Challenge Can You Solve These Math Challenges?

[QuantumQuest]

It's time for a basic math challenge! For more advanced problems you can check our other intermediate level math challenge thread!

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.
5) Mentors, advisors and homework helpers are kindly requested not to post solutions, not even in spoiler tags, for the challenge problems, until 16th of each month. This gives the opportunity to other people including but not limited to students to feel more comfortable in dealing with / solving the challenge problems. In case of an inadvertent posting of a solution the post will be deleted by @fresh_42$1.$ Finite Field $\mathbb{F}_8$
a) (solved by @lpetrich ) Find a minimal polynomial to determine the factor ring which is isomorphic to $\mathbb{F} _8$ .
b) (resolved in post #83) From there determine a basis of $\mathbb{F}_8$ over $\mathbb{F}_2$ and write down its multiplication and addition laws.
c) (solved by @lpetrich ) Why does the algebraic closure of a finite field have to be infinite?$\space$ (by @fresh_42)

$2.$ (solved by @nuuskur ) If $a$ is an odd integer show that $a^{2^n} \equiv 1(\mod{2^{n+2}})$ for all $n \in \mathbb{N} - \{0\}$ $\space$ (by @QuantumQuest)

$3.$ (solved by @Zafa Pi ) A carnival has a 3 sided coin with outcomes $\big \{\text{Heads, Tails, Other}\big\}$ and respective probabilities of $\big\{\frac{2}{5}, \frac{2}{5}, \frac{1}{5}\big\}$.

Rules of play:

Heads: $+1$ as a payoff
Tails: $+3$ as a payoff
Other: game over, lose all accrued winnings.

Otherwise the player may stop at any time and keep the accrued winnings.

question:
How much should a risk neutral player be willing to pay in order to play this game?

(For avoidance of doubt, this refers to playing one 'full' game, which is complete upon termination, and termination occurs on the first occurrence of (a) result of coin toss equals $\text{Other}$ or (b) the player elects to stop.)

optional:
How many rounds would it take on average for the game to terminate? (You may assume a mild preference for shorter vs longer games in the event of any tie breaking concerns.)

Now suppose the player doesn't care about the score and just loves flipping coins -- how long will the game take to terminate, on average, in this case? $\space$ (by @StoneTemplePython)

$4.$ (solved by @Hiero ) Calculate the volume $\mu(A)$ of

$A =\{(x,y,z)\in \mathbb{R}^3\,: \,x,y,z \ge 0\; , \;x+y+z \leq \sqrt{2}\; , \;x^2+y^2 \leq 1\,\}$ $\space$ (by @fresh_42)

$5.$ (solved by @nuuskur ) Determine the open balls with radius $3$ around $(2,1) \in \mathbb{R}^2$ w.r.t.
a) the French Railway metric with Paris at the origin $P$ and Reims at $R=(2,1)$ .
b) the Manhattan metric.
c) the maximum metric. $\space$ (by @fresh_42)

$6.$(solved by @Zafa Pi ) Two urns contain the same total number of balls, some black and some white in each. From each urn are drawn $\big(n \geq 3\big)$ balls with replacement. Find the number of drawings and the composition of the two urns so that the probability that all white balls are drawn from the first urn is equal to the probability that the drawing from the second is either all white or all black. $\space$ (by @StoneTemplePython)

$7.$ (solved by @lpetrich ) Show that if for some complex number $z$, ${\lvert \sin z\rvert}^2 + {\lvert \cos z\rvert}^2 = 1$ then $z \in \mathbb{R}$ $\space$ (by @QuantumQuest)

$8.$ (solved by @dRic2 ) Show that of all triangles with given base and given area, the one with the least perimeter is isosceles $\space$ (by @QuantumQuest)

$9.$ (solved by @lpetrich ) Solve $\int_{\Gamma} \omega$ with the curve $\Gamma = \gamma([0,1])$ given by

$\gamma : \mathbb{R} \longrightarrow \mathbb{R}^3\; , \;\gamma(t)=(t^2,2t,1)\text{ and }\omega = z^2 dx +2ydy+xzdz$

and compute the exterior derivative $\nu = d\omega$ . As such, the result is an exact $2-$ form. Is it also closed? Show this by direct calculation. $\space$ (by @fresh_42)

$10.$ (solved by @lpetrich ) Prove that $\lim_{n\to\infty} (n!e - \lfloor n!e \rfloor) = 0$ $\space$ (by @QuantumQuest)

 

[jedishrfu]

Nice set of problems. I like the spread even if I don't know how to do some of them.

 

[Hiero]

Wow... if this is basic then I must be really poorly versed in math, because I don’t even understand the language of many of these questions.

Is the answer to #4

Spoiler

\int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt2-x-y)dydx = \sqrt2\pi/4-2/3

?

(I did do it by hand, I just don’t want to type the intermediate steps.)

 

[fresh_42]

Wow... if this is basic then I must be really poorly versed in math, because I don’t even understand the language of many of these questions.

Is the answer to #4

Spoiler

\int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt2-x-y)dydx = \sqrt2\pi/4-2/3

?

(I did do it by hand, I just don’t want to type the intermediate steps.)

Yes, but the procedure belongs to the answer. You should at least roughly tell us what you did.

 

[Hiero]

Yes, but the procedure belongs to the answer. You should at least roughly tell us what you did.

The condition x,y,z ≥ 0 means we are in the positive octant. The next two conditions describe (under) a plane and (inside) a cylinder . So the integral should be of a quarter-cylinder under a plane.

Volume is given by (one way at least) ∫∫z dydx, so I put in the z from the plane equation and chose the integral boundaries to describe the quarter-circle and voila.
(The integral never gets messy, the worst part is the integral of cos^2(a)da which can be done by parts.)I am sorry I know this is a weak explanation but it’s more of a visual thought-process than a verbal thought-process.

 

[fresh_42]

I am sorry I know this is a weak explanation but it’s more of a visual thought-process than a verbal thought-process.

Your visual thought is right and it is exactly what we do formally. The process is called reduction of variables:
1.) $A =\{(x,y,z)\in \mathbb{R}^3\,: \,x,y,z \ge 0\; , \;x+y+z \leq \sqrt{2}\; , \;x^2+y^2 \leq 1\,\}$
2.) $A'=\{\{(x,y)\in \mathbb{R}^2\,: \,x,y \ge 0\; , \;x^2+y^2 \leq 1\,\}\; ; \; A_{(x,y)}= [0,\sqrt{2}-x-y]$
3.) $A''=[0,1]\; ; \;A'_x=[0,\sqrt{1-x^2}]$
We thus have
\begin{align*}
\mu(A)&=\int_A d\mu \\
&= \int_{A'} \left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right) \,d\mu_{A'} \\
&= \int_{A''} \left( \,\int_0^{\sqrt{1-x^2}}\,\left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right)\,dy \,\right) \,d\mu_{A''}\\
&= \int_{0}^1 \left( \,\int_0^{\sqrt{1-x^2}}\,\left( \, \int_0^{\sqrt{2}-x-y} \,dz \,\right)\,dy \,\right) \,dx \\
\end{align*}

 

[lpetrich]

Solution to problem #7:

Spoiler

Set z = x + i*y where x and y are real. Then,
$\cos z = \cos x \cosh y - i \sin x \sinh y$
and
$\sin z = \sin x \cosh y + i \cos x \sinh y$.
Substituting in,
$|\cos z|^2 + |\sin z|^2 = \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y + \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y$
The trigonometric functions in x drop out because of a familiar trig identity, giving
$\cosh^2 y + \sinh^2 y = 2 \sinh^2 y + 1$
Since y is real, this quantity only equals 1 if y = 0, meaning that z must be real.

 

[QuantumQuest]

@lpetrich

Your first two identities i.e. $\cos z = \cos x \cosh y - i \sin x \sinh y$ and $\sin z = \sin x \cosh y + i \cos x \sinh y$ are taught as a proposition so they need to be proved.

 

[lpetrich]

OK, I will prove them.

Spoiler

First, Euler's formula for real x: $e^{i x} = \cos x + i \sin x$
Next, the definitions of the hyperbolic functions $\cosh x = \frac12 (e^x + e^{-x})$ and $\sinh x = \frac12 (e^x - e^{-x})$.
One can easily rearrange those definitions to get $e^x = \cosh x + \sinh x$ and $e^{-x} = \cosh x - \sinh x$.

Generalizing Euler's formula to complex args gives $\cos z = \frac12 (e^{iz} + e^{-iz})$ and $\sin z = \frac{1}{2i} (e^{iz} - e^{-iz})$.

Setting z = x + i*y with both x and y real,
$\cos z = \frac12 (e^{ix -y} + e^{-ix+y}) = \frac12[(\cos x + i \sin x) (\cosh y - \sinh y) + (\cos x - i \sin x) (\cosh y + \sinh y)] = \cos x \cosh y - i \sin x \sinh y$
$\sin z = \frac1{2i} (e^{ix -y} - e^{-ix+y}) = \frac1{2i}[(\cos x + i \sin x) (\cosh y - \sinh y) - (\cos x - i \sin x) (\cosh y + \sinh y)] = \sin x \cosh y + i \cos x \sinh y$

Results that I'd used earlier.

 

[dRic2]

I think I came up with the worst proof one could ever imagine for problem #8

Spoiler

If base and area are given then also the height is given ($h = 2A/b$). Now consider the following image:

Now I can express the other two sides ($l_1$ and $l_2$) as a function of $x$ and the two constants ($b$ and $h$). So:

$l_1 = \sqrt{x^2+h^2}$
$l_2 = \sqrt{(b-x)^2+h^2}$

Then the perimeter is

$P = b + l_1 + l_2 = b + \sqrt{x^2+h^2} + \sqrt{(b-x)^2+h^2}$

Then I'm going to find the minimum for this function:

$\dot P = \frac x {\sqrt{x^2+h^2}} + \frac {x-b}{\sqrt{(b-x)^2+h^2}} = 0$

It's evident that $x = \frac b 2$ is a solution (and it is also the minimum). I don't know how to prove it rigorously, though. If you need a rigorous proof I will try to upload it tomorrow (it's gettin late here).

Finally we observe that for $x = b/2$ it happens $l_1 = l_2$.

 

[Sergio Rodriguez]

I have the proof for the 8)

As the second derivative of P is too long to do it, It's easier to proof that slope of the curve p(x) with x < b/2 is negative and the slope of the curve
with x > b/2 is positive. I have do it and there is a minimum in x = b/2

 

[lpetrich]

I will now take on problem #8. I will try to do it in a simpler fashion.

Spoiler

Consider a triangle ABC where AB is the base, with length 2b, and C is at perpendicular distance h and perpendicular-line intersection location at point D. That is, D is on AB and CD is perpendicular to AB. Let C have distance x toward A.

The area A = (1/2)*((2b)*h) = b*h. Since by the problem statement, both A and b are fixed, that means that only h is fixed.

The perimeter $L = 2b + \sqrt{(b-x)^2 + h^2} + \sqrt{(b+x)^2 + h^2}$. One finds the minimum by taking the derivative with respect to the remaining free parameter, x:
$$ \frac{dL}{dx} = - \frac{b-x}{\sqrt{(b-x)^2 + h^2}} + \frac{b+x}{\sqrt{(b+x)^2 + h^2}} $$
For solving (dL/dx) = 0, move the right term to the opposite side of the equation:
$$ - \frac{b-x}{\sqrt{(b-x)^2 + h^2}} = - \frac{b+x}{\sqrt{(b+x)^2 + h^2}} $$
Square it and take the reciprocal:
$$ \frac{(b-x)^2 + h^2}{(b-x)^2} = \frac{(b+x)^2 + h^2}{(b+x)^2} $$
Subtract 1 and take the reciprocal again:
$$ \frac{(b - x)^2}{h^2} = \frac{(b + x)^2}{h^2} $$
Multiply by h^2 and subtract the right term from the left term, giving -4*b*x = 0. This has solution x = 0 as its only solution.

Now consider the curvature of L at x = 0. We must take the derivative again:
$$\frac{d^2 L}{dx^2} = \frac{h^2}{(\sqrt{(b-x)^2 + h^2})^3} + \frac{h^2}{(\sqrt{(b+x)^2 + h^2})^3} $$
This is always positive, meaning that the zero-slope point x = 0 gives the minimum length of perimeter.

For x = 0, the distances AD = DB = b, making the triangle isosceles.

 

[QuantumQuest]

@lpetrich well done!

 

[QuantumQuest]

@lpetrich well done for problem $8$ but the solution of dRic2 came first ;)

 

[Biker]

I was going to post 8 and 10 but I was late :c

Spoiler: Question 10

Question 10:
The limit problem simply asks to prove that the fractional part of n!e tends to equal zero.
$n! e = n! (\frac{1}{1!} + \frac{1}{2!} ...)$

Thus the fractional part call it $K$ equals
$K = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + ...$

If you take n as a common factor (from each () ) you can show that the limit of this series tends to zero, Thus proving the limit in the question

I am interested in the probabilities questions but they same a bit vague for me. In Question 3, A risk neutral player would want to pay as much as he expects to earn?

 

[StoneTemplePython]

I am interested in the probabilities questions but they same a bit vague for me. In Question 3, A risk neutral player would want to pay as much as he expects to earn?

Right. Or equivalently, at what pricing would it be a "fair game"?

 

[QuantumQuest]

@Biker

I can see the points you make but it is not a full solution. In order to have a full solution you must show what the problem asks to be proved, using things that explicitly lead there.

 

[lekh2003]

If this is basic math, I am absolutely useless at math. Can we have an extra basic math problem set which us non-undergrads might be able to do?

 

[fresh_42]

If this is basic math, I am absolutely useless at math. Can we have an extra basic math problem set which us non-undergrads might be able to do?

Yes. We are learning, too. The answers themselves are not so difficult, more their setting - and some of these concepts can be looked up on Wiki, nLab, WolframAlpha etc.

But, you're right. The more as it makes our lives easier to find good questions. And because you support my personal opinion, that it might be a good idea to reverse the normal order: instead of answering randomly posed questions, let us take a subject and ask some exemplary questions around it, e.g. continuity, convergence, series, etc.

As said, we're learning, too.

 

[nuuskur]

It is basic in the sense it is mostly routine i.e what we've been doing in various tutorials of algebra, analysis etc. At the same time, it is instructive. Definitions can be looked up by doing some search around the web.

 

[Suyash Singh]

question 10 solution guys,
Using principle of mathematical induction,
when n=1
lim_n->inf(e-e)=0
Proved true for n=1
Now assume true for n=k,
lim_n->inf(k!e-⌊k!e⌋)=0 ------1
for n=k+1
lim_n->inf((k+1)!e-⌊(k+1)!e⌋)
lim_n->inf(k!e+e-⌊k!e⌋)-⌊(K+1)e⌋)
lim_n->inf(k!e+e-⌊k!e⌋-⌊(k-1)!e⌋-⌊ke⌋-⌊e⌋)
lim_n->inf(e-⌊(k!e⌋-2) using equation 1
again rearrange equation 1 and apply above to get,
lim_n->inf(e-k!e-2)
I am stuck here now

 

[fresh_42]

question 10 solution guys,
Using principle of mathematical induction,
when n=1
lim_n->inf(e-e)=0
Proved true for n=1
Now assume true for n=k,
lim_n->inf(k!e-⌊k!e⌋)=0 ------1
for n=k+1
lim_n->inf((k+1)!e-⌊(k+1)!e⌋)
lim_n->inf(k!e+e-⌊k!e⌋)-⌊(K+1)e⌋)
lim_n->inf(k!e+e-⌊k!e⌋-⌊(k-1)!e⌋-⌊ke⌋-⌊e⌋)
lim_n->inf(e-⌊(k!e⌋-2) using equation 1
again rearrange equation 1 and apply above to get,
lim_n->inf(e-k!e-2)
I am stuck here now

Your induction anchor, in case you separate the two occurrences and roles of $n$, should be $\lim_{n \to \infty} (1! \cdot e - \lfloor 1! \cdot e \rfloor) =0$ which is wrong. However, we can't separate the two, so the induction in this form doesn't make sense. @Biker in post #15 has been on the right track, he just didn't work out his idea.

 

[Suyash Singh]

Your induction anchor, in case you separate the two occurrences and roles of $n$, should be $\lim_{n \to \infty} (1! \cdot e - \lfloor 1! \cdot e \rfloor) =0$ which is wrong. However, we can't separate the two, so the induction in this form doesn't make sense. @Biker in post #15 has been on the right track, he just didn't work out his idea.

I guess i need to study "exponential and logarithmic series" again:(
I forgot everything

 

[fresh_42]

I guess i need to study "exponential and logarithmic series" again:(
I forgot everything

I'm not quite sure, whether @Biker's method actually works, I haven't checked the solution to this problem, but I guess the series for $e$ and maybe a good remainder term estimation of the Taylor series could be helpful.

 

[Adam Kohnle]

Can someone translate this into English? Just kidding but this is not basic unless I know less than I thought. Is the answer... 4? Just kidding but this is like Greek to me so I need to do some more studying then

 

[fresh_42]

Can someone translate this into English? Just kidding but this is not basic unless I know less than I thought. Is the answer... 4? Just kidding but this is like Greek to me so I need to do some more studying then

If you have specific questions, I mean those which Wiki doesn't answer (or PF hasn't already), feel free to ask, either here or if you're really interested in greater details, in a separate thread. The technical language often sounds like more than it actually is. The Manhattan metric, e.g. is nothing else as counting blocks instead of diagonals, and the French Railway metric means: all distances have to include Paris.

 

[Biker]

@Biker

I can see the points you make but it is not a full solution. In order to have a full solution you must show what the problem asks to be proved, using things that explicitly lead there.

What do you mean by showing what the problem asks to be proved? What methods do you want me to use?
It is asking to see whether the reminder tends to zero as n gets larger or no.
What inside the limit is just the number minus its integer value which is the reminder.

 

[QuantumQuest]

What do you mean by showing what the problem asks to be proved? What methods do you want me to use?
It is asking to see whether the reminder tends to zero as n gets larger or no.
What inside the limit is just the number minus its integer value which is the reminder.

You can pick any method you like to solve the problem provided that you'll not leave anything to be surmised by anyone who is reading your solution.
So, I'll ask it another way in order to be clear. Do you regard this as a full solution?

... Thus the fractional part call it $K$ equals
$K = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + ...$

If you take n as a common factor (from each () ) you can show that the limit of this series tends to zero, Thus proving the limit in the question

I don't say that you don't have a general understanding of the problem but according to rule #1 of the challenge

1) In order for a solution to count, a full derivation or proof must be given.

and I don't want to be unfair to anyone. The solution of this problem is not a one liner, no matter what way(s) you pick to solve it. On the other hand, I don't say it is difficult and that's why it is in a basic math challenge. In any case and talking about your solution, if you write it in a way that makes it complete I'll happily accept it.

 

[QuantumQuest]

I have the proof for the 8)

As the second derivative of P is too long to do it, It's easier to proof that slope of the curve p(x) with x < b/2 is negative and the slope of the curve
with x > b/2 is positive. I have do it and there is a minimum in x = b/2

I can see that you put some efforts in the right way but there are things missing so this can't be regarded as a full solution. Try to work it out in a complete way.

 

[QuantumQuest]

question 10 solution guys,
Using principle of mathematical induction,
when n=1
limn->inf(e-e)=0
Proved true for n=1
Now assume true for n=k,
limn->inf(k!e-⌊k!e⌋)=0 ------1
for n=k+1
limn->inf((k+1)!e-⌊(k+1)!e⌋)
limn->inf(k!e+e-⌊k!e⌋)-⌊(K+1)e⌋)
limn->inf(k!e+e-⌊k!e⌋-⌊(k-1)!e⌋-⌊ke⌋-⌊e⌋)
limn->inf(e-⌊(k!e⌋-2) using equation 1
again rearrange equation 1 and apply above to get,
limn->inf(e-k!e-2)
I am stuck here now

Besides what @fresh_42 has already pointed out there is no reason to invoke a cannon to shoot a sparrow. In other words, I appreciate your efforts but think about the problem in a simpler way i.e. by using sequences. You can mix and match some of them to your advantage.

 

[nuuskur]

The expression $a\equiv b\pmod{n}$ is equivalent to $n\mid (a-b)$. I will be working with the latter.

As previous attempts had failed, it's time for some good ol' induction.

Spoiler: Problem 2

Proof by induction on $n$.
Let $a=:2k+1$ be an arbitrary odd number. For $n=1$ it holds that
<br /> 8\mid (2k+1)^2 -1 \quad\mbox{i.e}\quad 8\mid 4k(k+1)\qquad (k\in\mathbb Z)<br />
Indeed, if $8\mid 4k(k+1)$, then $8\mid \left (4k(k+1) + 8(k+1) \right )$ i.e $8\mid 4(k+1)(k+2)$.Suppose for some $n>1$ it holds that $2^{n+1} \mid (2k+1)^{2^{n-1}}-1, k\in\mathbb Z$. We need to show
<br /> 2^{n+2} \mid (2k+1)^{2^{n}}-1,\quad k\in\mathbb Z<br />
We have
<br /> (2k+1)^{2^{n}}-1 = \left [(2k+1)^{2^{n-1}}\right ]^2 -1 = \left [(2k+1)^{2^{n-1}}-1\right ]\left [(2k+1)^{2^{n-1}}+1\right ]<br />
The quantity $2^{n+1}$ divides the left factor of the RHS by inductive assumption. The right factor of the RHS is even, therefore divisible by $2$. Therefore, for every $k\in\mathbb Z$ it holds that
<br /> 2^{n+2} \mid (2k+1)^{2^{n}}-1\qquad (n\in\mathbb N).<br />
(corrections suggested by @fresh_42 included)

Normally, I avoid induction like the plague. I use it as a last resort, since it often makes the result feel..cheap. It's some kind of silly superstition. (graph theoretic induction schemes are the work of the devil, but I digress...)

Spoiler: Previous idea

Sketch: proof by contrapositive. Assume $1\leq GCD(2^{n+2}, a^{2^n}-1) =:d <2^{n+2}$. Suffices to show the number $a$ is even. If $d=1$, the result follows immediately due to for some $u,v\in\mathbb Z$ the equality $u2^{n+2} + v\left (a^{2^n}-1\right )=1$ holds. If, however, $d$ is a power of $2$, I would instead get that $a^{2^n}$ is odd, therefore $a$ is odd.

Does anybody see a way to exclude $d>1$ ?

 

[fresh_42]

Well done, @nuuskur. I don't share your sentiments towards induction, the more as this proof well-nigh screams for an induction. I even met people, who would instantly call for an induction rather than to use anything with a negation. They would feel exact the opposite: Induction is algorithmic, constructive, and ergo good. Negation is a logical trick, and ergo evil.

Just to improve your writing, not to criticize your proof:

It suffices to show

That's misleading. As it is exactly what we have to prove, why is it "only" sufficient? Better would have been something like: "Thus we must show" or similar. "Sufficient" let's your readers search where you've restricted generality, but there is no such place.

The quantity $2^{n+1}$ divides the left term

factor on the RHS

by inductive assumption. The right term

factor

is even, therefore divisible by $2$.

A "left term" is sought on the left, not on the right side of an equation.

As said, not a big deal, but such small inaccuracies make it harder to read. E.g. while scratching your proof that I could better follow your steps, my induction assumption looked like: $a^{2^{n-1}}-1 =(2k+1)^{2^{n-1}} - 1 = 2^{n+1}\cdot N$ and I used this $N$ again in the last line to immediately see the divisibility: $a^{2^{n}}-1 = \ldots = (2^{n+1}\cdot N)\cdot (2\cdot M)$.
So with a little effort, I removed all steps which required to remember the steps in between (means to look up again and again: what was it here, n, n+1, n-1?).

I know, that is nitpicking and not really necessary to mention. I just wanted to demonstrate w.r.t. my previous criticism, how little helpers can make an entire proof easier to read.

 

[Adam Kohnle]

If you have specific questions, I mean those which Wiki doesn't answer (or PF hasn't already), feel free to ask, either here or if you're really interested in greater details, in a separate thread. The technical language often sounds like more than it actually is. The Manhattan metric, e.g. is nothing else as counting blocks instead of diagonals, and the French Railway metric means: all distances have to include Paris.

I appreciate that and if I have questions then I will ask y'all. For now I am going to worry about learning other things since I am beginning college and I still am not sure what my second major should be. I have to do one major as secondary education and then one major in the STEM field so I can teach STEM but I'm not sure which to pick. I might make a thread about it if it is allowed.

 

[fresh_42]

I appreciate that and if I have questions then I will ask y'all. For now I am going to worry about learning other things since I am beginning college and I still am not sure what my second major should be. I have to do one major as secondary education and then one major in the STEM field so I can teach STEM but I'm not sure which to pick. I might make a thread about it if it is allowed.

Sure, you're welcome. We have an extra forum for those kind of questions: Academic Guidance. Not all forums are read by all members, so to place a question in the right forum addresses the right people, here often current or former teachers and professors. And, of course, the better you pose your question, like telling what your major is, the better the answers are which you will receive.

 

[Adam Kohnle]

Sure, you're welcome. We have an extra forum for those kind of questions: Academic Guidance. Not all forums are read by all members, so to place a question in the right forum addresses the right people, here often current or former teachers and professors. And, of course, the better you pose your question, like telling what your major is, the better the answers are which you will receive.

Thanks so much! I will go and do that!

 

[QuantumQuest]

As previous attempts had failed, it's time for some good ol' induction.

@nuuskur you did well. Just try to be somewhat more straightforward and structured in your solutions and you'll be really fine. This is no kind of criticism or some negative comment; I just tell you for your own good.

 

[archaic]

A try at question 10.

Spoiler

Let x = ⌊n!e⌋
⟺ (property) n!e − 1 < x ≤ n!e
⟺ (multiply by -1) - n!e ≤ - x < 1 - n!e
⟺ (add n!e to both sides) 0 ≤ n!e - x < 1
$$\lim_{n\to\infty} 0 ≤ \lim_{n\to\infty} (n!e - \lfloor n!e \rfloor) ≤ \lim_{n\to\infty} 1$$ ?

 

[nuuskur]

@archaic
When you take the limit, strict inequalities become non-strict.

 

[archaic]

@archaic
When you take the limit, strict inequalities become non-strict.

Thank you

 

[nuuskur]

Number 5 seems like a fun exercise. I'll assume we already know these mappings are metrics.

Spoiler: Problem 5

French Metro. For every $a,b\in\mathbb R^2$
<br /> \rho (a,b) := \begin{cases} \lvert a-b\rvert, &amp;\exists k: ka=b \\ |a|+|b|, &amp;\mbox{otherwise}\end{cases}<br />
where $|a|$ denotes the Euclidean distance (from Paris). Describe the open ball $B((2,1),3)$.
If one observes $R(2,1)$ as a bound vector (on Paris), multiplying it by some number, we just stretch this vector so whenever $\rho (a,b) = |a-b|$, they are situated on the straight line determined by Paris and Reims. Then we have the condition $\rho (X,R)<3$ i.e
<br /> |(2k,k) - (2,1)| = |(2(k-1), k-1)| \overset{*}= |k-1||(2,1)| = |k-1|\sqrt{5}&lt;3<br />
(*) the mapping $|\cdot|$ is actually a norm, so I can use its homogeneity property
therefore $-\frac{3}{\sqrt{5}} +1 < k < \frac{3}{\sqrt{5}} +1$.
On the other hand, if $X$ does not lie on the line, the minimum distance between Reims and $X$ is at least the distance between Paris and Reims. (Even if the other city is like 5 km north, you have to take the train back to Paris and only then get to your destination, unless you don't mind walking )
More specifically we have $\rho (X,R)<3$ i.e
<br /> |(x,y)| + |(2,1)| = \sqrt{x^2 + y^2} + \sqrt{5} &lt; 3 \implies x^2+y^2 &lt; \left (3-\sqrt{5}\right )^2<br />
It's a circle around Paris with radius $3-\sqrt{5}$, boundary excluded.
The open ball consists of the segment with specified $k$ and the circle. (think lollipop)
---------------------------------------------------------------------------------------------------------------------------
Manhattan metric. For every $(a,b),(c,d)\in\mathbb R^2$
<br /> \rho ((a,b),(c,d)) := |a-c| + |b-d|.<br />
Suppose $\rho ((x,y),(2,1))<3$ i.e
<br /> |x-2| + |y-1| &lt;3<br />
Since we are adding two non-negative quantities they are both bounded by $3$. This means $-1<x<5$ or $-2<y<4$. (so the open ball definitely lives inside of this open rectangle)

Firstly, suppose $y\geq 1$. We have the condition $|x-2| <3 - (y-1) = 4-y$.

1. If $x\geq 2$, then $x+y<6$ (strictly under the line $y = -x+6$)
2. If $x<2$, then $y-x<2$ (strictly under the line $y=x+2$)

Secondly, suppose $y<1$. We have the condition $|x-2| < 3 +y-1 = y+2$.

1. If $x\geq 2$, then $x-y<4$ (strictly above the line $y=x-4$)
2. If $x<2$, then $x+y>0$ (strictly above the line $y=-x$).

Edit: the shape is a special case of a parallelogram with vertices lying on the intersections of the lines. The vertices are $(-1,1), (2,4), (5,1), (2,-2)$. The boundary is excluded.
---------------------------------------------------------------------------------------------------------------------------
Maximum metric. For every $(a,b),(c,d)\in\mathbb R^2$
<br /> \rho ((a,b),(c,d)) := \max \lbrace |a-c|, |b-d|\rbrace<br />
So, suppose $\rho ((x,y),(2,1))<3$ i.e
<br /> \max \lbrace |x-2|, |y-1|\rbrace &lt; 3<br />
This is the rectangle I was talking about in Manhattan exercise: $-1<x<5$ or $-2<y<4$

Edit: It's a square, sorry. With vertices $(-1,4), (5,4), (5,-2), (-1,-2)$. The boundary is excluded.

The balls in general metric spaces exhibit some very "unball"-like characteristics

 

[fresh_42]

The balls in general metric spaces exhibit some very "unball"-like characteristics

That's been the idea behind the question: a different metric is a different way of measuring, not just a different scale like meter and yards.

In the French Railway metric, we can easily travel to Luxembourg but won't reach - and that is the good news - Saint Quentin, although it is "closer" (in the ordinary sense). The ball is shaped like a hammer in a hammer throw competition: a rope of some fixed length and everything inside the Paris highway circle.
Your description and numbers are correct.

For the other two metrics, where a ball is shaped like a rectangular, can you tell the name of the shapes and their vertices?

 

[nuuskur]

For the other two metrics, where a ball is shaped like a rectangular, can you tell the name of the shapes and their vertices?

I've edited #40 with this information.

 

[fresh_42]

I've edited #40 with this information.

Thanks. Btw. the "ball" in the Manhattan metric is called a rhombus, a square standing on one of its corners. Here's a quick illustration:

 

[lpetrich]

I'll do Problem 10.

Spoiler

For that, we must start with the series
$$e = \sum_{k=0}^\infty \frac{1}{k!}$$
Multiply by n!:
$$n! e = \sum_{k=0}^\infty \frac{n!}{k!}$$
Split the series in two, after k = n: $n! e = e_1 + e_2$ where
$$ e_1 = \sum_{k=0}^n \frac{n!}{k!} ,\ e_2 = \sum_{k=n+1}^\infty \frac{n!}{k!} = \sum_{k=1}^\infty \frac{n!}{(n+k)!} $$
with k redefined for e₂.

For e₁, each of the terms is (k+1)*(k+2)*...*n, and is thus an integer. Therefore, e₁ is an integer.

For e₂, however, each of the terms is the reciprocal of (n+1)*(n+2)*...*(n+k), and each part of that term has lower limit n+1. That reciprocal thus has the lower limit (n+1)^k. Thus,
$$ e_2 = \sum_{k=1}^\infty \frac{1}{(n+1)(n+2) \cdots (n+k)} < \sum_{k=1}^\infty \frac{1}{(n+1)^k} = 1/n $$
using the familiar formula for the sum of a geometric series.

This means that e₂ is always between 0 and 1/n for n >= 1, and thus that it is always between 0 and 1. That means that (n!*e) has integer part e₁ and fractional part e₂ for n >= 1. Thus, $n!e - \lfloor n!e \rfloor = e_2$

Though e₂ is positive, it can be made arbitrarily small with some suitable selection of n, and thus $\lim_{n\to\infty} e_2 = 0$. Thus proving that
$$\lim_{n\to\infty}(n!e - \lfloor n!e \rfloor) = 0$$

 

[QuantumQuest]

A try at question 10.

I appreciate your efforts but can this way lead to any safe conclusion?Can you come up with some other way?

 

[QuantumQuest]

@lpetrich well done!

 

[lpetrich]

I'll try problem 9.

Spoiler

For a space with coordinates xⁱ, the one-form $\omega = \omega_i dx^i$, assuming summation over dummy indices like i here. Integrating over curve C with parameter t, with $x^i = x^i(t)$,
$$\int_C \omega = \int_C \omega_i \frac{dx^i}{dt} dt$$
Using Mathematica to do the mathematical grunt work, I find that this problem's integral has the value 5.

The exterior derivative of ω is
$$\nu = d\omega = \frac12 \left( \frac{d\omega_j}{dx^i} - \frac{d\omega_i}{dx^j}\right) dx^i \wedge dx^j$$
where the wedge denotes antisymmetry. For this problem, $\nu = z dx \wedge dz$.

Taking a further exterior derivative, I find $d\nu = dz \wedge dx \wedge dz = 0$, and ν is thus closed. This result can be proved more generally:

d(d(any n-form)) = 0

as a consequence of the antisymmetry of the exterior derivative.

 

[fresh_42]

I'll try problem 9.

Spoiler

For a space with coordinates xⁱ, the one-form $\omega = \omega_i dx^i$, assuming summation over dummy indices like i here. Integrating over curve C with parameter t, with $x^i = x^i(t)$,
$$\int_C \omega = \int_C \omega_i \frac{dx^i}{dt} dt$$
Using Mathematica to do the mathematical grunt work, I find that this problem's integral has the value 5.

The exterior derivative of ω is
$$\nu = d\omega = \frac12 \left( \frac{d\omega_j}{dx^i} - \frac{d\omega_i}{dx^j}\right) dx^i \wedge dx^j$$
where the wedge denotes antisymmetry. For this problem, $\nu = z dx \wedge dz$.

Taking a further exterior derivative, I find $d\nu = dz \wedge dx \wedge dz = 0$, and ν is thus closed. This result can be proved more generally:

d(d(any n-form)) = 0

as a consequence of the antisymmetry of the exterior derivative.

The results are correct, but I want to see the steps. E.g. there are only six steps to get 5 and only 4 to calculate $\nu$. Shouldn't be too many to write out.

 

[lpetrich]

Here are some evaluations.

Spoiler

For doing the integral, the integrand is
$$ z^2 \frac{dx}{dt} + 2y \frac{dy}{dt} + x z \frac{dz}{dt} = 1^2 \frac{d(t^2)}{dt} + 2(2t) \frac{d(2t)}{dt} + t^2 \cdot 1 \frac{d(1)}{dt} = 2t + 8t + 0 = 10t $$
Integrating it is easy: $\int (10 t) dt = 5 t^2$, and plugging in the limits of integration gives $5(1^2) - 5(0^2) = 5$.

For taking the derivative, I do
$$ d(z^2 (dx) + 2y (dy) + x z (dz)) = 2z (dz \wedge dx) + 2 (dy \wedge dy) + z (dx \wedge dz) + x (dz \wedge dz) = - 2z (dx \wedge dz) + z (dx \wedge dz) = z (dx \wedge dz) $$

 

[fresh_42]

Here are some evaluations.

Spoiler

For doing the integral, the integrand is
$$ z^2 \frac{dx}{dt} + 2y \frac{dy}{dt} + x z \frac{dz}{dt} = 1^2 \frac{d(t^2)}{dt} + 2(2t) \frac{d(2t)}{dt} + t^2 \cdot 1 \frac{d(1)}{dt} = 2t + 8t + 0 = 10t $$
Integrating it is easy: $\int (10 t) dt = 5 t^2$, and plugging in the limits of integration gives $5(1^2) - 5(0^2) = 5$.

For taking the derivative, I do
$$ d(z^2 (dx) + 2y (dy) + x z (dz)) = 2z (dz \wedge dx) + 2 (dy \wedge dy) + z (dx \wedge dz) + x (dz \wedge dz) = - 2z (dx \wedge dz) + z (dx \wedge dz) = z (dx \wedge dz) $$

Yes, and for sake of completeness, the initial steps are formally:
$$\int_{\Gamma} \omega = \int_{[0,1]} \gamma^*(\omega)=\int_{[0,1]} \omega(d\gamma)=\int_{[0,1]} (z^2 dx +2ydy+xzdz)d\gamma $$