Math Challenge - November 2018

[Infrared]

@julian You're on the right track, but you need to make your argument rigorous.

We also have:
So write the polynomial as:

$p(x) = a_n (x-a)^n + a_{n-1} (x-a)^{n-1} + \dots + a_4 (x-a)^4 + a_3 (x-a)^3 + a_0$

where $a_0 \not= 0$. We have an undulation point at $x=a$ if $a_3 = 0$ and the first non-zero term after it has even power in $(x-a)$. Say this first non-zero term is $a_{2m} (x-a)^{2m}$. For $x$ close to $x=a$ we have

$p (x) \approx a_{2m} (x-a)^{2m} + a_0$

so the corresponding trough or peak is not intersecting the $x-$axis and again we must have complex roots?

Can you give a proof for why the fact that this approximation to $p$ having no real roots implies that $p$ has a corresponding non-real root?

Your count in the case that $a$ is an inflection point also confuses me. Can you give a more complete argument for why $p(x)$ has fewer than $n$ real roots.

 

[timetraveller123]

hi i am new to this i am not sure this will work
this is for question 2 of basics

assume the opposite that such a polynomial exists
hence
$P(x) = \sum_i (x - \alpha_i)$
where $\alpha_i$ is real
hence
$P'(x) = P(x)\sum_i \frac{1}{x-\alpha_i}\\ P'(a) = P(a) \sum_i \frac{1}{x - \alpha_i} = 0\\ \sum_i \frac{1}{a - \alpha_i} = 0\\ P''(x) = P'(x)\sum_i \frac{1}{x-\alpha_i} + P(x) \sum_i \frac{-1}{(x-\alpha_i)^2}\\ P''(a) = P(a) \sum_i \frac{-1}{(a-\alpha_i)^2} = 0\\ \sum_i \frac{1}{(a-\alpha_i)^2} = 0$
if each term were to be real then all term would have to be equally zero else they would have to be complex would this work

 

[Delta2]

hi i am new to this i am not sure this will work
this is for question 2 of basics

assume the opposite that such a polynomial exists
hence
$P(x) = \sum_i (x - \alpha_i)$
where $\alpha_i$ is real
hence
$P'(x) = P(x)\sum_i \frac{1}{x-\alpha_i}\\ P'(a) = P(a) \sum_i \frac{1}{x - \alpha_i} = 0\\ \sum_i \frac{1}{a - \alpha_i} = 0\\ P''(x) = P'(x)\sum_i \frac{1}{x-\alpha_i} + P(x) \sum_i \frac{-1}{(x-\alpha_i)^2}\\ P''(a) = P(a) \sum_i \frac{-1}{(a-\alpha_i)^2} = 0\\ \sum_i \frac{1}{(a-\alpha_i)^2} = 0$
if each term were to be real then all term would have to be equally zero else they would have to be complex would this work

Ah @timetraveller123 I was on the exact same line of thinking but when I decided to start writing you had already post it.
Just a small typo you have at first line, you mean $P(x)=\prod_i (x-a_i)$

 

[timetraveller123]

oh you that's what i meant

 

[bhobba]

I didn't use de L'Hôpital but let me just add my solution for the sake of readability:

Yes yours is simpler to read. I personally do not usually resort to explicitly using L-Hopital either because as you used any polynomial divided by e^αx goes to zero as a simple corollary of it anyway. I remember one of my professors listed a number of them when he proved the theorem.

Thanks
Bill

 

[Infrared]

Nice job @timetraveller123 !

It is also possible to solve this problem by counting roots, as some other posters tried. Anyone should feel free to post a root-counting solution even though the problem is already solved.

Also, the basic-level questions were originally intended for high school students. If you are much past this, maybe try the other problems instead :)

 

[timetraveller123]

i am high school

 

[Infrared]

Sorry, I didn't mean you in particular. Not all of the solutions for the B-level problems came from high school students.

 

[Buzz Bloom]

Here is my (hopefully) final solution try for #7. I have the value for N, the number of times between midnight and midnight the following day that the clock will give an ambiguous reading.
ADDED
I had the needed insight to complete the calculation.
ALSO A LATER MINOR EDIT
ALSO Added a later edit taking into account the comments in post #47.
ALSO Added some more edits to make corrections and also to make my analysis clearer.
ALSO Added one more correction to my solution.
ALSO Added still one more correction.
ALSO Added one (hopefully) final minor edit.

Spoiler

The assumption is that hour and minute hands of a clock are completely indistinguishable from each other. This includes the detail that one can not see which of the two hands is on top of the other at the axis of the clock. I also assume that the hands move continuously without jumps. It is also assumed that if the time t₁ looks exactly like a different time t₂, this is considered to a confusion of times if and only if t₁ and t₂ are both within the same of two 12 hour periods: (a) between midnight and noon, or (b) between noon and midnight. The problem is to find the value of N, where N is the number of times in a 24 hour day (between midnight one day and midnight the following day) that the configuration of hour and minute hands of a clock at a time t₁ is exactly the same a different time t₂ with the hour and minute hands reversed, AND also that t₁ and t₂ are within the same 12 hour period (a) or (b).

I note that N = 2 M where M is the number of time confusions that occur in period (a) and also the number of time confusions that happen in period (b).

My solution is N = 264.

The solution requires determining the number of times, M, between 12am and 12 pm that actual time t₁ and will be confused with some false time t₂.

Let the angle in degrees between the actual hour hand at time t₁ and its position at either 12am or 12 pm be h₁, and let the corresponding angle for the actual minute hand be m₁.

m₁ = 12 h₁ - 360 INT(12 h₁/360)​

The false time t2 has corresponding angles for the hands, h2 and m2

m₂ = 12 h₂ - 360 INT(12 h₂/360)​

The following are the equations for the time confusion between t1 and t2.

[1] h1 = m2 = 12 h₂ - 360 INT(12 h₂/360)
[2] h2 = m1 = 12 h₁ - 360 INT(12 h₁/360)​

Let
A = INT(12 h₁/360), and
B = INT(12 h₂/360).
Then

h₁ = 12 h₂ - 360 A, and
h₂ = 12 h₁ - 360 B.​

solving:

h₁ = 144 h₁ - 360 (A+12 B) ->
143 h₁ = 360 (A+12 B). - >
h₁ = 360 (A+12 B) / 143​

Similarly

h₂ = 360 (B+12 A) / 143​

Therefore, since

0 <= A < 12, and
0 <= B < 12.​

there are 144 pairs of values of A and B, and therefore also 144 equations for h₁ and h₂. However, this value fails to take into account that when

h₁ = m₁ = h₂ = m₂​

then

t₁ = t₂,​

so there is no time confusion. This occurs when A = B. Therefore there are only

M=11×12=132​

pairs of possible solutions, each pair having both a before noon and a post noon value.

To get the actual count, one must solve the 132 pairs of equations and determine for each pair of A and B values that both h₁ and h₂ satisfy the constrains

0 < h₁ < 360, and
0 < h₂ < 360.​

Therefore

A + 12 B < 143, and
B + 12 A < 143.​

Both of these constraints are satisfied for all 132 acceptable combinations of A and B.

So there are M=132 occurrences when there is a confusion between midnight and noon. There are also M=132 confusions between noon and midnight, making

N=2×M=264​

confusions in a 24 hour day.

 

[eys_physics]

10a) By the change of variables, $$t=\log(x),$$
$$\int_{1}^{\infty}\frac{\log(x)}{x^3}dx=\int_{0}^{\infty}t\exp(-2t)dt=\frac{1}{4}\int_{0}^{\infty}t\exp(-t)dt=\frac{1}{4}\Gamma(2)=\frac{1}{4}$$.

 

[fresh_42]

10a) By the change of variables, $$t=\log(x),$$
$$\int_{1}^{\infty}\frac{\log(x)}{x^3}dx=\int_{0}^{\infty}t\exp(-2t)dt=\frac{1}{4}\int_{0}^{\infty}t\exp(-t)dt=\frac{1}{4}\Gamma(2)=\frac{1}{4}$$.

You could have been a bit more detailed, or at least used $s=2t$ in the second step, but yes, it's correct.

For all others: this can also be shown without the use of the $\Gamma-$function and integration by parts.

 

[StoneTemplePython]

Here is my partial solution try for #7. I have an upper limit for the number of times between midnight and noon that the clock will give an ambiguous reading...

The solution requires specifying the number of times, N, between 12am and 12 pm that actual time t1 and will be confused with some false time t2.

to be clear the question asks "How many times a day is..." so the final answer needs to be stated over an entire day-- going from midnight to noon only is cutting it a bit short.

The maximum of these two functions of A and B is when A = B = 11, yielding a maximum value of 143. Thus, A=11 and B=11 is not allowed, so there are 143 occurrences when there is a confusion.

The other issue: these numbers look quite familiar to me but some more adjustments are needed here. It's subtle, but shouldn't the length of the hands not matter on occasion?

- - - -
remark: its interesting how many different approaches there are to the problem. There is a simple gedanken physics / thought experiment styled approach that can get to the answer in only a few sentences. Many other approaches exist, of course.

 

[Buzz Bloom]

How many times a day is..." so the final answer needs to be stated over an entire day

Hi StoneTemplePython:

I confess I was fooled by the way the problem was phrased. It would have helped to use the phrase "times a 24 hour day". Now that I understand this, a see that the answer should be 286=2×143 since each of the 143 confused pairs between midnight and noon is exactly the same as another confused pair between noon and midnight.

It's subtle, but shouldn't the length of the hands not matter on occasion?

I do not understand why the length of hands is relevant to anything. I have assumed throughout that the hands are the same length, and in all other ways indistinguishable. I note here that noon and midnight are two occasions when the hour hand and minute hands are in the same place, and this is not a case of confusion since reversing the role of the two hands doesn't change the time.

Regards,
Buzz

 

[StoneTemplePython]

Hi StoneTemplePython:
I confess I was fooled by the way the problem was phrased. It would have helped to use the phrase "times a 24 hour day". Now that I understand this, a see that the answer should be 286=2×143 since each of the 143 confused pairs between midnight and noon is exactly the same as another confused pair between noon and midnight.

The puzzle was written by a famous mathematician -- I don't think your wordsmithing is needed -- evidently neither did he. As always people are free to ask questions to clarify.

Overall its a decent attempt but your answer is still wrong.

I do not understand why the length of hands is relevant to anything. I have assumed throughout that the hands are the same length, and in all other ways indistinguishable. I note here that noon and midnight are two occasions when the hour hand and minute hands are in the same place, and this is not a case of confusion since reversing the role of the two hands doesn't change the time.

I was trying to give you a subtle hint... but using the hint is of course optional. Honestly, playing around with an old school clock can be useful, which reminds me:

Oh.So do you mean that the clock ticks from one minute to another minute,rather than gradually changing all the time?I know both kinds of non digital clock exist.
The ticking one:
https://tenor.com/view/design-time-clock-tick-tock-gif-3428153

for avoidance of doubt here: the "ticking one" does not have a ticking minute hand. It has a ticking seconds hand. The minutes and hours hands don't move at all.

edit:
for extra avoidance of doubt: what I said in post 16 still stands. My point here was to correct errors made in post 15. The "ticking one" is a reference to the above link but I wanted to point out that it is the seconds hand that jumps, not the minute hand, and in that link the minute hand doesn't move at all. I've never referred to the clocks in this
problem as "ticking ones".

For this problem: the clock's hands have no jumps!

 

[Buzz Bloom]

It has a ticking seconds hand.

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.

The puzzle was written by a famous mathematician

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.

the "ticking one" does not have a ticking minute hand. It has a ticking seconds hand. The minutes and hours hands don't move at all.

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

 

[StoneTemplePython]

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]

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]

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.

 

[Delta2]

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]

@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]

@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!

 

[Buzz Bloom]

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]

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]

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.

 

[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]

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]

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]

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.

 

[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]

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]

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]

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.

 

[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.