"Proving" that $0^0 = 1$ I know that $0^0$ is one of the seven common indeterminate forms of limits, and I found on wikipedia two very simple examples in which one limit equates to 1, and the other to 0. I also saw here: Prove that $0^0 = 1$ using binomial theorem
that you can define $0^0$ as 1 if you'd like.

Even so, I was curious, so I did some work and seemingly demonstrated that $0^0$ always equals 1.

My Work:

$$y=\lim_{x\rightarrow0^+}{(x^x)}$$

$$\ln{y} = \lim_{x\rightarrow0^+}{(x\ln{x})} $$

$$\ln{y}= \lim_{x\rightarrow0^+}{\frac{\ln{x}}{x^{-1}}} = -\frac{∞}{∞} $$

$\implies$ Use L'Hôpital's Rule

$$\ln{y}=\lim_{x\rightarrow0^+}\frac{x^{-1}}{-x^{-2}} $$
$$\ln{y}=\lim_{x\rightarrow0^+} -x = 0$$
$$y = e^{0} = 1$$

What is wrong with this work? Does it have something to do with using $x^x$ rather than $f(x)^{g(x)}$? Or does it have something to do with using operations inside limits? If not, why is $0^0$ considered indeterminate at all?

Someone said that $0^0=1$ is correct, and got a flood of downvotes and a comment saying it was simply wrong. I think that someone, me for example, should point out that while saying $0^0=1$ is correct is an exaggeration, calling that "simply wrong" isn't quite right either. There are many contexts in which $0^0=1$ is the standard convention.

Two examples. First, power series. If we say $f(t)=\sum_{n=0}^\infty a_nt^n$ that's supposed to entail that $f(0)=a_0$. But $f(0)=a_0$ depends on the convention that $0^0=1$.

Second, elementary set theory: Say $|A|$ is the cardinality of $A$. The cardinality of the set off all functions from $A$ to $B$ should be $|B|^{|A|}$. Now what if $A=B=\emptyset$? There as well we want to say $0^0=1$; otherwise we could just say the cardinality of the set of all maps was $|B|^{|A|}$ unless $A$ and $B$ are both empty.

(Yes, there is exactly one function $f:\emptyset\to\emptyset$...)

Edit: Seems to be a popular answer, but I just realized that it really doesn't address what the OP said. For the record, of course the OP is nonetheless wrong in claiming to have proved that $0^0=1$. It's often left undefined, and in any case one does not prove definitions...

In here, I wrote a detailed explanation on why $0^0=1$ is convenient to use, and why $0^0$ is undefined.

Copying the text, we have

Write a polynomial, for example $x^3-5x^2+7x+2$.
If we write this polynomial as $x^3-5x^2+7x^1+2x^0$, it will give consistency and make it easier to deal with certain notations.
We have that $\lim_{x\to 0+}x^x = 1$
This can be proven easily by taking the natural log and applying L'Hopital's Rule.
From $\frac{d}{dx}\left(x^n\right) = nx^{n-1}$, set $n=1$.
This gives that $1 = x^{0}$. Now plug in $x=0$, giving $0^0=1$.
How many functions $f : X \rightarrow Y$ are there if $X = Y = \phi$?
There is only one such function. Also the number of functions is given by $0^0$ as well, which means that $0^0=1$.

Now for reasons that $0^0$ is still undefined, copying the text again,

Cauchy, who is a pioneer of analysis, wrote in 1821 that $0^0$ is undefined.
Libri, in 1830, wrote a paper that "proved" $0^0=1$, which was very flawed.
Few years later, Mobius wrote a paper in support of Libri, and claimed the following.

If $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} g(x) = 0$, we have $\lim_{x\to 0^+}f(x)^{g(x)} = 1$.

This was disproved by a mathematician, with its counterexample being $f(x)=a^{-\frac{1}{x}}, g(x)=x$ for $a>1$.

Again, as noted here, it is convention to leave $0^0$ as undefined.

Of course, in combinatorics, it's very useful to leave (and define) $0^0=1$.


Coefficient in a polynomial The coefficient of $x_1^{r_1}...x_k^{r_k}$ in $(x_1 +x_2+...x_k)^n$ is ${n \choose r_1r_2\dots r_k}$.

How would I go about finding the coefficient of $x_1^{r_1}...x_k^{r_k}$ in $(x_1^2+\dots+x^2_k)^n$.

Thanks!

That's easy : substitute your $x^2_i,\,i\in\{1,2,\dots,k\}$ like this :
$$ \forall 1\leqslant i\leqslant k : x^2_i=y_i $$
Then :
$$ [y_1^{\nu_1}y_2^{\nu_2}\dots y_k^{\nu_k}](y_1+y_2+\dots+y_k)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}}, $$
where $[m^2](1-m)^3$ is the coefficient of $m$ in the development of $(1-m)^3$ (that is : $-3$), and $\sum_1^k \nu_i = n$.

Substitute back, and the coefficient of $x_1^{r_1}\dots x_k^{r_k}$ in the expansion of your polynomial is :

$$[x_1^{r_1}x_2^{r_2}\dots x_k^{r_k}](x_1^2+x_2^2+\dots+x_k^2)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}},$$
where, for all $i\in\{1,2,\dots,k\}$, we have : $r_i=2\nu_i$


Given $\{\log_ab \mid a,b\in \mathbb N, \mathrm{gcd}(a,b)=1,a,b≥3\}$ does the sum of any two $\log_ab$ form an irrational or rational number? Given $\{\log_ab \mid a,b\in \mathbb N, \mathrm{gcd}(a,b)=1,a,b≥3\}$ does the sum of any two $\log_ab$ form an irrational or rational number?

I know that $\log_ab$ is irrational, but does the sum of any two $\log_ab$ form an irrational or rational number?

e.g. Is $\{\log_3(5)+log_7(11) \in \mathbb Q\}$ or is $\{\log_3(5)+\log_7(11) \in \mathbb R \setminus \mathbb Q\}$.

Hint What is the identity of this operation? Is it an element in your set?


Finding the equation of the sphere Find the equation of the sphere which touches the three coordinate planes and the plane 2x+y+2z=6.
Please help me to solve this problem. What does three coordinate planes mean?

A sphere that touches all the coordinate planes has center of the form $(\pm r, \pm r, \pm r)$. If the sphere lies in the first octant, then we have
$$\left|\frac{2r+r+2r-6}{3}\right| = r$$ and hence $$4r^2 - 15r + 9 = 0$$ or $r = 3/4$ or $r=3$. Thus the spheres are
$$(x-3)^2 + (y-3)^2 + (z-3)^2 = 3^2$$
and
$$(x-3/4)^2 + (y-3/4)^2 + (z-3/4)^2 = (3/4)^2$$
Suppose that the center is $(r, -r, r)$. Then we must have
$$\left|\frac{2r-r+2r-6}{3}\right| = r$$
and hence $(r-2)^2 = r^2$ and $r=1$. Thus the sphere
$$(x-1)^2 + (y+1)^2 + (z-1)^2 = 1$$
also satisfies the conditions. Similarly we can discuss the other cases.

I get $7$ solutions in all, considering all octants.
Here is a picture
(it doesn't want to load here, being over 2.7MB).

The centre of your sphere is a point $(\pm r, \pm r, \pm r)$, so that it is a distance $r > 0$ from each of the coordinate planes $x=0$, $y=0$, $z=0$, and
is also at distance $r$ from the plane $2x + y + 2z = 6$.


Local minimum global Let $f:(a,b)\to\Bbb R$ be continuous. Assume that $f$ has a local minimum at some point $x_0$. Further assume that this is the only point where $f$ has a local extremum. Does it follow that $f$ has a global minimum at $x_0$.
Thanks

The definition of a local extremum uses "soft" ineqaulities. For example for all $x$ in a neigborhood of your $x_0$ we have $f(x) \ge f(x_0)$.

Then the answer to your question would be yes, I think.

We know there is a neighborhood around $x_0$ such that $f(x) \ge f(x_0)$ for $x$ there. Choose a point from in there, $x_1 \ne x_0$, then actually $f(x_1) > f(x_0)$ strictly, because otherwise $f$ would have a local minimum (of the same size) at $x_1$ too.

Now, we want to prove that this minimum at $x_0$ is global. Suppose it was not. Then some $x_\star$ exists with $f(x_\star) < f(x_0)$ strictly. Without loss of generality, $x_\star$ lies on the same side of $x_0$ as does $x_1$. Then look at $f$ restricted to the compact interval between $x_\star$ and $x_0$. The image of this compact interval contains its supremum. In other words $f$ restricted to the compact interval has a maximum. By construction this maximum is attained in the interior of the interval. So there is a local maximum, which contradicts our hypothesis.

Let's try this again more carefully. (I'll eventually get it right.)

Assume, by way of contradiction, that $x_0$ is not a global minimum. So there exists $x_1 \in (a,b)$ such that $f(x_0) > f(x_1)$. WLOG, we may assume that $x_0 < x_1$. Since $x_0$ is a local minimum, there exists $\varepsilon>0$ such that for all $x\in(x_0-\varepsilon,x_0+\varepsilon)$ we have $f(x) \geq f(x_0)$. Then we can take the interval $[x_0,x_1]$, which is compact, and consider the image under $f$. Since $f$ in continuous, there must be a local maximum and a local minimum in the interval. Since $f(x_0) > f(x_1)$, $x_0$ is not the local minimum. Since there is some $x_2 \in (x_0, x+\varepsilon) \subset [x_0,x_1]$ with $f(x_2) \geq f(x_0)$, $x_0$ is not the local maximum. Hence one of the local extrema occurs in the interval $(x_0,x_1)$. But this gives a contradiction since it implies there is a second local extremum. Therefore $x_0$ is a global minimum.


General Solution of Linear Homogenous Differential Equation The function $y_1=x^\frac{1}{2}$ is a solution of $4x^2y''+y=0$. What is the general solution of the linear homogenous differential equation on the interval $(0,\infty)$.

Can someone help me out. I am preparing for my test and don't know how to do this.

We have
\begin{align}
4x^2y''\left(x\right)+y\left(x\right)=0.\tag{1}
\end{align}
This is a classic example of a Cauchy-Euler equidimensional equation. Substitute $y\left(x\right)=x^r$. Then
\begin{align}
y\left(x\right)=x^r,\\
y'\left(x\right)=rx^{r}x^{-1},\\
y''\left(x\right)=r\left(r-1\right)x^{r}x^{-2}.
\end{align}
This gives us
\begin{align}
0&=4x^2\left(r\left(r-1\right)x^{r}x^{-2}\right)+\left(x^{r}\right)\\
&=4r\left(r-1\right)x^r+x^r\\
&=4r\left(r-1\right)+1.\tag{2}
\end{align}
Now solve this characteristic equation for $r$:
\begin{align}
r&=\frac{4\pm \sqrt{16-4\left(4\right)\left(1\right)}}{8}\\
&=\frac{4\pm 0}{8}=\frac{1}{2}.\tag{3}
\end{align}
Therefore we have $y\left(x\right)=C_0 x^{1/2}$ as a general solution, but one of repeated roots. To find the second solution you make the observation that $\frac{\partial }{\partial r}\left(x^r\right)=x^r\log\left(x\right)$. Therefore, the most "general" solution is
\begin{align}
y\left(x\right)=C_0x^{1/2}+C_1x^{1/2}\log\left(x\right),\;\;x\gt 0.\tag{4}
\end{align}


How can I find these two limits? How can I find these two limits? I've no idea how to improve or continue now. Can someone give me a hint?

1)$$\lim_{x \to 0^+}\left(\frac{\cos^{\pi}(25x)} {\tan^3(x)}\right)=\lim_{x \to 0^+} (\cos^{\pi} (25x)\cot^3(x))$$

2)$$\lim_{x \to 0^-}\left(\frac{\cos^{\pi}(25x)} {\tan^3(x)}\right)=\lim_{x \to 0^-} (\cos^{\pi} (25x)\cot^3(x))$$

Thanks a lot!


Compute $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$ I want to calculate $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$.
No idea in my mind. Any help?
Context
I want to calculate the expected value of bits per symbols in adaptive arithmetic coding when the number of symbols goes to infinity.

Very nice!

I have the same idea the final result is:
$$F(1,1,-n,0)=\left(\frac{G(-n)G(-n-2)}{G(-n-1)^2}\right)=\left(\frac{\pi}{-\sin(n+1)\pi}\right)n!\left(\frac{\pi}{(-\sin(n+3)\pi}\right)\\(n+1)!\left(\frac{\pi}{(-\sin(n+2)\pi}\right)(n+1)!\left(\frac{-2(n+1)}{(n+2)}\right)$$

(There is a line break between the two parts of the equation, for those wondering)


A question about finite field extension of a finite field Let $K$ be a finite extension field of a finite field $F$. Show that there is an element $a\in K$ s.t. $K = F(a)$.

My attempt:

$K$ is a finite field and $char(K) = char(F) := p$. I know that for a given prime $p$ and $n\in N$, there exists a unique finite field upto isomorphism of order $p^n$.

1) But how can I show that $K \simeq GF(p^n)$ where $p = char(K)$?

Once I have that, $K\setminus \{0\}$ is a cyclic group under multiplication and thus $\exists a\in K$, s.t. $K\setminus \{0\} = <a>\subset F(a)$. Thus $K\subset F(a)$.

2) Also, how can I show that $F(a)\subset K$ so that $F(a) = K$? $K$ is an finite field extension of $F$ but I can't understand why that would imply $F(a)\subset K$??

Let $a$ be a generator of the multiplicative group $K^\times$. The elements $a,a^2,...,a^{|K|-1}$ are all different elements in $F(a)$ and hence $|F(a)|\geq |K|-1$. On the other hand $F(a)\subseteq K$ and hence $|F(a)|\leq |K|$. Since the order of $F(a)$ must be a power of $p$ (and $|K|-1$ is not a power of $p$) we conclude that $|F(a)|=|K|$. Since $F(a)\subseteq K$ and both fields are finite this implies $F(a)=K$.

Since the finite extension of a finite field is also a finite field, so there are at most finitely many middle fields $E_i$ such that $F\subset E_i\subset E$. so $E/F$ is a simplicial extension.


In how many ways can 10 cards be drawn so that the 10th card is the repetition of a previous draw? Suppose that we draw a card from a deck of 52 cards and replace it before the next draw. In how many ways can 10 cards be drawn so that the 10th card is the repetition of a previous draw ?

Solution:

No of ways the 9 cards can be chosen $ = 52^9$

And for the 10th card that should be repetition of 9th card should have only $1$ possibility (i.e. Same as 9th Card)

So the answer should  be $52^9$ but the answer is $52^{10} - (51)^9.52$

Can anyone explain how it is calculated ?

Does the previous draw means all the previous draws ? If that's the case then its correct.

Be careful about the distinction between the words the and a.  If you see the word the, it means only one.  The word a signifies that there may be more than one.

The problem says that the tenth card is a repetition of a previous draw, meaning that the tenth card can match any of the previous nine draws.

The justification for the answer is that we are subtracting those cases in which the tenth cards fails to match any of the previous draws from the $52^{10}$ ways of drawing ten cards from a standard deck of $52$ cards with replacement.  There are $52$ possible selections for the tenth card.  If none of the previous draws match that card, they must all have been selected from the other $51$ cards in the deck, which can occur in $51^9$ ways.  Hence, there are $51^9 \cdot 52$ ways for the tenth card not to match any of the previous nine draws.  Thus, there are $52^{10} - 51^9 \cdot 52$ ways for the tenth card to match at least one of the previous nine cards.

It implicitly says it is with replacement. But in that case the previous draws are irrelevant, the probability is just getting the same card in two draws, i.e. repeat whatever card you get the ninth draw in the tenth. If the last card is drawn at random (note that even how the ninth draw is done is irrelevant!), this is just $1/n$ for $n$ cards.


If $n$ is coprime to 10, then $1/n$ produces a repeating decimal. EDIT: I have changed the title (twice). I am getting my terms and symbols jumbled up in my head. I hope that I have asked a clear question now.

Here are three statements that I believe to be true, but have only been able to prove them in one direction.

CASE 1: Iff $n = 2^x5^y$ then $1/n$ produces a terminating decimal.
CASE 2: Iff n is coprime to 10, 1/n produces a purely repeating decimal
CASE 3: Iff $n = 2^x5^yn'$, where $n'$ is coprime to 10, then 1/n produces a repeating decimal after a non-repeating section.

I already proved case 1 both ways, but I'm stuck on case 2. I think if I can prove case 2, then case 3 would be very easy.

Question:
Are the above statements true? If so, what are some hints to help me solve case 2?

My work so far:

Assume that $n$ is coprime to 10.

$\implies 10^k \bmod n = m$, where $m \in \{1, \dots, 9\}, \forall k \in \mathbb{N}$

$\implies$ I'm really not sure where to go next...

I should add that I never studied number theory before. This is for fun, on my own time, so simple explanations are better for me :)

Yes, all three statements are true.

Terminating expansions: A positive real number $x$ has a terminating decimal expansion iff there exists $k$ such that $10^k x \in \mathbb N$. Now suppose $x = 1/n$. Then
$$\begin{align*}
&\exists k \in \mathbb N : 10^k \frac{1}n \in \mathbb N \\
\iff &\exists k, \ell \in \mathbb N : 10^k = n \ell
\end{align*}$$
That is, $1/n$ has a terminating decimal expansion iff it divides $10^k$ for some $k$. That is, iff its only prime divisors are $2$ and $5$. This proves the first claim.

Periodic expansions. A positive real number $x<1$ has a periodic decimal expansion iff there exists $k>0$ such that $10^k x - x \in \mathbb N$, i.e. $(10^k-1) x \in \mathbb N$. Now take $x = 1/n$, with $n > 1$. Then this is equivalent to saying that there exists $k>0$ such that $n$ divides $10^k - 1$. We want to show that this is equivalent to $10$ being coprime to $n$.

$n$ coprime to $10$ $\implies$ periodic expansion: Consider the numbers $10^0, 10^1, 10^2, \ldots$ and their remainders upon division by $n$. Because there are infinitely many numbers and only finitely many possible remainders, ...

   by the pigeonhole principle, there exist $p, q \in \mathbb N$ with $p < q$ and such that $10^p$ and $10^q$ have the same remainder upon division by $n$.

That is, ...

   $n$ divides $10^q - 10^p$.

We have $10^q - 10^p = 10^{p} (10^{q-p}-1)$.
Because $n$ divides this number, and $n$ is coprime to $10$, it must be that ...

   $n$ divides $10^{q-p} - 1$.

Thus $1/n$ has periodic expansion.

Periodic expansion $\implies$ $n$ coprime to $10$: By assumption, there exist $k, \ell \in \mathbb N$ with $10^k -1 = n \ell$, and $k \geq 1$. Suppose $n$ and $10$ are not coprime. Then there exists $d > 1$ that divides both $10$ and $n$. But then $d$ divides ...

  $$10^k - n \cdot \ell = 1\,,$$

... a contradiction with $d > 1$. Hence $n$ is coprime to $10$.

Eventually periodic expansions. The decimal expansion of any rational number is finite or eventually periodic, so it must be that the natural numbers $n$ not covered by the previous two cases are precisely those such that $1/n$ has an infinite, not immediately periodic decimal expansion.

Counterexample to the title:

Prove that, in general if $1/n$ produces a repeating decimal, then $n = 2^x5^y$.

Let $n=7$.  $$\frac{1}{7} = 0.\overline{142857}$$

The number $7$ is prime, so $7 \ne 2^x 5^y$.

UPDATE

Hey, the title changed.  This actually was a valid response at the time!


Lower bound for monochromatic triangles in $K_n$ Say $K_n$ is a complete graph of $n$ nodes, and every edge is either blue or red. I'm trying to find $T_n$, which is the lower bound for the number of monochromatic triangles in $K_n$ (monochromatic meaning triangles whose 3 edges are of the same color).

I'm trying this approaches:

A. Let $v_i$ a vertex in $K_n$. This vertex belongs to $\binom{n-1}{2} $ different triangles. Let $r$ be the number of red edges that $v_i$ belongs $b$ the number of blue edges. Of course,
$$r + b = n-1$$
So, the number of non-monochromatic (dual/stereo-chromatic?) triangles is $r \cdot b$, and so our bound would be (for vertex $V_i$)

$$T_n \le \binom{n-1}{2} - r \cdot b$$

Question is - how do I know when $r \cdot b$ is the maximal possible number?
Also, this would give me the upper bound, how do I find the lower bound?

Thanks in advance!

Here's a proof, from Wikipedia, that $K_6$ has at least 2 monochromatic triangles. You may find that this method of proof generalizes to give you the lower bound you want.

"An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices $x, y, z$ such that the edge $(xy)$ is red and the edge $(yz)$ is blue. Firstly, any given vertex will be the middle of either $0\times5=0$ (all edges from the vertex are the same color), $1\times4=4$ (four are the same color, one is the other color), or $2\times3=6$ (three are the same color, two are the other color) such triples. Therefore there are at most $6\times6=36$ such triples. Secondly, for any non-monochromatic triangle $(xyz)$, there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the $K_6$ are monochromatic."

On the other hand, if you want all the work done for you, there is an answer in A W Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783, also in Frank Harary, The two-triangle case of the acquaintance graph, Math. Mag. 45 (1972) 130-135.

If you take $G = K_n$, $n\geq 3$ and colour each edge red with probability 1/2 and blue otherwise. Define $X$ to be the random variable counting the number of monochromatic triangles. Then

$E[X] = \binom{\binom{n}{2}}{3}\left(\frac{1}{2}\right)^3$

Hence, there is a colouring of $G$ such that there are least this many.


Integration - finding the area between two polar curves I have attempted to solve a polar curves question and I keep getting the wrong answer.

The question is:
Find the area enclosed by between the curves

$$
r = 3 - 3\cos\theta
$$
and
$$
r =4\cos\theta
$$

So far I have forked out the angle that they interest at so:

$$
4cos\theta = 3 - 3cos\theta
$$
$$
\therefore \theta = 1.127885 radians
$$
Then I integrating with the upper bound as 1.127885 and the lower bound as zero.
$$
\int_{0}^{1.127885}{\frac{1}{2}*(4\cos\theta)^2}d\theta = 6.06037
$$

$$
\int_{0}^{1.127885}{\frac{1}{2}*(3-3\cos\theta)^2} d\theta  = 0.3527
$$

Then the difference in area is 6.06037 - 0.3527 = 5.70767.

Then by symmetry the total area is 5.70567*2 = 11.4

The answer I get is 11.4 but apparently the answer 1.15. Could you help me with this?

Well, that first integral is certainly suspect. Observe that when $t=0$, $r=4$, so you’ve computing the area of most of the upper black semicircle in the diagram below:

Your first integral computes the area of that semicircle less the segment bounded by the chord from the origin to point $A$ (the dotted line). Your value is plausible for this area—it’s a bit under $2\pi$, the semicircle’s area. The second integral computes the area between the blue curve and the dotted chord, so when you subtract its value, you take another little bite out of the semicircle. All in all, you’ve computed an area outside of the blue cardioid instead of inside it. What you need to do instead is compute the area of that segment of the circle bounded by the chord $OA$ and add that to the area between the cardioid and $OA$, which you’re already computing correctly.

grapg

I think it's will help you
You need to calc the diff between this areas


Reciprocal sum equal to $1$ Given the equation $$1/v + 1/w + 1/x + 1/y + 1/z = 1,$$ where all of $v,w,x,y,z$ are positive integers. One solution is the trivial $1/5 + 1/5 + 1/5 + 1/5 + 1/5$. For distinct solutions with $v<w<x<y<z$ find the minimum of $v+w+x+y+z$.

What is the minimum of $s:=x+y+z+u+v$ subject to $x<y<z<u<v$ and $\frac1x+\frac1y+\frac1z+\frac1u+\frac1v=1$?
From $\frac12+\frac14+\frac17+\frac1{14}+\frac1{28}=1$, we know that $s\le 55$. Hence this is a finite problem.
A quick brute force check of all possibilities with $x+y+z+u+v<55$ reveals that
$$\tag{38} \frac13+\frac14+\frac15+\frac16+\frac1{20}=1$$
with $s=38$ is the optimum.
The next solutions with small sums are:
$$\tag{43} \frac12+\frac14+\frac1{10}+\frac1{12}+\frac1{15}=1 $$
$$\tag{45}  \frac12+\frac14+\frac1{9}+\frac1{12}+\frac1{18}= \frac12+\frac15+\frac1{6}+\frac1{12}+\frac1{20}=1  $$
$$\tag{50} \frac12+\frac14+\frac1{8}+\frac1{12}+\frac1{24}=1 $$

I suspect $2,4,8,12,24$ with sum $50$ will be hard to beat, but don't have a proof (yet).


My proof that $4$ either divides $n^2$ or $n^2 − 1$ I am attempting to prove the following statement

Prove that if n is any integer then 4 either divides n^2 or n^2 − 1

I have started with the case of n = 2k
Consider the case n = 2k

  n = 2k
  n^2 = 4k^2
⇒ n = 4k

∴ 4 divides n^2 as there is some integer k in which n = k

Would this be considered as a correct proof for the first case here? Is there any additions I should make?
I also attempted case 2, where n = 2k+1, however I am less sure of the direction I have taken this and is incomplete, so some advice on this would also be appreciated.
Consider the case n = 2k+1

n       = 2k+1
n^2     = (2k+1)^2
n^2     = 4k^2 + 1
n^2 - 1 = 4k^2

Consider the case $n=2k$, then $$n^2=4k^2$$
$$\therefore \space 4|n^2 \space \text{that is if $n$ is even then $n^2$ is divisible by $4$}$$
For the case $n=2k+1$ we have $$n^2-1=(2k+1)^2-1=(2k)^2+2k+2k+1-1=4k^2+4k=4(k^2+k)$$
$$\therefore \space 4|(n^2-1) \space \text{that is if $n$ is odd then $n^2-1$ is divisible by $4$}$$
Since all integers are of the form $2k$ or $2k+1$ for some $k\in \mathbb Z$, we are done.

Here's another simple proof using parity. $[$Actually the same thing what @Ares did, but not taking the form of $2k$ or $(2k + 1)$$]$
Suppose $n$ is even, then $2|n$, which implies $4|n^2$, so $n$ has to be odd.
Now if $n$ is odd, then  note that $(n^2 - 1) = (n + 1)(n - 1)$ . Here in this case both $(n + 1)$ and $(n - 1)$ will be even as $n$ is odd.
So $2|(n+1)$ , $2|(n-1)$ which implies that :-  $$(2*2)|(n+1)(n-1)$$
$$\rightarrow 4|(n^2 - 1)$$
In other words, $4|n^2$ if $n$ is even, else $4|(n^2 - 1)$ if $n$ is odd.


condition for roots of quartic equation to be purely imaginary (a) Show that the roots of equation $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ where $a_1, a_2, a_3, a_4$ are real constants different from zero, has a pure imaginary root if $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$
  (b) Is converse of (a) true?

I tried factoring $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = (z^2 + b_1 z + b_2)(z^2 + c_1 z + c_2)$ and compared (equating both $b_1 = c_1 = 0$) but couldn't get the required results (that would get $a_1 = 0 $). Did I make any mistake on assumptions? I am beginning to doubt this question? Any suggestions, hints?

We will assume the problem means that the polynomial has a purely imaginary root, not that all the roots are purely imaginary.

The converse is not true unless you also know that $\frac{a_3}{a_1}>0$.

To do this, show that if $$a_3^2 + a_1^2 a_4 = a_1 a_2a_3$$ then $$z^4+a_1z^3+a_2z^2+a_3z+a_4 = (z^2+\frac{a_3}{a_1})(z^2+a_1z+\frac{a_4a_1}{a_3})$$

The coefficients for $x^3$ is $a_1$ on both side, the coefficient for $z$ is $a_3$ on both sides, and the constant is $a_4$ on both sides. So you only need to prove that the coefficient for $z^2$ is $a_2$ on both sides. On the right, that coefficient is $$\frac{a_4a_1}{a_3} + \frac{a_3}{a_1} = \frac{a_4a_1^2+ a_3^2}{a_1a_3}=\frac{a_1a_2a_3}{a_1a_3} = a_2$$

So the two polynomials are equal. But $z^2+\frac{a_3}{a_1}$ has a purely imaginary root when $\frac{a_3}{a_1}>0$, but does not when $\frac{a_3}{a_1}<0$. (The case of equal zero depends on purely imaginary, but since we know that $a_3\neq 0$, this is irrelevant.)

On the other hand, the other part $z^2+a_1z + \frac{a_4a_1}{a_3}$ never has purely imaginary roots when $a_1\neq 0$. So we can say that when we have real $a_1,a_2,a_3,a_4\neq 0$, then the polynomial $$z^4+a_1z^3+a_2z^2+a_3z+a_4$$ has a purely imaginary root if and only if $a_3$ and $a_1$ have the same sign and $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$.

Note, the image from the book is actually the opposite problem, and it is an error on the part of the publisher/authors. It asks you to prove that the polynomial has a pure imaginary root if $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$, which is not the case. For a simple example, let $(a_1,a_2,a_3,a_4)=(-1,1,1,-2)$. Then $$z^4+a_1z^3+a_2z^2+a_3z+a_4=(z-1)(z+1)(z^2-z+2)$$ which does not have a pure imaginary root.

let me come again descriptively.

let $z=ix$ (purely imaginary)

After substituting the value of $z$ at given equation we get a polynomial in terms of $x$. (mark it 1).

Again $z=-ix$ (conjugate of $ix$) will also satisfy the polynomial. (mark it 2)

Subtract 1 from 2, and we'll get $x=\pm\sqrt{a_3/a_1}$.
Use this value of $x$ in any of the equation (1 or 2). And we'll get the desired condition of having a purely imaginary root.
(suggestion: have a pen and paper)


Why can $5^{2n+3} + 3^{n+3} \cdot 2^{n} \quad \forall n \in \mathbb{N}$ never be a prime number? It seems to be true for the first thousand $n$ but I really can't think of a way to  prove this statement.

Any kind of help will be appreciated!

Hint $ $ Write it as $\,f_n.\ $ $\gcd(f_0,f_1) = 19\ $ so we check if the factor $19$ persists for all $\,n.$

$\qquad\ \  {\rm mod}\,\ 19\!:\,\ f_n\equiv 5^3\, 25^n + 3^3\, 6^n \equiv 11\cdot 6^n + 27\cdot 6^n \equiv 38\cdot  6^n \equiv 0\ \Rightarrow\ 19\mid f_n$

Remark $\ $ The reason that the common factor of  $\,f_0,f_1\,$ persists for all $\,n\,$ is that $\,f_n\,$ satisfies a $2$nd order recurrrence $\, f_{n} = 31f_{n-1} - 150 f_{n-2}.\,$ So $\,19\mid\color{#c00}{f_0,f_1}\,\Rightarrow\, 19\mid f_2 = 31\color{#c00}{f_1} - 150 \color{#c00}{f_0}.\,$ In the same way $\,19\mid f_1,f_2\,\Rightarrow\, 19\mid f_3.\,$ This shows the way to give an alternative (strong) induction proof that $\,19\mid f_n\,$ for all $\,n\ge 0.\,$ Or you could use the recurrence and strong induction to show there are integers $\,j,k\,$ such that $\, f_{n} = j f_0 + k f_1,\,$ for all $\,n\ge 0,\,$ because such integral linear combinations of $\,f_0,f_1\,$  are closed under integer scalings and addition.

This is how I would think about it:

$(5)^{2n+3}$ is going to be an odd, since a product of all odd integers is always going to be odd (if you want me to prove that I'd be more than happy to).

So then, think of the second part of the statement as
$$
A = [(2)(3)]^n \text{ and } B = (3)^3 = 27
$$
so we can write the second part as:
$$
A*B
$$
Since A is even, and B is odd, the product of $A$ and $B$ is going to be odd as well (again, if that doesn't makes sense, I can prove that as well).

So now we have the sum of 2 odds, which is even. And other than 2, even numbers can't be prime! Hope that helps!

-ZZ


2-dimesional cell complexes with fundamental group isomorphic to the following. I have been asked to give examples of 2-dimensional cell complexes whose fundamental group isomorphic to the following

$$ \Bbb Z_4 * \Bbb Z_5$$and
$$\Bbb Z_4\times \Bbb Z5$$

I know in the first case I am looking for the wedge of 2 spaces whose fundamental groups are $\Bbb Z_4$ and $\Bbb Z_5$ respectively and for the second I'm looking for the the cross product of two spaces with the same fundamental groups.

However I haven't come across any examples of spaces whose fundamental groups are anything but $\Bbb Z $ either crossed with itself or the free product of $\Bbb Z $ with itself.

Can anyone give examples of spaces whose fundamental groups are different?

Here are some hints. Most of what follows is already contained in your question and the comments.

As suggested by Dan in the comments, you should start out by finding two connected $2$-complexes $X$ and $Y$ that have $\Bbb Z/4$ and $\Bbb Z/5$ as fundamental groups. One way of constructing a connected $2$-dimensional CW complex with fundamental group $\Bbb Z/n$ (for some integer $n\geq 2)$ is to start with the circle $S^1$, and to attach a $2$-cell $D^2$ with the attaching map $\partial D^2\to S^1, z\mapsto z^n$.

To obtain the first space you want, just choose base points in $X$ and $Y$ and consider $X\vee Y$. This will give the desired result since the basepoint in $X\vee Y$ has a contractible neighborhood, and using van Kampen's theorem.

For the second construction, you proposed the idea of using $X\times Y$. While this breaks the dimensional requirement, it can be salvaged: just consider the $2$-skeleton of the product complex $X\times Y$. It is a general fact that, for any integers $0\leq k\leq n$, and any CW-complex $X$, the $k$-skeleton inclusion $X_n\hookrightarrow X$ induces isomorphisms on the homotopy groups $\pi_k(X_n)\to\pi_k(X)$ while $k<n$ (and a surjection $\pi_n(X_n)\to\pi_n(X)$). Hence, the $2$-skeleton $Z=(X\times Y)_2$ of the $4$-dimensional product complex $X\times Y$ has its fundamental group isomorphic to that of $X\times Y$, i.e. $\pi_1(Z)\simeq\Bbb Z/4\times\Bbb Z/5$.

Of course, it would be simpler to remember that $\Bbb Z/4\times\Bbb Z/5\simeq\Bbb Z/20$ since $4$ and $5$ are coprime, and apply the first construction.

A cell complex with 1 vertex, one edge (both ends attached to that vertex, forming a circle that I'll treat as $S^1$, and one 2-cell, attached to the 1-skeleton via the map
$$
\theta \mapsto 2\theta
$$
in which the boundary of the 2-cell double-covers the 1-skeleton, has $H_0 = Z$ and $H_1 = Z/2Z$. You should confirm this (and read up on $RP^2$ if you cannot do so).

Generalize to find a complex with $H_1 = Z/nZ$.

See what you can do from there.

Hint: for the second part, you're going to need a generator $a$ for the first cyclic group and a generator $b$ for the second to commute. You can get that by attaching a 2-cell along $aba^{-1}b^{-1}$. If this doesn't make sense, look at the torus and compute its homology to see why the latitude and meridian generators commute.


Solve Pell's equation Bonjour, How can I solve a pell's equation $x^2-181y^2=180$? I try continued fraction method but it is too long.


Percentage relations between x and y.(Relation between two variables) I would like to know easiest method to solve following:

Que 1:

If 10% of x=y, then y% of 30 is same as what % of x?

My Approach:

Step 1: (10 * x)/100 = (10 * y)/100

Step 2: (30 * y)/100 = (3 * y)/10

Don't know further.

Que 2:

What percentage of numberes from 1 to 89 are divisible by 5?

My Approach:

Step 1: Total numberes divisible by 5 from 1-89 = 17

Step 2: Considering 90 as 100%

Step 3: So 17 is what % of that

Which i solve like

(17 * 100)/90 = 18.888%

I get ans: 18.8888%, but i think i am wrong, help here.

Thank You!

For question $1$, you have: $\dfrac{x}{10} = y \Rightarrow \dfrac{y}{100}\cdot 30 = \dfrac{k}{100}\cdot x=\dfrac{k}{100}\cdot 10y\Rightarrow k = ?$

For question $2$, you have $17$ numbers that are multiple of $5$ out of $89$ numbers. Thus the percentage is $\dfrac{17}{89}\cdot 100\text{%} \approx 19 \text{%}$

For the 1st one: $10\%$ of $x=y$, then $y\%$ of $30$ is same as what $\%$ of $x$?

Consider "of" as multiplication, so
$10/100=1/10=y\%$

Now $y\%$ of $30$ means,
$1/10*30=3$.
So the answer is $3$.


What is the meaning of this sentence? "The right side of $a\equiv b \pmod c$ is less than or equal to the left side." Let us consider a congruence of the form
$$a\equiv b \pmod c$$
What is the meaning of the following sentence?
$$\text{The right side of the congruence is less than or equal to the left side.}$$

For context, the Jones, et.al, American Mathematical Monthly article "Diophantine Representation of the Set of Prime Numbers" (PDF link via umd.edu) has the following (Lemma 2.4, pg 452; emphasis added):

For any numbers $p$, $n$ and $a\geq 1$ we have the congruence
$$\chi_a(n) \equiv p^n+\psi_a(n)(a-p) \pmod{2ap-p^2-1}$$
Furthermore, when $0< p^n<a$, the right side of the congruence is less than or equal to the left side.

I am not an expert on the topic, but I think you can use the definition of congruence modulo which is that $\displaystyle a\equiv b\ (\bmod c) \ \Leftrightarrow a-b=ck$, so I would think you can replace the equality sign with a greater than or equal to. It would then be $\displaystyle a-b\geqslant ck$. I am not sure, however.


Cardinality of a set A is strictly less than the cardinality of the power set of A I am trying to prove the following statement but have trouble  comprehending/going forward with some parts! Here is the statement:

  If $A$ is any set, then  $|A|$ $<$ $|P(A)|$

Here is what I have so far:

We need to show that there is an injection from $A$ to $P(A)$ but not a surjection.

A natural choice for an injection is the function $ f(x)$ $=$ $\{x \}$, which in plain English, takes any element $x$ (that is in $A$) and sends it to the one-element set $\{x \}$. Thus $f(x)$ is injective!

To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from $P(A)$) to elements (from $A$) right?

If the above is correct, my problem arises here. I am not sure how to prove that $f$ is not surjective. Unfortunately, I am easily confused by notation so please explain in English.
Thank you in advance!! :)

What you want here is the so-called diagonal argument. The idea is to show that no matter what function $f:A\to\wp(A)$ you look at, you can find a subset $S_f$ of $A$ that is not in the range of $f$. If you can do that, you’ve shown that there is no map of $A$ onto $\wp(A)$ and therefore certainly no bijection from $A$ to $\wp(A)$.

To build the set $S_f$, imagine that you could somehow go through the set $A$ one element at a time. You look at an element $a\in A$, and one of two things must be true: either $a\in f(a)$, or $a\notin f(a)$. (Remember, $f(a)$ is some subset of $A$, so it’s meaningful to ask whether that subset contains $a$.) Since we’re building the set $S_f$ to suit ourselves, we get to decide whether $a\in S_f$ or not, and we’ll decide in exactly the opposite way from the function $f$: if $a\notin f(a)$, we’ll put $a$ into $S_f$, and if $a\in f(a)$, we won’t put $a$ into $S_f$. After we’ve done this for each $a\in A$, our set $S_f$ will contain exactly those $a\in A$ such that $a\notin f(a)$:

$$S_f=\{a\in A:a\notin f(a)\}\;.$$

For each $a\in A$, therefore, the sets $S_f$ and $f(a)$ differ in how they treat $a$: if $a\in f(a)$, then $a\notin S_f$, and if $a\notin f(a)$, then $a\in S_f$.

That’s almost the entire argument: all you have to do to finish it off is explain why this ensures that for $S_f$ is not the set $f(a)$ for any $a\in A$ and why this implies that $S_f$ is not in the range of $f$ and hence that $f$ is not a surjection.

Suppose that we had such a matching. Then for every element a ∈ A, there
is a subset Sa of A matched to it, so that every subset of A occurs as Sa for some a.
We have to show that this is impossible, by producing a set which is missed out.
We follow Bertrand Russell’s idea (see the supplementary material for Chapter
2). Let B consist of all those elements a of A which do not lie in the subset
they label:
B = {a ∈ A : a ∈/ Sa}.
Now, if we have a complete matching, then the subset B must occur in the matching,
that is, B = Sb for some element b ∈ A. But this is impossible. For, if b ∈ Sb,
then by definition b ∈/ B, while if b ∈/ Sb, then b ∈ B; so the sets B and Sb differ
at the element b (one contains it, the other doesn’t), and cannot be equal. So the
proposed matching is not complete


Is the blue area greater than the red area? Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?

Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since  the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that $$\begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}$$

Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.

Is this true? If so, is there another way of proving the assertion?

Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.

Source:
The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$.

The four numbered areas are congruent.

[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

The two areas are equal.  On the diagram with the red area draw the vertical and horizontal lines that define the blue area.  The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area.  Those two triangles are congruent.


estimate the perimeter of the island I'm assigned a task involving solving a problem that can be described as follows: Suppose I'm driving a car around a lake. In the lake there is an island of irregular shape. I have a GPS with me in the car so I know how far I've driven and every turns I've made. Now suppose I also have a camera that takes picture of the island 30 times a second, so I know how long sidewise the island appears to me all the time. Also assume I know the straight line distance between me and the island all the time. Given these conditions, if I drive around the lake for one full circle, will I be able to estimate the perimeter of the island? If yes, how? Thanks.

Assume the island $\Omega$ is convex. Parametrize the observed width $w$ of $\Omega$ with the direction from which it is observed. So we have a function $\phi\to w(\phi)$ of period $\pi$. The average observed width is $$\bar w:={1\over 2\pi}\int_0^{2\pi }w(\phi)\ d\phi\ .$$
Consider a line element $ds$ of $\partial\Omega$. Its contribution to $\bar w$ is half its average projection length over all directions $\phi$ (here we use the convexity of $\partial\Omega$), so it is $\ \lambda\thinspace\thinspace ds\ $ for some universal constant $\lambda>0$ yet to be determined. In any case we have  a formula of the following kind:

$$\bar w = \lambda \int_{\partial\Omega} ds =\lambda\ L(\partial\Omega)\ .$$

When $\Omega$ is the unit disk then $\bar w=2$ and $L(\partial\Omega)=2\pi$. It follows that $\lambda={1\over\pi}$, and solving for $L(\partial\Omega)$ we get

$$L(\partial\Omega)=\pi\ \bar w={1\over 2}\int_0^{2\pi} w(\phi)\ d\phi\ .$$

I am going to take a stab at this, although I would like to hear a flaw in my argument.

If you have a reference point on the island where your camera is always pointed to then, since you know your exact travel path and distance from your path to the edge of the island, you can plot the shape of the island, then, finding the area can be done in multiple ways (integration being one of the ways).

If you do not have a reference point that the camera(your eyes) can always point to, then I think the problem is tricky, and I believe there is no solution.


Number of 6-character passwords with at least 1 digit I have two methods for calculating the answer to a problem, and I'm unable to figure out where one of them goes wrong (they yield different values).

My question is, why is this the case? Am I doing something wrong in my calculation, or my logic?

In my class today the professor discussed how to calculate the number of password combinations under the following conditions:

The password is 6 characters long
The password contains only upper-case letters and digits
The must be at least 1 digit in the password

His solution was to take the total combinations of 6 letters and digits, and subtract the number of cases where there are only letters in the password:

combinations = $36^6 - 26^6$ (approx 1.868 million)

I tried solving the same problem with the following approach, getting a different value:

Take all the combinations of 5 letter-or-digit characters
Multiply that by the number of possible positions for the mandatory digit (6)
Multiply that by the number of values the mandatory digit could have (10)

combinations = $36^5 \cdot 6 \cdot 10$ (approx 3.628 million)

You counted several things twice in your example. For example, take a look at the password AAAA11.

You counted that password as the password you get when you put the mandatory digit on the sixth place, and write AAAA1 on the other places, and also as the password you get when you put the mandatory digit on the fifth place, and write AAAA1 on the other places.

You will use this password to access your account. Enter a combination of at least six numbers, letter and punctuation marks.


Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.

The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?

$$
\begin{align}
n(n+1)(n+5)
&=n(n+1)(n+2)+3n(n+1)\\[6pt]
&=6\binom{n+2}{3}+6\binom{n+1}{2}
\end{align}
$$

Binomial Coefficient Basics

If, instead of Pascal's Triangle, we define the binomial coefficients as
$$
\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}
$$
Then we have
$$
\begin{align}
\binom{n-1}{k}+\binom{n-1}{k-1}
&=\binom{n-1}{k-1}\frac{n-k}k+\binom{n-1}{k-1}\\
&=\binom{n-1}{k-1}\frac nk\\
&=\binom{n}{k}
\end{align}
$$
Thus, if $\binom{n-1}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$, then $\binom{n}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$.

You just needed one more step.
$$\begin{align}n(n+1)(n+5) = & (n+1)(n+2)(n+6) \\ = & n^3+9 n^2+20 n+12 \\ = & (n^3+6n^2+5n)+3n^2+15n+12 \\ = & n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$

$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.

Of course you could have done this immediately.

$n(n+1)$ is divisible by $2$.
If neither $n$ nor $n+1$ are divisible by $3$, then $n+2$ and $n+5$ will be.

exactly one of the three factors, $n$, $n+1$, $n+5$ is divisible by $3$.

Therefore $n(n+1)(n+5)$ is divisible by $2$ and $3$, and thus by $6$.


$3$ indistinguishable dice are rolled twice. What is the probability that both throws have the same configuration? $3$ indistinguishable dice are rolled twice, independently. If each face of each die has the equal probability to appear uppermost, what is the probability that both the throws have the same configuration?

What is the sample space of this experiment? For distinguishable I know that the sample space has size $6^6$. What will happen for indistinguishable case?

Please help me in this regard.

Thank you very much.

The first throw has $6^3$ outcomes (some are equivalent as to configuration).

There are $\binom{6}{1}=6$ outcomes where all dices are the same, so in the next throw chances are only $1$ in $6^3$ you'll get the same configuration.

There are $\binom{6}{3}3!=120$ outcomes where all dices are distinct, and the next throw has $3!$ outcomes with the same configuration (differently permuted which is ignored/indistinguishable).

The remaining $6^3-6-120=90$ outcomes have the form $(a,a,b)$ or $(a,b,b)$. In each such case the next throw has $3!/2!=3$ possible outcomes with the same configuration.

The weighted average of the probabilities in each case is

$$\frac{6\dfrac{1}{6^3}+120\dfrac{3!}{6^3}+90\dfrac{3}{6^3}}{6^3}=\frac{83}{3888}\approx 2.135\%$$

Could anyone please give help me for clarification?

If three indistinguishable dice are rolled for the first time, probability that any configuration will be formed is 1 (for in the question no specific conf is asked). Since the dice are indistinguishable the possible number of different configurations is 56. So that in the next throw, only 1 of the 56 configurations can be formed. The probability should then be 1/56.
If the dice are distinguishable, possible number of different confs are 216. First throw gives probability 1, and second throw gives 1/216, like the same principle above.

What is the error in this solution?


Prove that if $n \in \Bbb{N}$ and $n > 1$ is not prime, then $\exists p$ prime such that $p \mid n$ and $p \leq \sqrt{n}$ Not really sure how to do this question

this is what I have so far

$n = a \cdot b$

$(a \leq b)$

$a > \sqrt{n}$

$b > \sqrt{n}$

$ab > n$

You actually proved that if $n$ is not prime, it has a divisor $a<\sqrt n$.

What remains to prove is that the smallest $a>1$ that divides  $n$  is prime.

Indeed, suppose $a$ has a non-trivial factorisation: write $a=bc$, $\enspace1<b,c<a$. Then $n$ has divisors $>1$ that are strictly less than $a$, so that $a$ is not the smallest divisor $>1$ of $n$. Contradiction.

As $n$ is not prime it will at least have $2$ prime factors (may be repeated) and both of them can't be greater than $\sqrt n$, otherwise their product will exceed $n$. Hence at least of them will be $\leq \sqrt n$.


Finding the characteristic ODE from a nonlinear PDE I am studying for a PDE exam on Tuesday, and I am getting pretty confused about one specific type of problem and I am thinking that perhaps I am misinterpreting the correct procedure to follow. The problem is as follows: solve the equation $$y(\frac{\partial{u}}{\partial{x}})+x( \frac{\partial{u}}{\partial{y}}) = 0 $$ $$u(0,y) = e^{-y^2} $$

The way I wanted to do it is as follows:
$$
\frac{\partial{u}}{\partial{x}} + \frac{x}{y} (\frac{\partial{u}}{\partial{y}} )= 0
$$
$$
\frac{\partial{y}}{\partial{x}} + \frac{x}{y} = 0
$$
$$
\frac{\partial{y}}{\partial{x}} = - \frac{x}{y}
$$
$$
y \space \partial{y} = -x \space \partial{x}
$$
Integrating both sides yields
$$
\frac{y^2}{2} + \frac{x^2}{2} = C
$$

However, on the following solution (pictured below), there are two things I don't understand: how can one possibly arrive at $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$??? I have seen "slightly wrong" steps like this on several problems of this nature, and so I suspect I am simply misunderstanding the step that is being taken. Of course if $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$ is correct then it is indeed possible to arrive at the equation $x^2 - y^2 = C$ but I am truly at a loss as to how you can just introduce a minus sign to only one term in an equation. Is the author not actually solving the PDE but doing something else? I see that the exercise asks to find the characteristics; is this something related?

Also, is this PDE nonlinear? I thought that nonlinear meant that there were terms where partial derivatives of a function with respect to one of its independent variables were multiplied by other order derivatives of the same function (e.g. $u(x,y)(\frac{\partial{u(x,y)}}{\partial{y}})$) or when partial derivatives of a function were multiplied by the function itself, or independent variables of the function (e.g. $y\frac{\partial{u(x,y)}}{\partial{y}}$).

Denote the PDE as

$$a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} = 0, \tag{1}$$

where

$$a(x, y) = y \text{ and } b(x, y) = x. \tag{2}$$

Is this PDE nonlinear?

A PDE is nonlinear if the solution space of the homogeneous equation (with RHS = 0) is not a vector space. Suppose that $u$ and $v$ are solutions of the PDE (1-2), and that $k \in \mathbb{R}$ is an arbitrary constant. It can be shown that $k u$ and $u + v$ are solutions of the PDE (which is a homogeneous equation). Therefore the solution space of (1-2) is a vector space; therefore (1-2) is a linear PDE.

How can one possibly arrive at $\frac{d y}{d x} = \frac{x}{y}$?

Using the method of characteristics (MoC). I will give a simplified description of the MoC that is applicable to your current problem. As I recall, Strauss discusses the MoC in more generality.

Consider a curve $\Gamma$ given by the parametrization $\Gamma(s) = \begin{bmatrix} x(s), & y(s) \end{bmatrix}$, $s \in I$ ($I$ is an arbitrary interval). Moreover, suppose that $x(s)$ and $y(s)$ satisfy the following system of ODEs:

$$\frac{d x}{d s} = a(x, y) \text{ and } \frac{d y}{d s} = b(x, y). \tag{3}$$.

Then (1) becomes

\begin{align}
0 = a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} &= \frac{d x}{d s} \frac{\partial u}{\partial x} + \frac{d y}{d s} \frac{\partial u}{\partial y}. \tag{4}
\end{align}

You may recognize the last expression in (4) as the derivative $\frac{d u(x, y)}{d s}$ due to the chain rule:

$$\frac{d u(x, y)}{d s} = \frac{\partial u}{\partial x} \frac{d x}{d s} + \frac{\partial u}{\partial y} \frac{d y}{d s}.$$

Therefore the PDE (1) has been reduced to the following ODE along the curve $\Gamma(s)$:

$$\frac{d u}{d s} = 0. \tag{5}$$

From Eqn (5), we know something about the solution of the PDE (1). Along the curve $\Gamma(s)$ (defined by the system of ODEs (3)), the solution $u$ of the PDE (1) is constant.

The system of ODEs (3) for the particular PDE (1-2) are

$$\frac{d x}{d s} = y \text{ and } \frac{d y}{d s} = x,$$

alternatively

\begin{align}
\frac{\frac{d y}{d s}}{\frac{d x}{d s}} &= \frac{x}{y} \\
\frac{d y}{d s} \frac{d s}{d x} &= \frac{x}{y} \\
\frac{d y}{d x} &= \frac{x}{y}.
\end{align}

By separation of variables we find that $x^2 - y^2 = \text{constant}$.

$\newcommand{\+}{^{\dagger}}%
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 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\yy}{\Longleftrightarrow}$

$\dot{x} = y$ and $\dot{y} = x\quad$ yields $\quad\dd y/\dd x = x/y\quad$ which means $x^{2} - y^{2} = $ constant. Then ${\rm u}\pars{x, y} = {\rm f}\pars{x^{2} - y^{2}}$:
$$
{\rm u}\pars{0,y} = \expo{-y^{2}}\quad\imp\quad
{\rm f}\pars{-y^{2}} = \expo{-y^{2}}.\quad
\mbox{So,}\quad {\rm f}\pars{x} = \expo{x}
$$

and
$${\large
{\rm u}\pars{x,y} = \expo{x^{2} - y^{2}}}
$$
Let's checked
$$
y\,\partiald{u}{x} = 2xy\expo{x^{2} - y^{2}}\,,
\qquad
x\,\partiald{u}{y} = -2xy\expo{x^{2} - y^{2}}
$$


Algebraic solution to $2x + \ln x=2$ The answer to $2x+\ln x=2$ is $x=1$, which can be easily guessed after a failed attempt at solving it. I've tried to manipulate this into some inverse function of $2x+\ln x$, more specifically if $f(x)=2x+\ln x$, then I've tried to find $f^{-1}(x)$. But all I've gotten to is $xe^{2x}=e^2$, and I'm not sure how to simplify further.
Is there an algebraic solution to this? It seems like the product logarithm might be used, but I've only seen that function from Wolfram Alpha, and I'm not entirely sure how to use it.

Reiterating comments and complaints.
The way I read your question, you are asking for some function or method to algebraically/analytically solve the equation.  If we were told for instance that $x^2=4$ and were asked to find the positive value of $x$ satisfying this, sure we could say "Oh, $2$ is the answer because look here... $2^2$ does in fact equal $4$" but if we were to be asked about a similar question such as $x^2=3$ we would be hard-pressed to say "Oh, $1.732050807568877\dots$ is the answer because $(1.732050807568877\dots)^2=3$."
For the question of solving $x^2=3$ for $x$, we were able to introduce a notation and concept... the square root... which allows us to answer this easily.  $x^2=3\implies x=\sqrt{3}$.  The way I read your question, you are looking for the analogous "operation" which lets us extract the value of $x$ from the above equation which uses this operation rather than just spotting the solution.  Spotting the solution for a particular case and proving it is indeed the solution is all well and good however it does nothing to help us find a solution in a similar problem with slightly changed values.
The answer to whether or not there is such an operation to solve for $x$ is that hoping for such an operation which solves a transcendental equation such as this is futile.  Any such operation will only be as useful as saying the solution is "the solution" and will be non-elementary.  In some related problems you will not have any conveniently named function to use.  In other problems you may have a function such as the Lambert-W function (a.k.a. the product-log function as you alluded to in your post) but actually using the Lambert-W in practice will generally require numerical methods to evaluate.
For this specific problem, looking at the generalized equation of $2x+\ln(x)=a$ you can indeed use the Lambert-W function.  Modifying the equation a bit, we have $e^{2x+\ln(x)}=e^a$ and $xe^{2x}=e^a$ and so $(2x)e^{(2x)}=2e^a$
This is of the form $we^w=z$ whose solution is $w=W_k(z)$ for some integer $k$ where $W_k$ is the Lambert-W function on the $k$'th branch.  That is, replacing the values of $w$ and $z$ for what we had $2x = W_k(2e^a)$ giving us
$$2x+\ln(x)=a\iff x = \frac{W_k(2e^a)}{2}$$
In the specific case of $a=2$ this does simplify to $W_k(2e^2)/2$ which simplifies to $1$, again using that $we^w=z\iff w=W_k(z)$
In the specific case of $a=3$ (so the equation $2x+\ln(x)=3$) this would not simplify well and would just be written as $W_k(2e^3)/2$ which one can find via numerical methods is approximately $1.3499618380\dots$

As noticed in the comments, in general we can't find elementary algebraic solutions to this kind of equations.
To proceed by elementary inequalities, in this special case we have that
$$2x + \ln x=2\iff \ln (x)=2(1-x)$$
and since

for $x<1 \quad \ln (x)<0 \,\land\, 2(1-x)>0$
for $x>1 \quad \ln (x)>0 \,\land\, 2(1-x)<0$
for $x=1 \quad \ln (x)=2(1-x)=0$

$x=1$ is the unique solution for the given equation.


Optimal path around an invisible wall The Problem

  On an infinite plane there are two points, $A$ and $B$, a unit distance apart.  There is a $50\%$ probability that there is an invisible wall somewhere between the two points.  The wall extends $1/3$ unit in each direction perpendicular to the line between $A$ and $B$, and its position is uniformly distributed between the two points.  You don't know if the wall is present or where it is until you run into it.

  What strategy minimizes the expected distance to travel to point $B$ from point $A$?

(This problem originally came from Puzzling.SE but was closed for being "too mathy.")

My Solution

I use an $x$/$y$ coordinate system on the plane with the origin at $A$, and $B$ on the positive $x$-axis.

Once the position of the wall is known (i.e. when you run into the wall) the best strategy is to walk straight along the wall to the nearest end, then travel straight to $B$.  The best strategy should therefore be to follow a fixed path from $A$ to $B$ until you reach $B$ or hit the wall; if you hit the wall, follow the previously stated path around the wall.

I break up the fixed part of the path into a number of small segments.  A given segment covers a horizontal distance $\delta x$ and vertical distance $\delta y$.  The length of each segment is weighted by the probability it gets traveled, which is $\frac{1}{2} + \frac{1}{2}\left(1-x\right)$.  We also consider the probability that you have to walk around the wall, which for a given segment is $\frac{1}{2}\delta x$.  The distance it takes to get around the wall is just $(H-y) + \sqrt{H^2 + (1-x)^2}$, where $H=\frac{1}{3}$, the half-length of the wall.

Therefore the expected distance $E$ should be equal to:

$$
E = \sum_{i=1}^N\frac{2-x}{2}\sqrt{\delta x_i^2 + \delta y_i^2} + \frac{\delta x_i}{2}\left((H-y_i) + \sqrt{H^2 + (1 - x_i)^2}\right)
$$

If we treat $y$ as a function of $x$ (assuming that the optimal path doesn't travel backward) we can factor out $\delta x$, since $\delta y = y'(x)\delta x$; then taking the limit as $\delta x\to 0$ we obtain an integral:

$$
E = \frac{1}{2}\int_{0}^{1}(2 - x)\sqrt{1 + y'(x)^2} + H - y(x) + \sqrt{H^2 + (1-x)^2}~\mathrm{d}x
$$

Now I try to use the Euler-Lagrange equations to find the path $y(x)$ that minimizes $E$:

$$
E = \int \mathcal{L}(x, y(x), y'(x))~\mathrm{d}x \\
\mathcal{L}(x, y(x), y'(x)) = \frac{1}{2}\left((2 - x)\sqrt{1 + y'(x)^2} + H - y(x) + \sqrt{H^2 + (1-x)^2}\right) \\
\frac{\partial \mathcal{L}}{\partial y(x)} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial \mathcal{L}}{\partial y'(x)} = 0 \\
\left[-\frac{1}{2}\right]-\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{(2-x)y'(x)}{2\sqrt{1+y'(x)^2}}\right]=0 \\
1 = \frac{y'(x)(1 + y'(x)^2) - (2-x)y''(x)}{\left(1+y'(x)^2\right)^{3/2}} \\
y''(x) = \left(1+y'(x)^2\right)\frac{y'(x)-\sqrt{1+y'(x)^2}}{2-x}
$$

A solution to this differential equation should be a minimum-distance path, as long as it satisfies $y(x) \le H$ for all $x\in[0,1]$.  The boundary conditions are $y(0)=y(1)=0$.  Mathematica finds an explicit solution, but it it is extremely complex.  The simplest form I can get it in is:

$$
y(x) = \frac{168 \left(9 \sqrt{6}+\sqrt{38}\right) \xi -\sqrt{131-9 \sqrt{57}} \xi^3 - 448 \sqrt{3710+378\sqrt{57}}}{2688 \left(27+\sqrt{57}\right)} \\
\xi = \sqrt{7 \left(43+\sqrt{57}\right)-8 \left(27+\sqrt{57}\right) x} \\
y(x) \approx 0.3929 x\sqrt{1.2802-x} +0.3454 \sqrt{1.2802-x}-0.3908
$$

The solution looks like this:

The Question

Is this approach valid, and are my calculations correct?

One approach is via dynamic programming (which works the solution backwards). For simplicity, let's chop the $x$-axis into $n$ segments of size $\delta_x = 1/n$, and suppose the wall is on one of these chopping points (including point 0, but not point 1).  Chop the $y$-axis into $n$ segments of length $\delta_y = (1/3)/n$.  Define the value function $J(x,y)$ as the expected remaining travel distance to the destination under the optimal policy, given we are at a horizontal distance $x$, a vertical distance $y$, and none of our previous moves encountered a wall. The $J(x,y)$ function is defined over:
$$ x \in \mathcal{X} = \{0, \delta_x, 2\delta_x, ..., 1\} \quad , \quad y \in \mathcal{Y}=\{0, \delta_y, 2\delta_y, ..., 1/3\}$$

Suppose we start at $(x_0,y_0)=(0,0)$. Working backwards, and assuming there is no wall at $x=1$, we get:

$$J(1,y) =  y \quad \forall y \in \mathcal{Y} $$

Now assume $J(k\delta_x, y)$ is known for all $y$ and for some $k$.  Then:

\begin{align}
J((k-1)\delta_x, y) &= P[\mbox{wall is here | have not yet seen it}]\left[1/3-y + \sqrt{1/9+ (1-(k-1)\delta_x)^2}\right]\\
&+P[\mbox{wall is not here | have not yet seen it}]\min_{v\in\mathcal{Y}}\left[J(k\delta_x,v)+\sqrt{\delta_x^2 +(y-v)^2}  \right]
\end{align}
Since we have discretized the problem, if a wall exists then it is located uniformly over one of the the $n$ points $\{0, \delta_x, ..., (n-1)\delta_x\}$.
Thus, if we are at location $(k-1)\delta_x$ and have not yet seen a wall at any of the locations $\{0, \delta_x, ..., (k-2)\delta_x\}$, we get:
$$ P[\mbox{Wall here | not before}]=\frac{\frac{1}{2n} }{\frac{1}{2}+\frac{1}{2}\left(\frac{n-(k-1)}{n}\right)} $$

My solution to this problem (which I originally posted on the puzzle site), would be to aim for the midpoint up the wall assuming it is midway between A and B.  After that, I would lower my path back towards the centerline of A and B.  I would have to use a computer to try many different paths and lowering angles.  For example, if I don't hit the wall by midway between A and B, I could try going directly at B but this seems non optimal cuz think of the worst case where the wall is slightly before B.  You then get diverted quite a bit just to finish a small distance to B.  I think a simple and near optimal solution would be after not seeing the wall by the midpoint of the direct AB path, plot a few more discrete points that the wall could be at (such as $5/8$th of the way, $3/4$s of the way, $7/8$ths of the way), optimize just those paths, then find a formula that passes thru those $3$ points (called curve smoothing I think?). Not sure if that will be optimal for all wall positions past the halfway point but it is a good start to try to beat.


length of train which has passed through tunnel A train is travelling at a speed of 96 km/hr. It takes 3 s to enter a tunnel and 30 s more to pass through it completely. What is the length of the train?

HINT:

Entering the tunnel means crossing own length

Crossing the tunnel mean travelling own length + that of the tunnel

96 Km/hr equivalent to (80/3) m/s.
It takes 3s for a train to enter into the tunnel.

Hence, length of the train = (80/3) *3

Therefore, 80m is the length.

The second part stating 30s is not really required for calculating the length.


Quantitative aptitude-profit and loss A merchant earns 25% profit in general
Once his 25% consigment was abducted forever by some goondas. Trying to compensate his loss he sold the rest amount by increasing his selling price by 20%.What is the new percentage profit or loss ?
A.10% loss
B.12.5% loss
C.12.5% profit
D.11.11% loss

I took Initial CP as 100,so selling price after profit is 125
After 25% abduction,75 is available whose CP is 100 and SP is 125+50= 150
So 150-100/100=50%
But the answer is wrong.
What mistake have I done ?

Let the merchant had =100 product with total CP =100
Whose SP = 1.25 *100 = 125
In 2nd Case the price of 75 product = 125/100 *75 =375/4
New SP = 1.20*375/4 = 112.5
So profit = 112.5 - 100 = 12.5
profit percentage = 12.5/100 = 12.5%

Assume that the merchant was purchasing 100 goods at 100 Rs. each. He was selling them at 25% profit for 125 Rs. each.

His total profit = Total Selling Price - Total Cost Price = 100*125 - 100*100 = 2500 Rs.

Now, 25% of his good have been abducted. So, he is left with 100 - 25 = 75 goods

He increased his selling price by 20%. New selling price = 1.2 * 125 = 150 Rs.

Total money he makes now = 75*150 = Rs. 11250

His new profit = 11250 - 10000 = Rs. 1250

New percentage profit = (1250 / 10000) * 100 = 12.5 % (Option C)

You can check out some basic concepts of profit and loss here.


Prove $\sqrt{n+1} < \sqrt{n} + 1 $ Prove $\sqrt{n+1} < \sqrt{n} + 1 $ for all $n \ge 1$.
I have proven the base step for $n = 1$. $\sqrt{2}$ is less than $2$. The inductive hypothesis is $\sqrt{n+1} < \sqrt{n} + 1$. From here, I am unsure of how to proceed. How can I show that the inequality $\sqrt{(k+1)+ 1} < \sqrt{k+1} + 1$? After simplifying the expression $\sqrt{(k+1)+1}$, it becomes $\sqrt{k+2}$. Perhaps I can use this somehow to come up with an expression? The right hand side does not have a radical over the 1, so I do not think it is a duplicate question that has an answer in the suggested question.

A geometric proof:

Consider two segments of lengths $1$ and $\sqrt{n}$. If you attach these segments by their endpoints at an angle of $90^\circ$ (so they become two legs of a right triangle), then by Pythagorean theorem the hypotenuse of this right triangle is $\sqrt{n+1}$.

Now increase the angle between the two "leg" segments. By the law of cosines, the distance between the "loose" endpoints also increases, until the angle becomes $180^\circ$. But now at $180^\circ$, the distance between the "loose" endpoints is just the sum of the two segments $\sqrt{n}+1$; therefore
$$\sqrt{n+1} < \sqrt{n}+1.
$$
Q.E.D.

To prove by induction:
Consider n=1, $\sqrt {1+1} \lt \sqrt{1} + 1$
$$\sqrt{2} \lt 2$$
Assume n=k true:
$$\sqrt{k+1}\lt \sqrt{k} + 1$$
Consider n=k+1:
If true, this would mean that:
$$\sqrt{k+2}\lt\sqrt{k+1} + 1$$
We know that $\sqrt{k+1}\lt \sqrt{k} + 1$, so:
$$\sqrt{k+1} + 1 \lt \sqrt{k} + 2$$
Hence:
$$\sqrt{k+2}\lt\sqrt{k+1} + 1\lt\sqrt{k} + 2$$
Squaring both sides:
$$k+2\lt k + 4 + 4\sqrt{k}$$
This is indeed true, so n=1 true, and n=k true implies that n=k+1 true hence true for all natural numbers (integers greater than or equal to 1).


Does $(\pm 3, \pm 4)$ refer to 2 points or 4? I've seen the notation $\mp$ to contrast with $\pm$ to indicate "Yes there are only two cases and these guys must be opposite signs".  So something like $(\pm2, \mp3)$ would just be $\{(2, -3), (-2, 3) \}$.  But if I write $(\pm 3, \pm 4)$, does this refer to $\{(3,4), (-3, -4)\}$, or $\{(3,4), (-3, -4), (3, -4), (-3, 4)\}$?

This is obviously an abbreviation of four-point writing. For example we can write that the only solution in integers of the system

$$\begin{cases}x^4+y^4=17\\x\lt y\end{cases}$$ is given by $(x,y)=(\pm1,\pm2)$.


In how many ways $5$ different rings can be worn on $4$ fingers? In how many ways $5$ different rings can be worn on $4$ fingers ?

Although these is similar question here In how many ways $3$ different rings can be worn in $4$ fingers with at most one in each finger?

But I want to answer it in different way which is like take $4$ fingers like $a,b,c,d$. Now $a$ can be filled in $5$ ways and so are others. So total ways are $5^4$ but answer is $4^5$. What is wrong in my reasoning? Which cases have I left out?

I'm assuming that the four fingers are given, e.g., the fingers of one hand without the thumb. An arrangement of five distinguishable rings on the four labeled fingers amounts to a linear arrangement of $1$, $2$, $3$, $4$, $5$, and three indistinguishable zeros as separators. There are ${8!\over3!}=6720$ such arrangements. Note that a green ring and a blue ring on the index finger can be worn in $2$ ways.

In your argument that finger a can be filled in $5$ ways is wrong. This means one ring can be worn in more than one finger which is not possible.

You must argue that the first ring can be worn in any of the fingers, so it has $4$ choices.
Similarly, each ring has $4$ choices.
Total ways$=4.4.4.4.4=4^5$.

Second Question: Select $3$ rings from given $4$ rings and arrange the $3$ rings.
Answer=$^4C_3\cdot 3!=24$.


Show that $a_n<\sqrt{2}$ for every $n\in\mathbb{N}$. We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.

I want to prove that $a_n$ converges. But first I'd like to show that $a_n<\sqrt{2}$ for every $n\in\mathbb{N}$.

I try to use induction. For a fixed $n$, I suppose that $a_n<\sqrt{2}$.

Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)<\sqrt{2}$?

Thanks.

We have:
$$ 2 a_n a_{n+1} = a_n^2 + 2, $$
from which it follows that:
$$ a_{n+1}^2-2 = (a_{n+1}-a_n)^2 = \left(\frac{1}{a_n}-\frac{a_n}{2}\right)^2 = \left(\frac{a_n^2-2}{2a_n}\right)^2 > 0, $$
so $a_n > \sqrt{2}$ for every $n > 1$. If we set $b_n = a_{n+1}^2-2$, we have $b_1=\frac{1}{4},b_n>0$ and:
$$ b_n = \frac{b_{n-1}^2}{4a_n^2} = \frac{b_{n-1}^2}{4(b_{n-1}+2)} \leq \frac{b_{n-1}^2}{8}, $$
from which it follows that the sequence $\{b_n\}_{n\geq 1}$ decreases very fast towards zero. For instance, $b_1=\frac{1}{4}$ and the last inequality imply, by induction:
$$ b_n \leq \frac{1}{2^{5\cdot 2^{n-1}-3}}.$$

In a general case :
$$a + b - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \geq 0,$$
so just put $a = a_n, b=\frac{2}{a_n}$ in your case.


Field with 2 elements Let $ A $ be a ring with $ x^{2}+y^{2}+z^{2}=xy+yz+zx+xyz+1,\forall x,y,z\in A^{*} $.
Prove that $ A $ is a field with 2 elements.

If we put $ x=y=z=1 $ we obtain that $ 1+1=0 $.
If we put $y=z=1 $ we have that $ x^2=x $, which means $ A $ is a boolean ring.
That's all I did so far.

Hint: Try setting $x=y=z$.

(By the way, it is not correct to conclude that $|A|=2^n$ as you have done, since $A$ might be infinite.  Also, the statement you are trying to prove is slightly incorrect, since the condition also vacuously holds if $A=\{0\}$.  Your conclusion should be that $A$ is either $\{0\}$ or a field with two elements.)


Proving there are no periodic points I have the following function

$$f(x) = x^3-x$$

By inspection, I found that the only periodic points are $0, \pm \sqrt{2}$ with period $1$ (i.e. fixed points).

Nevertheless, I don't know how to prove that there's no periodic points other than the fixed points I found.

Let $f(x)=x^3-x$

$a^3-a=b^3-b$ where $a\ne b$

$\implies0=a^3-b^3-(a-b)=(a-b)(a^2+ab+b^2-1)$

So, we need $a^2+ab+b^2-1=0\implies a=\dfrac{-b\pm\sqrt{b^2-4(b^2-1)}}2=\dfrac{-b\pm\sqrt{4-3b^2}}2$

For real $a,$ we need $4-3b^2\ge0\iff3b^2\le4$

In that case $f(a)=f(b)$ right?


closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...

I tried the direct approach of pluggin in y=1/x into the distance formula but it leads to an order-4 polynomial...

That is not a mistake. The ellipse has two maxima and two minima of the distance to $P$ for most locations of point $P$.  This leads to a degree $4$ polynomial when solving for the coordinates of the extrema.  There is no special geometric relationship between the distance extrema that might make the polynomial easier to solve than the general quartic equation.

The slope of $y = 1/x$ at $x$ is $-1/x^2$,
so the tangent at $(p, 1/p)$ is
$(y-1/p)/(x-p) = -1/p^2$
or $y = 1/p - (x-p)/p^2$.
At $y = 0$, $(x-p)/p^2 = 1/p$
or $x = 2p$;
at $x = 0$, $y = 1/p + p/p^2 = 2/p$.
The intercept points are thus
$(0, 2/p)$ and $(2p, 0)$.

I will minimize the distance by making
the line from $(a, b)$ to
$(p, 1/p)$ perpendicular to this tangent
at $(p, 1/p)$
and do this by making their
dot product zero.

The dot product of the vector
from $(a, b)$ to $(p, 1/p)$
with the tangent is

$\begin{align}
(p-a, 1/p-b)\cdot (2p, -2/p)
&= 2p(p-a) - 2/p^2+2b/p \\
&= (2p^3(p-a) - 2 + 2pb)/p^2 \\
&= (2p^4 - 2ap^3 + 2bp-2)/p^2 \\
&= 2(p^4 - ap^3 + bp-1)/p^2
\end{align}
$

and this is zero when
$f(p; a, b) = p^4 - ap^3 + bp-1 = 0$.

I will consider the special case $a = b$
to check this;
there should be a root at $p = 1$.

If $a = b$,

$\begin{align}
f(p; a, a) &= p^4 - ap^3 + ap-1 \\
&= (p^4-1) - a(p^3-1) \\
&= (p-1)(p^3+p^2+p^1 - a(p^2+p+1)) \\
&= (p-1)(p^3 + (1-a)(p^2+p+1))
\end{align}
$

which has a root at $p = 1$
(as expected) and,
if $a > 1$
(so $(a, a)$ is to the right and above the hyperbola),
 another positive real root
since $p^3 + (1-a)(p^2+p+1) < 0$
for small $p$
and $p^3 + (1-a)(p^2+p+1) > 0$
for large $p$.

If $a < 1$,
so $(a, a)$ is between the hyperbola and the origin,
there are no other positive real roots
since $p^3 + (1-a)(p^2+p+1) > 0$
for all $p \ge 0$.

I could argue further about
where the real roots
of $f(p; a, b)$
are using
$f(1/p; a, b) = (1/p)^4 - a(1/p)^3 + b/p-1
= (1 - ap + bp^3 - p^4)/p^4
= -f(p; b, a)/p^4
$
so the roots of
$f(1/p, a, b)$ are the same as the roots of
$f(p; b, a)$,
but that's enough for now for me.

In general, I find it much easier
algebraically to make two vectors
orthogonal by making their dot product zero
than minimizing their distance
using the Pythagorean theorem.


Prove that the value of $\Delta$ is an integer for the given determinant $\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$

taking $\sqrt 6$ out from the first column and performing the following operations

$$R_2\rightarrow R_2 -\sqrt 2 R_1$$

$$R_3 \rightarrow R_3-\sqrt 3 R_1$$

$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$

What should I do next?

From here

$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$

Pull out a factor of $\sqrt{3}$ from the second row and a factor of $\sqrt{2}$ from the third to get

$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix}$$

Now you can compute it easily by expanding along the first row or by subtracting the second row from the third. Lets subtract.

$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix} = 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 0 & -1 \end {vmatrix}$$

The determinant is just the product of the diagonal entries, so $6 \cdot 1 \cdot 1 \cdot -1 = -6$.

Expand along the first column. Thanks to the two zeroes there, you will only have $\sqrt{6}\cdot1\cdot(2\times2\text{ determinant})$. And you can calculate that small determinate directly.


Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$. Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$.

It's obvious that if  $x\gt 1$ then  $x+1\gt 2$ , also $x^5\gt0 \;as \;x\gt1\gt0$, thus $x^5+x+1\gt 2$.

Is it fine ?

I can't understand what do they mean by direct proof?

Since the derivative $f'(x)=5x^4+1$ is always positive, the function $f(x)=x^5+x+1$ is increasing over $\mathbb{R}$ and $x>1$ implies $f(x)>f(1)=3$.

The function is monotone increasing for positive values. This is because addition and multiplication are increasing in either argument. Moreover the value at 1 is 3. The theorem follows.


If $f$ has a zero and $|f''|\leq M$, then $f$ is monotone on $(-h,h)$, where $h=\sqrt{2|f(0)|/3M}$ Let $f$ be twice differentiable on $\mathbb R$ and let $M$ be a bound
  of $f''$, $|f''|\leq M$ on $\mathbb R$. Assume $f(0)\neq0$ and define
  $h=\sqrt{\frac{2|f(0)|}{3M}}$. Prove that if $f$ has a zero in $(-h,h)$, it's
  monotone in $(-h,h)$.

So let's assume there's a zero $a\in (-h,h)$ such that $f(a)=0$ and also there's $b\in (-h,h)$ such that $f'(b)=0,\ f''(b)\neq0$. We need to get a contradiction.

This question is under the Taylor Expansion chapter although I can't really get anything out of $f(x)=f(0)+f'(0)x+\frac{f''(\xi_x)}{2}x^2$ nor by the expansions near $a,b$. I get that in $(-h,h)$ we have $\frac{M}{2}x^2<\frac{|f(0)|}{3}$ which may be related to the 2nd derivative term though.

Suppose that $f'(x_0)=0$ with $|x_0|<h.$ Then $|f'(x)|\leq M|x-x_0|< M(|x|+h)$ for all $|x|<h.$ In particular for $0<x<h$ we have $$|f(x)-f(0)|\leq \int_0^h |f'(x)|dx\leq \int_0^h M(x+h)dx<\tfrac32 Mh^2<|f(0)|$$ with the same bound holding for $-h<x<0.$ This means that $f$ cannot have a zero in $(-h,h).$

This is not true.
Consider $ f(x) = exp (-x^2) -0.1$

It's not monotone on $(-h,h)$ for any given $h>0$, yet it's second derivative is bounded by a finite $M$ and $f(0)=1$, and it has a root.


How to integrate $\cos(z)~dz$? How to find: $$\int\cos(z)~dz$$ where $z$ is complex? I know the result is $\sin(z)+C$ but I don't understand why I think I'mgetting confused on how we calculate the complex integrals can someone help explain?

Firstly, since your integral has no bounds, what the symbol "$\int f(z) \ dz$" represents is the antiderivative of $f(z)$, meaning,  it's the family of functions that when differentiated result in $f(z)$. So now the question becomes "How do we define derivatives for complex functions?".
As you can see in this answer, the derivative definition for complex functions is analogous to the derivative for real functions as it's given by
$$
f'(z) =\lim_{h \to 0} \frac{f(z+h) - f(z)}{h}
$$
Notice that because we use the same derivative definition as we do for real functions, it makes sense for a complex function to have the same derivative as its real counterpart.
Another question that might cause trouble is "How do we make sense of $\sin(z)$ and $\cos(z)$ when $z$ is complex?". We know that for real numbers a basic definition of the trig functions can be given as the $x$ and $y$ coordinates on points of a circle, but this doesn't make sense when the input is complex. So what do we do? Well, we remember that for real values we can express these trig functions as Taylor series given by
$$
\sin (x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}  \qquad \cos (x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}
$$
The important thing to notice is that if we substitute $x$ for a complex number $z$ in the above equations we can make sense of $\sin(z)$ and $\cos(z)$ because in the series part we would just be multiplying and then adding complex numbers,  which are both things we do know how to do! Using this definition combined with the definition of the derivative established earlier, you can show that for complex numbers it is also true that
$$
\frac{d}{dz} \sin(z) + C = \cos(z)
$$

The tl;dr version of the answer is that the antiderivative is the same as for real-valued functions because the complex definition for derivatives is designed to be a natural extension of the real derivative, and so, the behavior of the derivatives of the real function and its complex counterpart are analogous.
Lastly, if the concept of adding up "infinitely many complex numbers" by defining sine and cosine in terms of infinite series sounds a bit weird, I recommend watching this video by 3Blue1Brown that has some very nice explanations and visualizations on that very subject.

$z=\theta$ which is just an angle so whether it is real or complex its integration would be $\sin(z)+c$ also all real numbers are subset of complex numbers or as we are used to x replace $z=x$


showing an iequality without using a calculator During solving a problem, I faced the inequality $$\frac{37^{37}}{36^{36}}<100$$.
Is it useful to consider  the function $f(x)=x^x$ or something like that? Or is it obtained by algebraic manipulations? The calculator says that the two sides differ by less than $0.8$.
Please give me a hint to prove the inequality. Thanks.

Edit. If we take logarithms of both sides, we may consider the function $f(x)=x\log x$ and apply the mean value theorem on the interval $[36,37]$. However, this does not help as we dont know the $c$ obtained by the theorem satisfies $f'(c)<2$. Any other ideas?

I suggest rewriting ${37}^{37}$ as ${(36 + 1)}^{37}$, expand using the binomial theorem, then divide each successive term by ${36}^{36}$.  You should see the terms drop off in a suggestive pattern.


$T^2\times S^n$ is parallelizable This is taken from a UCLA Geometry/Topology qualifying exam.

How would one prove that $T^2\times S^n$ is parallelizable for all $n\geq 1$? Is there a way to find $n+2$ linearly independent vector fields? I am trying to think of the simplest case $n=2$ where $S^2$ is not parallelizable, but $T^2\times S^2$ has to be in some way. I would appreciate a general strategy to treat such problems.

There is a similar post a few days ago asking why
$$
T(\mathbb{S}^{2}\times \mathbb{S}^{1})
$$
is trivial. The idea is to use the extra dimension coming from $T(\mathbb{S}^{1})$ to fill in the place of the normal bundle. For your question I think it is similar but more subtle, as you have to consider carefully how the pull back bundle splits and how to place the tangent vectors. I am not sure if a long exact sequence+Whitehead's thoerem type of argument would be useful, but you can also give a try.

I may elaborate the reason. Since $\mathbb{S}^1$ has trivial tangent bundle. Denote $\xi$ to be one-dimensional trivial bundle, and thus
$$T_{\mathbb{T}^2}=T_{\mathbb{S}^2}=\pi_1^* T_{S^1}\oplus\pi_2^* T_{S^1}=\pi_1^*\xi\oplus\pi_2^*\xi=\xi^2.$$
Also use the same formula, note that the tangent bundle $\tau_n$ os $\mathbb{S}^n$ we have $\tau_n\oplus \xi =\xi^{n+1}$. Thus
$$T_{\mathbb{S}^n\times\mathbb{T}^2}=\pi_1^*T_{\mathbb{S}^n}\oplus\pi_2^*\xi\oplus\pi_3^*\xi=\pi_1^*(\tau_n\oplus\xi)\oplus\xi=\pi_1^* \xi^{n+1}\oplus\xi=\xi^{n+2}.$$


How many times can these two polynomial intersect with each other? Suppose $g(x)=a_1x+a_2x^2+...+a_kx^k$ and $f(x)=b_jx^j$ where $a_1,a_2...a_k>0$ , $j\in \{1,2.....,k-1\}$, $b_j >0$ and $x\geq0$.

Intuitively, I think they can have at most two intersections. Is that correct?

Well, the answer is it has two positive roots by Descartes' rule of signs. Thanks for your help!

The general answer to this question is given by Bezout's theorem. It states that two polynomials of degree $n$ and $m$ have precisely $n\cdot m$ intersection points provided that you

are over an algebraically closed field
count including multiplicities
also count possible intersections at infinity.

This is overkill for your situation but when applied to polynomials over the reals it still means you can have at most $m\cdot n$ intersections. One can find examples where you actually have that many real intersections.


Comparing big powers Which of the following is the largest?
A. $1^{200}$
B. $2^{400}$
C.$4^{80}$
D. $6^{300}$
E. $10^{250}$

I'm stuck trying to solve this. Obviously A and C are wrong ($4^{80}$ is less than $2^{400}$ and 1 to any power is always 1). And cancelling $2^{200}$ from each of the remaining choices, I can also eliminate B. However, I don't really know how to compare D and E... Any hints or helps?


Does a non-empty locally closed subset of a $k$-scheme of finite type always contain a closed point? Let $k$ be a field.
Let $X$ be a scheme of finite type over $k$.
We denote by $X_0$ the set of closed points of $X$.
Let $Y$ be a locally closed subset of $X$.
Is $Y \cap X_0$ non-empty whenever $Y$ is non-empty?

Motivation
See my comment to Avi Steiner's answer to this question.

Write $Y=Z \cap U$ with $Z \subseteq X$ closed and $U \subseteq X$ open. Now we may view $Y$ as an open subset of the closed subset $Z$ of $X$. We make $Z$ into a closed subscheme of $X$. Then it is again of finite type over $k$, and we have $Z_0 = X_0 \cap Z$. Thereby we reduce the problem to open subsets. But then it follows from a corollary of Hilbert's Nullstellensatz, stating that the closed points are dense.

As Martin Brandenburg's answer shows that we may suppose that $Y$ is a non-empty open subset of $X$.
Then $Y \cap X_0$ is non-empty because $X_0$ is dense in $X$ by the proposition of my answer to this question.


Can $p^{q-1}\equiv 1 \pmod {q^3}$ for primes $p<q$? For prime $q$ can it be that
$$
p^{q-1}\equiv 1 \pmod{q^k}
$$
for some prime $p<q$ and for $k\ge 3$?

There doesn't seem to be a case with $k=3$ and $q<90000$, and I also checked for  small solutions with $3<k\le 20$ and found none.

If we remove the condition $p<q$ then there are always solutions, e.g. $15441^{16}\equiv 1 \pmod{17^5}$. Also for $k=2$ there are many, e.g. $71^{330} \equiv 1 \pmod {331^2}$.

Let $w>1$ be any integer and let $q$ be an odd prime and $w^{q-1}$  $\equiv 1 \pmod {q^3}$. Let v be a primitive root mod $q^3$ where $v^h$  $\equiv w \pmod {q^3}$. So $v^{h(q-1)}$  $\equiv 1\pmod {q^3}$. Therefore h=$q^2 k$ ; k >= 1. Assume k> 1 , then $w^{(q-1)/k}$ $\equiv 1\pmod {q^3}$ ; $v^{q^2 k-k}$ $\equiv(w/v^k) \pmod {q^3}$ , so $v^{(q^2 k -k)(q^2)}$ $\equiv 1 \pmod {q^3}$ ,therefore $(w/v^k)^{q^2}$  $\equiv 1\pmod {q^3}$. If the order of w mod $q^3$ is M then given $(w/v)^{q^2 M} $ $ \equiv 1 \pmod {q^3}$ ; $v^{q^2 M}$ $\equiv 1 \pmod {q^3}$. Yet this implies M = (q-1).  Then the order of w mod $q^3$ is  not <(q-1) contradiction. So k = 1. And $v^{q^2}$ $\equiv w \pmod q^3$. The order of w mod $q^3$ is (q-1). If w = p a prime < q then $p^{q-1}$ $\equiv 1 \pmod {q^3}$ where (q-1) is the order of p. p = (q-v);  $(q-v)^q$ $\equiv(q-v)\pmod {q^3}$. So ($q^2$ $v^{q-1}$ -$v^q$) $\equiv(q-v)\pmod {q^3}$. Therefore $v^{q-1}$ ($q^2$-v) $\equiv (q-v)\pmod{q^3}$ ;  $(-v q)\equiv (q^2-v q)\pmod{q^3}$ ; $q^2 \equiv 0 \pmod{q^3}$ Contradiction , so if p < q the order of p mod $q^3$ can not be (q-1)


Generating Functions of Binary Strings I'm trying to learn generating functions and integer partitions and having particularly difficult time with this question:

"This question is about binary strings where each block of 0s is followed by a block of the same number of 1s."
1. Consider strings of this form which begin with 0 and have only one block of 0s. a) Write out all such strings of size at most 6. b) What is the generating function for these strings?

a) 01, 0011, 000111

b) Issue is that I have no idea how to represent this sequence as a generating function. Can anyone show me how to convert such sequences into generating functions?

Thank you.

Let $f(n)$ be the number of such strings of size $n$.  Then the (ordinary) generating function associated with this sequence is $$F(x):=\sum\limits_{n \geq 0} f(n) x^n. $$ Based on your description, $f(n) = 1$ if $n$ is even and $f(n) = 0$ if $n$ is odd.  The generating function is $$F(x) = \sum\limits_{n \geq 0}f(n) x^n = 1 + x^2 + x^4 + \cdots = \frac{1}{1 - x^2}.$$

Additionally, if we want the generating function for the number of such strings of size at most $n$, then we have $$\sum\limits_{n \geq 0 } \left(\sum\limits_{k = 0}^n f(k)\right) x^n = \frac{F(x)}{1 - x} = \frac{1}{(1 - x^2)(1 - x)}.$$

Think about how you'd come up with a decomposition for the set of binary strings you want. Intuitively, what you want is some number of zeros, the same number of ones and stick them together. And then you want to repeat that an arbitrary number of times. For convenience, lets define the set:

$$ A = 01 \cup 0011 \cup 000111 \cup ....$$

With respect to the weight function being the length of the string, the generating function for $A$ is:

$$ A(x) = x^2 + x^4 + x^6 + x^8 + ... \ = \frac {x^2} {1-x^2}$$

So this covers one repetition of the pattern you want. You want to allow this to happen an arbitrary number of times. So the set of all strings you're really interested in is this:

$$ S = \epsilon \cup A \cup A^2 \cup A^3 \cup ... $$

where $\epsilon$ is the empty string

Then the generating function for S is simply, (using the product and sum rules):

$$ S(x) = 1 + A(x) + A(x)^2 + A(x)^3 + ... $$
$$ = \frac {1} {1-A(x)}$$
$$ = \frac {1} {1 - \frac {x^2} {1-x^2}} $$

Which simplifies to:

$$ \frac {1 - x^2} {1 - 2x^2} $$

In general, the hardest and most important part is finding a clean decomposition of these strings -- an enumeration of the set you're interesting in counting properties of. It takes some practice, but I hope this outlines the general principle.


Qualitatively describing level surfaces from equations alone Describe level surfaces:
  (i) $f(x,y,z)=z-x^2-z^2$ (ii) $f(x,y,z)= 3x-y+4z-1$

Not sure think (i) is a type of paraboloid.
Think (ii) is a plane not sure what else I can describe in both cases any thoughts would be appreciated.

The first family of level sets are NOT paraboloids.

Here is a simple way to obtain them:

$z-x^2-z^2=k$ or  $z^2 -z+x^2=-k$ can be written $x^2+(z-\frac{1}{2})^2-\frac{1}{4}=-k$ of the form $Z^2+X^2=a$ with $a=0.25-k$.

Thus there are 2 cases, constituting a family of co-axial cylinders (if non void!):

$k \leq 0.25$; a cylinder with radius $\sqrt{0.25-k}$ and axis parallel to $y$-axis passing at $z=1/2$. The case $k=-0.25$ gives a degenerate cylinder reduced to its axis.
$k>0.25$: void sets.

The second equation is much simpler : $3x-y+4z-1=k$ is the equation of a plane for any $k$. All these planes share a common normal vector $(3,-1,4)$. They are thus all parallel.

You're correct in both cases. One way to see why is to set $f(x,y,z)=0$ (or more generally any constant $c$) and solve for $z$.


Separability of $A \subseteq X$ Let $(X, \tau)$ be a separable topological space. Let $A \subseteq X$. There exists a subset $D \subseteq A$ such that $D$ is dense over $A$ and $D$ is countable?
$X$ is separable if it contains a countable dense subset.
I supposed $D$ dense, and I would like to prove $D$ is not countable.
Thank you.

If I understand correctly, you’re asking whether a subset of a separable space is necessarily separable; the answer is no. Here’s one example.
Let $Y$ be the Sorgenfrey line, also known as the reals with the lower limit topology: the underlying set is $\Bbb R$, and the half-open intervals of the form $[a,b)$ are a base for the topology. $Y$ is separable, because $\Bbb Q$ is a countable dense subset of $Y$. Let $X=Y\times Y$, the Sorgenfrey plane, with the product topology. $X$ is also separable, because $\Bbb Q\times\Bbb Q$ is a countable dense subset of $X$. Now let $A=\{\langle x,-x\rangle:x\in\Bbb R\}$; $A$ is clearly uncountable, and it is also a discrete subset of $X$, since
for each $x\in\Bbb R$, the set $[x,x+1)\times[-x,-x+1)$ is an open nbhd of $\langle x,-x\rangle$ that contains no other point of $A$. (You can easily check that it is also a closed subset.) Clearly $A$ cannot be separable.
Another example can be constructed from $\Bbb R$ in a different way. We construct the new topology as follows. Points of $\Bbb Q$ are isolated. If $x\in\Bbb R\setminus\Bbb Q$, the sets of the form $\{x\}\cup\big((a,b)\cap\Bbb Q\big)$, where $a<x<b$, form an open nbhd base at $x$. In other words, each rational is an isolated point, and basic nbhds of an irrational consist of that irrational and all of the rationals in an open interval around it. Call this space $X$. The rationals are clearly dense in $X$, so $X$ is separable. However, the irrationals are an uncountable closed discrete subset of $X$, so the set of irrationals is not separable in $X$.
A space in which every subspace is separable is said to be hereditarily separable; all second countable spaces, and hence all separable metric spaces, are hereditarily separable.

Yes, since X is separable, there exist a countable set, Y, that is dense in X.
What can you say about $A\cap Y$?


is there any solution for $x^2 +x + 2 = e^x$ by using algebra? I know this can be solved by numerical methods but I would like to know whether this can be solved using logs or something similar.

Thanks

No, such equations have not an algebraic answer.
However, you can solve this by using log.

just $Log_e$ both side, simplify and divide you'll have log(x^2+x+2) = x  etc it should be after that.
in addition, remember the log addition rules!


Why is this method wrong on calculating this multivariable limit? So, I was studying Apostol's book while studying on the site "Brilliant" methods of calculating multivariable limits...
In particular, in $R^2$ we have polar coordinates to switch on and we have:
$\lim_{{(x,y)}\to(0,0)}f(x,y) = L$ iff $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = L$ since the statement $0\lt\sqrt{x^2+y^2}\lt\delta$ can be translated into $0\lt r \lt \delta$ from the $\epsilon-\delta$ definition of the limit while $x = r\cos(\theta)$ and $y = r\sin(\theta)$ (so the limit exists iff the limit exists in polar coordinates and it's $\theta-independent$) (taken from Brilliant)
But then Apostol came with the following function: $f(x,y) = \frac{xy^2}{x^2+y^4}$ if $x\neq 0$ and $f(0,y) = 0$ and things got messy in my mind because, if we switch to polar coordinates, it becomes $f(r\cos(\theta),r\sin(\theta)) = \frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+\sin^4(\theta)}$ if $r$ is different from $0$ and if we make $r\to0$ we'd have have $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = 0$
But, if you choose the curve $x = y^2$, we have $f(y^2,y) = \frac{1}{2}$ and so if we approach the origin by that curve we'd have $\lim_{y\to0}f(y^2,y) = \frac{1}{2}$ and by such we'd have the limit approaching $2$ different values which would mean the limit actually doesn't exist
So my doubt is about what is wrong about the procedure using polar coordinates instead of trying different curves, why the polar coordinate method didn't show me that the limit is "angle dependent" (and it doesn't exist in practice)? Did I make any mistakes in the procedure?

Your very first statement

In particular, in $\Bbb{R}^2$ we have polar coordinates to switch on and we have:
$\lim_{{(x,y)}\to(0,0)}f(x,y) = L$ iff $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = L$

is not very meaningful yet, because you haven't put a quantifier over $\theta$. I guess you meant the following:

In particular, in $\Bbb{R}^2$ we have polar coordinates to switch on and we have:
$\lim\limits_{{(x,y)}\to(0,0)}f(x,y) = L$ iff for every $\theta\in \Bbb{R}$, $\lim\limits_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = L$.

Even if this is what you meant, it is false, and this is a very common misconception (unfortunately there are several notes which promote the use of polar coordinates for solving limits, without carefully explaining the subtleties).
The implication $\implies$ is true, while the reverse implication is false. This is because if you fix a value of $\theta$, then $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta))$ is taking a limit of a single-variable function along a certain straight half-line (i.e it is a one-sided limit along a straight line) which is clearly a much weaker condition than what is actually required ($\lim_{(x,y)\to (0,0)}f(x,y)$ requires the limit to exist regardless of how you approach the origin: straight line, curvy line, zig-zag/criss-cross/oscillatory, whatever).
In fact your function is a perfect example, because it shows that along EVERY straight line to the origin, the limit of the function is $0$, yet despite this the multivariable limit $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist.

Just to drive the point home, let's write out in terms of quantifiers what each statement means:

$\lim_{(x,y)\to (0,0)}f(x,y) = L$ means:

for every $\epsilon>0$, there is a $\delta>0$ such that for all $(x,y) \in \Bbb{R}^2$, if $0< \lVert (x,y)\rVert<\delta$ then $|f(x,y) - L| < \epsilon$.

For every $\theta\in \Bbb{R}$, $\lim_{r\to 0^+} f(r\cos\theta, r \sin \theta) = L$ means:

For every $\theta\in \Bbb{R}$, for every $\epsilon>0$, there is a $\delta>0$ such that for all $r>0$, if $0<r<\delta$ then $|f(r\cos\theta, r \sin \theta) - L|< \epsilon$.

$\lim_{r\to 0^+}f(r\cos\theta, r \sin\theta) = L$, uniformly in $\theta$, means:

For every $\epsilon>0$ there is a $\delta>0$ such that for all $r>0$ and $\theta\in \Bbb{R}$, if $0<r<\delta$ then $|f(r\cos\theta, r \sin \theta) - L|< \epsilon$.

Note the differences in the statements, especially between 2 and 3 in terms of the quantifiers. We have $(1)\iff (3)$, and $(1)\implies (2)$ (so trivially $(3)\implies (2)$) but your specifc function shows that $(2)\nRightarrow (1)$.
In (1) and (3), the $\delta$ depends only on $\epsilon$, while in $(2)$, the $\delta$ depends on $\theta$ and $\epsilon$ (which is why order of quantifiers matters). Also, if you've seen the concept of uniform continuity, then you'll observe that it is a similar switch in the order of quantifiers which distingusihes between $(2)$ and $(3)$.

The moral of this: use Lagrange Multipliers to confirm the extreme values of your fraction as the denominator shrinks.
At some point you will learn Lagrange Multipliers. With a denominator such as $x^2 + y^4,$ positive definite but no homogeneous, you can take the denominator as equal to a (positive) number you name $c.$ So, the gradient of numerator $x y^2$   as a row vector is
$$ ( y^2, 2xy )  $$
and denominator gives
$$ ( 2x, 4 y^3) $$
more in a minute.......  If either of $x,y$ is zero (not both) the fraction is zero. The With nonzero variables, the two gradients are parallel when the cross products match,
$$ y^2 4 y^3 = 2xy 2x , $$
$$ y^5 = y x^2 $$
$$ y^4 = x^2 , $$
$$  x = \pm y^2 $$
When $x=y^2$ the fraction is $1/2.$ When $x = - y^2$ the fraction is $-1/2.$  Note that there is no explicit value needed for $c,$ in this problem it disappears. In other Lagrange multiplier problems it may be necessary to use terms that depend on the $c.$