Challenge Math Challenge - March 2020 (Part II)

[fresh_42]

I am unsure if I understood the question correctly

You did. The example should show that the quotient criterion for convergence and divergence of series needs all its strong restrictions.

 

[StoneTemplePython]

There are easier ways, e.g. with Stirling's formula, although the above solution seems to use the proof of Stirling. However, it can be used directly with appropriate error terms. Another possibility is the use of Wallis' series. So the problem still allows attempts of solutions!

for problem (6a)
there's actually a very nice probabilistic proof of this that completely abstracts away from binomial coefficients and works for just about any symmetric walk on the integers (where position at time n is given by $\sum_{i=1} ^n X_i$ where $X_i$ are iid and symmetric and $E\big[\vert X_i\vert^3 \big]\lt \infty$). The technique uses characteristic functions to get the so called 'local central limit theorem'. I'm pretty sure this is what Polya used to originally prove recurrence of simple random walk in 2-d and transcience in higher dimensions. There's a lot of interlocking parts but I'm going to try to write a high level coherent walk-though without getting bogged down in every detail. It's actually a really nice use of symmetry and basic complex analysis.

 

[mathwonk]

HINT: Re #4, the only solution I know seems a bit sophisticated, and shows further that there is no (finite dimensional) real division algebra whose dimension has an odd factor. I will not give it, in hopes some young person will find an original solution of the problem as asked. But the idea is that for an algebra to be a division algebra, equivalently to have no zero divisors, requires a certain system of homogeneous equations to have no non - zero solution. On the other hand there is a generalization of the basic calculus result that all (non constant) polynomials of odd degree have real solutions, that says that all real systems of n homogeneous polynomials in n+1 variables, in which all equations have odd degree, does have a non zero real solution. Even with this theorem, some clever tweaks are needed to get the result, but maybe the weaker result as asked follows without these tweaks.

 

[Infrared]

@mathwonk There are some easy topological arguments too. A division algebra structure on $\mathbb{R}^n$ gives a topological group structure on the unit sphere $S^{n-1}$ by $a\odot b= ab/|ab|$, and this can only happen in odd dimension by degree considerations.

I can confirm that @fresh_42 has an elementary solution in mind though ;)

 

[Not anonymous]

14. For natural numbers $1\leq k\leq 2n$ show that
$$
{{2n+1}\choose{k−1}} + {{2n+1}\choose{k+1}} ≥ 2⋅\dfrac{n+1} {n+2}⋅ {{2n+1}\choose{k}}
$$

Spoiler: Question 14

Let $S$ denote the sum on LHS of the inequality.
$$
S = {{2n+1}\choose{k−1}} + {{2n+1}\choose{k+1}} = \dfrac{(2n+1)!}{(k-1)!(2n-k+2)!} + \dfrac{(2n+1)!}{(k+1)!(2n-k)!} = \dfrac{(2n+1)!}{k!(2n-k+1)!} . \left( \dfrac{k}{2n-k+2} + \dfrac{2n-k+1}{k+1} \right) \\
= {{2n+1}\choose{k}} . \left( \dfrac{k}{2n-k+2} + \dfrac{2n-k+1}{k+1} \right)
$$

The inequality in the question is therefore equivalent to $s = \dfrac {S}{{2n+1}\choose{k}} \geq 2⋅\dfrac{n+1} {n+2}$

$s = \dfrac {S}{{2n+1}\choose{k}} = \left( \dfrac{k}{2n-k+2} + \dfrac{2n-k+1}{k+1} \right)$

For a fixed value of positive integer $n$, $s$ can be viewed as a function of $k$, i.e. $s(k)$. We may temporarily drop the question's requirement that $k$ be an integer since $s$ is defined for all real values of $k \in [0,2n]$.

$$
\frac {ds}{dk} = \frac{1}{2n-k+2} + \frac{k}{(2n-k+2)^2} - \frac{1}{k+1} - \frac{2n-k+1}{(k+1)^2} = \\
(2n+2) . \left( \frac{1}{(2n-k+2)^2} - \frac{1}{(k+1)^2} \right)
$$

It is easy to see that
$$
\frac {ds}{dk} \text{ is } \begin{cases}\\
= 0 & \text{if } k = n+\frac{1}{2} \\
\lt 0 & \text{if } k < n+\frac{1}{2} \\
\gt 0 & \text{if } k > n+\frac{1}{2} \\
\end{cases}
$$

Therefore, $s$ is strictly decreasing for $0 \leq k \lt (n+\frac{1}{2})$ and strictly increasing for $(n+\frac{1}{2}) \lt k \lt 2n$ and attains it minimum at $k=n+\frac{1}{2}$. This and the fact that $s(k)$ is symmetric around $n+\frac{1}{2}$ (it is easy to see that i.e. $s(k) = s(2n+1-k)$) implies that if we consider only integer values of $k \in [0, 2n]$, $s(k)$ reaches its minimum at $k=n$ and $k=n+1$. This minimum value is $s_{min} = s(n) = \left( \dfrac{n}{n+2} + \dfrac{n+1}{n+1} \right) = \dfrac{2n+2}{n+2} = 2 . \dfrac{n+1}{n+2}$. It follows that $s \geq \dfrac{2n+2}{n+2} = 2 . \dfrac{n+1}{n+2}$ for natural numbers $1 \leq k \leq 2n$ proving the inequality in the question

 

[wrobel]

Spoiler: Question 7

A set $X=\{u\in H^1(0,1)\mid u(0)=0\}$ being endowed with an inner product
$$(u,v)=\int_0^1u'(x)v'(x)dx$$ is a Hilbert space. By the Riesz representation theorem,
there exists a unique $w\in X$ such that $\int_0^1f(x)dx=(w,f).$ It remains to calculate $w$ (use integration by parts).
$w=x-x^2/2,\quad A=\|w\|_X=1/\sqrt 3$
The function on which the equality is attained $w/\|w\|_X$

 

[Not anonymous]

15. The year on the Earth-like planet Trappist-1e has $365$ days divided into months of $28,30,31$ days. How many months does its year have and how many months with (i) $28$, (ii) $30$, (iii) $31$ days?

Spoiler: Question 15

My understanding is that the question is just asking for positive integer only solutions to the equation $28a + 30b + 31c = 365$, where $a,b,c$ denote number of months of duration 28, 30 and 31 days respectively. If that is right, then 2 solutions exist if there must be at least 1 month of each duration (28, 30 and 31 days) and 1 more solution exists if a year is allowed to have no months of certain duration.

Solutions where $a, b, c$ are all non-zero:
(i) $a=1, b=4, c=7$. This means 12 months in a year, like a non-leap year in our calendar.
(ii) $a=2, b=1, c=9$. This too means 12 months in a year

There is just one solution where not all months of each of the 3 types need to be present:
(iii) $a=0, b=7, c=5$. This too means 12 months in a year

 

[Infrared]

@wrobel Nice, although I'll just remark that you don't actually use Riesz representation (since you directly verified that your $w$ represents the functional $f\mapsto\int_0^1 f$)

 

[fresh_42]

Spoiler: Question 15

My understanding is that the question is just asking for positive integer only solutions to the equation $28a + 30b + 31c = 365$, where $a,b,c$ denote number of months of duration 28, 30 and 31 days respectively. If that is right, then 2 solutions exist if there must be at least 1 month of each duration (28, 30 and 31 days) and 1 more solution exists if a year is allowed to have no months of certain duration.

Solutions where $a, b, c$ are all non-zero:
(i) $a=1, b=4, c=7$. This means 12 months in a year, like a non-leap year in our calendar.
(ii) $a=2, b=1, c=9$. This too means 12 months in a year

There is just one solution where not all months of each of the 3 types need to be present:
(iii) $a=0, b=7, c=5$. This too means 12 months in a year

Send me your crystal ball so that I can check ...

 

[mathwonk]

#5: Hint:

Unusual as its definition may seem, this is a rather nice and uncomplicated ring, i.e. all ideals are principal, indeed all are powers of the one unique maximal ideal. Perhaps you know something about the structure of an arbitrary finitely generated module in this situation, but if not, you might use the usual tool for dealing with finitely generated objects, i.e. induction on the number of generators.

 

[Not anonymous]

Send me your crystal ball so that I can check ...

I wrote a small script to find solutions to that equation since I thought the question was a non-proof-type, just solution-needed type puzzle.

Spoiler: Question 15

# Maximum number of months (in one year) of duration 28, 30 and 31 days cannot be more than 13, 12 and 11 respectively since 365/28 = 13.03.. , 365/30 = 12.16.., 365/31= 11.77..
for i in range(0,14):
   for j in range(0,13):
     for k in range(0,12):
       if (28*i + 30*j + 31*k == 365):
         print("a=", i, ", b=", j, "c=", k)

If this was meant to be a proof-type question, then I apologize for having posted the solution without math proof

 

[fresh_42]

I wrote a small script to find solutions to that equation since I thought the question was a non-proof-type, just solution-needed type puzzle.

Spoiler: Question 15

# Maximum number of months (in one year) of duration 28, 30 and 31 days cannot be more than 13, 12 and 11 respectively since 365/28 = 13.03.. , 365/30 = 12.16.., 365/31= 11.77..
for i in range(0,14):
   for j in range(0,13):
     for k in range(0,12):
       if (28*i + 30*j + 31*k == 365):
         print("a=", i, ", b=", j, "c=", k)

If this was meant to be a proof-type question, then I apologize for having posted the solution without math proof

We live in times where a script should be allowed, so I will accept it. However, the problem has been meant to practice an important trick: the pigeon hole principle.

E.g. one could start with the observation that $c>1$ and rewrite the equation as $28a+30b+31c'=334$ and consider $28(a+b+c')\leq 334 \leq 31(a+b+c')$ etc.

 

[mathwonk]

#9: SPOILER (partial hint):To rule out the case where y is odd, prove first that an odd square is congruent to 1 mod 8.

 

[Infrared]

@mathwonk I'm not sure if it's possible to solve with just modular arithmetic, but please do post an argument if you find one. It follows quickly from computing that the class group of $\mathbb{Q}(\sqrt{-21})$ is $C_2\times C_2$, but I posed the problem wondering if anyone here would find a more elementary approach.

 

[mathwonk]

I assumed you wanted an elementary approach, but I do not have a similarly easy one for y even. (Are you using Fueter's criterion for non existence of rational points? No, I guess not. At least not obviously. I am a novice here, having never studied algebraic number theory.)

 

[Infrared]

An elementary approach would be great if you could find one!

Maybe "follows quickly" was a bit optimistic. But you can view $x^3=y^2+21=(y+\sqrt{-21})(y-\sqrt{-21})$ as an equality of ideals in the ring of integers $\mathbb{Z}[\sqrt{-21}]$ and use the fact that the two ideals on the RHS are coprime to conclude that they are cubes. They have to be cubes of principal ideals by the computation of the class group, which you can directly check is impossible.

I also don't know much about algebraic number theory, and in particular am not familiar with Fueter's criterion. I didn't find a relevant fact with that name with a google search- do you have a reference to look at?

 

[mathwonk]

Thanks!

I meant this reference:

https://www.jstor.org/stable/2047452?seq=1

and you have inspired me to read mike artin's chapter on factorization in his algebra book, where apparently I can learn something of this topic of algebraic number theory!

 

[benorin]

I did problem 6 because it didn't say solved next to the problem, but I believe someone else solved already. I used Stirling's Approximation $n! \sim n^n e^{-n}\sqrt{2\pi n}$ hence

$$\tfrac{\sqrt{\pi n}}{2^{2n}}\cdot\tfrac{(2n)!}{(n!)^2} \sim \tfrac{\sqrt{\pi n}}{2^{2n}}\cdot\tfrac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{(n^n e^{-n}\sqrt{2\pi n})^2}\rightarrow 1 \text{ as }n\to\infty$$

 

[member 587159]

I did problem 6 because it didn't say solved next to the problem, but I believe someone else solved already. I used Stirling's Approximation $n! \sim n^n e^{-n}\sqrt{2\pi n}$ hence

$$\tfrac{\sqrt{\pi n}}{2^{2n}}\cdot\tfrac{(2n)!}{(n!)^2} \sim \tfrac{\sqrt{\pi n}}{2^{2n}}\cdot\tfrac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{(n^n e^{-n}\sqrt{2\pi n})^2}\rightarrow 1 \text{ as }n\to\infty$$

Doesn't the question ask to not use $\sim$?

 

[benorin]

6. Calculate (FR)
$$
\lim_{n \to \infty}\dfrac{\sqrt{n\pi}}{2^{2n}}\cdot \binom{2n}{n}
$$

Doesn't the question ask to not use $\sim$?

Nope.

 

[member 587159]

Nope.

b.) by using Stirling's formula, with accurate remainder terms, i.e. not simply $\sim$.

is the version that's still unsolved.

 

[benorin]

Oh I totally didn't see the a) and b), my bad.