Challenge Math Challenge - November 2018

[Buzz Bloom]

I am a bit confused by problem (12. a).

I am describing my confusion in a "SPOILER" box because I am not sure if it might unintentionally help someone solve the problem.

Spoiler

The area where f(x,y) ≠ 0 is a parallelogram bounded by the following four line segments connecting pairs of (x,y) pints.

[(0,0), (1,0)]
[(1,0), (1,3)]
[(1,3), (2,0)]
[(2,0), (0,0)]​

This parallelogram is divided into two regions, TOP and BOTTOM. The dividing line L separating TOP and BOTTOM is

L = [(0,1), (1,2)].​

The area of TOP is the same as the area of BOTTOM. Both areas are 1 square unit.
In the TOP area, f(x,y) = -1.
In the BOTTOM area, f(x,y) = +1.
(Along the line L f(x,y) = -1 but this does not influence the value of the integration.)
Therefore the integral
∫_R ∫_R f(x,y) dx dy = ∫_R ∫_R f(x,y) dy dx = 0.

What confused me is: Why it is necessary to bring Fubini's theorem into the discussion?

If Fubini's theorem must be discussed, then I offer the following.

Since the order of the two integration variables does not change the zero result, this demonstrates that the order does not matter, which is what Fubini's theorem proves.

 

[fresh_42]

I am a bit confused by problem (12. a).

I am describing my confusion in a "SPOILER" box because I am not sure if it might unintentionally help someone solve the problem.

Spoiler

The area where f(x,y) ≠ 0 is a parallelogram bounded by the following four line segments connecting pairs of (x,y) pints.

[(0,0), (1,0)]
[(1,0), (1,3)]
[(1,3), (2,0)]
[(2,0), (0,0)]​

This parallelogram is divided into two regions, TOP and BOTTOM. The dividing line L separating TOP and BOTTOM is

L = [(0,1), (1,2)].​

The area of TOP is the same as the area of BOTTOM. Both areas are 1 square unit.
In the TOP area, f(x,y) = -1.
In the BOTTOM area, f(x,y) = +1.
(Along the line L f(x,y) = -1 but this does not influence the value of the integration.)
Therefore the integral
∫_R ∫_R f(x,y) dx dy = ∫_R ∫_R f(x,y) dy dx = 0.

What confused me is: Why it is necessary to bring Fubini's theorem into the discussion?

If Fubini's theorem must be discussed, then I offer the following.

Since the order of the two integration variables does not change the zero result, this demonstrates that the order does not matter, which is what Fubini's theorem proves.

The integral isn't zero.

 

[Buzz Bloom]

The integral isn't zero.

Hi @fresh_42:
(12. a) Solution completed 7:10 pm EST 12/28.
FIXED some careless typos 1:21 pm EST 12/30.

Spoiler

I do now see I made a mistake in visualizing the problem. The non-zero domain is a infinite subset of

[(0,0),(0,inf)] x [(0,0,(inf,0)].​

However, I do not see why the integral with respect to y is not zero.
For every value of x, the non-zero range of y is [x,x+2). The value of y is +1 in [x,x+1) and -1 in[x+1,x+2).
Therefore, I see no reason why the following is not correct:

∫_Rf(x,y) dy = G(x) = 0.​

If this is correct then

∫_RG(x)dx = 0.​

OK. So the hint you are telling me is that

∫_Rf(x,y) dx = H(y)≠ 0.​

If fact

(a) for 0<y ≤ 1, H(y) = y,
(b) for 1< y≤2, H(y) = 1-(y-1) = 2-y
(c) for 2 < y, H (y) = 0​

I now need to integrate with respect to y. I can that see it will be > 0.

∫_RH(y)dy = [y²/2]₀¹

+ [2y-y²/2]₁²​

= (1/2) + (4-2-2+(1/2)) = 1​

Now: why isn't this a contradiction to Fubini's theorem?
For this I will need to do some reading:
https://en.wikipedia.org/wiki/Fubini's_theorem#The_Fubini–Tonelli_theorem.
The theorem says:

if X and Y are σ-finite measure spaces​

then the order of integration can be reversed with no change in the resulting double integral value.

However, since the space over which the integration takes place is infinite, this is not a contradiction to the theorem.

Regards,
Buzz

 

[lpetrich]

I will attempt to solve problem 18 again, using a different approach.

Spoiler

There is an important simplification of the problem that we can perform, since $\mathbb{Q}$ is a field. The multiplication identity for the value function yields a division identity. For any x and nonzero y,
$$ |x| = \left|\frac{x}{y} y \right| = \left| \frac{x}{y} \right| |y| $$ giving $$ \left| \frac{x}{y} \right| = \frac{|x|}{|y|} $$
The Archimedean inequality, $|na| > |b|$ thus reduces to $|n| > |b|/|a| = |b/a|$, thus giving the Archimedean condition this form: for all a in $\mathbb{Q}$, there is some nonnegative integer n that satisfies
$$ |n| > |a| $$
The non-Archimedean case has the negation of the Archimedean condition. There is some a in $\mathbb{Q}$ such that for every nonnegative-integer n,
$$ |n| \leq |a| $$
The values of a that satisfy that inequality I will call anti-Archimedean values. It is evident that if a is anyone of them, then |a| is an upper limit for all the possible values of |n|. In fact, the upper bound is the minimum of the |a| values for all the anti-Archimedean a's. It is evident that 0 is not anti-Archimedean, because nonzero n gives nonzero |n|, and that is greater than 0.

Our next task is to find out what kinds of possible values of |n| will be bounded. We can do this with $|n^m| = |n|^m$, something easy to prove with mathematical induction. $|n^m \cdot n| = |n^{m+1}| = |n|^{m+1}$, also $|n^m \cdot n| = |n^m||n| = |n|^{m+1}$. This means that |n| will be bounded only if $|n| \leq 1$.

There is a further constraint that comes from a special value of n. Since $1^2 = 1$, |1| = 1. This also means that 1 is anti-Archimedean.

So in conclusion, |0| = 0, |1| = 1, and for n > 1, |n| <= 1.

One can proceed further by using the prime factorization of each n, and that gives the value of each |n| in terms of the values of |p| for n's prime factors p. In the non-Archimedean case, |p| <= 1 for every prime p.

That also gives the value of |a| for all a in $\mathbb{Q}$, since every rational number is the ratio of two integers. One ought to be able to proceed further with the help of the value function's triangle inequality, $|x + y| \leq |x| + |y|$, but I am unable to do so.

 

[fresh_42]

I will attempt to solve problem 18 again, using a different approach...

This is a good route to go! Just don't restrict yourself too early.

As someone has mentioned earlier, the proof isn't easy. Let me give you some hints.
a) There is a smallest (by its usual ordering) natural number $|n_0| < 1$.
b) $n_0$ has to be prime.
c) Show $|a+b| \leq \max\{|a|,|b|\}$. Hint: Consider $|a+1|^m$.
d) Use the Euclidean division of a number $m$ by $n_0=p$.

You have proven the first statement. The second is an easy consequence. d) will finish the proof, but you need c) for the details in d).

 

[epenguin]

@epenguin I think you have the right idea, but it's a bit hard for me to follow what you've written.

If p(x)=x^3, then p&#039;(0)=p&#039;&#039;(0)=0 and 0 is a turning point according to your definition.I don't understand this. If p(x)=x^4-1, then the point 0 is 'such a point' (p&#039;(0)=p&#039;&#039;(0)=0), but the only zero of p&#039; is zero. Do you mean to say that there are at least two roots of p&#039; counted with multiplicity (because the point is already a zero for p of multiplicity at least 2). Anyways, what if the point a where p&#039;(a)=p&#039;&#039;(a) and p(a)\neq 0 is not in the interval between two roots (is smaller than the smallest root or larger than the largest root)?

Where are you using the assumption p(a)\neq 0? The problem statement is false if you don't assume this.

Sorry, I have hardly been able to be at the site for various reasons, but let me not be like certain of our students who never come back.
I agree my argument was badly expressed and hard to follow, I did not like it.
Now I think we could just say:

Between any two real roots of p there must be a turning point. (1)
Every turning point is a root of p' - but not every root of p'is a turning point, in particular double root of p' is not a turning point.
p has a total of n roots. To show it has two nonreal roots it suffices to show that with the given conditions it cannot have n real roots.
So since p' has a maximum of (n - 1) real roots but two of them are not turning points, it must have maximum of (n - 3) turning points.
Therefore by (1) p has a maximum of (n -2) real roots.
Therefore p has at least two nonreal roots.

In the case that p as well as its first and second derivative are zero at x = a, we have of course a triple real root and can have a total of n real roots and no nonreal ones. We can say this possibility is allowed for in the argument because in this case there is no 'between' as required by (1).

Still a bit of a mouthful. However more obvious and memorable so the way I would want to think I would say than time traveller123's.

 

[Infrared]

@epenguin This is definitely better, but I think you need to be a bit more careful when multiple roots are involved. For example, f(x)=x^3 has no turning points but still has 3 roots with multiplicity.

 

[epenguin]

Well I think it was taken into account in what I said above about between, but maybe the first sentence of the proof could be modified to say "Between any two distinct real roots of p there must be a turning point. (1)".

I don't know how the textbooks exactly say it because I haven't got any beside me at the moment.

In the present exercise since p(a) ≠ 0 , x = a is not a triple (or any other kind of) root of p.

 

[julian]

Is there a typo in question 16? According to the equation given in the question we would have:

$\varphi (\alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3) . x = - x \alpha_3 - y (i \alpha_1 + \alpha_2)$

Should we have:

$\varphi (\alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3) . x = - x \alpha_3 + y (i \alpha_1 + \alpha_2)$

instead?

If I use the former then $[\varphi (\alpha_1 , \alpha_2 , \alpha_3) , \varphi (\alpha_1' , \alpha_2' , \alpha_3')] . x \not= \varphi ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]) . x$ and the total angular momentum-squared operator's eigenvalue is negative. I get $[\varphi (\alpha_1 , \alpha_2 , \alpha_3) , \varphi (\alpha_1' , \alpha_2' , \alpha_3')] . x = \varphi ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]) . x$ and the correct eigenvalue if I use the latter.

 

[fresh_42]

Is there a typo in question 16? According to the equation given in the question we would have:

$\varphi (\alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3) . x = - x \alpha_3 - y (i \alpha_1 + \alpha_2)$

Should we have:

$\varphi (\alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3) . x = - x \alpha_3 + y (i \alpha_1 + \alpha_2)$

instead?

If I use the former then $[\varphi (\alpha_1 , \alpha_2 , \alpha_3) , \varphi (\alpha_1' , \alpha_2' , \alpha_3')] . x \not= \varphi ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]) . x$ and the total angular momentum-squared operator's eigenvalue is negative. I get $[\varphi (\alpha_1 , \alpha_2 , \alpha_3) , \varphi (\alpha_1' , \alpha_2' , \alpha_3')] . x = \varphi ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]) . x$ and the correct eigenvalue if I use the latter.

You have to adjust or modify $\varphi$, simply because the $\sigma_i$ are not elements of the Lie algebra $\mathfrak{su}(2,\mathbb{C})$. We need skew Hermitian matrices, whereas the $\sigma_i$ are not. To get an appropriate homomorphism, we have to consider $\varphi \, : \, \langle (i\sigma_1),(i\sigma_2),(i\sigma_3) \rangle \longrightarrow \mathfrak{gl}(\mathbb{C}_2[x,y])$.

\begin{align*}
\varphi(\alpha_1(i\sigma_1) ,\alpha_2(i\sigma_2),\alpha_3(i\sigma_3))&.(a_0+a_1x+a_2y+a_3x^2+a_4xy+a_5y^2)= \\
&= 1\cdot 0+\\
&+ x\cdot(\alpha_1 a_2 +i\alpha_2 a_2 + i\alpha_3 a_1 )+\\
&+ y\cdot(\alpha_1 a_1 -i\alpha_2 a_1 - i\alpha_3 a_2 )+\\
&+ x^2\cdot(-2\alpha_1 a_4 +2i \alpha_2 a_4 + 2i\alpha_3 a_3 )+\\
&+ xy\cdot(\alpha_1 a_3+\alpha_1 a_5 +i\alpha_2 a_3 -i\alpha_2 a_5 )+\\
&+ y^2\cdot(-2\alpha_1 a_4 -2i\alpha_2 a_4 -2i\alpha_3 a_5 )
\end{align*}

 

[julian]

You have to adjust or modify $\varphi$, simply because the $\sigma_i$ are not elements of the Lie algebra $\mathfrak{su}(2,\mathbb{C})$. We need skew Hermitian matrices, whereas the $\sigma_i$ are not. To get an appropriate homomorphism, we have to consider $\varphi \, : \, \langle (i\sigma_1),(i\sigma_2),(i\sigma_3) \rangle \longrightarrow \mathfrak{gl}(\mathbb{C}_2[x,y])$.

Physicists and mathematicians have different conventions. Physicists write a group element as $g = \exp (i \beta_i T_i)$ where $\beta_i$'s are real parameters and $T_i$ are traceless and hermitian. And say the basis of $\mathfrak{su}(2,\mathbb{C})$ is

$T_1 = {1 \over 2} \sigma_1 , \quad T_2 = {1 \over 2} \sigma_2, \quad T_3 = {1 \over 2} \sigma_3$

with commutation relations

$[T_i , T_j] = i \epsilon_{ijk} T_k .$

Whereas mathematicians write a group element as $g = \exp (\tilde{\beta}_i u_i)$ where $\tilde{\beta}_i$'s are real parameters and $u_i$ are traceless and skew-hermitian. And say the basis of $\mathfrak{su}(2,\mathbb{C})$ is

$u_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} , \quad u_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} , \quad u_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$

with commutation relations

$[u_1 , u_2] = 2 u_3 , \quad [u_2 , u_3] = 2 u_1 , \quad [u_3 , u_1] = 2 u_2 .$

You are using a slightly different basis in your question, you're using:

$u_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

but that doesn't really matter.

\begin{align*}
\varphi(\alpha_1(i\sigma_1) ,\alpha_2(i\sigma_2),\alpha_3(i\sigma_3))&.(a_0+a_1x+a_2y+a_3x^2+a_4xy+a_5y^2)= \\
&= 1\cdot 0+\\
&+ x\cdot(\alpha_1 a_2 +i\alpha_2 a_2 + i\alpha_3 a_1 )+\\
&+ y\cdot(\alpha_1 a_1 -i\alpha_2 a_1 - i\alpha_3 a_2 )+\\
&+ x^2\cdot(-2\alpha_1 a_4 +2i \alpha_2 a_4 + 2i\alpha_3 a_3 )+\\
&+ xy\cdot(\alpha_1 a_3+\alpha_1 a_5 +i\alpha_2 a_3 -i\alpha_2 a_5 )+\\
&+ y^2\cdot(-2\alpha_1 a_4 -2i\alpha_2 a_4 -2i\alpha_3 a_5 )
\end{align*}

You've changed the ordering of the terms. Can I stick to the original ordering: $a_0 + a_1 x + a_2 x^2 + a_3 y + a_4 y^2 + a_5 xy$ as I've already written up stuff? So I should write:

\begin{align*}
\varphi(\alpha_1 (i \sigma_1) +\alpha_2 (i \sigma_2)+\alpha_3 (i \sigma_3))&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
= \varphi( (i \alpha_1) \sigma_1 + ( i\alpha_2) \sigma_2+ (i \alpha_3) \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= x(\alpha_1 a_3 +i\alpha_2 a_3 - i\alpha_3 a_1 )+\\
&+ x^2(-2\alpha_1 a_5 +2 i\alpha_2 a_5 + 2i\alpha_3 a_2 )+\\
&+ y(\alpha_1 a_1 -i\alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ y^2(-2\alpha_1 a_5 -2i\alpha_2 a_5 -2i\alpha_3 a_4 )+\\
&+ xy(\alpha_1 a_2 +\alpha_1 a_4 +i\alpha_2 a_2 -i\alpha_2 a_4 )
\end{align*}

Is this right?

 

[fresh_42]

Physicists and mathematicians have different conventions.

I know. But physicists are not precise here, as their usage is outside the tangent space by a factor i.

You've changed the ordering of the terms.

IIRC I used the numbering of Pauli matrices on Wikipedia, and multiplied them by $i$ to get the actual Lie algebra vectors in order to avoid anti-isomorphisms.

I had thought making the replacement $\alpha_i \rightarrow i \alpha_i$ at the end of the calculations would have a trivial effect but now thinking about it it doesn't.

I have different signs and order such that I get a Lie algebra homomorphism and can use known results about $\mathfrak{sl}(2)$ representations with $H=\sigma_3\,(CSA)\; , \;X=-\dfrac{i}{2}(i\sigma_1)+\dfrac{1}{2}(i\sigma_2)\; , \;Y=-\dfrac{i}{2}(i\sigma_1)-\dfrac{1}{2}(i\sigma_2)$.

Btw., your $u_3$ looks strange.

The proper signs are necessary to actually have Lie algebras and Lie algebra homomorphisms instead of "anti's". It should not lead to a qualitatively different result, but in a way a wrong one. It's a nasty calculation anyway, easy but a bit of work. In case you do not use $[\varphi(X),\varphi(Y)]=\varphi([X,Y])$ you should at least say what you have instead, probably $[\varphi(X),\varphi(Y)]=\pm i \varphi([X,Y])$. But this is strictly speaking no Lie algebra representation. Maybe that's the hidden reason why physicists speak of "generators" instead. This way they avoid referring to tangents and left invariant vector fields.

 

[julian]

I've edited my post to correct $u_3$.

When I said you've changed the ordering of terms I meant the ordering of $1,x,x^2,y,y^2,xy$. In post #1 you wrote

$\varphi( \alpha_1 \sigma_1 +\alpha_2 \sigma_2+ \alpha_3 \sigma_3).(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)$
$\vdots$

but in post #130 you wrote

$\varphi( \alpha_1 (i \sigma_1) + \alpha_2 (i \sigma_2)+ \alpha_3 (i \sigma_3)).(a_0+a_1 x+a_2 y+a_3 x^2+a_4 xy+a_5 y^2)$
$\vdots$

 

[lpetrich]

Referring back to post 125 in this thread, by fresh_42, I was able to do (a) and (b) without much trouble, but I am totally stumped on (c) and (d). I've tried the hints for both of them, but I get nowhere.

 

[bahamagreen]

There is still a problem with the way #7 is proposed.

"How many times a day..." is typically meant to mean how many time per day, through some indefinite series of days, but the replies to questions seem to be stating that the problem period is one day, as if "times a day" meant "how many times in a day" or "how many times in one day".

The difference is that if the answer is a count of events occurring within only one day, that day is longer than if the answer is a rate (count per day) of a series of days... and the answer converges as the series of days increases.

For just one single day the day starts at 12:00 AM and ends at 12:00 AM 24 full hours later.

In a series of days each day's individual "full day" period in the sequence must have one of it's end points excluded.
If you adopt the convention to keep the starting point of time for each day (12:00 AM), then that day must not include the 12:00 AM 24 hours later because that is the starting point of the next day in the series.

This makes a difference in what will be the answer for this problem.

 

[mfb]

as if "times a day" meant "how many times in a day" or "how many times in one day".

What else can it mean?
How many times a day is it 15:00? Once. What is ambiguous about that?

Midnight is not ambiguous on the clock, so we don't have a problem with that time either.

 

[StoneTemplePython]

There is still a problem with the way #7 is proposed.

"How many times a day..." is typically meant to mean how many time per day, through some indefinite series of days, but the replies to questions seem to be stating that the problem period is one day, as if "times a day" meant "how many times in a day" or "how many times in one day".

How many times per day can be interpretted as being the frequency of these 'special times' in one day or the number of distinct 'special times' in one day-- however the problem statement clearly says that we can distinguish between am and pm, so both interpretations map to the same thing.
- - - -
for problem 7: I've been away but I think post 82 has it -- it is set up as a thought experiment, which I like, and the final answer is correct.

 

[fresh_42]

I've edited my post to correct $u_3$.

When I said you've changed the ordering of terms I meant the ordering of $1,x,x^2,y,y^2,xy$. In post #1 you wrote

$\varphi( \alpha_1 \sigma_1 +\alpha_2 \sigma_2+ \alpha_3 \sigma_3).(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)$
$\vdots$

but in post #130 you wrote

$\varphi( \alpha_1 (i \sigma_1) + \alpha_2 (i \sigma_2)+ \alpha_3 (i \sigma_3)).(a_0+a_1 x+a_2 y+a_3 x^2+a_4 xy+a_5 y^2)$
$\vdots$

It's been quite some time that I did the calculations. I guess I've reordered them in a way, that allows me to get a block structure for $\varphi$, i.e. grouped by the basis vectors of the irreducible components.

 

[fresh_42]

Referring back to post 125 in this thread, by fresh_42, I was able to do (a) and (b) without much trouble, but I am totally stumped on (c) and (d). I've tried the hints for both of them, but I get nowhere.

c) Show $|a+b| \leq \max\{|a|,|b|\}.$ Hint: Consider $|a+1|^m.$

$|a+b| \leq \max\{|a|,|b|\}$ is equivalent to $|a+1| \leq \max\{|a|,1\}$ and $|a+1|^m \leq (m+1)\max\{|a|,1\}$ from which the statement follows for $m \to \infty$.

d) Use the Euclidean division of a number $m$ by $n_0=p.$

We write $m=kp+r$, conclude $|m|=1$ for all $m$ coprime to $p$ and finally investigate all other numbers $m=p^s\cdot m'$ and $\dfrac{m}{n}\,.$

 

[julian]

It's been quite some time that I did the calculations. I guess I've reordered them in a way, that allows me to get a block structure for $\varphi$, i.e. grouped by the basis vectors of the irreducible components.

I guess when you originally set the question you didn't want to give any clues away that the irreducible components are organised as $\{1 \}$, $\{x , y \}$, and $\{ x^2 , xy , y^2 \}$ as that was question 16 b).

I'll be adjusting the calculations I did before (by adjusting $\varphi$) to get the appropriate homomorphism later when I'm less preoccupied.

 

[fresh_42]

I guess when you originally set the question you didn't want to give any clues away that the irreducible components are organised as $\{1 \}$, $\{x , y \}$, and $\{ x^2 , xy , y^2 \}$ as that was question 16 b).

I'll be adjusting the calculations I did before (by adjusting $\varphi$) to get the appropriate homomorphism later when I'm less preoccupied.

I actually calculated from behind: took some irreps of $\mathfrak{sl}_2$, wrote them as polynomials, and changed the basis to $\mathfrak{su}_2$ and finally choose the Pauli-matrices ($\sigma_j$) instead of proper Lie algebra vectors $(i\sigma_j$), mixed the basis vectors, in order - yes - not to give away the solution for free, et voilà: the standard example in every book about Lie algebras became a standard example of what physicists use instead. I thought this is an interesting (and simple) example to demonstrate what physicists call "ladder-operators".

 

[TGVF]

I think I found solution for Basics 4:
Let the bug altitude in the evening of day i be noted $z_i$ (in meters).
The tree height in the evening of day n is $L_0 +ir$ with r being the daily rate of tree growth (0.20 m/day).
Then the relative bug position w.r.t. tree top in the evening of day i is $\lambda _i = \frac {z_i} {L_0+ir}$
On the next day morning (day i+1), the bug altitude becomes: $z_i + d = \lambda _i *(L_0+ir) +d$, with d being the nighty rate of climb (0.10 m/night)
On the next day evening (day i+1), the relative bug position will be the same as in the morning: $\lambda _{i+1} = \frac {z_i +d} {L_0+ir} = \lambda _i +\frac {d} {L_0+ir}$. Obviously, its absolute altitude will be higher to uniform tree growth.
The bug will reach the tree top as soon as $\lambda _{i} \geq 1$. Although this can take a long time, we can be sure the bug will succeed because if we consider i forming a continuum (changing it from integer to real), then the relative position is a real function with logarithmic growth: $\lambda(x)= \int \frac {d} {L_0+(x-1)r} \, dx = \frac {d} {r} Ln(1+ \frac {(x-1)r} {L_0})$ which is increasing with x without upper bound.
The number of days for reaching the tree top is obtained by solving the equation $\lambda(x)= 1$. That is: $x= 1+ \frac {L_0}{r}*(e^{\frac {r}{d}} -1)$.
Numerically: $x=3,195.53$ days that is for an integer number of days: 3,196 days. However, the precise solution obtained by discrete summation is only 3,192 days. The actual result is the lower Darboux stepwise sum, as compared to the continuous logarithmic integral which is consequently too large.

 

[lpetrich]

I'll post proofs of parts a and b of fresh_42's proof outline for Problem 18 in post #125.

Spoiler

(a): I had earlier proved that for every positive integer $n > 1$, $|n|$ is at most 1. If for all such $n$, $|n| = 1$, then that is the trivial case. So to be nontrivial, there must be some such $n$ where $|n| < 1$.

Since the positive integers are bounded from below, there must be some minimum value of $n$ that satisfies $|n| < 1$, and since $|1| = 1$, this minimum value must be greater than 1.

(b): To prove that this minimum $n$ must be a prime number, I consider the case of composite $n$, $n = ab$, where both $a$ and $b$ are greater than 1 and less than $n$. If $n$ is this minimum $n$, then $|a|$ and $|b|$ must both equal 1. But $|n| = |ab| = |a| |b| = 1$, which contradicts this premise. Therefore, this minimum $n$ must be a prime number.

 

[lpetrich]

I continue to be stumped by parts c and d of fresh_42's proof outline in post #125, even with his hints in post #139. In particular, I find that the inequality of part c is a very weak constraint, not much stronger than the triangle inequality in the original definition of the value function. It appears to be too weak to prove anything in part d.

 

[fresh_42]

I continue to be stumped by parts c and d of fresh_42's proof outline in post #125, even with his hints in post #139. In particular, I find that the inequality of part c is a very weak constraint, not much stronger than the triangle inequality in the original definition of the value function. It appears to be too weak to prove anything in part d.

You know that there is a minimal prime with $|p|<1$ and that all natural numbers have $|n| \leq 1\,.$ The goal is to find out which numbers are of exactly norm $1\,.$

 

[lpetrich]

You know that there is a minimal prime with $|p|<1$ and that all natural numbers have $|n| \leq 1\,.$ The goal is to find out which numbers are of exactly norm $1\,.$

I can easily prove that for all $n$ with prime factors less than $p$, $|n| = 1$. But if $n$ has prime factors larger than $p$ and is not divisible by $p$, then I am stuck. Expressing $n$ in the form $kp+m$ with $0 < m < p$, I find
$|kp + m| \leq \max(|kp|,|m|) = \max(|kp|,1) = 1$,
and that does not add anything to the non-Archimedeanness of the value function.

 

[fresh_42]

I can easily prove that for all $n$ with prime factors less than $p$, $|n| = 1$. But if $n$ has prime factors larger than $p$ and is not divisible by $p$, then I am stuck. Expressing $n$ in the form $kp+m$ with $0 < m < p$, I find
$|kp + m| \leq \max(|kp|,|m|) = \max(|kp|,1) = 1$,
and that does not add anything to the non-Archimedeanness of the value function.

Have I missed that you have already shown the maximum formula, because you use it?

Anyway, if you have that all numbers are of norm one, which are coprime to $p$, then you also know all others for which $|n|<1$ and even their norm in dependence of $p$. With that, the rest is only a bit technical stuff.

 

[lpetrich]

Have I missed that you have already shown the maximum formula, because you use it?

I don't know what you have in mind, but I must note that $|kp+m| \leq 1$ is not equivalent to $|kp+m| = 1$. Or is there something that I was missing somewhere?

 

[fresh_42]

I don't know what you have in mind, but I must note that $|kp+m| \leq 1$ is not equivalent to $|kp+m| = 1$. Or is there something that I was missing somewhere?

We know $|p| < 1$ and $|n| \leq 1$ and assume for the moment that $|a+b| \leq \operatorname{max}\{\,|a|\, , \,|b|\,\}\,.$

Next we show that $|a+b|=\operatorname{max}\{\,|a|,|b|\,\}$:
Let $|a|<|b|\,.$ Then $|a|<|b|=|(a+b)-a| \leq \operatorname{max}\{\,|a+b|,|a|\,\} =|a+b| \leq \operatorname{max}\{\,|a|,|b|\,\} = |b|\,.$

Thus we have $|m|=1$ for all $m$ which are coprime to $p$. All other numbers are of the form $n=p^r\cdot m$ with $|n|=|p|^r$.

 

[lpetrich]

But if $|a + b| = \max(|a|,|b|)$ for all integers a and b, then |0| = |1 - 1| = max(|1|,|-1|) = max(1,1) = 1, which is contrary to our definition of this value function or norm. So there must be some flaw in the proof of this statement.