Probability of traversing a network of bridges This is problem 32 from this wonderful site here: https://www.probabilitycourse.com/chapter1/1_5_0_chapter1_problems.php :

You would like to go from point $A$ to point $B$. There are $5$ bridges on different branches of the river.

Bridge $i$ is open with probability $P_i$. Let $A$ be the event that there is a path from $A$ to $B$ and let $B_k$ be the event that $k$-th bridge is open.

Find $P(A)$
Find $P(B_3\ |\ A)$

My approach so far is this: $P(A)=P(A\ |\ B_3)P_3 + P(A\ |\ B_3^c)(1-P_3)$. Focusing on $P(A\ |\ B_3^c)$, we see that it's possible to cross from $A$ to $B$ only via the upper or lower route, so both bridges alone either or both routes should be open. Event $A$ becomes $(B_1\cap B_4)\cup(B_2\cap B_5)$, so

$$P(A\ |\ B_3^c)=P((B_1, B_4)\cup(B_2, B_5))=P(B_1, B_4)+P(B_2, B_5)-P(B_1, B_4, B_2, B_5)
\\=P_1P_4+P_2P_5-P_1P_2P_4P_5
$$

If bridge $3$ is open, then either of bridge $1$ or $2$ should be open, and either of bridge $4$ or $5$ should be open, so

$$P(A\ |\ B_3)=P((B_1\cup B_2)\cap(B_4\cup B_5))=P(B_1\cup B_2)+P(B_4\cup B_5)-P(B_1\cup B_2\cup B_4\cup B_5)
\\=P(B_1)+P(B_2)-P(B_1,B_2)+P(B_4)+P(B_5)-P(B_4,B_5)-[P(B_1)+P(B_2)+P(B_4)+P(B_5)-P(B_1,B_2)-P(B_1,B_4)-P(B_1,B_5)-P(B_2,B_4)-P(B_2,B_5)-P(B_4,B_5)+P(B_1,B_2,B_4)+P(B_2,B_4,B_5)+P(B_1,B_4,B_5)+P(B_1,B_2,B_5)-P(B_1,B_2,B_4,B_5)]
\\=P_1P_4+P_2P_4+P_1P_5+P_2P_5+P_1P_2P_4P_5\bigg(1-\frac{1}{P_1}-\frac{1}{P_2}-\frac{1}{P_4}-\frac{1}{P_5}\bigg)$$

I could plug these values back into the first equation to get $P(A)$, but this method seems extremely cumbersome and I've probably either made a mistake, or I'm missing a way to simplify this. There are ways to go from A to B that are common to both when bridge $3$ is open or closed. Have I overcounted somehow?

There are five bridges and there are 32 states of the bridges.

The probabilities of the 32 states can be easily calculated.

The probability of event $A$ is the sum of those probabilities that belong to states that make it possible to get through.

The conditional probability can be calculated the same way -- listing states...


What is the smallest unknown natural number? There are several unknown numbers in mathematics, such as optimal constants in some inequalities.
Often it is enough to some estimates for these numbers from above and below, but finding the exact values is also interesting.
There are situations where such unknown numbers are necessarily natural numbers, for example in Ramsey theory.
For example, we know that there is a smallest integer $n$ such that any graph with $n$ vertices contains a complete or an independent subgraph of 10 vertices, but we don't know the exact value of $n$.

What kinds of unknown small (less than 100, say) integers are there?
What are the smallest unknown constants which are known to be integers?
Or, more rigorously, what is the smallest upper bound for an unknown but definable number that is known to be an integer?

I know that asking for the smallest unknown integer is ill-defined since we do not know the exact values.
The more rigorous version of the question is well-posed, but I do not want to keep anyone from offering interesting examples even if they are clearly not going to win the race for the lowest upper bound.

An answer should contain a definition of an integer quantity (or a family of them) and known lower and upper bounds (both of which should be integers, not infinite).
Conjectures about the actual value are also welcome.
I have given one example below to give an idea of what I'm looking for.

The smallest infinitely often occurring prime gap, or

$$\liminf_{n\to\infty}\; (p_{n+1} - p_n)$$

is unknown as of now. Most likely, it is $2$, but the twin prime conjecture has not yet been settled.

Due to the work of Yitang Zhang and subsequent improvements by others (notably James Maynard and Terence Tao), we know that some prime gaps occur infinitely often. Zhang proved that gaps not larger than 70 million occur infinitely often, and the improvements lowered the bound to $246$ (perhaps there have been recent further improvements I'm not aware of).

The number $\pi$ has a very significant history and could be the ideal object of many problems about the first position in its decimal expansion in which a given property occurs:

In what position the following property occurs first?

1) two consecutive twin numbers of two digits (six possible couples)

2) the number  5 appears 5 consecutive times

3) the first square of ten digits

4) Nothing is known about which digit appears countless times however it is clear that at least one digit does.  What is this digit?

Many other similar questions can be formulated. Note that, excepting 4), it is not obvious that there are corresponding answers but it is obvious also that their probability of existence is nonzero. A way to be sure of the existence of an answer is as follows:

The famous BBP Formula

gives the value of any position in the expansion of $\pi$ so one can looking for some interesting property in an advanced position and ask about the minimal position at which this occurs (it can be the discovered position, of course).


If $X$ is well-ordered then $X\cup \left\{ \infty \right\}$ is well-ordered. Note that $\infty$ is a new element not in $X$, by putting the element $\infty$ to the very end of $X$ we obtain a new set $X\cup \left\{ \infty \right\}$.

Lemma. If $X$ is well-ordered then $X\cup \left\{ \infty \right\}$ is well-ordered.

Proof-trying. Assume $X$ is well-ordered. Let $A$ be a nonempty subset of $X$. Then, since $X$ is well-ordered then $A$ has a least element, say $a_1$. Since $A\subseteq X\cup\left\{ \infty \right\}$ then $ X\cup\left\{ \infty \right\}$ has a least element,that is $a_1$. Hence, $ X\cup\left\{ \infty \right\}$ is well-ordered.

Can you check my proof-trying?

The proof is sort of correct. I mean, you are on the right track, but you haven't proved that any subset of $X\cup\{\infty\}$ has a least element. Let's make it fully correct.
Recall that a totally ordered set $X$ is well ordered if every nonempty subset of $X$ has a least element.
Pick any nonempty subset $A\subseteq X\cup\{\infty\}$. Then there are two options:

$\infty\not\in A$. Then $A\subseteq X$ so $A$ has a least element because $X$ is well ordered.
$\infty\in A$. Consider $B=A\backslash\{\infty\}$. If $B=\emptyset$ then $A=\{\infty\}$ contains exactly one element so $\infty$ is the least element of $A$. So assume that $B\neq\emptyset$. Since $B\subseteq X$ then $B$ has the least element. Now by definition $x<\infty$ for any $x\in X$. Obviously the least element of $B$ also satisfies this so it is a least element of $A$ as well since $A=B\cup\{\infty\}$.

And that's pretty much it.

Without loss of generality, X is a von-Neumann ordinal and infinity is X. X U {X}=X+1 is an ordinal, thus well-ordered.


A Conjecture Sharper than Cramér's and Firoozbakht's Notation: $\lfloor\cdot\rfloor$ is floor function; and $\pi(x)$ is the prime-counting function up to $x$. $g_k := p_{k+1} - p_k$ .

OEIS sequence A267549 is "Primes prime(k) such that floor( (prime(k)/k)^2 ) <= prime(k+1) - prime(k)" https://oeis.org/A267549, or stated differently if $p_k \neq 3, 5, 7, 13, 23, 113$ and $2 \le p_k  < 10^{s}$, with $s$ currently at $12$, then $g_k < \lfloor\left(\frac{p_k}{k}\right)^2\rfloor$ holds. Define
$$a_k := \left(\frac{p_k}{k}\right)^2.$$
It is conjectured that
$$a_k > g_k, \text{ for all }p_k > 113.\qquad \qquad(1)$$

Pierre Dusart proved in 2010:

$$ \frac{x}{\log{(x)} - 1} < \pi(x) \text{ for }x \ge 5393.\qquad \qquad(2)$$

By rearranging, taking $x = p_k$, and noting $\pi(p_k) = k$ we get

$$ \frac{p_k}{k} < \log{(p_k)} - 1 < \log{(p_k)} \text{ for }p_k \ge 5393.\qquad \qquad(3)$$

Clearly, the square of (3) is $$\left(\frac{p_k}{k}\right)^2 < (\log{(p_k)} - 1)^2 < (\log{(p_k)})^2\text{ for }p_k \ge 5393.\qquad \qquad(4)$$
One can also take (3) multiply by $\frac{p_k}{k}$ and have $$\left(\frac{p_k}{k}\right)^2 < \frac{p_k}{k}(\log{(p_k)} - 1) < \frac{p_k}{k}(\log{(p_k)}).\qquad \qquad(5)$$

Which is to say (1) is a stronger upper bound than both Cramér's conjecture, right-side of (4), and Firoozbakht's conjecture, right-side of (5) is the same as the right-side  of (*).

Note for the 64-th record maximal gap
$$p_k = 1693182318746371, k = 49749629143526, g_k = p_{k+1} - p_k = 1132 $$

$$a_k = (1693182318746371/49749629143526)^2 = 1158.3178633021 $$

$$a_k - g_k = 1158.3178633021 - 1132 = 26.3178633021 $$

$$\frac{g_k}{a_k}  = 1132 / 1158.3178633021 = 0.977279239 < 1.$$
One can deduce that the gaps get arbitrarily smaller in proportion to the primes: the quotient
$$\lim_{k\to\infty}\frac{g_k}{p_k}=0.\qquad \qquad(6)$$
In 1931, E. Westzynthius proved that prime gaps grow more than logarithmically. That is,
$$\limsup_{k\to\infty}\frac{g_k}{\log p_k}=\infty.\qquad \qquad(7)$$

We expect that there is constant $C$ such that,
$$\limsup_{k\to\infty}\frac{g_k}{C (\log p_k)^2}=1.\qquad \qquad(8)$$

This, (6) and (7), explains why Charles R Greathouse IV wrote the commented on OEIS sequence A267549:

$\qquad$"Andrew Granville conjectures that lim sup (prime(n+1)-prime(n))/log(prime(n))^2 >= 2/e^gamma = 1.1229189.... If so (or at least if the lim sup is greater than 1) then this sequence is infinite."

So, if there is constant $G$ such that:

$$\limsup_{k\to\infty}\frac{g_k}{G a_k} = 1$$
then:
If $G < 1$ then (1) holds. If $G > 1$ then (1) fails. As for $G = 1$, a bound above must be found so that $\frac{g_k}{a_k} < 1$ holds.

But we do not know $G$ value or hints of it past $4*10^{18}$. Or do we? What information have I left off? What is needed to prove $113$ is the largest counterexample of (1) or the value of the counterexample?

(*) See Section 4, page 4, (12) of arXiv:1506.03042 (Journal of Integer Sequences, 18, 2015, article 15.11.2, Alexei Kourbatov) Upper bounds for prime gaps related to Firoozbakht's conjecture.

In 2016, R. Farhadian conjectured that if $p_n$ be the $n$-th prime and $\psi_n= \left(\frac{p_{n+1}}{p_n}\right)^n$, then

$$p_n^{\psi_n} < n^{p_n}, \quad \text{ for all } n>4$$

The Farhadian's conjecture is more stronger than some other conjectures as Cramer's, Firoozbakht's, Granville's, Nicholson's conjectures [1, 2].

Furthermore, we know that the sequence $\psi_n$  is directly related to the Firoozbakht's and Nicholson's conjectures, since the Firoozbakht's conjecture states that
 $$\psi_n < p_n, \quad \text{ for all } n\geq 1$$
and the Nicholson's conjecture states that
$$\psi_n < n\log(n), \quad \text{ for all } n>4$$
Therefore, we believe that studying the sequence $\psi_n$ is useful for examining these conjectures.

[1] C. Rivera, ed., Conjecture 78: p(n)^(p(n+1)/p(n))^n<=n^p(n), 2016, Available at http://www.primepuzzles.net/conjectures/.

[2] R. Farhadian, R. Jakimczuk, On a New Conjecture of Prime Numbers, 12 (2017), 559 - 564.


Where i'm doing wrong in the solution? I'm learning Progression and Series with myself.At current about Arithmetic Progression.I'm stuck in a problem from sometime.I don't know why my answer is coming different from book.

Problem-

  The sum of n terms of two A.P's are in the ratio of 7n+1:4n+27 , find the ratio of their 11th term.

My solution-
To find the ratio of 11th term i used formula of finding Tn= Sn-Sn-1 .First I calculated ratio of sum of 11th term and 10th term and calculated the difference.

Answer in the book :- 4/3


Equilateral triangle whose vertices are lattice points? Is it possible to construct an equilateral triangle with vertices on lattice points?

I think the answer is no, but how can I prove this?

I started with a triangle with coordinates $(0,0)$ $(a,b)$ and $(c,d)$.
Equating the size of the 3 sides, I get

$a^{2}+b^{2}=c^{2}+d^{2}=2ab+2cd$

How should I continue?

I see there are solutions based on the fact that the angle between two edges can not be 60°. Is it possible to have a solution based on the fact that the length of the edges can not be the same?

Let the vertices of our triangle be $(0,0)$, $(a,b)$, and $(c,d)$, where $a$, $b$, $c$, and $d$ are integers. If all edge lengths are the same, then
$$a^2+b^2=c^2+d^2=(a-c)^2+(b-d)^2.$$
Minor manipulation turns this into
$$a^2+b^2=c^2+d^2=2ac+2bd.$$

Now we use my favourite identity, which was known more than a millenium ago in India, and even earlier by Diophantus, and so has often been called the Fermat Identity:
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.\qquad\qquad(\ast)$$
This identity can be easily verified by expanding both sides, or more conceptually by noting that the norm of the product of two complex numbers is the product of the norms.

Let $N=a^2+b^2=c^2+d^2=**2(ac+bd)**$. Then $ac+bd=N/2$. The identity $(\ast)$ now gives
$$N^2=\frac{N^2}{4}+(ad-bc)^2$$
or equivalently
$$3N^2=4(ad-bc)^2.$$
This is impossible, since $3$ times the perfect square $N^2$ cannot be a square unless $N=0$, which gives a very tiny triangle.

You can start like this. Without loss of generality, let the three points be $(0,0), (0,a)$ and $(b,a/2)$. You can do this because you can always rotate and translate the axis to get these points. Now for all the points to be integral, you need $a$ to be even. This is the first constraint. Secondly, from the basic trigonometry,

$$ \tan \theta = \frac{2b}{a} = \sqrt{3}. $$ From this, you get $b = a\sqrt{3}/2$, which is irrational.


Why does Euclid write "Prime numbers are more than any assigned multitude of prime numbers." In Euclid's Elements Book XI proposition 20 (http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html), Euclid proves that:

Prime numbers are more than any assigned multitude of prime numbers.

I know that this is supposed to say something similar as there are infinitely many primes, but I don't really see this from this wording.

In my mind, this sentence means something like:

There are more prime numbers than any amount of prime numbers.

You can see Aristotle and Mathematics and Actual infinity.

According to the Aristotelian philosophy, we cannot legitimately "handle" actual infinity; i.e. we have no experience of an infinite "collection" but only of an unlimited iterative process (the potential infinity).

Euclid's statement must be understood in this context : we never have a "complete" infinite set of prime numbers, but we have a procedure that, for a finite collection of prime numbers whatever, can "produce" a new prime which is not in the collection.

I disagree with your interpretation at the end, of what he meant to say. I think if you interpret his statement, you should come up with something similar to this idea. "Prime numbers" changes to "there exists prime numbers", "are more" should be read as "greater than", and finally "than any assigned multitude of prime numbers" meaning then any prime number that actually takes a non variable value. In full my interpretation is this.

"There exists prime numbers greater than any prime number that actually takes on a non variable form."

This is what I 'beleive' to myself he said, further I wonder if part of what he actually said has been removed? Reguardless I think it shows knowledge of the idea of prime numbers as a variable for representing all of them. I think it also is clear evidence of the understanding that there are infinite number of prime numbers.


About the equation $\frac{\partial }{\partial t}x(n,t)=x(n+1,t)$ For the equation $\frac{\partial }{\partial t}x(n,t)=x(n+1,t)$,

What is the general solution?
How to represent $x(n,t+1)$ using $\{x(n+k,t)\}_{k\in \mathbb{Z}}$?

This equation seems easy, but I have no idea to solve that. Besides, the second question is the discretization about $t$.

Thanks

Quoting the beginning of a review, on page 268 in the May 1950 issue of Bulletin of the American Mathematical Society,  of An essay toward a unified theory of special functions by C. Truesdell:

  It is very difficult to draw the line between mathematical physics
  and applied mathematics but this book shows that there does exist a
  definite and important difference between them. In mathematical
  physics many special functions such as the Legendre, Hermite, or
  Lauguerre polynomials, Bessel or hypergeometric functions are used
  as tools to solve particular problems. As a consequence many
  properties of these fucntions and connections between them have been
  established. The author, as an applied mathematician, has posed the
  question of finding a unified approach to these different special
  functions so that from it most of the known properties could be found
  directly. The monograph under review gives an answer to this question.

  The author found that many of the special fuunctions, not only those
  previously mentioned but also such as the generalized Riemann zeta
  function, the incomplete gamma function, or the Poisson-Charlier
  polynomials, can be transformed into solutions of the equation
  $$ \tag{1}  \frac{ \partial F(z,\alpha)} {\partial z} = F(z,\alpha+1). $$

The equation $\frac\partial{\partial t} x(n,t)=x(n+1,t)$ says that the function $t\mapsto x(n+1,t)$ is the derivative of the function $t\mapsto x(n,t)$, nothing more, nothing less. So for any integer $k>0$,
$x(n+k,t)=\frac{\partial^k}{\partial t^k}x(n,t)$ is the $k$-th derivative, and $x(n-k,\cdot)$ is a $k$-th antiderivative of $x(n,\cdot)$.

That describes the general solution of the equation---note $x(n,\cdot)$ can be any smooth function. Without more information, the value of $x(n,t+1)$ cannot be determined from the values of $x(n+k,t)$. If, though, $x(n,\cdot)$ is analytic and its Taylor series has radius of convergence larger than 1, then of course $$x(n,t+1) = \sum_{k=0}^\infty \frac{x(n+k,t)}{k!}.$$

As elementary as this is, it is curious that the similar-looking system $$\beta_n'(t)= n\beta_n(t)-n\beta_{n+1}(t)$$ appears to be significant in this work on zero-free regions of the Riemann zeta function.


Take away, subtraction or nim? During most of my life my math skills were an absolute joke mostly due to bad teatchers (since 1st grade till 11th grade) but at some point, though late, of my life fortunatelly I was blessed with a bit of understanding of math, and what a beautiful world it is. I'm saying this in advance in order to humbly ask to whoever can to please help me in the most simple way possible in order for me to understand because I know that I'm no Cauchy.
Here's the problem: In a bar a guy comes to me with a game. The rules are as follow:
1- There was a matrix 3*5 of coins on the tabble.
2- Two players required to play.
3- Sets of {1},{2} or {3} coins could be drawed at time.
4- The player who wipes the tabble clean wins.
Now here's the twist, that fellow droped out of school very early so his mathematical skills are very limited let's say, so I'm guessing that there must be a simple way to understand how to win. Never the less he aced this game with such a confidence and he really won every single time no matter the opponent. I searched for an answer but so far only got confused.
Of my refection I figyred out that:
1- If it's my turn and there are 5, 6 or 7 coins it's a win to me.
2- The only way to assure a win is to let the other player with 8 coins on the tabble thus leadding to 100% of chance for me to win.
The question is what's the strattegy to win and what happens when both players kwon it?
thank you.

You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.

What is a kernel y directed graph theory? it is an independent set of vertices  $N$ such that for every vertex $v\notin N$ there exists an arrow from $v$ to a vertex in $N$.

Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w \leq 3$.

So the set of vertices is $\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$ and the
 arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= \{0,4,8,12\}$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.

Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.


I have a hard time understanding why $\ln e=1$ I have a hard time understanding why $\ln e=1$

Can someone explain to me why the natural logarithm of e is exactly equal to the first nonzero but positive integer?

It depends on how you define the natural logarithm; but, let's do it this way:

By definition, $\ln(x)$ is the unique number $y$ such that $e^y=x$.  In other words, the natural logarithm $g(x)=\ln(x)$ is the inverse function for the exponential function $f(x)=e^x$.

So, $\ln(e)=1$ because $e^1=e$.

A logarithm (being a function) along with a base $b$, when given an input $x$, could be interpreted as "What is $y$ such that $x=b^y$".
So if $b=e$, when evaluating $\log_e(e)=\ln(e)$, we actually ask "What is $y$ such that $e=e^y$". Therefore $\ln(e)=y=1$ must hold..


How to integrate $\int \frac{\cos^m x}{\cos nx} \,\mathrm{d}x$ How to integrate $$\int \frac{\cos^m x}{\cos nx} \,\mathrm{d}x$$

Motivation: $\int \cos mx \sec nx \,\mathrm{d}x $ and $ \int \cos mx\sec^n x\,\mathrm{d}x $ and  $\int \cos^m x \cos nx \,\mathrm{d}x $ and $\int \cos mx \cos nx \,\mathrm{d}x$ are all integrals with closed forms (including summations) I've solved myself or read elsewhere.

Plugging various $m,n$ into wolfram shows the antiderivative always has a combination of trig functions and $\tanh^{-1}$. Nothing I've tried has worked or even simplified it.

HINT

Let $y=\cos(x),$ then
$$\frac{\cos^m x}{\cos nx} = \frac{(\cos x)^m}{T_n(\cos x)} = f(\cos x),$$
where $T_n(y)$ are the Chebyshev's polynomials of the first kind.

If $m\ge n$ it is necessary to single out an integer part of the ratio of polynomials.
Let
$$g_{n-1}(y) = y^m\bmod T_n(y),\quad h(y) = f(y) - g_{n-1}(y),$$
then
$$f(y) = r(y) + h(y),$$
where
$$r(y) = \frac{g_{n-1}(y)}{T_n(y)},$$
$h(y)$ are polynomial of $y$ or zero, and $g_{n-1}(y)$ is polynomial of order $(n-1)$ or less.

Then,
$$T_n(y) = 2^n\prod_{k=1}^n (y-y_{n,k}),$$
where
$$y_{n,k} = \cos\left|\frac{2k-1}{2n}\right|.$$
So
$$r(y) = \frac {g_{n-1}(y)}{2^n \prod_{k=1}^n (y-y_{n,k})} = \sum_\limits{k=1}^n \frac{a_k}{y-y_{n,k}},$$
and
$$a_k = \lim_{y\to y_{n,k}}r(y)(y-y_{n,k}).$$
In this way,
$$\int\frac{\cos^m(x)}{\cos nx}\,dx = \sum_{k=1}^n a_k\int\frac{dx}{\cos x - y_{n,k}} + \int h(\cos x)\,dx,$$
where $h(y)$ is the polynomial, and the integrals under the sum can be calculated using the universal trig substitution.

You may wanna check out Liouville's theorem (Hamiltonian); I guess it can't be integrated.


Gardner riddle on mathemagicians A cute riddle (but maybe not so easy!) from Gardner:

At a gathering of mathemagicians, the Grand Master and his 8 disciples are seated at a round table. The Grand Master will judge each of his displices on their newest trick. After he has seen all the performances, the Grand Master hands to each of his disciples a card with their score on it (the score is some integer number of points greater than $0$). In return, The Grand Master then performs the trick himself and allows his disciples to judge his own perforamce. The disciples agree on a score and give the Grand Master a card with his score on it. It turns out that each mathemagician at the table has a different score.

The Grand Master then remarks: "I can think of a number that divides the product of my own score and the score of anyone seated at this table, other than the two people sitting beside me.". Each of the disciples look at their own scorecard and then look around the table, and each discplie remarks: "I can also think of such a number!". All at once, everyone seated at the table announces the number they have in mind. Incredibly, they all say same the same number! "Now that is some trick!" the Grand Master laughs.

What is the smallest possible number the mathemagicians could have announced?

The Grand Master top-scores with a mere $3^2 \times 5^2 \times 7^2 \times 11 = 121275$ and the other scores are:
$$\begin{array}{|c|c|}
\hline \text{Disciple} & \text{Score} & \text{Composition} \\
\hline 1 & 80850 & 3 \times 5 \times 5 \times 7 \times 7 \times 11 \times 2  \\
\hline 2 & 53900 & 5 \times 5 \times 7 \times 7 \times 11 \times 2 \times 2  \\
\hline 3 & 32340 & 5 \times 7 \times 7 \times 11 \times 2 \times 2 \times 3  \\
\hline 4 & 19404 & 7 \times 7 \times 11 \times 2 \times 2 \times 3 \times 3  \\
\hline 5 & 13860 & 7 \times 11 \times 2 \times 2 \times 3 \times 3 \times 5  \\
\hline 6 & 9900 & 11 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5  \\
\hline 7 & 6300 & 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 7  \\
\hline 8 & 22050 & 2 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7  \\
\hline \end{array}
$$
and the new marvellous number is $485100$
(thanks to Prajanan Pate for the improvement)

original answer
So the Grand Master scores highly, of course, getting $5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 = 37182145$
The remaining disciples, in order round the table, get:
$$\begin{array}{|c|c|}
\hline \text{Disciple} & \text{Score} & \text{Composition} \\
\hline 1 & 14872858 & 2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \\
\hline 2 & 6374082 & 2 \times 3 \times 11 \times 13 \times 17 \times 19 \times 23 \\
\hline 3 & 2897310 & 2 \times 3 \times 5 \times 13 \times 17 \times 19 \times 23 \\
\hline 4 & 1560090 & 2 \times 3 \times 5 \times 7 \times 17 \times 19 \times 23 \\
\hline 5 & 1009470 & 2 \times 3 \times 5 \times 7 \times 11 \times 19 \times 23 \\
\hline 6 & 690690 & 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 23 \\
\hline 7 & 510510 & 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \\
\hline 8 & 4849845 & 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \\
\hline \end{array}
$$
and the "incredible" number is $23\#=223092870$.
... and the field is open for a lower number!


Prime and maximal ideals of this ring? $R=\bigl\{\frac{a}{b}\  |  \gcd(3, b) = 1,\ a, b ∈ \Bbb Z\bigr\}$

I am trying to find all of the ideals in this ring. So far I have $\bigl\{0\bigr\}$, $R$, and $S=\bigl\{\frac{a}{b}\  |  \gcd(3, b) = 1, \gcd(3,a)=3 \ a, b ∈ \Bbb Z\bigr\}$ . I'm not sure how to find any more or, if there are no more, show that this is the case. I am then interested in which of the ideals are prime and maximal.

This is the ring of fractions with denominator not divisible by $3$. In other words, it is the localisation of $\mathbf Z$ at the prime ideal $3\mathbf Z$, usually denoted as $\mathbf Z_{(3)}$.

It is known the only non-trivial ideals of this ring are generated by the powers of $3$, so the only prime ideals are $(0)$ and $3R$ (which is the single maximal ideal of $R$).


We flip 100 fair coins 6 times each, how many coins will end up being heads all 6 times or tails all six times? I've solved for a single coin flipped 6 times and the probability that we get 6 flips as all heads or tails as 1/64 --> 1.56%
But I'm stuck on where to go from there.

The binomial probability that a given fair coin will be only heads in 6 trials, or only tails in 6 trials, is $2^{-6}$.  The probability that in 6 trials it is only heads or only talks is thus $2^{-5}$.
If 100 fair coins are flipped 6 times each, the expected number that come up only heads or only tails is therefore $100 \times 2^{-5}=3.125$.
This is equivalent to a non-rounded version of one of the solution proposed by @DontWorry: after 1 roll, you expect $100/2$ coins to be only heads; after 2 rolls, the expectation is $100/2^2$, and after 6 rolls, the expectation is $100/2^6$, so for both only heads or only tails, the expected number is $100/2^5$.

Assuming that
Pr[tails] = 0.5, Pr[heads] = 0.5
Then if we get 100 people flip the coins at the same time and we eliminate the ones with heads we have
100 in the beginning
50 after 1st flip
25 after 2nd flip
12 after 3rd flip (rounding downward/floor)
6 after 4th flip
3 after 5th flip (rounding downward)
1 after 6th flip
So we have in total 1 (or 2 because we rounded downward)  person with tails 6 times in a row
The same for heads
Finally, 1 coin will heads all 6 times and 1 coin will be tails all 6 times.


continuous function: preimage of closure of a set is closure of preimage i am struggeling with the following problem: let $f:X\to Y$ be a continuous mapping bet ween two metric spaces.

Does it hold that $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$ and $f^{-1}(int(A))=int(f^{-1}(A))$, where the overline denotes the closure of a set and $int()$ the interior of a set.

One direction is obvious: $f^{-1}(A)\subset f^{-1}(\overline{A})$ and by continuity of $f$ the set$f^{-1}(\overline{A})$ is closed. So we proved $\overline{f^{-1}(A)}\subset f^{-1}(\overline{A}$. Any idea or reference where I can find the proof or an counter example?

Let it be that $X$ and $Y$ both have $\mathbb R$ as underlying set, that the topology on $X$ is discrete and the topology on $Y$ is the usual one.

Both topologies are metrizable and $f:X\to Y$ prescribed by $x\mapsto x$ is continuous.

If $A\subset Y$ is not closed then: $$\overline{f^{-1}(A)}^X=f^{-1}(A)=A\varsubsetneq\overline{A}^Y=f^{-1}(\overline{A}^Y)$$

If $B\subset Y$ is not open then: $$f^{-1}\left(\text{int}_Y(B)\right)=\text{int}_YB\varsubsetneq B=f^{-1}(B)=\text{int}_X(f^{-1}\left(B\right))$$

You can, on the other hand, prove that a map $f: X\rightarrow Y$ is closed if and only if $\forall A\subseteq Y$ you have $f^{-1}(\overline{A})\subseteq\overline{f^{-1}(A)}$.

So $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$ if and only if f is continuous and closed.

(you get analogous results for the interior and open maps)


Calculate $\oint_{\gamma}\frac {\sin z}{z(z-2i)}dz$ on $|z| =3$ in trigonometric sense and on the inverse trigonometric sense. $\oint_{\gamma}\frac {\sin z}{z(z-2i)}dz$ on $|z| =3$ in trigonometric sense and on the inverse trigonometric sense.

On the trigonometric sense, we can apply the Residue theorem, we have :

$z_1 = 0$ apparent singular point because $\lim_{z\to0} f(z)= -\frac 1{2i}.$ Which is different from infinity and it clearly exists.
And also we have: $z_0=2i$ simple pole.

If we check that $|z_0|<3$ is true then $z_0\in Int\gamma.$ Then by the Residue theorem:

$$I = 2\pi i Res(f, 2i)$$

But $Res(f, 2i)= \frac {\sin(2i)}{2i} \implies I=\pi\sin(2i)=i\pi\frac {e^2-e^{-2}}{2}$.

And on the inverse trigonometric sense, I thought that i have to check if $|z_0|>3$ so then that is false then $I=0.$ Am i correct?

Ofcourse, with the notations: $I$ being the integral, and $f(z)$ the function under the integration sign.

$$\int_{\gamma}\dfrac{\frac{\sin z}{z}}{z-2i}\mathrm dz=2\pi i\mathrm{Res}_{z=2i}=2\pi i\cdot\lim_{z\to 2i}\left[(z-2i)\cdot\dfrac{\frac{\sin z}{z}}{z-2i}\right]$$

Aliter:

Another way to tackle this is to use Cauchy Integral formula that gives you the following thereby confirming what we obtained in the first part:$$2\pi if(2i)=\int_\gamma\dfrac{{\frac{\sin z}{z}}}{z-2i}\mathrm dz \ : f(z)=\dfrac{\sin z}{z}$$

Another way for anticlockwise (or trigonometric) (yes, I read the question which mentions Residue theorem) which I believe is easier to grasp, you can rewrite
$$\int\limits_{|z|=3}\frac {\sin z}{z(z-2i)}dz=
\frac{1}{2 i}\int\limits_{|z|=3}\sin{z} \left(\frac{1}{z-2i}-\frac{1}{z}\right)dz=\\
\frac{1}{2 i}\int\limits_{|z|=3}\frac{\sin{z}}{z-2i}dz-\frac{1}{2 i}\int\limits_{|z|=3}\frac{\sin{z}}{z}dz=...$$
and now use Cauchy's integral formula
$$f^{(n)}(a)=\frac{n!}{2\pi i}\int\limits_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz \tag{1}$$

In this case, both $0$ and $2i$ are "inside" $|z|=3$ and $f(z)=\sin{z}$, thus
$$...=\pi f(2i)-\pi f(0)=\pi\sin{(2i)}$$

For clockwise, just flip the sign of the result above.


Definition of connected graph The definition of a connected graph states that:
A graph $G$ is called connected provided for each pair $a,b$ with $a\neq b$ of vertices $\exists$ a walk joining a and b.(equivalently a chain joining $a$ and $b$)...

What does the definition mean by (equivalently a chain joining $a$ and $b$) .Please help...

A chain is simply a sequence of edges, forming a path.

A path is a walk without repeated vertices.

If there is a walk between two vertices a and b, there is also a path connecting them.

An undirected graph is connected when there is a path between every pair of vertices. In a connected graph, there are no unreachable vertices. A graph that is not connected is disconnected. An undirected graph G is said to be disconnected if there exist two nodes in G such that no path in G has those nodes as endpoints.
A graph with just one vertex is connected. An edgeless graph with two or more vertices is disconnected


Find minimum number of customers that must have visited the bakery that day? A bakery sells three kinds of pastries -pineapple,
  choclate and black forest. On a particular day,
  the bakery owner sold the following number of
  pastries : $90$ pineapple , $120$ chocolate and
  $150$ black forest. If none of the customers bought
   more than $2$ pastries of each type, what is the
  minimum number of customers that must have
  visited the bakery that day ?

$a.)\ 80 \\
\color{green}{b.)\ 75} \\
c.)\ 60 \\
d.)\ 90 $

I have drawn the venn diagram and concluded that the visitors are $210$.

But that is not in options.

I look for a short and a simple way.

I have studied maths up to $12$th grade.

You haven’t enough information to fill in the Venn diagram. (And what you’ve done is inconsistent with the data, since you have $120$ pineapple pastries.)

To get the smallest possible number of customers, you want each customer to buy as many pastries as possible consistent with the totals given. $150$ black forest pastries were bought; if each customer bought $2$, there would have been at least $\frac{150}2=75$ customers, and there can’t possibly be any fewer customers than that. Is $75$ customers enough to account for the other sales? Yes: for instance, it could be that $60$ of the customers also bought $2$ chocolate pastries each, and $45$ of them also bought $2$ pineapple pastries each. (Many other arrangements are also possible.)

It’s the black forest pastries that determine the lower bound on the number of customers, simply because more black forest pastries were sold than any other type.

As $~90~$ is min.

So $~\frac{90}{2}=45~$ should be ans.

But it is not in options.

If we consider $~60~$ single pastries out of $~90~$ pastries are bought by $~60~$ customers and remaining $~30~$ pastries  are bought by $~15~$  customers each $~2~$. i.e.  $~(15*2=30)~$.

Min customers $~60+15=75~$.


Lie algebra, maximal toral sub-algebra Is there a relation between number of roots of a finite dimensional semi-simple Lie algebra L and  dimension of the maximal toral sub-algebra H(Cartan sub-algebra) of L?

Thanks!

There is the relation
$$
\dim (H)+\mid \Phi\mid =\dim (L),
$$
where $\mid \Phi\mid$ denotes the cardinality of the root system of $L$. For example, with $L$ of type $A_n$ we have $\dim (H)=n$, $\mid \Phi\mid=n^2+n$ and $\dim(L)=n^2+2n$.

With notations as in the question,
2 dim(H) <= no. of roots.
This is true because roots span H*(Dual space of H) and roots always occur in pairs.(I mean x is a root iff -x is a root.)

I do not hope there is any other relation, in general. But, may be, if we consider some special classes of L, we may have some better results!


Eigenvalues and powers Is this statement true?

  $L:V\to V$ is a linear map with eigenvalue (not necessarily the only one) $a$. Suppose $(L-aI)^{m+1}(v)=0$ where $m$ is the power of the $(x-aI)$ term in the minimal polynomial of $L$. Then $(L-aI)^m(v)=0$ also.

Some thoughts:

So this is essentially saying that $(L-aI)[(L-aI)^m(v)]=0\implies (L-aI)^m(v)=0$. In other words the nullspace of $ (L-aI)$ is $\{0\}$. Therefore I don't think the statement is necessarily true. Is there a way of constructing an explicit example to disprove this?

Ah wait, I haven't used the fact that $m$ is the power of the term in the minimal polynomial. So this may yet be true...

The statement is true. In general you have
$$Ker(L-aI)\subseteq Ker(L-aI)^2\subseteq\cdots $$
This sequence stabilises, i.e. there is an $m$ such that
$$Ker(L-aI)^i=Ker(L-aI)^{i+1}$$
for all $i\geq m$. This $m$ is precisely the $m$ which is the exponent of $x-a$ in the minimal polynomial. In particular you have
$$Ker(L-aI)^m=Ker(L-aI)^{m+1}.$$

You write that the kernel of $L-aI$ restricted to $Ker(L-aI)^m$ is zero. Indeed it is, but for trivial reasons, since the whole source is the zero space.

Edit: The sequence stabilises since the dimension is bounded from above by the dimension of $V$. To see that the $m$ for which this happens is precisely the power in the minimal polynomial, one has to carefully analise the proof of the Caley Hamilton Theorem. This is done in any basic book on linear algebra. The key idea is that there is a squence of spaces with decreasing dimension which finally will be the image of the characteristic polynomial evaluated at $L$. This shows that the image is trivial. One observes carefully that we may reduce the power of the linear coefficients precisely to $m$ to obtain the similar result for the minimal polynomial.

But $\,L-aI\,$ cannot have nullspace equal to zero as it is a singular map: $$\exists\,0\neq v\in V\,\,s.t.\,\,Lv=av\Longleftrightarrow (L-aI)(v)=0\,$$ and this seems to disprove your idea.

Now, if $\,Tv=0\,$ for some lin. transf., then $\,T^n(v)=0\,,\,\forall n\in\mathbb{N}\,$ , so both $\,(L-aI)^m\,\,,\,\,(L-aI)^{m+1}\,$ vanish at $\,v\,$, but not that from one we can deduce the other.


How to replace 8-sided dice with other dice The exact question is:
You need an 8-sided die for a game. You only have a coin, two four-sided and one 10-sided dice. How can you replace the 8-sided die?
Re-rolls are not allowed.

There are several solutions to this I've been told, I found one, but my solution wasn't one of the expected solutions. What solutions can you think of?

My solution was:

Roll 10-sided and 2x 4-sided dice, sum up the result and roll and substract 2-sided die (the coin) which you divide by 2 at the end to get the result (round up)

We need to not only generate numbers between 1 and 8, but also to make sure they are uniformly distributed.

Your solution does not produce uniformly distributed results (at least according to MJD, in the comments).

However, this procedure does: you can roll a 4-sided die for a value between 1 and 4, and then toss a coin: if heads, add 4 to the result.

It is easy to see that each number from 1 to 8 is produced by exactly one outcome: for example, a result of 3 requires a roll of 3 and a toss of tails, while a result of 5 requires a roll of 1 and a toss of heads.

To keep the uniform repartition, here are some proposals :

Roll the 10-face dice, re-roll on 9 and 10.
Roll the 4-face dice, then flip the coin. Add 4 on heads.
Roll the 4-face dice and multiply the result by 2. Flip a coin, and substract 1 on heads.

There are probably a number of other solutions, the important part is to keep the repartition uniform.


An Infinite Cyclic Group has Exactly Two Generators: Is My Proof Correct? I have completed a proof of this that I am inclined to believe is correct, or at least on the right track. I would like to ask if it is indeed correct, or if I need a nudge in the right direction. Perhaps there is a better way to go about this?

My Proof:

(1) I will show that the existence of a generator for an infinite cyclic group implies the existence of a second which is the inverse of the former.

If $G$ is an infinite cyclic group, then every element of $G$ can be written as $a^n$ for $n\in\mathbb{Z}$ and $a$ some element of $G$. It is clear, then, that $a^{-1}$ and $a$ are both generators for $G$ since, given $n\in\mathbb{Z}$, we can choose $m\in\mathbb{Z}$ such that $(a^{-1})^n=(a)^m$ by taking $m=-n$. Moreover, we are guaranteed that $a^{-1}\ne a$ since this would imply $a=a^{-1}=e_G$ leaving us with $G$ the trivial group. So given one generator, there must be another, namely, the inverse of the given generator.

(2) Now I will show that $G$ cannot have more than two generators.

Now suppose for the sake of contradiction that $n>2$ and $G=\langle{a_1}\rangle=\ldots=\langle{a_n}\rangle$ (which is to say that any one of these $n$ elements and its inverses generate $G$), where $a_i=a_j\iff i=j$. Then, by invoking the pigeonhole principle, at least one of the members of $\{a_1,\ldots,a_n\}$ is not an inverse of one of the others. Let us call this element $a_i\ne a_1,a_{1}^{-1}$. Thus, if $a_1$ generates $G$, then there exists $b\in\mathbb{Z}$ for which $a_{1}^{b}=a_i$. Hence, if these both generate $G$, then $\forall{c\in\mathbb{Z}},\exists{j\in\mathbb{Z}}$ such that $a_{1}^{c}=a_{i}^{j}$, but $a_{i}=a_{1}^{b}$, so $a_{1}^{c}=a_{1}^{bj}$. But if $b\ne{\pm1}$, this is cannot be true, as this would state (for $c=b-1$) that $b-1=bj\implies j=1-\frac{1}{b}$ and then both $j$ and $\frac{1}{b}$ cannot be an integer for any value other than $b=\pm{1}$, which contradicts the assumption that $a_{i}$ is neither the inverse of $a_{1}$ nor equal to $a_1$.

Combining statements (1) and (2),it follows that if $G$ has one generator and it is an infinite cyclic group, then this generator is not the identity element and $G$ must have two generators. Additionally, if $G$ is an infinite cyclic group, it cannot have $n>2$ generators. These two statements mean that $G$ must have precisely two generators.

You can simplify it greatly, and also do part (2) without contradiction (i.e. directly).

(1) If $G = \langle a \rangle$, then $G$ also equals $\langle a^{-1} \rangle$, because every element $a^n$ of $\langle a \rangle$ is also equal to $(a^{-1})^{-n}$.

(2) If $G = \langle a \rangle = \langle b \rangle$, then $b = a^n$ for some integer $n$, and $a = b^m$ for some integer $m$. These imply that $a = b^m = (a^n)^m = a^{nm}$; since $G$ is infinite cyclic, $nm$ must equal $1$. As an equation in the integers, $nm = 1$ has only two solutions, $(1,1)$ and $(-1,-1)$. These give $b = a$ or $b = a^{-1}$.

One can simply use isomorphisms.We can apply the fact that any infinite cyclic group say G is isomorphic to Z. Now ,any isomorphism maps a generator(of the domain group) to a generator of the co domain group,since Z has precisely 2 generators this implies G has also 2 generators

PS: I concede that this is not the best solution as isomorphisms have been used ,instead of concept of cyclic groups only


If $N = 2^{(2^2)}$, find real $x$ such that $N^{N^N}=2^{(2^x)}$. Understanding the solution. Harvard-MIT Math Tournament 2019 Algebra Problem:

Let $N = 2^{(2^2)}$ and $x$ be a real number such that $N^{N^N}=2^{(2^{x})}$. Find $x$.

Solution:
$$N^{(N^N)}=16^{16^{16}} = 2^{4\cdot2^{^{4\cdot2^{{4}}}}} = 2^{2^{2^{6}+2}} = 2^{2^{66}}$$
I don't understand how to get from (the previous step) $2^{4\cdot2^{^{4\cdot2^{{4}}}}}$ to (the final answer) $2^{2^{66}}$

To my eye, the difficulty lies in the typography: It's hard to see which superscripts are at what level.* I recommend taking logarithms instead.  Using the abbreviation $\lg$ for $\log_2$, we have
$$\begin{align}
x&=\lg(\lg(2^{2^x}))\\
&=\lg(\lg(N^{N^N}))\\
&=\lg(N^N\lg(N))\\
&=\lg(N^N)+\lg(\lg(N))\\
&=N\lg(N)+\lg(\lg(N))\\
&=16\lg(16)+\lg(\lg(16))\\
&=16\cdot4+\lg(4)\\
&=64+2\\
&=66
\end{align}$$
*In fact there is a typo in the OP's display: "2^{4\cdot^{2^{4\cdot^{2^{4}}}}}" should be "2^{4\cdot{2^{4\cdot{2^{4}}}}}" (without the spurious ^'s after the \cdot's), i.e
$$2^{4\cdot{2^{4\cdot{2^{4}}}}}\qquad\text{instead of}\qquad2^{4\cdot^{2^{4\cdot^{2^{4}}}}}$$
but even when TeXed correctly, I find it hard to see which superscript is at which level.


How Many Ways are There to Climb 10 Stairs Using 1, 2, ..., or an Unlimited Number of Steps? In how many ways can one climb a staircase composed of 10 stairs when each step covers either one stair or two stairs?

This question is very similar to one I have seen with 6 stairs. The solution was to find how many possibilities are there to take one stair at a time, then 2, then 3, and so on. I need help because the number is larger.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  The number of $\ds{n}$-steps configurations is given by

\begin{align}
&\sum_{s_{1} = 1}^{2}\ldots\sum_{s_{n} = 1}^{2}
\bracks{s_{1} + \cdots + s_{n} = 10} =
\sum_{s_{1} = 1}^{2}\ldots\sum_{s_{n} = 1}^{2}
\oint_{\verts{z} = 1}{1 \over z^{11 - s_{1} - \cdots - s_{n}}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z} = 1}{1 \over z^{11}}\pars{\sum_{s = 1}^{2}z^{s}}^{n}
\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{11 - n}}\,{\dd z \over 2\pi\ic} =
{n \choose 10 - n}
\end{align}
which is non zero whenever
$\ds{\pars{~n \geq 10 - n \geq 0 \implies 5 \leq n \leq 10~}}$.

  The total number of configurations is given by

$$
\sum_{n = 5}^{10}{n \choose 10 - n} =
\sum_{n = 0}^{5}{n + 5 \choose 5 - n} = \bbx{\ds{89}}
$$


Can a number have infinitely many digits before the decimal point? I asked my teacher if a number can have infinitely many digits before the decimal point. He said that this isn't possible, even though there are numbers with infinitely many digits after the decimal point. I asked why and he said that if you keep adding digits to the left of a number it will eventually approach infinity which is not a number and you could no longer distinguish two numbers from one another.
Now this is the part of his reasoning I don't understand: why can we distinguish two numbers with infinitely many digits after the point but not before it? Or is there a simpler explanation for this?

The formal way to understand this is, of course, using the definition of real numbers.
A real number is "allowed" to have infinite digits after the decimal point, but only a finite number of digits before. (http://en.wikipedia.org/wiki/Real_number)

(if it interests you, there are numbers that have infinite digits before the decimal point, and only a finite number after. Take a look at http://en.wikipedia.org/wiki/P-adic_number . just to have some fun, know that the $10$-adic expansion of $-1$ is $\color{red}{\dots 99999} = -1$)

If you want to get some intuition about this, first think that, as your teacher said, said number would approach infinity, which is not a real number. This is reason enough.

About the comparing two numbers part: if I give you $$1234.983...$$ and $$1234.981...$$
you know which one is bigger, it does not matter what the other digits are.

But with $$...321.99$$, $$...221.99$$ you don't, because the information relies in the "first" digit. Of course nobody know what the first digit is, since there is no first digit.

But as I said before, this is to gain some intuition; the correct way to think about this is using the definition (which is not trivial)

It's a matter of semantics and definition.  Technically, since integers are a subset of real numbers and there are an infinite number of them, you can have an infinite number of digits; however, for the sake of defining properties, and defining numeric fields, you limit the number of digits to n... which can be as large as you like.

So, yes, you CAN have an infinite number of digits - because you can always make your number larger by multiplying it by 10, say - but, for the sake of definition and math theory, it is NOT possible, by nature of its definition.


Transformation of $f(x)$ We have $f(x)$. If we want to make up $f(ax+b)$ :

1.We should first create $f(ax)$ and then $f(ax+b)$

2 Firstly create $f(x+b)$ and then $f(ax+b)$ ?

In the other words , I'm in doubt when we create $f(ax)$ the coefficient of $x$ only is multiplied by $x$ and make up $f(ax+b)$ or multiple by $x+b$ and make up $f(ax + ab)$ and therefore second way is wrong.
If someone provide several examples like $f(x) = x^2$ and create $f(ax+b) = (ax+b)^2$ using function transformations is helpful.

Example : Consider $f(x) = sin(x)$ . I used Mathematica for drawing functions.  We want to create $f(3x-1)$. We should draw $sin(3x)$ and then $sin(3x-\frac{1}{3})$ but I think we have to draw $sin(3x-1)$ in last step. Am I wrong ? Why?

Going from $f(x)$ to $g(x) = f(ax + b)$ requires two $2$ function transformations.

A horizontal movement of $-b$ by substitution $x$ by $x+b$. $(g(x))$
A horizontal multiplication of $\frac{1}{a}$ by substition $x$ by $ax$. $(h(x))$

Example: Let $f(x) = x^2$. Then

$$g(x) = (x+b)^2$$ and $$h(x) = (ax+b)^2.$$ Question: Does the order matter, that is, can we first do the second and then the first transformation?


Why is the Fibonacci ratio though a decreasing function, it is alternating and decreasing? I tried to find the ratio of consecutive terms of the Fibonacci series and found that it is a decreasing function and it converges .
I tried it though a small code piece in python so that I can have a lot of data points to analyze.

import math

def F(n):
    return ((1+math.sqrt(5))**n-(1-math.sqrt(5))**n)/(2**n*math.sqrt(5))

for i in range(53):
    a = F(i)
    b = F(i+1)

print str(b) + '/' + str(a) + '=>' + str(b/a)

And it gave values which suggested that it is a decreasing function and it converges at around 1.618. But I found that, though it was decreasing, it was not a continuously decreasing function, it in fact increases and decreases alternatively. But overall it was decreasing and converging.

And I tried to plot the values to prove this graphically.

(I expanded the graph a bit by plotting for $n*3$, just to make the view clear)

My doubt is this - What is so special in this series which makes it behave alternating in nature?

I tried to do it for consecutive numbers ratio and found that it was a increasing but not alternating like the Fibonacci ratios.

Thanks

Consider the formula you used for calculating the values: $$F_n = \frac{1}{\sqrt{5}}\left(\phi^n - \psi^n\right)$$ where $\phi=\frac{1}{2}\left(1+\sqrt{5}\right)$ and $\psi=\frac{1}{2}\left(1-\sqrt{5}\right)$.

The ratio of consecutive terms can then be expressed as $$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-\psi^{n+1}}{\phi^n - \psi^n} = \phi + \frac{(\phi-\psi)\psi^n}{\phi^n -\psi^n}$$

There are is a couple of observations we can now make:

We have $|\phi|>1$ and $|\psi|<1$. Thus, with increasing $n$, the value of $\psi^n$ will get closer and closer to zero, while $\phi^n$ will grow further and further. This already implies the ratio $(F_{n+1}/F_n)$ converges to $\phi$, since the other term converges to zero (its numerator is going to zero and denominator to infinity).
$\psi<0$, so $\psi^n$ flips its sign whenever we increase $n$ by one. Therefore, the second term oscillates from positives to negatives; making the ratio alternate between being greater than $\phi$ and smaller than $\phi$.

The first observation can be generalized to any sequence which can be expressed as a linear combination of exponentials, the greatest of which has base greater (in absolute value) than 1. The ratio of consecutive terms will converge to this greatest base.

Let $F_i$ be the $i$-th Fibonacci number. Put $F_i$ people at one table, and deliver $F_{i-1}$ pizzas to them. Put $F_{i+1}$ people at a second table and deliver $F_i$ pizzas to them. At the first table, there are $F_{i-1}/F_i$ pizzas per person, and at the second table there are $F_i/F_{i+1}$ pizzas per person.

Now have everyone at the first table move to the second table, and bring their pizzas along. In total, there are $F_i + F_{i+1} = F_{i+2}$ people and $F_{i-1} + F_i = F_{i+1}$ pizzas, so we have $F_{i+1}/F_{i+2}$ pizzas per person. If there were more pizzas per person at the first table than at the second (in other words, if $F_{i-1}/F_i > F_i/F_{i+1}$), then bringing over the people from the first table increased the number of pizzas per person at the second table. In other words, $F_{i+1}/F_{i+2} > F_i/F_{i+1}$. If there were more pizzas per person at the second table ($F_{i-1}/F_i < F_i/F_{i+1}$), then bringing over the people from the first table decreased the number of pizzas per person, and $F_{i+1}/F_{i+2} < F_i/F_{i+1}$. This shows the oscillation. Note that either way, $F_{i+1}/F_{i+2}$ sits in between $F_{i-1}/F_i$ and $F_i/F_{i+1}$. The sequence of ratios converges by the alternating series test.

I phrased this in terms of ratios smaller than $1$ to avoid the appearance of gluttony, but obviously you can do the same thing with ratios greater than $1$ ($F_{i+1}/F_i$), or just take reciprocals.


Prove $\exists$ a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$. If $f$ is continuous function ($f:\mathbb{R}\rightarrow\mathbb{R}$), then there exists a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$.

Question: Since $K$ is compact, there exists some interval $[a,b]$, such that $K$ is contained in. By Weierstrass, for all $\epsilon>0$, there exits a sequence of polynomials of $P_n$, such that $P_n$ converges to $f$ uniformly, i.e. $\exists N\in\mathbb{N}$ $\forall n\geq N$, $|P_n(x)-f(x)|<\epsilon$, for all $x\in [a,b]$, therefore, certainly in $K$. So we are done. Correct?

Your hypothesis is not true!

We can find continuous real function $f_{0}$ such that $f_{0}$ not differentiable everywhere!
But as $f_{0}$ be the limit of a real polynomials at such interval $[a,b]$ then $f_{0}$ will be expanded by Taylor expansion, so $f_{0}$ is infinitely differentiable over $[a,b]$. Which contradict with the first property of $f_{0}$ which is: $f_{0}$ not differentiable everywhere.

  Now let us tray to set a continuous real function $f_{0}$, such that $f_{0}$ not differentiable everywhere:

let us define $g$ as following:
\begin{equation}
g(x)=x-E(x) \text{ if } E(x) \text{ is even}\\
g(x)=E(x)-x+1 \text{ if } E(x) \text{ is odd}\\
\text{where } E \text{ is the integer part function}
\end{equation}

We can easily show that $g$ is continuous, periodic with period 2, and bounded, moreover norm of $g$ less than $1$.

Take the following function sequence:
\begin{equation}
g_{n}(x) =g(2^{n}x+1/2)/2^{n} \text{ for all } x\text{ in } \mathbb{R} \\
\text{for any positive integer } n
\end{equation}
 We can easily show that $g_{n}$ is continuous, periodic with period 2^{-n+1}, and bounded, moreover norm of $g_{n}$ less than $1/2^{n}$, for every positive integer $n$.

Take the following function series:
\begin{equation}
S_{n}(x) =\sum_{i=0}^{i=n}g_{i}(x) \text{ for all } x\text{ in } \mathbb{R} \\
\text{for any positive integer } n
\end{equation}
We can easily show that $S_{n}$ is continuous for every positive integer $n$.

Therefore this series is converges by norm to a real continuous function S.

  Now we just need to show that S is not differentiable every where.

Let us find $B$ the set of all points where S is not differentiable Using the following steps:

$\mathbb{Z}$ is the set of all points where $g$ is not differentiable,see the definition of $g$.

let $A_{n}$ be the set of all points where $g_{n}$ is not differentiable.
$g_{n}$ is not differentiable at the points $x$ such that $2^{n}x+1/2$ in $\mathbb{Z}$, which means $A_{n}=2^{-n}\mathbb{Z}-2^{-n-1}$. The must important think is $A_{i}\cap A_{j}=\phi$ for each two different positive integers $i,j$.

let $B_{n}$ be the set of all points where $S_{n}$ is not differentiable.
\begin{equation}
\text{let } x\in \bigcup_{u=0}^{u=n}A_{u}\\
x\in \bigcup_{u=0}^{u=n}A_{u}
 \Rightarrow x\in A_{u} \text{ for some } 0\leq u\leq n\\
 \Rightarrow g_{u}\text{ is not differentiable at } x \text{ for some } 0\leq u\leq n
\end{equation}
since $A_{i}\cap A_{j}=\phi$ for each two different positive integers $i,j$: so we have $x$ doesn't belong to every $A_{v}$ where $0\leq v\leq n$ and $v\neq u$, that means  $g_{v}$ differentiable at $x$ for every $0\leq v\leq n$ and $v\neq u$. Which gives:
\begin{equation}
\sum_{i=0}^{i=n}g_{i} \text{ is not differentiable at } x \text{, hence } S_{n} \text{ is not differentiable at } x.\\
S_{n} \text{ is not differentiable at } x \Rightarrow x\in B_{n}
\end{equation}
therefore: $B_{n}=\bigcup_{u=0}^{u=n}A_{u}$ ($\bigcup_{u=0}^{u=n}A_{u}\subseteq B_{n} $ has been proved; $B_{n}\subseteq \bigcup_{u=0}^{u=n}A_{u}$ is trivial inclusion).
\begin{equation}
\text{Then } B=\bigcup_{u=0}^{u=\infty}A_{u}.
\end{equation}

In order to show that S is not differentiable every where we will show that:
\begin{equation}
(\forall \epsilon>0)(\forall a \in \mathbb{R})(S \text{ is not diff. over } ]a-\epsilon,a+\epsilon[)
\end{equation}

Let us choose $a \in \mathbb{R}$. for each fixed $\epsilon>0$ the following is true:

We will choose $n_{0} \in \mathbb{N}^{*}$ such that $1/2^{n_{0}}<\epsilon$. Suppose $g_{n_{0}}$ is differentiable over $]a-\epsilon,a+\epsilon[$, then $g_{n_{0}}$ is differentiable over $[a-\epsilon/2,a+\epsilon/2]$, then $g_{n_{0}}$ is differentiable over $[a-\epsilon/2,a+\epsilon/2]+m/2^{n_{0}}$ for all integer m (since $1/2^{n_{0}}$ is the period of $g_{n_{0}}$), then $g_{n_{0}}$ is differentiable over $\bigcup_{m=0}^{m=\infty} ([a-\epsilon/2,a+\epsilon/2]+m/2^{n_{0}})$, then $g_{n_{0}}$ is differentiable over $\mathbb{R}$ which isn't true. Therefore $g_{n_{0}}$ isn't differentiable over $]a-\epsilon,a+\epsilon[$.

but: $g_{n_{0}}$ is not differentiable over $]a-\epsilon,a+\epsilon[$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $g_{n_{0}}$ is not differentiable at $b$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $b \in A_{n_{0}}$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $b \in B$, equivalent to say $\exists b \in ]a-\epsilon,a+\epsilon[$ such that $S$ is not differentiable at $b$, equivalent to say $S$ is not differentiable over $]a-\epsilon,a+\epsilon[$.

Finally $S$ isn't differentiable every where, $S$ is an example for $f_{0}$ continuous function not differentiable every where.


The product of any three consecutive natural numbers is divisible by 9.prove or find a counterexample can anyone give me feedback on my solution
false counterexample:
n(n+1)(n+2)
let n= 1
1(2)(3)
=6 is not divisible by 9

is this correct?

Indeed, $1\cdot2\cdot3=6$ is not divisible by $9$, so your counterexample works perfectly well.


Two sets of polynomials with distinct roots build the ring of polynomials. Definitions:

$i \in K$
$U_{i}:=\{f\in K[X] |f(i)=0 \}$
$K[X]$ is the ring of polynomials

HINTS:

K[X] is a vector space
Every $U_{i}$ is a vector subspace of $K[X]$

Question:

(i) With $s \neq t$ show, that $U_{s} + U_{t} = K[X]$

(ii) Is this sum direct?

I tried:

I thought it is easy to show, that $U_{s} + U_{t} \subseteq K[X]$ because every $U_{i}$ is allready a subset of $K[X]$ their sum ends up being in $K[X]$ as well.  My Problems are in the other direction... $"\supseteq"$

How can I show that $U_{s} + U_{t}$ build the complete vectorspace $K[X]$ ?

I tried to think of an $N:=\{U_{i}|i \in K\}$ and wanted to show that $dim_{N}(K[X])=2$. This would proof that there always exists a linear combination of $u, v \in N$ with $u \neq v$ with $a_{1}v + a_{2}u = K[X]$. But I am at the stage of confusion.

Hint
For $f\in K[X]$ chose $g\in K[X]$ such that $g(s) = f(s)$ and $g(t) = -f(t)$. Now look at
$$f = \frac12 (f + g) + \frac12(f - g)$$
Here $\frac12 := (1+1)^{-1}$ in the field $K$ (We must thus assume $\mathrm{char}(K) \ne 2$).
If $K = \mathbb F_2$ (up to isomorphism the only field with characteristic $2$), it should be very easy for you since $K[X]$ is particularly easy.


In a game of Minesweeper, a number on a square denotes the number of mines that share atleast one vertex with that square. In a game of Minesweeper, a number on a square denotes the number of mines that share atleast one vertex with that square.A square with a number may not have a mine,and the blank squares are undetermined.In how many ways can the mines be placed in the given configuration on the blank squares.

$(A)120\hspace{1cm}(B)105\hspace{1cm}(C)95\hspace{1cm}(D)100$

Number on the middle square is $1$,so there is one mine which shares atleast one vertex with this square.So there are 8 ways to put the mine adjacent to square numbered $1$.But then i got stuck.Please help me.

Hint: First, think about where the mine causing the middle number $1$ can be. How many possible positions are there?

Then, given the position of that mine how many mines does the left $2$ need, and in how many places can you place them?

From Brian M. Scott's hint: Answer $= \text{C}^3_1\times\text{C}^5_1\times\text{C}^2_2 + \text{C}^3_1\times\text{C}^5_2\times\text{C}^2_1 + \text{C}^2_1\times\text{C}^5_2\times\text{C}^2_2$

$= 3\times5\times1 + 3\times10\times2 + 2\times10\times1 = 15 + 60 + 20 = 95$


Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$ How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.

You know the limit of $1+\frac12 +\frac14+\frac18+\ldots$, don't you?

Your sequence is just like that, but without the $1$ and the $\frac12$.

I will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.

We have that $\sum 2^n \delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\Delta 2^n=2^{n+1}-2^n=2^n$).

In your sum, doesnt taking limits, we have that

$$\sum 2^{-n-1}\delta n=-2^{-n}$$

If we take limits then

$$\sum_{n=1}^{\infty} 2^{-n-1}=-2^{-n}\bigg\lvert_{1}^{\infty}=-2^{-\infty}+2^{-1}=\frac{1}{2}$$

Take care with notation because

$$\sum_{k\ge 0}2^k\ne\sum_{k\ge 0}2^n$$


Are there any open mathematical puzzles? Are there any (mathematical) puzzles that are still unresolved? I only mean questions that are accessible to and understandable by the complete layman and which have not been solved, despite serious efforts, by mathematicians (or laymen for that matter)?

My question does not ask for puzzles that have been shown to have either no solution or multiple solutions (or have been shown to be ambiguously formulated).

The sofa problem.

From Wikipedia:

  It asks for the rigid two-dimensional shape of largest area $A$ that
  can be maneuvered through an L-shaped planar region with legs of unit
  width. The area $A$ thus obtained is referred to as the sofa constant.
  The exact value of the sofa constant is an open problem.

Author of the picture: Claudio Rocchini, see this link

Two envelopes problem

From wikpedia:
The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle, or paradox in logic, philosophy, probability, and recreational mathematics. It is of special interest in decision theory, and for the Bayesian interpretation of probability theory. Historically, it arose as a variant of the necktie paradox.

The problem:

  You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.

It can be argued that it is to your advantage to swap envelopes by showing that your expected return on swapping exceeds the sum in your envelope. This leads to the paradoxical conclusion that it is beneficial to continue to swap envelopes indefinitely.

Example -
Assume the amount in my selected envelope is \$20. If I happened to have selected the larger of the two envelopes, that would mean that the amount in my envelope is twice the amount in the other envelope. So in this case the amount in the other envelope would be \$10. However if I happened to have selected the smaller of the two envelopes, that would mean that the amount in the other envelope is twice the amount in my envelope. So in this second scenario the amount in the other envelope would be \$40. The probability of either of these scenarios is one half, since there is a 50% chance that I initially happened to select the larger envelope and a 50% chance that I initially happened to select the smaller envelope. The expected value calculation for how much money is in the other envelope would be the amount in the first scenario times the probability of the first scenario plus the amount in the second scenario times the probability of the second scenario, which is \$10 * 1/2 + \$40 * 1/2. The result of this calculation is that the expected value of money in the other envelope is \$25. Since this is greater than my selected envelope, it would appear to my advantage to always switch envelopes.

A large number of solutions have been proposed. The usual scenario is that one writer proposes a solution that solves the problem as stated, but then another writer discovers that altering the problem slightly revives the paradox. In this way, a family of closely related formulations of the problem have been created, which are discussed in the literature.
No proposed solution is widely accepted as correct. Despite this it is common for authors to claim that the solution to the problem is easy, even elementary. However, when investigating these elementary solutions they often differ from one author to the next. In the last two decades, several new papers have been published every year.


product of six consecutive integers being a perfect square A 1939 paper of Erdos (Note on Products of Consecutive Integers, J. London Math. Soc. 14 (1939), 194–198) shows that a product of consecutive positive integers cannot be a perfect square. He cites a 1917 paper by Narumi which proves that a product of at most 202 consecutive positive integers cannot be a perfect square. I cannot seem to easily find Narumi's paper.

Although this result is known, I am curious about self-contained elementary proofs of special cases.

It's not too difficult to come up with fairly quick proofs for two, three, four, five, or seven consecutive integers.

Is there a short self-contained elementary proof that the product of six consecutive positive integers cannot be a perfect square? Or is it perhaps fair to say that this is the first "tricky" case?

I am not sure how elementary you want your proof to be, but here is a proof that uses elliptic curves...

Suppose that there are $x,y\in\mathbb{Z}$ such that $x>0$ and

$$y^2=x(x+1)(x+2)(x+3)(x+4)(x+5).$$

If we put $t=x+2+\frac{1}{2}$, then we have

$$y^2=(t-5/2)(t-3/2)(t-1/2)(t+1/2)(t+3/2)(t+5/2)=\left(t^2-\frac{1}{4}\right)\left(t^2-\frac{9}{4}\right)\left(t^2-\frac{25}{4}\right),$$

or, equivalently,

$$4^3y^2 = (4t^2-1)(4t^2-9)(4t^2-25).$$

If we put $U=2^3y$ and $V=4t^2$, then we have a solution for the equation

$$U^2=(V-1)(V-9)(V-25)=V^3 - 35V^2 + 259V - 225.$$

This defines an elliptic curve $E/\mathbb{Q}$, and we can use standard techniques to calculate the rank of the group of rational points $E(\mathbb{Q})$. This method ($2$-descent) shows that the rank of the curve is $0$, and one can easily separately show that the torsion subgroup is $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. It follows that the only points on $E(\mathbb{Q})$ are the trivial points $(V,U)=(1,0)$, $(9,0)$ and $(25,0)$, plus the point ``at infinity'' on the curve. These correspond to $t$-values $t=\pm 1/2$, $\pm 3/2$ and $\pm 5/2$, and therefore do not give any integer values of $x$ with $x> 0$. Hence, there are no integer solutions to our original equation.

There is probably some elementary argument that shows that $U^2=(V-1)(V-9)(V-25)$ only has $3$ solutions, but I can't think of one right away.

There's an elementary answer actually. You have n, n+1, n+2, n+3, n+4, n+5. Every two numbers there will be a multiple of two. This means, because there are six numbers in all, that either the first xor the last number will be a multiple of two. There will be three multiples of two. Every three numbers will be a multiple of three. Because three is odd, if the first multiple of three is even, the second will be three plus that and so will be odd. If the first multiple of three is odd, the second is three plus an odd number so is even. So no matter what, one of the multiples of two is also a multiple of three, and there are only two multiples of three in our list. There will be one or two multiples of five in the list. If there are two, then it must be the first and last numbers. Since one of the first and last numbers is already a multiple of two, we only gain information about one "new" number, unless the first is a multiple of two and the last is a multiple of three, or vice versa, in which case we gain information about no "new" numbers. Now if the five appears once in the list only we then gain information about one "new" number at max. So, we have found all the places a 2, 3, or 5 may be a factor -- but this only accounts for five unique numbers (three multiples of 2, one extra multiple of three [because one overlaps with the multiples of two], and one (or none) extra multiple of 5 [because if there are two fives one overlaps with the multiples of two]. That means we have five numbers which have prime factors 2, 3, or 5 in our list, and one which does not. But a list of 6 consecutive positive integers can not repeat any prime factor more than 5. So our unknown number must have a prime factor that is only once in the list, and thus the product of the list cannot be a square. The only exception is the number without any prime factors, 1. If our unknown number is 1, then the list will be 1, 2, 3, 4, 5, 6. But five is a prime factor only once there in that list, so that isn't a square either.

Edit: This is wrong because the number without factors 2, 3, or 5 could be a square itself.


Finding the Equation of an Ellipse given the Length of the Latus Rectum and the Distance between the Foci For a National Board Exam Review:

  Find the equation of the ellipse having a length of latus rectum of ${ \frac{3}{2} }$ and the distance between the foci is ${ 2\sqrt{13} }$

Answer is ${ \frac{x^2}{16} + \frac{y^2}{3} = 1 }$

So I try:

$${ LR = \frac{2b^2}{a} = \frac{3}{2} }$$

$${ a^2 - b^2 = c^2 }$$
$${ a^2 - b^2 = ( 2\sqrt{13} )^2 }$$

Solve two equations; I get a = 7.59...

From there I'm stuck. I cant use the variable for the answer... What is wrong with my method?

$c=13$,
$b^2=\frac{3a}4$

Substituting in $a^2-b^2=c^2$,

$$a^2-\frac{3a}{4}=13$$

Multiply by $4$ to cancel fraction and make it whole.

$$4a^2-3a-52=0$$

$a=4$ and $a=\frac{-13}{4}$

Since we need positive use $4$ to substitute in $$\tag{1}b^2=\frac{3a}4$$

$b^2=3$ and substitute the value of $b$ in $(1)$. You can have $a^2=16$.

Therefore;
$$\frac{x^2}{16}+\frac{y^2}{3}=1$$ or $$3x^2+16y^2=48$$


Finding the derivative of $y^x = e^y$ I have a problem that I've been working on and I don't understand how to start:

  If $y = y(x)$ satisfies $y^x = e^y$, then when $(x, y) = (2\sqrt{e}, \sqrt{e})$, the derivative $\frac{dy}{dx} = \ ...$

I can't understand what $y = y(x)$ satisfies $y^x = e^y$ means. Any help or hints are greatly appreciated! Thanks!

Let us use implicit differentiation

$$ y^x = e^y \implies x \ln y = y \implies \ln y + x  \frac{y'}{y} = y'$$

$$ \implies \ln y = y'(1 - \frac{x}{y}) \implies \frac{dy}{dx} = \frac{\ln y }{1 - \frac{x}{y}}$$

$$ \therefore \; \text{at}\ \ (2 \sqrt{e}, \sqrt{e} ) ,\ \frac{dy}{dx} = \frac{\frac{1}{2}}{1 - 2} = - \frac{1}{2}$$

Rewrite it as $y^{1/y}=e^{1/x}$, look up Lambert W function and ask Wolfram...


If $aRb \land bRc$ never occurs, is a relation considered transitive? Studying relation properties. My definition of a transitive relation is as follows:

  A relation is transitive if and only if $\forall a,b,c \in A [aRb \land bRc \implies aRc]$

My question is: if $aRb \land bRc$ never occurs in the first place, is the relation considered transitive?

I ask this because when I read $\forall a,b,c \in A [aRb \land bRc \implies aRc]$ I read it as follows: "when $a$ relates with $b$ and $b$ relates with $c$, $a$ relates to $c$". But since this "when" never happens, there is no condition to evaluate to decide if it is transitive or not.

The same issue occurs with antisymmetry, where $\forall a,b \in A[aRb \land bRa \implies a = b]$ - what if $aRb \land bRa$ never occurs in the first place?

EDIT

I just remembered that $F_0\implies whatever$ is always $V_0$.... I guess this answers my question... kinda.

The answer is yes in both cases; these are examples of statements that are vacuously true, to use the usual expression. To see why we would want to use this interpretation, ask yourself what it would take in order to show that $R$ is not transitive: you’d need to find elements $a,b,c\in A$ such that $a~R~b$, $b~R~c$, and yet $a~\not R~c$. But if there are no elements $a,b,c\in A$ such that $a~R~b$ and $b~R~c$, you can’t possibly find such a counterexample. In short, there is no counterexample to transitivity of $R$, so by default $R$ must be transitive.

Another way to put it: transitivity says that when certain conditions are met, a certain thing has to happen. It says nothing about cases in which those conditions are never met: it requires nothing in those cases, so it cannot possibly fail.

On the set $\{-1,1\}$ with $R$ defined $xRy$ if $y=-x$, we see that $R$ is transitive by vacuity.


Compute the integral $\int_C (z^2-1)^\frac{1}{2} dz$ where $R>1$ Let C be the circle of Radius $R>1$, centered at the origin, in the complex plane. Compute the integral

$\int_C (z^2-1)^\frac{1}{2} dz$

where we employ a branch of the integrand defined by a straight branch cut connecting $z=1$ and $z=-1$, and $(z^2-1)^\frac{1}{2} > 0 $ on the line $y=0$, $x>1$. Note that the singularities are not isolated in this case. One way to do this integral is to expand the integrand in a Laurent series valid for $|{z}|> 1$ and integrate term by term

Consider the contour $\Gamma$ below.

This consists of two pieces: the circle $z=R$, $R \gt 1$, minus the dogbone inside the circle, consisting of two small circular arcs at $z=\pm 1$ of radius $\epsilon$, and joined by line segments above and below the real axis.

Now consider $f(z) = \sqrt{(z-1)(z+1)}$.  On the upper line segment, $\arg{(z+1)} = 0$ and $\arg{(z-1)} = \pi$.  On the lower line segment, $\arg{(z+1)} = 0$, but $\arg{(z-1)} = -\pi$.  Thus, the contour integral about $\Gamma$ is, in the limit as $\epsilon \to 0$,

$$\oint_{\Gamma} dz \, \sqrt{z^2-1} = \int_{|z|=R} dz \, \sqrt{z^2-1} + i \int_1^{-1} dx \, \sqrt{1-x^2} - i \int_{-1}^1 dx \, \sqrt{1-x^2} $$

By Cauchy's theorem, the integral about $\Gamma$ is zero.  Thus,

$$ \int_{|z|=R} dz \, \sqrt{z^2-1}  = i 2 \int_{-1}^1 dx \, \sqrt{1-x^2} = i \pi$$

Hint: The value is independent of $R$ (by Cauchy's integral theorem). Let $R \to \infty$ and use the "residue at $\infty$" or compute the limit in another way (for example using that $(z^2-1)^{1/2} \approx z$ for $|z|$ large). You need to precise that estimate of course.


Average number of guesses for simple game The guess-a-number game starts with one player (the chooser) picking a number
between 1 and 100 (inclusive) and secretly writing it down. The other player
(the guesser) attempts to guess the number. After each guess, the chooser responds
with “correct” (the guesser guessed the number and the game is over),
“higher” (the actual number is higher than the guess), or “lower” (the actual
number is lower than the guess).

What is the average number of guesses the guesser will need to find the number given that he uses the optimal, most efficient strategy?

I found that it will be 6 guesses $29/32$ times and 7 guesses $3/32$ times using the most clumsy method ever which I won't post. It can be no more than 7 since $2^7=124>100$.

It was like

$(100-1)/2 = 49.5, \\(49-1)/2 = 24,  \\(50-1)/2 = 24.5 \\-> 24, 25 $

with 24 $3/4$ of the time and 25 $1/4$ of the time, etc etc etc.

What is a better way (since the method had so many opportunities for error I'm probably wrong anyway)?

If you want to split the interval, your first guess should be $\frac {100+1}2=50.5$ You can be a bit more efficient by making your guesses whole numbers, giving a chance of getting "correct". early.  So if your first guess is $50$, you get away with $1$ guess $1/100$ of the time, while if you guess $50.5$ you lose the chance to be lucky.  You have $2/100$ chance of winning in two guesses, $4/100$ chance in three, and so on.  Other than that, binary search cannot be beaten.

The average number of guesses would be ~5.80905. As the analytical approach to solving this would be rather extensive, the answer could be estimated with a simple simulation, in which the program follows the optimal strategy of cutting the set of values in half by selecting the mid point, and then we average the number of trials after a large number of iterations.
n <- 100
count<-c()

for (i in 1:100000){

  s <- round(runif(1,0.5,n+0.5),0)
  iter <- 1
  guess <- trunc(n /2)
  iter_par <-trunc(guess/2)

  while (guess!= s){

    ifelse(guess<s,guess<-guess+iter_par,guess<-guess-iter_par)
    iter_par<-round(iter_par/2,0)
    ifelse(iter_par<1,iter_par<-1,iter_par<-iter_par)
    iter<-iter+1
  }

  count<-append(count,iter)

}

mean(count)


Check whether $36^{36}+41^{41}$ a multiple of $77$ Let $$a=\frac{36^{36}+41^{41}}{77}.$$

Is $a$ an integer?

I know that:

The last digit ofr $41^{41}$ is $1$.
The last digit of $36^{36}=6^{72}$ is $6$.

How can I use this information to answer my question, or do I need another thing to help me?


How many maps of this set into itself question? Suppose we have a set $(1, 2, ... n)$, which has $2^n$ subsets. I am taking an introduction to analysis course, and I don't seem to understand the following two questions about this set:

How many maps of this set into itself?
How many maps of this set onto itself?

I understand the definitions of "into" and "onto" in this context, but I'm struggling with understanding what exactly they're asking. Any help?

Call the set $S$. By "into" they mean any map $f\colon S\rightarrow S$. By "onto" they mean a map $f\colon S\rightarrow S$ whose image is $S$ itself (i.e. $f(S)=S$). That is, each element in the codomain is mapped to by some element in the domain.

There are $n^n$ maps from an $n$ element set to itself.

A surjection from a finite set to itself is also an injection, hence a bijection. (Proof: If $f\colon X=\{1, \dotsc n\}\to X$ is a surjection, then $\{f^{-1}(x) \mid x\in X\}$ is a partitions $X$ into $n$ nonempty disjoint sets, so every $f^{-1}(x)$ must have exactly one element.) Thus, the number of surjections from $X$ to itself equals $n!$, as that is the number of bijections from $X$ to itself.

Note that in the general case, computing the number of surjections from a set of $n$ elements to a set of $k$ elements is more involved, and is answered here and here.


Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?

By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.

There is a theorem (by Riemann, I think) which states that:
"A convergent series which is not absolutely convergent
can be rearranged to sum to any number we choose"

So all we need to do to show that what the OP asks is possible, is to take
an arbitrary convergent series which is not absolutely convergent,
and which consists of irrational numbers only.

Then by applying the theorem, it means we can rearrange its elements
and get any rational (and also any irrational) number we want.

An example of such a series is, let's say:

$\sum_\limits{n=1}^\infty \frac{(-1)^n}{n\sqrt5}$


Using differential equations to determine the number of rolls on a roll of toilet paper Puzzle: A role of toilet paper has $180$ sheets on it. The outside is covered with exactly two sheets. The inside around the cardboard cylinder is covered by exactly one. Question of the puzzle: how many layers of toilet paper are on the roll of toilet paper?

The "Given" Solution: One way to solve this is by saying that the average round is covered by $1.5$ sheets, so therefore the answer is $120$ (I have no source for whether this is actually correct)

I Tried: I tried to solve it with a differential equation, but ultimately failed:  Let $S$ be the number of toilet paper sheets on the roll, and $n$ the number of rotations.

I think the number of sheets per rotation depends linearly on the number of rotations at a given point, because with every rotation the toilet role becomes more thick, so:
$$\frac{\mathrm{d}S}{\mathrm{d}n}=kn.$$

This equation is separable, so
$$dS=(kn)\,\mathrm{d}n.$$

Integrate to get
$$S=\frac{1}{2}kn^2+C.$$

Now we need to find the values of constants $k$ and $C$: we know that when $S=1$ then $n=1$ and also when $S=180$ then $\frac{dS}{dn}=2$.

But now I am stuck.

My Question: What would be the correct way of solving this problem using differential equations?

This question is also linked to this question: Toilet paper puzzle (question 2)

Use the "sheets" as units of length, so the total length is $L=180$, the "inner" radius is $r_i = 1/(2\pi)$, the "external" one is $r_e = 2/(2\pi)$. Assume that the paper draws a spiral and makes $n$ loops (not necessarily an integer).
Then, we parametrize this spiral in terms of the angle $t \in [0, 2 n \pi]$ (I use rad for angles, so that $n$ is the number of layers).
The radius is slowly increasing from $r_i$ to $r_e$, say linearly:
$$
r(t) = r_i +(r_e-r_i) \frac{t}{2 n \pi}
$$
We have
$$
L = \int_0^{2 \pi n } dt  \sqrt{ r(t)^2 + r'(t)^2} \approx \int_0^{2 \pi n } dt \,     r(t) =   n \pi (r_e + r_i)
$$
This gives that the number of loops (or layers) is
$$
n \approx \frac{L}{\pi (r_e + r_i)} = 2L/3 = 120
$$
Note: the approximation is valid if the spiral is "spiralling out" slowly. Otherwise the line of reasoning is the same, but you have to deal with more complex calculations (i.e. taking the full square root term in the integral).

PS: it is not explicitly with differential equations, but there is an integral. Solving the integral above is equivalent to "integrate" the ODE
$$
\frac{dL(n)}{dn}  \approx 2 \pi \, \frac{r_e + r_i}{2}
$$
You have to integrate this simple ODE with the initial condition $L(0) = 0$ till you touch the value $L(n) =180$.

The following is simple lateral area calculation:

Let outer,inner radii be  (R,r) respectively.
Since circumferences $2 \pi r , 2 \pi R $ proportional to $(r,R)$are given double,

$$ \dfrac{R}{r}=2 \tag1$$
It is given that total toilet paper length when unrolled
$$ L= 180\, (2 \pi r)\tag2$$

Let effective toilet paper thickness be $t$. The lateral visible layered packed area of paper roll cylinder between radii $(R,r)$ is
$$ L\cdot t= \pi (R^2-r^2) \tag3$$
Eliminate L between (2) and (3) and simplify and  factor it
$$ 360 r= \dfrac {(R-r)(R+r)}{t} \tag4$$
Plug in from(1) and simplify to eliminate R
$$ 120 =\dfrac{r}{t} \tag5$$
Number of layers by tight roll winding the layers
$$n =  \dfrac{R-r}{t} =  \dfrac{r}{t}  \tag6 $$
and lastly from (5) and (6)
$$ n= 120 \tag7$$

And yes, it comes to the same result by integrating linear variation to second degree with smaller radius of roll boundary condition:
$$S=\frac{1}{2}kR^2-\frac{1}{2}kr^2,\,k= 2 \pi, $$
in other words effectively same as (3) with (2).


$\int f(x)\,\mathrm{d}x = \left(\int_0^x f(t) \, \mathrm{d}t\right) + C$ If $f(x)$ is a continuous function on $\mathbb{R}$ and I am asked to find
$\int f(x) \, dx$, what is the problem with the following answer:

$$\int f(x)\,\mathrm{d}x = \left(\int_0^x f(t) \, \mathrm{d}t\right) + C$$

The problem with your answer is the problem with your answer.

And I hope the problem with my answer will tell you something about the problem with your answer.

There are cases when it's appropriate do this:

Someone says

  $e^{-x^2}$ is not integrable

So we say

  Of course it is integrable.  All continuous functions are integrable.  The integral is $\int_0^x e^{-t^2}\;dt$.  What you mean is: the integral is not an elementary function.


Manipulation of an expression with derivatives: which sign is more suitable, "d" or "∂"? I am writing a scientific paper about Mechanics of Beams. At a given point, I discuss the quadratic strain in function of $u(x,t)$ and $w(x,t)$, which refer to the beam's motion in function of position $x$ and time $t$. The quadratic strain reads:
\begin{equation}
    \varepsilon_q = \dfrac{\left(du+dx\right)^2+dw^2-dx^2}{dx^2} \cong \dfrac{\partial{u}}{\partial{x}}+\dfrac{1}{2}\left(\dfrac{\partial{w}}{\partial{x}}\right)^2.
\end{equation}
As you can see, on the right side I use " $\partial$ ", which is suitable to the fact that $u(x,t)$ and $w(x,t)$ depend on both $x$ and $t$. However, on the left side I use " $d$ ", which is suitable to the fact that $du = u(x+dx)-u(x)$ and $dw = w(x+dx)-w(x)$. I wonder if it would not be more accurate to write:
\begin{equation}
    \varepsilon_q = \dfrac{\left({\partial}u+{\partial}x\right)^2+{\partial}w^2-{\partial}x^2}{{\partial}x^2} \cong \dfrac{\partial{u}}{\partial{x}}+\dfrac{1}{2}\left(\dfrac{\partial{w}}{\partial{x}}\right)^2,
\end{equation}
even though I have neever seen $\partial{x}$, $\partial{u}$ or $\partial{w}$ written aside - let us say out of a "derivative fraction" - as we commonly see in math or engineering texts for $dx$, $du$ and $dw$. For example, I have often seen the expression $du = u(x+dx)-u(x)$; but never $\partial{u} = u(x+\partial{x})-u(x)$.
That being said, which expression do you think I should choose?

To address your concern directly, it would not be more accurate to replace $du$ with $\partial u$. The partial derivative notation $\partial u$ is meaningless without a corresponding $\partial x$ indicating which variable is meant to vary.
It can only be made meaningful if the function $u$ is interpreted as a one-variable function anyway. And that's precisely what's happening here - the time dependence of $u$ is being implicitly suppressed, and the $du$ refers to only the variable $x$.
Anyone who fails to understand the $du$ notation would certainly fail to understand the $\partial u$ notation.

The standard notation for partial derivatives is terrible, as the denominator contains information about the numerator, and, because of that, they can never be split.
A notation I have suggested is to subscript partial differentials with the variables that were allowed to change.  So, for $f(x, y)$:
$$ df = \partial_x f + \partial_y f $$
In other words, the total differential is the sum of all of the partials.
So, what is normally thought of as $\frac{\partial f}{\partial x}$ is actually more explicitly written as $\frac{\partial_x f}{dx}$.  This allows the differentials to be treated algebraically.
Using the notation above, you can be more clear as to whether the differentials being considered are partial or total, and, if partial, which one was allowed to vary.


Probability of first and second drawn balls of the same color, without replacement I have an urn with 10 balls: 4 red and 6 white. What is the probability that the first two drawn balls have the same color?

Approach No. 1:

$|U| = 10! = 3628800$

White balls on first and second draw: on the first and second draw, I can pick any of the 6 white balls on the first draw, and 5 on the second draw. After that, I have 8 balls left, of which I don't care what order they are in. This will become $6\times5\times8\times7\times6\times5\times4\times3\times2\times1=1209600$.

$\left(\frac{1209600}{3628800} \right)=\frac{1}{3}$

If we have red balls on the first and second draw, then: I can pick any of the 4 red balls on the first draw, and on the second draw any of the left over 3. This will be$4\times3\times8\times7\times6\times5\times4\times3\times2\times1=483840$.

$\left(\frac{483840}{3628800} \right)=\frac{2}{15}$

The sum of these two will be: $\frac{1}{3}+\frac{2}{15}=\frac{7}{15}$

Approach number two:

Chance of a red ball on first draw: $\frac{4}{10}$. Chances of a red ball on second draw: $\frac{3}{9}.$

$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$.

Chance of a white ball on first draw: $\frac{6}{10}$. Chances of a white ball on second draw: $\frac{5}{9}$.

$\frac{6}{10}\times\frac{5}{9}=\frac{5}{15}$.

$\frac{5}{15}+\frac{2}{15}=\frac{7}{15}$

What I want to know, is this answer correct? Somehow I can't get it straight in my head that this is the right answer, since the $8!$ seems a bit weird to me. I mean don't the chances of the other red balls in the case of red balls on first and second draw count?

Yes this is correct.

With conditional probability, you could formulate it this way:
    P(2 balls of the same color) = P(2 balls are white OR 2 balls are red)
      = P(2 balls are white) + P (2 balls are red) [since the two conditions are mutually exclusive]

P(2 balls are white) = P(first ball is white AND (second ball is white given first ball is white))
  = P(first ball is white) * P(second ball is white | first ball is white)
  = 6/10 * 5/9, as you mentioned.

In the same way, P(2 balls are red) = 4/10 * 3/9.

The two rules that we use are described here: http://people.richland.edu/james/lecture/m170/ch05-rul.html

It is 1) the addition of mutually exclusive events and 2) dependent events and contitional probabilities.

S = {Set of all possible combination of 2 balls from the available 4 Red and 6 White Balls}

n(S) = $^{10} P_{2}$ ways = $10 \times 9 = 90$ (Since, 2 balls can be selected from $10$ in $^{10} P_{2}$ ways)

Now,

E = Both are of same colour

n(E) = $2$ Red balls can be selected from $4$ in $^4 P_{2}$ ways AND $2$ White balls can be selected from $6$ in $^6 P_{2}$ ways)    Total = $^4 P_{2} + ^6 P_{2} =(12 + 30) =42$ ways.

P(E) = $\frac{n(E)}{n(S)} = \frac{42}{90} = \frac{7}{15}$.


Putnam 2012 B3 - Tournament combinatorics A round-robin tournament among $2n$ teams lasted for $2n-1$ days, as follows. On each day, every team played one game against another team, with one team winning and one team losing in each of the $n$ games. Over the course of the tournament, each team played every other team exactly once. Can one necessarily choose one winning team from each day without choosing any team more than once?

There is a solution by applying Hall's theorem, but I was wondering if there exists an elementary solution that someone without knowledge of Hall's theorem could find during the test.

This question was simple enough, and did not require Hall's theorem at all. Consider the question. We need to choose a distinct winning team for each of $2n-1$ days. Thus, in order to fail this condition, we must only have $2n-2$ teams with a win, or put another way, only $2$ of the $2n$ teams without a win. This is clearly impossible because at some point those two teams must have played one another, and one team would have won the game, and thus not be winless. Therefore, it is impossible to have less than $2n-1$ teams with a win, and thus it is always possible to select a distinct winning team for each of $2n-1$ days of the tournament.


Vector multiplication with scalars I'm a little confused as to the results of multiplying vectors and scalars. If I multiply a vector and a scalar, is it a vector? And if I multiply a vector and another vector, is the results also a vector? I looked at http://mathworld.wolfram.com/VectorMultiplication.html but still haven't figured it out.

Ok, this is a math question really, but a 'physics answer' could go as follows:

Velocity is a vector: it has a size (speed) and a direction. If you multiply a velocity vector by - say - a factor of two, you go twice as fast in the same direction. So multiplying a (velocity) vector with a scalar gives again a (velocity) vector.

Now the multiplication of two vectors. There are several meaningful definitions of vector multiplications. Out of these, the inner product multiplication (or dot product) is most abundant in physics. We take force vectors and displacement vectors as examples. Dot multiplying the force vector exerted on a particle with its displacement vector yields the amount of work done. This amount of work is a scalar (a number without further attributes such as direction). So the dot product between two vectors gives a scalar.

As I mentioned, other definitions of products between vectors are possible. For instance, the direct product between two vectors yields a (rank two) tensor. No need to be bothered about this in basic physics though.

God the answers above are mere showing off. Vector multiplied by a scalar is a scalar. vector multiplied by another vector can take two forms, known as scalar product (inner product) or cross product. The former (inner) is a scalar.


Any volume form on a smooth $n$-dimensional manifold is locally a pullback of the standard volume form $dx_1\wedge ...\wedge dx_n$ on $\mathbb R^n$? Let $M$ be an $n$-dimensional smooth manifold equipped with a volume form $\omega$ , and let $\omega_0:=dx_1\wedge ...\wedge dx_n$ be the standard volume form on $\mathbb R^n$ , then is it true that for every $a \in M$ , there exists an open set $U$ containing $a$ in $M$ and a diffeomorphism $g:U \to \mathbb R^n$ such that $\omega = g^*\omega_0$ ?

The following is a variation of Pedro's answer (which doesn't work in my opinion, as explained in Ted's comment).

Take a coordinate chart around $a$ and write $\omega=fdx^1\wedge\ldots\wedge dx^n$. Try new coordinates as follows: $y_1=gx_1,y_2=x_2,\ldots,y_n=x_n$, where $g$ is some smooth function. Then$$\begin{align}dy^1&=\left(\frac{\partial g}{\partial x^1}x_1+g\right)dx^1+\frac{\partial(gx_1)}{\partial x^2}dx^2+\ldots+\frac{\partial(gx_1)}{\partial x^n}dx^n,\end{align}$$and so,$$\begin{align}dy^1\wedge\ldots\wedge dy^n=\left(\frac{\partial g}{\partial x^1}x_1+g\right)dx^1\wedge\ldots\wedge dx^n.\end{align}$$Hence, in order for the $y$'s to be good coordinates, we need to solve $$\frac{\partial g}{\partial x^1}x_1+g=f.$$ For convenience, we may assume that $x_1$ does not vanish, and then the above first order linear equation is easy to solve.

Edit: In second thought, the computation is even easier when guessing a coordinate change of the form $y_1=g,y_2=x_2,\ldots,y_n=x_n$, where $g$ is a smooth function (here, $g$ replaces $gx_1$). Now we have $$dy^1\wedge\ldots\wedge dy^n=dg\wedge dx^2\wedge\ldots\wedge dx^n=\frac{\partial g}{\partial x^1}dx^1\wedge\ldots\wedge dx^n,$$and so, the function $g$ only has to satisfy $\frac{\partial g}{\partial x^1}=f.$

It's true. If you choose coordinates $x_1, \ldots, x_n$ in $M$, the volume form will have the form $f dx_1 \wedge \ldots \wedge dx_n$, with $f$ non-vanishing. So just make the coordinate change $y_1 = x_1/f$ and $y_2 = x_2, \ldots, y_n = x_n$. Then in these coordinates the volume form is $dy_1 \wedge \ldots \wedge dy_n$ as you wish.

In fancier language: orientations are really $\text{GL}^{+}$-structures, and these are always integrable (as the argument above shows).


How do I show $(f_1(x), f_2(x), ..., x-a) = (f_1(a), ..., f_r(a), x-a)$ where $a \in R$, $R$ is a commutative ring, $f_i(x) \in R[x]$ Let $R$ be a commutative ring, $a \in R$, and $\forall i = 1, ...,r \ \ f_i(x) \in R[x]$.
Prove the equality of ideals
$(f_1(x), ..., f_r(x), x-a ) = (f_1(a), ...f_r(a), x-a)$.

We want to show that $$\forall g_1(x), ..., g_{r+1}(x) \in R[x] \ \ \exists y_1(x), ..., y_{r+1}(x) \in R[x]: \ \ g_1(x)f_1(x) + ... + g_r(x)f_r(x) + g_{r+1}(x)(x-a) = y_1(x)f_1(a) + ... + y_r(x)f_r(a) + y_{r+1}(x)(x-a)$$
and vice versa.

$(f_i(x)-f_i(a))(a)=0$. Write $f_i(x)-f_i(a)=q(x)(x-a)+c$ where $c\in R$, you have $f_i(a)-f_i(a)=q(a)(a-a)+c=0$, implies that $c=0$, thus $f_i(x)\in (f_1(a),...,f_r(a),x-a)$.

On the other hand, $f_i(x)=q(x)(x-a)+c$ this implies that $c=f_i(a)=f_i(x)-q(x)(x-a)$ thus $f_i(a)\in (f_1(x),...,f_r(x),x-a)$.

as  the coefficient dominant of $x-a$ is 1 ivertible in R the euclidien division is ok  for any polynomial by $x-a$. so $f_i(x)=g(x)+b $  for some constante $b$ the equality with x=a come $f(a)=b$ and then as reauired  the ideal  $(f_1(x),...,f_r(x),(x-a))$ is the ideal $(f_1(a),...,f_r(x),(x-a))$


If $f \le g$ on $[a,b]$ and both are bounded, then the upper integral of $f$ is less than that of $g$ How to prove the following:

  If $f \le g, \forall x \in [a,b]$, f and g are both bounded on [a,b], then upper integral of f is less then or equal to upper integral of g.

It is easy to show this when f and g have the same partition. But when they have the different partition, I think maybe we can use a new partition, which can refine both of them, but I do not know how to set up the inequality.

The upper integral of $h$ is
$$
                 \overline{\int_{a}^{b}}h(x)dx = \inf_{\mathcal{P}}\overline{S}_{\mathcal{P}}(h),
$$
where $\overline{S}_{\mathcal{P}}(h)$ is the supremum of all Riemann sums over a fixed partition $\mathcal{P}$ obtained by allowing the evaluation points to vary. For the same evaluation points, $S_{\mathcal{P}}(f)\le S_{\mathcal{P}}(g)$. Therefore,
$$
  S_{\mathcal{P}}(f)
            \le S_{\mathcal{P}}(g)
            \le \overline{S}_{\mathcal{P}}(g).
$$
The far right side is independent of the evaluation points of $\mathcal{P}$. Therefore, $S_{P}(f)\le \overline{S}_{\mathcal{P}}(g)$, and the right side does not depend on evaluation points of $\mathcal{P}$. Taking the supremum over all evaluation points then gives $\overline{S}_{\mathcal{P}}(f) \le \overline{S}_{\mathcal{P}}(g)$ and
$$
        \overline{\int_{a}^{b}}fdx \le \overline{S}_{\mathcal{P}}(f) \le \overline{S}_{\mathcal{P}}(g).
$$
Therefore,
$$
              \overline{\int_{a}^{b}}fdx \le \overline{S}_{\mathcal{P}}(g),
$$
and taking the infimum over all $\mathcal{P}$ gives
$$
             \overline{\int_{a}^{b}}fdx \le \overline{\int_{a}^{b}}gdx.
$$

From $f<g$ follows $0<g-f$ Hence,

Now integrate with same limits of integration:
$$\int 0 dx=0 < \int (g-f) dx = \int g dx - \int f dx$$

Or

$$\int f dx < \int g dx$$


Is the sequence $\{(-1)^n 1/n\}$ divergent? Is the sequence $\{(-1)^n{1\over n}\}$ divergent or convergent?

It's convergent to $0$ because for every $\varepsilon$, exists $1/N < \varepsilon$ by archimedean property, so $|1/n|< \varepsilon$ for all $n>N$?

Hint: $-1\le(-1)^n\le1$. Either that, or split it up into two subsequences, depending on the parity of n, and prove that both converge to the same limit.

The sequence converges by the alternating series test.
$$ (-1)^n \over n $$
See a tutorial about alternating series test
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

This fits the criteria for convergence by the alternating series test.