Challenge Math Challenge - September 2021

[fresh_42]

Yes, in most of the books I have see $U,V$ for open sets. I am less familiar with the notation for closed sets.

My suspicion is that it has historical reasons. Topology was first developed and researched by Russian and German mathematicians. One could even say that Cantor and Dedekind kind of started the field:
https://www.physicsforums.com/threa...for-the-particular-names.962018/#post-6102165

A neighborhood is usually open, and Umgebung is the German word for neighborhood. Closed is abgeschlossen in German. I guess the abbreviations $U$ and $A$ came in handy for a quick distinction.

 

[jbergman]

My suspicion is that it has historical reasons. Topology was first developed and researched by Russian and German mathematicians. One could even say that Cantor and Dedekind kind of started the field:
https://www.physicsforums.com/threa...for-the-particular-names.962018/#post-6102165

A neighborhood is usually open, and Umgebung is the German word for neighborhood. Closed is abgeschlossen in German. I guess the abbreviations $U$ and $A$ came in handy for a quick distinction.

What is the meaning of $\trianglelefteq$ in 5.4?

 

[fresh_42]

What is the meaning of $\trianglelefteq$ in 5.4?

$K\trianglelefteq G$ means that $K$ is a normal subgroup of $G,$ i.e. $gKg^{-1}\subseteq K$ for all $g\in G.$ Normality is crucial, because it makes
$$
\varphi \, : \,G/K \times G/K \longrightarrow G/K\, , \,\varphi (gK,hK)=gh^{-1}K
$$
a well-defined function.

Hint: You could start with an open set $\overline{V} \subseteq G/K$.

 

[jbergman]

$H$ is obviously a topological group because restriction of a continuous map to a subset with the subspace topology is continuous.

Let $m$ and $m'$ represent the multiplication maps on $G$ and $G/K$ respectively.

Then,

$$\pi(m(g, h))=m'(\pi(g),\pi(h))$$

Sine the left is the composition of continuous maps, the right is continuous. $\pi$ is surjective onto $G/K$. Therefore $m'$ must be continuous, because if $m'^{-1}(U)$ mapped into a closed set in $G/K\times G/K$, $\pi^{-1}\times\pi^{-1}$ would also map to a closed subset since $G/K$ has the quotient topology.

Still thinking about how to show that $\pi$ is open. It must have something to do with the interplay of $H$. But, I'm not quite clear on how subgroups and a normal group interact.

 

[Not anonymous]

11. 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 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.

Spoiler: Question 11

Proof by contradiction: Suppose neither of $T, U$ is closed under multiplication. Then there must exist $t_1, t_2 \in T$ (not necessarily distinct) such that $x \equiv t_1 t_2 \notin T \Rightarrow x \in U$ (since $T \cup U = S$ and $x \in S$) and similarly there must exist $u_1, u_2 \in U$ such that $y \equiv u_1 u_2 \notin U \Rightarrow y \in T$. Now consider the product $xy$.

Since $xy = t_1 t_2 y$ and $t_1, t_2, y \in T$ and it is given that product of any three elements of $T$ is in $T$, we must have $xy \in T$. (1)

But the product $xy$ is also expressible as $xy = x u_1 u_2$. Since $x, u_1, u_2 \in U$ and it is given that product of any three elements of $U$ is in $U$, we must have $xy \in U$. (2)

Conditions (1) and (2) are contradictory since $T$ and $U$ are disjoint subsets and hence $xy$ cannot belong to both the sets. The contradiction arises due to the assumption neither $T$ nor $U$ is closed under multiplication. Hence it must be the case that at least one of $T, U$ is closed under multiplication.

Addendum: While at least one of $T, U$ will be closed under multiplication given the conditions stated in the question, it is not always necessary that both will be closed under multiplication. For example, if $S$ is the set of all integers, $T$ is the set of all negative integers and $U$ is the set of all non-negative integers, then both $T, U$ meet the condition that the product of any three elements from the set is also in the same set, but only $U$ and not $T$ is closed under multiplication.

 

[fresh_42]

$H$ is obviously a topological group because restriction of a continuous map to a subset with the subspace topology is continuous.

Let $m$ and $m'$ represent the multiplication maps on $G$ and $G/K$ respectively.

Then,

$$\pi(m(g, h))=m'(\pi(g),\pi(h))$$

Sine the left is the composition of continuous maps,

Why is $\pi$ continuous? It is an algebraic function, not a topological.

the right is continuous. $\pi$ is surjective onto $G/K$. Therefore $m'$ must be continuous, because if $m'^{-1}(U)$ mapped into a closed set in $G/K\times G/K$, $\pi^{-1}\times\pi^{-1}$ would also map to a closed subset since $G/K$ has the quotient topology.

What is $U$? And closed is not the opposite of open!

Still thinking about how to show that $\pi$ is open. It must have something to do with the interplay of $H$. But, I'm not quite clear on how subgroups and a normal group interact.

 

[jbergman]

Why is $\pi$ continuous? It is an algebraic function, not a topological.

A quotient topology requires a surjective map from $f: X \mapsto Y$. Then the open sets $U \in Y$ are the sets such that $f^{-1}(U)$ is open in $X$. In our case $\pi$ is the map used to define the quotient topology on $G/K$.

What is $U$? And closed is not the opposite of open!

$U$ is in an open set in $G/K$.

 

[fresh_42]

A quotient topology requires a surjective map from $f: X \mapsto Y$. Then the open sets $U \in Y$ are the sets such that $f^{-1}(U)$ is open in $X$. In our case $\pi$ is the map used to define the quotient topology on $G/K$.

$U$ is in an open set in $G/K$.

Ok, right so far. Now simply "calculate" what $\pi (U)$ is and use the fact that arbitrary unions of open sets are open. You do not need something closed.

 

[jbergman]

Ok, right so far. Now simply "calculate" what $\pi (U)$ is and use the fact that arbitrary unions of open sets are open. You do not need something closed.

Can't I just replace closed with not open in my argument?

To restate, my argument is the following.

We know that
$m'(\pi(g), \pi(h))$ is continuous from my argument above, where $m' : G/K \times G/K \mapsto G/K$

For an open set $U \in G/K$, $m'^{-1}(U)$ maps to a set $X \times Y \in G/K \times G/K$.

If $X$ or $Y$ is not open, then $\pi^{-1}(X)$ or $\pi^{-1}(Y)$ because of the definition of the quotient topology. But we know that, $\pi^{-1}(X) \times \pi^{-1}(Y)$ is open since $m'(\pi(g), \pi(h))$ is continuous. Hence, $X$ and $Y$ must have been open.

I know it's stated a bit clumsy, but I am not versed in the working with products.

 

[fresh_42]

Can't I just replace closed with not open in my argument?

To restate, my argument is the following.

We know that
$m'(\pi(g), \pi(h))$ is continuous from my argument above, where $m' : G/K \times G/K \mapsto G/K$

For an open set $U \in G/K$, $m'^{-1}(U)$ maps to a set $X \times Y \in G/K \times G/K$.

If $X$ or $Y$ is not open, then $\pi^{-1}(X)$ or $\pi^{-1}(Y)$ because of the definition of the quotient topology. But we know that, $\pi^{-1}(X) \times \pi^{-1}(Y)$ is open since $m'(\pi(g), \pi(h))$ is continuous. Hence, $X$ and $Y$ must have been open.

I know it's stated a bit clumsy, but I am not versed in the working with products.

You were done when you showed that $m'$ is continuous. This means that $m'^{-1}(U)$ is open. It is the definition of a continuous function.

To show that $\pi$ is open, take an open set $V$ and calculate $\pi(V)=VK=\ldots .$

We have defined a topological group by continuity of $(g,h)\mapsto gh^{-1}$. So you have to show that $\varphi$ as defined in post #63 is continuous.

 

[jbergman]

You were done when you showed that $m'$ is continuous. This means that $m'^{-1}(U)$ is open. It is the definition of a continuous function.

To show that $\pi$ is open, take an open set $V$ and calculate $\pi(V)=VK=\ldots .$

Ahh, I finally get it!
$$\pi^{-1}(\pi(V)) = \pi^{-1}(VK) \bigcup_{k_i \in K} k_iV$$

Which is just the union of open sets, again using left multplication is a homeomorphism.

We have defined a topological group by continuity of $(g,h)\mapsto gh^{-1}$. So you have to show that $\varphi$ as defined in post #63 is continuous.

Well, my argument to show that $m'$ is continuous is essentially the same for the inverse map $i'$ and then from 5.1 we know that is equivalent to showing that $\phi$ is continuous.

Briefly,

$\pi(i(g)) = i'(\pi(g))$ and again the left is continuous so the right side must be, and since $\pi$ is surjective and continous, $i'$ must also be.

 

[fresh_42]

Ahh, I finally get it!
$$\pi^{-1}(\pi(V)) = \pi^{-1}(VK) \bigcup_{k_i \in K} k_iV$$

Which is just the union of open sets, again using left multplication is a homeomorphism.

Well, my argument to show that $m'$ is continuous is essentially the same for the inverse map $i'$ and then from 5.1 we know that is equivalent to showing that $\phi$ is continuous.

Briefly,

$\pi(i(g)) = i'(\pi(g))$ and again the left is continuous so the right side must be, and since $\pi$ is surjective and continous, $i'$ must also be.

Modulo some typos, yes.

$\pi(V)=VK=\displaystyle{\bigcup_{k\in K}Vk}=\pi(V)\subseteq G/K$ and the $Vk$ are open.

Here is the version without using part (a):

Let $\overline{V}\subseteq G/K$ be an open set, and $(gK,hK)\in \varphi^{-1}(\overline{V}).$ Since $(g,h)\longmapsto \pi(gh^{-1})=gh^{-1}K$ is continuous, there are open neighborhoods $V_g,V_h\subseteq G$ of $g,h$ such that $V_gV_h^{-1}\subseteq \pi^{-1}(\overline{V}).$ Since $\pi$ is open, $\pi(V_g)\times \pi(V_h)=V_gK\times V_hK\subseteq G/K\times G/K$ is an open neighborhood of $(gK,hK)\in G/K\times G/K.$ This proves that $\varphi$ is continuous since $V_gK\times V_hK\subseteq \varphi^{-1}(\overline{V}).$

 

[Not anonymous]

13. Let $d := d_1d_2...d_9$ be a number with not necessarily distinct nine decimal digits. A number $e := e_1e_2...e_9$ 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 $f := f_1f_2...f_9$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 $7$.

Spoiler: Question 13

Let $D_{j}$ denote the number obtained by replacing $d_j$ by $e_j$ in $d$ for all $1 \leq j \leq 9$. So, for example, $D_1$ will have the decimal representation $e_1 d_2...d_9$. Similarly let $E_{j}$ denote the number obtained by replacing $e_j$ by $f_j$ in $e$ for all $1 \leq j \leq 9$.

The numeric values of $D_j, E_j$ are given by:
$D_j = (\sum_{i=1}^9 10^{9-i}d_i) - 10^{9-j}d_j + 10^{9-j}e_j = d - 10^{9-j}d_j + 10^{9-j}e_j$
$E_j = (\sum_{i=1}^9 10^{9-i}e_i) - 10^{9-j}e_j + 10^{9-j}f_j = e - 10^{9-j}e_j + 10^{9-j}f_j$
for all $1 \leq j \leq 9$.

Since $D_j \equiv 0 \, \text{mod 7}$, we get $(d - 10^{9-j}d_j + 10^{9-j}e_j)\, \text{mod 7} = 0 \Rightarrow$

$10^{9-j}e_j \, \text{mod 7} \equiv 10^{9-j}d_j\, \text{mod 7} - d \, \text{mod 7}$ (Eq. 1)

Similarly, since $E_j$ is divisible for 7 for $j = 1, 2, ...,9$, we have $E_j \equiv 0 \, \text{mod 7}$ and therefore $(e - 10^{9-j}e_j + 10^{9-j}f_j)\, \text{mod 7} = 0 \Rightarrow$

$10^{9-j}f_j \, \text{mod 7} \equiv 10^{9-j}e_j\, \text{mod 7} - e \, \text{mod 7}$ (Eq. 2)

Substituting based on (Eq.1) in (Eq. 2) gives:
$10^{9-j}f_j \, \text{mod 7} \equiv 10^{9-j}d_j\, \text{mod 7} - d \, \text{mod 7} - e \, \text{mod 7} \Rightarrow$
$10^{9-j}(d_j - f_j) \, \text{mod 7} \equiv d \, \text{mod 7} + e \, \text{mod 7}$ (Eq. 3)

Now $e = \sum_{i=1}^{9} 10^{9-i}e_i$. Using (Eq. 1), we get $$
e \, \text{mod 7} \equiv \sum_{i=1}^{9} (10^{9-i}d_i \, \text{mod 7} - d \, \text{mod 7}) \equiv (\sum_{i=1}^{9} (10^{9-i}d_i \, \text{mod 7}) - 9d \, \text{mod 7} \equiv -8d \, \text{mod 7}$$ (Eq. 4)

Applying (Eq. 4) in (Eq. 3) gives $10^{9-j}(d_j - f_j) \, \text{mod 7} \equiv d \, \text{mod 7} - 8d \, \text{mod 7} \equiv -7d \, \text{mod 7} = 0 \, \text{mod 7}$. Since $10^{9-j} \, \text{mod 7} \neq 0$ for any integer valued $j$, it follows that $(d_j - f_j) \equiv 0 \, \text{mod 7}$, i.e. Hence proved that $d_j - f_j$ must be divisible by 7 for all allowed values of $j$.

 

[fresh_42]

Spoiler: Question 13

Let $D_{j}$ denote the number obtained by replacing $d_j$ by $e_j$ in $d$ for all $1 \leq j \leq 9$. So, for example, $D_1$ will have the decimal representation $e_1 d_2...d_9$. Similarly let $E_{j}$ denote the number obtained by replacing $e_j$ by $f_j$ in $e$ for all $1 \leq j \leq 9$.

The numeric values of $D_j, E_j$ are given by:
$D_j = (\sum_{i=1}^9 10^{9-i}d_i) - 10^{9-j}d_j + 10^{9-j}e_j = d - 10^{9-j}d_j + 10^{9-j}e_j$
$E_j = (\sum_{i=1}^9 10^{9-i}e_i) - 10^{9-j}e_j + 10^{9-j}f_j = e - 10^{9-j}e_j + 10^{9-j}f_j$
for all $1 \leq j \leq 9$.

Since $D_j \equiv 0 \, \text{mod 7}$, we get $(d - 10^{9-j}d_j + 10^{9-j}e_j)\, \text{mod 7} = 0 \Rightarrow$

$10^{9-j}e_j \, \text{mod 7} \equiv 10^{9-j}d_j\, \text{mod 7} - d \, \text{mod 7}$ (Eq. 1)

Similarly, since $E_j$ is divisible for 7 for $j = 1, 2, ...,9$, we have $E_j \equiv 0 \, \text{mod 7}$ and therefore $(e - 10^{9-j}e_j + 10^{9-j}f_j)\, \text{mod 7} = 0 \Rightarrow$

$10^{9-j}f_j \, \text{mod 7} \equiv 10^{9-j}e_j\, \text{mod 7} - e \, \text{mod 7}$ (Eq. 2)

Substituting based on (Eq.1) in (Eq. 2) gives:
$10^{9-j}f_j \, \text{mod 7} \equiv 10^{9-j}d_j\, \text{mod 7} - d \, \text{mod 7} - e \, \text{mod 7} \Rightarrow$
$10^{9-j}(d_j - f_j) \, \text{mod 7} \equiv d \, \text{mod 7} + e \, \text{mod 7}$ (Eq. 3)

Now $e = \sum_{i=1}^{9} 10^{9-i}e_i$. Using (Eq. 1), we get $$
e \, \text{mod 7} \equiv \sum_{i=1}^{9} (10^{9-i}d_i \, \text{mod 7} - d \, \text{mod 7}) \equiv (\sum_{i=1}^{9} (10^{9-i}d_i \, \text{mod 7}) - 9d \, \text{mod 7} \equiv -8d \, \text{mod 7}$$ (Eq. 4)

Applying (Eq. 4) in (Eq. 3) gives $10^{9-j}(d_j - f_j) \, \text{mod 7} \equiv d \, \text{mod 7} - 8d \, \text{mod 7} \equiv -7d \, \text{mod 7} = 0 \, \text{mod 7}$. Since $10^{9-j} \, \text{mod 7} \neq 0$ for any integer valued $j$, it follows that $(d_j - f_j) \equiv 0 \, \text{mod 7}$, i.e. Hence proved that $d_j - f_j$ must be divisible by 7 for all allowed values of $j$.

Well, one can write it a bit shorter, but the ideas are the same.

We are given that for all $1\leq j\leq 9$
$$
(e_j-d_j)10^{9-j}+d \equiv 0\equiv (f_j-e_j)10^{9-j}+e \mod 7\quad (*)
$$
Thus $\sum_{j=1}^9(e_j-d_j)10^{9-j}+d=e-d+9d\equiv e+d\equiv 0\mod 7.$ Now add the first and second relation from $(*)$ for any particular value $j$ and get
$$
0 \equiv (f_j-d_j)10^{9-j}+e+d \equiv (f_j-d_j)10^{9-j} \mod 7
$$
Because $7$ is prime and $7\,\nmid\,10^{9-j}$ this implies $7\,|\,(d_j-f_j).$

 

[jbergman]

Modulo some typos, yes.

$\pi(V)=VK=\displaystyle{\bigcup_{k\in K}Vk}=\pi(V)\subseteq G/K$ and the $Vk$ are open.

Here is the version without using part (a):

Let $\overline{V}\subseteq G/K$ be an open set, and $(gK,hK)\in \varphi^{-1}(\overline{V}).$ Since $(g,h)\longmapsto \pi(gh^{-1})=gh^{-1}K$ is continuous, there are open neighborhoods $V_g,V_h\subseteq G$ of $g,h$ such that $V_gV_h^{-1}\subseteq \pi^{-1}(\overline{V}).$ Since $\pi$ is open, $\pi(V_g)\times \pi(V_h)=V_gK\times V_hK\subseteq G/K\times G/K$ is an open neighborhood of $(gK,hK)\in G/K\times G/K.$ This proves that $\varphi$ is continuous since $V_gK\times V_hK\subseteq \varphi^{-1}(\overline{V}).$

One surprising fact to me is the fact that $\pi$ is open implies that if there is an open set $U \in gK$, then the entire coset is open, since $\pi^{-1}(\pi(U)) = gK$. This must be related to 5.5.

 

[fresh_42]

One surprising fact to me is the fact that $\pi$ is open implies that if there is an open set $U \in gK$, then the entire coset is open, since $\pi^{-1}(\pi(U)) = gK$. This must be related to 5.5.

$\pi^{-1}(\pi(U)) =U \neq_{i.g} gK$

$\pi$ is open means that it maps open sets to open sets, so there are open neighborhoods around the images of those points.

 

[Not anonymous]

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?

Partial or incomplete answer, but I am posting this because I think the approach is not too far off from the correct, complete answer and so would like to know if and what mistake, if any, exists in my attempted solution. Since I don't get simple, closed-form expressions relating $a, b$ and $c$, I suspect there might be a mistake in at least one of the intermediate equations, but I am unable to find where the mistake arises.

Spoiler: Question 14

The ellipse's circumference must be equal to the length one period of the mentioned curve (a sinusoidal wave) if the ellipse completes exactly one revolution when rolls over one period of the wave.

The circumference of the ellipse, assuming $a$ is the semi-major axis and $b$ is the the semi-minor axis, is $$C = 4a \int_{0}^{\pi / 2} \sqrt {1 - (1 - b^2/a^2) \sin^2{θ}} \, dθ$$

The length of a curve $y = f(x)$ between $x=x_1$ and $x=x_2$, assuming that $f(x)$ is differentiable in the range $[x_1, x_2]$, is given by $\int_{x_1}^{x_2} \sqrt {(1 + (f'(x))^2} \, dx$. For $y = c \sin(x/a)$, $y' = f'(x) = \dfrac {c} {a} \cos (x/a)$ and therefore the length $L$ of one period of the curve is given by $$
L = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {1 + \dfrac {c^2} {a^2} \cos^2 (x/a)} \, dx = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {1 + \dfrac {c^2} {a^2} (1 - \sin^2 (x/a))} \, dx = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {\dfrac{a^2+c^2}{a^2}} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 (x/a)} \, dx
$$

Due to the symmetry of sine wave's shape, the length of the curve between $\frac{x}{a} = 0$ and $\frac{x}{a} = 2π$ equals 4 times the length of the curve between $\frac{x}{a} = 0$ and $\frac{x}{a} = π/2$. Using this property, and defining $z = x/a$ and substituting for $x, dx$ accordingly in the above equation for $L$, we get $$
L = 4 \sqrt {a^2+c^2} \int_{z = 0}^{π/2} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 z} \, dz
$$

Since $L = C$, we get $$
\dfrac {4 \sqrt {a^2+c^2}} {4a} = \dfrac {\int_{0}^{\pi / 2} \sqrt {1 - (1 - b^2/a^2) \sin^2{θ}} \, dθ} {\int_{0}^{π/2} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 z} \, dz}
$$

When I first attempted to solve this, I had wrongly used $\pi$ instead of $2\pi$ as the upper bound for one of the integrals and that led to neat and simple relations between $a, b, c$ and I believed that a similar simple relations would be found even after correcting that error, but that didn't happen. I thought I could try equating the constants outside the integrals in expressions for $L$ and $C$ and similarly equate the multipliers of the $\sin^2(.)$ expression inside the 2 integrands to get at least one of the possibly many valid relations between the $a, b, c$, but equating $4\sqrt {a^2+c^2}$ to $4a$ leads to $c=0$ and that cannot be a valid solution. Where is the mistake?

Thanks in advance!

 

[jbergman]

$\pi^{-1}(\pi(U)) =U \neq_{i.g} gK$

$\pi$ is open means that it maps open sets to open sets, so there are open neighborhoods around the images of those points.

This doesn't make sense to me...

If $U \subset \bigcup_{k_i \in K} gk_i$ is open in $G$, then
$\pi(U) = gK$ which must be open in $G/K$.

Therefore, $\pi^{-1}(gK) = \bigcup_{k_i \in K} gk_i$, which must be open in $G$.

Where is the above logic incorrect?

 

[fresh_42]

Partial or incomplete answer, but I am posting this because I think the approach is not too far off from the correct, complete answer and so would like to know if and what mistake, if any, exists in my attempted solution. Since I don't get simple, closed-form expressions relating $a, b$ and $c$, I suspect there might be a mistake in at least one of the intermediate equations, but I am unable to find where the mistake arises.

Spoiler: Question 14

The ellipse's circumference must be equal to the length one period of the mentioned curve (a sinusoidal wave) if the ellipse completes exactly one revolution when rolls over one period of the wave.

The circumference of the ellipse, assuming $a$ is the semi-major axis and $b$ is the the semi-minor axis, is $$C = 4a \int_{0}^{\pi / 2} \sqrt {1 - (1 - b^2/a^2) \sin^2{θ}} \, dθ$$

The length of a curve $y = f(x)$ between $x=x_1$ and $x=x_2$, assuming that $f(x)$ is differentiable in the range $[x_1, x_2]$, is given by $\int_{x_1}^{x_2} \sqrt {(1 + (f'(x))^2} \, dx$. For $y = c \sin(x/a)$, $y' = f'(x) = \dfrac {c} {a} \cos (x/a)$ and therefore the length $L$ of one period of the curve is given by $$
L = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {1 + \dfrac {c^2} {a^2} \cos^2 (x/a)} \, dx = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {1 + \dfrac {c^2} {a^2} (1 - \sin^2 (x/a))} \, dx = \int_{\frac{x}{a} = 0}^{2 \pi} \sqrt {\dfrac{a^2+c^2}{a^2}} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 (x/a)} \, dx
$$

Due to the symmetry of sine wave's shape, the length of the curve between $\frac{x}{a} = 0$ and $\frac{x}{a} = 2π$ equals 4 times the length of the curve between $\frac{x}{a} = 0$ and $\frac{x}{a} = π/2$. Using this property, and defining $z = x/a$ and substituting for $x, dx$ accordingly in the above equation for $L$, we get $$
L = 4 \sqrt {a^2+c^2} \int_{z = 0}^{π/2} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 z} \, dz
$$

Since $L = C$, we get $$
\dfrac {4 \sqrt {a^2+c^2}} {4a} = \dfrac {\int_{0}^{\pi / 2} \sqrt {1 - (1 - b^2/a^2) \sin^2{θ}} \, dθ} {\int_{0}^{π/2} \sqrt {1 - \dfrac {c^2} {a^2 + c^2} \sin^2 z} \, dz}
$$

When I first attempted to solve this, I had wrongly used $\pi$ instead of $2\pi$ as the upper bound for one of the integrals and that led to neat and simple relations between $a, b, c$ and I believed that a similar simple relations would be found even after correcting that error, but that didn't happen. I thought I could try equating the constants outside the integrals in expressions for $L$ and $C$ and similarly equate the multipliers of the $\sin^2(.)$ expression inside the 2 integrands to get at least one of the possibly many valid relations between the $a, b, c$, but equating $4\sqrt {a^2+c^2}$ to $4a$ leads to $c=0$ and that cannot be a valid solution. Where is the mistake?

Thanks in advance!

If I should check your calculations, then you will have to write way more lines of calculation.

The solution is actually very easy and short. Write the circumference of the ellipse as
$C=\int_{0}^{2\pi}\sqrt{(-a\sin \left(\theta\right))^2+(b\cos\left(\theta\right))^2}\,d\theta$ and keep the polar coordinates for $L$! Next simply bring those integrals in a comparable form for $C=L$ and conclude the solution.

 

[fresh_42]

This doesn't make sense to me...

If $U \subset \bigcup_{k_i \in K} gk_i$ is open in $G$, then
$\pi(U) = gK$ which must be open in $G/K$.

Therefore, $\pi^{-1}(gK) = \bigcup_{k_i \in K} gk_i$, which must be open in $G$.

Where is the above logic incorrect?

And the construction $gk_i$ in $G$ doesn't make any sense to me. It confuses the distinction between elements of $G$ and elements of $G/K$.

Why do you look at $U\subseteq \{g\cdot k\,|\,k\in K\}$ at all? Yes, you get $\pi(U)=gK$, so what?

$\pi^{-1}(gK)=\{g\}$ not $U$.

 

[jbergman]

And the construction $gk_i$ in $G$ doesn't make any sense to me. It confuses the distinction between elements of $G$ and elements of $G/K$.

I am not sure the confusion. A coset in $G$ consists of the set of elements, $gK = \bigcup g_i \exists k \in K s.t. g_i k = g$. In $G/K$ that set becomes a single element.

Why do you look at $U\subseteq \{g\cdot k\,|\,k\in K\}$ at all? Yes, you get $\pi(U)=gK$, so what?

$\pi^{-1}(gK)=\{g\}$ not $U$.

I know it isn't $U$. I never claimed it was. $U$ isn't a saturated open subset unless it is the entire coset. It also isn't a single element $g$ since all elements of the coset in $G$ map to the same element $gK$.

Either I have a giant misunderstanding or we are having some miscommunication issue here.

 

[fresh_42]

Either I have a giant misunderstanding or we are having some miscommunication issue here.

I guess misunderstanding.

One surprising fact to me is the fact that $\pi$ is open implies that if there is an open set $U \in gK$, then the entire coset is open, since $\pi^{-1}(\pi(U)) = gK$. This must be related to 5.5.

I did not understand $U\in gK$. This is a mixture of $G$ and $G/K$. If you say open, then you have to say open in what.

 

[jbergman]

I guess misunderstanding.

I did not understand $U\in gK$. This is a mixture of $G$ and $G/K$. If you say open, then you have to say open in what.

Sorry for the confusion. I am not sure how people typically notate these things.

 

[fresh_42]

Sorry for the confusion. I am not sure how people typically notate these things.

One convention is to write cosets with an overline: $gK=\overline{g}$. If you write $gK$ then it is unclear whether you mean $gK=\{gk\,|\,k\in K\}\subseteq G$ or $gK\in G/K$.

So $U\subseteq G$ and $\overline{U}\subseteq G/K$ is a possibility to distinguish $G$ from $G/K$. And, of course, we could also write everything in $G/K$ as images of $\pi\, : \,G \twoheadrightarrow G/K$ namely $gK=\overline{g}=\pi(g) \in G/K\, , \,\overline{U}=\pi(U)\subseteq G/K.$

So $gK \subseteq G$ is actually the misleading notation. It would be better to write $\pi^{-1}(\overline{g}).$ This has also the advantage, that it is immediately obvious that $gK\subseteq G$ is closed, as it is the pre-image of a closed set $(\overline{g}\in G/K$ is a singleton and therefore closed, at least in $T_1$ topologies) under a continuous function.

 

[julian]

Problem #7:

Spoiler

Write

\begin{align*}
f (x) = x - (1+x) \log (1+x) + \frac{3x^2}{2 (3+x)}
\end{align*}

We must show that $f (x) \leq 0$ for $x > -1$.

The first derivative is

\begin{align*}
\frac{d}{dx} f (x) = - \log (1+x) + \frac{3 x (x + 6)}{2(3+x)^2}
\end{align*}

This has a zero at $x=0$, with $f (0) = 0$.

The second derivative is

\begin{align*}
\frac{d^2}{dx^2} f (x) = - \frac{x^2 (x + 9)}{(1+x) (3+x)^3}
\end{align*}

which is zero at $x = 0$, so we may have a maximum, a minimum or an inflection.

As

\begin{align*}
\frac{d}{dx} f (x) = - \int_0^x \frac{y^2 (y + 9)}{(1+y) (3+y)^3} dy
\end{align*}

we have $\frac{d}{dx} f (x)$ is positive for all $-1 < x < 0$ and we have $\frac{d}{dx} f (x)$ is negative for all $x > 0$.

Which means the function $f (x)$ is monotonically increasing for $-1 < x < 0$ and monotonically decreasing for $x > 0$. So $f(x)$ has a global maximum of zero at $x = 0$ for $x > -1$.

ADDITONAL: The third derivative, $f''' (0)$, is zero but the fourth derivative, $f'''' (0)$, is negative, so by the higher derivative test the function $f(x)$ has a local maximum at $x = 0$.

 

[julian]

Problem #9 parts 5 and 6.

Spoiler

5.

We prove $\frac{2^{2n-1}}{n} \leq \binom{2n}{n}$ by induction. The base case is trivial $\frac{2^1}{1} \leq \binom{2}{1}$.

We assume

\begin{align*}
\frac{2^{2n-1}}{n} \leq \binom{2n}{n} .
\end{align*}

This together with $4n \leq 2(2n+1)$ implies

\begin{align*}
4n \cdot \frac{2^{2n-1}}{n} \leq 2 (2n+1) \binom{2n}{n}
\end{align*}

or

\begin{align*}
\frac{1}{n+1} \cdot 2^{2n+1} \leq \frac{(2n + 2) (2n+1)}{(n+1)^2} \binom{2n}{n}
\end{align*}

or

\begin{align*}
\frac{2^{2(n+1)-1}}{n + 1} \leq \binom{2(n+1)}{n+1}
\end{align*}

completing the inductive argument.

We prove $\binom{2n}{n} \leq 2^{2n-1}$ by induction. The base case is trivial $\binom{2}{1} \leq 2^1$.

We assume

\begin{align*}
\binom{2n}{n} \leq 2^{2n-1} .
\end{align*}

This together with $2n+1 \leq 2 (n+1)$ implies

\begin{align*}
2 \frac{2n+1}{n+1} \binom{2n}{n} \leq 4 \cdot 2^{2n-1}
\end{align*}

which easily becomes

\begin{align*}
\binom{2(n+1)}{n+1} \leq 2^{2(n+1)-1}
\end{align*}

completing the inductive argument.
6.

We prove $\prod_{p \leq n} p < 4^n$ by induction. The base cases $n = 2$ obviously works (I guess the for the case $n = 1$ you use the convention that $\prod_{p \leq 1} p = 0$ as 1 is not a prime).

Assume $\prod_{p \leq m} p < 4^m$ for all $m < n$, we wish to prove the result $\prod_{p \leq n} p < 4^n$.

This is easy if $n$ is not prime because then

\begin{align*}
\prod_{p \leq n} p = \prod_{p \leq n-1} p < 4^{n-1} < 4^n .
\end{align*}

We now consider the case where $n$ is prime. As we are considering primes greater than 2, we can write $n = 2k+1$.

Consider

\begin{align*}
\binom{2k+1}{k} = \frac{(2k+1) \cdots (q) \cdots (k+2)}{k!} = (2k+1) \cdots \frac{(q)}{k!} \cdots (k+2)
\end{align*}

where $q$ is a prime. As none of the factors in $k!$ can divide $q$ the $q$ cannot be cancelled. Hence $q$ divides $\binom{2k+1}{k}$. Hence, $\binom{2k+1}{k}$ is divisible by all prime numbers greater than or equal to $k + 2$ and less than or equal to $2k + 1$, that is, it is divisible by

\begin{align*}
Q = \prod_{k+2 \leq p \leq 2k+1} p = \dfrac{\prod_{p \leq 2k+1} p}{\prod_{p \leq k+1} p}
\end{align*}

So that $Q \leq \binom{2k+1}{k}$. But

\begin{align*}
\binom{2k+1}{k} & < \binom{2k+1}{0} + \binom{2k+1}{1} + \cdots + \binom{2k+1}{k - 1} + \binom{2k+1}{k}
\nonumber \\
& = \frac{1}{2} \left( \binom{2k+1}{0} + \binom{2k+1}{1} + \cdots + \binom{2k+1}{2k} + \binom{2k+1}{2k + 1} \right)
\nonumber \\
& = \frac{1}{2} (1+1)^{2k + 1} = 4^k .
\end{align*}

Hence $Q < 4^k$, or in other words

\begin{align*}
\dfrac{\prod_{p \leq 2k+1} p}{\prod_{p \leq k+1} p} < 4^k
\end{align*}

which by the inductive hypothesis implies

\begin{align*}
\prod_{p \leq 2k+1} p < 4^k \prod_{p \leq k+1} p < 4^{2k + 1} .
\end{align*}

Completing the inductive argument.

 

[fresh_42]

(I guess the for the case $n=1$ you use the convention that $\prod_{p\leq 1}p=0$ as $1$ is not a prime).

No. The usual convention for an empty product is, to set it equal to $1$. This is because $0$ isn't part of multiplicative groups, and $1$ is their neutral element. It is the same reason why empty sums are considered to be $0.$ And it preserves the important functional equation $\exp(0+0)=\exp(0)\cdot\exp(0).$

 

[julian]

Problem #9 part 4.

Spoiler

By definition of $\text{ord}_p (N)$ it is the exponent of $p$ in the prime number decomposition of the number $N!$. Using part 1 of the question we have that the exponent of $p$ in the prime number decomposition of $\binom{2n}{n}$, denote it $\nu_p (n)$, is given by

\begin{align*}
\nu_p (n) & = \text{ord}_p (2n) - 2 \text{ord}_p (n)
\nonumber \\
& = \left( \left\lfloor \frac{2n}{p} \right\rfloor - 2 \left\lfloor \frac{n}{p} \right\rfloor \right) + \left( \left\lfloor \frac{2n}{p^2} \right\rfloor - 2 \left\lfloor \frac{n}{p^2} \right\rfloor \right) + \cdots \quad (*)
\end{align*}

When $p^k >2n$ we have that $\left( \left\lfloor \frac{2n}{p^k} \right\rfloor - 2 \left\lfloor \frac{n}{p^k} \right\rfloor \right) = 0$. Otherwise each such term in $(*)$ is either 0 or 1 as $\left\lfloor 2a \right\rfloor - 2 \left\lfloor a \right\rfloor$ is either 0 or 1. Let $k_{max}$ be the largest integer such that $p^{k_{max}} \leq 2n$. Then $\nu_p (n) \leq k_{max}$.

So if $p^r | \binom{2n}{n}$ then

\begin{align*}
p^r \leq p^{\nu_p (n)} \leq p^{k_{max}} \leq 2n .
\end{align*}

 

[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?

Who's grading this month, @fresh_42 ? I think I should also get credit for #1) bc of the quoted post, the Bohr-Mollerup Theorem:

It is trivial from this theorem's definition of the Gamma function to obtain the form of it given in the problem (by existence + uniqueness of the Gamma function), but if you like I can furnish the leg work.

 

[MathematicalPhysicist]

Who's grading this month, @fresh_42 ? I think I should also get credit for #1) bc of the quoted post, the Bohr-Mollerup Theorem:

View attachment 289735

It is trivial from this theorem's definition of the Gamma function to obtain the form of it given in the problem (by existence + uniqueness of the Gamma function), but if you like I can furnish the leg work.

Is there a prize for the highest grades?
Otherwise I wouldn't care if I got graded or not.