Solve the equation $3x+7=6$ in $\mathbb{Z}_{13}$ The title is exercise 2.2 in The Fundamental Theorem of Algebra.

The hint for the problem is: Find the value of $\frac{1}{3}$ in $\mathbb{Z}_{13}$

  (please realize that my knowledge of the subject is what I read in Chapter 2)

I have gotten that $x = -\frac{1}{3}$.

I know that $\mathbb{Z}_{13}$ is the integers modulo 13.  Thus $x \equiv n \pmod{13}$? for some integer n,  $0\le n < 13$.

Thus for $x – n = km$ for some integer $k$, with $m \neq 0$ and an integer.

How does m divide $(x-n)$, when $(x-n)$ is not an integer, if $x = -\frac{1}{3}$????

The notation can indeed be confusing the first few times one encounters in. When one says $x=-1/3$ in the context of $\mathbb{Z}_{13}$, it is simply short-hand for saying that $x$ is some number from $0$ to $12$ such that $-3x \equiv 1 \pmod{13}.$ So remember that $x$ is not the usual real number $-1/3$, but we use this notation because it behaves like $-1/3$ would, in that $-3x=1$ in $\mathbb{Z}_{13}.$

Hint:  what is $-1 \pmod {13}$ in the normal set of representatives $0 \le n \lt 13$?


Help with commutative property of matrices problem Given matrices $A$ and $B$, where that $AB = A + B$, prove $AB = BA$.
I keep coming up with AB = AB. It seems like basic algebra, but for the life of me, I'm getting nowhere :/. Someone help please?

Hint: Consider $(A-I)(B-I)$, where $I$ is the identity matrix.

The following must apply simultaneously
$$AB+A+B=0$$
$$BA+B+A=0$$
Subtract the two equations and get the assumption.


If $x$ is a set then $\cap x$ is a set Consider the following problem:

(*) If $A$ is a non-empty set, then $\cap A$ is a set.

But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $$\{x \in A \mid \forall y (y \in A ) \rightarrow (x \in y)\}$$ This is a set by the separation axiom and it is the intersection set $\cap A$, isn't it?

What did I do wrong?

Let $A=\{\{a\}\}$, say. Then $\bigcap A=\{a\}$ but $a\notin A$, so your proposed
definition gives $\emptyset$.

Axiom: $\cup A=\{x:\exists y\in A\,(x\in y)\}$ exists. So let $B=\{x\in \cup A: \forall y\in A\,(x\in y)\},$ which exists by Comprehension (Separation). If $A=\emptyset$ then $\cup A=\emptyset$ so $B=\emptyset.$ If $A\ne \emptyset$ then $B=\{x:\forall y\in A\,(x\in y)\}=\cap A$.


motion on the surface of a 3-sphere I am trying to use motion on the surface of a 3-sphere to represent constrained, autocorrelated points in 4D.  The phenomena I'm trying to model is with the autocorrelation is fairly cyclic -- if a particular dimension is increasing, it tends to increase for awhile then decrease and vice versa.

Mechanically, this works fine by choosing an initial $(\phi_1,\phi_2,\phi_3)$, angular velocities and then doing a series of steps $\phi_1^{(n+1)}=\phi_1^{(n)}+v_1^{(n)}$, etc.  I also do multi-normal noising to the angular velocity on each step.  I simulate motion on the hypersphere surface for many steps, then translate the angles back into cartesian coordinates.

This generates values that comply with constraint to the surface, but when I check the distribution of points in the 4D space, I find that ${x_1}^2$ average is roughly 1/2, ${x_2}^2$ is roughly 1/4, and ${x_3}^2$ and ${x_4}^2$ are both roughly 1/8.

I need the averages to be roughly equal.  Looking at the equations for converting between coordinate systems, this outcome seems obvious in retrospect -- each cosine and sine of $\phi_i$ terms would have the same average value, and each subsequent $x_j$ multiplies by another cosine or sine term, and movement in the angular dimensions is independent.

Is there a straightforward way to modify this general approach to yield the correct average $x$ behavior?

Thanks to @IvanNeretin's comments, I got the right perspective.

My dynamic model results in uniform distributions for all of the $\phi_i$ terms, which is what I thought I wanted.

But when I thought in terms of randomly distributing points on the surface, I realized that uniformly distributed points only result in uniformly distributed $\phi_{n}$ for an $n$-sphere.  Thinking about the regular sphere (a 2-sphere), the azimuthal angle ($\phi_2$) is uniformly distributed, but the polar angle is not - this is easy to realize by recognizing the relative area strips associated with projecting $d\phi_1$ at different $\phi_1$.  Near the poles, the strip area is small, while near the equator, it is large.  The relative difference is a factor of $\sin\phi_1$.  So a uniformly distributed sample of $\phi_1$ on $(0,\pi)$ needs to be re-distributed to be a $\sin\phi_1$ weighted sample, which can be accomplished with the CDF method.  We know the associated CDF would be

$$
CDF(x) = \frac{\int_0^{x}\sin\phi d\phi}{\int_0^\pi \sin\phi d\phi}
$$

which can be solved, inverted, and applied to uniformly sampled $\phi_1/\pi$

For each additional sphere dimension, there is an extra power of $\sin\phi$ in the leading $\phi_1$.  So for the the 3-sphere, the distribution of $\phi_1$ needs to become $\sin^2\phi_1$, $\phi_2$ becomes $\sin\phi_2$ and $\phi_3$ remains uniform.

By squishing the $\phi_i$ dimensions to $\sin^{n-i}\phi_i$ distributions, I am able to keep the boundary constraint, get the autocorrelation, and meet the mean balance of the $x_i$s.


Time between 3 and 4 that the hour and minute hands overlap each other? This problem is on the practice problems for my math club, but I'm not sure how to solve it.

What's the exact time that the hour and minute hands overlap each other between 3 and 4 o clock on an analog clock?

Thanks!

Let $x$ be the number of minutes after 3:00 that the hour and minute hands overlap.
Since the hour hand moves $360^\circ$ twice over a $24$-hour period, so it moves $\frac{360^\circ}{12\times60}=0.5^\circ$ per minute.  The hour hand would be at $90^\circ$ at time 3:00, therefore, it would be at $90^\circ+0.5^\circ\times x$ at the time that the hands overlap.
For the minute hand:  it moves $360^\circ$ every hour, so it moves $\frac{360^\circ}{60}=6^\circ$ every minute.  And since it starts at an angle of $0^\circ$ at 3:00, the minute hand should be at the angle of $6^\circ\times x$ when they overlap.
Now, set them equal to each other, you get:
\begin{align*}
90+0.5x&=6x\\
180+x&=12x\\
180&=11x\\
x&=\frac{180}{11}
\end{align*}
$\therefore\text{The hour and minute hands overlap each other }180/11\text{ minutes after 3:00}.$

big hand moves 12 times as fast as little hand so minutes after 3:15 (when minute hand will be on top of hour hand) equals $X$. Therefore $$12X - X = 11X;$$ $$11X = 15;$$ so $$X = \frac{15}{11};$$ so time will be 3:16 and $\frac{4}{11}$.


Does every vector space have an inner product? I have a question: Does every vector space have an inner product?

I think, yes. But I failed to find an essential reason. If it does not exist, then give me a counterexample. Thanks.

Take a vector space $X$ over $\mathbb R$ (or $\mathbb C$) and fix a basis $\{v_i\}_{i\in I}$.

If $x,y\in X$, then they can be uniquely expressed as
$$
x=\sum_{i\in I}c_iv_i,\quad y=\sum_{i\in I}d_iv_i,
$$
with each of the sum above finite.

Define
$$\langle u,v\rangle=\sum_{i\in I}c_id_i.
$$
This is an inner product in $X$.

In the case of $\mathbb C$, the definition is
$$\langle u,v\rangle=\sum_{i\in I}c_i\overline{d_i}.
$$

Inner products are defined for spaces over $\mathbb{R}$ or over $\mathbb{C}$ only.

The reason is that in the definition we require $$\langle u,v \rangle=\overline{\langle v,u\rangle}$$

and what is the meaning of $$\overline{\langle v,u\rangle}$$

when $\langle v,u \rangle$ is not complex ?


First order ODE with Dirac delta funtcion I am looking for a direct method to solve this first order ODE with Dirac delta funtcion

$$\frac{dU(t)}{dt}+k^2U(t)=\frac{1}{\sqrt{2\pi}}\delta(t)$$

with the initial condition $U(0)=\frac{1}{\sqrt{2\pi}}$.

The solution to this problem is $$U(t)=\frac{1}{\sqrt{2\pi}}e^{-k^2t}$$

My try

The integrating factor for this ode is
$$I=e^{\int k^2 dt }=e^{ k^2 t  }$$

then multiplying both sides of the differential equation by  $\,\,e^{ k^2 t  }$, we get

$$\frac{d }{d t}\left(e^{ k^2 t  } U(t)\right)=\frac{1}{\sqrt{2\pi}}e^{ k^2 t  } \delta(t)$$
Integrating both sides, we have

$$e^{ k^2 t  } U(t)=\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t  } \delta(t)}dt+C$$

From here on, I am lost. Any suggestions?

Back ground of this problem

The above ode we got after applying Fourier transform to the following PDE
$$u_{t}=u_{xx}+\delta{(x)}\delta{(t)}$$
with $u(x,0)=\delta(x)$.

This maybe an answer to my question but needs your conformation/validation.

Continuing from where I left
$$e^{ k^2 t  } U(t)=\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t  } \delta(t)}dt+C$$
$$\Rightarrow U(t)=\bigg[\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t  } \delta(t)}dt+C\bigg]e^{ -k^2 t  }$$
Now recalling some properties of the Dirac delta function

$$\int_{-\infty}^{\infty}f(t)\delta(t)dt=f(0)$$
$$\int_{a}^{b}f(t)\delta(t)dt=
\begin{cases}
f(0) & \text{if } a<t<b\\
0 & \text{if } x=0
\end{cases}
$$
Now since we are dealing with an IVP, where $t\geq0$, then adopting the above properties "we can have"
$$\int{e^{ k^2 t  } \delta(t)}dt=e^{0}=1,$$

which gives
$$U(t)=\bigg[\frac{1}{\sqrt{2\pi}}+C\bigg]e^{ -k^2 t  }$$
Finally utilizing the initial condition $U(0)=\frac{1}{\sqrt{2\pi}}$, gives $C=0$, and hence
$$U(t)=\frac{1}{\sqrt{2\pi}}e^{ -k^2 t  }$$

Is this correct?

I wonder if you have a redundant initial condition? What is normally seen is the following:

U’+k^2*U=dirac(t)/sqrt(2pi)

U(0)=0

This system is equivalent to the following:

U’+k^2*U=0, t>0

U(0)=1/sqrt(2pi)

So I guess your ODE should be instead:

U’+k^2*U=dirac(t)/sqrt(2pi)

U(0)=0


Traditional Marriage GS female pessimality In the context of stable matchings I am currently trying to understand the Gale-Shapely Algorithm (traditional marriage algorithm).

I have proved it s correctness and I have shown that its male optimal in the sense that each boy  will end up in the best stable match. (i.e. any matching where he would be with a girl he prefers would not be stable), however I am struggling a little to come up with a good strategy to show the algorithm will return a female pessimal matching. (i.e. among all stable matchings  each female will be matched to the least preferred boy)

A hint to get me started would be very appreciated.

Let $M_1$ be the male-optimal matching, and suppose that $M_1$ was not female-pessimal. By definition this would mean that there is some girl (call her Alice) who could be worse off with some other stable matching $M_2$. Let's say that $M_1$ pairs Alice with Bob, and $M_2$ pairs Alice with Charles. By assumption Alice likes Bob better than she likes Charles.

However, since $M_2$ pairs Alice to Charles, it must pair Bob with some other girl, Denise. However, since $M_1$ is assumed male-optimal, Alice is the girl Bob likes best among those he has any chance with. We can conclude that Bob likes Alice better than he likes Denise. But that means that under matching $M_2$, Alice and Bob ought to elope together rather than stay with Charles and Denise. In other words, $M_2$ cannot be stable after all.

I think that if Alice is paired to another boy, Bob has to find another girl. Then it isn't male-optimal for Bob since Alice is the optimal girl.


If $a0$ then $a\leq b$ Show that if $a,b\in\mathbb{R}$ and $a<b + r$ for every positive rational $r$, then $a\leq b$
I think it might be useful to use the density of $\mathbb{Q}$ in $\mathbb{R}$, but I don't really know how to approach this problem

Suppose that $a>b$. Then we can find a rational number $r$ such that $0 < r < a-b$, and then $a<b+r<a$, which is a contradiction.

another idea is to take $r=\frac{1}{n}$ so $$a<b+\frac 1n$$ how to read inequality when $n$ tends to $+\infty$ ?


How to prove that it is a surjective homomorphism?? Ergodic Theory, Symbolic Dynamics and Hyperbolic Spaces, Chapter 1, Page 12.

Let $\mathcal{M}$ be the group of all Möbius transformations. It is well known that the map
$$A=\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)\mapsto g_A,\ g_A(z)=\frac{az+b}{cz+d}$$
is a homomorphism of $SL(2,\mathbb{C})$ onto $\mathcal{M}$ [...]

How to prove that this homomorphism is surjective?
In $SL(2,\mathbb{C})$ we have $ad-bc=1$ where in $\mathcal{M}$ we have $ad-bc\not=0$.

So $\mathcal{M} = \{ g :\hat{\mathbb{C}} \to \hat{\mathbb{C}}, \exists(a,b,c,d) \in \mathbb{C}^4, ad-bc \ne0, g(z)=\frac{az+b}{cz+d}  \} $
So take any $(a,b,c,d) \in \mathbb{C}^4$ with $ad-bc \ne 0$, and let called $\mathcal{M} \ni g:z \mapsto\frac{az+b}{cz+d}$.
Let $t$ be one of the two square root of $ad-bc$ then
$$A = \begin{pmatrix}a/t & b/t \\ c/t & d/t
 \end{pmatrix} \in SL(2,\mathbb{C})$$
and $g_A(z)=\frac{za/t +b/t}{zc/t+d/t}=\frac{az+b}{cz+d}=g(z)$
So we find a preimage of $g$ and the map is surjective.

The inverse of a matrix with determinant $1$ also has determinant $1$.
So $g_{A^{-1}}$ is also in $SL(2,\mathbb{C})$, which shows that $g_A$ is both onto and one-to-one.
Indeed, $$g_{A^{-1}}\circ g_A=g_A\circ g_{A^{-1}}={\rm id}.$$


Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$ In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian.

My approach was the following:

Let $a,b\in G$

Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.

Now, $$\begin{align}
a^3b^3=b^3a^3 \\
\implies aaabbb=&bbbaaa\\
\implies a\cdot a^2\cdot b^2&\cdot b=b\cdot b^2\cdot a^2\cdot a\\
\implies a\cdot b^2\cdot a^2&\cdot b =b\cdot a^2\cdot b^2\cdot a\\
\implies ab \cdot ba \cdot a&b=ba\cdot ab\cdot ba
\end{align}$$

I am unable to proceed further.

What I need is a simple proof using simple theorems on groups, better if could be done using elementary properties.

Hint: We can show that
$$a^{6}b=ba^{6}$$
for all $a, b\in G$. With assumptions, we have
$$a^{2}=b^{-2}a^{2}b^{2}\ \ \text{and}\ \ a^{3}=b^{-3}a^{3}b^{3},$$
and we get
$$a^{6}=b^{-2}a^{6}b^{2}\ \ \text{and}\ \ a^{6}=b^{-3}a^{6}b^{3},$$
thus
$$a^{6}=b^{-2}a^{6}b^{2}=b^{-3}a^{6}b^{3}$$
and the above relation implies that  $a^{6}b=ba^{6}$ for all $a, b\in G$.

Proof. The center of a group $G$, denoted $Z(G)$ is the set of elements that commute with every element of $G$, which is subgroup and $Z(G)=G$ iff $G$ Abelian group.. With the above hint we get for any $a\in G$, $a^{6}\in Z(G)$, now  $a^{8}b^{6}=b^{6}a^{8}$ and so
$$aa^{7}b^{6}=b^{6}a^{7}a;$$
therefore $a^{7}b^{6}\in Z(G)$ and since $Z(G)$ is a subgroup thus $a^{7}, a^{6}\in Z(G)$ (note that $a^{6}\in Z(G)$) and so $a^{7-6}=a\in Z(G)$ and this implies that $Z(G)=G$ i.e., $G$ is Abelian group.

Hint: If you can somehow show that $a^2b^3=b^3a^2$. First take $(a^2b^3)^2$ and then prove it equal to $(b^3a^2)^2$. After this show that $a^2b^3=b^3a^2$ implies the group to be abelian


diophantine equation with squares over 3 variables I am trying to find solutions for this diophantine equation

$$x^2+y^2+x^2y^2=4z^2$$

I am looking for advice on a procedure to find all positive integer solutions for this equations.

by considering x,y are both even and let $x^2=s ,y^2=r$

then $sr+s+r=(2z)^2$ this equation is equivalent to:

$(2s+r+1)^2=(r-1)^2+(2s)^2+(4z)^2$  you can check that
and the positive solutions for the last equation are given by the dimensions and the length of the diagonal of a rectangular box which is a related problem to Pythagorean Triple.

so $s=a , b=a+1 , r=(4z^2-a)/b$

& b is a divisor of $a^2+4z^2  ,b<√(a^2+4z^2 )   ,   1<2z^2$

one can solve this equation over the positive integers if he just know the value of  $a$


Examining convergence of the series. Examine convergence of the series $$1-\frac{1}{2}+\frac{1\cdot3}{2\cdot4}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+\cdots$$

Let's denote $a_n=\frac{1 \cdot 3 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot \cdots \cdot (2n)}$ and consider:
$$\frac{a_n}{a_{n+1}}=1+\frac{1}{2n+1}$$
So by Gauss test we have absolute divergence and conditional convergence by Leibniz test and for limit 0


Why isn't the set of all $m\times n$ matrices over $F$ denoted ${F^m}^n$? Is there some technical reason that this notation is not used, or is it just a convention to use $\mathrm{M}_{m\times n}(F)$ instead?

It’s not hard to prove $F^{mn} \cong M_{m\times n}(F)$ as vector spaces, or rings (in the case $m=n$); it's only a matter of notation. However, notation is a matter of convenience and things should be written clearly.

The first one is not very clarifying: you could add a matrix ring structure to $\mathbb R^4$, but reading it as $M_{2\times2}(\mathbb R)$ makes it easier to think about the set with operations as matrices, since the notation speaks about the underlying algebraic structures. If you're talking about vectors, you shouldn't use $M_{2\times2}(\mathbb R)$, and if you're talking about matrices, it doesn't make sense to use $\mathbb R^4$.

Actually both is correct it is just matter of convension which we follow in whole mathamatics.
Thanks for a2a.


If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not? In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?

I tried by assuming a set is equal to its interior points and contains its limit points.

A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?

Anyway, it must then be unbounded.

If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$.

Suppose $X \subset \mathbb{R}$ is nonempty, open and closed. Let $x_0 \in X$. Finally suppose that $X \neq \mathbb{R}$; then there is some $y \not\in X$; WLOG we can assume that $y > x_0$.

Then the set $Z = \{ x \in \mathbb{R} : x > x_0, x \not\in X \}$ is bounded below (by $x_0$) and nonempty ($y \in Z$). Therefore $\inf Z = z$ exists.

Suppose $z \in X$. Then since $X$ is open, it contains an open neighborhood $(z - \epsilon, z + \epsilon)$. This contradicts the definition of $z = \inf Z$, because there would be a sequence $z_n > z$, $|z - z_n| < \frac{1}{n}$, $z_n \in Z \Rightarrow z_n \not \in X$. This is not possible, because $[z, z + \epsilon) \subset X$.
Suppose $z \not \in X$. Then since $X$ is closed, its complement is open, therefore there is an open neighborhood $(z - \epsilon, z + \epsilon)$ contained in $\mathbb{R} \setminus X$. Then $z - \frac{\epsilon}{2}$ contradicts the $\inf$ definition of $z$.

It follows that $X = \mathbb{R}$.

If $A$ is open, $A$ is ($\emptyset$ or) a disjoint union of a countable collection of open intervals. If $A$ is the latter, then $A$ is closed if and only $A=\mathbb R$ because otherwise, $A$ would have nonempty border, violating the rule that Set is Clopen iff Boundary is Empty.

Note: This is more in line with my thinking that $A$ has something to do with intervals, but unfortunately the first sentence of this post was not proven (or stated) in one of my elementary real analysis textbooks, namely the one by Trench. Luckily: 1 - it is an exercise in the other textbook, namely the one by Lay. 2 - it is a proposition in my advanced real analysis textbook by Royden and Fitzpatrick.


Proving $\, a^b = (a\bmod n)^b\, \pmod{\!n}\ $ [Congruence Power Rule] I am working on a problem I am pretty close to solving but I can't figure out the last part. I used some algebraic manipluation to break the problem down. The problem is:

  Show that the following congruence is true

  a^b = ((a mod n)^b) (mod n)

  for any positive integers a, b, and n.

  Hint: Number a can be written as a = q*n + r, where q = floor brackets a/n, and r = a mod n.

My attempt it:

  (q*n +r)^b = ((a mod n)^b) * (mod n)

  (q*n)^b + (a mod n)^b = ((a mod n) ^b) * (mod n)

  (a)^b -(r)^b = ((a mod n) ^b) * (mod n)

  (a)^b = modn  ----> divided by ((a mod n)^b)

Am I doing this correctly and if so how can I prove this is true? Is there a trick I am missing? I think that by simplifying it further I would just keep going in a loop.

Thanks!

Let $\ A = (a\ {\rm mod}\ n).\,$  Then $\ A \equiv a \,\Rightarrow\,  A^b \equiv a^b\pmod n\,$ by the Congruence Power Rule below.

Congruence Sum Rule $\rm\qquad\quad  A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \  B\equiv b\ \Rightarrow\ \color{blue}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{blue}{AB - ab} $

Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \  (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then  $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$  By induction $\,g(A)\equiv g(a)\,$ so $\, \color{#0a0}{A g(A)\equiv a g(a)}\,$ by the Product Rule. Hence $\,f(A) = f(0)+\color{#0a0}{Ag(A)}\equiv f(0)+\color{#0a0}{ag(a)} = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g.
the exponential analog of above $\rm A^B\equiv\, a^b$ is not generally true (unless $\rm B = b,\,$ when it follows by applying the Polynomial Rule with $\,f(x) = x^{\rm b}).$

We have
$$a = c \pmod{n}$$ This means $n \mid (a-c)$, i.e., $a-c = nd$. Now consider $a^b-c^b$. We have
$$a^b-c^b = (a-c) \left(a^{b-1} + a^{b-2}c + a^{b-3}c^2 + \cdots + ac^{b-2} + c^{b-1} \right) = nd\left(a^{b-1} + a^{b-2}c + a^{b-3}c^2 + \cdots + ac^{b-2} + c^{b-1} \right)$$
Hence, $$n \mid \left(a^b-c^b \right) \implies a^b \equiv c^b \pmod{n}$$


Are there any practical applications for higher dimensional geometry? My current understanding of higher dimensions is that we can take geometric properties from 2D or 3D, and simply extend them further theoretically. So in the same way that 3D objects can be projected onto a 2D space, 4D objects can be projected onto a 3D space.

But do higher dimensions "exist" in any sense outside of this type of theory? Is it just that we humans can't sense them, like we can't see infrared or hear ultrasound? And can we apply this theory to solve practical problems?

Higher dimensions, and in particular, four dimensional spaces "exist" in the physical sense you describe. The study of such spaces is general relativity, which seeks to explain relationships between space, time, gravity, mass, and energy. General relativity explains the universe as a four dimensional space with three spatial dimensions and a time dimension. There are also other theories of physics which rely on the reality of ten or more dimensions, but I don't know too much about that.

But higher dimensional spaces are practical in other ways. One of my favorite examples is parameter space. Suppose you were designing a new cell phone. You might give it parameters, or numbers that describe it. For instance, you could let $w$ represent the width, $l$ the length, and $t$ the thickness. You could have $c$ represent the camera resolution, $m$ the memory, and $p$ the processor speed. We could go on, but that is sufficient for this example. With these numbers, we could represent the space of all possible cell phones by the ordered 6-element tuple $(w,l,t,c,m,p)$. Thus, these cell phones exist in a six dimensional space. While true that we cannot see this space, hold it in our hands, or draw it in a nice way, it is still a very real space. And it is far less abstract than most mathematical objects--ever element represents a cell phone one could build, hold, and use. Of course, cell phones aren't the only application. I know that biologists model certain processes that occur in cells in up to at least forty-something dimensional space. I once saw an analysis of how electricity was generated, moderated and distributed which employed a space with over one hundred parameters. My own research involves infinite dimensional spaces which nevertheless have real life application to the shapes of materials under stress. Thus, higher dimensional spaces can real and practical.

The question is asking about four dimensional space, specifically referencing tesseracts, and is not a general inquiry about higher dimensions. Although spacetime is clearly a way of understanding four dimensions and higher dimensions in general, this is not the question at hand.

It's absolutely necessary to point out this conflation, as Alex S. begins his answer saying that four dimensional space exists in a physical sense. I know he simply misunderstood the question, but this answer is so egregiously wrong that it can't stand.

Humanity has developed many ways of measuring physical quantities that we cannot or have difficulty perceiving, and these generally can be used to understand complex phenomena. Space, time, and temperature are at least five of the necessary dimensions for you to model the rate at which the heater on your stove warms up, for example. Alex mentions other examples where higher dimensions are useful.

On the contrary, four dimensional space is but a theoretical modelling abstraction, and there is no evidence that it exists in a physical sense.
Even if we cannot perceive it like large portions of the electromagnetic spectrum, we can't measure it either. That doesn't mean that it isn't practically useful, but I'm not the right person for that. I've asked a friend to comment here on her work.


solve the Following Differential equation: $y''=2yy'$ Solve the following equation: $y''=2yy'$

My attempt:

$y''=2yy'$

integrate on both sides:

$y'=\int(2yy')$

Let y'=s

We have $\int 2y \dfrac{ds}{dy}dy = 2 \int yds = y^2$

Is that a way to integrate $2yy'$?

I fthat's the case don't we get $y=\dfrac{1}{3}y^5$ which is false

Indeed,

$$\frac{dy'}{dx}=2y\frac{dy}{dx}$$
can be integrated by

$$\int dy'=\int2y\,dy,$$

$$y'=y^2+C.$$

Then

$$\frac{dy}{dx}=y^2+c^2,$$

$$\int\frac{dy}{y^2+c^2}=dx,$$

$$\frac1c\arctan\left(\frac yc\right)=x+c',$$

$$y=c\tan(cx+c').$$

Note that the constant was denoted as $c^2$ to ensure positiveness. A similar solution holds for a negative constant, involving the hyperbolic tangent.

Yeah. More generally if you have a 2nd degree nonlinear autonomous equation (no independent variable) do the following substitutions

y = y

y' = s

y'' = s*ds/dy

This reduces the order.


Can you provide examples of fractal coordinates for points? On a 1D line there is one real number needed to uniquely locate a point by coordinates.
On a 2D plane, a pair of real numbers are needed.
But what if a fractal has dimension d, between 1 and 2 ($1<d<2$), how can points on the fractal be specified with more than one, but less than 2 coordinates?
Can you provide examples of 1.xD coordinates for points?
Edit: I'm asking for a numerical example, so this is not the same question tan this one.
A point in the plane can be specified by a single number in base 4
if each one of the symbols represents one quadrant
O=0
→=1
↑=i
↗=1+i
. is a decimal separator

a point with coordinates 1.5+2.5i can be written ↑→.↗
meaning (the dot is a decimal separator)
$\uparrow \rightarrow . \nearrow = i*2^1+1*2^0+(1+i)*2^{-1} $
That representation system can be rewritten as pair of numbers in decimal system.
On the same way, a point on sierpinski triangle can be written in base 3 using a symbol for each of the 3 subtriangles
Any point on the fractal can be specified with arbitrary precision with a string like
←→↑←↑↑←←→.←↑←→←←↑
where each position is multiplied by a power of 2 and
←=0
→=1
↑=0.5+(3/4)^½i
. is a decimal separator

So, I expect that number to be decomposed into 1.D coordinates as a real number, and something else.
On information theory, a character may represent a non integer number of bits, and that is solved by combining multiple characters, so I expected a similar solution for expressing coordinates.
Note that on the real line, if base N is used, the characters are multiplied by power of 4, but on other dimensions, it looks like the dimension depend on the difference between the base N and the power used as positional weight.

When we talk of "dimensions" of fractals we mean something entirely different that the number of independant spatial coordinates to determine a point.
So the question of how to indicate a point doesn't apply or even make sense.
So what does the dimensions of a fractal mean?
In euclidean geometry, if we increase the radius of an $n$dimensional object in $\mathbb R^n$ by a factor of $k$, then the hyper-volume of the object increase by a factor of $k^n$.  And it is the linear increase causes an exponential growth in volume that we mean by "dimension" and that has nothing to do with coordinates.
So the area of a fraction increases in comparison to an increase in linear size by a non-integer factor.
So in that definition it is fractional dimensional.  But it has nothing to do with coordinates.
https://medium.com/@cosinekitty/understanding-fractional-dimensions-f2ed2e4e1600
(Sometimes we teach by analogy.  But then analogy breaks.  The classic example if this is $b^{n} = \underbrace{b\times b\times b......\times b}_{n\text{ times}}$.  And as just by counting we must have $b^{n+m} = b^nb^m$.  But what if $x\not \in \mathbb Z$? What does $b^x$ mean?  Well, knowing that $b^{x+y} = b^xb^y$ we come up with another definition based on... other stuff.  So $2^\pi = 8.824977827076287623856429604208$......  Punch it into a calculator! That's what you get.  But does that mean $2^\pi = \underbrace{2\times2\times 2\times ....}_{\pi \text{ times}} =  8.824977827076287623856429604208$????  How the heck do you multiply a number by itself $\pi$ times?  Does that make any sense?  The answer is "no" it doesn't make sense but that is not what $b^x$ means any more.  It means something else.)

You're confusing the topological dimension of an embedding with the intrinsic dimension.  You need as many (integer) coordinates as the topological dimension.


If a series converges uniformly on $[a+\varepsilon,b) $ for all $\varepsilon>0$ and pointwise on $[a,b)$, is then the convergence uniform on $[a,b) $? The question arises from the specific series $\sum_{k=0}^\infty \dfrac{k (ex)^k}{(k+1)^2} $, which converges pointwise on $[-1/e,1/e) $ and uniformly on $[-1/e,1/e-\varepsilon] $ for every $\varepsilon>0$.

I found the intwrval of pointwise convergengence and I proved uniform convergence on $[-1/e+\varepsilon,1/e-\varepsilon] $, is this enough to conclude the interval of uniform convergence is $[-1/e,1/e-\varepsilon] $?

In general the above statement is false.
Consider, for example, $\sum_{n=0}^\infty x^n$, that converges pointwise to $f(x) = \frac{1}{1-x}$ for $x\in (-1,1)$ and uniformly on every compact interval $[a,b]\subset (-1,1)$.

On the other hand, you have $s_n(x) := \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}$, and this sequence does not converge uniformly to $f$ in $(-1,1)$.


Probability when arranging $N$ different numbers in a line we would like to arrange the numbers: $ 1, 2, \ldots , N$ in a line.

what is the probability that digits $1, 2$ are not arranged next to each other?
what is the probability that digits $1, 2$ are not  arranged next to each other, and also the digits $3, 4$ are arranged next to each other?

my attempt:

This is mostly a combinatorics question, as the "Probability", meaning: $$\Omega = N!$$

I treated $1, 2$ as a unit, looking for the number of different ways to arrange them next to each other - which would be the Complementary event.
$$P(A)^C = \frac{(N-1)! \cdot 2 }{N!}$$
where there's in fact $N-1$ needed to be arranged, ans 2 is the number of permutations.

so that $$P(A) = 1- \frac{(N-1)! \cdot 2 }{N!} = 1 - \frac{2}{N}$$

for the second sections - I'm not sure.
suppose I'd be looking the number of different ways to arrange  $1, 2, \ldots N$ in a line so that the digits $1, 2$ and $ 3, 4$ are always arranged next to each other, then that will be:
$$\frac{(N-2)! \cdot 2 \cdot 2}{N!}$$ where 4 is the number of permutations (2 for each unit of digits)
but does that mean that there's $$\left(N! - {(N-2)! \cdot 2 \cdot 2}\right) \cdot \left((N-2)! \cdot 2 \right)$$ ways to arrange the numbers?

that doesn't seem right.

The number of ways that $1$ and $2$ appear next to each other can be calculated by considering both the numbers as a single unit. So the packets will be $$ (1,2),(3),(4),(5) ....... (N) $$
Number of ways these can be arranged are $(N-1)!$ as there are $(N-1)$ packets. To consider the backward and forward arrangement of 1 and 2 inside the packet we need to multiply by two. So,
Total ways for 1 and 2 to be arranged together is $2(N-1)!$

So the numerator will be (Total ways-Undesired ways).  $$N! - 2(N-1)!$$.
Divide this with the total ways we get $$\frac{N! - 2(N-1)!}{N!}  = 1 - \frac 2N$$

For the second part we consider one packet with 1 and 2. The other packet with 3 and 4. So the packets will be,$$ (1,2),(3,4),(5),(6),(7) ....... (N) $$

Arrangements in which 1 and 2 are together(like the first part) are $P(X)=2.(N-1)!$
Arrangements in which 3 and 4 are together are $P(Y)=2.(N-1)!$
Arrangements in which both groups (1 and 2) and (3 and 4) are together are $P(Z)=2.2.(N-2)!$. The two packets have an internal arrangement of two ways and the total count of packets is $(N-2)!$

The numerator in the second part is $N! - P(X) - P(Y) + P(Z)$.
The denominator will be $N!$

Regardless of the other numbers in the list, there are $2(N-1)$ ways to position $1$ and $2$ next to each other in a list of size $N$. Assume they take the first position, then you have to sort the other $N-2$ values into a list of size $N-2$. Thus we have that the total ways of satisfying the question is:
$$2(N-1)\cdot (N-2)!=2(N-1)! $$

The probability follows nicely from this, it is $$\frac{2(N-1)!}{N!}=\frac 2N $$

This approach should also work for (b).


Let u4 be a vector that is not a linear combination of {u1,u2,u3,}. Select the best statement.

A) span{u1, u2, u3} = span{u1, u2, u3, u4}

B) There is no obvious relationship between span{u1, u2, u3} and span{u1, u2, u3, u4}

C) span{u1, u2, u3} is a proper subset of span{u1, u2, u3, u4}

D) We only know that span{u1, u2, u3} ⊆ span{u1, u2, u3, u4}

E) span{u1, u2, u3} = span{u1, u2, u3, u4} when none of  {u1, u2, u3, }  is a linear combination of the others.

F) We only know that span{u1, u2, u3, u4} = span{u1, u2, u3}

G) None of the above.

I have tried B and G thinking that there is not an explicit relationship between the two sets, however, both those choices are incorrect.

For any $u_1, u_2, u_3, u_4$, we always have
$$\text{span}\{u_1, u_2, u_3\} \subseteq \text{span}\{u_1, u_2, u_3, u_4\}.$$
This is because any linear combination of $u_1, u_2, u_3$ is automatically a linear combination of $u_1, u_2, u_3, u_4$. (Explicitly, $c_1 u_1 + c_2 u_2 + c_3 u_3 = c_1 u_1 + c_2 u_2 + c_3 u_3 + 0 \cdot u_4$.

Now the question is whether this containment is strict containment $\subsetneq$ or equality $=$. If $u_4$ is not a linear combination of $u_1, u_2, u_3$, then it does not lie in $\text{span}\{u_1, u_2, u_3\}$. Therefore the containment must be strict $\subsetneq$.

Assuming there is no additional information you have not told us, D is correct.

A) "$span(x_1, x_2, x_3)= span(x_1, x_2, x_3, x_4)$" is true only if $x_4$ is a linear combination of $x_1$, $x_2$, and $x_3$.

B) It's not clear what "obvious relationship" means here but we can certainly say that
$span(x_1, x_2, x_3)$ is a subspace of $span(x_1, x_2, x_3, x_4)$.  I would call that an "obvious relationship"!

C)  I just said it was a subspace but it is not necessarily a "proper" subspace.

D)  This is what being a subspace means.

E)  Again not necessarily true because $x_4$ is not necessarily a linear combination of the others.

F)  This is just a restatement of (A) and (E)!


Prove that $x^2+1$ cannot be a perfect square for any positive integer x? I started this problem by trying proof by contradiction.

I first noted that the problem stated that $x$ had to be a positive integer, and thus $x=0$ could not be a solution. I then assumed that $x^2+1=n^2$ for some integer $n$ other than $1$. From here I have tried various methods, to no avail:

Factoring:

$n^2-x^2=1\implies(n+x)(n-x)=1$. It would be nice if I could say $(n+x)=(n-x)=1$ and $(n+x)=(n-x)=-1$. However, the issue here is that $(n+x)$ and $(n-x)$ could take on any value. For example, $(n+x)=2$ and $(n-x)=\frac{1}{2}$, or $(n+x)=3$ and $(n-x)=\frac{1}{3}$. Thus, I ruled out factoring.

For any number $n$, $n\equiv 0 \pmod4$ or $n\equiv 1 \pmod 4$.

The trouble with this is that I would have to prove the above statement, so I ruled this out.

Does anyone have any tips on how to continue? I feel like this should be an easy proof, but no solutions are coming to me, without having to prove something else.

For any $n$, the square is $n^2$.

The following integer is $n+1$, and its square is $n^2 + 2n + 1$.

Since (assuming $x>0$) $$x^2 < x^2+1 < x^2 + 2x + 1$$

it also true that  $$x = \sqrt{x^2}  < \sqrt{x^2+1} < \sqrt{x^2 + 2x + 1} = x+1$$ but there are no integers between $x$ and $x+1$.

You could try proving by contradiction by assuming that $x^2+1$ is a perfect square for some $x∈\mathbb{Z}$ then show how this cannot be possible


Integral $ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx$ Hello there I am trying to calculate
$$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx
$$
NOT using mathematica, matlab, etc.  We are given that $\sigma, \omega$ are complex.  Note, the integral should have different values for $|\sigma \omega^{-1/2}| < 1$ and $|\sigma \omega^{-1/2}| > 1.$  I am stuck now and not sure how to approach it.
Note this integral is useful since in the limit $\sigma \to \sqrt{\omega}$ and using $Li_2(-1)=-\pi^2/12$ we obtain
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^2)}{x^3}dx=\frac{\pi}{2}.
$$
We also know that
$$
\ln(1+x)=-\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}, \ |x|\leq 1.
$$
Thanks

One may adopt the approach as in Pranav Arora's comment. But this approach involves a double integral whose calculation seems painful. So here is an indirect approach that makes calculation slightly easier (at least to me):

Let us consider the following integral: for $\alpha, \beta \in \Bbb{C}\setminus(-\infty, 0]$ and $0 < s < 1$,

$$ I = I(s,\alpha,\beta) := \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta x)}{x^{2+s}} \, dx, $$

Differentiating w.r.t. $\alpha$ and $\beta$, we have

$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \int_{0}^{\infty} \frac{dx}{x^{s}(1+\alpha x)(1+\beta x)}. $$

Using standard complex analysis technique (you man use keyhole contour), it follows that

$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \frac{\pi}{\sin \pi s} \frac{\beta^{s} - \alpha^{s}}{\beta - \alpha}
\quad \Longrightarrow \quad
I = \frac{\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{x^{s} - y^{s}}{x - y} \, dxdy. \tag{1} $$

Replace $\beta$ by $i\beta$ (with $\beta > 0$). Then (1) yields

$$ 2I(s, \alpha, i\beta) = \frac{2\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{i^{s}x^{s} - y^{s}}{x + iy} \, dxdy. $$

Now assume that $\alpha, \beta > 0$. Taking real parts of the identity above and taking $s \to 1^{-}$, it follows that

\begin{align*}
\tilde{I}(\alpha, \beta)
&:= \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta^{2}x^{2})}{x^{3}} \, dx \\
&= \int_{0}^{\alpha}\int_{0}^{\beta} \frac{2xy \log(y/x) + \pi x^{2}}{x^{2}+y^{2}} \, dxdy. \tag{2}
\end{align*}

In particular, when $\beta = \alpha$, by symmetry we retrieve the following formula

$$ \tilde{I}(\alpha, \alpha)
= \pi \int_{0}^{\alpha}\int_{0}^{\alpha} \frac{x^{2}}{x^{2}+y^{2}} \, dxdy
= \frac{\pi}{2} \int_{0}^{\alpha}\int_{0}^{\alpha} dxdy
= \frac{\pi}{2}\alpha^{2}. $$

which also follows from the formula in OP's posting. In general, using polar coordinates shows that we have

$$ \tilde{I}(\alpha, \beta) = \beta^{2}J(\alpha/\beta) - \alpha^{2}J(\beta/\alpha) + \frac{\pi \alpha \beta}{2} + \frac{\pi^{2}\beta^{2}}{4} - \frac{\pi(\alpha^{2}+\beta^{2})}{2}\arctan(\beta/\alpha), \tag{3} $$

where $J$ is defined by

$$ J(x) =  \int_{0}^{x} \frac{t \log t}{1+t^{2}} \, dt. $$

This function can be written in terms of elementary functions and dilogarithm.

Remark. Though we have derived this formula for positive $\alpha, \beta$, by the principle of analytic continuation (3) continues to hold on the region containing $(0, \infty)^{2}$ where both sides of (3) are holomorphic.

why not use the $\Gamma$ function ? it seems that this question is the special case!

$\int_0^\infty \frac{\ln(1+ x)\ln(1+x^2)}{x^3}dx$$=\frac{1}{2}(\int_0^\infty \frac{e^{-x}}{\sqrt{x}}dx)^{2}$

$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx
$$=$
$\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{x^3}dx
$

$\Longrightarrow$$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{(\sigma{x})^3}dx$$=$$\frac{\sigma^{2}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$

$\Longrightarrow$
$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{{x}^3}dx$$=$$\frac{\sigma^{5}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$

$=\frac{1}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{x}}dx)^{2}$$\cdot{\sigma^{4}}$

therefore, it seems that you just need to multiply $\sigma^{3}$

and the positive and negative of your equation depends on :

$|\sigma \omega^{-1/2}| < 1$ or $|\sigma \omega^{-1/2}| > 1$


Finding $\lim_{x\to 0} \frac {2\sin x-\sin 2x}{x-\sin x}$ geometrically While looking at this question, I noticed an interesting geometric interpretation of the limit the OP was trying to evaluate.  His limit came to twice the value of the limit

$$\lim_{x\to 0}\frac{\sin x (1-\cos(x))}{x-\sin x}$$

which can be interpreted as the limit of the ratio of the area of $\triangle DBC$ and the area bounded between the line $CB$ and the arc $CB$. (See the image below).

I know that the limit $\lim_{x\to 0} \frac{\sin x}{x} = 1$ is often derived through a geometric method.  Can this the limit I've given be evaluated geometrically?

Not sure how convincing this is, but consider:

That point in the middle of trangle $DBC$ is its centroid, so that all three subtriangles have the same area. Does the area of the upper right subtriangle approximate the area of the arc sliver to a high enough degree as the angle approaches $0$? If so, the ratio is of course $3$.

I think this limit is impossible to evaluate geometrically, at least using the squeeze theorem. First, note that the area of $∆ACB=sinx/2$ because the area of a triangle is $bh/2$ with $b=1$ and $sinx=h$ in this case. Also, $∆ADC=sin(x)cos(x)/2$ and the area of the sector is $x/2$. Thus, to use the Squeeze Theorem, we would need this inequality:

$sinxcosx/2 < sinx/2 < x/2$

The problem is that this inequality only holds true in quadrant 1 where everything is positive, but in order to take the limit as x goes to 0, we need to evaluate the limit from quadrant 4 because we take the limit from left to right. But because sinx is negative in quad 4, the inequality no longer holds true. Obviously, we would need to majorly rewrite the inequality before we can even take the limit, but I strongly suspect that we'd still have the same problem. In fact, I suspect that no matter how we set up the inequality, this problem still remains. So I doubt you can evaluate this limit geometrically, using basic methods anyway.


$\left|\sin(x)\right|$ derivative at $\pi$ Let $f$ be a function, $f:\mathbb{R}\to\mathbb{R}$

$$f(x)=\begin{cases}
      e^{x^2}-2 & x< 0 \\
x^3+x-1 & 0\leq x \leq 1 \\
     \left|\sin(x)\right|  & x>1.
   \end{cases}$$

Check if $f$ is continuous and differentiable at $a$, when $a=0,1,\frac{\pi}{2}, \pi$. If $f$ is differentiable at $a$, find $f'(a)$.

What I've been doing:

I found that:

$f$ is continuous at $0$ but not differentiable.
$f$ is not continuous at $1$ so it's not differentiable.

And then I thought that $f$ was differentiable at $\pi$ and $\frac{\pi}{2}$ because $f$ is continuous at $(1, +\infty)$ ($\left|\sin(x)\right|$), so I looked for:

$f'(\frac{\pi}{2})=\left|\cos(\frac{\pi}{2})\right|=0$ (by the solution my prof gave me this is correct).
$f'(\pi)=\left|\cos(\pi)\right|=-1$ Now this is wrong. The solution they gave me says that $f$ is not differentiable at $\pi$, and I'm really lost. Why is it differentiable at $\frac{\pi}{2}$ and not $\pi$?

$f$ behaves like $\sin(x)$ around $\frac{\pi}{2}$. It is differentiable.
$f$ behaves like $\sin(x)$ on the left of $\pi$ but it behaves like $-\sin(x)$ on the right.  We have to be more careful.
$f(\pi)=0$.
$\lim_{x \to \pi^-} \frac{f(x)-f(\pi)}{x-\pi}=\lim_{x \to \pi^-} \frac{f(x)-0}{x-\pi}=\lim_{x \to \pi^-} \frac{\sin(x)-\sin(\pi)}{x-\pi}=\cos(\pi)=-1$
$\lim_{x \to \pi^+} \frac{f(x)-f(\pi)}{x-\pi}=\lim_{x \to \pi^+} \frac{-\sin x-0}{x-\pi}=-\lim_{x \to \pi^+} \frac{\sin(x)}{x-\pi}=-\lim_{x \to \pi^+} \frac{\sin(x)-\sin(\pi)}{x-\pi}=-\cos(\pi)=1$

Hopefully this picture can help you, notice where the kink occurs.

The function can be treated as piecewise function when there is absolute function involved.
$$f(x)=\begin{cases}
      \sin x & 1<x\le\pi\\
-\sin x & \pi< x \leq 2\pi
   \end{cases}$$
So
$$\lim_{x→\pi^-}f^\prime(x)→-1$$ while $$\lim_{x→\pi^+}f^\prime(x)→1$$ left gradient does not equal to right gradient, so it is not differentiable at $\pi$.


Finding a number knowing the modular multiplicative inverse Let a = 44 and m = 67
The modular multiplicative inverse of 'a' and 'm' is 32

i = inverseModFunction( a, m )
i = inverseModFunction( 44, 67 )
i = 32

It can be calculated here:
https://www.xarg.org/tools/modular-inverse-calculator/

Is there a way to find 'a' knowing the inverse modulo 'i' and 'm'?

i = 32
m = 67

a = function( i , m )
a = function( 32 , 67 )
a = ?

Modular inverse relation is symmetric, so if i is the modular inverse of a, then a is the modular inverse of i and can be found by inverseModFunction(i,m).


Number of Solutions for $p^2 - q^2$ = n? Number of Solutions for $p^2 - q^2 = n$? how do I find it? It was asked in aptitude test of a company , couldn't figure it out. What is the answer of it?

Edit:

once again i faced similar question for :$p^2 - q^2 = 388$

options : 1 , 2 , 3 , More than 3.

what is answer ?

$n = p^2 - q^2 = (p+q)(p-q)$  So every such decomposition gives you a factorization and vice versa.

However there is a gotcha.  If only one of $p+q$ and $p-q$ is odd, then $p$ is not an integer.  So it is the number of factors where the parity of the two factors is the same.

Given the prime factorization of $n$ it is not hard to figure out how many such factorizations exist.

Write the given equation as $p^2-q^2=388$.
Now prime factorize R.H.S.
Click this to view the entire solution.


Suppose that not and exist $s,t$ divisors of $n$ satisfying that $\frac{b^n-1}{b^s-1}$ and $ \frac{b^n-1}{b^t-1}$ are coprimes and divisors of $b^n-1$... My solution was wrong


Let be $G=(V,E)$ a graph and let $U$ be a set containing circles in $G$. Is $(U, E)$ a matroid? Let be $G=(V,E)$ a graph and let be $U$ a set containing circles in $G$. Is $(U, E)$ a matroid?
I draw an easy graph (undirected) and checked this property.  It does not hold for my example.

My example was like a graph with nodes $V=\{1,2,3,4\}$ and edges $E=\{a,b,c,d,e\}$ with the circles $\{a,b,d\}$ , $\{a,b,c,e\}$ and $\{d,e,c\} $ and this one would be a matroid if  the empty set would not be missing. So the statement is not true or am I missing something? Any Help is much appreciated.

A subset $C$ of circles (which I will call cycles here) of a graph $G$ cannot form the independent sets of a matroid. The reason for this is that the empty set is always an independent set (see the first axiom here) of a matroid while it is always (trivially) acyclic in $G$.

What happens if we append the empty set to $C$? Well, if $C' := C \cup \{\emptyset\}$ are the independent sets of a matroid then they form a simplicial complex (every subset of set in $C'$ is also in $C'$). In particular, for every edge $e$ in some cycle in $C'$ we have $\{e\} \in C'$. But since $C'$ is a collection of cycles of $G$ it follows that $e$ is a loop in $G$. But this implies that every cycle in $C'$ is a loop. So any matroid whose independent sets are of the form $C'$ is the trivial matroid (whose only basis is the empty set) together with some loops, that is, the uniform matroid $\mathcal{U}_{0,n}$ where $n$ is the number of loops.

On the other hand, the cycles of any graph $G$ do yield a matroid (the cycle matroid of $G$) whose circuits are the cycles of $G$ and whose independent sets (respectively, bases) are the forests (spanning forests) of $G$.

From definition of matroid,

  Every subset of member of independent set is also an independent set.

So, if $U$ are set of independent sets, then for $A \in U$, one has $\{e\} \subset A$ is also independent set, but it is obvious not true because $\{e\}$ contains no circle.


Center of a group vs centralizer vs conjugacy classes I am trying to fully understand the connection between a centralizer to conjugacy class.

Starting with the definitions:

Center: $Z(G)=\{z\in G:\forall g\in G, zg=gz\}$ which are all the elements in the group that commute

Centralizer: $C_G(S)=\{g\in G:gs=sg \text{   for all     } s\in S\}$ which are all the elements of $g$ that commute with all of the elements  subset of $S$

Conjugacy class: $Conj(x)=\{h\in G:\exists g\in G: gxg^{-1}=h  \}=\{gxg^{-1}: g \in G \}$ can not explain which elements are in the conjugacy class

Center VS Centralizer: the center takes a group and returns all the elements in the group that commute whereas the Centralizer take a subset of the group and return only that elemnts of the subset which commute

So If for example all the subset commute , let say that the group is cyclic and there for every subset commute, we will have $Z(G)=C_G(G)$

Did I get that right? what are Conjugacy classes and how they differ from the centralizer and the center? is there a simple example for these definitions?

Saying that the center of $G$ are the elements of $G$ that commute makes no sense. Commute with what? The center of $G$ is the set of those elements of $G$ that commute with all other elements of $G$.

Your definition of centralizer is correct. Note that the centralizer of $G$ is precisely the center of $G$ (for every group). And that a group is commutative if and only if its center is the whole group.

As for the conjugacy classes: the conjugacy class of $x$ are all those elements of $G$ of the form $hxh^{-1}$. For instance, if $G$ is commutative, then the conjugacy class of each $x\in G$ is $x$ itself (because then $hxh^{-1}=hh^{-1}x=ex=x$). But if $G$ is the group $S_3$ (the group of the permutations of $\{1,2,3\}$), then you can check that the conjugacy class of $(1\ \ 2\ \ 3)$ is $\{(1\ \ 2\ \ 3),(1\ \ 3\ \ 2)\}$. Note that $x$ always belongs to the conjugacy class of $x$ (because $x=exe^{-1}$).

By definition You have $Z(G)=C_G(G)$. If an element is in the center, it commutes with all elements in $G$ as You correctly said and that means for $x\in Z(G)$ that $gxg^{-1}=x$ for all $g\in G$. The mappings $\phi_g:G\rightarrow G,x\mapsto gxg^{-1}$ are called the inner automorphisms of the group $G$ and make the group operate on itself. The conjugacy classes are the orbits of elements under this operation and the elements in the center are exactly those that have orbits of length $1$.


Show that: $a·c \equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$ Show that $a·c\equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$

I've seen many similar examples, but can't seem to find a step by step explanation or solution to any of them (like an actual proof).

Here is a counter example. $a=b+1$ , $c=m$

$2\cdot2\equiv 2\cdot5(mod\ 2)$ but $2\not\equiv 5(mod\ 2)$


Set theory - what are these questions asking? For my Discrete structures class in school, we have the following homework problem:

Preface: I naturally associate the word and with the intersection operator $\cap$, for example: $x \in A \text{ and } x \in B \iff x \in A\cap B$.

a.)

To me a.) can be translated to the following:

$$A\cap B  = \emptyset \iff B\cap A  = \emptyset$$

I'm inclined to make my proof say that this implies $A\cap B = B\cap A$ which we can prove by stating the commutative law of set theory. I can't tell if it wants me to go beyond that statement and actually prove the commutative law though?

b.)

I thought this problem was going to have me show the distributive law of set theory which in this case would be $(A\cup B)\cap C = (A\cap C)\cup(B\cap C)$ HOWEVER I translate b.) to the following:

$$(A\cup B)\cap C = \emptyset \iff A\cap  C = \emptyset \land  B\cap C = \emptyset$$ Now, because statement 1 and statement 2 must be true I think I may be able to simplify it down to $\iff (A\cap C)\cap(B\cap C) = \emptyset$ but logically that does not make sense to me. We're trying to prove that C has no elements in common with A and C has no elements in common with B which you could just as easily do by simplifying it to $(A\cap C)\cup(B\cap C) = \emptyset$. The statement I was able to simplify it to ($(A\cap C)\cap(B\cap C) = \emptyset$) looks like it is more proving that $A$ and $B$ and $C$ have no common elements, namely $A\cap B\cap C = \emptyset$. Did I interpret it correct, and the problem is just kinda weird?

If my original interpretation is correct: $(A\cap C)\cap(B\cap C) = \emptyset$ then I believe the sets $A = \{1, 2, 3\}, B = \{4, 5, 6\}, C = \{3, 4\}$ proves their statement wrong.

c.)

I translate the c.) to:

$$(A\cap B)\cap C = \emptyset \iff (A\cap C)\cap(B\cap C) = \emptyset$$

Can I just use the distributive law here?

Am I interpreting these even close to correctly?

(a) seems okay, Except that you do not need to prove the commutative law, just use it as a justification.

For (b) note that: Intersection $A\cap C$ is empty and intersection $B\cap C$ is empty if and only if the union of both these intersections is empty.

$$(A\cap C=\emptyset)\wedge (B\cap C=\emptyset) \iff (A\cap C)\cup (B\cap C) = \emptyset$$

Because, basically, deMorgan's Laws:
$$\begin{align} (P=\emptyset) ~\wedge~ (Q=\emptyset)
~\iff~ & \forall x~(x\notin P) ~\wedge~ \forall x~(x\notin Q)
\\[0.5ex] ~\iff~& \forall x~(x\notin P ~\wedge~ x\notin Q)
\\[0.5ex] ~\iff~& \forall x~\neg(x\in P~\vee~ x\in Q)
\\[0.5ex] ~\iff~& \forall x~\neg(x\in P\cup Q)
\\[0.5ex] ~\iff~& \forall x~(x\notin P \cup Q)
\\[0.5ex] ~\iff~& (P \cup Q = \emptyset)
\end{align}$$

For (c) you want to disprove : $(A\cap B)\cap C=\emptyset \iff (A\cap C)=\emptyset\wedge(B\cap C)=\emptyset$

Because, $(A\cap B)\cap C=\emptyset \nLeftrightarrow (A\cap B)\cup C=\emptyset$

For a) I am with you up to here:

  I'm inclined to make my proof say that this implies $A\cap B = B\cap A$ which we can prove by stating the commutative law of set theory.

Where you say implies you probably mean is implied by or follows from.  And as far as “proving” $A \cap B = B \cap A$ from the commutative law of set theory, that's really the only content to the law.  So there's nothing to prove there—you can just mention it parenthetically.

For c), I'm not aware of a distributive law for intersection over itself.  But you could invoke the associative (and commutative) law for intersection and the fact that $C\cap C= C$:
$$
    (A \cap B) \cap C = (A \cap B) \cap (C \cap C) = A \cap B \cap C \cap C = A \cap C \cap B \cap C = (A \cap C) \cap (B \cap C)
$$

For b) I agree with Graham.

As for your question, “Do I need to prove the commutative law?”, that depends on the context of the assignment.  Is the law stated in your textbook before this exercise, or has it been covered in class?  If so, you are probably free to use it.


A change of variable for $\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$ Question 1

  I would like to understand the following change of variable

for $\phi(x)=nx,$ $\ d\phi=n\cdot dx,$ and $\chi_{E}(y)$ the usual characteristic function of a set $E$:

$$\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx= \int_0^{+\infty}\frac{\phi(x)}{n^{2/3}+n^{1/6}\phi(x)}\chi_{(0,n)}(\phi(x)) \ d\phi$$

Question 2

Is there a more slick way to evaluate the integral

$$\lim_{n\to +\infty}\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$$

I can answer your question 1. (I don't understand your question 2, in a similar vein to @joriki's comment, and I don't understand your reply to that comment. Maybe if you elaobrate further in the comment, I will understand.)

If you make the substitution $u=nx$, then you would have
$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}dx=\int_{u=0}^{u=n}\frac{n^{1/3}u}{1+n^{-1/2}u^3}\frac{du}{n}=\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}du$$

Now introduce that function $\chi$. Its output is $1$ if the input is in $[0,n]$, and $0$ otherwise.

$$\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$

That's fine because we just multiplied by $1$. But also we can extend the upper limit of the integral to $\infty$ because over $(n,\infty)$ this would be multiplying by $0$, and therefore not changing the value of the integral.

$$\int_{u=0}^{u=\infty}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$

The posted version uses $\phi(x)$ in place of $u$, which is fine. Except I wonder if you left out an exponent of $3$ in the denominator when typing up the question.

As you've noted in the comments, this can be easily solved with DCT. However, to follow what you've tried to gone through with, we could substitute $u=nx$ to get $x=u/n$ and $\mathrm dx=\mathrm du/n$, which give

$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}~\mathrm dx=\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n$$

Expanding out the denominator:

$$1+n^{5/2}(n^{-1}u)^3=1+n^{5/2}n^{-3}u^3=1+n^{-1/2}u^3$$

and the numerator:

$$n^{4/3}n^{-1}u=n^{1/3}u$$

which, put together and dividing by $n$ becomes

$$\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n=\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du$$

or equivalently,

$$\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du=\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du$$

We can factor $n^{1/6}$ out from the denominator and bound the integral by

$$\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du\le\int_0^\infty\frac{n^{-1/6}u}{1+u^3}~\mathrm du$$

where $\displaystyle\int_0^\infty\frac{u~\mathrm du}{1+u^3}$ converges by seeing it is bounded on $[0,1]$ and comparing the rest to $\displaystyle\int_1^\infty\frac{u~\mathrm du}{u^3}$.


Give examples of the following rings (concerning identities) Disclaimer: I do not assume rings to be unital (otherwise the question wouldn't make sense).

  Give examples of a ring $R$ with subring $S$ such that

  (1) $R$ has an identity but $S$ has not.

  (2) $S$ has an identity but $R$ has not.

Attempt:

(1) Take $R = \mathbb{Z}$ and $S = 2 \mathbb{Z} = (2)$

Suppose that $S$ has an identity $e$, then we must have (in $S$)

$$2e = 2$$

This is also true in $R$, and thus we can conclude $e = 1$. But $1 \notin S$. Contradiction. Thus $S$ has no idenity.

(2) Take $R$ the compactly supported functions $\mathbb{R} \to \mathbb{R}$. This is easily seen to be a subring of the set of all functions $\mathbb{R} \to \mathbb{R}$ (and in fact thus also gives an example for (1), as this ring has  identity $f = 1$). Suppose $R$ has an identity $g$. Take a function $l$ that is non-zero everywhere. Then we must have

$$lg = l \implies l(g-1) = 0 \implies \forall x \in \mathbb{R}: g(x) = 1 \implies g = 1$$

But $g \notin R$. Thus $R$ has no identity.

Define a function $k$ as the indicatorfunction $k= I_{[0,1]}$ which is $1$ on $[0,1]$ and $0$ elsewhere. This function is compactly supported and an idempotent. Hence, the ideal $(k) = kR$ is a subring with identity $k$ (since $(kr)k = k^2r = kr$).

Observation: If $R$ has an identity that also lives in $S$, this is also the identity for $S$. Thus the only way $S$ can be non-unital when $R$ is unital is that the idenity of $R$ does not live in $S$.

Is this correct?

Yes, both examples and the observation are correct.

Note for (2), that any idempotent element $e$ (even in a nonnecessarily commutative ring) gives rise to a subring with identity, namely $eRe$.

I think these examples are sufficient.

R has an identity but S has not.
R={0,1,2,3} under addition and multiplication modulo 4.
S={0,2} under the same operations
S  has an identity but R has not.
R={0,2,4,6,8,10} under addition and multiplication modulo 12.
S={0,4,8} under the same operations.(S is even a field)


Radius of convergence and analyticity If the radius of convergence of a function $f(z)$ is $\infty$, is it always true that $f(z)$ is entire?

I always thought that was obvious since any analytic function can be represented with a Taylor series and the radius of convergence being infinite should mean that it is entire. But I've come across this function that is a bit tricky:

$$
 f(z) = \frac{e^{z^2}-1}{z^2}
$$
$$
 \frac{e^{z^2}-1}{z^2}=1+\frac{1}{2!}z^2+\frac{1}{3!}z^4+\frac{1}{4!}z^6+\cdots
 =\sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!}
$$

The ratio test tells me that $f(z)$ has the radius of convergence infinity.

However, $f(z)$ doesn't seem to be defined when $z=0$, although according to the series representation $f(0)=1$.

How can this be explained?

A power series has a radius of convergence, not a function.

In your case, $f(z) = \frac{e^{z^2}-1}{z^2}$ is defined on $\Bbb C \setminus \{ 0 \}$, and
$$
f(z) = \sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!} \quad \text{for all }
z \in \Bbb C \setminus \{ 0 \}
$$
The radius of convergence of the power series on the right-hand side is $\infty$,
therefore
$$
g(z) := \sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!} \quad (z \in \Bbb C)
$$
is an entire function.

So $f$ is the restriction of an entire function to $\Bbb C \setminus \{ 0 \}$. In other words, $f$ has a removable singularity at $z=0$.
This would also follow directly from the existence of the limit
$$
\lim_{z \to 0} f(z) = 1
$$
due to
Riemann's theorem.

You can extend the definition of $f$ to $\Bbb C$ by setting $f(0) := 1$. Strictly speaking, this is a different function (because it is
defined on a different domain). But loosely speaking, the extended
function is sometimes called $f$ again, and this would be an entire
function.

You need to check the convergence with ratio test
$$\require{cancel}L=\lim\limits_{n \to \infty} \left|{\frac{f(n+1)}{f(n)}}\right|$$
That results in
$$L=\lim\limits_{n \to \infty} \left|{{z^{{2n}+2}\over (n+2)!}\over {z^{2n}\over {(n+1)!}}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2+\cancel{2n}}\over (n+2)\cdot\cancel{(n+1)!}}\over {z^\cancel{2n}\over \cancel{(n+1)!}}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2}\over (n+2)}\over {1\over 1}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2}\over (n+2)}\over {1\over 1}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{z^2\over (n+2)}\right|$$
$$L=|z^2|\lim\limits_{n \to \infty} \left|\cancelto{0}{1\over (n+2)}\right|$$
$$L=0$$
So for convergence, $L < 1$. You got $0$ so lets check

$$0 < 1$$
$0$ will always be less than $1$, so you can say that is absolute convergence, that result in convergence and is for all $Z$ values


groupoids and localization in categories I've been learning homology (and hence Category Theory) and have been interested in groupoids for some time.  I've gotten to localization in categories, and that has sparked the following question:  Is the localization of a category a groupoid?  So going further, isn't localization a functor from Cat (the category of categories) to groupoids (which can be seen as a sub-category of Cat)?

No. When you localize you're inverting some of the morphisms, not all of the morphisms. There is a special localization where you try to invert all of the morphisms; this is (up to size issues) the left adjoint to the inclusion from groupoids into categories.

Let C be a category, and W a class of maps of C which satisfies the 2-3 property. This means that if two of the maps $f:X\to Y, g:Y\to Z, g\circ f$  is in W so is the third. Then other some circumstances, there is a notion of localization of W.

This is the purpose of the theory of closed models of Quillen: you need a class of cofibrations cof(C) and a class of fibrations fib(C) such that $(cof(C)\cap W,fib(C))$ and $(cof(C),fib(C)\cap W)$ are weak factorization systems. There exists a homotopy category which is a localization of W.

In general there are set theoretical problems which prevent to localize any class of maps in a given category.


Computing a transformation of a bivariate random variable Given the bivariate random variable (X,Y) with PDF:
$   \begin{cases}
   \frac{2}{7}(2x+5y) & 0<x<1,\ 0<y<1 \\
   0 & otherwise
   \end{cases}
    $

$ Let\ Z=\frac{Y}{X}. Compute\ the\ probability\ density\ function\ of\ Z. $
$Compute \ P(X>\frac{1}{3}|Z=z) for \ z \in(0,\infty) $

I tried computing the Jacobian matrix but am not sure what to do then.
Thanks

Using delta function approach ...

$$p(z')dz' = \delta(z'-z(x,y))p(x,y)dx dy$$

and hence

$$P(z''< z)=\int_0^z{p(z')}dz'$$


Calculating an Angle from $2$ points in space Given two points $p_1$ , $p_2$ around the origin $(0,0)$ in $2D$ space, how would you calculate the angle from $p_1$ to $p_2$?

How would this change in $3D$ space?

Assuming this is relative to the origin (as John pointed out): Given two position vectors $\vec p_1$ and $\vec p_2$, their dot product is:

$$\vec p_1\cdot \vec p_2 = |\vec p_1| \cdot |\vec p_2| \cdot \cos \theta$$

Solving for $\theta$, we get:

$$\theta = \arccos\left(\frac{\vec p_1 \cdot \vec p_2}{|\vec p_1| \cdot |\vec p_2|}\right)$$

In a 2D space this equals:

$$v = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$

And extended for 3D space:

$$v = \arccos\left(\frac{x_1x_2 + y_1y_2 + z_1z_2}{\sqrt{(x_1^2+y_1^2+z_1^2) \cdot (x_2^2+y_2^2+z_2^2)}}\right)$$

Let the two points be $(x_1,y_1)$ and $(x_2,y_2)$ then angle $\theta$ between them $=$
$$\theta=Tan^{-1}{\frac{y_2}{x_2}}-Tan^{-1}{\frac{y_1}{x_1}}$$


Can anyone clarify the meaning of zero content? I am having hard time understanding the definition of zero content. The following are the definitions of zero content in $\mathbb{R}$ and $\mathbb{R}^2.$

A set $Z \subset \mathbb{R}$ is said to have zero content if $\forall \epsilon > 0$ there is a collection of intervals $I_1, \ldots, I_L$ such that
(i) $Z \subset \bigcup_1^L I_l,$ and
(ii) the sum of the lengths of the $I_l$ is less than $\epsilon.$
A set $Z \subset \mathbb{R}^2$ is said to have zero content if $\forall \epsilon > 0$ there is a finite collection of rectangles $R_i$ such that
(i) $Z \subset \bigcup_1^M R_i$ and
(ii) the sum of areas of the $R_i$ is less than $\epsilon.$

Thanks!

Update: the answer below is not correct, see comments.

Essentially, that means that Lebesgue measure of $Z$ (length in $\Bbb R$, area in $\Bbb R^2$ etc.) is zero. Measures are monotone, i.e. $\lambda(Z)\leq \lambda (\cup I_l)$ since $Z\subseteq \cup I_l$. Measures are additive, that is $\lambda (\cup I_l) = \sum \lambda(I_l) \leq \epsilon$. Hence, you get $\lambda (Z)\leq \epsilon$ for all $\epsilon>0$ and so $\lambda(Z) = 0$.

Measure theory is a bit abstract, and formal definition of $\lambda$ is rather technical. Yet, in early applications in analysis you often just need to know that something has measure of $0$, and you don't care much what is the measure whenever it is greater than $0$ - for example, you want boundaries of nice sets to have measure of $0$. For this reason, instead of giving the whole definition of measures, you simply define what does it mean to be of zero measure, or of zero content in your terminology.


What's the difference between tuples and sequences? Both are ordered collections that can have repeated elements. Is there a difference? Are there other terms that are used for similar concepts, and how are these terms different?

A sequence requires each element to be of the same type.
A tuple can have elements with different types.

You can compare it with arrays and structs in C (respectively).

For example, let's consider the DFA $(Q,F,s,\delta,\Sigma)$ quintuple. It's a tuple with the full set of states $Q$; a set of final states $F$; a starting state $s$; the state transition function $\delta$ and the alphabet $\Sigma$.

Sequences' elements are indexed with natural numbers N.

Tuples' elements are indexed with positive integers N*.

Sequence is use to define relations in general
(i.e. where the cardinality of arbitrary sets is not clearly identified).

Tuple is use to define finitary relations in particular
(i.e. where the cardinality of finite sets can be directly identified).

"While the cardinality of a finite set is just the number of its elements, extending the notion to infinite sets usually starts with defining the notion of comparison of arbitrary (in particular infinite) sets." From Wikipedia, the free encyclopedia.


Why is the order of $\mathbb{Z}[i]/\langle 3+i\rangle$ equal to 10? The reasoning my book uses is as follows:

Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.

I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.

Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.

You might find clearer the following way of viewing the proof. Write $\,A = \Bbb Z[i]/(3+i)$.
Note $\, h\!: \Bbb Z \to A\,\color{#0a0}{ \ {\rm is\ surjective\  (onto)}}\,$ by $\bmod\, 3\!+\!i\!:\ \, i\equiv -3\,\Rightarrow\, a\!+\!bi\equiv a\!-\!3b\in\Bbb Z$
$\color{#c00}{I := \ker h = 10\,\Bbb Z}\ $ follows immediately by means of $\,\rm\color{#90F}{rationalizing}\,$ a denominator
$$ n\in I\iff 3\!+\!i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \dfrac{n}{3\!+\!i}\in \Bbb Z[i]\!\!\color{#90f}{\overset{\large \rm\ rat}\iff}\! \dfrac{n(3\!-\!i)}{10}\in\Bbb Z[i]\iff \color{#c00}{10\mid n}\ $$
Thus $\, \color{#0a0}{A = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{10\,\Bbb Z}\ $ by the First Isomorphism Theorem.
Remark $ $ Another view you might find illuminating arises from rewriting the ideal as a module in Hermite normal form: $\ I =
 (3\!+\!i) = (10,3\!+\!i) = 10\Bbb Z + (3\!+\!i)\Bbb Z.\,$ But it is trivial to test module membership given such a triangularized basis, namely
$$\begin{align} a\!+\!bi = a\!-\!3b +b(3\!+\!i)&\in I = 10\Bbb Z + (3\!+\!i)\Bbb Z\\
\iff\ a\!-\!3b&\in I\\
\iff\ a\!-\!3b &\in 10\Bbb Z \iff  10\mid a\!-\!3b
\end{align}$$
Further this shows that $\, a\!+\!bi\bmod I\, =\, a\!-\!3b\bmod 10.\ $
The criterion generalizes to an ideal test for modules $\rm\,[a,b\!+\!c\:\!\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 Franz Lemmermeyer's notes linked here..
This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.

Prop 1) $10 \equiv 0 + <3+i>$.

Pf:  $a \equiv b + <3+i>$ means there is a $k \in \mathbb Z[i]$ so that $a = b + k(3+i)$.

$10 = 0 + (3-i)(3+i)$ so $10 \equiv 0 + <3_i>$.

Prop 2) $i \equiv -3 + <3+i>$

Pf:  $i = -3 + (3+i)$ so $i \equiv -3 +<3+i>$.

Prop 3) If $a = j + mi; j,m \in \mathbb Z$ Then $a \equiv j - 3m + <3+i>$.

proof: $j+mi = j - 3m + m(3 + i)$.

Prop 4) If $a = j+ mi$ then $a \equiv r + <3+i>$ for some $0 \le r < 10$.

Pf: We can find some $v$ so that $j-3m = 10*v + r$ where $0 \le r < 10$.  That is basic division.

So  $a \equiv j - 3m  \equiv r + 10v + <3 + i>\equiv r + <3+i>$.

ANd that's more or less it.  Any element of $\mathbb Z[i]$ is equivalent to an integer between 0 and 9.  So the set of equivalence classe $\mathbb Z[i]/<3 + i> = \{k + <3+i>| k = 0...9\}$

The only thing to show is if $j,k \in \{0...9\}$ and $j\ne k$ then $j \not \equiv k + < 3+i>$.


How many values of c in the equation $x^2 - 5x + c$ result in rational roots which are integers? This question was asked in the e-litmus exam.

Hint: $(x-a)(x-b) = x^2 - (a+b) x + ab$.  How many integer solutions does $a+b=5$ have?

Answer should be $6$.

In order for $x$ to be an integer, numerator must be even which would be divisible by $2$.
 $d= \sqrt{25-4c}$

$x=(5\pm\sqrt{25-4c}) /2$

Now solve inside the brackets first such that d becomes odd so that the numerator $5\pm$(any odd integer) is divisible by $2$ and and hence $x$ is an integer.

The values of $c$ could be $24/5,-24/5,16/5,-16/5,0$ (though even)


Newtonian Mechanics - Differential equation If we combine Newton's second law of motion i.e. $F=m\ddot{x}$ and Newton's law of gravity i.e.,
$$
F=G\frac{mM}{x^2},
$$
where $x$ is distance, we obtain the following equation:

$$\ddot{x}=MG\frac{1}{x^2}$$ My question is how do we solve this differential equation for $x$ ? I see no ordinary way of doing it. I have tried using Laplace transformation but apparently there exists no transformation for $\frac{1}{x^2}$ so I'm left with nothing to use.

Hint:  Set $p(x)=x'(t)$ so you'll have $x''(t)=p(t)p'(t)$ so $$x''=\frac{MC}{x^2}\to pp'=\frac{MC}{x^2}$$ and so $$\int pdp=\int \frac{MC}{x^2}dx$$

(double dot χ)*(χ^2)= MG
But (double dot χ) equals the acceleration a, which equals ΔV/t.
This implies that;
(χ squered)= MGt/ΔV.
Hence
χ= Squere root(MGt/ΔV).
I hope this is helpful, as it is simple since it does not involve the trouble of integration constants.


Determine the convergence of the following series $\sum_{n=2}^{\infty} (-1)^n \frac{1}{\sqrt{n} + (-1)^n}$ $\sum_{n=2}^{\infty} (-1)^n \frac{1}{\sqrt{n} + (-1)^n}$

Now i know that this is alternating series which means that i should determine the  absolute convergence $|a_n|=\frac{1}{\sqrt{n} + (-1)^n}$ but i don't know how to do it, d'Alembert's test isn't working Cauchy tests can't help here here either so i am out of ideas now.

$$\begin{align}
\frac{(-1)^n}{\sqrt{n}+(-1)^n}&=\frac{(-1)^n}{\sqrt{n}}+\Bigl(\frac{(-1)^n}{\sqrt{n}+(-1)^n}-\frac{(-1)^n}{\sqrt{n}}\Bigr)\\
&=\frac{(-1)^n}{\sqrt{n}}-\frac{1}{\sqrt{n}\,\bigl(\sqrt{n}+(-1)^n\bigr)}.
\end{align}$$
Now:

$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt{n}}$ converges by Leibniz's criterion.
$\displaystyle\sum_{n=2}^\infty\frac{1}{\sqrt{n}(\sqrt{n}+(-1)^n)}$ diverges.

The original series diverges.

Another way of seeing it is to observe that the sum of two consecutive terms behaves like $1/n$, so that the even partial sums do not converge.

Compare the first two terms, $ \frac{1}{\sqrt{2} +1}, \frac{1}{\sqrt{3}-1}$: the sequence is not monotone decreasing, so the alternating series test does not apply


Principale value, how can we consider it? How can we consider principal value where as, it's something that "doesn't exist". For example,
$$\int_{-\infty }^\infty \frac{1}{x}\mathrm d x$$
doesn't exist, but the principal value is nulle. What is the motivation between the principale value ?

Let $f:\mathbb R^2\longrightarrow \mathbb R$ a function. You can have that $$\lim_{(x,y)\to (0,0)}f(x,y)$$
doesn't exist, but that $$\lim_{x\to 0}f(x,g(x))$$
exist for some function $g:\mathbb R\to \mathbb R$ such that $$g(x)\underset{x\to 0}{\longrightarrow }0.$$

Now, the (Cauchy) principale value is the same idea. Indeed, let says that $x<0$ and $y>0$ and $$f(x,y)=\int_{-\infty }^x g(t)\mathrm d t+\int_{y}^\infty g(t)\mathrm d t.$$
Then $$\lim_{(x,y)\to 0}f(x,y)$$
may not exist, but we can get interested to $$\lim_{x\to 0}f(-x,x)$$
that is well defined if it exist. In other word, you define a function $$h(x)=f(-x,x),$$
and compute $$\lim_{h\to 0}h(x).$$

Sice $\frac{1}{x}$ is an odd function, the value of the integral must be zero. An for calculation you can calculate it in this way
$$I=\lim_{a\to \infty}\int _{-a}^{a} \frac{1}{x}dx=0$$

Hope this will be helpful !


I need help with this graph proof please Let $G$ be a self-complementary graph in where $n = 4k+1$ . Prove that there is an odd number of vertices of degree $(n-1)/2$.

I dont even know where to start with this. I really need help.

Thanks!

As it is isomorphic to its complement we can say that

$$\sum_{i=1}^n deg(v_i)=\sum_{i=1}^n 4k-deg(v_i) $$

Then we have $$2\sum_{i=1}^n deg(v_i)=\sum_{i=1}^n 4k=4k(4k+1) $$

$$\sum_{i=1}^n deg(v_i)=2k(4k+1)$$

Notice that if $r=deg(v_i)$ then $4k-r$ is also a degree of another vertex. And if $r\neq 2k$ ,it is clear that it is another vertex. Set $s$ be number of vertex with degree $2k$ and $l$ be the number of vertex with degree less than $2k$.

$$s2k+l(4k-r+r)=2k(4k+1)$$
$$s+2l=4k+1$$
$$s\equiv1 \text{ } mod(2)$$

Hint: $v$ has degree $\frac{n-1}{2}$ in $G$ if and only if $v$ has degree $\frac{n-1}{2}$ in its complement...

Can you  pair the vertices which don't have degree $\frac{n-1}{2}$?


What does it mean for a function to be a solution of a differential equation? What does it mean for a function to be a solution of a differential equation?  I think I understand, but I still don't have a simple or intuitive understanding.

Solving $x+1=3$ means finding a value for $x$ that satisfies the equation $x+1=3$. In this case, $x=2$ does the job because $2+1=3$.

Now, given a differential equation such as:

$\frac{dy}{dx}= 2x$            ..........(1)

a solution to this equation is a function (call it $y(x)$). Of course not any function will do. The correct function must satisfy the equation we have in (1) above.

for this particular case, y(x) can be equal to $x^2$.

Why? Because the derivative of $x^2$ with respect to $x$ is $\frac{dy}{dx}= 2x$ which is what we have in our equation.

But wait, what about $y(x)=x^2$+5?

In fact, this is another solution. As you can see, in this case we end up with many solutions all of the form:

$y(x)=x^2+k$

where k is a constant. This is because, all of such functions satisfy our differential equation.

When you write an algebraic equation you are trying to generalise a calculation for say. In a way you are saying what will be the end result if you give certain input or vice-versa.

Similarly when you write a function you are generalising the notion of the end result itself. After defining a function you can opt for any result your wish and find the input required to get that desired result.

In the same way when you define a differential equation you are actually trying to emphasis that in what way the rate of change of a function and its value will merge together to give the result. This kind of approach is useful in study of complex systems where the rate of changes can be measured with time and which helps in understanding the nature of system mathematically.


Noetherian, Artinian on graded ring and localization There's an exercise in Tom Marley's Graded Rings and Modules making me confused, stating

  $R$ is a nonnegatively graded local ring with $R_0$ being local. Let $M$ be the unique homogeneous maximal ideal Prove $R_M$ Artinian (Noetherian) implies $R$ Artinian (Noetherian).

This may be an easy question. But I can't solve it since I still don't see the connection between localization and Artinity of $R$. Any ideas

For Artinian case, It is result of Exercises 4.7, and 2.10:
 Take a descending chain of homogeneous ideals in $R$ and localize at the homogeneous maximal ideal, $M$. Now use 2.10.

For Noetherian case replace 4.7 with 4.3


Calculate the expansion of  $(x+y+z)^n$ The question that I have to solve is an answer on the question "How many terms are in the expansion?".

Depending on how you define "term" you can become two different formulas to calculate the terms in the expansion of $(x+y+z)^n$.

Working with binomial coefficients I found that the general relation is $\binom{n+2}{n}$. However I'm having some difficulty providing proof for my statement.

The other way of seeing "term" is just simply as the amount of combinations you can take out of $(x+y+z)^n$ which would result into $3^n$.

Depending on what is the right interpretation, how can I provide proof for it?

For the non-trivial interpretation, you're looking for non-negative solutions of $a + b + c = n$ (each of these corresponds to a term $x^a y^b z^c$). Code each of these solutions as $1^a 0 1^b 0 1^c$, for example $(2,3,5)$ would be coded as $$110111011111.$$ Now it should be easy to see why the answer is $\binom{n+2}{n}$.

It can never be $3^n$. As you can see for $(a+b)^n$ contains just $n+1$ terms. Note that we have to keep the sum of powers in each of the combinations of $x,y,z$ to $n$, so it will be reduced.

Now replace $a$ and $b$ by $x$ and $(y+z)$ respectively. So total number of terms should be $1+2+3+\cdots+(n+1)=\dfrac{(n+1)(n+2)}{2}$.


If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$ I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?

Hint$\;$ from my $ $ sci.math post $ $ 2006/4/4 $ $ here or here (see there for much motivation)

$$\rm\quad\begin{eqnarray}
  p-1 \!\!\! &&\mid\rm \color{#c00}{n-1} \\
 \rm p\!\!\! &&\mid\rm \color{#0a0}{a+n} && \\
\end{eqnarray}\! \Rightarrow\,\
p\mid \overbrace{a^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!a+\color{#0a0}{a\!+\!n}}^{\Large a^n+n}\mid \overbrace{b^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!b+\color{blue}{b\!-\!a}+\color{#0a0}{a\!+\!n}}^{\Large b^n +n} \ \,\Rightarrow\,\ p\mid \color{blue}{b-a}\qquad\qquad\quad  $$

UPDATE: the proof seems incorrect as b=4, a=2, fail when n=4 (check the comments below).

First, there is a counter example when n=1, where a=6 and b=13 (as $ (6+1) | (13+1) $ ). However I shall continue for all n>1.

If $(a^n+n) | (b^n+n)$ then there is a natural number k $\geq 1$ such that $b^n+n=k(a^n+n)$. Clearly $a=b \mbox{ iff } k=1$. We shall proceed to prove that k is always 1 for all n>1.

Continuing from $b^n+n=k(a^n+n)$,

$b^n+n \equiv 0 \quad (\mbox{mod } k)$

If we allow n to vary for the same a and b, then the only value for k that will make that equation hold for all n>1 is k=1, which implies that a=b.


What are the invariant subspaces I'm trying to find all subspaces of $\Bbb F^n$ (where $\Bbb F$ can be $\Bbb R$ or $\Bbb C$) invariant under $\Phi$ where $\Phi:\Bbb F^n\to \Bbb F^n$ is given by

$$\Phi(x_1,\dots, x_n) = (x_1,2x_2, \dots, nx_n)$$

I've already figured out that eigenspaces have the form $\operatorname{span}(0,\dots, 0,1,0,\dots, 0)$ where the $1$ is in the $i$th position for all $i\in \{1,\dots, n\}$.  So these are all the $1$-dimensional invariant subspaces.  Beyond these and the two trivial invariant subspaces, every sum of the $1$-dimensional subspaces should be an invariant subspace as well. For instance, the subspace $$\operatorname{span}(1,0,\dots, 0) + \operatorname{span}(0,1,0,\dots, 0) = \text{the $xy$-plane}$$ should be invariant as well.

So can I conclude that every subspace of $\Bbb F^n$ is an invariant subspace under $\Phi$?  Or are there subspaces of $\Bbb F^n$ that can't be written as a sum of the spans of the standard basis vectors?  And if I am forgetting some (as I suspect I am), how can be either be sure I've gotten all of the invariant subspaces or find the ones I've missed?

Thanks.

Edit: As I learned from Mark in the comments, clearly I can't conclude that every subspace of $\Bbb F^n$ is an invariant subspace under $\Phi$.  So how can I either rule out any subspaces that aren't a sum of spans of the basis vectors or find the subspaces that I'm missing so far?

Edit 2: I have only just gotten to eigenspaces/ invariant subspaces.  I don't know what a minimal polynomial is or what a diagonalisable map is.

Suppose $U\leqslant \mathbb{F}^n$ is $\Phi$-invariant, and consider the restriction of $\Phi$ to $U$, denoted say by $\Psi$. Then the minimal polynomial of $\Psi$ divides the minimal polynomial of $\Phi$; hence is also a product of distinct linear factors, so that $\Psi$ is diagonalisable. That is, $U$ is the direct sum of eigenspaces of $\Psi$; but each eigenspace of $\Psi$ is an eigenspace of $\Phi$. Hence $U$ is one of the $2^n$ possible sums of the $n$ eigenspaces of $\Phi$.

You've already figured out all the subspace spanned by the eigenvectors are invariant.  And go one step further, take any two eigenvectors, the subspace spanned by any two eigenvectors are also invariant.  And in general, the subspace spanned by any number of eigenvectors are invariant.


Show that every Cauchy sequence can have at most one limit point. Show that every Cauchy sequence can have at most one limit point.

My solution

If a Cauchy sequence $x_{n}$ admits a limit point $L$, then there is a subsequence of $x_{n}$ which converges to $L$. Therefore $x_{n}$ converges to $L$. Since the limit of a sequence is unique in a metric space, the result holds.

Can someone check if is it right? An answer based on $\varepsilon-\delta$ definitions would be nice.

Edit 1

My answer is based on the following result:

  Let $(X,d)$ be a metric space and $x_{n}$ is a Cauchy sequence in it. Then, if $x_{n}$ admits a subsequence which converges to $x_{0}$, then the Cauchy sequence is also convergent and $x_{n}\rightarrow x_{0}$.

Edit 2

Since my question has caused some confusion, I am going to explicit what I was looking for.

The proposition "if $L$ is a limit point of the sequence $x_{n}$, then there is a subsequence of $x_{n}$ which converges to $L$" has been proved earlier as well as the other two results involved in the solution.

Maybe I should have asked for an alternative solution involving $\varepsilon-\delta$, but as it has been pointed out it does not make sense since they are already involved in the assumptions used to solve the problem.

Anyway, thank you guys for the feedback.

Assume there are two subsequence $ x_{\phi(n)} $ and $ w_{\psi(n)} $ such that

$$x_{\phi(n)}\to L_1$$
and
$$x_{\psi(n)}\to L_2$$

Given $ \epsilon>0$,

we know that
$$\phi(n)\ge n ; \psi(n)\ge n$$

then there exists $N_1, N_2,N_3 \ge 0$ such that
$$n\ge \max(N_1,N_2,N_3) \implies$$

$$|L_1-L_2|=|L_1-x_{\phi(n)} +x_{\phi(n)}-x_{\psi(n)}+x_{\psi(n)}-L_2|\le \frac \epsilon 3+\frac \epsilon 3 +\frac \epsilon 3$$

We conclude that $$L_1=L_2$$

A Cauchy sequence is a convergent sequence so it has exactly 1 limit point, not at most 1.

Edit: If the sequence comes from real numbers.


Proving that a set is closed with respect to subtraction I have read that an equivalent definition for a subring is the following:

  B is a subring of A if and only if B is closed with respect to subtraction and multiplication

If I was trying to prove that a set is closed with respect to subtraction, would I just need to prove that for $a, b \in B$ that $a-b\in B$ or would I also need to prove that $b-a\in B$?

They are equivalent.

As $(a,b)$ ranges over all of $B^2$, $(b,a)$ also ranges over all of $B^2$.  So, by the time you have checked every $a-b$ you have also checked every $b-a$.

Proving that for any $a,b\in B$ we have that $a-b\in B$ would be sufficient. The key word here is "ANY".

Once you prove this property in general, and assuming this subset $B$ is nonempty, you can take any $a\in B$ and apply this property to $a$ and $a$, thus showing that $0=a-a\in B$. Then you can can apply it to $0$ and $a$ to show that $-a=0-a\in B$. This is why proving only one direction, i.e. only $a-b$, is sufficient, because containing additive inverses follows.


Finite set cardinality decreasing Say I have a finite set $A$ and know its cardinality. How can I prove that, by repeatedly applying some algorithm, which removes a known number of elements from $A$ each time, its cardinality will stop at some value? It should stop because the aforementioned algorithm requires a fixed number of elements present in the set.

I tried induction, but couldn't figure out a condition over $n \in \mathbb{N}$ that resembled this problem. I also thought about limits of sequences, but this would be a finite sequence.

Any hint is appreciated! Thanks in advance.

Try something like this: given a finite set $A$ of cardinality $c$, find an expression $f(n)$ in terms of $n$ (and involving $c$) such that if the algorithm is iterated $n$ times successfully (i.e. without terminating) starting with $A$ as input, the resulting finite set has cardinality at most $f(n)$. This is something you could prove by induction on $n$. Now if the function $f(n)$ is chosen well, you can show that it will be be negative for sufficiently large $n$, which would imply that for such large $n$, the algorithm cannot be iterated $n$ times (since it couldn't possible produce a set of negative cardinality!) That is, it will terminate before then.

Sets have finite cardinality if and only if there is a bijection between them and the set $\{0,\ldots,n-1\}$ for some $n\in\mathbb N$.

That is, the cardinality of a finite set is some natural number $n\in\mathbb N$.

If your algorithm removes finitely many elements at each stage, the cardinality cannot go up. It can stay the same or become smaller.

Now we can use a fact (or prove it, if you have not seen it before) that every non-increasing sequence of natural numbers stabilizes after finitely many steps. (If not, then at some finite point we have to go below $0$ which is impossible)

Also note that while "removing" seems to require existing, this is not true. Mathematically we can remove one apple from a bag of oranges. This may seem strange, but recall: $$A\setminus B=\{x\in A\mid x\notin B\}$$ Thus if your algorithm removes 10 elements each round, even if the set has 9 elements we can still remove 10. We are simply left with an empty set.


Show that if $m$ is prime then the product of two non-zero elements of $\mathbb Z/m\mathbb Z$ is again non-zero I am having some trouble concluding this.

I have the following:

$[a][b]=[ab]$ by the definition of multiplication in $\mathbb Z/m\mathbb Z$

By the remainder theorem, we know that $a=qm+r$ and $b=km+s$ where $q,k,r,s\  \epsilon\  \mathbb Z$

Additionally,  $0\lt r\lt m$ and  $0\lt s\lt m$

*It is greater than (not greater than or equal to) because at the onset of this problem, we have defined $[a]$ and $[b]$ to be non-zero

So, with those stated:

$[a][b]=[qm+r][km+s]=[(qm+r)(km+s)]$

which is equal to $[m^2qk+qms+mkr+rs]$

which is simply equal to $[rs]$

This is where I run into trouble. I am aware that I need to show that $[rs]$ is not equal to zero (i.e. $r*s$ should not be a divisor of $m$ or some multiple of $m$...say $\alpha m$.)

I thought that I could maybe setup a contradiction by stating something like:

"Assume $r*s=\alpha* m$" and then go on to show this can only be the case if $r$ or $s$ is not coprime with m (we know that $r$ and $s$ must be coprime with $m$ if $m$ is prime because $r$ and $s$ are both less than $m$ and greater than $0$...which would be a contradiction!)

Unfortunately, I cannot figure out how to explicitly demonstrate this. Any help would be appreciated!

edit: Using the recommended Euclid's Lemma, while retaining the structure of my argument, we have the following:

Assume $r*s=\alpha* m$

This means that $m|r*s$. Recall, $m$ is prime and therefore, by Euclid's Lemma, we have either $m|r$ or $m|s$. This produces a contradiction for the following reason:

$r$ and $s$ have both been defined to be between $0$ and $m$. A number cannot evenly divide a number that is smaller than itself but greater than $0$.

Therefore, $rs\neq \alpha*m$ which means that $[rs] \neq[0]$ (i.e. it is non-zero).

If $[rs]$ is zero then $m \mid rs$ but $m$ is prime so $m \mid r$ or $m \mid s$ (no loss of generality) let's assume that $m\mid r$ then $m \mid a = qm +r$, i. e., $[a]$ is zero, wich is a contradition.

The answer is much easier.

Take $~x~$ a non multiple of $~m~$. Then the additive cyclic group generated by $~[x]~$ is cyclic of a order dividing $~m~,~$ thus $~m~$ since $~m~$ is prime.

Therefore $~n[x]=[nx]=0=[m]~$ if and only if $~n~$ multiple of $~m~$.


Proof of $\lim_{x\to b}\int_a^x| f(t)|dt$ exists $\implies$ $ \lim_{x\to b}\int_a^x f(t)dt$ exists. Let $f:[a,b[\to \mathbb R$ continuous and integrable on $[a,b[$. I want to prove that $\displaystyle\lim_{x\to b}\int_a^x |f(t)|dt$ exists $\implies$ $\displaystyle\lim_{x\to b}\int_a^x f(t)dt$ exists. I set $F(x)=\int_a^x f(t)dt.$ In my course they proved that $F$ is uniformly continuous on $[a,b[$ and thus $\lim_{x\to b}F(x)$ exist.

Can I do as follow : I set $f^+(x)=\max\{f(x),0\}$ and $f^{-}(x)=-\min\{f(x),0\}$. Since $f^{+}$ and $f^-$ are continuous  $$0\leq  f^+\leq |f|\quad \text{and}\quad 0\leq f^-\leq |f|,$$
and $\int_a^b |f|$ exists, we have that $$\lim_{x\to b}\int_a^x f^+\quad \text{and}\quad \lim_{x\to b}\int_a^x f^-$$
exists. Therefore, $$\lim_{x\to b}\int_a^x f=\lim_{x\to b}\int_a^x f^+-\lim_{x\to b}\int_a^x f^-,$$
and thus it exists.

Does it work ?

It is enough to show that for every Cauchy sequence $x_n$, $\int_{a}^{x_n}f(t)dt$ is a Cauchy sequence. This results from the fact that $|\int_{x_n}^{x_m}f(t)dt|\leq \int_{x_n}^{x_m}|f(t)|dt$.


Rotating two vectors to point in the same direction I have two vectors, $v$ and $u$.  How do I rotate $u$ around the x-, y-, and z-axes (or one axis) so that it points in the same direction as $v$?

Two 3-vectors define a plane. Rotation in that plane (i.e. about normal vector of that plane) brings one vector to another. So the quick sketch for the solution would be:

find normal vector to the common plane (I think this is just the vector product $u \times v$)
find rotation angle $\theta$ using dot product ($\theta = \cos^{-1}(\frac{u \cdot v}{||u|| ||v||})$)
express the rotation using some axis-angle representation (axis is the normal vector from 1. and angle is the $\theta$ from 2.)

I am curious if someone expresses the straightforward formula here...

If $u,v\in\mathbb R^2$, Find the angle between $u$ and $v$ by $$\cos\theta=\frac{<u,v>}{|u|.|v|}$$
Now take matrix of rotation $A_\theta$ of angle $\theta$. Now take $A_\theta u$ or $A_{-\theta} u$. These will rotate $u$ to the direction of $v$.
$$A_\theta= \left(
  \begin{array}{cc}
  \cos\theta   &-\sin\theta   \\
    \sin\theta &\cos\theta   \\
  \end{array}
\right) $$

For $u,v\in \mathbb R^3$,
Write $v= (a,b,c)$ and $u= (x,y,z)$. If all x,y,z is non zero, then we want $T$ such that $Tu = v$ , define $T$ such that $$T(e_1)= \frac{a}{x} e_1$$
$$T(e_2)=\frac{b}{y} e_2$$
$$T(e_3)= \frac{c}{z}e_3$$  Then we have $T(x,y,z)= (a,b,c)$.  That is rotation matrix is matrix of $T$ that is $$\left(
  \begin{array}{ccc}
   \frac{a}{x } &0  & 0 \\
    0 &\frac{b}{y}  &0  \\
    0 & 0 &\frac{c}{z}  \\
  \end{array}
\right)$$

If some of x,y,z is zero, then case reduces to $\mathbb R^2$ keeping one axis fixed.

Example:   Assume $u= (1,2,3)$ and $v= (2,-5,6)$  Then rotation matrix which take $u$ to $v$ is $\left(
  \begin{array}{ccc}
   \frac{2}{1} &0  & 0 \\
    0 &\frac{-5}{2}  &0 \\
    0 & 0 &\frac{6}{3}\\
  \end{array}
\right)= $
$\left(
  \begin{array}{ccc}
   2 &0  & 0 \\
    0 &\frac{-5}{2}  &0  \\
    0 & 0 &2  \\
  \end{array}
\right)$

putting value of $a,b,c$ and $x,y,z $ you may have many more example.... If some of $x,y,z$ is what happen see and if didn't get comment it.. i will give example for that too.


Reflecting a golfball off a wall to a hole and compensating for the balls radius Problem: I'm struggling to compensate for the radius of a ball when reflecting it off a wall towards a target. (sorry I cannot yet post images)

What I want is to do this:
golf reflections

but this does not compensate for the size of the ball. Any guidance would be helpful.

Would love to know how to compensate for this. Thanks!

Offset the wall by the radius of the ball. By doing this, you compensate for the radius of the ball, so thereafter, you can consider the ball to be a point, and do the computations shown in your reference.

The point that is reflecting off a wall according to those rules is not the center of the ball but the point on the ball where it touches the wall. Thats how you account for it. If you have a rectangle, you only need to use 4 points on the ball plus its center.


Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$. Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$

Hint $1$ :

Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes:
$$y^2-3ty+(3t)^2-(t+1)^3=0 $$

a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square.

Hint $2$: $\Delta_y=(t-2)^2(4t+1)$

Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$

And it's your turn to do some work!

Edit

Because my answer is down-voted and I don't see the reasons, I will write a full solution based on my little hints below:

  Claim If $(x,y)$ are integer solutions to the equation  :
  $$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3\tag 1$$ then
  $(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}$

Proof

Let $(x,y)$ be a solution to the given equation, and let $t=\frac{x+y}{3}\in \Bbb Z$ so that:
$$y^2-3ty+(3t)^2-(t+1)^3=0 \tag 2$$
and this is a quadratic equation on $y$ so let's compute its discriminant:
$$\Delta_y=(3t)^2-4\left((3t)^2-(t+1)^3\right)=4(t+1)^3-27t^2=(t-2)^2(4t+1)\tag 3$$
so this equation has an integer solution $y$ only if $\Delta_y$ is a square of an integer, and this is equivalent to $4t+1$ is a square of an integer which implies that $t=a^2+a$ for some integer $a$. and because the equation $(2)$ is quadratic and we know the value of its discriminant we can find the two solutions on $y$:
$$\begin{align}y&=&\frac12\left(3t-(t-2)\sqrt{4t+1}\right)&=-a^3+3a+1\tag 4\\ y&=&\frac12\left(3t+(t-2)\sqrt{4t+1}\right)&=a^3+3a^2-1 \tag 5\end{align}$$
now that we found the value of $t$ and $y$ we can find the value of $x=3t-y$ which gives:
$$\left\{\begin{matrix}
 x=a^3+3a^2-1&\text{ and }& y=-a^3+3a+1 &\text{ or }\tag 6\\
 x=-a^3+3a+1&\text{ and } & y=a^3+3a^2-1
\end{matrix}\right.$$

Finally:
$$(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}\tag 7$$