u/Limp_Ordinary_3809

I live in Nicosia and I’ve been doing a lot of night walks and exploring the old town, buffer zone edges, Pedieos park, etc. I’m looking for the stuff that doesn’t show up in guides:

Strange or forgotten places

Weird little spots (a door, a courtyard, a wall, a view)

Unusual things you’ve witnessed in the city (events, encounters, odd scenes)

I’m less interested in typical tourist recommendations and more in: ‘One time I saw/found X in Nicosia and it stuck with me.’

If you’re comfortable, please mention roughly where (neighborhood or landmark) so I can check it out on foot.

I’m curious if anyone knows whether SNNs are actually used in practice on edge devices. I understand that they use much less power, but are mismatched to chips like esp32, so there is no way to use it in practice, and dedicated chips like loihi are not intended for prototyping or consumer products.

I’m entering Year 11 of a 13-year system, so I have two selection cycles before university applications and three before I age out. Based in Cyprus.

Where I’m starting from:

—Sat the national multiple-choice competition three times (25 questions, 60 minutes, negative marking). Peaked around 10 correct. Never entered the actual team-selection pipeline, which I only recently learned is separate.
—No olympiad-specific training. Some independent university-level maths: working through Spivak’s calculus, so real analysis proofs rather than competition technique.
—Aged out of the junior track, so I’d be up against people with years of prep.

The path here is: provincial competitions in November—> national competition in December (top ~10 per grade advance)—> four selection tests during a preparation course, five eliminated per test—> six-person team.

My plan is roughly 10 hours a week for two years — about 1,000–1,400 hours — using AoPS Volume 2, then Engel, Evan Chen’s geometry book, and past BMO and Greek olympiad papers, with weekly timed cold papers to fix a specific problem: I do fine untimed and collapse on contest day.

Questions:
1**.** Is 1,000–1,400 hours a realistic budget for team selection in a small country from this starting point, or am I without a chance?
2**.** For those who made a national team starting late what did the path actually look like? How long before scores improved?
3**.** Is the contest gap fixable in two years, and what specifically worked?
4.** **Is my book sequence right, or should I be somewhere else given the timeline?

I've been working through Calculus by Spivak to learn calculus, and I really enjoy doing the problems. I have finished Part II, and am eager to get on to Integrals and Derivatives, however currently I am working through the problems of chap 7 & 8. Henceforth, should I continue doing all of the problems? I enjoy them, but the starred ones can take me several days of work and iterative correction to solve. I try to contribute 4 hours of my day to these problems, but progress feels slow.

Theorem. 

There is no continuous function f, defined on ℝ, such that every value f attains is attained exactly twice.

PROOF

Setup

 f(0) is some value. By hypothesis it is attained exactly twice; call these points a, b with a < b, so f(a) = f(b).

Step 1 — Dichotomy on (a,b). For x ∈ (a,b), f(x) cannot equal f(a): that would be a third occurrence of f(a), contradiction. So f(x) is strictly above or below f(a) for each such x. Suppose both occurred — some x₁ ∈ (a,b) with f(x₁) > f(a), some x₂ ∈ (a,b), x₂ ≠ x₁, with f(x₂) < f(a). By IVT applied between x₁ and x₂, f equals f(a) at some point strictly between them, hence in (a,b) — a third occurrence, contradiction. So f is entirely above f(a) or entirely below it on (a,b).

WLOG f(x) > f(a) for all x ∈ (a,b) (otherwise replace f by −f, which is continuous and has the same "exactly twice" property).

Step 2 — Interior maximum. By EVT on [a,b], f attains a maximum M at some c ∈ (a,b). Since f > f(a) on the interior and f(a) = f(b) at the endpoints, M > f(a), and c is interior (not an endpoint).

Step 3 — Outside is forced below f(a). Suppose some x₄ ∉ [a,b] has f(x₄) ≥ f(a). It can't equal f(a) — third occurrence. So f(x₄) > f(a). Pick v with f(a) < v < min(f(x₄), M). By IVT on [a,c], f attains v at some q₁ ∈ (a,c); by IVT on [c,b], f attains v at some q₂ ∈ (c,b); q₁ ≠ q₂ since the intervals are disjoint. By IVT between x₄ and the nearer endpoint of [a,b] (where f = f(a) < v, while f(x₄) > v), f attains v at some third point outside [a,b]. Three occurrences of v — contradiction. Hence f(x) < f(a) for all x ∉ [a,b].

Step 4 — M is the global maximum. For x ∈ [a,b], f(x) ≤ M by definition of M. For x ∉ [a,b], f(x) < f(a) < M by Step 3. So f(x) ≤ M for all x ∈ ℝ.

Step 5 — M's second occurrence lies in (a,b). M is attained (at c), so by hypothesis it's attained exactly twice; let c′ ≠ c be the other point. By Step 3, c′ cannot lie outside [a,b] (there f < f(a) < M), and it can't be a or b (there f = f(a) < M). So c′ ∈ (a,b). WLOG c < c′.

Step 6 — Dichotomy at (c,c′). For x ∈ (c,c′), f(x) > M is impossible since M is the global maximum (Step 4); f(x) = M is impossible since c, c′ are M's only two occurrences. So f(x) < M for all x ∈ (c,c′); in particular, pick any x₅ ∈ (c,c′), with f(x₅) < M.

Step 7 — Triple occurrence, contradiction. Take v with max(f(x₅), f(a)) < v < M. By IVT:

• on [a,c]: f goes from f(a) up to M, so f = v at some p₁ ∈ (a,c);
• on [c,x₅]: f goes from M down to f(x₅), so f = v at some p₂ ∈ (c,x₅);
• on [x₅,c′]: f goes from f(x₅) back up to M, so f = v at some p₃ ∈ (x₅,c′).

These three intervals are pairwise disjoint, so p₁, p₂, p₃ are distinct. Thus v is attained at least three times. But v is attained, so by hypothesis it must be attained exactly twice. Contradiction.

Therefore no such f exists. ∎

QUESTION

a) Suppose f is continuous on [a,b] and let (x,0) be a point on the horizontal axis. Prove that there is a point on the graph of f which is closest to (x,0); that is, prove that there is some y∈[a,b] such that the distance from (x,0)to (y,f(y)) is less than or equal to the distance from (x,0) to (t,f(t)) for every t∈[a,b].

b) Show that this same assertion is not necessarily true if [a,b] is replaced by (a,b) throughout.

c) Show that the assertion is true if [a, b] is replaced by R throughout.

PROOF OF (c)

Fix x and define D(t)=sqrt{(x-t)^2+f(t)^2} on R. Since D(t)>=|x-t| and |x-t|—> infty as t—>infty, we have D(t)—>infty in both directions. Let M = D(x). Because D—>infty, there exists a<x<b with D(t)>M for all t∉[a,b]. On the closed interval [a,b], D is continuous (composition of continuous functions, f continuous), so by EVT it attains a minimum at some y ∈ [a,b]:D(y)<=D(z) ∀z ∈ [a,b]. Since x ∈ [a,b], in particular D(y)<=D(x)=M. Now, take any t ∈ R. If t ∈ [a,b] then D(y)<=D(t) by minimality. If t∉[a,b], then D(t)>M>=D(y). Either way, D(y)<=D(t), so (y,f(y)) is a closest point.

QUESTION

a) Suppose f is continuous on [a,b] and let (x,0) be a point on the horizontal axis. Prove that there is a point on the graph of f which is closest to (x,0); that is, prove that there is some y∈[a,b] such that the distance from (x,0)to (y,f(y)) is less than or equal to the distance from (x,0) to (t,f(t)) for every t∈[a,b].

b) Show that this same assertion is not necessarily true if [a,b] is replaced by (a,b) throughout.

PROOF (of (b))

Let f(x)=x on (0,1) and take the point (0,0). For t∈(0,1) D(t) = sqrt{(t-0)^2 + f(t)^2} = sqrt{2t^2} = t sqrt2, which is strictly increasing. Given any y∈(0,1), let t=y/2∈(0,1). Then D(t)=y/2*sqrt2<y*sqrt2=D(y), so y is not a closest point. Since y was arbitrary, no closest point exists.

THEOREM

Suppose f is continuous on [a,b], and let x be any number. Prove that there is a point on the graph of f which is closest to (x,0); in other words there is some y in [a,b] such that the distance from (x,0) to (y,f(y)) is ≤ the distance from (x,0) to (z,f(z)) for all z in [a,b].

PROOF

Let x be fixed and define g:[a,b]→R

g(t)=f(t)^2+(x−t)^2

the squared distance from (x,0)to (t,f(t))

Since f is continuous, g is continuous (sums and products of continuous functions). By the Extreme Value Theorem, g attains a minimum at some y∈[a,b], so g(y)≤g(z)for all z∈[a,b]. As sqrt is increasing, y also minimizes the actual distance. qed.

THEOREM:

Suppose that f is a continuous function with f (x) &gt; 0 for all x, and lim as x approaches infty f (x) = 0 = lim as x approaches -infty f (x). (Draw a picture.) Prove that there is some number y such that f(y) &gt;= f(x) for all x.

PROOF:

The picture looks like a lump, where the ends never meet the floor.

Let M=f(0)>0. Since f(x)→0 as x→±∞ & M>0, ∃ a<0<b | f(x)<M ∀x∉[a,b].

f is continuous on the compact [a,b]∋0, so by EVT ∃ y∈[a,b] with f(y)≥f(x) ∀x∈[a,b]. In particular f(y)≥f(0)=M.

Now ∀x: if x∈[a,b], f(x)≤f(y); if x∉[a,b], f(x)<M≤f(y). Either way f(x)≤f(y). ■

THEOREM: Let f be any polynomial function. Prove that there is some number y such

that lf(y)I <= lf(x)I for all x.

PROOF:
We will split this proof into two conditions.

1. Constant polynomials in the form f(x)=c.
2. Non constant polynomials with a degree of at least 1

As all polynomials fall into these two categories, this proof applies to all polynomials.

Let g(x)=|f(x)|.

Condition 1 — Constant polynomials
If f is constant, then f(x)=c, additionally, |f(x)|=|c|, for all x, thus there is a y which satisfies |f(y)|=|f(x)| for all x, which also satisfies |f(y)|≤|f(x)|, thus proving that this applies to condition 1.

Condition 2 — Non constant polynomials
Non constant polynomials are in the form f(x)=xⁿ+aₙ₋₁xⁿ⁻¹+...+a₀. Since we apply the absolute function, the limit as x approaches negative infinity, and positive infinity is infinity: lim_{x→∞} g = lim_{x→-∞} g = ∞.
Using this information, we will construct an interval and apply the Extreme Value Interval to find a local minimum, and thus a global minimum across the domain of the polynomial, which is R.
First, we can choose a c∈R such that g(c)=M. Since g approaches infinity on both sides, there is a cutoff point a, with a<c: ∀x<a, g(x)>M. Similarly, there is a point b with c<b: g(x)>M ∀x>b. g is continuous on [a,b] as [a,b]∈R and [a,b] is non empty as c∈[a,b], so we can apply the EVT on this interval, giving us a point y, such that ∀x∈[a,b], g(y)≤g(x).

Conclusion for part 2.
For x<a, g(x)>M. For x>b, g(x)>M. For x∈[a,b], g(y)≤g(x), and since c∈[a,b], g(y)≤g(c)=M.
Thus, across all three regions, the domain of |f(x)|, there is a minimum |f(y)|≤|f(x)| for all x.

Also, even if this is correct, if possible, can someone tell me if there is a cleaner way. It felt uncomfortable to split it into two conditions, but I couldn't find another way.

I'm working on a project involving LoRa, and I'd like to use an RMF95, and to experiment with it on a breadboard. Unfortunately, it seems that it is incompatible with a standard breadboard, as it's holes have 2mm spacings. I tried to find a breakout adapter board online so that it can be used on a breadboard, but with no luck. Are there any other methods to use an RMF95 on a breadboard?

I'm working on a project involving LoRa, and I'd like to use an RMF95, and to experiment with it on a breadboard. Unfortunately, it seems that it is incompatible with a standard breadboard, as it's holes have 2mm spacings. I tried to find a breakout adapter board online so that it can be used on a breadboard, but with no luck. Are there any other methods to use an RMF95 on a breadboard?

\documentclass{amsart}

\usepackage{amsmath, amssymb, amsthm}

\newtheorem{theorem}{Theorem}

\begin{document}

\begin{theorem}

Suppose $f$ is continuous on $(a,b)$, and

\[

\lim_{x \to a^+} f(x) = +\infty, \qquad \lim_{x \to b^-} f(x) = +\infty.

\]

Then $f$ attains a minimum on $(a,b)$; that is, there exists $x_0 \in (a,b)$ such that

$f(x_0) \leq f(x)$ for all $x \in (a,b)$.

\end{theorem}

\begin{proof}

Let $c \in (a,b)$ be any point, and set $M = f(c)$.

\medskip

\noindent\textbf{Choosing cut-off points.}

Because $\lim_{x \to a^+} f(x) = +\infty$, there exists $\alpha$ with $a < \alpha < c$ such that

\[

f(x) > M \quad \text{for all } x \in (a, \alpha].

\]

Because $\lim_{x \to b^-} f(x) = +\infty$, there exists $\beta$ with $c < \beta < b$ such that

\[

f(x) > M \quad \text{for all } x \in [\beta, b).

\]

\medskip

\noindent\textbf{Applying the Extreme Value Theorem.}

Since $f$ is continuous on $(a,b)$ it is in particular continuous on the closed interval

$[\alpha, \beta] \subset (a,b)$. Moreover $c \in [\alpha, \beta]$, so the interval is

non-empty. By the Extreme Value Theorem, $f$ attains its minimum on $[\alpha, \beta]$:

there exists $x_0 \in [\alpha, \beta]$ such that

\[

f(x_0) \leq f(x) \quad \text{for all } x \in [\alpha, \beta].

\]

In particular, $f(x_0) \leq f(c) = M$.

\medskip

\noindent\textbf{Conclusion.}

We verify that $x_0$ is a global minimum on $(a,b)$ by checking each sub-interval:

\begin{itemize}

\item For all $x \in (a, \alpha)$: $f(x) > M \geq f(x_0)$.

\item For all $x \in [\alpha, \beta]$: $f(x) \geq f(x_0)$ (by the choice of $x_0$).

\item For all $x \in (\beta, b)$: $f(x) > M \geq f(x_0)$.

\end{itemize}

Since $(a,b) = (a,\alpha) \cup [\alpha,\beta] \cup (\beta,b)$, we conclude that

$f(x_0) \leq f(x)$ for all $x \in (a,b)$. Hence $f$ has a minimum on $(a,b)$,

attained at $x_0$.

\end{proof}

\end{document}

Suppose\ that\ \phi\ is\ continuous\ and\ \displaystyle \lim_{x \to \infty }\phi(x)/x^{n} = 0 = \displaystyle \lim_{x \to -\infty } \phi(x)/x^{n}.

Prove\ that\ if\ n\ is\ even\, then there is a number y such that y^{n}+\phi(y)\leq x^{n} + \phi(x) for all x.

PROOF: Let f(x)= x^{n} + \phi(x). First, we will show that the x^{n} term dominates. Since \lim_{x \to \infty }\phi(x)/x^{n} = 0 = \displaystyle \lim_{x \to -\infty } \phi(x)/x^{n}, there is some \left | x\right |\geq M such that

\left | \phi(x)/x^{n}\right | \leq 1/2. This implies that x^{n}/2\leq f(x). Let b>0 be a number such that b^{n}\geq 2f(0) and b^{n}\geq M. Then, if we have x\geq b, we have

f(x)\geq x^{n}/2\geq b^{n}/2\geq f(0). Similarly, if x\leq -b, then f(x)\geq x^{n}/2\geq (-b)^{n}/2\geq f(0).

Now we apply the extreme value theorem on the function f on the interval [-b,b]. We conclude that there is a number y such that (1)if -b\leq x\leq b, f(y)\leq f(x). (2)f(y)\leq f(0), so f(x)

\geq f(0)\geq f(y). Putting 1 and 2 together, we find that f(x)\geq f(y) for all x. QED.

................ ................ ................ ................ ................ ................ ................

Suppose that ϕ is continuous and lim⁡x→∞ϕ(x)/x^n=0=lim⁡x→−∞ϕ(x)/x^n

Prove that if n is odd, then there is a number x such that x^n+ϕ(x)=0

PROOF: Let g(x)=x^n+ϕ(x)

For sufficiently large positive x, we can find ϕ(x)/x^n>-1. Thus, x^n+ϕ(x)>0, or g(x)>0.

For sufficiently large negative x, we can find ϕ(x)/x^n>-1. Thus, x^n+ϕ(x)<0, because x^n<0.

Since there exists a g(x)>0, and g(x)<0, by the Darboux property of continuous functions, there must exist some x such that g(x)=0, or x^n+ϕ(x)=0.

QED.

Also, is there a way to write notation on reddit? Its difficult and painful to read and write it in this form.

...................................

Hey,

I got frustrated that every AI tool for coding (Claude Code, Cursor, etc.) completely skips over Roblox developers. You can't point those tools at a Studio project — Studio is a closed ecosystem, scripts don't live on disk in a normal way, nothing external can reach in.

So I built RobloxClaw. It's a plugin that adds a chat panel to Studio. You talk to it, it reads your scripts, edits them, creates new ones, searches your codebase, and reads the output log when something breaks. No copy-pasting. No describing your file structure from memory. It actually knows your project.

What it can do:

• Read any script in your game
• Write and edit scripts (surgically — find and replace a block, not rewrite the whole file)
• Create new Scripts, LocalScripts, ModuleScripts
• Search across all your scripts for a function or variable
• Read the Output log to help debug errors
• Create/rename/delete any instance

Every change it makes is wrapped in undo history — Ctrl-Z works on everything the AI does.

How it works:

Two pieces — a Python server that runs on your computer and talks to an AI model (via OpenRouter, so you pick the model), and a Lua plugin inside Studio that executes the commands. Studio can't receive incoming connections so the plugin long-polls the server. The whole bridge is about 100 lines of code.

To use it:

1. Drop plugin.lua in your Plugins folder
2. Run python main.py in a terminal
3. Add your OpenRouter API key to a .env file
4. Start chatting

Works with any model on OpenRouter — Claude, GPT-4o, Gemini, whatever.

GitHub: https://github.com/nikolask11/scriptblox

Open source, MIT license. Would love feedback, bug reports, or contributions. Especially curious if anyone runs into issues with more complex projects.