问:
Measure, Integration & Real Analysis Sheldon Axler Section 2B Exercise 16
I apologize for the title in advance, but this is a long problem. So, here it is:

Suppose $\mathcal{S}$ is a $\sigma$-algebra on a set $X$ and $A \subset X$. Let $$\mathcal{S}_A = \{E \in \mathcal{S} : A \subset E \text{ or } A \cap E = \emptyset\}$$

In the first question in this exercise, we're asked to show that $\mathcal{S}_A$ is a $\sigma$-algebra, which I have done and we can certainly use it if necessary in the second question, which is the one I'm struggling with. Here is the entire question:

Suppose $f: X \rightarrow \mathbb{R}$ is a function. Prove that $f$ is measurable with respect to $\mathcal{S}_A$ if and only if $f$ is measurable with respect to $\mathcal{S}$ and $f$ is constant on $A$

I was able to show that if $f$ is measurable with respect to $\mathcal{S}_A$ then it would be measurable with respect to $\mathcal{S}$ simply because $f^{-1}(B) \in \mathcal{S}_A$ for any Borel set $B$, but by the way $\mathcal{S}_A$ is defined, that would mean $f^{-1}(B) \in \mathcal{S}$ as well. The part that I'm struggling with is to first show that not only $f$ would be measurable with respect to $\mathcal{S}$ but also $f$ would be constant on $A$, and second is to show the converse to conclude the equivalence that is required. So, would anyone be able to help, please?

答:
To show that $f$ is a constant on $A$ take $a,b \in A$ and suppose $f(a)=c \neq d=f(b)$. Then $f^{-1} (\{c\})$ and $f^{-1} (\{d\})$ belong to $\mathcal S _A$ but $A$ is neither contained in $f^{-1} (\{c\})$  nor is it disjoint from it. [ $b \in A$ and $b \notin f^{-1} (\{c\})$;  Also, $a \in f^{-1} (\{c\}) \cap A$].

This contradiction proves that we must have $c=d$ or $f(a)=f(b)$.

The converse part is really trivial. If $f$ is a constant on $A$ what is the inverse image of any Borel set?


问:
Supremum of $x^2y^2(x^2+y^2)$
Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is

$3n^2$

$2n^6$

$4n^6$

$n^6$

My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.

But calculation will be cumbersome. Is there any smart way?

Thanks in advance.

答:
This is a good opportunity to use Lagrange's multipliers method:

Let $g(x,y)=x+y-2n$. We have

$$\nabla g=\langle 1,1\rangle \textrm{ and }\nabla f = \langle 2xy^2(y^2+2x^2), 2yx^2(x^2+2y^2)\rangle$$

At a maximum, $\nabla f$ and $\nabla g$ are parallel, so $xy^2(y^2+2x^2)=yx^2(x^2+2y^2)$. If $x=0$ or $y=0$, we have $f(x,y)=0$, which his the minimum value, so we can assume that $x\neq 0$ and $y\neq 0$. Therefore,

$$ y(y^2+2x^2)=x(x^2+2y^2) \iff y^3+2yx^2=x^3+ 2xy^2 \iff y^3-x^3=2xy^2-2yx^2$$

so $$ (y-x)(y^2+xy+y^2)=2xy(y-x)$$

This means that $x=y$ or $y^2+x^2-xy=0 \iff (y-\frac{x}{2})^2+\frac{3}{4}x^2=0 \iff x=y=0$, which we excluded.

Therefore, $x=y$ at the maximum, so $(x,y)=(n,n)$. Substituting, the maximum value is $f(n,n)=2n^6$.

Note: to be totally rigorous, Lagrange's method only says that, if the maximum value exists, then $\nabla f$ and $\nabla g$ are parallel at the maximum. The maximum actually exists here because $\{ (x,y) | x\ge 0, y\ge 0, x+y=2n \}$ is compact.

ContentLicense: CC BY-SA 4.0

Title: Supremum of $x^2y^2(x^2+y^2)$


问:
Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor.
Is there any reference that contains this proof:

Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor.

Or any help in the proof will be appreciated.

We defined projective module as follows:

An $R$-module $P$ is projective if, whenever $A,B$ are $R$-modules, $f \in \operatorname{Hom}_R(P,B),$ and $g \in \operatorname{Hom}_R(A,B),$ is surjective, there exists $\tilde{f} \in \operatorname{Hom}(P,A)$ such that $f = g \tilde{f}.$

And I proved that the $\operatorname{Hom}_R(M, -)$ functor is a left exact functor. Also, I have proved if $P$ is free then $P$ is projective.

Any help will be appreciated.

EDIT:

I already solved this problem:

$(a)$ Let $P$ be an $R$-module. Show that the following three conditions are equivalent.\
1.$P$ is projective.\
2. Any surjective homomorphism $g: A \rightarrow P$ of $R$-modules has a section.\
3. $P$ is a direct summand of a free module, i.e.,there is a free $R$-module $F$ and submodule $Q$ such that $F = P \oplus_R Q.$\

$(b)$ Part $(a)$ shows that, if $P$ is free, then $P$ is projective. Give an example of a finitely generated projective module which is not free.\

答:
Projectivity is basically defined to be exactly right exactness of $\hom_R(M,-)$; suppose we have a short exact sequence

$$0\to A\to B\overset g\to C\to 0.$$

Now you want to know that

$$0\to\hom_R(M,A)\to\hom_R(M,B)\to\hom_R(M,C)\to0$$

is right exact (you've already checked left exactness). So we really just need surjectivity of $\hom_R(M,B)\to\hom_R(M,C)$. If you have an element $f\in\hom_R(M,C)$, then by projectivity of $M$ you know there exists some $\tilde f\in\hom_R(M,B)$ such that $f=g\tilde f$. Unraveling the definitions check this is really just saying that $\tilde f\mapsto f$ under $\hom_R(M,B)\to\hom_R(M,C)$.


问:
Are all finite simple groups essentially permutation groups?
Pardon what may be my stupidity, but I genuinely need some direct clarification on whether or not this is true. I know that by the classification of the finite simple groups that every finite simple group is either cyclic, alternating, a type of lie group, or maybe one of the exceptions called sporadics (side note, is this what they called the "ENORMOUS theorem"?)

I think cyclic and alternating groups are permutation groups. I believe lie groups to be the group formed by symmetries on particular figures (which sounds very permutation-y to my simple brain). I have no idea what sporadic groups are.

That being said, simple groups have a very general definition that don't seem to automatically imply it to be equivalent to a permutation group. I believe this definition to be a nontrivial group with only {e} and the whole set as its normal subgroups. Perhaps when you have any particular finite simple groups it is isomorphic to some group of permutations on some set?

Edit:
I forgot about Cayley's thoerem for a second as I was just focusing on finite simple groups in particular. So am I to assume that cyclic, alternating, lie groups, and sporadic groups are all actually permutation groups?

As an additional question, if they all are, why identify finite simple groups as permutation groups in particular. Is there sometime special about identifying groups with their respective permutation group?

答: Cayley's theorem says that every group is a permutation group, under the correct definition of "permutation group": for any cardinality $\kappa$ we can define $S_\kappa$ as being the group of permutations of a set of cardinality $\kappa$, where "permutation" is simply defined to be a bijective map. Since left-multiplication by a group element $g \in G$ is automatically bijective, being invertible by the map that left-multiplies by $g^{-1}$, this gives us a map $\rho: G \to S_G$, where $S_G$ again is the group of permutations on $G$, considered as a set. To reiterate, we have $g \mapsto \rho_g$ where $\rho_g(h) = gh$. This embeds $G$ as a subgroup of a group of permutations.


问:
Show $f: [0,1] \rightarrow [0, \infty)$ is Lebesgue measurable.

Let $X = [0,1]$. Let $f: X \rightarrow [0, \infty)$. Define $G(f) = \{ (x,y)\in X \times [0, \infty] : y \leq f(x) \}$. Show that $f$ is $m$-measurable iff $G(f)$ is $m^2$-measurable and that $$m^2(G(f))= \int f \,\mathrm{d}m.$$

I already proved the direction by defining $g(x,y)= f(x)-y$, showing $g$ is $m-$measurable and using Tonelli's Theorem, but I'm not sure how to do the backwards direction.

Any help is appreciated!

Thanks!

答:
Let $X = [0,1]$. Let $f: X \rightarrow [0, \infty)$. Define $G(f) = \{ (x,y)\in X \times [0, \infty] : y \leq f(x) \}$. Show that $f$ is $m$-measurable iff $G(f)$ is $m^2$-measurable and that $$m^2(G(f))= \int f \,\mathrm{d}m.$$

Proof:  $(\Rightarrow)$ Suppose $F$ is $m$-measurable (where $m$ is the Lebesgue measure). Then $g : X \times [0, \infty) \rightarrow \Bbb R$ defined by $g(x,y)= f(x)-y$ is a mesasurable function. So $ G(f) = g^{-1}([0, \infty))$  is a $m^2$ measurable set.

$(\Leftarrow)$ Suppose that $G(f)$ is $m^2$-measurable. Then,
for all $b \in [0, \infty)$, and all $n \in \Bbb N$, $n>0$, the function $h_{b,n}:  X \times [0, \infty) \rightarrow  X \times [0, \infty)$ defined by $h_{b,n}(x,y) = \left (x, \frac{1}{n}y+b \right)$ is Lebesgue-Lebesgue-measurable (because $h_{b,n}$ is an invertible affine transformation in $\Bbb R^2$, restricted to $ X \times [0, \infty)$).

So,
\begin{align*} h_{b,n}^{-1}(G(f)) & = \left \{ (x,y) \in X \times  [0, \infty) : \left (x, \frac{1}{n}y+b \right)\in G(f)\right \} = \\
& =\left \{ (x,y) \in X \times  [0, \infty) : f(x) \geqslant  \frac{1}{n}y+b \right \}
\end{align*}
is $m^2$-measurable.

So
\begin{align*}
h_{b,n}^{-1}(G(f)) \cap (X \times (0,\infty))  & =\left \{ (x,y) \in X \times  [0, \infty) : f(x) \geqslant  \frac{1}{n}y+b ,\: y>0\right \}
\end{align*}
is $m^2$-measurable.

So, for all $b \in [0, \infty)$,

\begin{align*} \bigcup_{n=1}^\infty \left (h_{b,n}^{-1}(G(f)) \cap (X \times (0,\infty)) \right ) & = \bigcup_{n=1}^\infty \left \{ (x,y) \in X \times  [0, \infty) : f(x) \geqslant  \frac{1}{n}y+b ,\: y>0\right \} =\\
& = \{(x,y)  \in X \times  [0, \infty): f(x)>b , \: y > 0 \} = \\
& = f^{-1}((b, \infty))\times (0, \infty)
\end{align*}

is $m^2$-measurable. So $f^{-1}((b, \infty))\times (0, \infty)$ is a $m^2$-measurable rectangle and $m((0, \infty)) >0$. So $f^{-1}((b, \infty))$ is $m$-measurable. So $f$ is $m$-measurable.

Finally if one of the two conditions hold (and so both holds) we have that $\chi_{G(f)}$ is a measurable non-negative function, and applying Tonelli's Theorem, we have
$$ m^2(G(f)) = \int \chi_{G(f)} dm^2 =\iint \chi_{G(f)}(x,y) \mathrm{d}m(y) \mathrm{d}m(x) = \int f(x) \,\mathrm{d}m(x) =  \int f \,\mathrm{d}m$$


问:
Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$
In my AI textbook there is this paragraph, without any explanation.

The sigmoid function is defined as follows

$$\sigma (x) = \frac{1}{1+e^{-x}}.$$

This function is easy to differentiate because

$$\frac{d\sigma (x)}{d(x)} = \sigma (x)\cdot (1-\sigma(x)).$$

It has been a long time since I've taken differential equations, so could anyone tell me how they got from the first equation to the second?

答:
Another approach using the quotient rule is as follows:

let $\sigma=\frac{1}{1+e^-x}$
upon rearranging ,

$\sigma=\frac{e^x}{1+e^x}$

using the quotient rule

$\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}$,

let $f(x)=e^x$ and $g(x)=1+e^x$,

we get
$\frac{d\sigma}{dx}=\frac{(e^x)(1+e^x)-e^xe^x}{(1+e^x)^2}$

upon rearraning, we get :

$\frac{d\sigma}{dx}=\frac{e^x}{1+e^x}\frac{(1+e^x)-e^x}{1+e^x}$

upon further rearrangement:

$\frac{d\sigma}{dx}=\frac{e^x}{1+e^x}(1-\frac{e^x}{1+e^x})$

$\frac{d\sigma}{dx}=\sigma(1-\sigma)$

ContentLicense: CC BY-SA 3.0


问:
Contraction Mapping and fixed point
Given $f: X \to \mathbb{R}$ ($X \subseteq \mathbb{R})$ such that $f$ is differentiable on the domain. Similarly, $x_0$ is a fixed point that lies in the interior of $X$. We are given that $f'$ is continuous at $x_0$ with $|f'(x_0)|< 1.$ Show that there exists a closed interval $K$ centered at $x_0$ such that $f$ is contraction on $K$.

I used the definition of continuity to start with:
Fix $\epsilon > 0.$ Then, there exists $\delta > 0$ such that whenever $x \in (x_0 - \delta , x + \delta), |f'(x) - f'(x)| < \epsilon.$ I think now I should try to show that for all $x \in (x_0 - \delta , x + \delta), |f'(x)| \leq r$ such that $|f'(x_0)| \leq r < 1.$ Then, I could use mean value theorem to find the interval needed. But I am not able to proceed ahead.

Does this look like a right direction?

答: Notice that there exists $ \epsilon >0$ such that $|f'(x_0)|<1-\epsilon$, for that $\epsilon > 0$ there exists $\delta_1 >0$ such that $|f'(x)-f'(x_0)|< \epsilon$ for all $x \in (x_0-\delta_1, x_0+\delta_1)$, then $|f'(x)|\leq |f'(x_0)|+|f'(x)-f'(x_0)|<|f'(x_0)|+\epsilon~~~(<1-\epsilon+\epsilon=1)$ for all $x \in (x_0-\delta_1, x_0+\delta_1)$. Let $\delta=\frac{1}{2}\delta_1$, for any $x_1<x_2 \in [x_0-\delta, x_0+\delta]$ and by MVT in $[x_1,x_2]$, $|f(x_1)-f(x_2)|=|f'(\xi)|\cdot|x_1-x_2|\leq(|f'(x_0)|+\epsilon)|x_1-x_2|$ since $\xi \in (x_0-\delta_1, x_0+\delta_1)$. Thus $f$ is a contraction on $[x_0-\delta, x_0+\delta]$.

Title: Contraction Mapping and fixed point


答:
We have the following fact.

Proposition. Let $(X,\mathcal{F})$ and $(Y,\mathcal{G})$ be measurable
spaces. Let $(X\times Y,\mathcal{F}\otimes\mathcal{G})$ be the usual
product. For any subset $G\subseteq X\times H$, $x\in X$, $y\in Y$,
define sections $G^{x}=\{y\mid(x,y)\in G\}$ and $G_{y}=\{x\mid(x,y)\in G\}$.
If $G\in\mathcal{F}\otimes\mathcal{G}$ and $x\in X$, $y\in Y$,
then $G^{x}\in\mathcal{G}$ and $G_{y}\in\mathcal{F}$.

Proof: Let $x\in X$ and $y\in Y$ be fixed. Define $\theta^{x}:Y\rightarrow X\times Y$
and $\theta_{y}:X\rightarrow X\times Y$ by $\theta^{x}(b)=(x,b)$
and $\theta_{y}(a)=(a,y)$. We go to show that $\theta^{x}$ is $\mathcal{G}/\mathcal{F}\otimes\mathcal{G}$-measurable
while $\theta_{y}$ is $\mathcal{F}/\mathcal{F}\otimes\mathcal{G}$-measurable.
Let $\pi_{X}:X\times Y\rightarrow X$ and $\pi_{Y}:X\times Y\rightarrow Y$
be canonical projections, i.e., $\pi_{X}(a,b)=a$ and $\pi_{Y}(a,b)=b$.
Recall that product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{G}$
has the universal property: For any measurable space $(Z,\mathcal{H})$
and map $g:Z\rightarrow\mathcal{F}\otimes\mathcal{G}$, $g$ is measurable
iff $\pi_{X}\circ g$ and $\pi_{Y}\circ g$ are measurable. Now, observe
that $\pi_{X}\circ\theta^{x}:Y\rightarrow X$ is the constant map
$t\mapsto x$ while $\pi_{Y}\circ\theta^{x}:Y\rightarrow Y$ is the
identity function $t\mapsto t$, and both of them are clearly measurable.
Therefore, $\theta^{x}$ is $\mathcal{G}/\mathcal{F}\otimes\mathcal{G}$-measurable.
Similarly $\theta_{y}$ is $\mathcal{F}/\mathcal{F}\otimes\mathcal{G}$-measurable.

Observe that $G^{x}=(\theta^{x})^{-1}(G)$ and hence $G^{x}\in\mathcal{G}$.
Similarly, $G_{y}=(\theta_{y})^{-1}(G)$ and hence $G_{y}\in\mathcal{F}$.

For your question. Suppose that $G(f)\in\mathcal{B}([0,1]\times[0,\infty))=\mathcal{B}([0,1])\otimes\mathcal{B}([0,\infty))$
(this is true because the usual topology on $\mathbb{R}$ is second
countable). Let $y\in[0,\infty)$ be arbitrary. By the above proposition,
$G(f)_{y}\in\mathcal{B}([0,1])$. However, $G(f)_{y}=\{x\in[0,1]\mid y\leq f(x)\}=f^{-1}([y,\infty)).$
This shows that $f$ is Borel.

Title: Show $f: [0,1] \rightarrow [0, \infty)$ is Lebesgue measurable.

Body:
Let $X = [0,1]$. Let $f: X \rightarrow [0, \infty)$. Define $G(f) = \{ (x,y)\in X \times [0, \infty] : y \leq f(x) \}$. Show that $f$ is $m$-measurable iff $G(f)$ is $m^2$-measurable and that $$m^2(G(f))= \int f \,\mathrm{d}m.$$

I already proved the direction by defining $g(x,y)= f(x)-y$, showing $g$ is $m-$measurable and using Tonelli's Theorem, but I'm not sure how to do the backwards direction.

Any help is appreciated!

Thanks!


问:
Let N1 and N2 are normal subgroups in the finite group G. Is it true that if N1≃N1 then G∖N1≃G∖N2.?
Let $N_1$ and $N_2$ are normal subgroups in the finite group $G$. Is it true that if $N_1 \simeq N_2$ then $G/ N_1 \simeq G/ N_2$? </p>

答:
No.

Think of $G = \langle a \rangle \times \langle b \rangle$, where $a$ has order $4$, and $b$ has order $2$.

The quotient of $G$ modulo $\langle b \rangle$ is cyclic of order $4$, while in the quotient modulo $\langle a^2 \rangle$ all non-identity elements have order $2$.

Still $\langle a^2 \rangle \cong \langle b \rangle$, as they are both cyclic of order $2$.


问:
Why linear independency of solutions of ordinary differential equations or functions is important?
There are so many methods for checking linear independency of functions. But why do we require this? I mean what is the use to find out whether functions or solutions are linear independent or not.</p>

答: Linear independence of solutions is useful for describing the entire solution space. Typically, with an ODE you have a finite-dimensional space of solutions, and to describe this solution space you want to supply a list of solutions $\{f_1,\ldots,f_n\}$ that will be sufficient to describe any solution to the equation via a linear combination $\alpha_1f_1 + \cdots \alpha_nf_n$. It's helpful in this process to ensure that the functions $f_1,\ldots,f_n$ are linearly independent, since otherwise we would effectively be throwing in redundant solutions that do us no good in this regard. Thus, ensuring that the set is linearly independent guarantees that we have exactly as many functions as we need and no more.


问:
Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$
When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)

Thanks

答:
Let the left side be $f(a,b),$ a polynomial in $a$ and $b.$  It is easily confirmed that  the coefficient of $b^3$ is zero. Thus, $$f(a,b)=c_0(a)+c_1(a)b+c_2(a)b^2.$$ It is easy to check that
$$ f(a,0)=f(a,a)=f(a,-1)=a^2+a.$$
Thus, by interpolation (considering $f(a,b)$ as a polynomial of degree two in $b$),
we conclude that $f(a,b)=a^2+a.$


答:
By the AM-GM inequality, we have
$$xy \leqslant \left(\frac{x+y}{2}\right) ^2 =n^2,$$
and
$$xy(x^2+y^2) \leqslant \frac{1}{2} \cdot \left(\frac{2xy+(x^2+y^2)}{2}\right)^2=\frac{1}{2} \cdot \left(\frac{(x+y)^2}{2}\right)^2 = 2n^4.$$
Therefore
$$x^2y^2(x^2+y^2) \leqslant 2n^6.$$

Title: Supremum of $x^2y^2(x^2+y^2)$

Body:
Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is

$3n^2$

$2n^6$

$4n^6$

$n^6$

My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.

But calculation will be cumbersome. Is there any smart way?

Thanks in advance.


问:
prove that $\forall k\geqslant1\land k\in\mathbb{Z}.\nexists q\in\mathbb{Z}.q^{2}=(k-i)(k+i)$
prove that $\forall k\geqslant1\land k\in\mathbb{Z}.\nexists q\in\mathbb{Z}.q^{2}=(k-i)(k+i)$.
My try: use gaussian integer to claim that we have ufd. the problem, is that k might not be irreducible in gaussian primes.
Edit: without using that non zero squares cannot differ by 1.

答: Hint: Two nonzero perfect squares in $\Bbb{Z}$ cannot differ by $1$.


问:
How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$?
The problem is as follows:

Which value should have $K$ in the expression from below so it makes an identity?

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$

I'm confused exactly how to get that particular value for $K$.

The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$

$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$

$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$

$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$

$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$

However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.

Could someone help me with an orderly method to solve this without much fuss?.

答:
If it is an identity, it will be true for any small angle $x$.

Using Taylor series, we have

$$\cos(nx)=1-\frac{n^2 }{2}x^2+O\left(x^4\right)$$ This makes the lhs to be
$$4 (1-32 x^2+1-18 x^2)(1 - 18 x^2+1-2 x^2)-1=15-560 x^2+O\left(x^4\right) \tag 1$$

Now, for the rhs
$$\sin(Kx)=K x-\frac{K^3 }{6}x^3+O\left(x^5\right)$$
$$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$$
$$\sin(Kx)\csc(x)=K+\frac{1}{6} \left(K-K^3\right) x^2+O\left(x^3\right)\tag 2$$

Compare $(1)$ and $(2)$. This gives $K=15$ and you can check that, for $K=15$
$$\frac{1}{6} \left(K-K^3\right)=-560$$


问:
Why is $F=\{0\}$ a field?
Context: I am new to this. I started my course yesterday.

I know that the operations $+$ and $\cdot$ are required to satisfy the field axioms. So how can $F=\{0\}$ be a field?
Recall the additive and multiplicative identity:
there exist 2 different elements $0$ and $1$ in $F$ such that $ a + 0 = a$ and $a \cdot  1 = a$

But there is only one element in $F$? Shouldn't that mean that this rule has not been satisfied and thus $F$ is not a field?

My teacher mentioned something about $0=1$ can someone explain that?

Update my lecture notes:

答: Under the usual axioms $\{0\}$ is not a field. E.g., if $F$ is s field and $0$ its additive neutral then $F\setminus\{0\}$ is a group under multiplication - but groups cannot be empty. Put differently, one often takes $0\ne 1$ as one of the axioms of a field.

Title: Why is $F=\{0\}$ a field?


答: No, take $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, with $H=\langle (1,0)\rangle$ and $K=\langle (0,2)\rangle$</p>

Title: Let N1 and N2 are normal subgroups in the finite group G. Is it true that if N1≃N1 then G∖N1≃G∖N2.?

Body: Let $N_1$ and $N_2$ are normal subgroups in the finite group $G$. Is it true that if $N_1 \simeq N_2$ then $G/ N_1 \simeq G/ N_2$? </p>


问:
$C^{k}_{a}(X) = a \forall X \in \mathbb{N}^{k}$ explanation.
I was reading about functions in "Discrete Mathematics and its Applications", and in the "Recursive Functions" section I came across initial functions. One of which is the idea of a constant function.

Now I understand the idea of a constant function. A constant function is a kind of function which no matter the domain, will always give the same output. For example the following is a constant function :-

$$f(x) = 9, \forall x \in \mathbb{R}$$

But the definition provided in the example in this book, that I could not understand. The definition is as follows :-

$$
\text{ For each } k \geq 0 \text{ and each } a \geq 0 \text{ ,the constant function } C^{k}_{a}:\mathbb{N}^{k} \to \mathbb{N} \text{ is defined by the formula } \\
C^{k}_{a}(X) = a \forall X \in \mathbb{N}^{k}
$$

Apart from the definition itself, I don't understand the notations $C^{k}_{a}$ and $\mathbb{N}^{k}$. Any help would be highly appreciated.

答:
It is saying that for every natural number $k$ and for every natural number $a$, for each element $X$ of $\mathbb{N}^k$ the function $C_k^a$ maps $X$ to $a$. The notation of the function $C_k^a$ provides you with information about the domain of the function, given by $k$, and the output of the function, given by $a$.

The notation $\mathbb{N}^k$ is notation for the cartesian product of $N$ with itself $k$ times. It’s elements are $k$-tupples of natural numbers.


问:
Extending injectivity of closed sets in injective immersions
I am trying to do the following exercise :

A $C^1$ immersion $f:M\rightarrow N$ which is injective on a closed subset $F\subset M$ is injective on a neighborhood of $F$. In fact $f$ has a neighborhood  $ V\subset C_S^1(M,N)$ and $F$ has a neighborhood $U\subset M$ such that every $g\in V$ is injective on $U$. If $K$ is compact $V$ can be taken in $C_W^1(M,N)$.

Focusing only on the first part of the problem that is the part where can extend injectivity of closed sets ininjective immersions.
I have been thinking about this for a while and there are two ideas, either I construct the neighborhood , I have tried this using the fact that $f$ is locally injective , or I go by contradiction but I can't get anytihing. Either way I got nowhere useful and was wondering if there is anything else that I can try or if there is something I am missing ?

Any advice on this is aprecciated. Thanks !

答:
Hint: Proper maps are closed maps; immersions are locally injective.
Then try to construct the neighborhood on a countable series of ordered compact subsets , with induction.</p>


答:
Every group $G$ is a group consisting in permutations, in the sense that, by Cayley's theorem, it can be imbedded in a symmetric group, the group of symmetries of $G$.

In case $G$ is finite, we have that $G$ is isomorphic to a subgroup of the symmetric group $S_n$, where $n=|G|$.

In the infinite case, we still have that $G\hookrightarrow\rm{Sym}G$, the group of symmetries of $G$.

Because of this anything that is true for subgroups of symmetric groups is true for all groups.

What I know is still fairly limited, even though I appear to be headed for gold badges in group theory and abstract algebra, eventually.

The classification of finite simple groups was very difficult and time consuming, and is considered one of the great intellectual achievements.

Furthermore, since every finite group has a composition series, it can be realized as a series of extensions of simple groups.  Thus a solution to the so-called "extension problem", would lead to a classification of finite groups.

Title: Are all finite simple groups essentially permutation groups?

Body:
Pardon what may be my stupidity, but I genuinely need some direct clarification on whether or not this is true. I know that by the classification of the finite simple groups that every finite simple group is either cyclic, alternating, a type of lie group, or maybe one of the exceptions called sporadics (side note, is this what they called the "ENORMOUS theorem"?)

I think cyclic and alternating groups are permutation groups. I believe lie groups to be the group formed by symmetries on particular figures (which sounds very permutation-y to my simple brain). I have no idea what sporadic groups are.

That being said, simple groups have a very general definition that don't seem to automatically imply it to be equivalent to a permutation group. I believe this definition to be a nontrivial group with only {e} and the whole set as its normal subgroups. Perhaps when you have any particular finite simple groups it is isomorphic to some group of permutations on some set?

Edit:
I forgot about Cayley's thoerem for a second as I was just focusing on finite simple groups in particular. So am I to assume that cyclic, alternating, lie groups, and sporadic groups are all actually permutation groups?

As an additional question, if they all are, why identify finite simple groups as permutation groups in particular. Is there sometime special about identifying groups with their respective permutation group?


问:
What is the other way to write this triple integral?
I need to find all $5$ ways to write this integral and so far I have found $3$ of the $5$ but I need some help to find the last $2$ which are $dy$ $dz$ $dx$ and $dy$ $dx$ $dz$. I assume one of them might require you to add $2$ triple integrals  together. Anyways when evaluated the answer needs to come out to about $12.669$ which can be checked via Desmos. Can anyone help?

$$\int_0^2\int_{e^y}^{e^2}\int_0^{e^\left(\frac{y}{2}\right)} f(x,y,z) \ dz \ dx \ dy$$

答:
$$\begin{matrix}\begin{matrix}2\log z<y<\log x\\z^2<x<e^2\\1<z<e\end{matrix}&\begin{matrix}0<y<\log x\\1<x<e^2\\0<z<1\end{matrix}\end{matrix}$$

So the integral is $\int_1^e\int_{z^2}^{e^2}\int_{2\log z}^{\log x} dydxdz+\int_0^1\int_1^{e^2}\int_0^{\log x} dydxdz$. It consists of a slice of a rectangular prism on the bottom and a twisted part on the top.


问:
Accessible Proof of Denjoy-Riesz Theorem?
The Denjoy-Riesz Theorem states that any compact zero-dimensional subset of the plane is contained in the image of an embedded copy of the arc $[0,1]$.  The only place I've seen this proved is in Kuratowski's Topology II, and it's proved as a trivial consequence of a huge body of theory on unicoherence, Janiszewski spaces, characterizations of the sphere etc.

The "original proof" by Riesz is known to have flaws.  Actually I've also seen it proved in a paper by Moore, but there he proves a generalization and it's atrocious to read as well.

Is there a more self-contained, modern proof accessible anywhere?

答:
I'll use the Schoenflies Theorem: If $J_1$ and $J_2$ are Jordan curves in the plane (or sphere) there is an ambient plane (resp. sphere) isotopy sending $J_1$ to $J_2$.  This is sometimes called the "strong" Schoenflies Theorem (it's a consequence of the Annulus Theorem).  Let $\text{Int}(J)$ and $\text{Ext}(J)$ denote the bounded and unbounded complementary regions of $J$, respectively.

If $J_1, \dots, J_n$ are a collection of mutually disjoint Jordan curves, then call the collection simple if no $J_i$ is contained in any $\text{Int}(J_k)$.  By induction and repeated use of the Schoenflies Theorem:

Observation 1: Any two simple collections of Jordan curves are ambient
isotopic if they have the same number of elements.

A continuum (plural continua) is a compact, connected metric space.  We need the following theorem: A nested intersection of continua is a continuum, as proved in the body of the question here.  Relevant is that since the Cantor Set is totally disconnected, it doesn't contain a non-degenerate continuum.

Let $B_\epsilon(x) = \lbrace y: |x-y| \leq \epsilon \rbrace$ and let $B_\epsilon(X) = \cup_{x \in X} B_\epsilon(x)$.  If $X \subset \mathbb{R}^2$ is a continuum then let $X_\circ$ denote the filled in continuum obtained by adding each bounded component of its complement (it's a continuum since its complement is open and a sufficiently large disc around $X$ has the boundary bumping property).

Lemma 1: Let $C_1$ and $C_2$ be disjoint Cantor sets in the plane (or sphere).
Then there is a simple pair of disjoint Jordan curves $J_1$ and $J_2$
with $C_i \subset \text{Int}(J_i)$.

Proof: By Observation 1, it's sufficient to show that there are simple collections $\mathcal{A}_1$ and $\mathcal{A}_2$ containing $C_1$ and $C_2$ respectively such that $\mathcal{A} = \mathcal{A}_1 \cup \mathcal{A}_2$ is also simple, since in this case it is explicitly constructible after suitable isotopy.  Since each $C_i$ is compact each $B_\epsilon(C_i)$ is a finite cover by Euclidean discs and thus each complementary component is a Jordan domain (apart from the unbounded one).

Letting $\epsilon$ be small enough that $B_\epsilon(C_1) \cap B_\epsilon(C_2) = \varnothing$, we get disjoint covers of each such that filling in their components produces covers by Jordan domains.  However, it's possible that for some components $D$ of $B_\epsilon(C_1)$ and $E$ of $B_\epsilon(C_2)$, we have $D_\circ \cap E \neq \varnothing$, without loss of generality.  In which case $E$ is contained in a complementary region of $D$ since it's disjoint from it.

In $D$, find a $\delta$ sufficiently small so that $E$ is no longer contained in any complementary component of $B_\delta(D \cap C_1)$.  If none such existed, then for $\delta_n = \frac{1}{n}$ we have that $E$ is contained in a bounded complementary component of each $B_{\delta_n}(D \cap C_1)$, say of component $D_n$ respectively which are nested by construction.  But this is impossible since it would imply the existence of a non-degenerate continuum in $C_1$, as $D_n$ cannot converge to a point while all separate $E$ from $\infty$ unless $D_n \rightarrow E$, but $D_n \subset D$ which is disjoint from $E$ and closed.

After doing this in each component of $B_\epsilon(C_1)$, by finite descent we have a simple collection $\mathcal{A}_1$ of Jordan curves with interior domains covering $C_1$ such that none intersect $C_2$, nor do their interiors.  Doing the same for $B_\epsilon(C_2)$ produces the desired family $\mathcal{A}$. QED

Let $C$ be the standard Cantor ternary set, viewed as a subset of the real line $\mathbb{R} \times \lbrace 0 \rbrace \subset \mathbb{R}^2$.  Let $E$ denote the set of endpoints of the (bounded) deleted intervals in the real line, and let $E_n$ denote the endpoints of the intervals with length $\frac{1}{3^n}$, called the $n$th level of $E$.  Note that $E$ is dense in $C$.  Call the removed intervals $U_j^n$ the $n$th level of removed intervals.  Call the $2^n$ neighborhoods $D_j^n$ complementary to the $n$th level of removed intervals the $n$th levels of $C$.  Note that the collection $\mathcal{D}_n = \lbrace D_j^n \rbrace$ is a covering of $C$ for every $n$, and each $D_j^n$ is a Cantor set that's clopen in $C$.

Let $I = [0,1]$ be the unit interval.  If $X, Y \subset \mathbb{R}^2$ and $f: X \rightarrow Y$ is a homeomorphism, we say that an ambient isotopy $F: \mathbb{R}^2 \times I \rightarrow \mathbb{R}^2$ extends $f$ if we have that $f(X) \equiv F(X \times \lbrace 1 \rbrace)$.

Theorem: If $C_1, C_2$ are Cantor sets in the plane and $f: C_1 \rightarrow C_2$ is a homeomorphism, there exists an ambient plane isotopy extending $f$.

Note that the proof below also works for the sphere.

Proof: Let $g$ be an arbitrary homeomorphism from $C_1$ onto $C$, the standard ternary Cantor set.  We produce an ambient isotopy from $C_1$ to $C$ extending $g$.  Then there also exists isotopies extending arbitrary homeomorphisms $C_2 \rightarrow C$, and homeomorphisms between compact metric spaces are quotient maps.  Thus we may factor $f$ through $g$ to obtain an $h^{-1}$ such that $f = h^{-1} \circ g: C_1 \rightarrow C_2$, a concatenation of ambient isotopies and thus an ambient isotopy itself.

By Lemma 1 and Observation 1 there is a simple pair of Jordan curves $J_1, J_2$ around $g^{-1}(D_1^1)$ and $g^{-1}(D_2^1)$, and these may be isotoped to narrow discs sitting above $D_1, D_2$ as shown below such that the preimages of the first level of $E$ share $x-$coordinates with their images.  The interior of each is homeomorphic to the plane, so by induction we may apply Lemma 1 and Observation 1 repeatedly to isotope inverses as necessary to approximate $g$ on $g^{-1}(E)$.  By concatenating isotopies which differ on uniformly smaller sets, we obtain a homotopy on the plane sending $C_1$ to $I$ and which is a homeomorphism for all $t < 1$.  We need to show it's a homeomorphism at $t = 1$ on $C_1$, as it is clearly bijective off of $C_1$.

Since $g$ is a homeomorphism, $g^{-1}(E)$ is dense in $C_1$ so it uniquely determines the function $F$ restricted at time $t = 1$.  Since both are compact metric spaces it's sufficient to show that $F$ is bijective on $C_1$ at $t = 1$.  But this is immediate by construction, since any two points in $C_1$ are contained in disjoint elements of some $n$th level of $C$ and these have images with distance bounded below by $\frac{1}{3^n}$.  QED

In higher dimensions the proof breaks down at the observation.

Anyone see how to continue to grab the Moore-Kline result?


答:
Let $x = \pi/4$. Then $\cos 2x = 0, \cos 6x = 0$, and $\cos 8x = 1$, which gives:

$$0 - 1 = \sin(K\pi/4) \cdot \csc \pi/4 \Rightarrow\sin(K\pi/4) = -\frac{1}{\sqrt2}$$
$$\Rightarrow K \pi/4 + 2\pi k= 5\pi/4, 7\pi/4$$

which only $K = 15$ satisfies from the given options.

Title: How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$?

Body:
The problem is as follows:

Which value should have $K$ in the expression from below so it makes an identity?

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$

I'm confused exactly how to get that particular value for $K$.

The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$

$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$

$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$

$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$

$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$

However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.

Could someone help me with an orderly method to solve this without much fuss?.


问:
Let $(x_n)$ be a convergent sequence in a metric space. Show that $\{x_n: n \in \Bbb N\} \cup \lim x_n$ is a compact subset.
Hello I am trying to show the following.

Let $(x_n)$ be a convergent sequence in a metric space. Show that $\{x_n: n \in \Bbb N\} \cup \{\lim x_n\}$ is a compact subset.

I am proving by Heinel-Borel, therefore I have to prove that the set is bounded and closed. I have already proved that the set is bounded, can anyone help me out with the close statement?

答: The Heine-Borel theorem does not hold in metric spaces in general. However, the result is easy to prove directly from the definition of compactness: start with an open cover $\mathscr{U}$ of the set in question, and show that it has a finite subcover. HINT: Some member of $U$ must contain the limit of the sequence.


问:
Evaluate this integral without using the residue theorem
I want to evaluate this integral

$$ I = \int_c \tan z + \frac{\csc z}{z} dz $$
$$ c :|z| = 1 $$

apparently tanz is analytic in this region so its integral equals to zero
now
$$ I = \int_c\frac{\csc z}{z} dz $$
okay I did the following

$$ \frac{\csc z}{z} = \frac{1}{z\sin z} =\frac{1}{z(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots)}  $$

$$ \frac{\csc z}{z} = \frac{1}{z^2(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots)} $$

letting
$$  g(z) = \frac{1}{(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots)} $$
then

$$ \frac{\csc z}{z} = \frac{g(z)}{z^2} $$

$$ I = \int_c\frac{\csc z}{z} dz = \int_c\frac{g(z)}{z^2} dz  $$

using cauchy's integral formula for derivatives
$$ \int_c\frac{g(z)}{z^2} dz = \frac{2\pi i }{1!}  g'(0) $$

well $$ g'(0) = 0 $$
and the whole integral equals to zero this means the function is analytic which doesn't seem to be true did I make a mistake?

答:
First of all, $\tan$ is only analytic away from its poles, but there aren't any inside the unit circle. Second, you would need to know $\int_C g(z)\,dz=0$ with $g$ continuous and for all circles $C$ to conclude $g$ is analytic.

One shortcut for you: $(\csc z)/z$ is an even function, so, without computation, its residue at $0$ must be $0$.


答:
As pointed out in the comments, Heine Borel does not apply.

Let $x := \lim x_n$ and $X$ be the set in question. Let $\cal U$ be an open cover of $X$. There exists some $U_0 \in \cal U$ containing $x$.

Since $U_0$ is open, there exists some $N \in \Bbb N$ such that $x_n \in U_0$ for all $n > N.$

Thus, $U_0 \setminus X$ has at most $N$ elements. For each element, you can pick open sets $U_1, \ldots, U_N$ from the cover $\cal U$.

Then, $\{U_0, \ldots, U_N\}$ is a desired finite subcover.

Title: Let $(x_n)$ be a convergent sequence in a metric space. Show that $\{x_n: n \in \Bbb N\} \cup \lim x_n$ is a compact subset.

Body:
Hello I am trying to show the following.

Let $(x_n)$ be a convergent sequence in a metric space. Show that $\{x_n: n \in \Bbb N\} \cup \{\lim x_n\}$ is a compact subset.

I am proving by Heinel-Borel, therefore I have to prove that the set is bounded and closed. I have already proved that the set is bounded, can anyone help me out with the close statement?


答: To address your additional question, perhaps it worths noting that if $G$ is a finite simple group of order $n$ and $H$ is a proper, nontrivial subgroup of $G$, then $G$ is isomorphic to a subgroup of $S_m$, where $m=[G:H]<n$, namely there's a "sharper" embedding of $G$ into a symmetric group than the one ensured by Cayley's theorem. In fact, the action of $G$ by left multiplication on the left quotient set $G/H$ has trivial kernel.


问:
How to get $A$ and $B$ from $A\csc 10^\circ+B=$ $\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$?
The problem is as follows:

Find $A+B$ from:

$A\csc 10^\circ+B=\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{5}\\
2.&\textrm{2}\\
3.&\textrm{1}\\
4.&\textrm{4}\\
\end{array}$

What I did to attempt solving this problem was as follows: Please note that what itis written below follows the right side of the equation so it can be equated properly.

$\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$

The curious thing which I noticed is that I could arrange them in the following way:

$\sin 10^\circ + \sin 30^\circ + \sin 50^\circ + \sin 70^\circ + \sin 90^\circ$

But that's it. Its just the part where I got stuck. Is there any trick here?. Can someone help me here?.

Since this problem was found in a section belonging to prosthaphaeresis topic on my workbook. Thus what could it be done here?.

What it confuses me the most is how to get to appear the cosecant of $10^\circ$. So all and all can someone help me with the appropiate method for this?.

答:
When angles are in arithmetic progression, the formula for sum of their sines (cosines) is given by

$$\sin A + \sin (A+D) + \sin (A+2D) + \ldots + \sin (A+(n-1)D) =$$
$$ \frac{\sin nD/2}{\sin D/2}\, \sin \left( \frac{2A+(n-1)D}{2} \right)$$

This should remind you of an usual arithmetic series sum
$$A+(A+D)+(A+2D)+\ldots+(A+(n-1)D)=$$
$$n \times \frac{\text{first term + last term}}{2}$$

In a memorable way,
$$\sum_{k=0}^{n-1} \sin(A+kD)= \frac{\sin n\tfrac{D}{2}}{\sin \tfrac{D}{2}}\, \sin \left( \frac{\text{first angle + last angle}}{2} \right)$$
$$\sum_{k=0}^{n-1} \cos(A+kD)= \frac{\sin n\tfrac{D}{2}}{\sin \tfrac{D}{2}}\, \cos \left( \frac{\text{first angle + last angle}}{2} \right)$$
Please note, in sum of cosines, only the second factor changes.

Now to solve the problem, we first use $\cos \theta = \sin (90-\theta)$ to get the angles in AP.
$$\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$$
$$=\cos 80^\circ+\cos 60^\circ+\cos 40^\circ+\cos 20^\circ+1$$
$$=1+\frac{\sin 4\cdot \tfrac{20^\circ}{2}}{\sin \tfrac{20^\circ}{2}}\, \cos \left( \frac{20^\circ+80^\circ}{2} \right)=1+\frac{\sin 40^\circ \cos 50^\circ}{\sin 10^\circ}$$
$$=1+\frac{2\sin^2 40^\circ}{2\sin 10^\circ}=1+\frac{1-\cos 80^\circ}{2\sin 10^\circ}=\frac{1+\sin 10^\circ}{2\sin 10^\circ}$$
$$=\frac{1}{2}+\frac{1}{2}\csc 10^\circ$$


问:
Homomorphism of $R$-modules gives a commutative diagram
I am working on Problem 2.10 of Atiyah & Macdonald's Introduction to Commutative Algebra.

In this problem, we have a commutative ring $A$ and a homomorphism $u : M \longrightarrow N$ of $A$-modules.

Let $a$ be an ideal of $R$. In the solution here, they claim that the following diagram is commutative :

It's clear that $u$ induces the quotient map $w$ and the restriction map $v$, but it's unclear to me how the author makes the claim that the diagram is commutative fairly quickly. How can I see this ?

Note : It is given that $N$ is finitely generated. It is also given that $a$ is an ideal contained in the Jacobson radical of $A$. However, I don't think that either of these assumptions affect the claim being made here.

Thanks!

答:
Let $\varphi : M \to M'$ be a module homomorphism, and let $N$ and $N'$ be submodules of $M$ and $M'$, respectively. If $\varphi(N) \subseteq N'$, then we can define two new homomorphisms:

The restriction of $\varphi$ on $N$, namely the map $\tilde\varphi : N \to N'$ that sends $n$ to $\varphi(n)$. Indeed, if $i : N \to M$ and $i' : N' \to M'$ are the inclusions, the true restriction of $\varphi$ on $N$ is the map $\varphi \circ i : N \to M'$, which is the same as $i' \circ \tilde\varphi$ (that is, $\tilde\varphi$ with extended codomain). In other words, $\tilde\varphi : N \to N'$ is the unique map that makes the diagram
$$
\require{AMScd}
\begin{CD}
N @>{\tilde\varphi}>> N' \\
@V{i}VV @VV{i'}V \\
M @>>{\varphi}> M'
\end{CD}
$$
commutative.

The induced map of $\varphi$ on the quotients, namely the map $\overline \varphi : M/N \to M'/N'$ that sends $m+N$ to $\varphi(m)+N'$. This is well-defined, for if $m_1+N=m_2+N$, then $m_1-m_2 \in N$ and then $\varphi(m_1)-\varphi(m_2) = \varphi(m_1-m_2) \in \varphi(N) \subseteq N'$, which means $\varphi(m_1)+N'=\varphi(m_2)+N'$. Now, if $\pi : M \to M/N$ and $\pi' : M' \to M'/N'$ are the canonical projections, we observe that $$\pi'(\varphi(m)) = \varphi(m)+N' = \overline\varphi(m+N) = \overline\varphi(\pi(m))$$ for all $m \in M$, in other words, $\overline \varphi : M/N \to M'/N'$ is the unique map that makes the diagram
$$
\require{AMScd}
\begin{CD}
M @>{\varphi}>> M' \\
@V{\pi}VV @VV{\pi'}V \\
M/N @>>{\overline\varphi}> M'/N'
\end{CD}
$$
commutative.


问:
Existence for $-\Delta u + u + f(u) =0$ when $f$ is a compact operator?
Let $\Omega$ be a smooth domain and consider on it
$$-\Delta u + u + f(u) =0 $$
with $u=0$ on $\partial\Omega$. Here $f\colon H^1(\Omega) \to H^1(\Omega)$ is a linear, compact operator, but I don't know if it is monotone and I don't have any sign information.

Is it possible to get existence for this equation (in eg. Sobolev spaces) without assuming that $\langle f(u), u \rangle$ can be "absorbed" into the lower order term $u$?

答: Since $f$ is linear, clearly $u=0$ is a solution.


问:
Prove $\lim_{n\to \infty}\frac{n}{n+1} = 1$ using epsilon delta
$\lim_{n\to \infty}\frac{n}{n+1} = 1$

Prove using epsilon delta.

答:
By your notation I believe you're talking about the sequence $(a_n)$ of elements of $\mathbb{R}$ defined by:

$$a_n =\frac{n}{n+1}$$

Now, limit for sequences has the following definition: "given $\varepsilon >0$ there's some $n_0 \in \mathbb{N}$ such that if $n > n_0$ then $|a_n - L|<\varepsilon$". So we want some $n_0$ such that:

$$\left|\frac{n}{n+1} - 1\right|<\varepsilon$$

Rewrite this as:

$$\left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right|$$

But $n$ is natural so that the thing inside of the module sign is already positive and so we want in truth that:

$$\frac{1}{n+1}<\varepsilon \Longrightarrow n>\frac{1-\varepsilon}{\varepsilon}$$

This part is the deduction part. Now we prove, we say: given $\varepsilon > 0$ take $n_0 =(1-\varepsilon)/\varepsilon$, then we have that for $n > n_0$:

$$\left|\frac{n}{n+1}-1\right| = \left|\frac{1}{n+1}\right| = \frac{1}{n+1}$$

But $n>n_0$ so that $1/(n+1) < 1/(n_0 + 1)$ and hence:

$$\left|\frac{n}{n+1} - 1\right| < \frac{1}{n_0 + 1} = \frac{1}{\frac{1-\varepsilon}{\varepsilon} + 1} = \varepsilon$$

The important part for you to note is that first we deduce which $n_0$ works, after that we throw this part away usually and just say: "take this $n_0$" so that we show that it really works as predicted.


问:
System of recurrence relations with different even and odd relations
I know the following:

if n is even, then: $A_{n+1}=A_n+B_n$ and $B_{n+1}=B_n$

if n is odd, then it is the reverse: $A_{n+1}=A_n$ and $B_{n+1}=A_n+B_n$

I am given the initial conditions that $A_1=A_2=B_1=1$ and $B_2=2$

I've solved recurrence relations before, but this has two going at once with each dependent on the other so I'm not sure what to do. I can code it out well enough, but I'm trying to get a closed form instead of a recurrence. I would appreciate a solution if anyone could demonstrate how to tackle this.

答:
Assume $n$ is even. Then, you have

$A_{n+2} = A_{n+1} = A_{n}+B_{n}$,

$B_{n+2} = A_{n+1}+B_{n+1} = A_{n}+2B_{n}$

Combining them, you can get $A_{n+4} = 3A_{n+2} - A_{n}$ and solve it as usual one.

Once you get closed form of $A_{n}$, you can do similar step for $B_{n}$ or substitute the closed form into $B_{n}=A_{n+2}-A_{n}$.

Then you can do similar step for odd $n$, or substitute the closed form of even cases into the reccurence relation.


答:
Several completely different answers, and now here's another completely different answer, using something I just now learned about! There's a trig identity related to the Dirichlet kernel:

$$1+ 2\sum^n_{k=1} \cos(kx) = \frac{\sin \left((2n+1)\frac{x}{2}\right)}{\sin \frac{x}{2}} \tag{1}$$

I noticed you were getting lots of different multiples inside your cosines and then saw this formula and thought it might work. Here's what we're gonna do. First, let $z = 2x$, because it lets us get easier numbers on the left of the original, and it lets us make the RHS of $(1)$ equal to $\frac{\sin (n+1)z}{\sin z}$, which looks a lot like the RHS of your problem, doesn't it? We'll just say $\csc 30° = 2$ because it's a red herring (similar replacement on the RHS of $K = 2n + 1$). Now we have:

$$4(\cos 4z+\cos 3z)(\cos 3z + \cos z)-1= \frac{\sin ((2n+1)z)}{\sin z}$$

We're gonna head allllll the way back to basic algebra and FOIL this baby, dropping the RHS for a bit:

$$4(\cos 4z \cos 3z + \cos 4z \cos z + \cos 3z \cos z + \cos 3z \cos 3z)-1$$

Those product to sum formulas you were using before were on the right general track. Let's use them on this. They all need a $2$ out front, and conveniently we can grab one away from that $4$ on the outside, getting us:

$$2\left[(\cos 7z + \cos z) + (\cos 5z + \cos 3z) + (\cos 4z + \cos 2z) + (\cos 6z + \cos 0z)\right]-1= \\ 2(\cos 7z + \cos 6z + \cos 5z + \cos 4z + \cos 3z + \cos 2z + \cos z  +1)-1$$

At this point I think you can see where we're headed.  We can take care of the constants and we have:

$$1 + 2 \cos z + 2 \cos 2z + \cdots +2 \cos 7z= \frac{\sin (2n+1)z}{\sin z}$$

This means the $n$ on the sum in $(1)$ is $7$, and $K = 2n+1 = 15$.

And having done that, and seeing the other three answers, I cannot imagine any single one of them having been something I'd have been expected to know in pre-calculus. This is crazy difficult.


问:
If $2+\log_2a=3+\log_3b=\log_6(a+b)$, find $\frac1a+\frac1b$ and $\log_{ab}\left(\frac1a+\frac1b\right)$
I have a kind of system of equations with logarithms.
How do I get to the expression that the problem asks for?
(Taken from the net.)

If $a>0$ and $b>0$ and it is given that
$$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}$$

find the value of

a) $\quad\dfrac{1}{a}+\dfrac{1}{b}$

b) $\quad\log_{ab}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}$

Edit 1
I have related the first with the third and the second with the third to form a system of equations, but the values that come out are not very comfortable to deal with, I do not know if I am wrong or if the problem is like that.

$$2+\log_{2}{a}=\log_{6}{(a+b)}$$
$$3+\log_{3}{b=}\log_{6}{(a+b)}$$

Edit 2 :I have reformulated the problem by splitting it in two

答:
A probably awful solution.

Rewrite the sustem as
$$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}\color{red}{=k}$$ Switching to natural logarithms
$$2+\frac{\log (a)}{\log (2)}=3+\frac{\log (b)}{\log (3)}=\frac{\log (a+b)}{\log (6)}=k$$
So
$$a=2^{k-2} \qquad \text{and} \qquad b=3^{k-3}$$ and we face the problem of solving now
$$2^{k-2}+3^{k-3}=6^k \qquad\text{or} \qquad \log(2^{k-2}+3^{k-3})=k\log(6)$$ which cannot show explicit solution. By inspection, the solution is close to $k=-1$ and a first iteration of Newton method will give
$$k\sim -1-\frac{89 \log \left(\frac{108}{89}\right)}{81 \log (3)+6\log (2)}$$ and then estimates of $a$ and $b$. Using them
$$\frac 1a+\frac 1b \approx 108.024$$
$$\log_{ab}{\left(\frac{1}{a}+\frac{1}{b}\right)}\approx -0.688551$$ which looks like $-\log(2)$.

Using Newton method for an almost exact solution, shows the following iterates
$$\left(
\begin{array}{cc}
 n & k_n \\
 0 & -1.0000000000000000000 \\
 1 & -1.1821700749166010948 \\
 2 & -1.1819646254300294942 \\
 3 & -1.1819646251791712419
\end{array}
\right)$$ Using this number
$$\frac 1a+\frac 1b = 108.0000000000000000000000\cdots$$ but this does not improve the second result.


问:
Let N1 and N2 are normal subgroups in the finite group G. Is it true that if N1≃N1 then G∖N1≃G∖N2.?
Let $N_1$ and $N_2$ are normal subgroups in the finite group $G$. Is it true that if $N_1 \simeq N_2$ then $G/ N_1 \simeq G/ N_2$? </p>

答: However you wanted $G$ to be finite and @Tobias remaked me, you can keep my small hint when the group is infinite. $$N_1=2\mathbb Z$$ and $$N_2=3\mathbb Z$$ while $G=\mathbb Z$.</p>


问:
Quaternions +Geometric (Clifford) Algebra: What Is the Proper Prerequisite Sequence Before Learning These Subjects
What is the systematic prerequisite sequences of learning that must be mastered before approaching the subject of learning Quaternions, and then Clifford Algebra? My ultimate goal is to learn and study Maxwell’s equations in their original Quaternion form, plus other obscure scientific papers that utilize Clifford Algebra and Octonians. But before I can approach that, I need the solid mathematical background that comes before studying and understanding Quaternions and Clifford (Geometric) Algebra, beginning with a recommended syllabus that starts with pre-algebra mathematics.

I know that there is a systematic approach to learning mathematics in general. For example: Multiplication and division at the foundation, then fractions, then pre-algebra, then algebra, then geometry, then trigonometry, and so on. With each division of mathematics itself having a systematic sequence of learning the subject from its fundamentals to their ultimate advanced topics. So to state my question differently, what I want to know is this:

What is the best sequence of mathematical prerequisites (the syllabus) to master before I begin studying Quaternions, and then advance to Clifford (Geometric) Algebra?

Where does Quaternion and Clifford (Geometric) algebra fit in the sequence of learning algebra, and what will be their prerequisites necessary to have been already mastered before entering these subjects?

Also, can someone recommend a track of material that would properly prepare me for the topic of Quaternions and Clifford (Geometric) Algebra from a pre-algebra starting point? i.e. Completed knowledge of addition/subtraction, multiplication/division, fractions and percentages.

答:
If Science is your destination, then there is no other noble subject beyond philosophy than mathematics itself to give you the foundation you need to study and learn the sciences. It is a commendable cause.

I give my answer to your question for the benefit of all on being appropriately prepared to understand Quaternions and Clifford (Geometric) Algebra (which utilizes Octonions). Because of the increase in traffic to mathematical forums for queries about quaternions due to a heightened advantageous awareness of them in computer science, and an increasing popularity for applications in physics (particularly their broader inclusion within Geometric Algebra), it now necessitates a consolidated forum page to answer these many queries. Thus, for that reason, this question has been carefully considered. For my answer, I draw upon the contributions of RobertTheTutor, CyclotomicField, J. W. Tanner, Rob Arthan, along with the results of my own extensive research to learn the answer to this very same question.

CONTENTS

I.   The Proper Prerequisite Learning Sequence.

II.  Recommended Sequential Learning Track Material.

III. Why Use Quaternion Based Mathematics? Benefits and Features.

IV. Historical Fun Facts About Quaternions and the Truth About Maxwell Theory.

V.  James Clerk Maxwell Original Work and References.

VI. Deriving the Maxwell Source Equations In Quaternion Space.

I. The Proper Prerequisite Learning Sequence.

To be appropriately prepared for the subjects of your inquiry, the proper prerequisite you will need is as follows:

Arithmetic: addition, subtraction, multiplication, division, fractions, decimals, percentages.

Pre-Algebra 1

Pre-Algebra 2

Algebra 1

Algebra 2

Geometry

Advanced Algebra

Including Matrices

Including Complex Numbers

Trigonometry

Pre-Calculus

Calculus 1

Linear Algebra; Basic + Advanced (very Important To Master!!!)

Quaternion Algebra

Which includes Versor Algebra

Abstract Algebra

Including Abstract Algebra: Group Theory

Including Abstract Algebra: Ring Theory

Geometric (Clifford) Algebra

This sequence will not only prepare you for entering into quaternions (and octonians of Clifford Algebra), but also give you the fundamental foundation you need to advance eventually into forms of Calculus that are based upon Geometric (Clifford) Algebras.

It may be that you have already obtained a particular level of mathematics already. If that is the case, then simply choose the point where you are in your mathematical level, and proceed forward from that point.

As a side note, if you come upon anyone speaking in a video lecture online about “Versor Algebra”, know that a versor is just a unit quaternion. In other words; in mathematics, a versor is a quaternion of norm one (a unit quaternion). Thus Versor Algebra is not a separate algebra, but relates precisely to a unit in quaternion mathematics. It was something that tripped me up when I first heard the term, but now it doesn’t. I just know that I have to learn quaternions to understand their online lecture.

II.  Recommended Sequential Learning Track Material.

I recommend this track of material to properly prepare you:

1.1 MathtutorDVD (MathandScience) courses. These fantastic video courses will take you by the hand, and patiently take you from arithmetic all the way through to mastery of advanced calculus if that is your desire. However, their courses from arithmetic through to engineering mathematics (which is their courses on leaner algebra) will be sufficient for your needs. This will cover all mathematics I have placed above in my proper prerequisite list from number 1 to number 11. However, instead of buying the courses separately, I recommend their package deal because by the time you buy all of these math courses separately, you may as well have bought their package deal of all courses.

1.2 If free is your thing, then you may prefer to check out the Real Not Complex website. Then after you master arithmetic through to pre-algebra, you can check out this full free Algebra Course posted to YouTube. Total length is 5:22:06. Personally, I think MathtutorDVD does a better job than these free resources. However, you can decide for yourself.

2.1 If you have mastered mathematics all the way through linear algebra, then you should be ready for the study of quaternions. Before you begin your study into quaterions, it is very important that you understand how complex numbers produce rotations in the complex plane first. If you have that down, then you can proceed to these next recommendations.

2.2 How To Visualize Quaternions, and Gain an Intuitive Grasp of Them? The best I have encountered (and learned of by RobertTheTutor, is Grant Sanderson and Ben Eater's Explorable Video Series on Visualsing Quaternions. This YouTube video is a good introduction and overview of their method of teaching: Visualizing Quaternions(4D Numbers) With Stereographic Projection. To get the full treatment, go to their website dedicated to interactive instruction on Quaternions.

2.3 John S. On YouTube also has a 3 part instruction on understanding Quaternions.

2.4 There is also a book I can recommend: Visualizing Quaternions by Andrew Hanson and Steve Cunningham.

2.5 When you are ready to plunge deeper into learning quaternions, Introduction to Quaternions and An Elementary Treatise On Quaternions by Philip Kelland and Peter Guthrie Tait will give you a deeper breadth of learning them from the standpoint of this mathematics's origins. These fellows were contemporaries with the inventor of quaternion mathematics.

2.6 To gain the full treatment on quaternions, what better place to turn than the author and originator of quaternion mathematics himself: Sir William Rowan Hamilton. Whom made his discovery in 1843, October 16. His three chief works on the subject are: Lectures of Quaternions, On Quaternions, and Elements of Quaternions. You can go to the embedded links to view or download them for free.

2.7 At this point, you may have follow up questions that have already been answered on this forum:

How do Quaternion represent rotations?

How can one intuitively think about Quaternion?

How do you construct the Quaternion and the multiplication rules, like Hamilton did?

Is there a geometric realization of Quaternion group?

Quaternions and Rotations. See also Wikipedia’s Article: Quaternions and Spacial Rotation. You may also like to check out Joan Sola's free turorial guide Quaternion Kinematics for the Error-State.

2.8 I save these last two books until later, for the simple fact that they have been frequently recommended, yet at the same time, comments about them indicate that they are better suited for someone whom has achieved junior or senior level of college mathematics. If you are a beginner in quaterion mathematics, then save these book until later, otherwise these two books are for you: Rotations, Quaternions, and Double Groups by Simon Altman, and On Quaternions and Octonians by John Horton Conway and Derek Smith.

2.9 Quaternions: Other Resources You May Want to Check Out:

Wolfram Mathword Website’s overview of Quaternions.

Euclidean Space Website - Introduction and Exploration of Quaternions.

Wikipedia’s Article on the Quaternion.

Sir William Rowan Hamilton Complete Mathematical Papers Collection.

When you are ready to apply Quaternions to physics, check out the book by J.B. Kuipers, Quaternions and Rotation Sequences, published by Princeton University Press (December 14, 1998).

3.1 If you have mastered Trigonometry, Pre-Calculus, Calculus 1, Leanear Algebra, and  Quaternion Algebra, then you should now be ready for the study of Abstract Algebra on your destined goal for mastery of Geometric (Clifford) Algebra.

3.2 I recommend starting off with Abstract Algebra for Beginners by Dr. Steve Warner PhD (Mathematics). It should give you a great overview study of the course to familiarize yourself of the deeper study to come. Meanwhile at the same time, it will get you up and running with the concepts presented. You may also like instead, A Book of Abstract Algebra, Second Edition, by Charles Pinter. Both books come with 4.5 star book review ratings. However, if a single book is what you are after, and jumping right into a path for full in depth mastery of Abstract Algebra from the standpoint of a beginner, then you may like instead of either of those former books I mentioned, Abstract Algebra, Third Edition, by David Dummit and Richard Foote.

4.1 If you have mastered everything through to Abstract Algebra in the proper learning sequence list above [Roman numeral (I) in the Table of Contents], you should now be ready for the study of Geometric (Clifford) Algebra.

4.2 Geometirc (Clifford) Algebra Video Series by Prof. Alan Macdonald. Note: He also publishes two books (and has a support website for those taking his course). His course starts with linear algebra and works up from there, including all the way up to advanced Calculus based upon Geometric (Clifford) Algebra.

4.2.1 A Swift Introduction to Geometric Algebra [44:22 minute YouTube video]. I give this entry as a good visual overview of the subject that may be of interest to you meanwhile you are checking out Professor Alan Macdonald’s videos and course.

4.2.2 Geometric Algebra YouTube Playlist by Mathoma. I give this entry merely as a supplemental to Professor Alan Macdonald’s video and course, in the event that you find a different angle on the subject useful.

4.3 Geometric (Clifford) Algebra: Other Resources You May Want to Check Out:

Wolfram Mathword Website’s overview of Geometric (Clifford) Algebra.

Euclidean Space Website - Introduction and Exploration of Geometric (Clifford) Algebra.

Wikipedia’s Article on Geometric (Clifford) Algebra.

Wikipedia’s Article on the inventor of Geometric Algebra, William Kingdon Clifford.

In the event that you should need to follow a course for Lie Algebras on your path in Geometric (Clifford) Algebra, you may want to start with Introduction to Lie Algebras by Erdmann and Wildon (be sure to do all the exercises in the book to establish a firm foundation. Next, move on to to Humphreys' Introduction to Lie Algebras and Representation Theory (the exercises in this book are more difficult, but it gets you deeper into the subject matter). Finally, if you should like to become a heavy hitting champ in this subject, then take up the book Lie Groups Beyond an Introduction by Anthony Knapp. The prerequisite for this book, however, is that you already have a good grasp of everything before this point in the recommended learning sequence, and have knowledge of smooth manifolds (topology and differential geometry). Other than this, the book come well recommended.

If electrical engineering will be your application of Geometric (Clifford) Algebra, then you may like to check out the book: Geometric Algebra for Electrical Engineers: Multivector Electromagnetism, by Peeter Joot. However, this book is for graduate level. So if you are just a beginner, or at intermediate level, then this text is not for you.

If physics will be your application of Geometric (Clifford) Algebra, and you want a good coverage of what Geometric Algebra is capable of doing when applied in modern physics, then you will want to follow the mathematical presentation given by Professor Anthony Lasenby of Cambridge University here. The total length of this YouTube video is 2:05:15 hours in length [Title: GAME2020 3. Professor Anthony Lasenby. A new language for physics]. After watching his presentation, you may be inspired to obtain his two textbooks on Geometric Algebra: Geometric Algebra for Physicists, and Space-Time Algebra Second Edition. However, both these Geometric Algebra textbooks presume an already existing understanding of concepts in physics such as quantum mechanics, relativity (special and general), and several other higher undergraduate physics topics. So that would be your prerequisite before approaching these textbooks by Professor Anthony Lasenby.

III. Why Use Quaternion Based Mathematics? Benefits and Features.

Native to problems consisting of four dimensional space and easily handles 3D and 4D problems.

Bypasses the common problem of 3D rotational locking in Euler’s angles (known as Gimbal Lock). It is often an inevitably incurred problem when using vector calculus [If that is the case, I can only imagine what errors could happen when trying to describe particles and field rotational variables in electromagnetism (just saying)].

Integrates easily into 3D programming routines for multi-physics problems.

Requires less computational power to implement than vector calculus in 3D and multi-physics simulation situations. Although not popularly known outside of game developers, 3D gaming graphics engines implement multi-physics simulations to accurately display user interactivity with physical events in the simulated environment. A majority of 3D games utilize Quaternions in their 3D physics engines and not Vector Calculus.

In comparison to rotation matrices, Quaternions use less computer memory, compose faster, and are naturally suited for efficient interpolation of rotations.

Meanwhile vector calculus often fails at interpolate orientations, quaternions excel exceedingly at them.

It helps to ground the practicing physicist utilizing them in describing the sciences in mathematical reality.

Finally, check out this YouTube video, Quaternions and 3D Rotation, Explained Interactively by Grant Sanderson and Ben Eater.

IV. Historical Fun Facts About Quaternions and the Truth About Maxwell Theory.

Oliver Heavside and his side-kick Gibbs back in the day called Quaternions, “pure evil”, and “the work of the devil”… no joke! True story! I reference Grant Sanderson and Ben Eater’s YouTube video. The reason why vector calculus won the day back in the late 19th century over Quaternions was not because vector calculus was a superior way to describe events in 3D and 4D space, but rather because Quaternions generally are not readily intuitive, seem contrary to convention of normal thinking (seem convoluted), and thus were difficult to learn and grasp. However, with the advent of modern instructional methods that can add interactive video graphics to the learning methodology (such as provided at Ben Eater’s website, this is no longer the case. Quaternions are now increasingly accessible to the person desiring to learn them and whom have completed at least tenth grade high school mathematics courses in linear algebra (in reference to the United States). With Quaternions and Octonians (used in Clifford/Geometric Algebra), it is very possible and easy to describe and simulate most aspects of physics (from the standpoint of computational power of computers) without venturing into vector calculus at all.

“Somehow quaternions are a fundamental building block of the Universe” - Sir William Rowan Hamilton, originator and developer of Quaternion Algebra.

From my investigation and what I have learned, I believe retrospectively, that James Clerk Maxwell was ultimately correct to use Quaternions in his equations to describe his electromagnetic theory, and Oliver Heavside was wrong to mutilate them with his truncated vector calculus translation of them. Has anyone ever heard the term: "lost in translation"?

Original Maxwell equations were written using quaternions and potentials. Quaternion is a combination of vector and scalar part. Electric and magnetic fields was defined as difference in potential. There was two kinds of potentials - electric and magnetic. Today's "Maxwell equations" are actually Heaviside equations, which are a truncated limited edition of the original Maxwell theory.

Meanwhile there is much that can be practically implemented with Heaveside's equations (like in any translation), there is much to be mined and extrapolated upon in the original and authentic theory of Maxwell and his equations described in QUATERNIONS... NOT in Vector Calculus (as is often now currently taught in Electrical Engineering Schools).

One may find it relevant to see a comparison study between original Maxwell’s Equations and Heavside’s Equations made by André Waser in his essay, On the Notation of Maxwell's Field Equations. Back up link here.

V.  James Clerk Maxwell Original Work and References.

Since Maxwell’s original quaternion equations were one of the goals of your question, you may find these original references helpful if you do not have them already. Once one has Quaternions mastered, one would do well to pay a visit upon Maxwell’s original work done in Quaternions, and thoroughly examine them. One may even discover their ingenious mathematical beauty and ditch Oliver Heavside’s equations for the original Maxwell’s Equations. Maxwell’s original unadulterated work is available here:

A Dynamical Theory Of The Electromagnetic Field - 1865. Maxwell's 1865 paper describing his 20 Equations in 20 Unknowns - Predecessor to the 1873 Treatise. Proceedings of the Royal Society Of London Vol XIII.

A Treatise on Electricity and Magnetism, Volume 1 – 1873.

A Treatise on Electricity and Magnetism, Volume 2 – 1873.

Five of Maxwell's Papers by James Clerk Maxwell (On Project Gutenberg Website).
These papers relates to light being integrated into the electromagnetic field.

On Physical Lines of Force, J.C. Maxwell, 1861. Back up link here.

There is much more than these reference works mentioned above. The full accumulation of his life’s work can be found in: The scientific letters and papers of James Clerk Maxwell published by Cambridge University Press.

PLEASE NOTE: The above reference mentioned is Volume 1 and 2 of a total 3 volume set of his complete collection containing all of his works and writings. This single set online containing volume 1 and 2 is 1064 pages long. Its free online, but registration is required to view it online. Google Books also has all three volumes online, but as I reviewed them, I saw that several hundred pages were censored out (although listed they cannot be read or accessed). I find it very interesting also that the hard back edition of volume two is no longer available in print, but has been substituted with a paperback edition. However, to get the full three volume set in print, I recommend looking up ISBN-13: 978-0521757942, or ISBN-10: 0521757940 (but this edition too appears to be missing several hundred pages when compared to the original hardback three volume editions by the same publisher – Cambridge University Press… very strange).

VI. Deriving the Maxwell Source Equations In Quaternion Space.

I recommend looking into these references if deriving the Maxwell source equations in Quaternion Space is your goal:

M. M. Acevedo, J. López-Bonilla, et al: Quaternions, Maxwell Equations and Lorentz Transformations.

Vic Chrisitianto and Florentin Smarandache, et al, A Derivation of Maxwell Equations in Quaternion Space.

Doug Sweetser’s YouTube Channel here apparently derives the Maxwell theory in quaternions, and addresses several applications of Quaternions to modern field equations (i.e. Relativity, Gravity, Quantum Mechanics, et al). However, this particular video series may be of greater relevance to your question:

GEM Blog 28: Deriving How to Derive Field Equations. (1/5)

GEM Blog 29: Deriving the Maxwell Source Equations Using Quaternions. (2/5)

GEM Blog 30: Deriving the Maxwell Homogeneous Equations Using Quaternions.  (3/5)

GEM Blog 31: Deriving the Hypercomplex Gravity Field Equations. (4/5)

GEM Blog 32: Deriving the GEM Field Equations. (5/5)

Using Quaternions for EM and GEM, Manifestly Covariant Form.


问:
Error in two approaches when finding the number of distinct integers between 100 and 1000
My first approach

there are 3 places A B C to be filled.

For the unit place C :-

There are 10 possible ways to fill it because possible outcomes equals 0 to 9.

For the tens place B :-

In order to be distinct number of ways to fill is 9.

For A :-

From 10 possible choices 0 is out of the equation and the two previous numbers so this leaves us with 7 ways. So Number of Ways = 630.

But there is a case when 0 is either in the middle of at C then only two numbers gets eliminated i.e. 0 and one more number because B AND C cannot repeat the same numbers.

After Finding that my answer was wrong I tried the same process in reverse order

2nd Approach

For A

0 is out of the equation so there are 9 ways to fill A.

For B

Now 0 is included so no of ways to fill = 9 again.

For C

Number of ways to fill = 8.

So total number of ways = 648 which is the right answer.

Why is this happening ? Am I repeating some numbers?

答:
This answer is in relation to the question suggested by the body of the OP rather than the title. Hence this post adresses the following problem:

How many integers with all three digits being distinct from one another are there between $100$ and $1000$?

Repair of 1st approach

If you start from C and progress, the trouble is that when you get to A, the number of available choices depends on the choices for the other two, namely if zero was included among B, C, you then have 8 choices for A.

To break that down:

Your original approach correctly stated what amoutns to: Choose C freely among $0..9$, then choose B not equal to C from the same set. This gives us $10\cdot 9$.

By excluding the digit $0$, we instead get $9\cdot 8$.

Hence the cases including the digit $0$ must be $10\cdot 9-9\cdot 8=2\cdot 9$. For those, the number of cases for A should be 8 instead of 7.

This leads to the figure:
$$
9\cdot 8\cdot 7+2\cdot 9\cdot 8=9\cdot 9\cdot 8
$$
which is the same figure now as your second approach.


问:
For $A$ a unital Banach Algebra and $B$ maximal ableian subalgebra, $Inv(B)=B\cap Inv(A)$
Let $A$ be a unital Banach Algebra. Let $B$ be a maximal Abelian Banach sub-algebra containing 1. I want to show that $Inv(B)=B\cap Inv(A)$, where $Inv(P)$ is the set of invertible elements in $P$.

One inclusion is trivial. For the other inclusion let $x\in B\cap Inv(A)$. We know that $b^{-1}$ exists in $A$, we need to show that it lies in $B$. I believe the idea should be to try and use $b^{-1}$ to generate an abelian sub-algebra bigger than $B$. This contradicts the maximality of $B$ and we are done.

I do not see how to generate such an sub algebra however, it seems to me that $B+b^{-1}\mathbb C$ is not a sub algebra of $A$.

答:
Let $b\in B$; then for any $a\in B$, $ab=ba$, so $$b^{-1}a=ab^{-1}$$
This implies that the set of polynomials generated by $B$ and $b^{-1}$ is an abelian subalgebra of $A$; each element takes the form $$\sum_{i=0}^na_ib^{-i}$$ By the above identity, the product of such elements remains in the subalgebra and commutes. Since this algebra contains $B$, but $B$ is maximal, it follows that $b^{-1}\in B$.


答:
In such problems it rarely hurts to gather some numerical evidence to see if an answer suggests itself, especially if you don’t see any better approach. It’s not hard to calculate the following table:

$$\begin{array}{ccc}
n&A_n&B_n\\\hline
1&1&1\\
2&1&2\\
3&3&2\\
4&3&5\\
5&8&5\\
6&8&13\\
7&21&13
\end{array}$$

The numbers $A_n$ and $B_n$ are easily recognizable as Fibonacci numbers. In fact, we can expand the table as follows:

$$\begin{array}{ccc}
n&A_n&B_n\\\hline
1&1=\color{red}{F_2}&1=\color{red}{F_1}\\
2&1=\color{red}{F_2}&2=F_3\\
3&3=F_4&2=F_3\\
4&3=F_4&5=F_5\\
5&8=F_6&5=F_5\\
6&8=F_6&13=F_7\\
7&21=F_8&13=F_7
\end{array}$$

The number $1$ is both $F_1$ and $F_2$; for the red entries above the choice between $F_1$ and $F_2$ was made to fit the obvious pattern of the unambiguous entries.

This evidence suggests the following conjecture:

$$A_n=\begin{cases}
F_n,&\text{if }n\text{ is even}\\
F_{n+1},&\text{if }n\text{ is odd}
\end{cases}\tag{1}$$

and

$$B_n=\begin{cases}
F_{n+1},&\text{if }n\text{ is even}\\
F_n,&\text{if }n\text{ is odd}\,.
\end{cases}\tag{2}$$

It’s easy to check that $(1)$ and $(2)$ give the correct values for $A_1,A_2,B_1$ and $B_2$, so if they satisfy the recurrences defining the sequences, they will yield the correct values of $A_n$ and $B_n$ for all $n\ge 1$.

For instance, if $(1)$ and $(2)$ hold for $1\le m\le n$, and $n$ is even, then

$$A_{n+1}=A_n+B_n=F_n+F_{n+1}=F_{n+2}\,,$$

and since in this case $n+1$ is odd, $F_{n+2}$ is exactly what $(1)$ says that $A_{n+1}$ should be. That leaves three cases to be checked: $B_{n+1}$ for even $n$, and $A_{n+1}$ and $B_{n+1}$ for odd $n$.

There are of course far more efficient ways to proceed, but sometimes even very simple tools are enough to get the job done.

Title: System of recurrence relations with different even and odd relations

Body:
I know the following:

if n is even, then: $A_{n+1}=A_n+B_n$ and $B_{n+1}=B_n$

if n is odd, then it is the reverse: $A_{n+1}=A_n$ and $B_{n+1}=A_n+B_n$

I am given the initial conditions that $A_1=A_2=B_1=1$ and $B_2=2$

I've solved recurrence relations before, but this has two going at once with each dependent on the other so I'm not sure what to do. I can code it out well enough, but I'm trying to get a closed form instead of a recurrence. I would appreciate a solution if anyone could demonstrate how to tackle this.


问:
If a set of differentiable functions S={$f1,…,fn$} with common domain $D=[a,b]$, is linearly independent, then so is S′={$f′1,f′2,…,f′n$}?
I have a (maybe too simple) question: I have to prove if different sets of equations are or aren't linearly independent. The case which is making trouble is ${sinx, cosx, 1}$. I have found out about the Wronskian, so I guess this implies that if a colection of functions (i.e. ${sinx, cosx, 1}$) is linearly independent, then so will be its derivative? Is that approach right? My idea was to derivate twice, and then try to solve the equations system for $a, b, c$ to prove that $a=b=c=0$, and thus the three functions are linearly independent. Is this also right?
Thank you very much.

答:
Your idea is good. But there is a simpler solution. Suppose that

$$a \sin x+b\cos x+c=0$$

for all $x$.

With $x=0$ we get

(1) $b+c=0.$

With $x=\pi/2$ we get

(2) $a+c=0.$

With $x=\pi$ we get

(3) $-b+c=0.$

From (1),(2),(3) it is easy to see that $a=b=c=0.$


问:
Given position vector of points A, B, find the equation of perpendicular bisector of AB in a vector form.
I am trying to solve the problem: A, B are given point. a is the position vector of A, and b is the position vector of B. I want to find the equation of perpendicular bisector of AB in a vector form.

I solved this problem in a regular form, which I first set A $(x_1,y_1)$ B $(x_2,y_2)$, then the equation of this line is
$$ (x_2-x_1)x+(y_2-y_1)y-\dfrac{x_2^2-x_1^2}{2}-\dfrac{y_2^2-y_1^2}{2}=0$$

But I don't know how to convert it in to something in the form of

$$\vec{r}=\vec{r}_0+t\vec{v}=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle $$

Would someone give me a hint? Thanks in advance.

答:
Note, for the question to be meaningful, we must be working in a plane (otherwise 'the' perpendicular bisector is not defined). Let $\hat{z}$ be a vector pointing out of the plane.

We want the equation of a line which passes through the midpoint of $\mathbf{a}$ and $\mathbf{b}$ and is perpendicular to $\mathbf{a}-\mathbf{b}$ and perpendicular to $\hat{z}$. Such a line is given by:

$$\frac12(\mathbf{a}+\mathbf{b})+\lambda (\mathbf{a}-\mathbf{b})\times \hat{z} $$


问:
Checking if given set forms basis for $P_3(\mathbb{R})$
The question says whether the polynomials $x^3+2x-4,\ x^3+x^2-3x+1, $and $x^3+5$ generate $P_3(\mathbb{R})$? I did the question in a sneaky way as follows. We already know $\{1,x,x^2,x^3\}$ forms the basis for $P_3(\mathbb{R})$, so it's dimension is $4$. Since the given set contains only $3$ elements, it won't generate $P_3(\mathbb{R})$. But I am not satisfied with my argument. How to find those polynomials which won't get generate by the span of the above set. Any hint. Thanks.

答:
One demonstration. Let

$A = x^3+2x-4,$

$B = x^3+x^2-3x+1,$
$C = x^3+5$

You can see the subspace Span{$A,B,C$}∩Span{$1,x,x^2$} is generated by

$D = B-A = x^2-5x+5$,

$E = B-C = x^2-3x-4$

Similarly, the subspace Span{$A,B,C$}$\cap$Span{$1,x$} is generated by

$F = E-D = 2x-9$,

which is clearly smaller than Span{$1,x$}


问:
Euclid's Algorithm - Finding positive solutions (x,y)
Ok, so maybe I'm missing some crucial steps here, but I just can't see it.

So, I have got $\gcd(144,252) = 36$

$144x+252y=36$

I need to find ALL positive solutions. I can find $x=2$ and $y=-1$ by reversing the algorithm. The answer is given as follows.

$36=144-108
  =144-(252-144)
  =2\times144-252     $

But then says

$\times 117$

$4212=118 \times2 \times144 -117 \times252$

solution of $4212 = 144x + 252y$
are $x= 234- 252t/36 = 243 - 7t$
   $ y= -117 + 144t/36 = -117 + 4t$

$x,y>0$ if $234/7 > t > 117/4$

$t=30,31,32,33$

$(x,y)=(24,3),(17,7),(10,11),(3,15)$

FIRSTLY - where on earth does the multiplier $117$ come from? this seems to have been plucked from thin air,

and secondly, please, how to convert the $t$ values into the $(x,y)$ values. ie, how does $30$ relate to $(24,3)$

Many thanks

答:
sIt isn't clear what problem you are trying to solve. Using multiples of $144=4 \cdot 36$ and $252=7 \cdot 36$ you can generate any multiple of $36$.

Note that (Fact A) $1008 =28 \cdot 36 = 7 \cdot144=4\cdot252.$

Now it looks as though your original question may have been to express $4212=144x+252y$ with both $x$ and $y$ being positive integers.

The first step in this is to check the gcd of $144$ and $252 = 36$ and then check that this is a factor of $4212$ - if it isn't, the problem cannot be solved. $4212=117\cdot 36$. So that is where the $117$ comes from.

If we have a convenient way of writing $36$ - in this case $36=2 \cdot 144-252$ we can multiply this by $117$ to get an expression for $4212=234\cdot144-117\cdot252$

The next thing we need is to adjust this so that the $-117$ is replaced by a positive integer. Using Fact A, we can increase $-117$ by $4t$ provided we reduce $234$ by $7t$ (the total remains the same). The smallest multiple of $4 \ge117$ is $120=4\cdot30$, corresponding to $t=30$. We can't allow $234$ to be replaced by a negative number, and we note that: $7\cdot33=231, 7\cdot34=238$, so $t \lt34$. This tells you which valued of $t$ will work.

Finally $(x,y)=(234-7t,-117+4t)$

I hope this helps you to understand the logic involved - you will need to work it through yourself.


问:
Prove $(1+2/x)^{ -1}= x/x+2 = x/2(1+x/2)^{ -1}$
Prove $(1+2/x)^{ -1}= x/x+2 = x/2(1+x/2)^{ -1}$ I am stuck on this question I do not know how these three are related and tried working it out but got no where i will attack some of my work.And none of my collages know how to solve it  Any help is much-appricated. thank you.

答:
Welcome to MSE!

Here are some hints:

$(1+2/x)^{-1} = \frac{1}{1+2/x} = \frac{1}{x/x+2/x} = \frac{1}{(x+2)/x} = \frac{x}{x+2}$ which is related to your 2nd expression (but in yours the parentheses are missing).

Here is the other equality:

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


问:
Notation of injective hull $(0 :_E p^k)$
If $R$  is a Noetherian ring and $p$  is a prime ideal, set $E = E(R/p)$  as the injective hull.
Now it is here
clamed that
the injective hull of $R/p$ over the Artinian ring  $R/p^k$  can be computed as the module
$(0 :_E p^k)$.

What does the notation mean? I know that $(J:I)$ denotes the  ideal quotient. What does the subnotation $_E$ mean inside $(0 :_E p^k)$?

答:
In general, if $M,N$ are submodules of some $R$-module $U$, and $I$ is an ideal of $R$, then
$$
(N:_MI):=\{m\in M\mid I\cdot m\subseteq N\}
$$
is the "colon module".  This is just the natural generalisation of the colon ideal $(J:I)$ where the $:$ would now have the subscript $R$ (which isn't doing anything so we drop it again).

Here, the subscript $E$ is the $E=E(R/\mathfrak{p})$, the injective hull of $R/\mathfrak{p}$ in the first sentence of the paragraph.  Then $(0:_E \mathfrak{p}^k)$ is the submodule of $E$ that is annihilated by $\mathfrak{p}^k$, and $(0:_E\mathfrak{p}^\infty)=\bigcup_d(0:_E\mathfrak{p}^d)$ is the submodule of $E$ consisting of everything in $E$ that get annihilated by some power of $\mathfrak{p}$.


Body: Prove $(1+2/x)^{ -1}= x/x+2 = x/2(1+x/2)^{ -1}$ I am stuck on this question I do not know how these three are related and tried working it out but got no where i will attack some of my work.And none of my collages know how to solve it  Any help is much-appricated. thank you.

Body:
$\forall x \in \mathbb{R} \setminus \{0,-2\},$
$$\frac{1}{1+\frac{2}{x}} = \frac{1}{1+\frac{2}{x}} \times \frac{x}{x} = \frac{x}{x+2}$$
$$\frac{x}{2} \times \frac{1}{1+\frac{x}{2}} = \frac{x}{2\times(1+\frac{x}{2})} = \frac{x}{2+x}$$


问:
How can I evaluate average with negative values, given I will apply this weight for both negative and positive values
I am more a programming guy than math one, so please be easy on me with math expressions

Before I start describing to make it clear: I have 2 types of scores: reddit score and sentiment score.
Reddit score can be of any value.
Sentiment score ranges between -1 and 1

So my task is to calculate weight for each sentiment score. Besides analyzing post for sentiments I want to give this post a weight to enhance evaluation of sentiment value it carries

Basically it would look like this

```<code>sentiment value = sentiment_analyzer(post) * weight
</code>```

I want to use average of reddit scores, which can be both negative and positive
Then apply this average to calculate weight the following way:

```<code>weight = reddit_score / average(reddit_scores)
</code>```

So here comes the problem, it is not typical but what happens if average will end up negative ?
On one hand we can leave it as is, hoping this will never happen, but if it will it would ruin all scoring (I guess ?) , on the other we can't just make it positive and ignore that fact it was originally calculated with negative values.

Maybe I am missing some basic Math concepts, but I really can remember this situation, when I studied Math.

As for method itself I understand it far from accurate, but I am also limited on functions, so I would be grateful if you tell me whether this weight estimation method makes sense at all. And if there is a better simple method. (Even quartile formula not available)

答:
You can apply the transformation $\exp$ to your reddit_scores. (Exponentiate everything, e.g. $e^{reddit\_scores}$.) This will make everything positive. Then really bad posts will have very little leverage. For example, if a post got a negative value it would be just a fraction. But if a post got a positive rating then it would be a larger number. If you do not have access to exponentiation, you can remove posts with negative votes because they are supposed to be weighed very little.

A sentiment score times a weight by itself carries little meaning because it will obscure the actual sentiment score that was received. For example, if something was slightly positive (say .5) and you had a weight of 1.5 you would get a weighed sentiment score of .75, which actually means something else on the sentiment scale. So don't actually use the weights until you have all the weights and want to take the average. Then you can do

```<code>average sentiment= Sum of (weight * sentiment_analyzer(post))
                   __________________________________________
                   Sum of weights
</code>```

Body:
You can apply the transformation $\exp$ to your reddit_scores. (Exponentiate everything, e.g. $e^{reddit\_scores}$.) This will make everything positive. Then really bad posts will have very little leverage. For example, if a post got a negative value it would be just a fraction. But if a post got a positive rating then it would be a larger number. If you do not have access to exponentiation, you can remove posts with negative votes because they are supposed to be weighed very little.

A sentiment score times a weight by itself carries little meaning because it will obscure the actual sentiment score that was received. For example, if something was slightly positive (say .5) and you had a weight of 1.5 you would get a weighed sentiment score of .75, which actually means something else on the sentiment scale. So don't actually use the weights until you have all the weights and want to take the average. Then you can do

average sentiment= Sum of (weight * sentiment_analyzer(post))
                   __________________________________________
                   Sum of weights


问:
Every function in $D'(\Omega)$ is the limit of a sequence in $D'(\Omega)$ with compact support.
How will we prove that for every $f\in D'(\Omega)$ there exists a sequence $(f_j)$ in $D'(\Omega)$ with compact support such that $f$ is the limit of $(f_j)$?</p>

答:
Let $K_n$ be a sequence of compacta with $K_n\subset K_{n+1}$ and $\bigcup_{n\in \mathbb{N}} K_n = \Omega$. Let $\chi_n\in C_c^\infty(\Omega)$ with $\chi_n(x)=1$ for $x\in K_n$ and $\operatorname{supp }\chi_n \subset K_{n+1}$. Then define $f_n=\chi_n f$.

For a test function $\phi$ you have $\operatorname{supp }\phi \subset K_m$ for some $m\in\mathbb{N}$. Then for $n>m$ you have
$$ f_n(\phi) = (\chi_n f)(\phi) = f(\chi_n\phi) = f(\phi).$$


答:
I think the question you have is the following:

If a set of differentiable functions  $S=\{f_1,\ldots,f_n\}$ with common domain $D=[a,b]$, is linearly independent, then so is
$S'=\{f_1',f_2',\ldots,f_n'\}?$

The answer is yes, provided no function in $S$ is the translation of another, ie you cannot have $x^2$ and $x^2+3$ in $S$ for example. Also, $S$ cannot contain constant functions or else you'd have the zero function in $S'$ and $S'$ automatically becomes linearly dependent.

So suppose otherwise. You know that you can write
$$f_i(x)=\int_{a}^{x}f_i'(t)dt,\;\;\text{ for all }x\in[a,b].$$
Therefore if $a_1,\ldots, a_n\in\mathbb{R}$ are scalars and $$a_1f_1'(t)+a_2f_2'(t)+\cdots+a_nf_n'(t)=0,\;\; \text{ for all } t\in[a,b]$$
then integrating the above equation yields
$$\int_a^x a_1f_1'(t)+a_2f_2'(t)+\cdots+a_nf_n'(t)dt=0\;\;\text{ for all }x\in [a,b].$$
so that
$$a_1f_1(x)+\cdots+a_nf_n(x)=0,\;\;\text{ for all }x\in[a,b]. $$
But the set $S$ is linearly independent and therefore $a_1,\ldots a_n$ are all zero. This implies that $S'$ is also linearly independent.

Title: If a set of differentiable functions S={$f1,…,fn$} with common domain $D=[a,b]$, is linearly independent, then so is S′={$f′1,f′2,…,f′n$}?

Body:
I have a (maybe too simple) question: I have to prove if different sets of equations are or aren't linearly independent. The case which is making trouble is ${sinx, cosx, 1}$. I have found out about the Wronskian, so I guess this implies that if a colection of functions (i.e. ${sinx, cosx, 1}$) is linearly independent, then so will be its derivative? Is that approach right? My idea was to derivate twice, and then try to solve the equations system for $a, b, c$ to prove that $a=b=c=0$, and thus the three functions are linearly independent. Is this also right?
Thank you very much.


问:
Looking for Advanced Single Variable Calculus textbooks (NOT REAL ANALYSIS)
Does anyone know any other good advanced calculus textbooks that have much more advanced or rigorous problems?

Textbooks like: Spivak, JEE Main, Advanced Calculus Explored, Problems in Mathematical Analysis (Demidovich), AoPS Calculus, Examples & Problems in Advanced Calculus, Advanced Calculus of a Single Variable, I.A. Maron, etc.

Textbooks that aren't what I'm looking for: Rudin, Apostol, Courant, Folland, G.H. Hardy, etc.

Again, I'm not looking for Real Analysis, I'm looking for like a very "honors" type calculus book without proofs. Please, no proofs, just a calculus book with challenging or advanced problems.

I tried looking through other common Mathexchange questions and they're just proofs and answer the same exact thing that I'm not looking for.

答:
I would like to recommend

Irresistible integrals

Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis(This does have some multivariable stuff but it is challenging book to solve, especially the challenge problems)

Inside interesting integrals


问:
$\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ adds up to 1
Question:

Let a discrete r.v be denoted $X_1,..,X_n$ denote the birthdays of $n$ people in a room. Assume that $X_1,..,X_n$ are mutually independent and that $X_i$ is a distribution such that $X_i \sim U(\left \{ 1,2,...,365 \right \})$ for all $i \in {(\left \{ 1,...,n\right \}}$

Find the joint pmf $p:\mathbb{R}^n \rightarrow \mathbb{R}$ for the vector $(X_1,..,X_n)$ and show that it is a valid joint pmf..

Answer:

Since the birthdays are independent you have $p(\mathbf{x}) $
$=p(x_1,x_2,\ldots,x_n) $ $= \mathbb P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n)
 $ $= \mathbb P(X_1=x_1)\mathbb P(X_2=x_2)\cdots\mathbb P(X_n=x_n) $ $=
 p_1(x_1)p_1(x_2)\cdots p_n(x_n)$.  This will be $\frac{1}{365^n}$ when
all of the $x_i \in \{1,2,3,\ldots,365\}$ i.e. when $\mathbf{x}\in
 \{1,2,3,\ldots,365\}^n$, and $0$ otherwise.  You can easily check that
it adds up to $1$.

Difficulties

I am doing the birthday problem but have to show that $\frac{1}{365^n}$  when all of the $x_i \in \{1,2,3,\ldots,365\}$ adds up to 1. This does not make sense for me as I am adding up powers of a already large denominator over 1. Is there any way to show this

Body:
Simply

$$\frac{1}{365^n}\underbrace{\sum_{i=1}^{365}1\dots\sum_{j=1}^{365}1}_{n\text{ times}}=\frac{1}{365^n}\times365^n=1$$