Micromass' big October challenge

[micromass]

Time for the october challenge! This time a lot of people sent me suggestions for challenges. I wish to thank them a lot! If you think of a good challenge that could be included here, don't hesitate to send me!

Ranking [and previous challenges] here: https://www.physicsforums.com/threads/micromass-big-challenge-ranking.879070/

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

ADVANCED CHALLENGES:

1. SOLVED BY mfb Thanks to Charles Link I use the notation from the Frenet-Serret frame for curves:
a) Find the trajectory of an object moving in the $x$ direction with velocity $v_0$ at time $t=0$ that experiences a centripetal force (in the $N$ direction) that increases linearly with time (so that acceleration $a=btN$).
b) Does the speed of the object change?

2. Thanks to Erland
Definitions:
The following functions from $\mathbb N^k$ to $\mathbb N$, with $k\ge 0$, are called basic functions:

B1) The zero constant function $Z:\mathbb N^0\to \mathbb N$, given by $$Z()=0$$(so $Z$ is actually the constant $0$).

B2) For all $n\ge 0$ and $i$ such that $1\le i\le n$, the i:th n-ary projection function $P_i^n:\mathbb N^k\to\mathbb N$, given by $$P_i^n(x_1, x_2,\dots, x_n)=x_i,$$ for all $x_1,x_2,\dots,x_n\in \mathbb N$.

B3) The successor function $S:\mathbb N^1\to \mathbb N$, given by $$S(x_1)=x_1+1,$$ for all $x_1\in \mathbb N$.

We obtain new functions from old ones by applying the following operations:

O1) Composition: For all $m,n\ge 0$ and all functions $g:\mathbb N^m\to \mathbb N$ and $h_1,h_2,\dots, h_m:\mathbb N^n\to \mathbb N$, we obtain the function $f:\mathbb N^n\to \mathbb N$, given by
$$f(x_1,x_2,\dots, x_n)=g(h_1(x_1,x_2,\dots,x_n),h_2(x_1,x_2,\dots,x_n),\dots,h_m(x_1,x_2,\dots,x_n)),$$
for all $x_1,x_2,\dots,x_n\in \mathbb N$.

O2) Primitive recursion: For all $n\ge 0$ and all functions $g:\mathbb N^n\to\mathbb N$ and $h:\mathbb N^{n+2}\to\mathbb N$, we obtain the (unique) function $f:\mathbb N^{n+1}\to\mathbb N$ which satisfies the following conditions:

i)
$$f(x_1,x_2,\dots,x_n,0)=g(x_1,x_2,\dots,x_n),$$
for all $x_1,x_2,\dots,x_n\in\mathbb N$,

ii)
$$f(x_1,x_2,\dots,x_n,x_{n+1}+1)=h(x_1,x_2,\dots, x_n, x_{n+1},f(x_1,x_2,\dots,x_n,x_{n+1})),$$
for all $x_1,x_2,\dots,x_n,x_{n+1}\in\mathbb N$.

A function $f:\mathbb N^n\to \mathbb N$, for any $n\ge 0$, is called primitive recursive if it can be obtained from the basic functions B1-B3 by finitely many applications of the operations O1 and O2.

Ackermann's function is the (unique) function $A:\mathbb N^2\to \mathbb N$ which satisfies the following conditions:

a) $A(0,n)=n+1$,
b) $A(m+1,0)=A(m,1)$,
c) $A(m+1,n+1)=A(m,A(m+1,n))$,

for all $m,n\in \mathbb N$.

a) Prove that the factorial function $n!$ is primitive recursive.
b) Prove that the "truncated subtraction" $f(a,b) = \max\{a-b,0\}$ is primitive recursive.
c) Prove that to each primitive recursive function $f:\mathbb N^n\to \mathbb N$, for any $n\ge 0$, there is an $m\in\mathbb N$ such that
$$f(x_1,x_2,\dots,x_n)\le A(m, \max_{1\le i\le n} x_i),$$
for all $x_1,x_2,\dots, x_n\in\mathbb N$.
d) Prove that $A$ is not primitive recursive.

3) Let $X$ denote the set of all bounded real-valued sequences. As was shown in last challenge, a generalized limit exists for any such sequence. A generalized limit is any function function $L:X\rightarrow \mathbb{R}$ such that
1) $L((x_n + y_n)_n) = L((x_n)_n) + L((y_n)_n)$
2) $L((\alpha x_n)_n) = \alpha L((x_n)_n)$
3) $\liminf_n x_n \leq L((x_n)_n)\leq \limsup_n x_n$
4) If $x_n\geq 0$ for all $n$, then $L((x_n)_n)\geq 0$.
5) If $y_n = x_{n+1}$, then $L((x_n)_n) = L((y_n)_n)$
6) If $x_n\rightarrow x$, then $L((x_n)_n) = x$.
Find (in ZFC) the number (= cardinality) of all generalized limits.

4) A pirate ship is at the origin of $\mathbb{R}^2$. A merchant vessel is at $(x_0,0)$ with $x_0\neq 0$ as is sailing vertically upwards with constant speed $v$. The pirate vessel is sailing with constant speed $V$ in such a way that at every instant it is always sailing directly towards the merchant vessel.
a) SOLVED BY Charles Link Find the curve in which the pirate vessel moves.
b) SOLVED BY Charles Link For $V>v$, find the total distance traveled by the pirate ship until they capture the merchant vessel.
c) SOLVED BY Chestermiller In the case $v=V$ show that the pirate ship will lag behind the merchant vessel endlessly. Find the asymptotic distance between pirate ship and merchant vessel.

5) SOLVED BY mfb Consider the recursion relation $S_{n+2} = 2(S_{n+1} - S_n)$ with $S_1 = a$ and $S_2 = b$. Find all $a$ and $b$ such that $S_n = 0$ for infinitely many $n$.

6) SOLVED BY Erland Suppose $X$ and $Y$ are independent random variables distributed according to a uniform distribution on $[0,1]$. Find the distribution of $X^Y$.

7) SOLVED BY TeethWhitener A laser is located $b$ units directly across from position $a$ on an infinite straight wall. An angle $\theta$ is chosen uniformly out of $(-\pi/2,\pi/2)$ and the laser is positioned such that it makes an angle $\theta$ when shining on the wall. Find the expected value and the variance of the random variable $X$ which is the (signed) distance from $a$.

8) SOLVED BY Fightfish On a table are $2016$ bells standing in a sequence. At every turn you have to pick $2$ bells, ring them and then exchange their place.
For example, if there were only $4$ bells, they stand initially as $A-B-C-D$. In turn $1$, you pick bells $A$ and $D$, ring them and exchange them to get $D-B-C-A$. In turn $2$, you pick bells $D$ and $B$, ring them and exchange them to get $B-D-C-A$.
The goal of the bell ringer is to take $n$ turns after which the sequence of bells is reversed. For example an easy way to reverse the order in $A-B-C-D$ is first to ring $A$ and $D$ to get $D-B-C-A$ and then to ring $B$ and $C$ to get $D-C-B-A$. We have reversed the bells in $2$ turns.
Show that it is impossible to reverse $2016$ bells in an odd number of turns.

9) SOLVED BY QuantumQuest Find all $10$-digit numbers such that
a) each digit $\{0,1,2,3,4,5,6,7,8,9\}$ is used exactly once
b) the first $n$ digits form a number divisible by $n$ ($n\in \{1,2,3,4,5,6,7,8,9,10\}$).
For example, if my number would be $1234567890$, then $1$ must be divisble by $1$, $12$ must be divisible by $2$, $123$ must be divisible by $3$ and so on.

10) SOLVED BY Fightfish Find all 10-digit numbers where the first digit is how many zeros are in the number, the second digit is how many 1s are in the number etc. until the tenth digit which is how many 9s are in the number.

PREVIOUS UNSOLVED ADVANCED CHALLENGES:

1) Show that it is impossible to find four distinct squares as the subsequent elements in an arithmetic progression.
The following sketch may be used (credit will be given for each step, and not just for the entire problem!)
STEP 1: Assume $\alpha$, $\beta$, $\gamma$ and $\delta$ are four distinct squares that are consecutive elements of an arithmetic progression.
Show that we may assume that they are relatively prime nonnegative integers.
Show that $\text{gcd}(\alpha,\beta) = \text{gcd}(\beta,\gamma) = \text{gcd}(\gamma,\delta) = 1$ and that $\alpha$, $\beta$, $\gamma$, $\delta$ are all odd.
Show that $\text{gcd}(\beta+\alpha, \beta - \alpha) = \text{gcd}(\gamma + \beta,\gamma - \beta) = \text{gcd}(\delta+\gamma,\delta - \gamma) = 2$.

STEP 2:
Define
$$2a = \text{gcd}(\beta+\alpha,\gamma + \beta)$$
$$2b = \text{gcd}(\beta + \alpha, \gamma - \beta)$$
$$2c = \text{gcd}(\beta - \alpha, \gamma + \beta)$$
$$2d = \text{gcd}(\beta-\alpha, \gamma-\beta)$$
Show
$$\beta +\alpha = 2ab,~\beta-\alpha = 2cd$$
$$\gamma + \beta = 2ac,~\gamma-\beta = 2bd$$
$$\delta + \gamma = 2bc,~\delta - \gamma = 2ad$$

STEP 3:
Show that $(a+d)b = (a-d)c$and $(c+d)a = (c-d)b$.
Show that $\text{gcd}(a+d, a-d)$ and $\text{gcd}(c+d,c-d)$ are both either $1$ or $2$.
Show that
$$a+d = mc,~a-d= mb$$
$$c+d = nb,~c-d=na$$
with $m,n\in \{1,2\}$.

STEP 4: Derive a contradiction.2) SOLVED BY TeethWhitener Let $p\neq 0$ be a real number. Let $x_1,...,x_n$ be positive real numbers, we define the $p$-mean as
$$M_p(x_1,...,x_n) = \sqrt[p]{\frac{1}{n}\sum_{i=1}^n x_i^p}$$
Note that $M_1(x_1,...,x_n)$ is the usual mean.
Prove that for all $p,q\in \mathbb{R}\cup \{- \infty,+\infty\}$ has that $p\leq q$ implies $M_p(x_1,...,x_n)\leq M_q(x_1,...,x_n)$.
Hint: Use Jensen's inequality with the function $x^{p/q}$.

3) SOLVED BY mfb Take a wire stretched between two posts, and have a large number of birds land on it at random. Take a bucket of yellow paint, and for each bird, paint the interval from it to its closest neighbour. The question is: what proportion of the wire will be painted. More strictly: as the number of birds goes to infinity, what is the limit of the expected value of the proportion of painted wire, assuming a uniform probability distribution of birds on the wire.
Note: an answer that comes from a simulation program will be accepted! Be sure to present both the answer and the program you used.CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) Let $A,B,C,D$ be complex numbers with length $1$. Prove that if $A+B+C+D=0$, then these four numbers form a rectangle.

2) On an arbitrary triangle, we produce on each side an equilateral triangle. Prove that the centroids of these three triangles forms an equilateral triangle.

3) SOLVED BY Biker Find the total area of the red spot below:

4) SOLVED BY MAGNIBORO A man starts at the origin of $\mathbb{R}^2$. He walks $1$ meter left and $1$ meter up. He walks then $1/2$ of the last distance right. He walks $1/3$ of the last distance down. He walks $1/4$ of the last distance left. Thus the man walks in a spiral. Where does he converge to?5). SOLVED BY MAGNIBORO Thanks to Erland Ackermann's function is the (unique) function $A:\mathbb N^2\to \mathbb N$ which satisfies the following conditions:

a) $A(0,n)=n+1$,
b) $A(m+1,0)=A(m,1)$,
c) $A(m+1,n+1)=A(m,A(m+1,n))$,

for all $m,n\in \mathbb N$.

a) Find (and prove) a closed form expression for $A(1,n)$.
b) Find (and prove) a closed form expression for $A(2,n)$.
c) Find (and prove) a closed form expression for $A(3,n)$.

6) Thanks to Math_QED Consider the integrals I and J.

$I = \int\limits_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{x+1}dx$
$J = \int\limits_0^{\pi}\frac{\cos x}{(x+2)^{2}}dx$

What is I in function of J?

7) Compute the integral $\int \sqrt{\tan(x)}dx$ (the final answer must be written with elementary functions only).

8) Thanks to mfb Find all bases $b>6$ where $5654$ is a prime power.

9) SOLVED BY MAGNIBORO Thanks to Math_QED Let $I_n = \int\limits_0^1 x^n\sqrt{1-x}dx$

Show that $I_n = \frac{2n}{2n+3}I_{n-1} \quad \forall n \in \mathbb{N}\backslash\{0\}$

10) Thanks to Math_QED Calculate:

$\int\limits_{-1}^1\frac{1}{(e^x+1)(x^2 +1)}dx$

11) Thanks to Math_QED Calculate:

$\int \frac{\sqrt[6]{\frac{x}{x-3}} - \sqrt[4]{\frac{x}{x-3}}}{x^2 -3x}dx$

12). Thanks to Math_QED Prove that every matrix $A \in M_{n,n}(\mathbb{R})$ can be written as the sum of a symmetric and an antisymmetrix matrix.PREVIOUS CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) Show that for $0<\lambda<1$ and $\alpha,\beta\geq 0$ holds
$$\alpha^\lambda \beta^{1-\lambda}\leq \lambda \alpha+(1-\lambda)\beta$$
Hint: Fix all but one of the variables and use derivatives.

 

[mfb]

With problems that long, some clearer separation between the problems would help.

1. Thanks to Charles Link I use the notation from the Frenet-Serret frame for curves:
a) Find the trajectory of an object moving in the $x$ direction with velocity $v_0$ at time $t=0$ that experiences a centripetal force (in the $N$ direction) that increases linearly with time (so that acceleration $a=btN$).
b) Does the speed of the object change?

As far as I can see, we have some ambiguity concerning the initial direction of N. Let it point in the y-direction, and introduce a complex coordinate, the real part is x and the imaginary part is y. The acceleration is then given by $\dot v(t) = i b t \frac{v(t)}{|v(t)|}$. Use the ansatz $v(t)=v_0e^{iwt^2}$. Then $\dot v(t) = 2iv_0wt e^{iwt^2}$. Plugging this into the differential equation leads to $2iv_0^2wt e^{iwt^2} = i b t v_0e^{iwt^2}$. Simplification shows that $2v_0w = b$. Therefore, $$v(t)=v_0 \exp(\frac{ibt^2}{2v_0})$$
As we can see, the speed does not change.

Splitting it into components again, $v_x = v_0 \cos(\frac{bt^2}{2v_0})$ and $v_y = v_0 \sin(\frac{bt^2}{2v_0})$. To check the method from above we can calculate the derivatives of those:
$a_x = - bt \sin(\frac{bt^2}{2v_0})$ and $a_y = bt \cos(\frac{bt^2}{2v_0})$. The magnitude is "bt" as expected and it is always orthogonal to the velocity as well.

To find the position as function of time, we have to integrate cos(x²) which does not have a solution in closed form, but of course it is possible finding the position using the Fresnel integral. There might be a function for the path (but not the position as function of time) that avoids this issue, I don't know.

5) Consider the recursion relation $S_{n+2} = 2(S_{n+1} - S_n)$ with $S_1 = a$ and $S_2 = b$. Find all $a$ and $b$ such that $S_n = 0$ for infinitely many $n$.

There is the trivial case of a=b=0 where all elements are zero. Apart from that:
Consider an arbitrary series that has a zero at some point: S_c = 0, S_c+1 = d for some d. Then S_c+2 = 2d, S_c+3 = 2d, S_c+4 = 0 and S_c+5 = -4d. As we can see, if a series has 0 at some point we get some pseudo-repetition with S_c+4 = -4S_c+0. The series is also unique in its backward evolution: If we know S_c=0 then S_c-4=0 with the same argument. A series has either zeros everywhere, zeros exactly every 4th step, or no zeros. That allows a classification of all series which have infinitely many zeros:

a=0 and b arbitrary
b=0 and a arbitrary
a=b
2a=b

 

[Fightfish]

Problem 10

Let $x_{n}$ denote the value of the $n$-th digit of the number. One immediate constraint that follows is that $\sum x_{n} = 10$ since $x_{n}$ represents the number of times the digit $n$ appears in the number. Suppose that $x_{0} = \alpha$, where $\alpha$ is obviously non-zero. Then there are $(9-\alpha)$ nonzero elements (one of which is $x_\alpha$ which must sum up to $(10-\alpha)$. It is thus clear that all of these nonzero elements must be unity except for one that has the value $2$. Thus, we have $x_{2} = 1$, which forces $x_{1} = 2$ and $x_{\alpha} = 1$. There can be no other non-zero elements, and so it must be the case that $9 - \alpha = 3$.
Hence, $\{x_{0} = 6, x_{1} = 2, x_{2} = 1, x_{6} = 1 \}$ with the rest being zero is the only solution.

 

[Fightfish]

Problem 8

Not sure how much detail you want here - but the answer seems to follow directly from the theorems pertaining to permutations/transpositions. In particular, that any particular permutation is of definite parity: if a permutation can be represented as an odd (even) number of transpositions, then it cannot be decomposed into an even (odd) number of transpositions. Here, we see that one possible decomposition involves $(1\,\,2016)(2\,\,2015) \cdots (1008\,\,1009)$, which is an even number of transpositions. Thus, by that theorem, there does not exist any other decomposition that contains an odd number of transpositions.

 

[Stella.Physics]

for UNSOLVED ADVANCED CHALLENGES #3. When you say proportion of the wire painted, you mean for an individual bird or for the sum?
do the birds occupy space on the wire or are they dimensionless?

 

[micromass]

for UNSOLVED ADVANCED CHALLENGES #3. When you say proportion of the wire painted, you mean for an individual bird or for the sum?

The total sum of the painted pieces.

do the birds occupy space on the wire or are they dimensionless?

They are pointlike.

 

[Stella.Physics]

The total sum of the painted pieces.
They are pointlike.

before writing down my calculations is the proportion $\frac{2(n-1)}{n}$ ?

 

[micromass]

before writing down my calculations is the proportion $\frac{2(n-1)}{n}$ ?

I'm not saying anything without solution. But I have no idea what $n$ is. I don't expect a dependence on a number $n$.

 

[Stella.Physics]

I'm not saying anything without solution. But I have no idea what $n$ is. I don't expect a dependence on a number $n$.

sorry my bad, n is the number of birds :P

 

[micromass]

sorry my bad, n is the number of birds :P

Read the problem carefully. The number of birds should not enter into the solution.

 

[Stella.Physics]

Read the problem carefully. The number of birds should not enter into the solution.

So there are 2 problems the first with random distribution and the second with a uniform distribution.
So the question to the first problem is: about the proportion that is painted, is this about the whole painted area from the first to the last bird or should I take into account the areas that would be double painted?

 

[micromass]

So there are 2 problems the first with random distribution and the second with a uniform distribution.
So the question to the first problem is: about the proportion that is painted, is this about the whole painted area from the first to the last bird or should I take into account the areas that would be double painted?

No, there is one problem. The birds have uniform distribution, and you are supposed to find the asymptotic proportion of the painted wire.

 

[Stella.Physics]

No, there is one problem. The birds have uniform distribution, and you are supposed to find the asymptotic proportion of the painted wire.

OK, now it's clear thank you. What about the double painted areas, can the proportion value be over 1? Meaning taking double painted areas into account.

 

[micromass]

OK, now it's clear thank you. What about the double painted areas, can the proportion value be over 1? Meaning taking double painted areas into account.

Double painted = painted. So proportion cannot be bigger than 1.

 

[Stella.Physics]

OK so this is what I have done so far, in simple thinking

n is the number of the birds
L is the length of the wire
P is the painted area
z is the distance between 2 birds In a homogeneous distribution
and z/2 is the distance from the end of the wire to the birds sitting at both ends of the distribution.

So P=(n-1)z
L = P+ 2 $\frac{z}{2}$ = (n-1)z+z=zn ⇒ z= $\frac{L}{n}$

so the proportion is $\frac{P}{L}$ = $\frac{(n-1)z}{zn}$ = $\frac{n-1}{n}$ = 1 - $\frac{1}{n}$

so of course if the number of birds goes to infinity the proportion becomes 1.

How come and the number of birds cannot affect the final painted proportion like you say?

 

[micromass]

That is not the correct answer.

 

[mfb]

Read the problem statement more carefully. A piece of wire is painted only if it is the connection to the closest neighbor for any bird.

 

[Stella.Physics]

Read the problem statement more carefully. A piece of wire is painted only if it is the connection to the closest neighbor for any bird.

The distribution is uniform meaning that each bird is the closest neighbor to the previous one and next one, as they are equally spaced. Except from the two birds sitting on the edges. And we want to know what part of the wire is painted, not the distance of yellow paint, so I left out the areas that where painted 2 times since they are between two neighbor birds.
So I think I've used this kind of thinking in my image.
This one has been puzzling me since last night :P

A second thought was that the first bird will sit on the middle (L/2) of the wire and then the others on the quarters (L/4) and so on...so I could use a Taylor series, but then the part "a large number of birds land on it at random" would be ruined :P

 

[micromass]

The distribution is uniform meaning that each bird is the closest neighbor to the previous one and next one, as they are equally spaced.

That is not what uniform distribution means

 

[aheight]

#5: (Derivation: just put the small piece in lower left corner in upper top corner, then half of the center red piece on the top to get a square 10x10. Then the area of the red is equal to 100 - the circle area and the small piece in lower left which is equal to the area in the bracket below. Use quadratic formula to find $(x_0,y_0)$.

red=$\left(100-25\pi\right)-\biggr[1/2(x_0 y_0)+\int_{x_0}^{5} \left(5+\sqrt{25-(x-5)^2}\right)dx\biggr]$
where the point $(x_0,y_0)$ is the solution to the simultaneous equations $y=1/2x$ and $y=5+\sqrt{25-(x-5)^2}$.

 

[Pepper Mint]

...3) Find the total area of the red spot below:

The area of red spots=S_OCD-S_OAB-S_circle
In my method, I have to find the area of OAB by 2 integrations (line segment OA and Ox plus arc segment AB and Ox) supposing that O's coordinates is (0,0).

• Find the equation of the line passes O and C
• Find the equation of the circle
• From both of these I can find A's coordinate.

 

[mfb]

I still don't find the mistake in my previous calculation, but I ran some simulations, and the fraction of painted wire is about 38.8% (+- 0.1%) based on 108,000 birds.

Simulated in Excel: Put the first bird at location 0, let the following bird have a separation of -LN(RAND()) from the previous one where RAND() has a uniform distribution over (0,1). Then, for each interval, see if it is painted, sum over the lengths of painted intervals, divide by the location of the last bird. Simulate 12,000 birds. Repeat a few times to get a feeling how large the statistical fluctuations are, ignore the tiny bias from the two border cases, average over the runs.

 

[micromass]

#5: (Derivation: just put the small piece in lower left corner in upper top corner, then half of the center red piece on the top to get a square 10x10. Then the area of the red is equal to 100 - the circle area and the small piece in lower left which is equal to the area in the bracket below. Use quadratic formula to find $(x_0,y_0)$.

red=$\left(100-25\pi\right)-\biggr[1/2(x_0 y_0)+\int_{x_0}^{5} \left(5+\sqrt{25-(x-5)^2}\right)dx\biggr]$
where the point $(x_0,y_0)$ is the solution to the simultaneous equations $y=1/2x$ and $y=5+\sqrt{25-(x-5)^2}$.

What question are you answering??

 

[micromass]

I still don't find the mistake in my previous calculation, but I ran some simulations, and the fraction of painted wire is about 38.8% (+- 0.1%) based on 108,000 birds.

Simulated in Excel: Put the first bird at location 0, let the following bird have a separation of -LN(RAND()) from the previous one where RAND() has a uniform distribution over (0,1). Then, for each interval, see if it is painted, sum over the lengths of painted intervals, divide by the location of the last bird. Simulate 12,000 birds. Repeat a few times to get a feeling how large the statistical fluctuations are, ignore the tiny bias from the two border cases, average over the runs.

That is remarkably close to the right answer, so I'll give you credit. Whoever finds an analytic solution in this month gets credit too, otherwise I'll post the solution.

 

[mfb]

Remarkably? Precision is just a matter of bird number. 1.5 million birds lead to 38.83% (+- 0.03%).
I don't see any obvious analytic expression. e/7=0.388326 fits, but 7 is a large number.

 

[aheight]

What question are you answering??

Sorry about that. Pepper Mint got it. I did not previously read the qualifications and find now I do not qualify. My apologies to Pepper Mint.

 

[Charles Link]

With problems that long, some clearer separation between the problems would help.

As far as I can see, we have some ambiguity concerning the initial direction of N. Let it point in the y-direction, and introduce a complex coordinate, the real part is x and the imaginary part is y. The acceleration is then given by $\dot v(t) = i b t \frac{v(t)}{|v(t)|}$. Use the ansatz $v(t)=v_0e^{iwt^2}$. Then $\dot v(t) = 2iv_0wt e^{iwt^2}$. Plugging this into the differential equation leads to $2iv_0^2wt e^{iwt^2} = i b t v_0e^{iwt^2}$. Simplification shows that $2v_0w = b$. Therefore, $$v(t)=v_0 \exp(\frac{ibt^2}{2v_0})$$
As we can see, the speed does not change.

Splitting it into components again, $v_x = v_0 \cos(\frac{bt^2}{2v_0})$ and $v_y = v_0 \sin(\frac{bt^2}{2v_0})$. To check the method from above we can calculate the derivatives of those:
$a_x = - bt \sin(\frac{bt^2}{2v_0})$ and $a_y = bt \cos(\frac{bt^2}{2v_0})$. The magnitude is "bt" as expected and it is always orthogonal to the velocity as well.

To find the position as function of time, we have to integrate cos(x²) which does not have a solution in closed form, but of course it is possible finding the position using the Fresnel integral. There might be a function for the path (but not the position as function of time) that avoids this issue, I don't know.

Very Good @mfb ! The integrals for the path (a scaled version) are found as a graphical solution called the Cornu Spiral that is actually presented in the Optics texts. The Cornu Spiral comes up in diffraction theory in Optics where sometimes the Fresnel integrals are tabulated, but they are also presented in graphical form. I will try to supply a "link" but I haven't used the "link" feature yet. And thank you @micromass for posting the problem (as part of your October challenge) that I forwarded to you that I had come up with (and solved) quite a few years ago after studying Purcell's calculus book which introduces the Frenet equations in two dimensions... $\\$ I solved it by writing $\vec{v}=(ds/dt) \hat{T}$ with $\hat{T}=cos(\phi) \hat{i} +sin(\phi) \hat{j}$ and writing for acceleration $\vec{a}=(d^2 s/dt^2) \hat{T}+(ds/dt) (d \hat{T}/dt )$. Now $d \hat{T} / dt=(d \hat{T} / d \phi ) (d \phi /ds) (ds/dt)=\hat{N} \kappa (ds/dt)$. (Note that $d \hat{T} / d \phi=-sin(\phi) \hat{i}+cos(\phi) \hat{j}=\hat{N}$ where $\hat{N}$ is perpendicular to $\hat{T}$, and $\kappa=d \phi /ds=1/r$ by definition where $r$ is the instantaneous radius of curvature.) This gives $\vec{a}=(d^2 s/dt^2) \hat{T}+\kappa (ds/dt)^2 \hat{N}$. (These equations are found in Purcell's calculus book). The next part is simply to compare the given acceleration to this acceleration and integrating twice, first to get the velocity and then to get the position as a function of time. Notice from the above equation that $d^2 s / dt^2=0$ so that $ds/dt$ must be a constant.

 

[Pepper Mint]

Sorry about that. Pepper Mint got it. I did not previously read the qualifications and find now I do not qualify. My apologies to Pepper Mint.

micromass is a strange person in my opinion. My sole purpose to post on PF is to share, nothing else really matters.

 

[micromass]

micromass is a strange person in my opinion.

Hmm, why?

 

[Biker]

Highschool 3:

:C you have to deal with the mess of my solution
First we find theta, Through tan-1(10/20) = 26.5650 degree
Next, Find the area of ADC sector of the circle.
$\frac{180-2\theta}{360} ~ \pi 5^2$
we substract the area of the triangle DAC from it to get the required space to remove.
$\text{Area of ADC sector} - (25 cos(\theta) ~ sin (\theta))$
And for the the other circle we just subtract the area we found above from the area of the circle.

What we are left with is that small area near the word theta.
So I formed FDG a triangle to find the area of FCG by using similar steps to the above. Once we get that, we can form a square FAGH find the area of it (5*5) subtract quarter of the circle, remove the part FCG and we are left with GCH

By getting the area of the triangle then removing all the parts we found we get that the area to be equal to $19.50413775 m^2$Unfortunately, I know the other geometry question and it looks like the bells problem was solved. I am left with the previous challenge :C

 

[QuantumQuest]

ADVANCED CHALLENGES: Problem 9

We will utilize letters a, b, c, d, e, f, g, h, i, j for digits. We'll begin with divisibility (i.e b) part) and take into account a) part along the way.

According to part b) constraint, the whole number denoted $abcdefghij$, must be divisible by 10, so $j = 0$. Now, we'll proceed to the first five digits. The reason for this, is that we can quickly conclude that $e$ is $0$ or $5$. Now, taking into account part a) constraint (no digit repetitions) and because $0$ is already taken, we conclude that $e = 5$.

At this point, it would be wise to divide remaining candidate digits into odd and even. So, numbers ending in $b$, $d$, $f$ and $h$ must be divisible by 2, 4, 6, 8 respectively, so the previous letters are even "spots". This also means that in accordance to part a), $a$, $c$, $g$, $i$ are odd "spots" (with the same previous meaning). Proceeding to the first four digits $abcd$, because 4 divides them, it also divides $cd$. But because $c$ is odd, $d$ must be 2 or 6$^{*}$. In a similar fashion, $abcdefgh$ is divisible by 8, so $gh$ is divisible by 4. Because $g$ is odd, this leaves us with $h$ to be 2 or 6. Taking into account all the above, we conclude that $b$ and $f$ can be 4 or 8.

Now, going to $abcdef$, this must be divisible by 6. But from the odd parts of the number, $abc$ must be divisible by 3, so $def$ must also be divisible by 3. We already know from the above that $d$ is 2 or 6, $e = 5$ and $f$ is 4 or 8. So, from the four numbers that are formed ($def$ must be divisible by 3), only 258 and 654 survive. Also, according to the above analysis, whatever values $d$ and $f$ take, the values of $h$ and $b$ are fixed accordingly.

Proceeding to $abcdefgh$, this must be divisible by 8, so from divisibility rules, $fgh$ must be divisible by 8. But as we stated earlier, $f$ is even, so it has to be that $gh$ is divisible by 8. Now, $g$ is odd (excluding of course the value 5) and $h$ can be 2 or 6 as we stated earlier, so from all the forming numbers for $gh$ only 4 numbers (16,32,72,96) survive.

From this point on, it would be a good idea to create some sketches for each situation.
So far:

$_4_25816_0$ or $_4_25896_0$ and $_8_65432_0$ or $_8_65472_0$ are the four candidate patterns.

Starting from the first two, we examine the first three digits. Now, we know that $a$ and $c$ can take all the odd values from 1 to 10 except the value 5. Also, $b = 4$ and $abc$ divisible by 3. Taking also into account the constraint a), we have twelve candidate numbers for $abc$, but only 147 and 741 survive the divisibility by 3. So:

$14725816-0$, $14725896-0$, Now, we turn our attention to the divisibility by 7 for $abcdefg$ and we see that it fails for both. So, we can safely exclude the first two candidate patterns.

Now, we examine the last two patterns. In the same fashion as before we examine the first three digits pattern: $abc$. Now, $a$ and $c$, can take the same values as in the previous case of the first two patterns, but $b = 8$ here. Also, taking into account divisibility by 3, constraint a) and the fact that $g$ can be only 3 or 7 (according to the form of the examined patterns), from the twelve possible values, only six survive: 183, 189, 381, 789, 981, 987.

Now, we examine each one of the formed patterns:

$18365432-0$ fails due to repetition of 3

$18965432-0$ fails to the divisibility of $1896543$ by 7

$38165432-0$ fails due to repetition of 3

$78965432-0$ fails to the divisibility of $7896543$ by 7

$98165432-0$ fails to the divisibility of $9816543$ by 7

$98765432-0$ fails to the divisibility of $9876543$ by 7

$18365472-0$, $18965472-0$, $78965472-0$, $98165472-0$, $98765472-0$ fail to the divisibility of the first seven digits by 7.

The only surviving pattern is $38165472-0$. We can now simply complete it with the missing digit 9, so the number is $3816547290$.
$^{*}$

This is trivial but for completeness, the candidate numbers for $cd$ (given the constraints) are 12, 14, 16, 18, 32, 34, 36, 38, 72, 74, 76, 78, 92, 94, 96, 98. From these, only 12, 32, 72, 92, 16, 36, 76, 96 survive the divisibility by 4, so $d$ can be 2 or 6. In the next, for brevity, I will omit showing similar cases.

 

[QuantumQuest]

I'll also give some missing explanation - sorry but my head spun around a bit during posting ;)

Starting from the first two, we examine the first three digits. Now, we know that $a$ and $c$ can take all the odd values from 1 to 10 except the value 5. Also, $b=4$ and $abc$ divisible by 3. Taking also into account the constraint a), we have twelve candidate numbers for $abc$, but only 147 and 741 survive the divisibility by 3. So:

$14725816−0$ $14725896−0$

741 survives the divisibility by 3, but if we insert it into the above pattern in position $abc$, both $74125816-0$ and $74125896-0$ fail to the divisibility by 7. So, we're left with the above two patterns beginning with 147 in $abc$.

 

[EnumaElish]

For the last question, "Show that for 0<λ<1 and α,β≥0 holds ...", isn't there a short proof using a well-known single-variabled function and its properties?

 

[micromass]

For the last question, "Show that for 0<λ<1 and α,β≥0 holds ...", isn't there a short proof using a well-known single-variabled function and its properties?

It's a question for high schoolers.

 

[ibkev]

Nice solution @Biker! I was working on finding the area of the red spots too but like the others took an algebraic approach. I got to the point where all I had left to solve was:

$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx = \int \sqrt {x(10 - x)} \ dx $$

This integral is not as easy as it looks... LOL!