# Challenge Basic Math Challenge - May 2018

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#### fresh_42

Mentor
2018 Award
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

Gold Member
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

Gold Member
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

Gold Member
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.
Proof by induction on $n$.
Let $a=:2k+1$ be an arbitrary odd number. For $n=1$ it holds that
$$8\mid (2k+1)^2 -1 \quad\mbox{i.e}\quad 8\mid 4k(k+1)\qquad (k\in\mathbb Z)$$
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
$$2^{n+2} \mid (2k+1)^{2^{n}}-1,\quad k\in\mathbb Z$$
We have
$$(2k+1)^{2^{n}}-1 = \left [(2k+1)^{2^{n-1}}\right ]^2 -1 = \left [(2k+1)^{2^{n-1}}-1\right ]\left [(2k+1)^{2^{n-1}}+1\right ]$$
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
$$2^{n+2} \mid (2k+1)^{2^{n}}-1\qquad (n\in\mathbb N).$$
(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...)
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 $d>1$ ?

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#### fresh_42

Mentor
2018 Award
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.

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" lets 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.

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

Mentor
2018 Award
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.

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

Gold Member
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.
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
$$\lim_{n\to\infty} 0 ≤ \lim_{n\to\infty} (n!e - \lfloor n!e \rfloor) ≤ \lim_{n\to\infty} 1$$ ?

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#### nuuskur

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

#### nuuskur

Number 5 seems like a fun exercise. I'll assume we already know these mappings are metrics.
French Metro. For every $a,b\in\mathbb R^2$
$$\rho (a,b) := \begin{cases} \lvert a-b\rvert, &\exists k: ka=b \\ |a|+|b|, &\mbox{otherwise}\end{cases}$$
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
$$|(2k,k) - (2,1)| = |(2(k-1), k-1)| \overset{*}= |k-1||(2,1)| = |k-1|\sqrt{5}<3$$
(*) 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
$$|(x,y)| + |(2,1)| = \sqrt{x^2 + y^2} + \sqrt{5} < 3 \implies x^2+y^2 < \left (3-\sqrt{5}\right )^2$$
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)
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Manhattan metric. For every $(a,b),(c,d)\in\mathbb R^2$
$$\rho ((a,b),(c,d)) := |a-c| + |b-d|.$$
Suppose $\rho ((x,y),(2,1))<3$ i.e
$$|x-2| + |y-1| <3$$
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.
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Maximum metric. For every $(a,b),(c,d)\in\mathbb R^2$
$$\rho ((a,b),(c,d)) := \max \lbrace |a-c|, |b-d|\rbrace$$
So, suppose $\rho ((x,y),(2,1))<3$ i.e
$$\max \lbrace |x-2|, |y-1|\rbrace < 3$$
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

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#### fresh_42

Mentor
2018 Award
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

Mentor
2018 Award
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:

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#### lpetrich

I'll do Problem 10.
$$e = \sum_{k=0}^\infty \frac{1}{k!}$$
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

Gold Member
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?

Gold Member

#### lpetrich

I'll try problem 9.
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

Mentor
2018 Award
I'll try problem 9.
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.
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

Mentor
2018 Award
Here are some evaluations.
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:
$$\int_{\Gamma} \omega = \int_{[0,1]} \gamma^*(\omega)=\int_{[0,1]} \omega(d\gamma)=\int_{[0,1]} (z^2 dx +2ydy+xzdz)d\gamma$$

"Basic Math Challenge - May 2018"

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