1,2,\dots,r$). But just for the Fibonacci example, we have initial conditions $a_0=0$ and $a_{-1}=1$. So I assume these are just off by 1. Therefore, substituting $a_{-1}=0$ and $a_{-2}=1$ gives the author's result above and using $a_{-2}=0$ and $a_{-1}=1$ gives my result above.


问:
Number of connected components preserved under counterimage by holomorphic embedding
Given $f:\Bbb C\to\Bbb C^2$ holomorphic embedding, is it true that if $U=\bigcup_{j=1}^NU_j$ is a disjoint union of sets, then $f^{-1}(U)$ is still a disjoint union of the same number of sets?

答: No. If $U$ is disjoint from the image of $f$, then $f^{-1}(U)$ is empty which shows that the number of connected components is not preserved under preimage.


问:
Is this linearisation of the uncapacitated facility location problem correct?

Assume we have $n$ factories with unlimited capacity that produce a certain product. The costs to build the factories are $c_1, \ldots, c_n$ respectively. There are $m$ clients whose demands for this product are $d_1, \ldots, d_m$ units respectively. We also have $e_{ij}$ is the cost to deliver $1$ unit from factory $i$ to client $j$. The total cost is the sum of building cost and delivery cost. Determine which factories should we build to minimize the total cost.

This is known as uncapacitated facility location problem. Could you please verify if my linearisation of the product $x_i y_{ij}$ of a binary variable $x_i$ and a continuous $y_i$ is correct?

Let $x_i = 1$ if the factory $i$ is built and $0$ otherwise. Let $y_{ij}$ be the proportion of $d_j$ supplied by factory $i$. To simplify notation, let $[m] := \{1, \ldots m\}$ and $[n] := \{1, \ldots n\}$. Then our problem becomes

$$
\begin{aligned}
\text{min} & \quad \sum_{i=1}^n c_i x_i + \sum_{j=1}^m \sum_{i=1}^n d_j e_{ij} x_i y_{ij}\\
\text{s.t} & \quad x_i \in \{0, 1\}  && \text{for all} \quad i \in [n] \\
           & \quad y_{ij} \in [0, 1]  && \text{for all} \quad (i, j) \in [n] \times [m] \\
           & \quad \sum_{i=1}^n x_i y_{ij}= 1 && \text{for all} \quad j \in [m]
\end{aligned}
$$

Then I want to linearise this optimisation problem as follows. Let $z_{ij} := x_i y_{ij}$. Then our problem becomes

$$
\begin{aligned}
\text{min} & \quad \sum_{i=1}^n c_i x_i + \sum_{j=1}^m \sum_{i=1}^n d_j e_{ij} z_{ij}\\
\text{s.t} & \quad x_i \in \{0, 1\}  && \text{for all} \quad i \in [n] \\
           & \quad z_{ij} \in [0, 1]  && \text{for all} \quad (i, j) \in [n] \times [m] \\
           & \quad \sum_{i=1}^n z_{ij}= 1 && \text{for all} \quad j \in [m]
\end{aligned}
$$

答:
Close, but not quite.

To see that your reformulation is not correct, note that if we have $c_i > 0$ for all $i$, then any optimal solution of your reformulated version will have $x_i=0$ for all $i$, hence the minimum possible value for your reformulated version will be less than the minimum possible value for the original version.

As a way to fix it, you can include the constraints $z_{ij}\le x_i$ for all $i,j$ in order to guarantee $z_{ij}=0$ whenever $x_i=0$.

But note: Since the variables $x_i$ are binary, you have a mixed-integer linear program, not a linear program.


问:
How to find $A$ such that $A^2$ is the zero matrix?

Let $\space t:\mathbb{R^3} \to \mathbb{R^3}$ such that $\space t(v)=0 \space \forall \space v \in \mathbb{R^3}$. The matrix form would be a $3$ by $3$, zero matrix.

If $\space t=f \circ f$, where $f:\mathbb{R^3} \to \mathbb{R^3}$ and $f\neq t$, what is the matrix form of $f$?

I know that $\space t=f \circ f= f\cdot f$ and if one let $A$ to be the matrix of $f$ and $\space T$ the matrix of $t$, then $T=A^2$. And so $A^2$ would be a zero matrix.

Is there a systematic way to solve this problem? Thanks for the hints.

答:
Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation satisfying $T^2 = 0$.

First of all, let us check that the kernel of $T$ has dimension at least 2.  Obviously the kernel cannot be trivial because that would mean $T$ is an isomorphism.  Let us suppose that the kernel has dimension exactly 1, generated by a vector $v \in \mathbb{R}^3$.  We can complete this to a basis $\{ v, v_1, v_2 \}$.  Since $T^2 = 0$, $T(v_1) = \lambda_1 v$ and $T(v_2) = \lambda_2 v$ for some $\lambda_1, \lambda_2 \in \mathbb{R} \setminus \{0\}$.  The problem is that $T(v_1-\frac{\lambda_1}{\lambda_2}v_2) = 0$ as well, even though $v_1-\frac{\lambda_1}{\lambda_2}v_2$ is not in the kernel!  So the kernel has dimension at least 2.

If the dimension of the kernel is 3, then the matrix for $T$ in any basis is the zero matrix.

If the dimension of the kernel is 2, then matrices for $T$ are all similar to a matrix which is zero everywhere except for a 1 at the upper-right corner.

Why is this true?  Since the kernel has dimension 2, we can find one vector $w$ which is not in the kernel.  However, $T(w)$ is in the kernel.  Now pick another vector $u$ in the kernel which is linearly independent from $T(w)$.  With respect to the basis $\{ T(w), u, w \}$, the matrix for $T$ is the special matrix we singled out.


Title: Matrices and images

Body:
Let T: $\mathbb{R}^4 \to \mathbb{R}^5$ defined by

$T (x_1, x_2, x_3, x_4) = (x_4, x_1, x_3, x_2, x_1 - x_3)$

A) Find the standard matrix for the linear transformation $T$

B) Find the image of $x = (1, 2, 1, 2)$ by $T$

Can anyone help me out with this question. I have been through my notes but can not understand how to do it. Hopefully, a detailed solution would be the best so I can get a better understanding.

Body:
First, it is important to see how matrices act on vectors. In our case, we are transforming vectors from $\mathbb{R}^4$ into vectors in $\mathbb{R}^5$ (i.e. T will transform a vector $x = (x_1,x_2,x_3,x_4)$ into $b =(b_1,b_2,b_3,b_4,b_5)$. If we write a general transformation T:
\begin{equation}
\begin{bmatrix}
t_{11} & t_{12} & t_{13} & t_{14}\\
t_{21} & t_{22} & t_{23} & t_{24}\\
t_{31} & t_{32} & t_{33} & t_{34}\\
t_{41} & t_{42} & t_{43} & t_{44}\\
t_{51} & t_{52} & t_{53} & t_{54}\\
\end{bmatrix} \begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
\end{bmatrix}=
\begin{bmatrix}
b_{1} \\
b_{2} \\
b_{3} \\
b_{4} \\
b_{5} \\
\end{bmatrix}
\end{equation}

where

\begin{cases}
 b_1 =  t_{11} x_1 + t_{12} x_2 + t_{13} x_3 +  t_{14} x_4\\
 b_2 =  t_{21} x_1 + t_{22} x_2 + t_{23} x_3 +  t_{24} x_4\\
 b_3 =  t_{31} x_1 + t_{32} x_2 + t_{33} x_3 +  t_{34} x_4\\
 b_4 =  t_{41} x_1 + t_{42} x_2 + t_{43} x_3 +  t_{44} x_4\\
 b_5 =  t_{51} x_1 + t_{52} x_2 + t_{53} x_3 +  t_{54} x_4
\end{cases}

Back to our exercise:

\begin{cases}
 b_1 & = x_4&=  0 \cdot x_1 &+ 0 \cdot x_2 &+ 0 \cdot x_3 &+  1 \cdot x_4 \\
 b_2 &=x_1&=  1 \cdot x_1 &+ 0\cdot x_2 &+ 0 \cdot x_3 &+  0 \cdot x_4  \\
 b_3 & = x_3&=  0 \cdot x_1 &+ 0 \cdot x_2 &+1 \cdot x_3 &+  0 \cdot x_4\\
 b_4 & = x_2&=  0 \cdot x_1 &+ 1 \cdot x_2 &+0 \cdot x_3 &+  0 \cdot x_4\\
 b_5 & = x_1-x_3&=  1 \cdot x_1 &+ 0 \cdot &+ (-1) \cdot x_3 &+  0 \cdot x_4
\end{cases}

Then,

\begin{equation}
T=
\begin{bmatrix}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
1 & 0 & -1 & 0\\
\end{bmatrix}
\end{equation}

For the image of $x = (1,2,1,2)$ just compute:

\begin{equation}
T (1,2,1,2) =
\begin{bmatrix}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
1 & 0 & -1 & 0\\
\end{bmatrix}
\begin{bmatrix}
1 \\
2 \\
1 \\
2 \\
\end{bmatrix} = \begin{bmatrix}
2 \\
1 \\
1 \\
2 \\
0 \\
\end{bmatrix}
\end{equation}


问:
Factor $9x^4-37x^2+4$

Factor $9x^4-37x^2+4$ using difference of cubes, sum of cubes,
difference of squares, substitution, grouping or common factoring,
whichever is simplest method.

Usually factoring isn't to difficult for me, but for some reason I'm stuck on this one. I would usually use the factor theorem but the question specifically states to use one of the methods listed. And the only one this equation kind of works with is substitution, at least to my eyes, but I'm having trouble with doing it that way and I can't figure any other way to do it besides the factor theorem.

答: You have an $x^4$, an $x^2$ and a constant term. Therefore substitute $u = x^2$ to get the quadratic $9u^2-37u+4$, which factors as $(9u - 1)(u - 4) = (9x^2 - 1)(x^2 - 4)$.

Title: Factor $9x^4-37x^2+4$


问:
why can we get $\frac{\partial q_t(x,y)}{\partial t}=\Delta q_t(x,y)?$
If for the transition density $p_n(x,y)=P^{x}(X_n=y)$ of discrete time random walk $X$ on a general garph $G$, we have
$$q_t(x,y)=\sum_{n=0}^{\infty}\frac{e^{-t}t^n}{n!}p_n(x,y), \quad t\ge0$$, why can we get
$$\frac{\partial q_t(x,y)}{\partial t}=\Delta q_t(x,y)?$$

If we differentiate both sides of $q_t$, we get
$$\frac{\partial q_t(x,y)}{\partial t}=-q_t(x,y)+\sum_{n=1}^{\infty}\frac{e^{-t}t^{n-1}}{(n-1)!}p_n(x,y).$$
Why the RHS would be $\Delta q_t(x,y)?$

答:
What is actually happening here is the usual decomposition of a continuous time Markov chain into a jump chain and holding times. In this case the average holding time at each state is $1$, and the $n$th term in the summation of $q_t(x,y)$ is the probability to go from $x$ to $y$ by making exactly $n$ jumps up to time $t$. The (suitably normalized) graph Laplacian appears as a result of the fact that as $\Delta t \to 0$ the probability that more than one jump occurs in time $[t,t+\Delta t]$ goes to zero asymptotically more rapidly than $\Delta t$.

In detail, if you want to differentiate at a later time, you have

$$q_{t+h}(x,y)=\sum_z q_t(x,z) \sum_{n=0}^\infty e^{-h} \frac{h^n}{n!} p_n(z,y).$$

This says the probability to go from $x$ to $y$ in time $t+h$ is the sum of the probability to go from $x$ to $z$ in time $t$ weighted by the probability to go from $z$ to $y$ in time $h$.

Now you differentiate that with respect to $h$ and set $h=0$. So you get

$$\frac{\partial q_t}{\partial t}(x,y)=\left. \sum_z q_t(x,z) \left ( -q_h(z,y) + \sum_{n=1}^\infty e^{-h} \frac{h^{n-1}}{(n-1)!} p_n(z,y) \right ) \right |_{h=0} \\
= \sum_z q_t(x,z) \left ( -q_0(z,y) + p_1(z,y) \right ).$$


问:
Circles and lines geometry problem

Here is my task: An arbitrary ray $OE$ crosses the circle at point $D$:
$$x ^ 2 + (y - a / 2) ^ 2 = (a / 2) ^ 2$$
and a tangent to it(at point $E$), passing through point $C$, diametrically opposite to $O$.

Through points $D$ and $E$, straight lines are drawn parallel to the axes $Ox$ and $Oy$, respectively, up to the intersection at point M.

Equate the line formed by the $M$ points and draw it (Agnesi's curl).

My attempt

I think that this problem need using Polar coordinates. But I have no idea how to start. Maybe if we assume that $M$ has coordinates $(x,y)$ we can find this point(using polar coordinates like $x=r\sin(q)$ or something like that) but how to equate the line? And of course I have already drawn it. After an hour i have idea if we find point $M$ can we use $M$-coordinates to get parametric equation of a line? If yes, how to make first step? Ok, after a few time i found $x$ like a coordinate of a point $M$ (i use as a parameter angle between $OE$ and $OC$ (q) so $x=tgqa/cosq$) how can find $OB$?

答:
Spoiler alert: this is not a hint, here is a solution using cartesian coordinates.

The calculation is based on similarity of triangles $OBD$ and $OCE$. First we consider the right side of the curve ($x \ge 0$) and then we note that the curve is symmetric. Our objective is to find $y_M$ as a function of $x_M$.

From the similarity of triangles $OBD$ and $OCE$ we have:
$$\frac{OB}{OC} = \frac{BD}{CE} $$
$$\frac{y_M}{a} = \frac{x_D}{x_M} $$
$$\therefore y_M = a \frac{x_D}{x_M} \qquad(1)$$
Now, to find $x_D$ we note that $D$ is on the given circle and that $y_D = y_M$ :
$${x_D}^2 + (y_M - \frac a2)^2 = (\frac a2)^2 $$
$$\therefore {x_D}^2 = y_M (a - y_M) \qquad(2)$$
By squaring equation (1) and using equation (2) we have:
$$y_M = a^2 \frac{a-y_M}{{x_M}^2} $$
which can be rewritten as
$$y_M = \frac{a^3}{a^2 + {x_M}^2} \qquad(3)$$
which is the equation of the loci of $M$ . Note that the curve of desired loci is symmetric, and equation (3) accounts for positive as well as negative values of $x$ .

Title: Circles and lines geometry problem


问:
How to show infinite $n$ exist for the following
If $x$ is cube free, it is not divisible by the cube of any prime $p$.

Show that there are infinitely many integers $x$ such that each of the numbers $x, x+1, x+2, x+3$ are not cube free.

How do you write an infinite proof like this?

答:
This is easiest to show via Chinese remainder theorem.

By CRT, there exists a solution $\mod (2\cdot3\cdot5\cdot7)^3$ to the system $$x\equiv0\bmod2^3$$$$x\equiv-1\bmod3^3$$$$x\equiv-2\bmod5^3$$$$x\equiv-3\bmod7^3$$Since there are infinitely many numbers congruent to $x\bmod (2\cdot3\cdot5\cdot7)^3$, we are done.

P.S. I chose $2,3,5,7$ arbitrarily. Any choice of $4$ integers $>1$ that are pairwise coprime would work just as well.


问:
Give an example of a function such that  '(0) exists but lim→0 () does not exist.
Give an example of a function such that   $f’(0)$ exists but $$\lim_{x \rightarrow 0}f(x)$$ does not exist.

Hello, I am struggling to find an example of this? Much help, thanks!

答:
Assuming the common rigorous definition of the derivative, differentiability implies continuity, and for a function continuous in 0, we have that $\lim_{x\to0}f(x) = f(0)$, in particular the limit exists.

Sometimes at a discontinuity there can be a left and a right limit of the derivative, and when they are equal their common vale is taken as the derivative. An example would be a function that is 0 up to and including 0, and 1 for $x>0$.


问:
Generating functions, the product between two formal power series with different denominators
I'm a bit lost in trying to take the product of a power series between two functions $f(x) = \frac{1}{(1-x)^k}$ and $g(x)=\frac{1}{(1-x^r)^k}$.

I know both can be expanded to,
\begin{align*}
\frac{1}{(1-x)^k} = \sum_{n=0}^{\infty}\binom{n+k-1}{k-1}x^n \\
\frac{1}{(1-x^r)^k} = \sum_{n=0}^{\infty}\binom{n+k-1}{k-1}x^{rn}.
\end{align*}

This is where I get a bit lost. Since the exponents for $x$ are different, it seems like the Cauchy product is not applicable. Is there a further step I'm missing?

答:
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {\left( \matrix{
  n + k - 1 \cr
  k - 1 \cr}  \right)x^{\,n} } \sum\limits_{0\, \le \,m} {\left( \matrix{
  m + k - 1 \cr
  k - 1 \cr}  \right)x^{r\,m} }  =   \cr
  &  = \sum\limits_{0\, \le \,m} {\sum\limits_{0\, \le \,n} {\left( \matrix{
  n + k - 1 \cr
  k - 1 \cr}  \right)\left( \matrix{
  m + k - 1 \cr
  k - 1 \cr}  \right)x^{r\,m + n} } }  =   \cr
  &  = \sum\limits_{0\, \le \,s} {\left( {\sum\limits_{0\, \le \,m\, \le \,\left\lfloor {s/r} \right\rfloor } {\left( \matrix{
  s - rm + k - 1 \cr
  k - 1 \cr}  \right)\left( \matrix{
  m + k - 1 \cr
  k - 1 \cr}  \right)} } \right)x^s }  =   \cr
  &  = \sum\limits_{0\, \le \,s} {\left( {\sum\limits_{0\, \le \,m\,\left( { \le \,\left\lfloor {s/r} \right\rfloor } \right)} {\left( \matrix{
  s - rm + k - 1 \cr
  s - rm \cr}  \right)\left( \matrix{
  m + k - 1 \cr
  m \cr}  \right)} } \right)x^s }  \cr}
$$


问:
Why do we have to use the exponents for this decay probelem.
"A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 milligrams of the material present and after two hours it is observed that the material has lost $10$ percent of its original mass, find ($a$) an expression for the mass of the material remaining at any time $t$"

I know the normal method would be to get the formula $y = 50e^{-kt}$ , then solve the two simultaneous equations to get $k$.

But if i use $dy = -ky\cdot dt$ with $dy$ as $0.1y$ and $dt$ as $2$, I don't get the exact same value for $k$.

I know the first method is the correct one, but, why don't I get the exact same value when i use that method?

答: Remember that dy is the differential on y, meaning that dy should be the difference between the initial ammount and the final amount after the two hours.</p>


问:
What is the smallest dimension a non-commutative C*-algebra can have?
What is the smallest dimension a non-commutative C-star-algebra can have? Let $d$ denote this dimension. Clearly, $d\leq 4$ as $M_{2}(\mathbb{C})$ is a $4$-dimensional non-commutative C-star-algebra. Also, $\{0\}$ and $\mathbb{C}$ are the only $0$- and $1$-dimensional C*-algebras, respectively, and they are both commutative, thus $d\geq2$. But are there $2$- or $3$-dimensional non-commutative C-star-algebras?

I can't find any examples, so I think $d=4$.

答: See Wikipedia.  A finite-dimensional $C^*$ algebra is isomorphic to a direct sum of full matrix algebras.  The smallest noncommutative full matrix algebra is $M(2,\mathbb C)$ which is $4$-dimensional.


问:
Find lim sup lim inf
How do I find $\lim \sup\text{ or } \lim \inf$ of $ \sin (\frac{n\pi}{5})$ ?

I know the $\sin$ function normally oscillates between $-1$ and $1$ but that obviously is not the answer for $\lim  \inf$ and $\lim \sup$.

答:
First, you need to know the definition of $lim sup$ and $lim inf$. Given a bounded sequence $ \sin (\frac{n\pi}{5})$. Let's name it $\{a_j\}$. We define
$$A_1=inf\{a_1,a_2,a_3,...\}$$
$$A_2=inf\{a_2,a_3,a_4,...\}$$
$$A_3=inf\{a_3,a_4,a_5,...\}$$
...
$$A_n=inf\{a_n,a_{n+1},...\}$$
It's easy to find the above lines shares the same infimum, because sin function is  periodical. But it's not obvious to see which is the infimum (here is also minimum.) Hence, I use R to help me find:

```<code>x <- seq(pi/5, 2*pi, by = pi/5)
min(sin(x))
</code>```

R gives me -0.9510565.
Hence,
$$lim inf{a_j}=limA_n=-0.9510565$$

Similarly,
$$lim sup{a_j}=limB_n=0.9510565$$, where $B_n=sup\{a_n,a_{n+1},...\}$

Title: Find lim sup lim inf

Body:
First, you need to know the definition of $lim sup$ and $lim inf$. Given a bounded sequence $ \sin (\frac{n\pi}{5})$. Let's name it $\{a_j\}$. We define
$$A_1=inf\{a_1,a_2,a_3,...\}$$
$$A_2=inf\{a_2,a_3,a_4,...\}$$
$$A_3=inf\{a_3,a_4,a_5,...\}$$
...
$$A_n=inf\{a_n,a_{n+1},...\}$$
It's easy to find the above lines shares the same infimum, because sin function is  periodical. But it's not obvious to see which is the infimum (here is also minimum.) Hence, I use R to help me find:

x <- seq(pi/5, 2*pi, by = pi/5)
min(sin(x))

R gives me -0.9510565.
Hence,
$$lim inf{a_j}=limA_n=-0.9510565$$

Similarly,
$$lim sup{a_j}=limB_n=0.9510565$$, where $B_n=sup\{a_n,a_{n+1},...\}$


问:
Solve Limit Tend To $\pi$, From Length Of Circle
Computing the length of a semicircle of radius 1. It's divided in $2^x$ equal arcs, I find the following formula for the total length of the chords created:

$$L=2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

When $x \to \infty$ we see, geometrically, that this expression tends to be  the length of the circle:

$$2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

My goal is to define $\pi$ as the value of that expression when $x \to \infty$ (the limit). But how can I rigorously prove that this limit exists and that the expression tends to be the length of the circle? In other words, how can I write rigorously the geometric ideas?

Note: The number  $\frac{45}{2^{x-1}}$ is in degrees.

答:
Well, if you define $45^{\circ}$ as equal to $\displaystyle\frac{\pi}{4}$ radians (equivalent to defining $\pi$ by $\text{circumference} = \pi d$), the limit's existence and value follows:

$$\lim_{n\to\infty}2^{n+1}\sin\bigg(\frac{45^{\circ}}{2^{n-1}}\bigg)=\lim_{n\to\infty}2^{n+1}\sin\bigg(\frac{\pi}{4\cdot2^{n-1}}\bigg)=\lim_{n\to\infty}\pi\cdot\frac{\displaystyle\sin\big(\frac{\pi}{2^{n+1}}\big)}{\displaystyle\frac{\pi}{2^{n+1}}}$$
$$=\pi\lim_{\frac{\pi}{2^{n+1}}\to 0^{+}}\frac{\displaystyle\sin\big(\frac{\pi}{2^{n+1}}\big)}{\displaystyle\frac{\pi}{2^{n+1}}}=\boxed{\pi}$$


问:
Examples of two inequivalent definitions which were once thought to be equivalent
I once thought two definitions that I made in proof theory were equivalent, but then I discovered a counterexample. So I am now led to wonder, have there ever been cases in the history of mathematics when people made two inequivalent definitions and thought they were equivalent, but later a counterexample was discovered?</p>

答:
It was once thought that the solution set of $x^2 + 1 = 0$ was $\emptyset$, until the notion of the complex numbers became widespread. Or is that not the kind of thing you want?

In a similar vein, nobody knew for a good long while how to define infinitesimals properly; the best thing we had until Robinson's nonstandard analysis in the 1960's was simply that infinitesimals are zero, end of story.

There are any number of "equivalent" statements in elementary analysis, which are actually inequivalent when you relax Choice to a greater or lesser extent, or you go through the various stages in reverse mathematics to discover exactly which ones are equivalent assuming an extremely weak background logic. For example, "a continuous function on a closed bounded interval is bounded" is equivalent to WKL0 (see https://en.wikipedia.org/wiki/Reverse_mathematics), so it's not equivalent to "every nonempty bounded set of reals has a least upper bound" which is equivalent instead to ACA0. There's also a truly wonderful paper by Propp which examines just what properties of the reals do indeed characterise the reals.


问:
How to compute a complex integral of $x \: dz$?
I'm working through chapter 4 of Alfohrs, and the first exercise is computing the integral $\int_\gamma x \: dz$ where $\gamma$ is the directed line segment from $0$ to $1+i$. What, exactly, does this mean?

I know that by definition, $\int_\gamma f(z)\:dz = \int_a^b f(z(t)) z'(t) \:dt$; however, I'm not quite sure what to do with this definition here. One thing that seems to be the case is $f$ is probably implied to be in the form $x + iy$ (is this a fair assumption?) If so, I could use the expansion $\int_\gamma f(z)\:dz = \int_\gamma (u dx - v dy) + i \int_\gamma (u dy + v dx)$, which would after substitution become $\int_\gamma x dx + i \int_\gamma x dy$.

If I can indeed have it in this form, I need to somehow parametrize $\gamma$, correct? My attempt here is saying that $\gamma(t) = t + it$ for $0 \leq t \leq 1$. Then I get that $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + (t^2/2)\mid_0^1 = (t^2)\mid_0^1 = 1^2=0^2=1$.

Is any of this reasoning correct?

Edit

Fixed my last calculation in response to mrsamy's answer: it's now $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + i(t^2/2)\mid_0^1 = 1/2 + 1/2 \cdot i$.

答:
You are correct that the first step is parametrizing the curve $\gamma$.

Instead of $\gamma(t)$, you write $z(t)=t+it$, $0\le t\le 1$. So $z'(t)=1+i$ and thus
$$
\int_\gamma f(z)dz=\int_0^1 f(z(t))z'(t)dt=\int_0^1t(1+i)dt=(1+i)\int_0^1tdt=\frac12(1+i).
$$

You miss an $i$ in your calculation.


问:
Simplifying $\nabla\times(Ax)$ by using index notation
I'm completely new to Einstein's summation notation/index notion and had a go at simplying the following expression:
$$\nabla\times(Ax)$$
For a 3x3 matrix A and a vector x.

My attempt:

I know
$$Ax=a_{ij}x_j$$
$$\nabla\times v=\epsilon_{ijk}\partial_jv_k$$
$$\partial_ix_j=\delta_{ij}$$
$$a_{ij}\delta_{jk}=a_{ik}$$
If we define a vector $c=Ax$ then $c_k=a_{kj}x_j$

Therefore
$$\nabla\times(Ax)=\epsilon_{ijk}\partial_j(a_{kl}x_l)$$
since they commute then
$$=\epsilon_{ijk}a_{kl}(\partial_jx_l)$$
By applying the third equation
$$=\epsilon_{ijk}(a_{kl}\delta_{jl})$$
Now apply the forth equation
$$=\epsilon_{ijk}a_{kj}$$
Thats where I got stuck on, if I didn't misunderstood anything, since j and k are the summation index, this term can be futher simplified but I don't know what to do next apart from brute force numeration. If you can lend a hand, thank you!

答:
When you say ‘they commute’, you're effectively assuming that the matrix $ A $ is constant and only the vector $ x $ is varying.  I'll assume that that's what you intended.

As @Arthur said in a comment, you can't always simplify indices away.  In index notation, the final answer is $ \epsilon _ { i j k } a _ { k j } $; you can't simplify it any further.  (Although if you'd like the indices to come in a consistent order, you can use the antisymmetry of $ \epsilon $ to make it $ - \epsilon _ { i k j } a _ { k j } $ or $ \epsilon _ { k i j } a _ { k j } $ and then relabel the indices to $ - \epsilon _ { i j k } a _ { j k } $ or $ \epsilon _ { i j k } a _ { i k } $, in case either of those looks nicer to you.)

Or if you want to translate the answer back into vector/matrix notation, then you can look through the known operations from matrices to vectors and see if you can find one that matches this.  In this case, I don't recognize it.  But that's the beauty of index notation: you don't have to worry about whether anybody has thought of your operation before; you have the notation for it regardless.

ETA:  And for the record, I checked your calculation, and it's correct (assuming that $ A $ is a constant matrix and $ x $ is the vector whose components are the coordinates).

ETA again:  I've thought of something that's similar to this operation that takes a matrix (in $ 3 $ dimensions) with components $ a _ { i j } $ to a vector with components $ \epsilon _ { i j k } a _ { j k } $.  If you have two vectors with components $ v _ i $ and $ w _ i $, then their dot product is $ v _ i w _ i $ and their cross product has components $ \epsilon _ { i j k } v _ j w _ k $.  But there's also the tensor product, a matrix with components $ v _ i w _ j $.  This contains all of the information in both the dot and cross products; for example, the dot product is the trace of the tensor product.  And how do you get the cross product from the tensor product?  It's this mystery operation!  So I still don't know what you call it or any symbol for it in vector algebra, but it does come up.  (Although note this technicality: the answer to your question is actually the opposite of the operation that takes a tensor product to a cross product, because of the extra minus sign.)


问:
Find all $n!$ that are products of two Fibonacci numbers
Find all positive integers $n$ such that $n!$ is the product of two Fibonacci numbers

I found this discussion, but it doesn't quite help my problem.
Check if the number is a result of multiplying two fibonacci numbers

答:
I might be overly complicating the stuff but I'll give it a try:

A first step might be to determine all Fibonacci numbers that are equal to some factorial.

from https://mathoverflow.net/questions/114850/is-every-fibonacci-number-fibonacci-prime

Carmichael's theorem says that for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for (positive integer) $k \lt n$

the second statement is equivalent to a sort of counterpart going in the other direction:
for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for $k \gt n$ save only for the obvious exceptions where $k$ is a multiple of $n$ (since $F_{n}\mid F_{k}$ when $n \mid k$).

As suggested in Doug M's comment, let us prove that for $i>12$, $F_i$ has a prime factor greater or equal than $i-1$.

Assume that $F_i$ has all its prime factors smaller that $i-1$, then since there exists a prime divisor of $i$ which doesn't divide any $F_k$ with $k<i$, and $f(p)\leq p+1$, the only possibility is that this factor is $i-1$ or $i$ (if one of them is prime)[this actually happens, take for instance $F_{23}$].

We want to know when some prime $p$ divides a Fibonacci number,  its done here, https://www.fq.math.ca/Scanned/1-2/vinson.pdf, where it is proven that, if $f(m)$ is the first time that $F_n$ is divisible by $m$, then $F_i \equiv 0 \:(m) \Longleftrightarrow f(m) | i$.

Notice that $f(p)\leq p+1$ ; $f(2^\ell) = 3.2^{\ell-2}$ ; $f(5)=5$ ; and if $m$ and $n$ are coprime, then $f(m.n)=f(m)f(n)$.

Now take $n$ such that $2^\ell\le n < 2^{\ell+1}$ and let $p$ be the largest prime in $n$'s decomposition. We're looking for $i$ such that $F_i = n!\:$. Now that means that $i\leq p+1$ (otherwise $F_i$ would have a prime factor greater than $p$). Meanwhile, if $n\geq 5$, then $2^\ell$ and $5$ divide $n!$, hence $f(2^\ell)f(5) = 3.2^{\ell-2} .5 |i$. But $3.2^{\ell-2} .5> 2^{\ell+1} >n $, this is absurd.

This means that the only possible solutions to $F_i=n!$ are for $n=0,1$ or $2$ (see also Peter comment)

Step 2 is to do the same with $n!=F_iF_j$. A different, probably easier, approach is to say that $n! \geq \sqrt{2\pi n}\left( {n \over e}\right)^n$ while $F_iF_j \leq F_{\max i,j}^2 \leq F_{n+1}^2$ (using the fact that $F_n$ has prime factor of size a least $n-1$ as we proved before).

Hence we want $n! \geq \sqrt{2\pi n}\left( {n \over e}\right)^n \geq 2^{2(n+1)} > F_{n+1}^2 \geq F_iF_j $ this is true for $n \geq 12$ (and again, we conclude using Peter's computations)

disclamer: this might need a bit of cleaning and checking at some point, sorry if it's a bit messy


问:
Probability of Poisson arrival in a vanishingly small interval
Consider a homogeneous Poisson arrival process, and we look at the time interval $(t,t+\varepsilon]$ for a vanishingly small $\varepsilon>0$.

Why is it the case that the probability of more than 1 arrival in $(t,t+\varepsilon]$ is $o(\varepsilon)$?

答:
The probability to have more than one arrival is then, assuming a rate $\lambda>0$ (seen as a constant),
$$\begin{align}
\mathbb{P}\{ N(t,t+\varepsilon] > 1 \}
&= 1 - \mathbb{P}\{ N(t,t+\varepsilon] = 0 \} - \mathbb{P}\{ N(t,t+\varepsilon] = 1 \} \\
&= 1 - e^{-\lambda \varepsilon} - e^{-\lambda\varepsilon}(\lambda\varepsilon) \\
&= \frac{\lambda^2\varepsilon^2}{2} + o(\varepsilon^2)
\end{align}$$
which is $o(\varepsilon)$.


问:
Convergence integral Riemann zeta
In the paper of Riemann and the book of Edwards I encountered the following representation for $\zeta$:
$$
\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx=\Pi(s-1)\zeta(s)
$$
which due to the fast growth of $e^x$ converges at infinity, for $x$ tending to $0$ the integrand grows like $x^{s-2}$, and therefore due to this pole the integral converges for $Re(s)>1$. Then he considers the following integral,taken over the Hankel contour from infinity to along the real axis anti-clockwise around the pole and back to infinity.
\begin{align*}
& \int_\infty^\infty\frac{(-x)^{s-1}}{e^x-1} \, dx \\[8pt]
= {} & (e^{i\pi s}-e^{-i\pi s}) \int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx \\[8pt]
= {} & 2i\sin(\pi s) \Pi(s-1)\zeta(s)
\end{align*}
It is claimed that the left integral converges for all $s.$ However, I was unable to see this convergence. I suspect by going around $z=0$ we omitted the pole, and therefore we only have to look at the convergence at infinity. But around $z=0$, a limiting process still takes place.

How can we rigorously prove the convergence of the left integral?

答:
I wasn't clear exactly where you were stuck in the proof, so I posted a (more-or-less) full proof.

We assume one easily proved lemma:

lemma 1: For $|z| < 1/2$, $\left|e^z -1\right| \geq |z|/2$.

Next, we define a version of the Hankel contour.  For small positive $\delta$ select two points in the complex plane:
\begin{align*}
p_1&=\delta e^{i\delta} = \delta_x + i\delta_y\quad\text{where}\quad \delta_x = \delta\cos(\delta),
\delta_y = \delta\sin(\delta)\\
p_2&=\delta e^{- i\delta} = \overline{\delta e^{i\delta}} = \delta_x  - i\delta_y
\end{align*}
Our contour $C$ consists of three piecewise smooth curves  $C_1+C_2+C_3$ defined as follows:

$C_1$ follows a path parallel to (and just above) the real axis from $\infty$ to $p_1$.

$C_2$ follows a counterclockwise circular path from $p_1$ to $p_2$, tracing the boundary of a circle that is centered at 0 with radius    $\delta$. (Making just barely less than a full circle).

$C_3$ follows a path parallel to (and just below) the real axis from $p_2$ to $\infty$.

Lemma 2
Let $ z\in\mathbb{C} \backslash [0,\infty)$ and $Re(s) > 1$.  Then
$\displaystyle \lim_{\delta \to 0^+}\int_{|z|=\delta}\frac{-z^{s-1}}{e^z -1}\, dz =0$.

Proof - Lemma 2
Fix $s$.  Let $A = | Re(s) - 1|$ and let $B = Im(s)$.  We use the ML Inequality to obtain an upper bound for our integral. We also assume that $\delta < 1/2$ and apply lemma 1:
\begin{equation*}
\int_{|z|=\delta}\frac{-z^{s-1}}{e^z -1}\, dz \leq 2\pi\delta \cdot \max_{|z| = \delta}
\left|  \frac{-z^{s-1}}{e^z -1} \right| \leq
2\pi\delta \cdot \max_{|z| = \delta} \left| \frac{-z^{s-1}}{\delta /2}  \right|
= 4\pi \cdot \max_{|z| = \delta} \left|-z^{s-1}\right|.
\end{equation*}
So it only remains to show that
\begin{equation*}
\lim_{\delta \to 0^+} \max_{|z| = \delta} \left|-z^{s-1}\right| = 0.
\end{equation*}
Using the principal value of the complex logarithm (for argument $\theta$: $-\pi < \theta \leq\pi$), we rewrite $-z = \delta e^{i\theta} = e^{\log(\delta) +i\theta}$, so that
\begin{align*}
 \left|-z^{s-1}\right|  = \left|  e^{ \left(  \log(\delta) +i\theta  \right) [ A + i B] } \right|
 = \left|  e^{A\log(\delta) - B\theta} e^{ i \left( B\log(\delta) +  A\theta  \right) } \right|
     = \left| e^{A \log(\delta)  - B\theta }   \right|
 = \delta^A  e^{- B\theta} \leq \delta^A e^{\pi \left| B \right| }.
\end{align*}
But $e^{\pi \left| B \right| }$ is fixed.  Because $A > 0$, we have $\lim_{\delta \to 0^+} \delta^A = 0$, completing the proof.

End of Proof - Lemma 2

Main Theorem
Let $s \in \mathbb{C}$.  We recall the traditional definition of $\zeta(s)$:
\begin{align*}
\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \prod_{p}\frac{1}{  \left(  1-p^{-s} \right)}
\quad\text{(where $p$ ranges over an ordered list of all the primes)},
\end{align*}
convergent for $Re(s) > 1$. For the contour of integration $C$ (a Hankel contour oriented in the positive direction defined above), we have that:
\begin{align}
\zeta(s)=\frac{\Gamma(1-s)}{ -2 \pi i}  \int_C \frac{-z^{s-1}}{e^z - 1} \, dz
\end{align}
extends $\zeta(s)$ to a meromorphic function on $\mathbb{C}$, holomorphic except for a simple pole at
$s = 1$.

Proof - Main Theorem.

Unless otherwise stated, we assume $Re(s) > 1$. We start with the Gamma function:
$$ \Gamma(s) = \int_0^\infty e^{-x} x^{(s-1)} \, dx. $$
Next, we replace $x$ by $nt$ in the integral (so that $dx=ndt$):
\begin{align*}
\Gamma(s) &= \int_0^\infty e^{-nt} (nt)^{(s-1)} n\, dt = \int_0^\infty e^{-nt}n^s t^{(s-1)} \, dt =  n^s \int_0^\infty e^{-nt} t^{(s-1)} \, dt  \\
n^{-s}\Gamma(s) &=\int_0^\infty e^{-nt} t^{(s-1)} \, dt  \\
\frac{1}{n^s} &= \frac{1}{\Gamma(s)}\int_0^\infty e^{-nt} t^{(s-1)} \, dt.
\end{align*}
Now sum both sides over $n$:
\begin{align}
\sum_{n=1}^\infty \frac{1}{n^s} &= \frac{1}{\Gamma(s)}\sum_{n=1}^\infty \int_0^\infty e^{-nt} t^{(s-1)} \, dt  \\
\zeta(s) &= \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=1}^\infty e^{-nt} t^{(s-1)} \, dt  \\
\zeta(s) &= \frac{1}{\Gamma(s)} \int_0^\infty t^{(s-1)}\sum_{n=1}^\infty e^{-nt}  \, dt.\tag{1}
\end{align}

Because we are assuming $Re(s) > 1$, the interchange of the sum and integral is justified by the uniform convergence of the sum.  Notice that the sum over $n$ in the last integral is just a geometric series with the $n=0$  term missing.  (We are justified in using a geometric series, because $0 < e^{-t} < 1$). So, we have:
$$
\sum_{n=1}^\infty e^{-nt} = \sum_{n=0}^\infty e^{-nt} - e^{-0t} = \sum_{n=0}^\infty (e^{-t})^n - 1= \frac{1}{1-e^t} -1 = \frac{e^t}{1-e^t}=\frac{1}{e^t-1}.
$$

But that means we can restate equation (1) (and while we are at it replace $dt$ with the slightly more standard $dx$), and have:
$$
\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty x^{(s-1)}\sum_{n=1}^\infty e^{-nx}  \, dx = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} \, dx.
$$

If we set $\displaystyle I(s)= \int_0^\infty \frac{x^{s-1}}{e^x - 1} \, dx$, then we have: $\displaystyle \zeta(s) = \frac{1}{\Gamma(s)} I(s) $.

Our next goal is to evaluate $I(s)$ in the complex plane and see where that takes us.  Riemann starts with a related but not identical integral, which we call  $I_0(s)$:
$$
I_0(s) = \int_C \frac{-z^{s-1}}{e^z - 1} \, dz,
$$
using the contour of integration $C$ described above.
This contour stays well away from the singularities of the integrand at $2\pi i \mathbb{Z}$.  We have:
$$
I_0(s) = \int_C \frac{-z^{s-1}}{e^z - 1} \, dz =  \int_\infty^{p_1} \frac{-z^{s-1}}{e^z - 1} \, dz + \int_{|z|=\delta} \frac{-z^{s-1}}{e^z - 1} \, dz + \int_{p_2}^\infty \frac{-z^{s-1}}{e^z - 1} \, dz.
$$
We now study $I_0(s)$ as $\delta\rightarrow 0$. By Lemma 2, the second integral can be disregarded so the contour includes only the horizontal lines $\infty\rightarrow p_1$ and $p_2\rightarrow\infty$.

Note that $(-z)^{(s-1)}=e^{(s-1)log(-z)}$ and that $log(-z)=log|z|+iarg(-z)$. We will use $Arg(z)$, the principal value of $arg(z)$, defined as the value $\theta$ satisfying $-\pi < \theta \leq \pi$. Recall that $z \mapsto Arg(z)$ is discontinuous at each point on the nonpositive real axis.  Let $z=x_0+iy$ for some fixed $x_0 < 0$. If $y\downarrow 0$ then $Arg(z) \rightarrow \pi$, whereas, if $y \uparrow 0$ then $Arg(z) \rightarrow -\pi$.

In the first integral, $-z=-\delta_x - i\delta_y$  approaches the nonpositive real axis from below, so (as $\delta\rightarrow 0$) we have
$Arg(-z) \rightarrow -\pi$. In the third integral, $-z=-\delta_x + i\delta_y$  approaches the nonpositive real axis from above, so we have $Arg(-z) \rightarrow \pi$.
\begin{align*}
I_0(s) &= %\int_C \frac{-z^{s-1}}{e^z - 1} \, dz =
  \int_{\infty}^{p_1} \frac{e^{(s-1)(\log |z| -\pi i)}}{e^z - 1} \, dz
+ \int_{p_2}^{\infty} \frac{e^{(s-1)(\log |z| +\pi i)}}{e^z - 1} \, dz \\
 &= -\int_{p_1}^{\infty} \frac{ e^{(s-1) \log |z| } e^{(s-1)(-\pi i)} }{e^z - 1} \, dz
+   \int_{p_2}^{\infty} \frac{ e^{(s-1) \log |z| } e^{(s-1)(\pi i)} }{e^z - 1} \, dz \\
 &= -e^{(s-1)(-\pi i)} \int_{p_1}^{\infty} \frac{ e^{(s-1) \log |z| }  }{e^z - 1} \, dz
+  e^{(s-1)(\pi i)} \int_{p_2}^{\infty} \frac{ e^{(s-1) \log |z| }  }{e^z - 1} \, dz.
\end{align*}
Taken to the limit, our integrals are over a horizontal line with $dy = 0$, so we can replace $dz$ with $dx$ (and $z$ with $x$), replace $p_1$ and $p_2$ with $0$,
and consider $e^{(s-1) \log |z|} = x^{(s-1)}$:
\begin{align*}
\lim_{\delta \to 0+} I_0(s)
&=-e^{(s-1)(-\pi i)} \int_{0}^{\infty} \frac{ e^{(s-1) \log |z| }  }{e^z - 1} \, dz
+  e^{(s-1)(\pi i)} \int_{0}^{\infty} \frac{ e^{(s-1) \log |z| }  }{e^z - 1} \, dz \\
&= \left[ e^{i\pi(s-1)} - e^{-i\pi(s-1)}  \right] \int_0^\infty \frac{x^{(s-1)}}{e^x - 1} \, dx.
\end{align*}
But we have: $e^{iw}-e^{-iw} = \cos(w) + i \sin(w) - \left[ \cos(w)-i\sin(w) \right] = 2i\sin(w)$.
So, the value in brackets is: $2i\sin(\pi (s-1)) = 2i\sin(\pi s - \pi) = -2i\sin(\pi s)$.
This gives our value for $I_0(s)$:
\begin{equation*}
\lim_{\delta \to 0+} I_0(s) =
-2i\sin(\pi s)\int_0^\infty \frac{x^{(s-1)}}{e^x - 1} \, dx = -2i\sin(\pi s) I(s).
\end{equation*}
Reviewing, we have:
\begin{equation*}
\zeta(s)= \frac{1}{\Gamma(s)} I(s)\quad\text{and}\quad I_0(s) = -2i\sin(\pi s) I(s)
\quad\text{so that}\quad \zeta(s)=\frac{1}{\Gamma(s)}\cdot\frac{1}{ -2i\sin(\pi s) } I_0(s).
\end{equation*}
Now use a rearrangement of Euler's Reflection Formula: $\displaystyle \frac{1}{\Gamma(s)} = \frac{\Gamma(1-s)\sin(\pi s)}{\pi}$, giving

\begin{equation*}
\zeta(s)= \left[  \frac{1}{\Gamma(s)} \right]   \frac{1}{ -2i\sin(\pi s) } I_0(s)
=  \left[    \frac{\Gamma(1-s)\sin(\pi s)}{\pi}  \right]    \frac{1}{ -2i\sin(\pi s) } I_0(s)
= \frac{\Gamma(1-s)}{ -2 \pi i} I_0(s).
\end{equation*}
We therefore have:
\begin{align}
\zeta(s)=\frac{\Gamma(1-s)}{ -2 \pi i}  \int_C \frac{-z^{s-1}}{e^z - 1} \, dz.
\end{align}

Up to this point, we have assumed $Re(s) > 1$.   But the integral is an entire function (it is uniformly convergent in any compact subset of $\mathbb{C}$  because $e^z$ grows faster than any power of $z$). That means that $\zeta(s)$ is analytic, except possibly at the positive integers where $\Gamma(1-s)$  has simple poles.  But we already know that $\zeta(s)$ is analytic for  $Re(s) > 1$, so the possible poles at $2,3,4...$ must be removable and cancel against zeros of the integral.  Thus, the only pole of $\zeta(s)$ is at $s= 1$.

End of Proof - Main Theorem.

Title: Convergence integral Riemann zeta


答:
Your reasoning seems to be mostly fine.

$\int_\gamma x\ dx + \int_\gamma x\ dy$

$x$ is the real part of the parameterization of $\gamma,$ y is the imaginary part.

$x = t, dx = dt\\
y = it, dy = i\ dt$

$\int_0^1 t\ dt + \int_0^1 it\ dt\\
\frac 12 + \frac 12 i$

Title: How to compute a complex integral of $x \: dz$?

Body:
I'm working through chapter 4 of Alfohrs, and the first exercise is computing the integral $\int_\gamma x \: dz$ where $\gamma$ is the directed line segment from $0$ to $1+i$. What, exactly, does this mean?

I know that by definition, $\int_\gamma f(z)\:dz = \int_a^b f(z(t)) z'(t) \:dt$; however, I'm not quite sure what to do with this definition here. One thing that seems to be the case is $f$ is probably implied to be in the form $x + iy$ (is this a fair assumption?) If so, I could use the expansion $\int_\gamma f(z)\:dz = \int_\gamma (u dx - v dy) + i \int_\gamma (u dy + v dx)$, which would after substitution become $\int_\gamma x dx + i \int_\gamma x dy$.

If I can indeed have it in this form, I need to somehow parametrize $\gamma$, correct? My attempt here is saying that $\gamma(t) = t + it$ for $0 \leq t \leq 1$. Then I get that $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + (t^2/2)\mid_0^1 = (t^2)\mid_0^1 = 1^2=0^2=1$.

Is any of this reasoning correct?

Edit

Fixed my last calculation in response to mrsamy's answer: it's now $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + i(t^2/2)\mid_0^1 = 1/2 + 1/2 \cdot i$.


问:
A theorem in Linear Algebra; linear dependence - Axler

I really am having trouble understanding the statement and the proof. Why does the theorem pick $v_1 \neq 0 $? Why not $v_2$? Also in proving (a), why do we consider the largest $j$?

I do not understand the statement Note all of $a_2, a_3, \dots, a_m$ can be $0$ (because $v_1 = 0 $) Can someone teach me by example?

答:
First, there's nothing special about $v_1$ that we set it not equal to $0$.  We're putting an arbitrary order to our list of basis vectors: the theorem would work the same if we switched the places of $v_1$ and $v_2$ (and changed the statement accordingly so that we could potentially express $v_2$ in terms of $v_1$ and require that $v_2 \ne 0$).

We have to require that $v_1 \ne 0$ because otherwise the singleton $\{ v_1 \}$ is linearly dependent, but it is nonsensical to express $v_1$ in terms of previous basis vectors (unless you consider $v_1 = 0$ an example of that).

We consider the largest $j$ with nonzero coefficient simply because we're expressing $v_j$ in terms of previous basis vectors.  If the coefficient of $v_k$ is $0$, then the statement says nothing about the vector $v_k$.  We want to be able to divide out the coefficient, and taking the largest $j$ vector allows us to do that.

Now to answer your last question, suppose all of $a_2 \dots a_m$ are $0$.  We're assuming that $a_1$ and $v_1$ are not zero, so $a_1 v_1 \ne 0$.  But $0 = a_1 v_1 + \dots a_m v_m = a_1 v_1$ if the other coefficients are all $0$.  This is a contradiction.

Hope that helped.


问:
Limit of the geometric sequence $\{r^n\}$, with $|r| < 1$, is $0$?

Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.

I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(

答:
Let $u=1-|r|$ and note that $|r|\lt1$ implies $0\lt u\le1$. This implies $0\le1-u^2\lt1$, which in turn implies $0\le1-u\lt1/(1+u)$, which in turn implies the first (strict) inequality in the string of assertions

$$|r^n-0|=|r|^n=(1-u)^n\lt{1\over(1+u)^n}\le{1\over1+nu}\lt{1\over nu}=\epsilon{1\over nu\epsilon}\le\epsilon{\lceil1/(u\epsilon)\rceil\over n}$$

for all $n\ge1$ and any $\epsilon\gt0$. So letting $N_\epsilon=\lceil1/(u\epsilon)\rceil$, we find that $n\ge N_\epsilon$ implies $|r^n-0|\lt\epsilon$, which tells us $\lim_{n\to\infty}r^n=0$.

Remark: This answer follows much the same logic as those of aes and Richard, differing mainly in style of presentation. Its one advantage, if any, is that it does not split out the trivial case of $r=0$. The key inequality, $(1+u)^n\ge1+nu$, can, if need be, be proved by induction. The other equalities and inequalities in the display are straightforward to verify; in particular, the equal sign preceding the sudden appearance of $\epsilon$ is justified by the fact that all we're doing there is multiplying and dividing by a nonzero quantity.


问:
Finding the Sum of Converging series when there is a range for $r$
The problem is: what values of $x$ will cause the following series to converge:

$\sum_{n=0}^\infty 2^n (x+1)^n$

I wasn't sure how to determine x but just took a few moments of "playing" with the series to see what values would make for an $|r|\lt1$.  I could tell I needed $-\frac 12\lt(x+1)\lt\frac 12$.

So, I would like to know the proper process for this rather than me just "playing with numbers".

Also, the next part of the problem said, find the sum of the series for those values of $x$.
I'm really stuck on this because it appears they want ONE sum.  Even if they want more than one sum, seems like the sum would have a range of values.  I say this because with different values of $x$ I get various values of $r$, all of which are $\lt1$, making the series converge.  But for $n=0$, the first term is always $1$, so it appears to me that I get a different sum for the series depending on what value of $x$ I use.  And since $-1.5\lt x \lt -.5$, I can't use those values to evaluate the sum since it's greater than or less than not equal to.

Stuck here.

答: This is a simple geometric series with common ratio $r=2(1+x)$. So it is convergent iff $|2(1+x)| <1$ or $-\frac  3 2 < x <-\frac 1  2$ and the sum  is $\frac  1 {1-r}=\frac 1 {1-2(1+x)} =-\frac 1  {1+2x}$.


问:
Equivalent sets proof
Let $A$ be a set. Show that $\cal P(A)$ - the power set of $A$ - is equivalent (same cardinal) to ${\{0, 1\}}^A$ - the set of all functions from $A$ to $\{0, 1\}$. Suppose $A$ was $\mathbb{N}$; would this hold? How about if $A$ was $\mathbb{N}^\mathbb{N}$ (the set of sequences with values in $\mathbb{N}$)?

My friend gave me the following proof:

It suffices to construct a bijective function.

Suppose that $f: A \rightarrow \{0, 1\}$, and let $A_f$ be the set of elements such that $a\in  A_f \iff f(a)=1$. Define the map $h(f) =A_f$ then we proceed to show injectivity and surjectivity:

Injectivity:

Suppose that for functions $f, g$, we have $f\ne g$. Then there is an $x\in A$ so $f(x)=1$ and $g(x)=0$ or vice versa. Then $A_f\ne A_g$. By the contrapositive argument, $h$ is injective.

Surjectivity:

Let $X\in{\cal P}(A)$ and then define a function $f$ as follows:

$$f(x)=\cases{1, if x\in B\cr 0, if x \notin B }$$

Then $X = h(f)$ and $h$ is surjective.

We have a bijection $h:{\{0, 1\}}^A \rightarrow {\cal P}(A)$ so, we may conclude that the two sets are equivalent.

Is this proof correct? Is there a more formal or more detailed way to express it? In addition, does this proof carry forward to $\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ defined in the problem above? Any assistance much appreciated.

答:
I'd be more direct.

Let $f\in \{0,1\}^A$.  Then $f$ is a function $f: A \to \{0,1\}$

Consider $\{a\in A| f(a) = 1\}$.  That's well defined subset of $A$.

And so $\phi:\{0,1\}^A\to \mathscr P(A)$ via $\phi(f) = \{a\in A| f(a) = 1\}$ is a well-defined function.

Now its just a matter of proving it is surjective and injective.

It is surjective.

Let $E\subset A$.  Define $g:A\to \{0,1\}$ via $g(a) =\begin{cases}1&a\in E\\0&a\not \in E\end{cases}$.  $g$ is a well defined function from $A\to \{0,1\}$. so $g \in \{0,1\}^A$.

Because that's what $\{0,1\}^A$ is:  Its the collection of all possible functions for $A\to \{0,1\}$ and $g$ certainly is a function that for every element of $A$ maps distinctly to either $0$ or $1$.

And $\phi(g)=\{a\in A|g(a)=1\} = \{a\in A|a \in E\} =E$.

SO for all $E\in \mathscr P(A)$ there a $g\in \{0,1\}^A$ so that $\phi(g) = E$. So $\phi$ is surjective.

It is injective.

$f, g\in \{0, 1\}^A$ and $\phi(f)=\phi(g)=E$.  Then for every $a \in A$ we have either $a\in E$ in which case $f(a)=g(a) = 1$; or $a \not \in E$ in which case $f(a)=g(a)=0$. In either case $f(a) =g(a)$ for all $a \in A$.

So $f = g$.  For any $f,g \in \{0,1\}$ we have $\phi(f)=\phi(g) \iff f=g$ so $\phi$ is injective.

....

ANd that's that.

....

This will hold of $A = \mathbb N$ or $A =\mathbb N^{\mathbb N}$ or for any set.

$\{0,1\}^{\mathbb N}$ is the set of all functions from $\mathbb N \to \{0,1\}$.  We can find a one-to-correspondence between any subset of the natural numbers to a function (and vice versa) by for each subset considering the function where $f(n)=1$ if $n \in A\subset \mathbb N$ and $f(m) = 0$ if $m\not \in A$.

......

COuld you try harder to put your finger on what in your gut doesn't feel right because this is utterly complete and formal and foolproof so far as I can tell.

Title: Equivalent sets proof


问:
Identify your two nonzero solutions to this homogenous system
Question: I am given a homogenous equation $A\vec{x}=\vec{0}$ where $\vec{x}$ is a vector and $\vec{0}$ is the $0$ vector. I have solved it and in parametric vector form it is:

$\begin{pmatrix}
x_1\\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{pmatrix}=x_2\begin{pmatrix}
1.5\\
1 \\
0 \\
4 \\
0
\end{pmatrix}+x_4\begin{pmatrix}
4\\
0 \\
-6\\
1\\
0
\end{pmatrix}$

The equations are $\:\: x_1 = 1.5 x_2 + 4 x_4;\:\: x_2=x_2 (\text{free});\:\: x_3=-6x_4;\:\: x_4=x_4 (\text{free});\:\: x_5=0$

The question I'm asked, and I can find no reference point in our study materials is to identify my two non-zero solutions to the system. Call them vector $\vec{u}$ and vector $\vec{v}$.

I'm also to determine whether $3 \cdot \vec{u}$ is a solution and whether the sum $\vec{u}+\vec{v}$ of $\vec{u}$ and $\vec{v}$ is a solution. I can't figure out the two solutions part. I'm sure I'll get everything else once I just understand what is being asked.

答: The two column vectors(let's call them $\vec{v}_1$ and $\vec{v}_2$) you found form a basis for the null space of $A$ which means they span the solution space of $Ax=0$ meaning that any linear combination belongs to the solution space. i.e: $k_1\vec{v}_1 + k_2\vec{v}_2$ is also a solution for all scalars $k_1,k_2$


问:
How to get $P_x(X_t=x, \tau<t)=P_{X_{\tau}}(X_{t-\tau}=x)\mathbb{E}_x\mathbb{1}_{\tau<t}$?
For a Markov process $X_t$, define a stopping time $\tau$ and $P_x(\cdot)=P(\cdot|X_0=x)$,
$$P_x(X_t=x, \tau<t)=P_x(X_t=x|\tau<t)P_x(\tau<t)=P_{X_{\tau}}(X_{t-\tau}=x)\mathbb{E}_x\mathbb{1}_{\tau<t}$$

My question is that how to the last equation.

Is the first term that using the strong Markov property and the second term is
$$P_x(\tau<t)=P(\tau<t|X_0=x)=\mathbb{E}(\mathbb{1}_{\tau<t}|X_0=x)?$$

答:
Showing that $\mathbb{P}_x(\tau<t)=\mathbb{E}_x[1_{\{ \tau<t \}}]$ is just a matter of expanding definitions.

The thing with the first factor is using the strong Markov property to restart the process at time $\tau$.


问:
Intuitions for ker of linear maps
I am doing some linear algebra ta for the first time, and while I can solve exercises and read through proofs and definition, I would like to offer my student some extra intuition...which I lack.

I'm currently reading about vector spaces and linear map and would like to find some intuition for the kernel of a linear map. I understand the definition, the kernel is a subset of vectors for which the map returns the zero vector of the target space.

What I would like is either:

an example where there is some geometric intuition to get

an example of how the concept itself can be used (e.g. In a proof)

答: My opinion –– which is biased probably –– is that the kernel is best understood geometrically using the standard projection map $\pi:\Bbb{R}^2\to \Bbb{R}$ given by $\pi(x,y)=x$. Then we can think of the fibre $\pi^{-1}(0)=\ker \pi$. You can also see that all of the elements of $\Bbb{R}^2$ are translates in the $x$ direction of an element of the fibre. Thinking of examples of this type in higher dimensions motivate the rank-nullity theorem also. Indeed, in this case it's quite clear that $\dim \ker \pi+\dim \operatorname{ran} \pi=\dim \Bbb{R}^2$.

Title: Intuitions for ker of linear maps


问:
Finding an integral submanifold of an involutive distribution
I've been given the following vector field on the manifold $M = \mathbb{R}^3 \setminus \{\text{$z$-axis}\}$ $$X = (x-2y)\,\frac{\partial}{\partial x} + (2x+y)\, \frac{\partial}{\partial y} + z\, \frac{\partial}{\partial z},\quad Y = -y\,\frac{\partial}{\partial x} + x\, \frac{\partial}{\partial y}$$
It was straightforward to show that they're linearly independent and commute, therefore they define an involutive rank $2$ distribution, say $D$. The question is find and describe its integral submanifolds. I tried the following methods:

I was able to figure out that the form $\omega = - xz\,dx - yz\,dy + (x^2 + y^2)\,dz$ annihilates $X$ and $Y$ and $d\omega = 3x\,dx\,dz + 3y\,dy\,dz$. Since our distribution is involutive, naturally $d\omega\wedge \omega = 0$. But I've been unable to figure out an $\alpha$ such that $d\omega = \omega\wedge\alpha$. This could then help me figure out the integral submanifolds

Alternatively, we can the flows of $X$ and $Y$, which I did after a painstaking computations but I'm not sure how to proceed after that.

Any help would be appreciated.

答:
Note that there is an easy integrating factor to guess for your $\omega$: Multiply through by $f=\dfrac1{(x^2+y^2)z}$ and the resulting form is exact. Indeed, $f\omega = d(-\log r + \log z)$, where $r=\sqrt{x^2+y^2}$. Integral manifolds are then $z=cr$ for various constants $c$.

Following Sophus Lie, there is a good way to look for such integrating factors. If you see an obvious one-parameter group leaving the distribution invariant (or sending $\omega$ to a functional multiple of $\omega$), let $V$ be its infinitesimal generator and let $f=1/\omega(V)$. Then you can check that $f\omega$ will be closed. In your case, the obvious one-parameter group is given by radial stretches, and $V=x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y}+z\dfrac{\partial}{\partial z}$. Unfortunately, the numbers work out to $\omega(V)=0$ — in other words, $\mathscr L_V\omega = 0$ rather than just a general functional multiple of $\omega$. But, still, we will see the appearance of $(x^2+y^2)z$ when we do this computation.

EDIT: A natural way to avoid the issue with $z=0$ about which you're worried is to introduce a homogenizing variable $u$ with $z=ur$, $r=\sqrt{x^2+y^2}$. Then
$$\omega = -zr\,dr + r^2\,dz = -ur^2\,dr + r^2(u\,dr + r\,du) = r^3\,du,$$
and so, naturally, $\dfrac 1{r^3}\omega = du$ is exact. (So, here, in the new coordinate system, $1/r^3$ does become the integrating factor!) Oh, and finally, the integral manifolds are patently obvious: They're given by $u=c$ for all constants $c$.


问:
Homotopy groups of non-contractible manifolds
Motivated by a proof in a differential geometry book and so far my lack of knowledge in algebraic topology I would like to know the following :

Is it possible to have a compact non-contractible manifold $M$ with finite fundamental group and trivial homotopy groups $\pi_k(M)$ for $k\geq 2$?

Any help or reference is appreciated, thanks in advance.

答:
This can also be achieved through Miller's theorem which states that the based mapping space $Map_*(BG,X)$ is weakly contractible if $G$ is a Lie group with a finite number of path components and $X$ is a nilpotent $CW$ complex with a finite number of cells.

In particular all maps $BG \rightarrow X$ are nullhomotopic.

Your $M$ is a $K(\pi_1(M),1)$ so also a $B\pi_1(M)$. Obviously $G = \pi_1(M)$ viewed as a discrete Lie group has finite number of path components and $M$ also has to be a finite $CW$ complex since it's compact. Thus the identity map $M =BG \rightarrow M$ is nullhomotopic and $M$ is contractible.


问:
The difference in finding a basis for the range of a matrix when using the columns versus the rows confuses me
Let $\{a_1,..,a_n\}$ be a set of vectors in $\mathbb{R}^m$.

In a first semester linear algebra course, two different but similar methods are introduced in order to find a basis for $span\{a_1,..,a_n\}$

Method 1: Consider $\{a_1,..,a_n\}$ as the rows of a matrix $A$ and reduce this matrix to echelon form, denoted $B$. Then the nonzero rows of $B$ will be a basis for $span\{a_1,..,a_n\}$

Method 2: Consider $\{a_1,..,a_n\}$ as the columns of a matrix $A$ and reduce this matrix to echelon form, denoted $B$. Then the columns of $A$ that correspond to the columns of $B$ that contain pivot positions form a basis for $span\{a_1,..,a_n\}$

I have two questions.

My first question is when using method $2$, wouldn't the columns of $B$ that contain pivot positions be a basis for $span\{a_1,..,a_n\}$? If so, why is this method presented as being dependent on going back to the columns of $A$?

My second question is, when using method $1$, would it be possible to use the rows of $A$ that correspond to rows of $B$ containing pivot positions to find a basis for $span\{a_1,..,a_n\}$? I suspect not. But what if we would have used column operations instead?

Anyway, I'm also just looking into general insight into understanding what is going on here, but my two questions would be a great place to start.

Thank goodness for MSE!!

答:
Elementary row operations do not change the row space of $A$. However, they can change the column space without altering the dependency of the columns. Meaning that if $\vec{v}_1,\vec{v}_2,\dots,\vec{v}_n$ are column vectors of $A$ and let $R$ be a row equivalent matrix to $A$ having column vectors $\vec{u}_1,\vec{u}_2,\dots,\vec{u}_n$ you can prove that the equation $k_1\vec{v}_1,k_2\vec{v}_2,\dots,k_n\vec{v}_n = 0$ has the same solutions as $k_1\vec{u}_1,k_2\vec{u}_2,\dots,k_n\vec{u}_n = 0$. It then follows that a set of columns of $A$ are linearly independent iff the corresponding columns of $R$ are linearly independent.

Your intuition for the second question is right. Row operations can alter the position of rows meaning that the rows on the original matrix in the same positions as the pivot rows may not form a basis for the row space.

Elementary column operations do not change the column space but alter the row space so you can basically swap the two methods if you use column operations.

Note that method 2 is beneficial if you also need to express the dependent vectors in the set as a linear combination of the basis vectors. Since the dependency between the columns is the same, you can solve the system using $R$(usually easier to compute) and the solution of scalars $\vec{k}$ is also the solution for the same system in $A$.


问:
Simple proof to show $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension
Let $(-2)^{\frac{1}{2}}$ be any element in $\overline{\mathbb Q_2}$ satisfying $x^2+2=0$, then is there a simple way to show that $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension of $\mathbb{Q}_2$?

What I know is that the polynomial $x^2+2$ is Eisenstein over $\mathbb{Q}_2$, hence irreducible, but I don't see how to continue.

This is a step in the proof of  local Kronecker-Weber theorem, so we can't use this theorem.

Thanks.

答:
$$\zeta_8= \frac{1+i}{\sqrt2}$$
(or $\frac{\pm 1\pm i}{\pm \sqrt2}$ if you prefer)

$\sqrt  {-2}= \zeta_8-\zeta_8^{-1}$