Math Challenge - November 2018

[Delta2]

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

 

[fresh_42]

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]

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 going to take it as having continuous positioning.

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

 

[Delta2]

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]

@Delta2 It's usually possible to determine the time. For example if one hand is pointing straight at the 12 and the other at the 3, then it must be 3:00. It can't be 12:15 since in that case the hour hand would be 1/4 of the way between the 12 and the 1, and not straight at the 12. (This is @StoneTemplePython's problem, but I'll comment because he's away and I know this is the right interpretation).

 

[fresh_42]

Interesting, my original line of thinking is that f should have some discontinuities in order for the two sides to be different. So you saying there is some f that is continuous in R yet the two sides are different?

Hmmmm,, trying to follow your hint.. does the function $f(x,y)=xe^{-x|y|}$ does the job?

I don't see what your absolute value of $y$ is good for. I had a slightly different function, such that differentiation will leave me a factor $x$ to make it zero at $x=0$ on one side.

 

[batboio]

Now I can't follow you anymore. Do you mean F/∼I\mathcal{F}/\sim_\mathcal{I} or F/{p}.\mathcal{F}/\{\,p\,\}\,. The latter isn't of interest, neither for the problem nor mathematically as G/{1}=GG/\{\,1\,\}=G for any group. So what do you mean by "the equivalence class of pp"?

OK I will be more explicit:
The equivalence class of the identity element (we will denote it by $[p]_{\sim_\mathcal{I}}$) is the set of all elements of $\mathcal{F}$ equivalent to $p$ when considering the relation $\sim_\mathcal{I}$. If we consider the quotient group $\mathcal{F}/\sim_\mathcal{I}$ (assuming for the moment that it is a quotient group), then $[p]_{\sim_\mathcal{I}}$ should be a normal subgroup of $\mathcal{F}$. But if I am not mistaken, by direct calculation one can see that $[p]_{\sim_\mathcal{I}}=\mathcal{I}$ and $\mathcal{I}$ is not a normal subgroup of $\mathcal{F}$.

 

[mfb]

Which implies only unstable (circular) orbits.

The 1/r² potential in the problem statement would arise in 4 dimensions for forces like gravity, while a 2-dimensional analog (ln(r)) leads to unbound potentials - making everything bound to everything. It is quite interesting that 3 dimensions is the only option that leads to proper orbits.

 

[Buzz Bloom]

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

We have 24 possible hours and 60 possible minutes So the total "times" is 24x60=1440.

Hi Delta2:

These two quotes are not consistent with each other. The first assumes continuous motion of the hands. The second assumes that only specific discrete positions are to be considered. BTW: I posted my solution in post #44 based on the continuous assumption. I made several errors, but with a few helpful hints by @StoneTemplePython I think I finally got it right.

Regards,
Buzz

 

[fresh_42]

... and $\mathcal{I}$ is not a normal subgroup of $\mathcal{F}$.

Proof?

 

[fresh_42]

Before we endlessly negotiate solutions instead of solving them, here is mine in detail:
https://www.physicsforums.com/threa...cannot-always-be-swapped.960617/#post-6092583

Please use https://www.physicsforums.com/threa...-integration-cannot-always-be-swapped.960617/ for further discussions on 10.e.).

 

[Jacob Nie]

It's my first time on the forums! Every other problem is like a foreign language to me so I will offer my solution to the clock problem:

Let us the define one "ruoh" to be the angular distance between two adjacent numbers on the clock (30 degrees). (The IBWM still hasn't responded to my proposal I sent for this new unit!) Let $m$ and $h$ be the angular distance, measured clockwise in ruoh, between {12, the minute hand} and {12, the hour hand}, respectively. A configuration of hour-hand and minute-hand is only "valid" if the following is satisfied:
$$m= 12\{h \}$$
where $\{x\}$ denotes the fractional part of x. A configuration will satisfy the problem condition if it is still a valid configuration when the minute and hour hands are switched. In other words,
$$m= 12\{ h\}\text{ and }h = 12\{m \}.$$
Now the easiest way to find the number of solutions is just to visualize its graph. (Or use a graphing calculator like desmos.com/calculator/) The graph of $y=\{x\}$ is simply a series of diagonal lines that slant upwards to the right, connecting (0,0) to (1,1), (1,0) to (2,1), (2,0) to (3,1), etc. So $m = 12\{ h\}$ graphed in the $h\text{-}m$ plane is really just that stretched by a factor of twelve vertically. Note that $h = 12\{ m \}$ is simply the reflection of $m = 12\{ h\}$ across $h=m.$

Suppose that $h = 12\{m \}$ is colored red and $m = 12\{h \}$ is colored blue. The relevant part of the graphs of both (the parts that contain intersections) consists of 12 red lines and 12 blue lines. Each red line intersects every single blue line exactly once, so the total number of intersections is 144. (Though this counts (12,12) which is a discontinuity for both equations.)

However, we count 12 points where $m=h.$ These points don't really satisfy the problem condition because this is the time where the hour and minute hands coincide. In this situation, though it cannot be determined whether one hand is an hour or minute hand, the time is still easily discernable. We check for the discontinuity we noticed earlier, and discover delightfully that we have killed two stones with one bird. Hence, between 12 am and 12 pm, there will be 132 times where the time cannot be determined on our unique clock. In one day, this will occur 264 times.

 

[Jacob Nie]

The beetle problem:

Let $t$ be the number of days since the beetle made his decision to climb. Let $b(t)$ be the height of the beetle, calculated at sunrise. Knowing $b(0)=0$, we can write the following recursive relation:
$$ b(t) = b(t-1) + 0.2\dfrac{b(t-1)}{100} + 0.1.$$
I am assuming that the tree only grows during the daytime. (Real trees have some sort of circadian rhythm that regulates this sort of thing.) When the tree grows is important because the amount of tree growth that the beetle is able to "take advantage of" is related to his height, which we know increases (very slightly) at night. The answer will be slightly different if we assumed the tree grows uniformly throughout the day. Maybe I will give this a shot later.

The height of the tree is given simply by $100 + .2 t.$ The first t for which $b(t)>100+0.2t$ is 965, so that will be my answer for now. My code: "b = 0; For[i = 1, b < 100 + .2 i, i++, b = 501/500 b + .1;]; i"

30 minutes later: well I tried it for a uniform growth. Here I make another assumption---the beetle is only climbing at night, which is 12 hours long. Using the same terminology as above, $b(t) = b(t-1) + 0.001 b(t-1) + \text{Night-time}$. $0.001b(t-1)$ is the height increase of the beetle during the daytime, which we assume to be half of one day. Calculating the beetle's progress at night requires a differential equation. In a time $dt$, the beetle will move by its own power at a rate of 0.1m per 12 hrs, and the tree will grow at a rate of 0.1m per 12 hrs. We can write
$$ db = \left(\dfrac{0.1}{12}+\dfrac{0.1}{12}\cdot\dfrac{b}{100}\right)dt.$$
This translates to
$$0.1 = \int_{1.001b(t-1)}^{b(t)} \dfrac{100}{100+b}db$$
which leaves us with the new recursive relation
$$ b(t) = 100(e^{0.001}-1) + 1.001e^{0.001}b(t-1).$$
Putting this back into our original program, we realize that the answer has changed to 964. An exercise in futility, but an exercise nonetheless!


Note: if the beetle moved 9.42 cm every night instead of 10 cm, it would be able to document its journey as a parody of a famous collection of Middle Eastern folk tales...

 

[fresh_42]

The beetle problem

So what is your solution? How many nights will it take for the beetle? I read it as if it will no succeed, which is wrong. Btw., the calculation is far easier.

 

[julian]

The beetle problem:

Let $t$ be the number of days since the beetle made his decision to climb. Let $b(t)$ be the height of the beetle, calculated at sunrise. Knowing $b(0)=0$, we can write the following recursive relation:
$$ b(t) = b(t-1) + 0.2\dfrac{b(t-1)}{100} + 0.1.$$
I am assuming that the tree only grows during the daytime. (Real trees have some sort of circadian rhythm that regulates this sort of thing.) When the tree grows is important because the amount of tree growth that the beetle is able to "take advantage of" is related to his height, which we know increases (very slightly) at night. The answer will be slightly different if we assumed the tree grows uniformly throughout the day. Maybe I will give this a shot later.

.

I made the same assumption that the tree only grows during the day. But I had a formula like:

$b(t) = b(t-1) + 0.2 {b(t-1) \over 100 + (t -2) 0.2} + 0.1$

because you have to take into account that the height of the tree is different at the beginning of every day. The solution to this can be converted into a sum/prod equation which crashed wolfram for reasonable values of $t$...

 

[fresh_42]

The solution to this can be converted into a sum/prod equation which crashed wolfram for reasonable values of $t$...

Calculate with $[t]= 1\, day$ and relative height instead of absolute.

 

[julian]

Hmm. Just to be clear with my notation and reasoning...

So at the first sunrise, $(t=1)$, we have

$b(1) = 0.1$

because the beetle walked a distance $0.1$ during the night. At this point the tree is still $100$.

Then during the following daytime the beetle gets carried a distance ${0.2 \over 100} b(1)$, during the next night he walks a distance $0.1$, so by the 2nd sunrise, $(t=2)$,the beetle is at height:

$b(2) = b(1) + {0.2 \over 100} b(1) + 0.1 .$

During this night the height of the tree is $100+0.2$.

Then during the following daytime the beetle gets carried a distance ${0.2 \over 100+0.2} b(2)$, during the next night he walks a distance $0.1$, so by the 3rd sunrise, $(t=3)$,the beetle is at height:

$b(3) = b(2) + {0.2 \over 100 + 0.2} b(2) + 0.1 .$

During this night the height of the tree is $100+ 2 \times 0.2$.

And so on.

So I will be wanting the value of $t$ for which $b(t) > 100 + (t-1) 0.2$.

 

[Buzz Bloom]

Here is my solution to problem #4.

Spoiler

On the evening of the 858th day tree tree has grown to a height of 271.6 m, and the beetle has reached this height.
My analysis was by means of a spread sheet.

 

[fresh_42]

Hmm. Just to be clear with my notation and reasoning...

In principle yes, but I have difficulties to read your various quotients. Especially the nominators look wrong, as if you used $\frac{a+b}{c+d}=\frac{a}{c}+\frac{b}{d}$. He always crawls a constant distance on an increasing track. Just add the portions.

 

[fresh_42]

Here is my solution to problem #4.

Spoiler

On the evening of the 858th day tree tree has grown to a height of 271.6 m, and the beetle has reached this height.
My analysis was by means of a spread sheet.
View attachment 234500
View attachment 234501
View attachment 234502

There is no need to distinguish night and day. Steps of 24 hours are sufficient. And he needs more than 5 years.

 

[julian]

In principle yes, but I have difficulties to read your various quotients. Especially the nominators look wrong, as if you used $\frac{a+b}{c+d}=\frac{a}{c}+\frac{b}{d}$. He always crawls a constant distance on an increasing track. Just add the portions.

I was assuming the tree only grows during the daytime, and that the beetle only walks during the night. What should I be assuming?

 

[Keith_McClary]

7. How many times a day is it impossible to determine what time it is, if you have a clock with same length (identical looking) hour and minutes hands, supposing that we always know if it's morning or evening (i.e. we know whether it's am or pm).

Say the clock has circumference $1$ and $H$ and $M$ are the distances of the of the hands from the top. Clockwise. Then legitimate positions are
$$M=12H+r$$
where $-r$ is the integer part of $12H$. Ambiguous times satisfy this and also
$$H=12M+s$$
where $s$ is an integer.
Eliminating $M$ we find that $143H$ must be an integer. The ambiguous times occur at times $12n/143$ hours for $n=0, ..., 142$.

 

[fresh_42]

I was assuming the tree only grows during the daytime, and that the beetle only walks during the night. What should I be assuming?

Day by day is sufficient: units of 24 hours: +20 cm for the tree, +10/L for the beetle.

 

[Buzz Bloom]

There is no need to distinguish night and day. Steps of 24 hours are sufficient. And he needs more than 5 years.

Hi fresh:

Thanks for your post #90. I understand that I could have combined the two motions (beetle and tree) in the same row, but for the purpose of making clear the non-simultaneous movements of beetle and tree, I decided to use a separate row in the spreadsheet for each distinct movement. I also found my careless error of reversing the speeds of the tree and the beetle. I will be editing the post with this correction soon.

Regards,
Buzz

 

[fresh_42]

I will be editing the post with this correction soon.

Better make a new one. To be honest, it would help a lot if an answer "xyz days" was at the end. Then I could either look for an error or confirm it.

 

[Buzz Bloom]

Hi @fresh_42:

I corrected my spreadsheet, but my post #88 no longer permits me to edit it. Can you allow me to make edits in this post, or should I post my corrected solution in a new post?

Regards,
Buzz

BTW: My corrected answer is 3191 days = 8.736483 years (1 year = 365.25 days).
SPREADSHEET IMAGES

Spoiler

 

[fresh_42]

Hi @fresh_42:

I corrected my spreadsheet, but my post #88 no longer permits me to edit it. Can you allow me to make edits in this post, or should I post my corrected solution in a new post?

Regards,
Buzz

BTW: My corrected answer with is 3191 days = 8.736483 years (1 year = 365.25 days).
SPREADSHEET IMAGES

Spoiler

https://www.physicsforums.com/attachments/234518 https://www.physicsforums.com/attachments/234519 https://www.physicsforums.com/attachments/234520

Can you show us how you derived this result - in mathematics, not in excel?

 

[Buzz Bloom]

Can you show us how you derived this result - in mathematics, not in excel?

Hi fresh:
I have made some partial progress with the math, but before I complete what I have started I would appreciate your comments regarding the approach I am taking.

Spoiler

I put the relationships for b_n into an equation:

(b_n+1 - b_n) - U_n b_n = V_n,
U_n = 20/(t₀+20n), and
V_n = 20n.​

.
Unfortunately, I am unskilled in difference equations, but I let

x = n,
dx = 1 (day),
y(x) = b_n = beetle's position on day n,
u(x) = U_n = 1/(1+Ax),
A = T₀/20,
T₀ = 10000cm, and
v(x) = 10x = V_n = 10n.​

This gives me a differential equation

y'(x) - y(x)/(1+Ax) = 10x, A = T₀/20.​

I also let

t(x) = T_n = 20n + T₀ = height of tree on day n.
​

To find an approximate number of days, I need to solve the following equation for x.

y(x) = w(x).​

Would this approach give me an acceptable solution for problem #4?

Regards,
Buzz

 

[fresh_42]

Hi fresh:
I have made some partial progress with the math, but before I complete what I have started I would appreciate your comments regarding the approach I am taking.

Spoiler

I put the relationships for b_n into an equation:

b_n+1 - b_n U_n = V_n​

Unfortunately, I am unskilled in difference equations, but I let

x = n,
dx = 1 (day),
y(x) = b_n,
u(x) = U_n, and
v(x) = V_n.​

This gives me a differential equation

y'(x) - u(x) y(x) = v(x).​

I also let

w(x) = t_n.​

To find an approximate number of days, I need to solve the following equation for x.

y(x) = w(x).​

Would this approach give me an acceptable solution for problem #4?

Regards,
Buzz

You have a good result, i.e. within acceptable error margins, but I have no idea what you are talking about. What are $b,x,y,u,v,U,V$ and $w$? I only see a letter salad.

You started with $b_{n+1} - b_nU_n=V_n$. Fine but what do $b,U,V$ stand for?
From there you get $y'(x)-y(x)u(x)=v(x)$. How? You can write $y'(x)=b_{n+1}-b_n$ but I think you set $y'(x)=b_{n+1}$ and where lucky by the form of the equation that this only costed you a day.

This puzzle really doesn't need a lot of formulas, one single sum to be exact. To solve it via a differential equation looks fun but isn't necessary. But that's not the point. The point is that you did not define your letters and how you achieved your initial recursion!

 

[Buzz Bloom]

The point is that you did not define your letters and how you achieved your initial recursion!

H fresh:

I apologize for skimping on the details. I have edited my post #98, fixing typos and adding details. I hope it is now clear.

Regards,
Buzz

 

[fresh_42]

H fresh:

I apologize for skimping on the details. I have edited my post #98, fixing typos and adding details. I hope it is now clear.

Regards,
Buzz

I still find that you haven't explained a lot, especially what your letters should mean, except $b_n$ and $y(x)$ which have already been the easiest to guess. I still don't know what $u,v,w,U,V$ are. But as your solution is close enough and I can see the right formulas shine through, and I do not want to ride this horse to death, I accept this as an answer and add 'my solution':

On the first night the beetle manages $10/10000$ of the tree trunk. On the second night he crawls $10/10020$ of the tree trunk. On the third night he crawls $10/10040$ of the tree trunk. He has reached the top when the sum of the track parts reaches $1$.
\begin{align*}
\sum_{n=0}^{N}\dfrac{1}{1000+2n}&\geq 1\\
\sum_{n=0}^{N}\dfrac{1}{500+n}&\geq 2\\
\sum_{n=1}^{N}\dfrac{1}{n} - \sum_{n=1}^{499}\dfrac{1}{n}& \geq 2\\
H_N - H_{499} \geq 2
\end{align*}
The closest estimation is $H_n=\gamma +\log(n)+\varepsilon$ with the Euler-Mascheroni constant $\gamma$ and a small error $\varepsilon$, which yields $N=3688$ and $3688-499=3189$ nights. The least lower bound are $3184$ nights, which I got by nested intervals and wolframalpha.

Correction: A close estimation is $H_n=\gamma +\log(n)+\varepsilon$ with the Euler-Mascheroni constant $\gamma$ and a small error $\varepsilon$. A numerical solution yields at least $3691-499=3192$ nights. (cp. post #108 f.)

 

[Delta2]

for 10 d), do we have to find an example for which the limits of both sides exist and are finite (yet they differ)?
I think I have an example where left side limit is +infinite while right side limit is zero. Does this count?

 

[Keith_McClary]

4.
Ant on a rubber rope - Wikipedia

 

[fresh_42]

for 10 d), do we have to find an example for which the limits of both sides exist and are finite (yet they differ)?
I think I have an example where left side limit is +infinite while right side limit is zero. Does this count?

Yes. Lebesgue integrability and finite limits.

 

[Delta2]

Yes. Lebesgue integrability and finite limits.

Something else, I see the domain of the functions $f_r$ is $\mathbb{R^{+}}$ while the domain of the integral is $\mathbb{R}$.. Is this a typo? The domain of the integral should be also $\mathbb{R^{+}}$ right?