Math Challenge - September 2021

[fresh_42]

Summary: Gamma function. Combinatorics. Stochastic. Semisimple Modules. Topological Groups. Metric spaces. Logarithmic inequality. Stochastic. Primes. Approximation theory.1. (solved by @julian and @benorin ) Let $f$ be a function defined on $(0,\infty)$ such that $f(x)>0$ for all $x>0.$ Suppose that $f$ has the following properties:

1. $\log f(x)$ is a convex function.
2. $f(x+1)=x\cdot f(x)$ for all $x>0.$
3. $f(1)=1.$

Then $f(x)=\displaystyle{\lim_{n \to \infty}\dfrac{n!n^x}{x(x+1)\cdots (x+n)}}=:\Gamma(x)$ for all $x>0.$2. (solved by @TeethWhitener ) Let $T = (x_1, x_2, \ldots , x_m)$ be a sequence of not necessarily distinct reals. For any positive $b$, define
$$
T_b := \{(x_i, x_j ) \,|\, 1 \leq i, j \leq m, |x_i-x_j | \leq b\}.
$$
Show that for any sequence $T$ and for every integer $r>1,$
$$|T_r| < (2r-1)|T_1|.$$

3. (solved by @Office_Shredder ) Let $X,Y$ be two independent identically distributed real random variables. For a positive $b,$ define $p_b:=\operatorname{prob}\left(|X-Y|\leq b\right).$ Then for every integer $r,$ $p_r\leq (2r-1)p_1.$ Thus
$$
\operatorname{prob}\left(|X-Y|\leq 2\right) \leq 3\cdot \operatorname{prob}\left(|X-Y|\leq 1\right)
$$

4. Let $\mathbb{F}$ be a field and $G$ a finite group, such that $\operatorname{char}\mathbb{F}\,\nmid\,|G|.$ Prove that $\mathbb{F}G$ is semisimple, and show that this is not true if $\operatorname{char}\mathbb{F}\,\mid\,|G|.$5. A group $G$ together with a topology, such that the mapping on $G\times G$ (equipped with the product topology) to $G$ given by $(x,y)\longmapsto xy^{-1}$ is continuous, is called a topological group (e.g. a Lie group). Prove

1. (solved by @jbergman ) $G$ is a topological group if and only if inversion and multiplication are continuous.
2. (solved by @jbergman ) The mappings $x\longmapsto xg$ and $x\longmapsto gx$ are homeomorphisms for each $g\in G.$
3. (solved by @jbergman ) Each open subgroup $U\leq G$ is closed, and each closed subgroup $U\leq G$ of finite index is open. If $G$ is compact, then each open subgroup is of finite index.
4. (solved by @jbergman ) Let $H \leq G$ be a subgroup equipped with the subspace topology, $K\trianglelefteq G$ a normal subgroup, and $G/K$ equipped with the quotient space topology. Then $H$ and $G/K$ are again topological groups and the projection $\pi\, : \,G \twoheadrightarrow G/K$ is open.
5. $G$ is Hausdorff if and only if $\{1\}$ is a closed set in $G.$ $G/K$ is Hausdorff for a normal subgroup $K\trianglelefteq G,$ if and only if $K$ is closed in $G.$ If $G$ is totally disconnected, then $G$ is Hausdorff.
6. Let $G$ be a compact topological group and $\{X_j \;(j\in I)\}\subseteq G$ a family of closed subsets such that for all $i ,j\in I$ there is a $k\in I$ with $X_k\subseteq X_i\cap X_j.$ Then we have for any closed subset $Y\subseteq G$
   $$Y\cdot \left(\bigcap_{i\in I}X_i\right) = \bigcap_{i\in I} YX_i $$

6. Let $(X_n,d_n)_{n\in \mathbb{N}_0}$ be a sequence of complete metric spaces, and let $(f_n)_{n\in \mathbb{N}_0}$ be a sequence of continuous functions $f_n:X_{n+1}\longrightarrow X_n$ such that the image $f_n(X_{n+1}) \subseteq (X_n,d_n)$ is dense for all $n\in \mathbb{N}_0$. Then
$$
M_0:= \left\{v_0\in X_0\,|\,\exists\,(v_n)_{n\in \mathbb{N}} \,\forall\,n\in \mathbb{N}\, : \,v_n\in X_n\,\wedge \,f_{n-1}(v_n)=v_{n-1} \right\}
$$
and
$$
M_0\subseteq M:= \bigcap_{n=0}^\infty(f_0\circ f_1\circ\ldots\circ f_n)(X_{n+1})
$$
are dense in $(X_0,d_0).$ In particular $M\neq \emptyset$ in case $X_0\neq \emptyset.$


7. (solved by @julian ) Prove for all $x>-1$
$$
x-(1+x)\log (1+x)\leq -\dfrac{3x^2}{2(x+3)}
$$

8. (solved by @julian ) Let $(\Omega, \mathcal{A},\mathcal{P})$ be a probability space, $B,T,\sigma$ positive real numbers, and $n\in \mathbb{N}.$ For independently distributed random variables $X_1,\ldots,X_n\, : \,\Omega \longrightarrow \mathbb{R}$ with expectation values $E(X_k)=0$ and $E(X_k^2)\leq \sigma^2$, and boundary $\|X_k\|_\infty \leq B$ for all $k=1,\ldots,n$ prove
$$
P\left(\dfrac{1}{n}\sum_{k=1}^nX_k \geq \sqrt{\dfrac{2\sigma^2 T}{n}}+\dfrac{2BT}{3n}\right) \leq e^{-T}.
$$

9. Let $\mathbb{P}$ be the set of all primes, $p\in \mathbb{P},$ and $n\in \mathbb{N}$ a positive integer. $\operatorname{ord}_p(N)$ denotes the number of primes $p$ which occur as divisor in $\{1,2,\ldots,N\}$ counted by multiplicity. E.g. $N=24=4!$ and $p=3$ yields in $\{3=3^1,6=3^1\cdot 2,9=3^2,12=3^1\cdot 4,15=3^1\cdot 5,18=3^2\cdot 2,21=3^1\cdot 7,24=3^1\cdot 8\}$
$$
\operatorname{ord}_3(24) = 1+1+2+1+1+2+1+1=10
$$
Prove

1. (solved by @TeethWhitener ) $\displaystyle{\operatorname{ord}_p(n)=\sum_{k\geq 1}\left\lfloor \dfrac{n}{p^k}\right\rfloor}$
2. (solved by @TeethWhitener )$2\left.\,\right|\binom{2n}{n}$ and $p\left.\,\right|\binom{2n}{n}$ for all $n<p\leq 2n$
3. (solved by @TeethWhitener ) $p\geq 3\,\wedge \,2n/3 <p\leq n \,\Longrightarrow \,p\,\nmid\,\binom{2n}{n}$
4. (solved by @julian ) $p^r\left.\,\right|\binom{2n}{n}\,\Longrightarrow \,p^r\leq 2n$
5. (solved by @julian ) $\dfrac{2^{2n-1}}{n}\leq \binom{2n}{n}\leq 2^{2n-1}$
6. (solved by @julian ) $\displaystyle{\prod_{p\leq n}p<4^n}$

10. Let $K$ be compact and $C(K):=\{f:K\to \mathbb{R} \text{ or }\mathbb{C}\,|\,f \text{ is continuous}.$ A $n-$dimensional subspace $M\subseteq C(K)$ is called Haar space, if all $f\in M-\{0\}$ have at most $n-1$ zeros. Linear independent functions $S:=\{\varphi_1,\ldots,\varphi_n \}\subseteq C(K)$ are called a Chebyshev- or Haar-system, if $\operatorname{span}(S)$ is a Haar space. We denote the (compact) unit circle $\mathbb{T} := \{e^{2\pi it} \,|\, t\in [0,1)\}.$ Let $K\subseteq \mathbb{R}$ be compact or $K=\mathbb{T}\, , \,f\in C(K).$

We call a point $\xi \in K$ with $f(\xi)=0$ a simple zero of $f$ if $\xi$ is either on the boundary of $K$ or $f$ changes sign in $\xi.$ If $f(\xi -t)f(\xi +t)>0$ in a neighborhood of $\xi,$ then we speak of a double zero.

1. A subspace $M\subseteq C(K)$ with $\dim M=n$ is a Haar space if and only if each $f\in M-\{0\}$ that has $j\in \mathbb{N}_0$ simple zeros and $k\in \mathbb{N}_0$ double zeros holdsSo each element $f\in M-\{0\}$ has at most $n-1$ different zeros.
   $$j+2k<n. $$ So each element $f\in M-\{0\}$ has at most $n-1$ different zeros.
2. The space of all real-valued trigonometric polynomials on $[0,1)$ of degree at most $n$ is a Haar space of dimension $2n+1.$
3. Let $n\in \mathbb{N}_0,\;p\in T_n,$ and $x\in \mathbb{T}.$ Then
   $$
   \left|p'(x)\right|\leq 2\pi n\sqrt{\|p\|^2_\infty -|p(x)|^2}.
   $$
   Remark: Consider the linear differential operator $D(p)=p'$ on $T_n.$ From $|D(\sin(2\pi nx))|=|2\pi n\cos(2\pi nx)|$ we conclude that
   $$
   2\pi n \leq \|D\|=\sup_{\|p\|_\infty \leq 1}\|p'\|_\infty=\sup_{\|p\|_\infty = 1}\sup_{x\in \mathbb{T}}|p'(x)| <\infty
   $$
   because $T_n$ is finite-dimensional.

High Schoolers only (until 26th)
11. (solved by @Not anonymous ) Let $S$ be a set of real numbers that is closed under multiplication (that is, if $a$ and $b$ are in $S$, then so is $ab$). Let $T$ and $U$ be disjoint subsets of $S$ whose union is $S$. Given that the product of any three (not necessarily distinct) elements of $T$ is in $T$ and that the product of any three elements of $U$ is in $U$, show that at least one of the two subsets $T,U$ is closed under multiplication.12. (solved by @Not anonymous ) Suppose we have a necklace of $n$ beads. Each bead is labeled with an integer and the sum of all these labels is $n-1.$ Prove that we can cut the necklace to form a string whose consecutive labels $x_1, x_2,\ldots, x_n$ satisfy
$$
\sum_{i=1}^k x_i \leq k-1 \quad (k=1,\ldots,n).
$$

13. (solved by @Not anonymous ) Let $d:=d_1d_2\ldots d_9$ be a number with not necessarily distinct nine decimal digits. A number $e:=e_1e_2\ldots e_9$ is such that each of the nine digit numbers formed by replacing just one of the digits $d_j$ by the corresponding digit $e_j$ is divisible by $7$ for all $1\leq j\leq 9.$ A number $f:=f_1f_2\ldots f_9$ is formed the same way by starting with $e,$ i.e. each of the nine numbers formed by replacing a $e_k$ by $f_k$ is divisible by $7.$ Example: If $d=20210901$ then $e_6\in \{0,7\}$ since $7\,|\,20210001$ and $7\,|\,20210701 \,.$
Show that, for each $j,$ $d_j-f_j$ is divisible by $7.$14. An ellipse, whose semi-axes have lengths $a$ and $b$, rolls without slipping on the curve $y=c\sin(x/a).$ How are $a,b,c$ related, given that the ellipse completes one revolution when it traverses one period of the curve?15. (solved by @Not anonymous ) For a partition $\pi$ of $N:=\{1,2,\ldots\, , \,9\}$, let $\pi(x)$ be the number of elements in the part containing $x.$ Prove that for any two partitions $\pi_1$ and $\pi_2,$ there are two distinct numbers $x$ and $y$ in $N$ such that $\pi_j(x)=\pi_j(y)$ for $j=1,2.$

 

[PeroK]

I was confused by number 15, but I think I understand it now.

 

[martinbn]

Small nitpicking: I don't agree that a topological group and a Lie group are the same thing.

 

[fresh_42]

Small nitpicking: I don't agree that a topological group and a Lie group are the same thing.

Me neither. I said for example, and I hope you agree that a Lie group is indeed a topological group.

 

[martinbn]

Me neither. I said for example, and I hope you agree that a Lie group is indeed a topological group.

I see.

 

[Chestermiller]

In question 7, did you mean log or ln?

 

[benorin]

Spoiler: #1

#1) "Um, I'll take what is the Bohr-Mollerup Theorem for 200 points, Alex."

I want you honest opinions here, what do you think of my new signature? Simple yet amazing truth but is it too long?

 

[fresh_42]

In question 7, did you mean log or ln?

I meant $\ln$. I just think that $\log$ has become common if no basis is noted.

 

[fresh_42]

Spoiler: #1

#1) "Um, I'll take what is the Bohr-Mollerup Theorem for 200 points, Alex."

I want you honest opinions here, what do you think of my new signature? Simple yet amazing truth but is it too long?

I take Gauß for 500. (He defined it that way.)

 

[benorin]

I meant $\ln$. I just think that $\log$ has become common if no basis is noted.

In Computer Science $\log := \log _2$, in algebra 2 $\log := \log _{10}$, in Analysis $\log := \ln$.

 

[LCSphysicist]

I am really puzzled about the exercise 7. I think the use of calculus can simplify a lot the demonstration of the inequalitty, but i ask myself (and of course it is more elegante if it exist) if there is a way to show it without calculus (not derivatives basically)...

 

[fresh_42]

I am really puzzled about the exercise 7. I think the use of calculus can simplify a lot the demonstration of the inequalitty, but i ask myself (and of course it is more elegante if it exist) if there is a way to show it without calculus (not derivatives basically)...

I don't know. My solution uses derivatives. At least it is rather short.

 

[TeethWhitener]

Here's my completely non-rigorous first stab at problem 2:

Spoiler: Problem 2

Thought process: try to make the right side as small as possible and the left side as big as possible and show the inequality still holds.

(NB--I'm pretty sure we can assume WLOG that $x_1\leq x_2\leq\dots\leq x_m$, so that's what I'll be doing)

For any given sequence $T = (x_1,x_2,\dots,x_m)$, we have $\min(|T_1|)=m$. Such a situation occurs, e.g., for $T=(2,4,6,8,10)$. Here the only members of $T_1$ are the pairs where $i=j$; that is: $T_1=\{(x_1,x_1),(x_2,x_2),\dots,(x_m,x_m)\}$. For this case to hold, it must be true that $\forall i \neq j, |x_i-x_j|>1$. Using the assumption above that the sequence $T$ is ordered low to high, we'll let $|x_i-x_{i+1}|=1+\epsilon_{i}, \epsilon_{i}>0$. This means that, in this case, $\max(|x_i-x_j|)=(m-1)(1+\max(\epsilon_i))$, because there are $m-1$ gaps between the $m$ members of the sequence, and each gap is less than or equal to $1+\max(\epsilon_i)$.

We also have $\max(|T_r|)=m^2$. This arises in the case where all possible pairs are present in $T_r$. However, for the maximum case to hold, it must be the case that $r\geq\max(|x_i-x_j|)$. Keeping with the theme of my thought process from earlier, we want to make $r$ and $|T_1|$ as small as possible. So we have $r\geq (m-1)(1+\max(\epsilon_i))$ and we choose to make $\epsilon_i\ll 1$ to get $r$ as small as possible. Now, we know that $r$ is an integer, and that $r>m-1$, so that means $\min(r)=m$.

If we then plug into the inequality $\max(|T_r|)=m^2$, $\min(r)=m$, and $\min(|T_1|)=m$, we get
$$m^2<(2m-1)m$$
or
$$m>1$$
To finish up, when $m=1$, we have $|T_r|=|T_1|=1$ for all $r$, and the inequality holds trivially.

I'm not sure this constitutes a full proof, because I'm not sure if it properly excludes some intermediate values of $|T_1|$, $r$, and $|T_r|$ that could mess things up.

 

[fresh_42]

Here's my completely non-rigorous first stab at problem 2:

Spoiler: Problem 2

Thought process: try to make the right side as small as possible and the left side as big as possible and show the inequality still holds.

(NB--I'm pretty sure we can assume WLOG that $x_1\leq x_2\leq\dots\leq x_m$, so that's what I'll be doing)

For any given sequence $T = (x_1,x_2,\dots,x_m)$, we have $\min(|T_1|)=m$. Such a situation occurs, e.g., for $T=(2,4,6,8,10)$. Here the only members of $T_1$ are the pairs where $i=j$; that is: $T_1=\{(x_1,x_1),(x_2,x_2),\dots,(x_m,x_m)\}$. For this case to hold, it must be true that $\forall i \neq j, |x_i-x_j|>1$. Using the assumption above that the sequence $T$ is ordered low to high, we'll let $|x_i-x_{i+1}|=1+\epsilon_{i}, \epsilon_{i}>0$. This means that, in this case, $\max(|x_i-x_j|)=(m-1)(1+\max(\epsilon_i))$, because there are $m-1$ gaps between the $m$ members of the sequence, and each gap is less than or equal to $1+\max(\epsilon_i)$.

We also have $\max(|T_r|)=m^2$. This arises in the case where all possible pairs are present in $T_r$. However, for the maximum case to hold, it must be the case that $r\geq\max(|x_i-x_j|)$. Keeping with the theme of my thought process from earlier, we want to make $r$ and $|T_1|$ as small as possible. So we have $r\geq (m-1)(1+\max(\epsilon_i))$ and we choose to make $\epsilon_i\ll 1$ to get $r$ as small as possible. Now, we know that $r$ is an integer, and that $r>m-1$, so that means $\min(r)=m$.

If we then plug into the inequality $\max(|T_r|)=m^2$, $\min(r)=m$, and $\min(|T_1|)=m$, we get
$$m^2<(2m-1)m$$
or
$$m>1$$
To finish up, when $m=1$, we have $|T_r|=|T_1|=1$ for all $r$, and the inequality holds trivially.

I'm not sure this constitutes a full proof, because I'm not sure if it properly excludes some intermediate values of $|T_1|$, $r$, and $|T_r|$ that could mess things up.

Looks ok to me. My solution uses induction on $m$ and intervals $[x_i-1,x_i+1].$

 

[TeethWhitener]

For problem 9 part 1, I think you meant $\displaystyle{\operatorname{ord}_p(n!)=\sum_{k\geq 1}\left\lfloor \dfrac{n!}{p^k}\right\rfloor}$? If so, I think I have a bead on parts 1 and 2--maybe.

Spoiler: Problem 9 Part 1

$\displaystyle{\operatorname{ord}}_p(N)$ counts the number of times a prime appears in the divisor list of $\{1,2,\dots,N\}$ along with the multiplicity. But let's consider a different operator $\overline{\displaystyle{\operatorname{ord}}}_m(N)$ that is the same as $\displaystyle{\operatorname{ord}}_p(N)$ except it doesn't count multiplicity, and $m\in\mathbb{N}$ doesn't have to be prime. If we look, e.g., at $\overline{\displaystyle{\operatorname{ord}}}_2(n!)$, we add 1 every time we pass an even number in $\{1,2,\dots,n!\}$, which means $\overline{\displaystyle{\operatorname{ord}}}_2(n!)=\left\lfloor\frac{n!}{2}\right\rfloor$. Similarly, $\overline{\displaystyle{\operatorname{ord}}}_3(n!)=\left\lfloor\frac{n!}{3}\right\rfloor$, $\overline{\displaystyle{\operatorname{ord}}}_4(n!)=\left\lfloor\frac{n!}{4}\right\rfloor$, etc., and generally
$$\overline{\displaystyle{\operatorname{ord}}}_m(n!)=\left\lfloor\frac{n!}{m}\right\rfloor$$
To go from $\overline{\displaystyle{\operatorname{ord}}}_m(n!)$ to $\displaystyle{\operatorname{ord}}_p(n!)$, we note that
$$\displaystyle{\operatorname{ord}}_p(n!)=\sum_{k\geq1}{\overline{\displaystyle{\operatorname{ord}}}_{p^k}(n!)}=\sum_{k\geq 1}{\left\lfloor \dfrac{n!}{p^k}\right\rfloor}$$

Spoiler: Problem 9 Part 2

$p|\binom{2n}{n}$ is really easy. Since $p>n$, there is no $p$ in the denominator of $\binom{2n}{n}$. But since $p\leq 2n$, there is a $p$ in the numerator. So $p|\binom{2n}{n}$.

$2|\binom{2n}{n}$ is trickier. I think the point is to use the result from part 1, but I'm pretty sure it's easier via induction. Denoting $n!=2^xa$ and $(2n)!=2^yb$, where $2\nmid a$ and $2\nmid b$, we have that
$$\binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{2^yb}{2^{2x}a}$$
(This is where I start to lose rigor and question my approach.)
We can map each even number in the set $\{1,2,\dots,n\}$ to its double in the set $\{1,2,\dots,2n\}$. We can map each multiple of 4 in the original set to its double in the larger set, and do the same for each multiple of 8, etc., or $2^k$ generally. This mapping let's us write $y\geq (x+\left\lfloor\frac{n}{2}\right\rfloor)$. But this mapping isn’t surjective: we haven’t accounted for the factors of 2 in $(2n)!$ of the form $4j+2$. When we do that, we get that $y\geq (x+n)$.

To show that $\binom{2n}{n}$ is divisible by 2, we have to show that $y>2x$, or equivalently, $x<n$. Edit: the following is all incorrect. We do this by induction by proving: $n!<2^na, \forall a\in\mathbb{N}$.
Base step:
$1!=1<2a, \forall a\in\mathbb{N}$
Inductive step:
Let $n!<2^na, \forall a\in\mathbb{N}$. Then $(n+1)!=(n+1)n!<(n+1)2^na$. Since $(n+1)2^n<2^{n+1}$, we have $(n+1)!<2^{n+1}a$.

Edit: crap, ignore. Found an error in problem 9 part 2. May work on it later.

 

[fresh_42]

For problem 9 part 1, I think you meant $\displaystyle{\operatorname{ord}_p(n!)=\sum_{k\geq 1}\left\lfloor \dfrac{n!}{p^k}\right\rfloor}$? If so, I think I have a bead on parts 1 and 2--maybe.

Uhm, no. Consider my example: $10=8+2+0$. $24!$ has many, real many more divisors $3^k.$

Edit: crap, ignore. Found an error in problem 9 part 2. May work on it later.

 

[TeethWhitener]

Uhm, no. Consider my example: $10=8+2+0$. $24!$ has many, real many more divisors $3^k.$

$\displaystyle{\operatorname{ord}_3(4!)=\sum_{k\geq 1}\left\lfloor \dfrac{4}{3^k}\right\rfloor}=1$? I think it has to be factorial, or else I'm really missing something.

 

[fresh_42]

$\displaystyle{\operatorname{ord}_3(4!)=\sum_{k\geq 1}\left\lfloor \dfrac{4}{3^k}\right\rfloor}=1$? I think it has to be factorial, or else I'm really missing something.

Oops, you're right. I confused $4!=24$ with $24!$, sorry. I will correct the OP.

 

[TeethWhitener]

Thanks for the clarification. I still think my problem 9 part 1 is right. I think half of problem 9 part 2 is right too, but I'll have to work on the other half (I've struck through the wrong part in post 15).

 

[fresh_42]

Thanks for the clarification. I still think my problem 9 part 1 is right. I think half of problem 9 part 2 is right too, but I'll have to work on the other half (I've struck through the wrong part in post 15).

It would be nice if you wrote it again for $n$ instead of the faculty, and if possible with $n$ and $p$ fixed. It's actually quite easy, which is why I gathered so many parts.

 

[Not anonymous]

12. Suppose we have a necklace of beads. Each bead is labeled with an integer and the sum of all these labels is $n-1$. Prove that we can cut the necklace to form a string whose consecutive labels $x_1, x_2, \ldots, x_n$ satisfy $$
\sum_{i=1}^k x_i \leq k-1 \, \, (k=1,...,n).
$$

Does "cut the necklace to form a string" mean any rearrangement of the individual beads is allowed, or does it only mean making at most one cut anywhere in the necklace and swapping the 2 resulting parts to form a new sequence (meaning, for e.g., (1, 2, 3, 4, 5) →(3, 4, 5, 1, 2) is allowed but not (5, 2, 1, 4, 3))? If any reordering of beads is allowed, here is my attempted solution.

Spoiler: Question 12

Let $A = (a_1, a_2, ..., a_n)$ be the original sequence of integer-valued labels of the necklace's beads. We create a new sequence $X = (x_1, x_2, ..., x_n)$ using only values from $A$ (also, if some value is repeated in $A$, that value will be repeated the same number of times in the new sequence too, but the order may differ, i.e. cardinality of values is retained) but arranged in ascending (more precisely, non-decreasing) order, i.e. $x_i \leq x_{i+1} \, \forall i \in \{1,...,n-1\}$.

Define $S_k = \sum_{i=1}^k x_k, \, k=(1,...,n)$. Since $X$ contains the same values as $A$ and with same cardinality, the sums of all values in $X$ must be the same as that for $A$. Hence $S_n = \sum_{i=1}^k a_k = n-1$. ∴ $S_k$ meets the condition $S_k \leq k-1$ when $k=n$, and this becomes the initial case for induction.

Inductive step: We now show that if $S_k \leq k-1$ for $k=j$ where integer $j > 1$, then it must be true for $k=j-1$ too.

Proof by contradiction: Suppose $S_{j-1} > j-1$. Since $S_j = S_{j-1} + x_j$, $S_{j-1} > j-1$ implies that $x_j = S_j - S_{j-1} < (j-1) - (j-1) \Rightarrow x_j < 0$. (C1)

But if $S_{j-1} > (j-1)$ as assumed, it follows that $S_{j-1}$ is a positive integer (since $j > 1$) and therefore there must be at least one positive integer in $(x_1, ..., x_{j-1})$ since $S_{j-1}$ is the sum of these values by definition. (C2)

Conditions (C1) and (C2) cause a contradiction since $x_j < 0$ and $(x_1, ..., x_{j-1})$ containing a positive value imply that the values in $(x_1, ..., x_{j-1}, x_{j})$ do not follow a non-decreasing order as required by the definition of the sequence $X$. Since this contradiction eventually arises from the assumption that $S_{j-1} > j-1$, it must be the case that this assumption is incorrect and we must have $S_{j-1} \leq j-1$ instead. Hence the inductive step is proved, and by the method of induction, it follows that $$
S_k = \sum_{i=1}^k x_i \leq k-1 \, \, \forall k \in \{1,..., n\}
$$. Hence, the beads can be rearranged as required.

 

[fresh_42]

Does "cut the necklace to form a string" mean any rearrangement of the individual beads is allowed, or does it only mean making at most one cut anywhere in the necklace and swapping the 2 resulting parts to form a new sequence (meaning, for e.g., (1, 2, 3, 4, 5) →(3, 4, 5, 1, 2) is allowed but not (5, 2, 1, 4, 3))? If any reordering of beads is allowed, here is my attempted solution.

Cut means one cut, i.e. we make a string out of the circle and can so determine $x_n$. Order and given labels remain unchanged.

 

[TeethWhitener]

Here it is rewritten without the factorial. I wasn't quite sure what you meant by n and p fixed.

Spoiler: Problem 9 Part 1

$\displaystyle{\operatorname{ord}}_p(N)$ counts the number of times a prime appears in the divisor list of $\{1,2,\dots,N\}$ along with the multiplicity. But let's consider a different operator $\overline{\displaystyle{\operatorname{ord}}}_m(N)$ that is the same as $\displaystyle{\operatorname{ord}}_p(N)$ except it doesn't count multiplicity, and $m\in\mathbb{N}$ doesn't have to be prime. If we look, e.g., at $\overline{\displaystyle{\operatorname{ord}}}_2(n)$, we add 1 every time we pass an even number in $\{1,2,\dots,n\}$, which means $\overline{\displaystyle{\operatorname{ord}}}_2(n)=\left\lfloor\frac{n}{2}\right\rfloor$. Similarly, $\overline{\displaystyle{\operatorname{ord}}}_3(n)=\left\lfloor\frac{n}{3}\right\rfloor$, $\overline{\displaystyle{\operatorname{ord}}}_4(n)=\left\lfloor\frac{n}{4}\right\rfloor$, etc., and generally
$$\overline{\displaystyle{\operatorname{ord}}}_m(n)=\left\lfloor\frac{n}{m}\right\rfloor$$
To take into account multiplicity and go from $\overline{\displaystyle{\operatorname{ord}}}_m(n)$ to $\displaystyle{\operatorname{ord}}_p(n)$, we note that
$$\displaystyle{\operatorname{ord}}_p(n)=\sum_{k\geq1}{\overline{\displaystyle{\operatorname{ord}}}_{p^k}(n)}=\sum_{k\geq 1}{\left\lfloor \dfrac{n}{p^k}\right\rfloor}$$

I also found another error in 9-2, but I haven't figured out how to complete the problem yet. I'll think on it some more.

 

[fresh_42]

Here it is rewritten without the factorial. I wasn't quite sure what you meant by n and p fixed.

Spoiler: Problem 9 Part 1

$\displaystyle{\operatorname{ord}}_p(N)$ counts the number of times a prime appears in the divisor list of $\{1,2,\dots,N\}$ along with the multiplicity. But let's consider a different operator $\overline{\displaystyle{\operatorname{ord}}}_m(N)$ that is the same as $\displaystyle{\operatorname{ord}}_p(N)$ except it doesn't count multiplicity, and $m\in\mathbb{N}$ doesn't have to be prime. If we look, e.g., at $\overline{\displaystyle{\operatorname{ord}}}_2(n)$, we add 1 every time we pass an even number in $\{1,2,\dots,n\}$, which means $\overline{\displaystyle{\operatorname{ord}}}_2(n)=\left\lfloor\frac{n}{2}\right\rfloor$. Similarly, $\overline{\displaystyle{\operatorname{ord}}}_3(n)=\left\lfloor\frac{n}{3}\right\rfloor$, $\overline{\displaystyle{\operatorname{ord}}}_4(n)=\left\lfloor\frac{n}{4}\right\rfloor$, etc., and generally
$$\overline{\displaystyle{\operatorname{ord}}}_m(n)=\left\lfloor\frac{n}{m}\right\rfloor$$
To take into account multiplicity and go from $\overline{\displaystyle{\operatorname{ord}}}_m(n)$ to $\displaystyle{\operatorname{ord}}_p(n)$, we note that
$$\displaystyle{\operatorname{ord}}_p(n)=\sum_{k\geq1}{\overline{\displaystyle{\operatorname{ord}}}_{p^k}(n)}=\sum_{k\geq 1}{\left\lfloor \dfrac{n}{p^k}\right\rfloor}$$

I also found another error in 9-2, but I haven't figured out how to complete the problem yet. I'll think on it some more.

Yes, but you had not to define $\overline{ord}$. The same argument at the end works with $\left\lfloor \dfrac{n}{p}\right\rfloor\, , \,\left\lfloor \dfrac{n}{p^2}\right\rfloor\, , \,\left\lfloor \dfrac{n}{p^3}\right\rfloor,\ldots$ from the start.

 

[TeethWhitener]

It occurred to me as I was falling asleep that I was going about this problem in the most roundabout way possible.

Spoiler: Problem 9 part 2…again

To show that $2\,|\binom{2n}{n}$, we proceed via induction, which we should’ve just done from the beginning. We’re trying to establish that $$\frac{(2n)!}{n!n!}$$ is divisible by 2.
Base step ($n=1$):
$$\frac{2!}{1!1!}=2$$
Inductive step:
Let $\frac{(2n)!}{n!n!}$ be divisible by 2. Then consider $$\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\frac{(2n)!}{n!n!}$$ We have two cases:
1) $n+1$ is odd. In this case, the denominator doesn’t acquire any additional factors of 2, and we assumed that $\frac{(2n)!}{n!n!}$ was divisible by 2, so the whole expression is divisible by 2.
2) $n+1$ is even. In this case, the denominator acquires a factor of 4 which must be canceled in the numerator for the whole expression to remain divisible by 2. We see that if $n+1$ is even, then $n=2k+1$ for some integer $k$, and the numerator can be written as $(4k+2)!(4k+3)(4k+4)$. Thus we have explicitly a factor of 4 in the numerator to cancel the 4 in the denominator and the inductive step goes through.

 

[TeethWhitener]

Spoiler: Problem 9 part 3

Since $n\geq p>\frac{n}{2}$, each $n!$ in the denominator of $\binom{2n}{n}=\frac{(2n)!}{n!n!}$ will have one factor of $p$. Since $p>\frac{2n}{3}$, the numerator $(2n)!$ will have a factor of $p$ and a factor of $2p$ and no other factors of $p$. So the binomial coefficient will have the form $$\frac{p^2a}{p^2b}$$ The $p$’s will cancel out and the binomial coefficient will not be divisible by $p$.

 

[fresh_42]

It occurred to me as I was falling asleep that I was going about this problem in the most roundabout way possible.

Spoiler: Problem 9 part 2…again

To show that $2\,|\binom{2n}{n}$, we proceed via induction, which we should’ve just done from the beginning. We’re trying to establish that $$\frac{(2n)!}{n!n!}$$ is divisible by 2.
Base step ($n=1$):
$$\frac{2!}{1!1!}=2$$
Inductive step:
Let $\frac{(2n)!}{n!n!}$ be divisible by 2. Then consider $$\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\frac{(2n)!}{n!n!}$$ We have two cases:
1) $n+1$ is odd. In this case, the denominator doesn’t acquire any additional factors of 2, and we assumed that $\frac{(2n)!}{n!n!}$ was divisible by 2, so the whole expression is divisible by 2.
2) $n+1$ is even. In this case, the denominator acquires a factor of 4 which must be canceled in the numerator for the whole expression to remain divisible by 2. We see that if $n+1$ is even, then $n=2k+1$ for some integer $k$, and the numerator can be written as $(4k+2)!(4k+3)(4k+4)$. Thus we have explicitly a factor of 4 in the numerator to cancel the 4 in the denominator and the inductive step goes through.

Spoiler: Problem 9 part 3

Since $n\geq p>\frac{n}{2}$, each $n!$ in the denominator of $\binom{2n}{n}=\frac{(2n)!}{n!n!}$ will have one factor of $p$. Since $p>\frac{2n}{3}$, the numerator $(2n)!$ will have a factor of $p$ and a factor of $2p$ and no other factors of $p$. So the binomial coefficient will have the form $$\frac{p^2a}{p^2b}$$ The $p$’s will cancel out and the binomial coefficient will not be divisible by $p$.

Well, yes, I think. What do you think about my solution?

$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}\,\Longrightarrow \,2\,|\,\binom{2n}{n}$

$\binom{2n}{n}=\dfrac{2n\cdot\ldots\cdot(n+1)}{1\cdot\ldots\cdot n}$ and a prime $n<p\leq 2n$ in the numerator doesn't cancel.

 

[TeethWhitener]

What do you think about my solution?

I think I learned a new binomial coefficient identity

 

[Not anonymous]

Cut means one cut, i.e. we make a string out of the circle and can so determine $x_n$. Order and given labels remain unchanged.

Spoiler: Question 12

Let $A=(a_1, ..., a_n)$ denote the sequence of integer-valued labels of the necklace's beads, with some arbitrary bead chosen as the first one in the sequence and subsequent beads being selected in the order in which they appear in the necklace when moving in a clockwise direction. Define $S_k(A) = \sum_{i=1}^k a_i$. If $S_k(A) \leq k-1 \, \forall k \in \{0,...,n\}$ (call this condition C_sum), then cutting the necklace between $a_n$ and $a_1$ is the solution as it gives us the string $A$ which meets the desired criterion. But if $A$ does not meet the condition C_sum, we show that it is possible to rotate the sequence one or more times to get another sequence that meets the condition C_sum. In other words, we can start at any arbitrary bead on the necklace, look at the $n$ different sums $S_1(A), S_2(A),...,S_n(A)$ to check if it meets the condition C_sum, and if doesn't meet the condition, we should be able to rotate it a finite number of times to reach an ordered sequence that meets C_sum and therefore defines an appropriate position where to make the cut. We only need t consider necklaces having 2 or more beads since the proof is trivial for $n=1$.

Claim: If $A=(a_1, ..., a_n)$ is a sequence of values formed by starting from any arbitrary bead on the necklace as explained above and if $S_k(A) \leq k-1 \, \forall k \in \{K, K+1, ..., n\}$ for some integer $K > 1$ but there exists some $1 \leq j < K$ for which $S_{j}(A) > j-1$. Then it is possible to rotate it to form a sequence $B = (b_1, ..., b_n)$ such that $S_k(B) \leq k-1 \, \forall k \in \{K', K'+1, ..., n\}$ for some integer $1 \leq K' < K$, i.e. $B$ has more $k$ values in its "tail" than $A$ for which the $S_k(.) < k$ condition holds true.
Proof: We prove by the method of induction. Regardless of which bead is chosen as the first bead to form the sequence, the sum of $n$ labels will always be the same, $n-1$. Hence $S_k(A)$ is obviously true for $k=n$ and hence we may choose $K = n$ for the base case. (P0)

Let $j \geq 1$ be the smallest index such that $S_{k}(A) > k-1 \Rightarrow S_{j} \geq j$. And since it is the smallest such index, we must also have $S_{k} \leq k-1$ if $k < j$. (P1)
Consider the rotated sequence $B = (b_1, b_2, ..., b_n) = (a_{j+1}, a_{j+2}, ..., a_n, a_1, a_2, ..., a_j)$, i.e. formed by rotating the necklace by $j$ beads w.r.t. start of sequence $A$.

$S_{n-j}(B) = \sum_{i=1}^{n-j} b_i = (a_{j+1} + a_{j+2} + ... + a_{n}) = S_{n}(A) - S_{j}(A) \leq (n-1) - j \equiv (n-j)-1$ (P2)

For $k > n-j$, $S_{k}(B) = S_{n-j}(B) + (a_1 + ... + a_{k-(n-j)})$. By property (P1), if $k-(n-j) < j$, we have $\sum_{i=1}^{k-(n-j)} a_i \leq k-(n-j)-1 \Rightarrow S_k(B) \leq (n-j-1) + (k-(n-j)-1)$

$\Rightarrow S_k(b) \leq k-2 < k-1$, and if $k-(n-j) =j$, i.e. $k=n$ (the maximum possible value of $k$), then $S_k(B) = S_n(B) = n-1$. Hence $S_k(B) \leq k-1 \, \forall k \in \{n-j, n-j+1, ..., n\}$. Thus we have $K' = (n-j) < K$ such that $S_k(B) \leq k-1 \, \forall k \in \{K', K'+1, ..., n\}$, thus proving the claim.

Thus, as long as there exists some a "violating" offset $j \geq 1$, i.e. one for which $S_j(A) > j-1$ for some sequence $A$ of the necklace with an upper bound $K-1$ for $j$, we can rotate it to obtain another sequence $A$ such that if it at all it too has a "violating" offset $j'$, it will have a smaller upper bound $K'-1 < K-1$. Thus by rotating the necklace we can keep reducing the "violating" offset till we reach a sequence where no violating offset is present (it is reduced to 0), i.e. this new sequence meets the condition C_sum. Say this sequence is $X=(x_1, x_2, ..., x_n) = (a_{r+1}, ..., a_n, a_1, a_2, ..., a_r)$, i.e. obtained by rotating the starting sequence $A$ by $r$ beads. Then cutting the necklace between $a_r$ and $a_{r+1}$ gives us the desired sequence.

 

[fresh_42]

I have really difficulties reading this. The first paragraph is basically only repeating what has to be shown. It can be removed.

12.

Let $A=(a_1, ..., a_n)$ denote the sequence of integer-valued labels of the necklace's beads, with some arbitrary bead chosen as the first one in the sequence and subsequent beads being selected in the order in which they appear in the necklace when moving in a clockwise direction. Define $S_k(A) = \sum_{i=1}^k a_i$. If $S_k(A) \leq k-1 \, \forall k \in \{0,...,n\}$ (call this condition C_sum), then cutting the necklace between $a_n$ and $a_1$ is the solution as it gives us the string $A$ which meets the desired criterion. But if $A$ does not meet the condition C_sum, we show that it is possible to rotate the sequence one or more times to get another sequence that meets the condition C_sum. In other words, we can start at any arbitrary bead on the necklace, look at the $n$ different sums $S_1(A), S_2(A),...,S_n(A)$ to check if it meets the condition C_sum, and if doesn't meet the condition, we should be able to rotate it a finite number of times to reach an ordered sequence that meets C_sum and therefore defines an appropriate position where to make the cut. We only need t consider necklaces having 2 or more beads since the proof is trivial for $n=1$.

Claim: If $A=(a_1, ..., a_n)$ is a sequence of values formed by starting from any arbitrary bead on the necklace as explained above and if $S_k(A) \leq k-1 \, \forall k \in \{K, K+1, ..., n\}$ for some integer $K > 1$ but there exists some $1 \leq j < K$ for which $S_{j}(A) > j-1$. Then it is possible to rotate it to form a sequence $B = (b_1, ..., b_n)$ such that $S_k(B) \leq k-1 \, \forall k \in \{K', K'+1, ..., n\}$ for some integer $1 \leq K' < K$, i.e. $B$ has more $k$ values in its "tail" than $A$ for which the $S_k(.) < k$ condition holds true.

And here I am getting lost. What is $K$ meant to be? As I see it, we can simply say $K=n$ and $j$ is a counterexample. It is neither the maximal nor the minimal index that violates the condition. You always say "some". That makes it difficult to handle sequences. You have a sequence $1,\ldots, K-1$ with "some" $K$ and "some" $j <K$. You do not say what the induction variable is. I assume $K$ since you say $K=n$ is your base case. Do you perform the induction backwards, or do you work with a minimal counterexample? But then you mentioned earlier $n=1$ as trivial case. What are you trying to do? I only see "some" and countless unspecified indices.

Here is a hint:
Consider $S_k= x_1 +\ldots+x_k - \dfrac{k(n-1)}{n}$ with $k=0,1,\ldots,n$ and $S_0=0.$

Proof: We prove by the method of induction. Regardless of which bead is chosen as the first bead to form the sequence, the sum of $n$ labels will always be the same, $n-1$. Hence $S_k(A)$ is obviously true for $k=n$ and hence we may choose $K = n$ for the base case. (P0)

Let $j \geq 1$ be the smallest index such that $S_{k}(A) > k-1 \Rightarrow S_{j} \geq j$. And since it is the smallest such index, we must also have $S_{k} \leq k-1$ if $k < j$. (P1)
Consider the rotated sequence $B = (b_1, b_2, ..., b_n) = (a_{j+1}, a_{j+2}, ..., a_n, a_1, a_2, ..., a_j)$, i.e. formed by rotating the necklace by $j$ beads w.r.t. start of sequence $A$.

$S_{n-j}(B) = \sum_{i=1}^{n-j} b_i = (a_{j+1} + a_{j+2} + ... + a_{n}) = S_{n}(A) - S_{j}(A) \leq (n-1) - j \equiv (n-j)-1$ (P2)

For $k > n-j$, $S_{k}(B) = S_{n-j}(B) + (a_1 + ... + a_{k-(n-j)})$. By property (P1), if $k-(n-j) < j$, we have $\sum_{i=1}^{k-(n-j)} a_i \leq k-(n-j)-1 \Rightarrow S_k(B) \leq (n-j-1) + (k-(n-j)-1)$

$\Rightarrow S_k(b) \leq k-2 < k-1$, and if $k-(n-j) =j$, i.e. $k=n$ (the maximum possible value of $k$), then $S_k(B) = S_n(B) = n-1$. Hence $S_k(B) \leq k-1 \, \forall k \in \{n-j, n-j+1, ..., n\}$. Thus we have $K' = (n-j) < K$ such that $S_k(B) \leq k-1 \, \forall k \in \{K', K'+1, ..., n\}$, thus proving the claim.

Thus, as long as there exists some a "violating" offset $j \geq 1$, i.e. one for which $S_j(A) > j-1$ for some sequence $A$ of the necklace with an upper bound $K-1$ for $j$, we can rotate it to obtain another sequence $A$ such that if it at all it too has a "violating" offset $j'$, it will have a smaller upper bound $K'-1 < K-1$. Thus by rotating the necklace we can keep reducing the "violating" offset till we reach a sequence where no violating offset is present (it is reduced to 0), i.e. this new sequence meets the condition C_sum. Say this sequence is $X=(x_1, x_2, ..., x_n) = (a_{r+1}, ..., a_n, a_1, a_2, ..., a_r)$, i.e. obtained by rotating the starting sequence $A$ by $r$ beads. Then cutting the necklace between $a_r$ and $a_{r+1}$ gives us the desired sequence.

 

[julian]

An idea of a proof of the Bohr-Mollerup theorem:

Spoiler

Let

\begin{align*}
\Gamma_n (x) = \frac{n! n^x}{x (x+1) \cdots (x + n)}
\end{align*}

\begin{align*}
\log \Gamma_n (x) = \log (n!) + x \log n - \sum_{m=0}^n \log (x + m)
\end{align*}

We have

\begin{align*}
\frac{d^2}{dx^2} \log \Gamma_n (x) = \sum_{m=0}^n \frac{1}{(x + m)^2} \geq 0 .
\end{align*}

meaning that it is a convex function. As the limit of a sequence of convex functions is convex we have that the log of $\Gamma (x)$ is convex. $\Gamma (0) = 1$. Also,

\begin{align*}
\Gamma (x + 1) = \lim_{n \rightarrow \infty} \frac{n! n^x n}{(x + 1) (x+2) \cdots (x + n + 1)} = \lim_{n \rightarrow \infty} \frac{n! n^x}{x (x+1) \cdots (x + n)} \cdot x \frac{n}{x + n + 1} = x \cdot \Gamma (x)
\end{align*}

So $\Gamma (x)$ satisfies the required conditions specified in the problem.

Note that

\begin{align*}
\lim_{x \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x) = 0 .
\end{align*}

To attempt to prove uniqueness consider

\begin{align*}
d (x) = \log G(x) - \log \Gamma (x)
\end{align*}

or

\begin{align*}
e^{d (x)} = G (x) / \Gamma (x)
\end{align*}

where $G (x)$ has the same properties as listed in the problem, which we just proved for $\Gamma (x)$.

We easily have that $d(0) = 0$. Also $d (x+1) = d (x)$ and so $d (x)$ is periodic.

Write

\begin{align*}
\frac{d^2}{dx^2} d (x) = \frac{d^2}{dx^2} \log G(x) - \frac{d^2}{dx^2} \log \Gamma (x) .
\end{align*}

Suppose that $d (x)$ is not constant. If we can then show that there exists an $x_0$ such that

\begin{align*}
\frac{d^2}{dx^2} d (x_0) < 0 ,
\end{align*}

then, using the periodicity of $d (x)$ we would have

\begin{align*}
\frac{d^2}{dx^2} d (x_0) & = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} d (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G (x_0 + n) - \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G(x + n) \geq 0
\end{align*}

producing a contradiction. We would then have that $d (x)$ is constant. Since $d(0) = 0$, $d (x) =0$ for all $x$. Implying $G (x) = \Gamma (x)$.

Say we always had that

\begin{align*}
\frac{d^2}{dx^2} d (x) \geq 0 ,
\end{align*}

then this would rule out periodic functions that are convex in places and concave in other places. But there are most likely going to be counter examples to these functions. I think I can think of some already.

I am very tired now. Tomorrow.

[SPLOILER]

 

[MathematicalPhysicist]

Spoiler: Problem 1

$$f(x)=f(x+1)/x=f(x+2)/(x(x+1))=\ldots = f(x+n)/(x(x+1)\ldots (x+n))$$

Spoiler: #1

#1) "Um, I'll take what is the Bohr-Mollerup Theorem for 200 points, Alex."

I want you honest opinions here, what do you think of my new signature? Simple yet amazing truth but is it too long?

Spoiler

I've seen a proof of this theorem in Askey's Special Functions red book.
And I am perplexed, why can you assume that $n$ is a positive integer and $0<x<1$?
It's on page: 35.

 

[julian]

I was hoping @fresh_42 would point out what periodic functions I'm not considering and maybe give a suggestion as to complete the proof I have started.

 

[fresh_42]

Spoiler: Problem 1

$$f(x)=f(x+1)/x=f(x+2)/(x(x+1))=\ldots = f(x+n)/(x(x+1)\ldots (x+n))$$

Spoiler

I've seen a proof of this theorem in Askey's Special Functions red book.
And I am perplexed, why can you assume that $n$ is a positive integer and $0<x<1$?
It's on page: 35.

$n>0$ can be assumed since we consider a limit with $n \to \infty$.
You first show $f(x+n)=(x+n-1)(x-n-2)\cdots (x+1)xf(x).$
Note that $f$ is defined by properties. And these properties are all we are allowed to use. Using too many $\Gamma 's$ makes the proof unreadable!

Next, we have to show with the help of this formula, that for any $y=x+N\in (N,N+1]$ holds $f(y)=\Gamma(y)$ where a) remember that $\Gamma(y)$ is defined as a limit here (see problem statement, and not as the gamma function) and which is b) the first essential part of the proof.

Then we can take log's and show (remember that $\Gamma$ is a limit): $L_n\leq f(x) \leq R_n$ with $L_n,R_n\to \Gamma(x)$.

This is at least the proof structure of my solution. I am skeptical if $\Gamma 's$ are used to make it misty because everybody has the faculty in mind.

The theorem says:
Gauß's definition of the gamma function as a limit is a necessary consequence of the functional requirements.

It's necessity which is the crucial point, not that it suffices.

 

[fresh_42]

I was hoping @fresh_42 would point out what periodic functions I'm not considering and maybe give a suggestion as to complete the proof I have started.

Well, I tried to provide hints in my previous post. I would be happy if you used less $G's$ and $\Gamma 's$ in your prove. This only creates confusion.