Challenge Math Challenge - November 2018

[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.

 

[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.