How many trees in a forest? Some time ago I met a forester. He told that there are only larches and spruces in his forest.
He also said that there are exactly $10$ spruces at the distance of exactly 1 km from each larch. Next, he argued that  there are more larches than spruces in his forest. Can he be right?

Nice number theoretic question in disguise! Not being much of a botanist I'll
call larches X and spruces Y, or I won't be able to tell them apart. Also I
take "argued" in the question to mean "asserted", since one can obviously not
argue that there must be more larches than spruces, but there can be.

Here is a solution, or rather a method to construct a solution. I'll take the
Y trees to form a large square grid, the distances between neigbours being 2
units (the unit of length is to be determined). In those units these trees
therefore occupy the points with both coordinates even, filling a large region
around the origin (we'll see about the boundary of the forest later).

There will be an X tree at $(2a+1,2b)$ and at $(2a,2b+1)$ for any integers
$a,b$ for which this does not come too close to the limit of the Y trees.

    0 0 0 0 0 0 0 0

    0 0*0*0*0*0*0 0
      * * * * * *
    0 0*0*0*0*0*0 0
      * * * * * *
    0 0*0*0*0*0*0 0
      * * * * * *
    0 0*0*0*0*0*0 0    X=*, Y=0
      * * * * * *
    0 0*0*0*0*0*0 0
      * * * * * *
    0 0*0*0*0*0*0 0

    0 0 0 0 0 0 0 0

Within the area occupied by X trees they are twice as numerous as the Y trees,
so we can make the X trees win out overall by making the forest large enough
(the advantage for X increases with the area, while the boundary, where the X
have stopped but not yet the Y, grows only with the circumference).

Now the challenge is to find a radius such that on a circle around each X
there are exactly 10 Y trees. If this works for the X tree at $(1,0)$, it will
work by translation and diagonal reflection symmetry for all X, provided we stop planting X when coming so close to the end of the forest that the circle protrudes outside.

Now we have our number theoretic question: for which integer $N$ (the square of
the radius) are there exactly 10 integer solutions $(x,y)$ to the equation
$(2x-1)^2+(2y)^2=N$ (giving a Y tree at $(2x,2y)$, at distance $\sqrt{N}$ of $(1,0)$)?
Obviously $N$ must be odd, and it is easy to see that dropping the parity
condition on the two squares summing to $N$ will double the amount of
solutions, every (odd,even) solution giving an (even,odd) solution by
interchange. So we need to have $N$ such that $k^2+l^2=N$ has exactly 20
integer solutions $(k,l)$. I claim the smallest such $N$ is $N=5^4=625$. That
it is indeed a solution is easy to see, with 5 solutions in the positive
quadrant (taken to include the the positive x-axis but not the y-axis):
$(25,0)$, $(24,7)$, $(20,15)$, $(15,20)$ and $(7,24)$, and the remaining
solutions are obtained from these by applying quarter turns.

To explain the "smallest such $N$" part, I applied to following theorem 2.3.12 from the course on arithmetic I gave last year (in French; I have no other reference at hand for the full statement, but its first part is well known, and the rest is easy once you know enough about Gaussian integers to prove the first part):

Theorem. Let $N$ be a positive integer, and $N=p_1^{m_1}\ldots p_l^{m_l}$ its
prime factorization. A necessary and sufficient condition for the existence of
integers $a,b$ with $N=a^2+b^2$ is that for every $p_i$ that is congruent to
$3$ modulo $4$ the corresponding exponent $m_i$ be even. Moreover the number
of solutions is given by
$$ 4 \prod_{\kern-14pt p_i\equiv1\pmod4\kern-14pt}(m_i+1).$$

So the number is always divisible by 4 (for obvious reasons) and for the other
factors only primes that are 1 modulo 4 are useful; for a factor 5 one needs
an exponent $m_1=4$, and the smallest appropriate prime is $p_1=5$, whence
$N=5^4=625$. The actual solutions are easily constucted using Gaussian integer
multiplication, from the elements $2\pm\mathbf{i}$ of norm $\sqrt5$.

All that remains is to choose the unit of measure to be $\frac1{25}\,\textrm{km}=40\,\textrm{m}$, and to find a size of the forest large enough that the X outnumber the Y even though they stop a kilometer before the the edge of the area planted with Y trees (left as exercise).

It may be argued that when trees are at least $40$ meters apart, one cannot reasonably talk about a forest. I would roughly guess a realistic distance between trees to be somewhere between $1$ and $5$ meters, which means we need $N$ to be larger, and in particular $\sqrt{N}$ (which is necessarily integer for a solution) in the range from $200$ to $1000$. This leaves us plenty of possibilities for $\sqrt{N}$, which can be any odd number whose set of prime factors $p$ with $p\equiv1\pmod4$ is a singleton, and which prime factor has multiplicity $2$. This leaves the possibilities $225$, $275$, $475$, $507$, $525$, $575$, $675$, $775$, $825$, $867$. Taking $867$ as an interesting value (giving a distance between trees of $\frac{1000}{867}\approx 1.15$ meter) the fundamental solutions of $k^2+l^2=N=867^2=751689$ are $(k,l)$ one of $(867,0)$, $(765,408)$ and $(720,483)$; the $17$ other solutions follow these by obvious symmetries.

Yes, he could. 10 spruces are in the middle of a formation, each point one kilometer away from the spruces. Larches could be planted around them,so there could be 20 larches sharing 10 spruces. Also the spruces could be in one metre square.


A group such that $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$ ($m$, $n$ coprime) is abelian? Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that
$$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$
$$\forall a,b \in G, \, a^nb^n=b^na^n.$$

Then how prove $G$ is an abelian group?

Some context: Some of these commutation relations often imply that $G$ is abelian, for example if $(ab)^i = a^i b^i$ for three consecutive integers $i$ then $G$ is abelian, or if $g^2 = e$ for all $g$ then $G$ is abelian. This looks like another example of this phenomenon, but the same techniques do not apply.

Let $M \subset G$ be the subgroup generated by all $m$-th powers and let $N \subset G$ be the subgroup generated by all $n$-th powers. These subgroups are clearly abelian normal subgroups. Since $m$ and $n$ are coprime, $G = MN$, and hence $M \cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] \subset M \cap N$ (since $M$ and $N$ are normal subgroups). Let $a \in M$ and $b \in N$.
Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.

As $\gcd(m, n) = 1$, by Bezóut's identity there are $u$, $v$ such that $u m + v n = 1$.
Take $a, b \in G$ arbitrary. By the conditions, as $a^u \in G$ for all $u \in \mathbb{Z}$, and as the conditions allow to commute certain powers:
$$
\begin{align*}
a^{u m} b^{u m} &= b^{u m} a^{u m} \\
(a^{u m} b^{u m})^{v n} &= a^{u m + v n} b^{u m + v n} \\
                        &= a b \\
(b^{u m} a^{u m})^{v n} &= b^{u m + v n} a^{u m + v n} \\
                        &= b a
\end{align*}
$$
This because, e.g.,
$(a^{u m} b^{u m})^2 = a^{u m} b^{u m} a^{u m} b^{u m}
                     = a^{u m} a^{u m} b^{u m} b^{u m}
                     = a^{2 u m} b^{2 u m},$
since the middle pair commutes by the conditions. Now use induction.
(Perhaps the switching around can be expressed in a more compact way).
Nice problem!


Recommendation on a rigorous and deep introductory logic textbook In this post, I don't mean any word by its somewhat "mathematical or logical" meaning but just "literally".

It's been three years since I started "formal" mathematics, and now I'm familiar with set theory and formal proof.

In the meantime, I have never studied "logic" before (it's nonsense to me), so now i think it's the time to start with it.

I have asked a similar question before, and people recommended me some texts. Almost all of them started with introducing "proposition logic". I guess authors intended to introduce a rather easier example at first. I don't think it's a good way to study logic rigorously. I felt like I'm not studying mathematics when I was reading those books, but I felt like I'm reading an philosophy article, which I felt extremely uncomfortable.

Frankly, to me, it's really hard to know what people mean by logic. I have searched wikipedia, but there are so many types of logics such as propositional logic, intuition logic(?), classical logic and etc. I even found some "logics" are subcategory of other!
What is logic exactly?

I don't want to start logic with 'handy and easy' examples. I want to study logic from its core so I could answer questions like: What is "proof"? What is "truth"?

Please... please recommend me a good precise logic textbook. I'm eager to learn logic precisely... Thank you in advance ! :)

There are two very different kinds of question here:

  What is logic exactly? ... What is "proof"? What is "truth"?

All good questions. But famously they do not have sharp, determinate, clear, uncontentious answers. Indeed, they are characteristically philosophical questions (that fall into the purview of what is often called "philosophical logic").

Of course, a technical logic text will introduce e.g. a sharp, technical, notion of a proof-in-a-given-formal-system (the fine print can be significantly different in different texts). But what is the relation between (1) the everyday notion of mathematical proof and (2) various notions of proof-in-a-given-formal-system which aim to model mathematical proof? This is up for (philosophical) debate. Similarly for the notion of truth, and indeed for the notion of a logic.

A "rigorous logic text" is therefore not the best place, really, to look for the discussion of the philosophical questions here. For those questions are (as it were) standing back from details in those rigorous texts and asking more general, philosophical, questions about them.

  Please recommend me a good precise logic textbook.

Still, if you do want pointers to formal logic textbooks then there are a lot of suggestions, at various levels, on various areas of logic, in the Guide you can find at http://www.logicmatters.net/tyl

According to Jim Nance texts Formal Logic is the science and art of reasoning well. He starts with deductive reasoning and the standard syllogism. He clearly defines the differences of truth and validity. If you are interested in a well built, straight forward approach, check out his Introductory AND Intermediate Logic texts. Get the Teachers manuals, as these house the student manual fused with the answer key, quizzes and texts. If you are a self taught learner, you will LOVE this format! There are also dvd videos he produced in which he teaches he lesson. He also has a facebookpage that he moderates and he has answered my questions with a positive and "mentor" minded demeanor. This text is a collegic entry level logic course/advanced highschool course. I teach/tutor claasically taught eighth graders both of these texts over the course of one year. Your time would not be wasted with a study as this and you could potentially finish both texts in one semester, depending on how motivated and determined you are. I have students who have found a LOVE for the structure of Logic through this examination approach of the standard syllogism. Good luck. And have fun! Please excuse my typos! I am using my phone, stumbled across your post, and felt led to share. I must run to teach classes now.


Find an example of a regular triangle-free $4$-chromatic graph Find an example of a regular triangle-free $4$-chromatic graph

I know that for every $k \geq 3$ there exists a triangle-free $k$-chromatic graph.

So if I can find a triangle-free graph $H$ such that $\chi(H)=3$, then I can use the Mycielski construction to obtain a triangle-free graph $G$ such that $\chi(G)=4$. However, the regular part keep getting me stuck. I try some odd cycle, I also tried the Petersen graph but still can't get a regular triangle-free $4$-chromatic graph.

I wonder if anyone can give me a hint, please.

If you're lazy you can check all the graphs on Wikipedia. The Hoffman–Singleton graph, for example, is strongly regular, has girth 5, and has chromatic number 4.

The Gewirtz graph, a strongly regular graph with parameters $(56,10,0,2)$, is another triangle-free graph with chromatic number $\chi=4$.
It is worth noting that Gewirtz graph is not isomorphic to Kneser graph $K(8,3)$. The Gewirtz graph has independence number $\alpha(G) = 16$ (i.e. the cardinality of a largest set of vertices pairwise nonadjacent, e.g. see Brouwer and Van Maldeghem' survey) whereas $\alpha(K(8,3)) = 21$ (the Erdős-Ko-Rado theorem  states that $\alpha(K(n,k)) = \binom{n-1}{k-1}, n\geq 2k $).


Material derivative of the dot product I was wondering if it is the case that:

$$\frac{d}{dt} \left( \vec{a}\cdot\vec{b} \right)= \vec{a} \cdot \frac{d}{dt} \vec{b} + \vec{b} \cdot \frac{d}{dt} \vec{a}$$

Where $\frac{d}{dt}$ indicates the material derivative $\frac{\partial}{\partial t} + \vec{u} \cdot \nabla$

If so, why is this the case?  I am having difficulty showing this to be true or false.

You really just have to expand out the material derivative. Consider the vectors $a = (a_1,\dots,a_n)$, $b = (b_1,\dots,b_n)$ and $u$ the velocity field. Then
\begin{align*}
\frac{d}{dt}\left(\sum_{j=1}^n a_jb_j\right) & = \frac{\partial}{\partial t}\left(\sum_{j=1}^n a_jb_j\right) + u\cdot\nabla\left(\sum_{j=1}^n a_jb_j\right) \\
& = \sum_{j=1}^n \left(\frac{\partial a_j}{\partial t}b_j + a_j\frac{\partial b_j}{\partial t}\right) + \sum_{k=1}^n u_k\frac{\partial}{\partial x_k}\left(\sum_{j=1}^n a_jb_j\right) \\
& = \sum_{j=1}^n \left(\frac{\partial a_j}{\partial t}b_j + a_j\frac{\partial b_j}{\partial t}\right) + \sum_{k=1}^n u_k\left(\sum_{j=1}^n \frac{\partial a_j}{\partial x_k}b_j + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial x_k}\right) \\
& = \sum_{j=1}^n \frac{\partial a_j}{\partial t}b_j + \sum_{k=1}^n u_k\left(\sum_{j=1}^n \frac{\partial a_j}{\partial x_k}b_j\right) + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial t} + \sum_{k=1}^n u_k\left(\sum_{j=1}^n a_j\frac{\partial b_j}{\partial x_k}\right)
\end{align*}
where we apply product rule. Interchanging sum for the second and fourth terms, we obtain
\begin{align*}
\frac{d}{dt}\left(\sum_{j=1}^n a_jb_j\right) & = \sum_{j=1}^n \frac{\partial a_j}{\partial t}b_j + \sum_{j=1}^n b_j\left(\sum_{k=1}^n u_k\frac{\partial a_j}{\partial x_k}\right) + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial t} + \sum_{j=1}^n a_j\left(\sum_{k=1}^n u_k\frac{\partial b_j}{\partial x_k}\right) \\
& = \sum_{j=1}^n b_j\left(\frac{\partial a_j}{\partial t} + \sum_{k=1}^n u_k\frac{\partial a_j}{\partial x_k}\right) + \sum_{j=1}^n a_j\left(\frac{\partial b_j}{\partial t} + \sum_{k=1}^n u_k\frac{\partial b_j}{\partial x_k}\right) \\
& = b\cdot\frac{da}{dt} + a\cdot\frac{db}{dt}.
\end{align*}

Consider $\vec a=[a_1\ \ a_2\cdots a_n]$ and $\vec b=[b_1 \ \ b_2\cdots b_n]$. Then it can simply be proved by applying scalar derivative property to their dot product
$$\vec a \cdot \vec b= a_1b_1+a_2b_2+\cdots +a_nb_n.$$


Is the set $U=\{ x \in \mathbb{R} : \sin (x) = \frac{1}{2} \}$ closed or open set? I tried to simplify the set definition by solving for different values of $x$'s and ended up with this set.
$U = \{ ... , -\pi-\frac{\pi}{6} , \frac{\pi}{6} , \pi-\frac{\pi}{6}, ... \}$
I know that a set is said to be closed if all of its cluster points belong to the set.
I don't know how to go ahead with the problem.
Thanks for the help in advance
Edit 1: From the comments although I don't know exactly why the set is closed but yes the set U is closed. Then π/6 is a cluster point i.e every neighborhood of π/6 has atleast one element in U leaving π/6 . How is that true?

$\sin(x): \mathbb{R} \to [-1,1]$ is a continuous function. Now since $\{\frac{1}{2} \}$ is closed in $[-1,1]$ and $U=\sin^{-1}\left(\frac{1}{2}\right)$ we get that $U$ is closed in $\mathbb{R}$.

Take any arbitrary $x$=nπ + $(-1)^n$$\frac{π}{6}$.
Now let $\epsilon$ < min {[(n+1)π + $(-1)^{(n+1)}$$\frac{π}{6}$] - [nπ + $(-1)^n$$\frac{π}{6}$], [nπ + $(-1)^n$$\frac{π}{6}$] - [(n-1)π + $(-1)^{(n-1)}$$\frac{π}{6}$] }.
Although it looks a bit complex but it's not at all complex. If you see the format of any three consecutive numbers in U, you'll understand what I have written in that minimum part. Now let's say $x_1$, $x$, $x_2$∈U such that $x_1$ < $x$ < $x_2$. Now what I have done is I have taken $\epsilon$<min{$x$-$x_1$, $x_2$-$x$}.(You can take equality but < makes the argument stronger and more explicit.) Then the neighbourhood N($x$,$\epsilon$) contains no point of U.This shows that there exists a neighbourhood of $x$ such that it doesn't contain any point of U. Hence, x is not a limit point. (Remember the definition of limit point). Since x is arbitrary, none of the points of U is a limit point. Hence, derived set U$'$ = {$\phi$} $\subset$ U and therefore, U is a closed set.


Finite number of subgroups $\Rightarrow$  finite group I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.

I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.

What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.

Is that correct?

Yes, it's correct, though perhaps somewhat disorganized.

The fact that a group is a union of its cyclic subgroups holds regardless of any hypothesis on the group. So you can state and prove that first.

Then: an infinite cyclic group has infinitely many (cyclic) subgroups. So, a group is infinite if and only if it has infinitely many cyclic subgroups: If the group is finite, then it has only finitely many subsets, so finitely many subgroups, so finitely many cyclic subgroups. If the group is infinite, then being a union of its cyclic subgroups, either one of the cyclic subgroups is infinite (and hence there will be infinitely many cyclic subgroups, since that subgroup has infinitely many cyclic subgroups), or else there are infinitely many finite cyclic subgroups in the union. Either way you are done.

That is your argument, just organized a bit more.

I might be oversimplifying, but it seems to me that if G has infinite elements, it has an infinite number of cyclic subgroups, one for each g in G.


Strictly Convex Implies Invertible Gradient? If $f:\mathbb R^n \rightarrow \mathbb R$ is strictly convex and continuously differentiable, does this imply that $\nabla f$ is a one-to-one mapping?

To be precise, can we say that $x, y \in \mathbb R^n$ and $\nabla f(x) = \nabla f(y)$ implies $x = y$?

Suppose that there exist $x, y \in \mathbb R^n$ such that $\nabla f(x) = \nabla f(y)$ and $x \neq y$.

Then, by the strict convexity of $f$, we can write
\begin{equation}
\nabla f(x) \cdot (y-x) < f(y) - f(x)
\end{equation}
and similarly
\begin{equation}
\nabla f(y) \cdot (x-y) < f(x) - f(y).
\end{equation}
Multiplying both sides of the latter inequality by $-1$ and substituting $\nabla f(x)$ in place of $\nabla f(y)$, we obtain
\begin{equation}
\nabla f(x) \cdot (y - x) > f(y) - f(x),
\end{equation}
which contradicts the first inequality.

Thus, if $x, y \in \mathbb R^n$ satisfy $\nabla f(x) = \nabla f(y)$, then $x =y$.


Show that $S^{-1}A = B$ for integral domains and $S = \{x\in A\setminus\{0\}: x^{-1}\in B\}$ Let $A \subset B$ be commutative integral domains with $\operatorname{Quot}(A) =  \operatorname{Quot}(B).$

Now consider the multiplicatively closed subset $S = \{x\in A\setminus\{0\}: x^{-1}\in B\}$.

I want to show that $S^{-1}A = B$. I would appreciate any help.

It suffices to prove $B\subseteq S^{-1}A$, since we already have $A\subseteq B\subseteq \mathrm{Frac}(A)$.
For any $x\in B$, $x=\frac{a}{b}$ for some $a,b\in A$. By definition, correct.


Angle and circle intersection, find the circular segment area Playing Kerbal Space Program, I found myself wondering about what a satellite would see of a planet depending on its field of view and its altitude. I tried attacking the problem from various angles but I always end up with a missing piece of information.

Speaking a bit more formally, I have a circle Planet with a radius r. I have a point Sat which is at a distance Orb from the center of Planet (for our case, Orb > r). The line Sat-Center(Planet) bisects A-Sat-B, with A and B being points on the circle.

I know Orb, r and the angle A-Sat-B, what ways do I have to calculate A-B? (See below for a diagram.)

From there I think the Circular segment will let me find the information I need but I'm stumped on the calculation.

So you have the angle $\theta$, the planet radius $r$ and the orbital distance ${\rm Orb}$

the solution arrives from two trigonometry  equations:

$$ \left.
\begin{aligned} {\rm Orb} & = r \cos \frac{\psi}{2} + \ell \cos \frac{\theta}{2} \\
 r \sin \frac{\psi}{2} & = \ell \sin \frac{\theta}{2} \end{aligned}
\right\}
\begin{aligned}  \frac{\psi}{2} & = \frac{\pi-\theta}{2} - \cos^{-1} \left(  \frac{{\rm Orb}}{r}\sin \frac{\theta}{2}\right)
\end{aligned}
 $$

With $h = r \sin \frac{\psi}{2} $  and ${\bf AB} = 2 h$

For example, when $r=1$, ${\rm Orb}=3.5$ and $\theta=25°$ then

$$ \frac{\psi}{2} = \frac{180°-25°}{2} - \cos^{-1} \left(  \frac{3.5}{1}\sin \frac{25°}{2}\right) =  36.74768°$$  which is same as measured $\beta$ angle in diagram above.

Then $h = 1 \sin (36.74768°) = 0.5982922 $ which also matches diagram above.

Also we can check $\ell = \frac{ r \sin \frac{\psi}{2} }{\sin\frac{\theta}{2}} = { r \sin(0.6413681) }{\sin(12.5°)} = 2.764245$

Diagram is made with GeoGebra 4

$\text{The answer can be as explained below -  }\\$

$\text{Join $A$ and $B$, Let $C$ be centre,}\\$
$\text{Now join $AC$ and $BC$, get $\theta$, the $\angle$ that lines $AC$ and $BC$ make at the centre,}\\$
$\text{find the length of the chord $AB$, which will be $\frac{1}{2}$ the $diameter$ if angle is $\frac {\pi}{2}$,}\\$
$\text{depending on the $\angle \theta $, whether it is $\pi$, $\frac{\pi}{2}$ respectively, and so on.}\\$

$\text{Now when you know $AB$,}\\$
$\text{You can also check (A-SAT) $or$ (SAT-B), by geometry finding the 2 equal}\\$
$\text{sides of a triangle by pythagorous theorem when base and height are known,}\\$
$\text{$or$ The Arc length $AB$ $=$ $r$ $\cdot$ $\theta$ & the }\\$
$$\text{$"Darkened Circular Segment Area"$} = \left[\frac {r^2}{2} \cdot \left(\frac{\pi \cdot \theta}{180} - \sin \theta \right)\right]$$


Cubic diophantine equation in 3 variables $(x+2y)(x-4y+k)(x-4y-k) - 28y^3 = 0$, $x,y,z \neq 0$ From research completely unrelated to Number Theory I stumbled onto the following equation:

$$
xyz = \frac{7}{16}\left(\frac{2x - y - z}{3}\right)^3
$$

for $x, y, z$ integers, $x,y,z \neq 0$(I forgot to put this condition in the original question). Are there integers that satisfy it?

Edit: Ok, I was missing the condition that all variables should be nonvanishing. Here is the original problem

$$
(x+2y)(x-4y+z)(x-4y-z) - 28y^3 = 0
$$

for $x,y,z$ integers and (all of them) nonvanishing.

The first equation can be obtained from this one by a trivial change of variables. Of course, solutions in the integers for the first one do not imply solutions in the integers for the second one but that is fine because rational solutions are also OK.

By expansion and simplification, your original problem can be written as
$$
  2(3x^2-2y^2+z^2)y=x(x-z)(x+z).
$$
Let $d = \gcd(x,y)$, say $(x,y)=(du,dv)$ for integers $(u,v)$ with $\gcd(u,v)=1$. Then
$$
  2\bigl((3u^2-2v^2)d^2+z^2\bigr)v=u(du-z)(du+z).
$$
Now re-simplifying in terms of $d$,
$$
d^2(u^3-6u^2v+4v^3) =(2v+u)z^2.
$$
Can you take it from there?

There are lots of easy solutions where one of $x,y,z$ is $0$.


chinese reminder theorem "A house wife spends 1770 Rupees for purchasing mangoes and apples. She pays 31 Rupees for each mango and 21 Rupees for each apple. If she buys more mangoes than apples, how many mangoes and apples she buys"
I know how to solve this using dolphins equation.
can it be solved using chinese reminder theorem.
this is what i think.
$1770 \equiv  x $ (mod $ 31)$
$1770 \equiv y $ (mod $ 21)$.
I know how to do the chinese reminder theorem for 1 variable.
please help me with this.

$1770=31M+21A$, so $3\equiv 21A\pmod{31}$.
To solve that congruence, you could use the Euclidean algorithm:
$31=1\times21+10$
$21=2\times10+1$
So $1=21-2\times10=21-2\times(31-21)=3\times21-2\times31$,
so $3\equiv9\times21\pmod{31}$.
Can you take it from here?


Prove that the distance between parallel planes $\vec{n}\cdot \vec{x} = d_1 $, $\vec{n}\cdot \vec{x}=d_2$ is $|d1-d2|/||\vec{n}||$ Prove that the distance between parallel planes with equations $\vec{n}\cdot \vec{x}  = d_1 $ and $\vec{n}\cdot \vec{x}=d_2$ is given by $\displaystyle\frac{\left|d_1-d_2\right|}{\left\|\vec{n}\right\|}$

Not sure how to go about this. Thought after taking a discrete class I would have better intuition towards proving such things.

I know the planes are both normal to $\vec{n}$ -- clearly the minimum distance is going to be on a line in the n direction.

Let $x_k$, $k=1,2$ be a point in each plane. Write $x_k = \alpha_k n + p_k$, where $p_k \bot n$.
Then
\begin{eqnarray}
\|x_1-x_2\|^2 &=& \| (\alpha_1-\alpha_2)n +(p_1-p_2)\|^2 \\
&=& (\alpha_1-\alpha_2)^2 \|n\|^2+\|p_1-p_2\|^2
\end{eqnarray}
Since each $x_k$ lies on each plane, we have $\langle x_k, n\rangle = d_k$, from which we get $\alpha_k = {d_k \over \|n\|^2}$, and so
$\|x_1-x_2\| = \sqrt{ ({d_1-d_2 \over \|n\|})^2+ \|p_1-p_2\|^2 }$.

It is easy to see that the distance is minimized if we choose $p_1=p_2$, in which case we get the minimum distance to be ${ |d_1-d_2| \over \|n\|}$

Same answer as "user183761" but with (perhaps) more explanation.

Given a point P(x1, y1, z1) on one plane and point Q(x2, y2, z2) on the other plane, the distance between the planes is the projection of vector PQ onto the normal vector n = (a, b, c) of the plane(s).  The vector PQ is given by (x2-x1, y2-y1, z2-z1).  By definition, for point P to lie on the plane it must satisfy the equation ax1 + by1 + cz1 = d1.  Likewise, for point Q, ax2 + by2 + cz2 = d2.

Thus, d2 - d1 = a(x2-x1) + b(y2-y1) + c(z2-z1) = (a, b, c)· (x2-x1, y2-y1, z2-z1) = n·PQ

So |d2 – d1|/||n|| = |n·PQ|/||n|| = |projnPQ|


Determine convergence or divergence of $a_n=\frac{4}{7}(1+\frac{1}{2n})^n$ $\sum_{n=1}^\infty a_n$ such that $a_1=1$, $a_n=\frac{4}{7}(1+\frac{1}{2n})^n$

I need to determine convergence or divergence of this series.

So I think the ratio test would work for this but i'm stuck when evaluating the limit.

$$\lim_{n\to \infty}\frac{(1+\frac{1}{2n+2})^{n+1}}{(1+\frac{1}{2n})^n}$$

Can anyone give me a hint as to how i should proceed?

The series $\sum_{n=1}^\infty a_n$ diverges since $a_n$ does not go to $0$ as $n\to\infty$. We have that
$$
a_n=\frac47\biggl(1+\frac1{2n}\biggr)^n=\frac47\sqrt{\biggl(1+\frac1{2n}\biggr)^{2n}}\to\frac{4\sqrt e}7
$$
as $n\to\infty$.

$$\lim _{x\rightarrow \infty} \left( 1+ \frac{1}{2x} \right) ^x \geq  \lim _{x\rightarrow \infty} 1^x = \lim _{x\rightarrow \infty} 1 = 1 \neq0 $$

since $a_n \not \rightarrow 0$ we have $\sum a_n = \infty$.


Fourier transform of compact supported function is entire Suppose $f\in L^1(\mathbb{R})$ has compact support, say $\operatorname{supp}f\subset[-r,r]$.
I want to show that its Fourier transform
$$ \hat{f}(z) = \int_{-a}^{a} f(t) e^{-2\pi itz}dt$$
for $z\in\mathbb{C}$ is entire, so $\hat{f}\in H(\mathbb{C})$. I think this follows from Morera's theorem, but I am not very familiar with contour integrals. Could someone help me out with the first steps?

"No, but I actually would like to learn how one could apply Morera's theorem here"
Well, for that we have to prove two things:
(a) $\hat f$ is continuous.
(b) The integral of $\hat f$ over each triangle is zero.
So, let's start with (a). Since for $z$ close to $z_0$,
$$
|e^{2\pi itz}-e^{2\pi itz_0}|\le 4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}|z-z_0|
$$
(that's at least what I could achieve), we have
$$
|\hat f(z)-\hat f(z_0)| = \left|\int_{-a}^af(t)\left(e^{2\pi itz} - e^{2\pi itz_0}\right)\,dt\right|\,\le\,4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}\|f\|_{L^1}|z-z_0|.
$$
Hence, $\hat f$ is locally Lipschitz. Now, let $\Delta$ be any triangle. Then with Fubini we get
$$
\int_\Delta\hat f(z)\,dz = \int_\Delta \int_{-a}^af(t)e^{2\pi itz}\,dt\,dz = \int_{-a}^af(t)\left(\int_\Delta e^{2\pi itz}\,dz\right)\,dt = 0.
$$

Here is a proof based on Moreras theorem that $\hat{f}(z) = \int_{-a}^a f(t) e^{-2 \pi i tz} dt$ is holomorphic.
Let $\gamma(t)$ be a simply closed curve in the complex plane.
Now, $\int_\gamma \hat{f}(z) dz=\int_\gamma \int_{-a}^a f(t) e^{-2 \pi i tz} dt dz = \int_{-a}^a \int_\gamma f(t) e^{-2 \pi i tz} dzdt$ by Fubinis theorem.
But the function $g(z)=f(t)e^{-2 \pi i tz}$ is holomorphic so $\int_\gamma g(z) dz = 0 $ by Cauchys theorem.
Hence, we have shown that for any simply closed curve $\gamma$, $\int_\gamma \hat{f}(z) dz =0$.  Moreras theorem, then, implies that $\hat{f}$ is holomorphic.


$(2x+y)\frac{dy}{dx} + x = 0$ Solve:
  $(2x+y)\frac{dy}{dx} + x = 0$ .

Only solving seperable equations and use of an integrating factor have been covered so far, and I can't see how to get it into a form such that it is solvable by these methods.

Try  $y = vx $

$\implies   y' = v'x + v$

The equation becomes
$$(2x+vx)(v'x+v) + x = 0 \\ (2+v)(v'x+v)+1 = 0$$

which is separable

That is :
$(x+y)dy + xdx=0$

Which is
$xdy + ydy + xdx = 0$

Integrate all parts:
$xy + \frac{y^2+x^2}{2}=c$


Probability that two randomly chosen people out of 10 have birthday in the same month There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?

My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility

I'm sure that your answer of 1/12 is correct. When you randomly choose 2 people, the probability of person 2 having a birthday in the same month as person 1 would just be 1/12, assuming that the 12 months have an equal chance at being someone's birth month.

Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.

For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.

There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.

In general the formula is: $P(k)=\frac{12!}{(12-k)!}\frac{1}{12^k}$ where $k$

is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$

So for 10 people the probability of two having the same birth month is 99.6132%.

This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.

The chances hit over 50% for any group over 4 members large.

If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.

So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.


characterization of the unipotent radical of a linear algebraic group Im having difficulties understanding a ceratain argument in the following proof, taken from T.A springer's linear algebraic groups.
In the following proof C is the intersection of the unipotent parts of all borel groups containing the maximal torus T, and C$\alpha$ is the intersection of the unipotent parts all such borel groups for wich $\alpha$ is a positive root.

I dont understand why should the borel groups be contained in TC or TC$\alpha$

The heuristic is as follows: The toral (and abelian) subgroup allows us to simultaneously diagonalize them, such that $C$ are all upper triangular matrices with ones on the diagonal (unipotent). Then $C_\alpha T=TC_\alpha$ is a Borel subgroup (maximal solvable). The fact that $\alpha$ is a positive root, fixes the root system and so the possible conjugates. Adding any other generator would mean to add a negative root and destroy solvability.


Functions $f:\mathbb R\to\mathbb R$ satisfying $f\left(\frac{x+y}r\right)=\frac{f(x)+f(y)}s$ Let $r$ and $s$ be distinct nonzero rational numbers. Find all functions $f:\mathbb R \to\mathbb R$ such that $f\left(\frac{x+y}r\right) = \frac{f(x)+f(y)}s$.

My attempt:
If we have $x=0$ and $y=0$, Then we have $f(0) = \frac{f(0)+f(0)}s \implies s=2$.
If we set $x=0$, we get $s\times f\left(\frac yr\right)-f(y)=f(0)$. $(1)$
If we set $y=0$, we get $s\times f\left(\frac xr\right)-f(x)=f(0)$. $(2)$
Setting $(1)$ and $(2)$ equal, $s\times f\left(\frac xr\right) - s\times f\left(\frac yr\right) = f(x) - f(y)$.
I don't know where this is going and need help.

You can show that the only solutions are constant functions, and in case $ s \ne 2 $, that constant must be zero (otherwise, any constant function will work).
Let $ y = 0 $ in
$$ f \left( \frac { x + y } r \right) = \frac { f ( x ) + f ( y ) } s \tag 0 \label 0 $$
to get
$$ f \left( \frac x r \right) = \frac { f ( x ) + f ( 0 ) } s \text . \tag 1 \label 1 $$
You can use \eqref{1} to rewrite the left-hand side of \eqref{0} and get
$$ f ( x + y ) + f ( 0 ) = f ( x ) + f ( y ) \text . \tag 2 \label 2 $$
Hence, if we define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = f ( x ) - f ( 0 ) $, \eqref{2} tells us that $ A $ is an additive function. Cauchy's functional equation then lets us deduce
$$ A \left( \frac x r \right) = \frac { A ( x ) } r \text , \tag 3 \label 3 $$
as $ r $ is rational. On the other hand, \eqref{1} can be rewritten as
$$ A \left( \frac x r \right) + f ( 0 ) = \frac { A ( x ) + 2 f ( 0 ) } s \text , $$
which together with \eqref{3} yields
$$ ( s - r ) A ( x ) + r ( s - 2 ) f ( 0 ) = 0 \text . \tag 4 \label 4 $$
As $ A ( 0 ) = 0 $, letting $ x = 0 $ in \eqref{4} we have $ r ( s - 2 ) f ( 0 ) = 0 $. Putting this back into \eqref{4} we get $ ( s - r ) A ( x ) = 0 $, which shows that $ A $ is the constant zero function, as we know that $ s \ne r $. This means that for all $ x \in \mathbb R $ we have $ f ( x ) = f ( 0 ) $; in other words, $ f $ must be constant. In case $ s \ne 2 $, using $ r ( s - 2 ) f ( 0 ) = 0 $ and $ r \ne 0 $, $ f $ must be constantly zero. In case $ s = 2 $, you can check that any constant function $ f $ satisfies \eqref{0}.

A starting point: Using $rx,ry$ for $x,y$ your functional equation reads as $f(x+y)=\frac1s f(rx)+\frac1s f(ry)$. With $g=h$ where $g(x):=\frac1s f(rx)$ this become a Pexider equation. Thus there is a function $a\colon\mathbb{R}\to\mathbb{R}$ such that $a(x+y)=a(x)+a(y)$ for all $x,y$ and constants $p,q$ such that $g(x)=a(x)+p$, $h(x)=a(x)+q$, and $f(x)=a(x)+p+q$ for all $x$.


let $a\sim b$ iff for some integer $k$, $a^k = b^k$ Let $G$ be a group,
 Let $a\sim b$ iff  for some integer $m$, $a^m = b^m$.

I am having a problem trying to figure out how to prove that the transitive property. I know that you start off by Assuming $a\sim b$ and $b\sim c$, such that $a^m = b^m$ and $b^n = c^n$ for some $m$, $n$ that exist in the integers. I am stuck and not sure where to go from here. I just need a nudge in the right direction.

Note that $a^{mn}=b^{mn}$ and $b^{nm}=c^{nm}$.

Remark: The relation is uninteresting if we allow exponents to be $0$.

Here's another approach. If $a^k = b^k$, then $a^k/b^k = (a/b)^k = 1$. Thus, $a\sim b$ iff $(a/b)^k$. Of course if an element has order dividing $k$, then so does its inverse, so $(b/a)^k = ((a/b)^{-1})^k = ((a/b)^k)^{-1} = 1^{-1} = 1$.


Rates of change question involving water leaking out of a hemispherical tank Water is leaking out the bottom of a hemispherical tank of radius 9 feet at a rate of 4 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height h is 8 feet? Note: the volume of a segment of height h in a hemisphere of radius r is $$πh^2[r− \frac {h}3]$$

The height at time $t$ is $h$=8ft. At a very small time interval after that $t+dt$, the new height is $h-dh$. Note the minus sign, since the height decreases. The volume change is $$dv=\pi h^2\left[r− \frac {h}3\right]-\pi (h-dh)^2\left[r− \frac {h-dh}3\right]$$ The rate of change in the volume is equal to 4ft$^3$/hr=$dv/dt$. If you ignore terms in $dh^2$ and $dh^3$, which are small, you get
$$\frac{dv}{dt}=-\pi h(h-2r)\frac{dh}{dt}$$ From here
$$\frac{dh}{dt}=\frac{1}{\pi(2r-h)h}\frac{dv}{dt}$$

Let the level of water after 1 hour be $h$.

so change in volume in one hour is :

$π8^2[9− \frac {8}3] - πh^2[9− \frac {h}3] = 4$,  (the difference in volume of water stored in tank)

you can take positive value of $h$ after solving the above equation and your change in level is $8-h$ feet per hour.


Tensor Product and its details I'm having a tough time trying to understand tensor product and how to compute it. I don't know what to do when I have $\alpha\in V^*$, $\vec v\in V$ and I want to compute $\alpha\otimes\vec v$ or $\vec v\otimes\alpha$... For example, if I have $\alpha = (e^3-4e^1)$ and $\vec v = (-2\vec e_1+\vec e_3)$, $\alpha\otimes\vec v$ and $\vec v\otimes\alpha$ look like what?

The definition that my professor has taught me is this:
given $\alpha,\beta\in V^*$ and $\vec u,\vec v\in V$ we have $(\alpha\otimes\beta)(\vec u,\vec v)\doteq \alpha(\vec u)\beta(\vec v)$. Thank you very much, folks.

The definition you wrote is for the tensor product of two linear functionals. You can consider the tensor product of a linear functional with a vector by "plugging in what makes sense", that is: $$(\alpha \otimes v)(w, \beta) \doteq \alpha(w)\beta(v),$$and so on. What you can do for now is write that as combination of "basic" tensors, using bilinearity of $\otimes$, as in: $$\begin{align} \alpha \otimes v &= (e^3 - 4e^1)\otimes(-2e_1+e_3)\\ &= -2 e^3\otimes e_1 + e^3\otimes e_3 + 8e^1\otimes e_1 - 4 e^1\otimes e_3.\end{align}$$If $f = f_i e^i$ and $w = w^j e_j$, you can see how $\alpha \otimes v$ acts explicitly by the above formula: $$(\alpha \otimes v)(w,f) = -2w^3f_1 + w^3f_3 + 8f^1w_1 - 4f^1w_3.$$

Probably you are using for aeV* and veV that aoxv is the bilinear form on VxV* where v acts on b* in V* by point evaluation - v (b*) = b*(v) (thus identifying V** with V so aoxv(w,b*) = a(w)b*(v) .This works as long as V is finite dimensional .Hope that helps .What book are you using ?


Find the ratio of the area inside the square but outside the circle to the area of the square in the figure. All I'm worried about it is (a); for now. Okay, so let's start off like this: I know what the question is asking, sort of. I know it wants the ratio of the inside corner pieces of the square to the square itself. But what??? I obviously know that $A_{circle}=\pi r^2$ and $A_{square}=wh$ and also that the piece they are looking for is $A_{circle}-A_{square}$ but the ratio? I don't understand that. Can someone explain to me what they mean by that and stay posted for I may have questions about (b)-(e). Thanks.

If $A_0$ is the total area of the four corner pieces, and $A_1$ is the area of the square, they want $\frac{A_0}{A_1}$: that fraction is by definition the ratio of $A_0$ to $A_1$. Note also that for the square you have $w=h=2r$.

For Q a)

The answer is $(4-\pi):4$ (or) $(4-\pi)r^2:4r^2$

First we have to find out the area of corner pieces. That we will get by subtracting the area of circle from the area of square.
I.e.,area of square of side $2r$ is $4r^2$ ( bz. They gave half of the side is $r$)
And area of circle with radius $r$ is  $\pi r^2$
So area of corner pieces is $4r^2-\pi r^2 = (4-\pi) r^2$
Finally the ratio of area of corner pieces to square is

$$(4-\pi)r^2:4r^2$$  (or)
                                                     $$(4-pi):4$$


Prove $(1 +\frac{ 1}{n}) ^ {n} \ge 2$ Using induction, I proved the base case and then proceeded to prove: $$(1 + \frac{1}{n+1}) ^ {n+1} \ge 2$$ given $$(1 + \frac{1}{n}) ^ n \ge 2$$ However, I'm stuck at this point and have no clue how to go about it. Other than induction, I tried simple algebraic transformations but couldn't prove this inequality. Any pointers on how to prove this will be appreciated.

[PS: This is my first question on stackexchange, so I'm sorry if there's anything wrong with this post and will be happy to edit if needed].

Hint Prove by induction the more general statement:

If $x >0$ then
$$(1+x)^n \geq 1+nx $$

Assuming $\left ( 1+\dfrac{1}{n}\right)^n \geq 2$, we want to prove $\left ( 1+\dfrac{1}{n+1} \right)^{n+1}\geq 2$.

You may start by saying: $\left ( 1+\dfrac{1}{n+1}\right)^{n+1} = \left ( 1+\dfrac{1}{n+1} \right)^n\left ( 1+\dfrac{1}{n+1}\right).$

But if $\left( 1+\dfrac{1}{n}\right)^n\geq 2, $ then $\left( 1+\dfrac{1}{n+1}\right)^n\geq 2$.

$\left( 1+\dfrac{1}{n+1}\right)^n\left ( 1+\dfrac{1}{n+1}\right)\geq 2\left ( 1+\dfrac{1}{n+1}\right)$.

It is easy to see that $\left ( 1+\dfrac{1}{n+1}\right) \geq 1$, therefore conclusion follows.


Show that an ordered 4-tuple is also an ordered $m$-tuple for every positive integer $m$ less than 4. I'm not really sure what this question is asking me to do. Can someone please explain this to me just a little bit more? Maybe give some examples.

This question was taken from Enderton's Elements of Set Theory.

Here is a hint:

Suppose we have the ordered $3$-tuple $\langle x_1, x_2, x_3\rangle=\langle\langle x_1, x_2\rangle, x_3\rangle$.

What happens if we define $y_1=\langle x_1, x_2\rangle$ and $y_2=x_3$?

A complete solution is:Take,$y_4=\langle x_1, x_2, x_3, x_4\rangle$ , $y_3=\langle x_1, x_2, x_3\rangle$  and $y_2=\langle x_1, x_2\rangle$ .
Now, $\langle x_1, x_2, x_3, x_4\rangle=\langle\langle x_1, x_2, x_3\rangle, x_4\rangle=\langle y_3, x_4\rangle$, which is a two tuple.
 Again,  $\langle x_1, x_2, x_3, x_4\rangle=\langle\langle x_1, x_2, x_3\rangle, x_4\rangle=\langle\langle \langle x_1, x_2\rangle, x_3\rangle, x_4\rangle=\langle\langle  y_2, x_3\rangle, x_4\rangle=\langle  y_2, x_3, x_4\rangle$ (by Definition), which is a triplet.
By definition,  $\langle y_4\rangle=y_4=\langle x_1, x_2, x_3, x_4\rangle$. Hence $\langle x_1, x_2, x_3, x_4\rangle=\langle y_4 \rangle$, which is a one tuple.
 Is this ans correct?


Area enclosed by polar curve I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer.  Can you assist in clarification?

Find the area of the region enclosed by $r=4cos(3 \theta)$.

I use $ \frac 12  \int_0^{2\pi} (16cos^2(3\theta) d\theta$.  For $cos^2(3\theta)$ I use the identity $\frac12[1+cos(6\theta)]$

This gives me $\frac{16}{4} \int_0^{2\pi} 1+cos(6\theta) d\theta$.

This gives me $4[\int_0^{2\pi}1 d\theta +\frac16\int_0^{12\pi} cos (u) du]$.

The integral of the cosine term is $0$, so I get $\theta $ evaluated from $0$ to $2\pi$.  This gives me $4(2\pi)=8\pi$.

When I use a symmetrical method A=$6\int_0^{\pi/6}\frac12(16 cos^2(3\theta)d\theta$ I get $4\pi$.   This is the text answer.

Don't understand why my 2 answers don't match.

Your first answer is twice the correct answer for the following reason: if you let $\theta$ range from $\theta=0$ to $\theta=2\pi$, the curve $r=4\cos(3\theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $\theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $\theta=\pi$; and then, from $\theta=\pi$ to $\theta=2\pi$ you retrace it once more.

In the second method, you find the area of a half of one petal, which you correctly determined to range from $\theta=0$ to $\theta=\dfrac{\pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $\theta=0$ to $\theta=\dfrac{\pi}{6}$ will produce the angle six times as wide, i.e. from $\theta=0$ to $\theta=\pi$, consistent with my explanation above.

To provide yet another approach, you can use Green's theorem on the differential form $x \,dy$ to transform the area integral over one petal into a line integral along the petal:

$$\frac13 A = \text{area of petal}  = \int\limits_{\text{petal}}\,1\,dA = \int\limits_{\text{boundary of petal}} x\,dy $$

Since $r = 4\cos(3\theta)$, the petal can be parameterized by
$$(x(\theta), y(\theta)) = (r(\theta)\cos\theta, r(\theta)\sin(\theta)) = (4\cos\theta\cos(3\theta), 4\sin\theta\cos(3\theta))$$

for $\theta \in \left[-\frac{\pi}6, \frac{\pi}6\right]$.

Differentiating gives $dy = (4\cos\theta\cos(3\theta) - 12\sin\theta\sin(3\theta))\,d\theta$ so

$$\frac13 A = \int_{-\frac{\pi}6}^{\frac{\pi}6}4\cos\theta\cos(3\theta) (4\cos\theta\cos(3\theta) - 12\sin\theta\sin(3\theta))\,d\theta = \frac{4\pi}3$$

so $A = 4\pi$.

The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.


Way to determine eigenvalues from 3. degree characteristic equation? I have derived the following characteristic equation for a matrix

$$a^3 - 3a^2 - a + 3 = 0$$ where $a = \lambda$.

I know that it's possible to find the roots (eigenvalues) by factorization, but I find this to be especially difficult with cubic equations and was wondering if there perhaps is an easier way to solve the problem.

Note that your cubic characteristic polynomial is always monic. If your matrix has integer coefficients, then ALL rational roots are (positive or negative) divisors of the constant term.

Therefore, the method that works in more than 90% of $3\times3$-exercise problems is: Try out the divisors starting with the small ones, split off the linear factor, solve the quadratic equation.

If your cubic polynomial with integer coefficients does NOT have rational solutions, the reasonable way to go is usually a numeric solution.

So, if it is an exam question and you have checked all divisors and you have not found a rational root, I recommend to search for the error in the calculation of the characteristic polynomial.

Personally, I hate dealing with polynomial equations to the third power. My escape from those is the following. Once you got a $3\times 3$ matrix, reduce it to a triangular matrix. According to the theorem from the book: "The eigenvalues of a triangular matrix are the entries on its main diagonal". For example, if the main diagonal entries in your triangular matrix are $1$, $3$, $4$, then the polynomial equation would be $(1-a)(3-a)(4-a)$. Thus your eigenvalues would be $1$, $3$ and $4$ respectively. Here we go, no synthetic division.


Show that for any real constants a and b, where b>0, (n + a)^b = θ(n^b) This problem is from "Introduction to Algorithms" Cormen 3rd edition. It's problem 3.1-2 from the book. I have ready many many websites, including this one, on how to solve this problem. My motivation is that I'm a grad student and taking "Introduction to Algorithms". I don't know how to solve this problem in a way that makes sense to me. I could loosely copy some elses work and hopefully not get docked on grading, but I want to get a doctorate in computer science after my master's degree. I'm not cutting corners, taking less classes to get straight A's and thoroughly understand everything in all my classes.
Here is the problem: Show that for any real constants a and b, where b>0, (n + a)^b = θ(n^b)
The way to set it up is using the definition of theta asymptotic notation, which is: θ(g(n)) = {f(n): there exists positive constants c1, c2 and n0 such that
0 <= c1g(n) <= f(n) <= c2g(n) for all n >= n0}.
So initially setting up the inequality is:
0 <= c1*(n^b) <= (n + a)^b <= c2*(n^b)
I don't know how to get from this point to the answer, which is c1=(1/2) and c2=2. I already have seen the answers to the first six pages that come up on Google by typing in the problem question. One of them was from this website. I thoroughly read all the solutions and some are done a lot different (or seem that way on the surface).
I need a detailed explanation. Here is an example. Changing (n+a) to (n + |a|). Why use the absolute value? Because a negative value is possible for a when n > 0? Why? Show examples. Explain each step in depth. I don't remember everything from every math class I took. I did read the appendices for the Algorithms book and read chapter 3 all the way through. So, yes I did study how to manipulate equations recently. I did my part and still don't know how to solve this problem all the way through making sense the entire way and can defend each step cogently.
Can anyone help?

\begin{align*}
(n + a)^b ≤ (n + |a|)^b & \text{ where } n > 0\\
          ≤ (n + n)^b & \text{ for } n ≥ |a|\\
          = (2n)^b\\
          = c1 · n^b & \text{ where } c1 = 2^b
\end{align*}
Thus
\begin{align*}
(n + a)^b = Ω(n^b)\\
(n + a)^b ≥ (n − |a|) b& \text{ where } n > 0\\
          ≥ (c′_2n)^b & \text{ for } c′_2 = 1/2 \text{ where }n ≥ 2|a|\\
\end{align*}
as  $n/2 ≤ n − |a|$, for $n ≥ 2|a|$.
Thus
$$(n + a)^b = \Omega(n^b)$$
The result follows from 1 and 2 with $c_1 = 2^b$, $c_2 = 2^{−b}$, and $n_0 ≥ 2|a|$.


Complex function on circle I have $f(z)= -\frac{z-i}{z+i}$ which, I believe, will map a circle into another circle. So if I start with a circle in the complex plane with centre $(x,y)$ of radius $r$ how do I determine the centre and radius after applying $f(z)$?

I can see that I substitute $x + iy$ for $z$ in the function $f(z)$ but then I have just an expression rather than an equation - don't I? (My university level maths is over $40$ years ago...).
Thanks.

Moebius transformations map lines and circles into lines and circles, see

A Möbius transformation maps circles and lines to circles and lines. What exactly does that mean?

Remark for a reference: Your map is (the negative of) the Cayley transform
$$
z\mapsto \frac{z-i}{z+i},
$$
which
maps the upper half plane biholomorphically to the open unit disk.  So $f(z)$ maps $\infty$ to $-1$, $1$ to $i$ and $-1$ to $-i$.

If I apply f to the center I get the new center,
f(x, y) -> c'

If I apply f to a point on the circle I get a point on the new circle. e.g.

f(x - r, y) -> p

and so, from c' and p, can determine  the radius.

Isn't this correct?


Suppose a group G has no proper subgroups (that is, the only subgroup of G is G itself and the trivial subgroup {e}. Show that G is cyclic. Suppose a group G has no proper subgroups (that is, the only subgroup of G is G itself and the trivial subgroup {e}. Show that G is cyclic.

how to solve the question?

Hint

If $G=\{e\}$ then we are done. Now assume that there's $G\ni g\ne e$. What's the subgroup of $G$ generated by $g$?

if you Let G be a group with no proper, nontrivial subgroups.let g be a non
-trivial element of G then  Consider < g >,the cyclic subgroup generated by g.
This is a subgroup of G, but since the only subgroups of G are < e >and
G, < g >=< e >or < g >= G. But we assumed g was nontrivial.
So< g > not equal< e > :Thus, < g >= G:But this means G is generated by
< g >


Is the book: A Course of Pure Mathematics (G H Hardy) still relevant today? A very intelligent family member of mine has recently decided to send me a large number of books and other resources on mathematics, include Hardy's book. Since this was published in 1921, is it still a useful textbook from which to learn pure mathematics?

Congratulations to such generous and wise members in your family. My answer is: Yes, of course. This text is great to read and will  help  to  gain  a deep and  proper understanding of the material.

Hint:  In order to better evaluate this treasure, the following might help to come to a decision by yourself:

Review 1 by Paramanand Singh, one of the members of this platform.

Review 2 from MAA.

Proposal: Select two, three sections from Hardy's book which you might find interesting, read them and compare it with corresponding sections from books you are familiar with.

It depends on your taste and interests. Many old books are fantastic, the age is not a problem. I personally dislike Hardy’s book and Hardy himself because I do not believe in “pure Mathematics”.  Hardy’s disregard for applications seems to me arrogant and childish. You can learn Math from many other books that include nice motivations from Physics, Engineering etc. The older texts by Hermite, Goursat etc. Are more interesting in my opinion.


Quadratic Diophantine equation in three variables How would one determine solutions to the following quadratic Diophantine equation in three variables:

$$x^2 + n^2y^2 \pm n^2y = z^2$$

where n is a known integer and $x$, $y$, and $z$ are unknown positive integers to be solved.
Ideally there would be a parametric solution for $x$, $y$, and $z$.

[Note that the expression $y^2 + y$ must be an integer from the series {2, 6, 12, 20, 30, 42 ...} and so can be written as either $y^2 + y$ or $y^2 - y$ (e.g., 12 = $3^2 + 3$ and 12 = $4^2 - 4$).  So I have written this as +/- in the equation above.]

Thanks,

Considering this as a quadratic in $y$, it is sufficient (and necessary) that

$n^4 - 4n^2(x^2-z^2)$ is a perfect square.

i.e.

$n^2 - 4(x^2 - z^2)$ is a perfect square, say $q^2$.

Rewriting gives us

$(n-q)(n+q) = 4 (x^2 - z^2)$

If $n=2k$ is even, then $q$ needs to be even too (say $2m$) and

$(k-m)(k+m) = (x-z)(x+z)$

If $n=2k+1$ is odd, then $q$ needs to be odd too (say $2m+1$) and

$(k-m)(k+m+1) = (x-z)(x+z)$

Thus you can pick any $m$, and try to factor the left hand side above (choosing the right one depending on the parity of $n$) into two terms of the same parity.

This is a good problem as it requires to write the (Diophantine) parametric representations of the variables.

This method can be extended to n variables; one way of stating this result is to say there is an algorithm to construct a n variable  quadratic system  and if our equation fits this construction then the solutions are read off trivial.

I will explain this later.


Existence of least squares solution to $Ax=b$ Does a least squares solution to $Ax=b$ always exist?

If you think at the least squares problem geometrically, the answer is obviously "yes", by definition.

Let me try to explain why. For the sake of simplicity, assume the number of rows of $A$ is greater or equal than the number of its columns and it has full rang (i.e., its columns are linearly independent vectors). Without these hypotheses the answer is still "yes", but the explanation is a little bit more involved.

If you have a system of linear equations

$$
Ax = b \ ,
$$

you can look at it as the following equivalent problem: does the vector $b$ belong to the span of the columns of $A$? That is,

$$
Ax = b \qquad \Longleftrightarrow \qquad \exists \ x_1, \dots , x_n \quad \text{such that }\quad x_1a_1 + \dots + x_na_n = b \ .
$$

Here, $a_1, \dots , a_n$ are the columns of $A$ and $x = (x_1, \dots , x_n)^t$. If the answer is "yes", then the system has a solution. Otherwise, it hasn't.

So, in this latter case, when $b\notin \mathrm{span }(a_1, \dots , a_n)$, that is, when your system hasn't a solution, you "change" your original system for another one which by definition has a solution. Namely, you change vector $b$ for the nearest vector $b' \in \mathrm{span }(a_1, \dots , a_n)$. This nearest vector $b'$ is the orthogonal projection of $b$ onto $\mathrm{span }(a_1, \dots , a_n)$. So the least squares solution to your system is, by definition, the solution of

$$
Ax = b' \ ,  \qquad\qquad\qquad   (1)
$$

and your original system, with this change and the aforementioned hypotheses, becomes

$$
A^t A x = A^tb  \ . \qquad\qquad\qquad    (2)
$$

EDIT. Formula (1) becomes formula (2) taking into account that the matrix of the orthogonal projection onto the span of columns of $A$ is

$$
P_A = A(A^tA)^{-1}A^t \ .
$$

(See Wikipedia.)

So, $b' = P_Ab$. And, if you put this into formula (1), you get

$$
Ax = A(A^tA)^{-1}A^tb \qquad \Longrightarrow \qquad A^tAx = A^tA(A^tA)^{-1}A^tb = A^tb \ .
$$

That is, formula (2).

I do not think it is true because solving finding a least squares solution amounts to solving A^{T}Ax=A^{T}b, and that A^{T}A might not always be invertible. If A^{T}A  is not invertible it follows, from the invertible matrix theorem, that the transformation it represents is neither onto nor one-to-one  .


Why must a normed space homeomorphic to a complete metric space be complete? Why must a normed space $X$ homeomorphic to a complete metric space $Y$ be complete?

I've solved this in the case where Y is a normed space (considering open balls gives Lipschitz equivalence), but am not sure how to proceed in general. Can we use an adapted open-balls argument even when Y isn't normed?

Maybe there's an equivalent condition for completeness that will come in useful? In a normed space we can use (absolutely convergent)$\implies$(convergent) for series, but this won't work in a metric space.

I've tried using an explicit homeomorphism $f:X\rightarrow Y$ and taking pullbacks and pushforwards along $f$ of a Cauchy sequence in $X$, but without success.

Many thanks for any help with this!

I don't know if this works, but some thoughts:

It is well known that a metric space $(X,d_X)$ can be homeomorphic to a complete metric space $(Y,d_Y)$ and the metric $d_X$ on $X$ is not complete. E.g. $X = (0,1)$ in the standard metric, compared with $Y = \mathbb{R}$ in the standard (complete) metric.
However, for such spaces $X$ it is the case that they are a $G_\delta$ in their completion (they are what is known as topologically complete).

Every normed space $(X,\|\cdot\|)$ has a normed completion $(\hat{X},\|\cdot\|)$ in which $X$ embeds isometrically. So we know from the above that $X$ is then a $G_\delta$ in $\hat{X}$. We want to show that these spaces are actually equal.

So maybe it is possible to show that a dense $G_\delta$ linear subspace of a Banach space is the whole space?

Added (after comments by Sean Eberhard)
This is indeed the case: $X$ is a dense $G_\delta$ in $\hat{X}$, so if $y \in \hat{X}\setminus X$, $y + X$ and $X$ are both dense $G_\delta$'s and their intersection is empty (otherwise $y$ would a difference of members of $X$ hence in $X$). This contradicts Baire's theorem, so such a $y$ cannot exist and $X = \hat{X}$, and thus is complete already.

This question is not true:
$\mathbb{R}$ is homeomorphic to $(0,1)$. $\mathbb{R}$ is a complete metric space whit Euclidean norm and $(0,1)$ is a normed vector space which isn't complete!


showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ I studied the cyclotomic extension using Fraleigh's text.

To prove that Galois group of the $n$th cyclotomic extension has order $\phi(n)$( $\phi$ is the Euler's phi function.), the writer assumed, without proof, that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$.

I know that for n=p, p is prime, $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion.

But I don't know how $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ when n is not prime.

The proof which follows is the one provided by Gauss and it uses modular arithmetic in very ingenious way. We will summarize the results needed as follows:

1) For a given prime $ p$, the numbers $ 0, 1, 2, \ldots, (p - 1)$ form a finite field under the operations of addition and multiplication modulo $ p$.

2) Since these numbers form a field, say $ F_{p}$, we can talk about polynomials $ f(z)$ whose coefficients are in $ F_{p}$. The set of all such polynomials, say $ F_{p}[z]$, has the unique factorization property i.e. any such polynomial can be factored as a product of irreducible polynomials in $ F_{p}[z]$ in a unique way apart from the order of the factors. The proof is same as that used for normal polynomials with rational coefficients.

3) If $ f(z)$ is a polynomial in $ F_{p}[z]$ then $ \{f(z)\}^{p} = f(z^{p})$. This is true for constant polynomials by Fermat's theorem which says that $ a^{p} \equiv a\,\,\text{mod} (p)$. For higher degree polynomials this is achieved by induction by writing $ f(z) = az^{n} + g(z)$ and using binomial theorem to raise both sides to power $ p$. In so doing we only need to note that the binomial coefficients involved are divisible by $ p$.

The proof of irreducibility of $ \Phi_{n}(z)$ is done in two stages:

Stage 1:

Let $ \zeta$ be a primitive $ n^{th}$ root of unity and let $ f(z)$ be its minimal polynomial i.e. $ f(z)$ is monic (leading coefficient $ 1$), has rational coefficients and irreducible and $ f(\zeta) = 0$. Since $ \zeta$ is also a root of $ z^{n} - 1 = 0$, it follows that $ f(z)$ divides $ (z^{n} - 1)$ and by Gauss Lemma $ f(z)$ has integer coefficients as well. We now establish the following:

If $ p$ is any prime which does not divide $ n$ then $ \zeta^{p}$ is a root of $ f(z) = 0$.

Proof: Since $ \zeta$ is also a root of $ \Phi_{n}(z) = 0$ it follows that $ f(z)$ divides $ \Phi_{n}(z)$. Thus we have $ \Phi_{n}(z) = f(z)g(z)$ where $ g(z)$ is also monic and has integer coefficients (by Gauss Lemma). Since $ p$ is coprime to $ n$ it follows that $ \zeta^{p}$ is also a primitive $ n^{th}$ root. And therefore $ \Phi_{n}(\zeta^{p}) = 0$.

Assuming that $ \zeta^{p}$ is not a root of $ f(z) = 0$ (otherwise there is nothing to prove), we see that it must be a root of $ g(z) = 0$. Therefore $ \zeta$ is a root of $ g(z^{p}) = 0$. Since $ f(z)$ is the minimal polynomial of $ \zeta$, it follows that $ f(z)$ divides $ g(z^{p})$ so that $ g(z^{p}) = f(z)h(z)$ where $ h(z)$ is monic with integer coefficients. Also since $ \Phi_{n}(z)$ is a factor of $ (z^n - 1)$ so that we have $ z^{n} - 1 = \Phi_{n}(z)d(z)$ where $ d(z)$ is again monic with integer coefficients. We thus have the following equations:
$$z^{n} - 1 = f(z)g(z)d(z)\tag{1}$$
$$g(z^{p}) = f(z)h(z)\tag{2}$$
We now apply the modulo $ p$ operation to each of the equations above, i.e. we replace each coefficient in the polynomials involved with its remainder when it is divided by $ p$. The resulting polynomials are all in $ F_{p}[z]$ and we will use the same letters to denote them. So the above equations are now to be interpreted as relations between some polynomials in $ F_{p}[z]$. The equation $(2)$ can now be equivalently written as
$$\{g(z)\}^{p} = f(z)h(z)\tag{3}$$
Let $ k(z)$ in $ F_{p}[z]$ be an irreducible factor of $ f(z)$. Then from the above equation $(3)$, $ k(z)$ divides $ \{g(z)\}^{p}$ and so divides $ g(z)$. Thus from equation $(1)$ the polynomial $ \{k(z)\}^{2}$ divides $ (z^{n} - 1)$. Thus $ (z^n - 1)$ has repeated factors and therefore $ (z^{n} - 1)$ and its derivative $ nz^{n - 1}$ must have a common factor. Since $ n$ is coprime to $ p$, therefore the derivative $ nz^{n - 1}$ is non-zero polynomial and it clearly does  not have any common factor with $ (z^{n} - 1)$.

We have reached a contradiction and therefore the initial assumption that $ \zeta^{p}$ is not the root of $ f(z) = 0$ is wrong. The result is now proved.

Stage 2:

We have thus established that if $ f(z)$ is the minimal polynomial for any primitive $ n^{th}$ root of unity then for any prime $ p$ not dividing $ n$, $ \zeta^{p}$ (which is again a primitive $ n^{th}$ root) is also a root of $ f(z) = 0$. And since $ f(z)$ is irreducible and monic, it will act as minimal polynomial for the primitive root $ \zeta^{p}$.

The same logic can be applied repeatedly and we will get the result that $ f(z)$ is the minimal polynomial for $ \zeta^{p_{1}p_{2}\ldots p_{m}}$ where $ p_{1}, p_{2}, \ldots, p_{m}$ are any primes not dividing $ n$. It follows that $ \zeta^{k}$ where $ k$ is coprime to $ n$ is also a root of $ f(z)$. Thus all the primitive $ n^{th}$ roots of unity are roots of $ f(z) = 0$. Hence $ \Phi_{n}(z)$ divides $ f(z)$. Since $ f(z)$ is irreducible it follows that $ f(z) = \Phi_{n}(z)$ (both $ f(z)$ and $ \Phi_{n}(z)$ are monic). We thus have established that $ \Phi_{n}(z)$ is irreducible.

Update: User Vik78 points out in comments that Gauss had proved the irreducibility of $\Phi_{n}(z)$ for prime values of $n$ only and it was Dedekind who established it for non-prime values of $n$. Note that the problem is far simpler when $n$ is prime (one easy proof is via Eisenstein's criterion of irreducibility) and the crux of the argument presented above is essentially due to Gauss. I came to know of this proof from the wonderful book Galois Theory of Algebraic Equations by Jean Pierre Tignol.

Tignol mentions in his book that Gauss proved the irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$ for non-prime values of $n$ in 1808 (see section 12.6 titled Irreducibility of cyclotomic polynomials on page 196 in the book mentioned in last paragraph). Dedekind on the other hand proved an even more powerful theorem:

  Theorem: Let $\zeta_{m}$ denote a primitive $m^{\text{th}}$ root of unity. If $m, n$ are relatively prime then $\Phi_{n}(z)$ is irreducible over the field $\mathbb{Q}(\zeta_{m})$.

And Gauss used this result implicitly (thinking that it could be proved in similar manner as irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$) in obtaining solution to $\Phi_{n}(z) = 0$ via radicals.

The simple proof, is to show that a span $\mathbb{S}$ of a finite set closed to multiplication, when intersected with $\mathbb{Q}$ gives $\mathbb{Z}$.  Since cyclotomic numbers are of this form, one can show that the span of such a set contains no fractions.

The proof is based on an attempt to construct a set $S$ whose span includes a fraction.  Since one can show that some $\frac 1{z^n}$ can not belong to $S$, then the intersection between $S$ and $Q$ is the same as between $S$ and $Z$.

Since the cyclotomic numbers form a span closed to multiplication, and the base set is finite, the intersection of any cyclotomic set and the rationals is the same as that of the integers.

Since also, one can show that the chords of a polygon $\{p\}$, which can be constructed from the cyclotomic numbers, are governed by this rule, and since this contains the chords of every rational angle, ditto.

If one takes the product of $a-\operatorname{cis}(x/n)$ for x=1 to n, it gives an algebraic equation of the form $a^n-1$, which has a unique factor for every $m \mid n$.  If the GCD of $x, n$, is greater than 1, then the particular root occurs at a lesser n.  It only gives the unique factor for $n$, if the GCD is 1.  Since the Euler totient gives the number of co-primes to n between $0$ and $n$, this is the order of the equation.

The second stage of the proof is to show that the equation is irreducable.  This is done by a process of automorphism: that the set $\operatorname{cis}(x/n)$ (where the gcd of x, n is 1), is identical to the set $\operatorname{cis}(xy,n)$.  This particular process has the effect that if a function $F(x) = F(xy)$, where F is some element in the span of $\operatorname{cis}(x,n)$, for all y, then F(x) must belong in the set $Z$.

The span is then a reduction of a sparse array in $\phi(n)$D onto  $2$D, which means that the equations are irreducable.


Probability of consecutive events This question refers to the formula by @leonbloy {Occurrence of 5 consecutive tails before occurrence of 2 consecutive heads}

Why is the formula asymmetric, i.e. the order of head/tail as arguments matters (e.g. 5 head + 2 tails vs. 5 tails + 2 heads), while $p_h$ (probability of head) is equal to $p_t$ (probability of tail) given the implicit assumption of a fair coin ?

Also what would it be if $p_t ≠ p_h$ ?

Your thoughts appreciated.

It might be better to say:

  The probability of getting $a$ runs of one particular side of the coin before getting $b$ runs of the other side of the coin is $$P(E) = \frac{2^b-1}{2^a + 2^b - 2}$$

The situation is symmetric in the sense that it doesn't matter what sides of the coin you're talking about --- heads or tails. But it does matter that the runs of $a$ come before the runs of $b$.

Intuitively, $a$ and $b$ are not symmetric because if you increase $a$ (the number of runs you have to get first), you expect the likelihood of success to go down. If you increase $b$ (the number of runs you have to avoid getting before succeeding at $a$), you expect the likelihood of success to go up:

It is rather likely to get a run of 2 before your first run of 1000. It is rather unlikely to get a run of 1000 before your first run of 2. This is true regardless of whether you're talking about heads/tails or tails/heads.

Let's generalize to weighted coins. Suppose you want to know the probability of getting $a$ runs of one side of the coin before getting $b$ runs of the other. Suppose the one side of the coin comes up with probability $\alpha$, and the other side of the coin comes up with probability $\beta = 1-\alpha$.

Then, following that post:

$$\begin{align*}
P(E) &= P(E|T)P(T) + P(E|T^C)P(T^C)\\
& = \frac{p_1}{1-q_1q_2}\alpha + \frac{p_1q_2}{1-q_1q_2}(1-\alpha)\\
& = [q_2 + (1-q_2)\alpha] \frac{p_1}{1-q_1q_2}
\end{align*}$$

and here $p_1$ and $q_1$ are the probability of succesfully/unsuccessfully getting a run of $a$, and similarly for $p_2, q_2$.

We have $p_1 = \alpha^{a-1}$, $p_2 = (1-\alpha)^{b-1}$, $q_1 = 1-p_1$, $q_2 = 1-p_2$.

$$\begin{align*}
P(E) & = [q_2 + (1-q_2)\alpha] \frac{p_1}{1-q_1q_2} \\
&= [1-(1-\alpha)^{b-1} + (1-\alpha)^{b-1}\alpha] \frac{p_1}{1-q_1q_2}\\
&= [1 + (\alpha-1)(1-\alpha)^{b-1}] \frac{p_1}{1-q_1q_2}\\
&= [1 - (1-\alpha)^{b}]\frac{\alpha^{a-1}}{1-(1-\alpha^{a-1})(1-(1-\alpha)^{b-1})}
\end{align*}$$


Estimating a certain row of Pascal's triangle I need to calculate all the numbers in a certain row of Pascal's triangle. Obviously, this is easy to do with combinatorics. However, what do you do when you need to estimate all the numbers in, say, the 100000th row of Pascal's triangle?

Is there any way to estimate the number so that the costly multiplications and divisions of binomials can be avoided? I'm already estimating factorials with Stirling's formula, but it still takes a number of seconds to calculate only one number - and I need about 100000/2 (since a row is symmetric).

Are you computing the factorials and then dividing?  Don't do that.  You can combine the estimates you get from Stirling's formula into

$${n+m \choose n} \approx \sqrt{ \frac{m+n}{2 \pi mn} } \frac{(m+n)^{m+n}}{m^m n^n}$$

and even then you can optimize, for example rewriting the second factor as

$$\left( 1 + \frac{n}{m} \right)^m \left( 1 + \frac{m}{n} \right)^n$$

and, depending on the relative sizes of $m$ and $n$, applying the approximation $\left( 1 + \frac{x}{n} \right)^n \approx e^x$.  This will work if one of $m$ and $n$ are large compared to the other and in the intermediate case the above powers should be straightforward to compute.  See also Exercises 1 and 2 at this blog post by Terence Tao on the subject.

Depending on what kind of precision you need you should consider just working with the logarithms and not with the numbers directly.

The entries in the $n$th row of Pascal's triangle (for large $n$) follow a Gaussian curve quite closely, as a consequence of the central limit theorem. See the Wikipedia article about the binomial distribution.


Find the contour integral around unit circle. Evaluate the below integral by turning it into a contour integral around a unit circle:

$$\int_{0}^{\pi}\frac{\cos2\phi}{1-2a \cos\phi + a^2} d\phi$$ $where\;a\neq \pm1$

Let $\gamma(a) = -\frac{1+a^2}{2a}$. Since the integral is even, we can write it as
$$
\int_0^{\pi}\frac{\cos(2\phi)}{1-2a\cos(\phi)+a^2}d\phi=\frac{-1}{4a}\int_0^{2\pi}\frac{\cos(2\phi)}{\cos(\phi)+\gamma}d\phi
$$
Recall that $\cos(\phi) = \frac{e^{i\phi}+e^{-i\phi}}{2}$. Then let $z = e^{i\phi}$ and $\cos(2\phi)=\Re\{e^{2i\phi}\}$ so $dz=ie^{i\phi}d\phi$ or $d\phi = \frac{dz}{iz}$. Therefore, we can write the integral now as
$$
\frac{-1}{2ai}\oint_C\frac{z^2}{z^2+2\gamma z+1}dz
$$
The poles occur when $z=-\gamma\pm\sqrt{\gamma^2-1}$. Plugging in $\gamma(a) = -\frac{1+a^2}{2a}$, we get that the poles are at $z=a,1/a$. By the Residue theorem,
\begin{align}
\int_0^{\pi}\frac{\cos(2\phi)}{1-2a\cos(\phi)+a^2}d\phi &=
\begin{cases}
\frac{-\pi}{a}\lim_{z\to a}(z-a)\frac{z^2}{z^2+2\gamma z+1},&\lvert a\rvert < 1\\
\frac{-\pi}{a}\lim_{z\to 1/a}(z-1/a)\frac{z^2}{z^2+2\gamma z+1},&\lvert a\rvert > 1
\end{cases}\\
&=
\begin{cases}
\Re\Bigl\{\frac{a^2\pi}{a^2-1}\Bigr\},&\lvert a\rvert < 1\\
\Re\Bigl\{\frac{\pi}{a^2(a^2-1)}\Bigr\},&\lvert a\rvert > 1
\end{cases}
\end{align}

First observe that by the cosines law (using the isomorphism between $\mathbb{R}^2$ and $\mathbb{C}$):
$$
\|1-ae^{ix}\|^2 = \|1\|^2 + \|a\|^2 -2\|1\|\cdot \|a\|\cos x =  1 + a^2 -2a\cos x.
$$
So the denominator is measuring the squared distance between the elements of the circle of radius $a$ centered at the origin and $1$.

On the other hand, the image of $2[0,\pi]$ is $2\pi$, so we only need to reparametrize the numerator.

So, parametrizing the unit circle as $\gamma(x)=e^{ix}$, we get that:
$$
I
= \int_0^{2\pi} \frac{\cos x}{\|1-ae^{i\frac{x}{2}}\|^2}\left(-ie^{-ix}\right)\left(ie^{ix}\right) dx
= -i\int_{S^1} \frac{\Re(z)}{z\|1-a\sqrt{z}\|^2} dz.
$$


Deriving $\sin{ix}=i\sinh{x}$ Derive $\sin{ix}=i\sinh{x}$ from $(5)$. What is $\sin{i}$?
  $$\cos{x}=\frac{1}{2}\left(e^{ix}+e^{-ix}\right)\quad\text{and}\quad\sin{x}=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag 5$$

We have
$$\begin{align}
\sin{ix}&=\frac{1}{2i}\left(e^{i^2x}-e^{-i^2x}\right)\\
&=\frac{1}{2i}\left(e^{-x}-e^x\right)\\
&=-\frac{\sinh{x}}{i}
\end{align}$$
It does not look like what we are deriving. What went wrong?

Multiply
 the denominator and the numerator by $i.$

We know that $sinh(ix)=i sin(x)$.
                          Put $i=ix$
                   $$Sinh(i.ix)=isin(ix)$$
                  $$ Sinh(-x)=i sin(ix)$$
                   $$-sinhx=i sin(ix)$$
                  $$i\times isinhx=i sin(ix)$$
                    $$i sinhx=sin(ix)$$


Limit problem, two cases, how to solve? $\lim_{x\to1}\frac{x + \sqrt{x}}{\sqrt{x-1}}$
$\lim_{x\to1}\frac{x - \sqrt{x}}{\sqrt{x-1}}$

Lately I've been trying to satisfy some curiosity about the nature of limits and I found this example, it's really bugging me that I can't solve it. It seems so simple, yet when I attempt to solve it, it's like I'm trapped in a loop. Everything I do, undos the previous action.

I'll take anything you can give, a solution, a good hint... I tried manipulating the expression without changing it, but I can't get through it.

The numerator tends toward $2$ while the denominator tends toward $0$; the form $2/0$ is not indeterminate at all, so no energy needs be spent beating it into shape when we can see for a fact just from this that the limit is $\infty$ (i.e. for any $N>0$ as large as you want there is a neighborhood of the argument $x=1$ for which the ratio is larger than $N$ throughout the entire neighborhood).

In the alternate situation, with a minus sign we can multiply/divide by the numerator's conjugate,

$$\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x-1}}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}\sqrt{x-1}}{\sqrt{x}+1}.$$

This allows for us to directly plug in $x=1$.

1+√1=2; however √1−1=0

2/0 -> infinity
This simple...


Rings $R/I \cong h(R)/h(I)\,$ for injective ring hom $h$ I still struggle with elementary knowledge about quotient rings.

For example, why can I just write down the second following isomorphism:

$$\frac{\mathbb{C}[x, y, z]}{(x^2 + y^2 - 1, y \pm iz)} \simeq \frac{\frac{\mathbb{C}[x, y, z]}{(y \pm iz)}}{\frac{(x^2 + y^2 - 1,y \pm iz)}{(y \pm iz)}} \simeq \frac{\mathbb{C}[x, y]}{(x^2 + y^2 - 1)} $$

I know the first isomorphism is right by the third isomorphism theorem. But why can I write down the second one?

It follows from the first isomorphism theorem for rings, like this:

  Suppose $\phi: R\to S$ is any injective ring homomorphism. Then $R/I\cong \phi(R)/\phi(I)$ as rings.

  Proof:
  Let $\psi: R\to \phi(R)/\phi(I)$ be given by $\psi(r)=\phi(r) + \phi(I)$. It can be checked that this is a ring homomorphism.

  Moreover, its kernel is precisely $I$. Just chase it down: $\psi(r)=0+\phi(I)$ iff $r-i\in \ker \phi$, and $\phi$ was injective so...

Now, it's fairly obvious that there is an isomorphism between the rings in the "numerators" of the latter half, and the denominator of the second quotient is the image of the denominator in the first quotient, so you can apply this lemma above.

Well, let $R$ be a commutative ring with ideals $J\subseteq I$. Then $R/I$ is isomorphic to $(R/J)/(I/J)$.

In the first isomorphism, you use $I=\langle x^2+y^2-1, y\pm iz\rangle$ and $J=\langle y\pm iz\rangle$. In the second isomorphism, you use the same result.


Finding a direction vector Let $u ∈ \mathbb{R}^2$ be the vector $\begin{pmatrix}3\\-4\end{pmatrix}$
. Let $l$ be the line with equation
$u · v = 50.$

Express $l$ in parametric form.

Solution: A direction vector for $l$ is $\begin{pmatrix}4\\3\end{pmatrix}$
. The point $\begin{pmatrix}6\\-8\end{pmatrix}$
lies on $l$. Hence $l$ consists of all points
of the form

$\begin{pmatrix}6\\-8\end{pmatrix}$ $+ λ\begin{pmatrix}4\\3\end{pmatrix}$.

What I want to know is how is it immediately known that $\begin{pmatrix}4\\3\end{pmatrix}$ is a direction vector for $l$? I thought you would have to find two points on $l$, say $x=\begin{pmatrix}6\\-8\end{pmatrix}$ and $y=\begin{pmatrix}18\\1\end{pmatrix}$ and compute $y-x=\begin{pmatrix}3\\9\end{pmatrix}$ Is this still correct?

Your longer way is correct, only notice that $y-x=\begin{pmatrix}12\\9\end{pmatrix}$.

But any vector orthogonal to $u$ will do, because if $x$ and $y$ both satisfy the equation of the line, then:
$$
u\cdot(x-y)= u\cdot x-u\cdot y =0.
$$
And if $u=\begin{pmatrix}a\\b\end{pmatrix}$, an obvious vector orthogonal to it is
$\begin{pmatrix}-b\\a\end{pmatrix}$.

Let $v = \begin{pmatrix}x\\ y\end{pmatrix}.$ Then, the line $l$ is described as follows:

$$u \cdot v = 50 \Rightarrow x - 4y = 50.$$

Hence, a parametric representation is:

$$\begin{cases}
x = 50 -4 k \\
y = k
\end{cases},
$$

with $k \in \mathbb{R}$.


Cosets and finite groups of orthogonal operators on the plane Can someone show me how to compute the left cosets of the subgroup $H=\{1, x^5\}$ in the Dihedral group $D_{10}$

I know that $D_{10}$ is generated by two elements $x$ and $y$ such that: $$x^{10}=1, y^2=1 \text{ and } yx=x^{-1}y$$

This is -essentially- a straight forward calculation, using the fact that for two elements $a, b\in D_{10}$ we have that $a\in bH$ if and only if $b^{−1}a\in H$. Each coset contains precisely two elements (why?)

For example, $xH=x^6H$, because $x^6\in xH$ as $x^{-1}x^6=x^5\in H$. Then because each coset contains two elements we have that $xH=\{x, x^6\}$.


Continuous function and Borel sets If $f:\mathbb{R}^p \rightarrow \mathbb{R}$ is continuous and $B\subset \mathbb{R}$ is Borel set, how to show that $f^{-1}(B)$ is also Borel set?

I was trying to construct a $\sigma$-ring of sets whose preimage is Borel, but they're not necessarily all borel in that sigma ring...

What would be the main idea for the proof here?

Hint: prove that $\mathcal A := \{B \in B(\Bbb R): f^{-1}(B) \in B(\Bbb R^p)\}$ is a sigma algebra containing the open sets, where $B(\Bbb R^d)$ is the Borel sigma algebra on $\Bbb R^d$.

Added:

This way, $\tau \subset \mathcal A$ (where $\tau$ is the usual topology of $\Bbb R$), "applying $\sigma$ both sides" we get $\sigma(\tau)\subset \cal A$, therefore $B(\Bbb R)\subset \cal A$. Which means that each $B \in B(\Bbb R)$ is in $\cal A$, i.e. $f^{-1}(B) \in B(\Bbb R^p)$.

A Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement.

As the inverse images of open sets under a continuous function are open sets and inverse images of a countable union is the countable union of the inverse images, same for countable intersection and relative complement. We get the the inverse image of a Borel set under a continuous function is a Borel set.


Why can't calculus be done on the rational numbers? I was once told that one must have a notion of the reals to take limits of functions. I don't see how this is true since it can be written for all functions from the rationals to the rationals, which I will denote $f$, that
$$\forall L,a,x:(L,a,x\in\mathbb{Q})\forall \epsilon:(\epsilon>0)\space \exists\delta:(\delta>0)$$
$$\lim_{x\rightarrow a}f(x)=L\leftrightarrow(\mid x-a\mid<\epsilon\leftrightarrow\mid f(x)-L\mid<\delta) $$
Since, as far as I know, functions like $\mid x\mid$ and relations like $<$ can be defined on the rationals. Is it true you couldn't do calculus on just the rational numbers? At the moment I can't think of any rational functions that differentiate to real functions. If it's true that it isn't formally constructible on the rationals, what about the algebraic numbers?

Edit

Thanks for all the help, but I haven't seen anyone explicitly address whether or not we could construct integrals with only algebraic numbers. Thanks in advance to anyone who explains why or why not this is possible.

$\newcommand{\QQ}{\mathbb{Q}}$
Derivatives don't really go wrong, it's antiderivatives.
(EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.)

Consider the function $f : \QQ \to \QQ$ given by
$$f(x) = \begin{cases} 0 & x < \pi \\ 1 & x > \pi \end{cases}$$

This function is continuous and differentiable everywhere in its domain. If $x < \pi$, then there's a neighborhood of $x$ in which $f$ is a constant $0$, and so it's continuous there, and $f'(x) = 0$. But if $x > \pi$, there's a neighborhood of $x$ in which $f$ is a constant $1$, so it's continuous there too, and $f'(x) = 0$ again.

So the antiderivatives of $0$ can look rather messy. By adding functions like this, you can construct arbitrarily "jagged" functions with zero derivative. As you can imagine, this completely destroys the Fundamental Theorem of Calculus, and any results that follow from it.

This can happen in the real line to some extent, but it's not nearly as bad. The traditional antiderivative of $1/x$ is $\ln|x| + C$. But so is the following function:
$$ g(x) = \begin{cases} \ln x + C_1 & x > 0 \\ \ln(-x) + C_2 & x < 0 \end{cases} $$

By changing $C_1$ and $C_2$, we can push the two halves of the real line around completely independently. This is only possible because $1/x$ isn't defined at $0$, and so we've "broken" the real line at that point.

If you like dumb physical metaphors, here's one:

The real line is kind of like an infinite stick. If you wiggle a section of it, the whole thing must move.

With the $1/x$ example, you've made a cut at $x = 0$, and now you have two half-sticks. They can be wiggled independently, but each half must still move as a unit.

The rational numbers are more like a line of sawdust. You can't really move one grain by itself, but you can certainly take an interval and move it around independent of its neighbors.

By completing the rationals, you're adding all the glue between the grains to form a stick again. (I hope no one from diy.stackexchange is reading this...)

On the rational numbers you can not really do Algebra, much less Calculus.

Since the rational numbers are not "complete", you can not for instance define the square root function. Similarly you can not define many other functions, e.g. power functions with non-integer exponent, ln(x), log(x), exp(x), sin(x), cos(x) etc.

Even if you restrict yourself only to functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$, you can not always take limits, derivatives, antiderivatives etc. as pointed out in other answers.

So, the problem is a lot more fundamental than "you'll get no theorems out".


Discontinuous limit function and uniform convergence The sequence I am working with is this, in the interval [-1,1]:
$f_n (x)= \begin{cases}-1,-1\leq x\leq-\frac{1}{n}\\[2ex]nx, -\frac{1}{n}<x<\frac{1}{n} \\[2ex]1, \frac{1}{n}\leq x \leq1\end{cases}$
We can show pretty easily that this converges pointwise to the function
$f (x)= \begin{cases}-1, x∈[-1,0[\\[2ex]0, x=0 \\[2ex]1, x∈]0,1]\end{cases}$
I would like to find the intervals in which this converges uniformly. It can't converge uniformly to a discontinuous function, so obviously we have no uniform convergence in the interval [-1,1]. However, when it comes to showing convergence in ]0,1], can I somehow use the uniform limit theorem here as well to show that the sequence cannot converge uniformly?

You’re right that you can use the uniform convergence theorem (in the contrapositive) to rule out uniform convergence. On all of $[-1,1]$, uniform convergence + continuity of the $f_n$ would imply continuity of $f$, which is false as you point out. To answer your question about $(0,1]$ specifically, yes the same argument works. Uniform convergence in $(0,1]$ plus ordinary convergence at $0$ would imply uniform convergence on $[0,1]$ (hopefully it’s clear why; adding a finite set does not affect the uniformity). But the limiting function is not continuous on $[0,1]$ so the convergence could not have been uniform.

hint
For $ x\in (0,1]$ and $ n $ great enough, we will have $$x>\frac 1n \;and \; f_n(x)=1$$
thus
$$f(x)=\lim_{n\to+\infty}f_n(x)=1$$
then
$$|f_n(x)-f(x)|=0 \; if \; x\ge \frac 1n $$
and
$$|f_n(x)-f(x)|=1-nx \; if \; 0<x<\frac 1n$$


For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$ For all $x$ which are real numbers, prove that
  $$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$

I know that

Let $\lfloor x\rfloor = n$

$n \leq x < n+1$

I am assuming $\lfloor x\rfloor = n$

Let's take two cases - one - the fractional part of $x$ is less than $0.5$  LHS will be $2n$ RHS will be $n+n$

case two - The fractional part of $x$ is greater than $0.5$LHS will be $2n+1$ RHS will be $n+n+1$

Hence, proved.


Prime Counting Function Difference If $\pi\left(x\right)$ is the prime counting function, is anything known about the number of x values satisfy
$$\pi\left(2x\right)-\pi\left(x\right)=1$$
Sorry if this is a stupid question, but I am just curious about the size of such $x$ values and perhaps some properties that they may have. Thanks.

As you know, Bertrand's postulate states that $\pi(2x) - \pi(x) \ge 1$ for all $x \ge 2$. A refinement due to Nagura in 1952 states that as long as $x \ge 25$ then $\pi(6x/5) - \pi(x) \ge 1$. But then $\pi(36x/25) - \pi(6x/5) \ge 1$ too, and since $36/25 < 2$ leads to $\pi(2x) - \pi(x) \ge 2$. This can be improved significantly, but as far as your question goes, the values of $x$ satisfying $\pi(2x) - \pi(x) = 1$ are very limited in number.

The prime counting function is defined for real numbers $x$.  The exact set of real numbers $x$ for which $\pi(2x) - \pi(x) = 1$ is the union of the following intervals:

$$[1,1.5) \cup [2,2.5) \cup [3,3.5) \cup [5,5.5).$$

You can see a sketch of this in the graph (Wolfram Alpha).


For any prime $p > 3$, why is $p^2-1$ always divisible by 24? Let $p>3$ be a prime. Prove that $24 \mid p^2-1$.

I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.

The most elementary proof I can think of, without explicitly mentioning any number theory: out of the three consecutive numbers $p - 1$, $p$, $p + 1$, one of them must be divisible by $3$; also, since the neighbours of p are consecutive even numbers, one of them must be divisible by $2$ and the other by $4$, so their product is divisible by $3 · 2 · 4 = 24$ — and of course, we can throw $p$ out since it's prime, and those factors cannot come from it.

As Unnikrihnan stated, every prime > 3 is of the form 6n + 1, or 6n − 1.

So, we could write the problem as ((6n ± 1)^2 - 1) /24.
Solving, we have: ((36n^2 ± 12n +1) – 1)/24;

Which, is equal to (36n^2 ± 12n)/24, and also equal to (3n^2 ± n) / 2, and to
n (3n  ± 1) / 2.

Since we are not interested on the numerical value of this equation, but that the result of the equation to be an integer, not a fraction, the n outside of the parenthesis is irrelevant.

So the equation could be written as, (3n ± 1)/2, and the result of this equation is always an integer for any odd or even value of n.


Semicircular paper and creasing of a chord A semicircular piece of paper with radius $2$ $cm$ is folded along a chord so that the arc is tangent to the diameter.If the contact point of the arc divides the diameter in the ratio $3:1$,determine the length of the crease.

Sorry I can't show any work as I can't develop any approach to solve this question. Any suggestions?

Picture:

Hint: Let the  semicircle be centered at the origin with radius 2, and lie above the x axis, so that the point $P=(1,0)$ cuts the diameter in a 3:1 ratio. Then consider another circle also of radius 2 but centered at $(1,2)$ so it would touch the x axis (diameter) at P so be tangent to the diameter there.

Figure out by algebra the two points where the new circle goes through the original semicircle, and get the length of the chord via distance formula.

Calculation has two parts. Finding the segment on outside circle and the angle  $ \alpha $ subtended between fold/diameter.

By creasing or folding T falls on t

$$ r = nK = 2 ( 2 cm) ;\, OK = k ;$$

$$(k+r) ( k-r ) = OT^2 = Ot^2 = ( k-r/2)^2 \rightarrow  k = 5 r /4 $$

$$ on = om = OK- nK= r/4 ; \,  OT = Ot= on + nt= r/2 + r/4 = 3r/4 $$

dot marked angle $= \alpha , $

$$  \tan 2 \alpha = \ tan TOK  = TK/OT = 1/(3/4) = 4/3 $$

$$ \cos 2 \alpha = 3/5 ;\, \cos \alpha = 2/\sqrt 5 ;\, $$

$$ OM = 5 r/4 \cdot 2/\sqrt 5 = \sqrt 5 r/2 ; $$

By Circle segment product property,

$$ OT^2 = OP \cdot OQ = (OM + MQ) (OM-MP) = OM^2 - x^2 $$

$$ (if \, MP = MQ = x) $$

$$ x^2 = OM^2 - OT^2  = 5 r^2/4 - (3 r/4)^2 ;\, \rightarrow x = \sqrt {11}/4 $$

$$ PQ = 2 x  = r \sqrt{11}/2 = \sqrt{11} cm \;  (Answer).\,$$


Contradictory equations Question: Find whether the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is a singleton.

My attempt: Rewriting the first equation will give us $x = \frac{1}{2}$.

The other two equations can be written as $x - y = 5$ and $x - y = 3$.

Now, the solution of these equations is $\emptyset$.
So we can say that the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is not singleton.

Kindly verify and throw some insights.

The equations you gave represent three lines in $\mathbb R^2$. When we have three (or more) lines in a plane, we can have  one of the following situations:

They have infinitely many common points.
They have one point in common.
They have no points in common.

The lines $2x=1$ and $y+5=x$ are clearly not parallel (can you see this?). So they intersect at one point, which means we are not dealing with case one. The lines will intersect at the point $(x,y)=\left(\frac12;-4\frac12\right)$.

Now the three lines will have one point in common if $\left(\frac12;-4\frac12\right)$ lies on the third line and no points in common when it doesn't. I will leave it to you to figure out if it is on the third line or not.

After edit from OP: We can solve this even quicker if we just look at the bottom two equations: $$\begin{cases} x=y+5\\ x=y+3\end{cases}$$ These lines are obviously parallel (and don't coincide). You could also argue that this implies $y+5=y+3$, which is a contradiction.

A vertical line ( first equation ) cuts two parallel lines ( last two ) transversely, creating two points of intersection or two solutions.

If there were three (non-concurrent) lines then there would be three points of intersection enclosing a triangle.


Three pumps filling up a tank "A large tank can be filled by 2 similar small pumps and 1 larger pump working together in 1 hour and 12 minutes. The larger pump takes one hour less than the smaller pump to fill up the tank alone. Find how long each pump takes, given that V/P = T where V = volume of the tank, P = Pump, T = Time."
These are the exact words from a mathematics exam question and I'm unsure how to interpret P (I assumed it was the rate). If someone could provide a solution, that would be great.

This reminds me of the painting a room word problem. Let's proceed as follows.
Suppose the large pump can fill the tank in $t_L$ hours working alone, and the small pump can fill the tank in $t_S$ hours working alone.
Then $t_L=t_S-1$ and $\frac{1}{t_S}+\frac{1}{t_S}+\frac{1}{t_L}=\frac{1}{1.2}$. Combining these equations and simplifying yields $\frac{2}{t_S}+\frac{1}{t_S-1}=\frac{1}{1.2}$ This implies $t_S=4$ and $t_L=3$. (Note: $t_S=3/5$ is extraneous).


How to prove that $(p-1)^2$ $\mid$ $(p-1)!$ when $p$ is a prime number and $p>5$? I say that $p-1$ $\mid$ $(p-1)!$ then I want to prove that $p-1$ $\mid$ $(p-2)!$.

I started by saying that $p-1$ is an even number so $2\mid (p-1)$ and that means that $\frac{p-1}{2}$ is an integer.

Then I just know that $\frac{p-1}{2}<p-1$ but I don't know what to do next.

Thanks in advance !

Since $(p-2)>\frac{p-1}{2}>2$ and $\frac{p-1}{2}\cdot 2=(p-1)$, $(p-1)$ divides:
$$ 1\cdot\color{red}{2}\cdot 3\cdot\ldots\cdot\color{red}{\frac{p-1}{2}}\cdot\frac{p+1}{2}\cdot\ldots\cdot(p-2)=(p-2)!.$$

In fact, if $m (m \ge 6)$ is an composite number, we have $(m - 1)! \equiv 0 \pmod m$.

Here, $m = p - 1$.


Remainders questions help If we divide a number by 3, 4 ,5 , 6 , we have  the remainders 2, 3 , 4 , 5. Is there any way to  get a pattern without guessing so many numbers and checking by 3, 4 ,5 ,6?

Hint: We are told that our number $x$ is congruent to $-1$ modulo $3$, $4$, $5$, and $6$.

This is first Modular Arithmetic problem I'm solving, I hope it's correct.
We have
$$-1\equiv3 \mod( y )$$
$$-1\equiv4\mod (y)$$
$$-1\equiv5\mod (y)$$
$$-1\equiv6\mod (y)$$

then
$$-1\equiv60\mod(y)$$

Any Number of the form $60p-1$, ( $p \in \mathbb N$) will have mentioned properties.


Duplicate - Proof by Ordinary Induction: $a^n-b^n \leq na^{n-1}(a-b)$ This question has already been asked:

Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$

However has not been answered properly. (Even thought the OP checked an answer)

The answers provided are direct proofs and not induction proofs.

Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:

$$a^n-b^n \leq na^{n-1}(a-b)$$

Anybody has ideas?

We do the induction step. Suppose that for a certain $k$ we have $a^k-b^k\le ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}\le (k+1)a^k(b-a)$.

We have
$$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).\tag{1}$$
Note that by the induction hypothesis $b(a^k-b^k)\le a(a^k-b^k)\le ka^k(a-b)$. So the sum on the right-hand side of (1) is $\le (k+1)a^k(a-b)$.

The proof proceeds by induction. Here we will need to make use of two base
cases, since it turns out that n = 1 needs to be treated separately.

(1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,

so the statement holds.

(2) Second base case (n = 2):
a^2 − b^2 = (a + b)(a − b)
< 2a(a − b)    since a > b
= na^(n−1)(a − b), for n = 2

(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 <
b < a. Then
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
< a^n(a − b) + ana^(n−1)(a − b) since b < a
= (n + 1)a^n(a − b)
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.


Understanding of what it means to say that a question is in NP Let's say the the function $f$ can be evaluated in polytime in the size of the input $x$.  Are the following problems in NP?

Is there an $x$ such that $f(x) = y$ for a particular value of $y$?

Find an $x$ such that $f(x) = y$ for a particular value of $y$?

Do either of these answers change when not all values of $y$ have a corresponding $x$?

It seems like the answer to both of these would be "yes" because your certificate is just that particular value of $x$ and then you can verify it by running $f$, but I'm not sure because the second question seems much more involved than the first.

I have since learned that the concept of NP applies to questions that have a yes or no answer, not questions that are requesting a particular number.  Therefore, my first question is in NP and the second is not.