u/Limp_Ordinary_3809

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.