# Challenge Math Challenge - November 2018

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#### StoneTemplePython

Gold Member
Hi StoneTemplePython:

Ah, another wrong assumption on my part. Since there was no specification about the hands moving abruptly second by second, rather than continuously, I assumed continuously.

I wonder if clocks moving discontinuously second by second were more common when the mathematician created this problem then analogue clocks are today. I confess I vaguely remember that when I was in elementary school in the 1940s, the classrooms had a clock something like that. I don't remember whether it moved incrementally second by second, or minute by minute.

I apologize for my denseness, but the quote above makes no sense to me in the context of the puzzle statement. Are you describing the puzzle clock as having three hands: hour, minute, second? If time is read by something other than the hour and minute hands, and "The minutes and hours hands don't move at all," then how is time read?

Regards,
Buzz
No man, Youngphysicst made an inaccurate statement about the links being in post 15 -- seemed to confuse the seconds and minutes hands. Did you look at them? I quoted one in particular in my response for a reason.

It seemed obvious to me that the minutes and hours hands in this problem must move yet they don't in the link from Youngphysicst -- another red flag. I was trying to clear up post 15, that's really all.

(I'll edit my post to make this extra clear)

#### PeroK

Homework Helper
Gold Member
2018 Award
23.
Example: Given a particle of mass $m$ in the potential $U(\vec{r})=\dfrac{U_0}{\vec{r\,}^{2}}$ with a constant $U_0$. At time $t=0$ the particle is at $\vec{r}_0$ with velocity $\dot{\vec{r}}_0\,.$

Hint: The Lagrange function with $\vec{r}=(x,y,z,t)=(x_1,x_2,x_3,t)$ of this problem is $$\mathcal{L}=T-U=\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}$$
a) Give a reason why the energy of the particle is conserved, and what is its energy?
b) Consider the following transformations with infinitesimal $\varepsilon$
$$\vec{r} \longmapsto \vec{r}\,^*=(1+\varepsilon)\,\vec{r}\,\, , \,\,t\longmapsto t^*=(1+\varepsilon)^2\,t$$
and verify the condition (*) to E. Noether's theorem.
c) Compute the corresponding Noether charge $Q$ and evaluate $Q$ for $t=0$.
I've just been learning about Noether's theorem, so I thought I'd try this one. The notation is a bit different from what I'm used to.

a) The Lagrangian is independent of time, so energy is conserved (Euler-Lagrange equation for time coordinate):

$\dot{H} = - \frac{\partial L}{\partial t} = 0$

Hence, the total energy $E$ is equal to the Hamiltonian:

$E = H = p_i \dot{x^i} - L = \frac{\partial L}{\partial \dot{x^i}}\dot{x^i} - L = m\dot{x^i}\dot{x^i} - L = \frac12 mv^2 + \frac{U_0}{r^2}$

Hence:

$E = \frac12 mv_0^2 + \frac{U_0}{r_0^2}$

b) We have the transformation:

$x^* = (1 + \epsilon)x; \ y^* = (1 + \epsilon)y; \ z^* = (1 + \epsilon)z; \ t^* = (1 + \epsilon)^2t$

Hence: $\psi^x = x; \ \psi^y = y; \ \psi^z = z; \ \phi = 2t$

The transformed Lagrangian is:

$L^* = \frac12 m(v^*)^2 - \frac{U_0}{(r^*)^2}$

Where:

$(v^*)^2 = (\frac{dx^*}{dt^*})^2 + \dots = (1+\epsilon)^{-2}(\frac{dx}{dt})^2 + \dots = (1+\epsilon)^{-2}v^2$

And:

$(r^*)^2 = (1+\epsilon)^{2}r^2$

Therefore, we have:

$L^* = L(1+\epsilon)^{-2}$

Hence:

$L^* \frac{dt^*}{dt} = L$

Which is the condition for invariance under the transformation.

c) The corresponding conserved Noether Charge is given by:

$Q = p_i\psi^i - H\phi = m(\dot{x}x + \dot{y}y + \dot{z}z) - 2Et = m\vec{v} \cdot \vec{r} - 2Et$

At $t=0$, we have $Q = m\vec{v_0} \cdot \vec{r_0}$, which leads to:

$m\frac{d}{dt}(r^2) = 2m\vec{v} \cdot \vec{r} = 2Q + 4Et$

And

$mr^2 = 2Et^2 + 2Qt + mr_0^2$

Which implies only unstable (circular) orbits.

• fresh_42

#### fresh_42

Mentor
2018 Award
I've just been learning about Noether's theorem, so I thought I'd try this one. The notation is a bit different from what I'm used to.

a) The Lagrangian is independent of time, so energy is conserved (Euler-Lagrange equation for time coordinate):

$\dot{H} = - \frac{\partial L}{\partial t} = 0$

Hence, the total energy $E$ is equal to the Hamiltonian:

$E = H = p_i \dot{x^i} - L = \frac{\partial L}{\partial \dot{x^i}}\dot{x^i} - L = m\dot{x^i}\dot{x^i} - L = \frac12 mv^2 + \frac{U_0}{r^2}$

Hence:

$E = \frac12 mv_0^2 + \frac{U_0}{r_0^2}$

b) We have the transformation:

$x^* = (1 + \epsilon)x; \ y^* = (1 + \epsilon)y; \ z^* = (1 + \epsilon)z; \ t^* = (1 + \epsilon)^2t$

Hence: $\psi^x = x; \ \psi^y = y; \ \psi^z = z; \ \phi = 2t$

The transformed Lagrangian is:

$L^* = \frac12 m(v^*)^2 - \frac{U_0}{(r^*)^2}$

Where:

$(v^*)^2 = (\frac{dx^*}{dt^*})^2 + \dots = (1+\epsilon)^{-2}(\frac{dx}{dt})^2 + \dots = (1+\epsilon)^{-2}v^2$

And:

$(r^*)^2 = (1+\epsilon)^{2}r^2$

Therefore, we have:

$L^* = L(1+\epsilon)^{-2}$

Hence:

$L^* \frac{dt^*}{dt} = L$

Which is the condition for invariance under the transformation.

c) The corresponding conserved Noether Charge is given by:

$Q = p_i\psi^i - H\phi = m(\dot{x}x + \dot{y}y + \dot{z}z) - 2Et = m\vec{v} \cdot \vec{r} - 2Et$

At $t=0$, we have $Q = m\vec{v_0} \cdot \vec{r_0}$, which leads to:

$m\frac{d}{dt}(r^2) = 2m\vec{v} \cdot \vec{r} = 2Q + 4Et$

And

$mr^2 = 2Et^2 + 2Qt + mr_0^2$

Which implies only unstable (circular) orbits.
Thank you for solving this! I already began to believe it would survive Emmy's centenary!
There is nothing to add to your solutions. Mine (taken from an exam for engineers!) is a bit closer in notation to the one given by the problem statement and the hints, so I will add it here - mostly because it took so long to solve this one, so people might choose what they find easier to grasp, so please don't interpret it as patronizing or so.

a)
(i) Energy is homogeneous in time, so we chose $\psi_i=0 , \varphi=1$
(ii) and check our equation by
\begin{equation*}
\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon = 0} \left(\mathcal{L}^*\,\cdot\,\dfrac{d}{dt}\,(t+\varepsilon )\right)=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon = 0} \left(\mathcal{L}^*\,\cdot\,1\right) = 0
\end{equation*}
since $\mathcal{L}^*$ doesn't depend on $t^*$ and thus not on $\varepsilon$, and calculate
(iii) the Noether charge as
\begin{align*}
Q(t,x,\dot{x})&=\mathcal{L}- \sum_{i=1}^N\dfrac{\partial \mathcal{L}}{\partial \dot{x}_i} \,\dot{x}_i\\
&=T-U-\dfrac{m}{2}\left( \dfrac{\partial}{\partial \dot{x}_i}\left( \sum_{i=1}^3 \dot{x}^2_i \right)\,\dot{x}_i \right)\\
&=\dfrac{m}{2}\, \dot{\vec{r\,}}^2 - U -m\,\dot{\vec{r\,}}^2\\
&=-T-U\\
&=-E\\
&=-\dfrac{m}{2}\, \dot{\vec{r\,}}^2- \dfrac{U}{\vec{r\,}^2}\\
&=-\dfrac{m}{2}\, \dot{\vec{r\,}}_0^2- \dfrac{U}{\vec{r\,}_0^2}
\end{align*}
by time invariance.

b) $\dot{\vec{r}}\,^*=\dfrac{d\vec{r}\,^*}{dt^*}=\dfrac{(1+\varepsilon)\,d\vec{r}}{(1+\varepsilon)^2\, dt }=\dfrac{1}{1+\varepsilon}\,\dot{\vec{r}}\,$ and thus $\,\mathcal{L}^*=\dfrac{1}{(1+\varepsilon)^2}\,\mathcal{L}\,$, i.e.
\begin{align*}
\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\left(\mathcal{L}\left(t^*,x^*,\dot{x}^*\right)\cdot \dfrac{dt^*}{dt} \right) &= \left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0} \mathcal{L}^*\,\dfrac{dt^*}{dt}\\ &=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0} \dfrac{\mathcal{L}}{(1+\varepsilon)^2}\cdot (1+\varepsilon)^2\\ &=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\mathcal{L} \\&= 0
\end{align*}
and the condition of Noether's theorem holds.

c) For the given transformations we have
\begin{align*}
x &\longmapsto x^* = (1+\varepsilon)x & \Longrightarrow \quad& \psi_x=x\\
y &\longmapsto y^* = (1+\varepsilon)y & \Longrightarrow \quad& \psi_y=y\\
z &\longmapsto z^* = (1+\varepsilon)z & \Longrightarrow \quad& \psi_z=z\\
t &\longmapsto t^* = (1+2\varepsilon)t & \Longrightarrow \quad& \varphi=2t\\
\end{align*}
and so the Noether charge is given by
\begin{align*}
Q(t,x,\dot{x})&= \sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\psi_i + \left(\mathcal{L}-\sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\dot{x}_i\right)\varphi\\
&=\sum_{i=1}^3 \dfrac{\partial}{\partial \dot{x}_i}\left(\dfrac{m}{2}\,\dot{\vec{r}\,}^2-\dfrac{U_0}{\vec{r\,}^{2}}\right)\,\psi_i \,+\\
&+ \left(\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}-\sum_{i=1}^3 \dfrac{\partial }{\partial \dot{x}_i}\,\left(\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}\right)\dot{x}_i\right)\varphi\\
&=m(\dot{x}x+\dot{y}y+\dot{z}z) \,+ \\
&+\left( \dfrac{m}{2}\dot{\vec{r\,}}^2-\dfrac{U_0}{\vec{r\,}^{2}}-m(\dot{x}^2+\dot{y}^2+\dot{z}^2)\right)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\,+\left( -\dfrac{m}{2}\dot{\vec{r\,}}^2-\dfrac{U_0}{\vec{r\,}^{2}} \right)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\, -(T+U)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\, -2Et\\
&\stackrel{t=0}{=} m\, \dot{\vec{r}}_0\,\vec{r}_0
\end{align*}
which shows that invariance under different transformations result in different conversation quantities.

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#### Delta2

Homework Helper
Gold Member
Is there some typo in the statement of 14.?

I don't seem to understand the meaning of the expression $I_A(f')^{-1}$. Is it meant to be $I_A\circ[(f')^{-1}]$??

#### Delta2

Homework Helper
Gold Member
@fresh_42 in post #53 in second line of b) at the end (after the $L^*$) did you omit a factor of $\frac{dt^*}{dt}$ or is my understanding wrong?

#### fresh_42

Mentor
2018 Award
@fresh_42 in post #53 in second line of b) at the end (after the $L^*$) did you omit a factor of $\frac{dt^*}{dt}$ or is my understanding wrong?
Your are right, I was a bit lost in the formula jungle. I corrected it now, and ... thanks for reading it!

• Delta2

#### Buzz Bloom

Gold Member
For this problem: the clock's hands have no jumps!
Hi StoneTemplePython:

I am still unclear about your meaning regarding the minute and hour hands. If these hands do not jump, then they must move continuously. However, you may mean something else: they each jump by one second intervals synchronized with the second hand jumps. That is, the minute hand jumps 1/10 of a degree every second, and the hour hand jumps 1/120 of a degree every second. Is this what you understand to be the nature of the clock in the puzzle statement?

My solution assumed that the minute and hour hands move continuously, and this apparently gave me the wrong answer. If the clock has the property of minute and hour hand jumps each second which I described above, I can then use that understanding to make another attempt to solve the puzzle.

Regards,
Buzz

#### StoneTemplePython

Gold Member
Hi StoneTemplePython:

I am still unclear about your meaning regarding the minute and hour hands. If these hands do not jump, then they must move continuously. However, you may mean something else: they each jump by one second intervals synchronized with the second hand jumps.
If you look through this thread, I've said multiple times now: no jumps! I meant it.

My solution assumed that the minute and hour hands move continuously, and this apparently gave me the wrong answer.
Continuity is fine. The issue is that there's a bug in your solution (no pun intended!)
- - - -
This problem has been out there for a few months now and oddly comes up when I'm about to travel somewhere. You may not hear back from me for a week or 2 but it'll keep.

#### QuantumQuest

Gold Member
Is there some typo in the statement of 14.?

I don't seem to understand the meaning of the expression $I_A(f')^{-1}$. Is it meant to be $I_A\circ[(f')^{-1}]$??
No, there is no typo. $I_A(f')^{-1}$ is a product.

• Delta2

#### Keith_McClary

10. c) Assuming the sum on the RHS means pointwise convergence.
Take $f_0 = 1$ on $[0,1]$ and $0$ elsewhere.
For n=1, ... , take $f_n = -1$ on $[n-1,n ]$, $f_n = 1$ on $[n,n+1 ]$ and $0$ elsewhere.
Then the LHS = $1$ and the RHS = $0$.

• fresh_42

#### fresh_42

Mentor
2018 Award
10. c) Assuming the sum on the RHS means pointwise convergence.
Take $f_0 = 1$ on $[0,1]$ and $0$ elsewhere.
For n=1, ... , take $f_n = -1$ on $[n-1,n ]$, $f_n = 1$ on $[n,n+1 ]$ and $0$ elsewhere.
Then the LHS = $1$ and the RHS = $0$.
That's correct, and I think it is called the vanishing bump or so, because what would equal the gap is shifted to infinity.

#### Keith_McClary

12. b)
$x^2+y > y$ and $x^2+y > x^2$ so
$$\int _0 ^1\frac{1}{x^2+y}dx < \int _0 ^{\sqrt y}\frac{1}{y}dx +\int_{\sqrt y} ^1\frac{1}{x^2}dx =\frac{1}{\sqrt y} -1 + \frac{1}{\sqrt y}$$ .
The integral over $y$ is finite.

#### fresh_42

Mentor
2018 Award
12. b)
$x^2+y > y$ and $x^2+y > x^2$ so
$$\int _0 ^1\frac{1}{x^2+y}dx < \int _0 ^{\sqrt y}\frac{1}{y}dx +\int_{\sqrt y} ^1\frac{1}{x^2}dx =\frac{1}{\sqrt y} -1 + \frac{1}{\sqrt y}$$ .
The integral over $y$ is finite.
But we are interested in $\int_0^1\int_0^1 \dfrac{1}{x^2+y}\,dy\,dx$.

#### Delta2

Homework Helper
Gold Member
ok thanks for the clarification @QuantumQuest and here is my attempt at 14.

This seems like a problem with a "scary statement" since it involves inverse of derivatives and derivatives of inverse functions but after working it out seems rather simple. Also I am not sure where the result for the second derivative is needed (at least as the hint says).

I ll use the function $g$ with $g(y)=(f')^{-1}(y)$ or simply put $g$ is the inverse of of $f'$ cause I believe this will make the symbols used less messy.

Using the product rule of differentiation (i differentiate with respect to y and not with x) we get that $(I_A(f')^{-1})'(y)=(I_A(y)g(y))'=(yg(y))'=g(y)+yg'(y)$ (1)

Using the chain rule of differentiation we get that $(f\circ g)'(y)=(f'\circ g)(y)g'(y)=f'(g(y))g'(y)$

But we can see that $f'(g(y))=f'([(f')^{-1}](y))=y$ hence $(f\circ g)'(y)=yg'(y)$ (2)

Subtracting (2) from (1) we get that
$((I_Ag)-(f\circ g))'(y)=g(y)$ which is simply what we wanted to prove.

• QuantumQuest

#### Keith_McClary

8.
Is "in same row or same column" the definition of "neighboring entries", or do you mean they must be adjacent?

#### fresh_42

Mentor
2018 Award
8.
Is "in same row or same column" the definition of "neighboring entries", or do you mean they must be adjacent?
Given $A = (a_{ij})\in \mathbb{M}_n(\mathbb{Z})$ with $\{\,a_{ij}\,\}=\{\,1,2,\ldots , n^2\,\}$ show that there exists a pair $(i,j)$ with $|a_{ij}-a_{i+1 j}|\geq n$ or for symmetry reasons there exists a pair $(i,j)$ with $|a_{ij}-a_{ij+1}|\geq n\,.$

#### batboio

Hello! I have a problem with problem 20...

If we have any quotient group, the equivalence class of the identity element should be a normal subgroup. But the equivalence class of $p$ (which is the identity element in $\mathcal{F}$) with respect to the relation $\sim_\mathcal{I}$ is the whole of $\mathcal{I}$ and it is not a normal subgroup of $\mathcal{F}$.

I will think about this more after I get some sleep and I am probably wrong, but I decided to post this anyway.

#### fresh_42

Mentor
2018 Award
I have difficulties to understand you.
Hello! I have a problem with problem 20...

If we have any quotient group
$G/H$
... the equivalence class of the identity element ...
$1\cdot H=H$
... should be a normal subgroup.
Yes. But only because you have written it the way you did.
But the equivalence class of $p$ (which is the identity element in $\mathcal{F}$) with respect to the relation $\sim_\mathcal{I}$ is the whole of $\mathcal{I}$ and it is not a normal subgroup of $\mathcal{F}$.
Now I can't follow you anymore. Do you mean $\mathcal{F}/\sim_\mathcal{I}$ or $\mathcal{F}/\{\,p\,\}\,.$ The latter isn't of interest, neither for the problem nor mathematically as $G/\{\,1\,\}=G$ for any group. So what do you mean by "the equivalence class of $p$"?
I will think about this more after I get some sleep and I am probably wrong, but I decided to post this anyway.

#### QuantumQuest

Gold Member
Well done @Delta2.

Also I am not sure where the result for the second derivative is needed (at least as the hint says).
The hints I gave pertain to a particular way to solve the problem but of course there are different ways too. If you follow the first hint then you will find an expression involving $f''(x)$. Next, following the second hint and combining you'll eventually reach the asked conclusion.

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• Delta2

#### Keith_McClary

But we are interested in $\int_0^1\int_0^1 \dfrac{1}{x^2+y}\,dy\,dx$.
12. b)
$x^2+y \ge y$ and $x^2+y \ge x^2$ so
$$\int_0^1 \frac 1{x^2+y} dy \le \int_0^{x^2} \frac 1{x^2}dy + \int_{x^2}^1 \frac 1{y}dy = 1 - 2 \ln x$$
The integral over $x$ is finite.

#### Delta2

Homework Helper
Gold Member
For 10. e) the domain of integration has to be the whole $\mathbb{R}$ or can it be some interval $[a,b]$ for a,b finite real numbers?

For 7 we can assume that the hour and minute hand have discrete positioning and can take only values that correspond to the positions of the minutes? Or perhaps the hour hand is moving instantaneously from hour 1(that correspond to position of minute 5) to hour 2 (that corresponds to position of minute 10) when it is 1:59 and then it goes 2:00

I believe in a real clock the hour and minute hands have some sort of continuous positioning as the seconds hand is moving around

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#### fresh_42

Mentor
2018 Award
For 10. e) the domain of integration has to be the whole $\mathbb{R}$ or can it be some interval $[a,b]$ for a,b finite real numbers?

For 7 we can assume that the hour and minute hand have discrete positioning and can take only values that correspond to the positions of the minutes? Or perhaps the hour hand is moving instantaneously from hour 1(that correspond to position of minute 5) to hour 2 (that corresponds to position of minute 10) when it is 1:59 and then it goes 2:00

I believe in a real clock the hour and minute hands have some sort of continuous positioning as the seconds hand is moving around
I don't see how these affect the solution. Take it as given.

#### Delta2

Homework Helper
Gold Member
Well for 10. e) I believe I have a function if the domain is say $[-1,1]$
Take $f(x,y)=\frac{e^{xy}}{y}$. Then the left hand side of 10.e) condition is not even defined (or it is infinite) while the right hand side is $\int_{-1}^{1}e^{xy}dy=\frac{e^x}{x}-\frac{e^{-x}}{x}$ . But this example doesn't work if the domain is $\mathbb{R}$ cause then in both sides we get infinite.

As for the 7. I was thinking that maybe we could understand which hand is the hour hand because it would take discrete positioning at the hour markings. But ok I think I see now, we gonna take it as having continuous positioning.

EDIT: For 10 e) the derivative of infinite is undefined or infinite?

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#### Delta2

Homework Helper
Gold Member
Ok well here is my attempt at #7 for which I am not sure at all.

We have 24 possible hours and 60 possible minutes So the total "times" is 24x60=1440.
It is impossible to determine what time is it unless the hour and the minute hand point at the same position which happens only for a total of 24=12+12 times (0:00, 1:05, 2:10,3:15,4:20,5:25,6:30,7:35,8:40,9:45,10:50,11:55,12:00 and so on). So the requested number is 24x59=1416.

#### Infrared

• 