Image credits: https://en.m.wikipedia.org/wiki/Proof_without_words
Francesco Polizzi
Un blog matematico
01 giugno 2024
06 maggio 2024
Proofs without words 5
For every natural number $n$, the quantity $n^2-n$ is divisible by $2$.
Credits: Benjamin Dickman's answer in MO6045702 aprile 2024
Proofs without words 4
Viviani's Theorem: In an equilateral triangle, the sum of the distances of an interior point to the three sides equals the altitude of the triangle.
Image from the web, attribution unknown |
(Image from the web, attribution unknow
06 gennaio 2024
An overkill proof of the divergence of the harmonic series
The following proof can be found in the MO thread [1].
Assume that the harmonic series $\sum_{n=1}^{+ \infty}\frac{1}{n}$ converges. Then the sequence of functions $\{f_n\}$ defined by $f_n = \frac{1}{n} \chi_{[0, \, n]}$ is dominated by the function $$g = \chi_{[0, \, 1]} + \frac{1}{2} \chi_{[1, \, 2]}+\frac{1}{3} \chi_{[2, \, 3]}+ \frac{1}{4} \chi_{[3, \, 4]}+\ldots,$$ which is by assumption absolutely integrable over $\mathbb{R}$.
By applying Lebesgue's Dominate Convergence Theorem [2] we get: $$1 = \lim_n \int_{\mathbb{R}} f_n(x) dx = \int_{\mathbb{R}} \lim_n f_n(x)dx = 0,$$ a contradiction. $\Box$
References
[1] https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts
[2] https://en.wikipedia.org/wiki/Dominated_convergence_theorem
04 dicembre 2023
Four colors suffice
[2] Wolfgang Haken (1928-2022), Memorial Tribute, Notices of the AMS 70, Number 9 (2023). DOI: https://doi.org/10.1090/noti2781
05 novembre 2023
Automorphisms of the real numbers
Theorem. The only field automorphism of $\mathbb{R}$ is the identity.
31 ottobre 2023
21 ottobre 2023
A proof of the Fundamental Theorem of Algebra based on Differential Topology
Theorem. Every non-constant polynomial $P \in \mathbb{C}[z]$ must have a zero.
Proof. Using the stereographic projection from the point $\infty=(0, \, 0, \, 1)$, we can extend the polynomial function $P \colon \mathbb{C} \to \mathbb{C}$ to a function $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ such that $f(\infty)=\infty$. A straightforward use of the change of coordinates $z=1/w$ shows that $f$ is smooth on the whole of $\mathbb{S}^2$, including the point $\infty$.
Let us remark that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ has a finite number of critical points (and so a finite number of critical values), corresponding to the zeros of the derivative $P’$ and possibly to the point $\infty.$
If $0\in \mathbb{C}$ corresponds to a critical value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ then, by definition, it has a preimage via $P \colon \mathbb{C}^2 \to \mathbb{C}^2$, and there is nothing to prove.
We may therefore assume that $0\in \mathbb{C}$ corresponds to a regular value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$. The locus of these regular values is the complement of finitely many points on the sphere, so it is connected.
From the Implicit Function Theorem for differentiable manifolds, it follows that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ is a local diffeomorphism when restricted to the locus of regular values. So the number of points which are preimages of a fixed one is locally constant on such locus, hence constant (being the locus connected).
Since $\deg(P) \geq 1$, there are surely (infinitely many) points $y \in \mathbb{S}^2$ such that $f^{-1}(y) $ is non-empty. Thus, the same must be true for the point corresponding to $0 \in \mathbb{C}.$
In other words, $0$ is in the image of $P$. $\square$
12 ottobre 2023
An overkill way to recover the Taylor expansion of $e^x$
Let us consider the linear operator of Banach spaces $$F \colon C[0, \, 1/2] \to C[0, \, 1/2], \quad F(f(x)):= \int_0^x f(t)dt.$$ Then the identity $\int_0^x e^tdt=e^x-1$ can be rewritten as $$(I-F)(e^x)=1. \quad (\sharp)$$ The operator $F$ is bounded and we have $$||F|| = \sup_{f \in C[0, \, 1/2]} \frac{||F(f)||}{||f||} \leq \frac{1/2 \, ||f||}{||f||} = \frac{1}{2} <1.$$ This implies that $I-F$ is invertible as a bounded linear operator and moreover $$(I-F)^{-1} = \sum_{k=0}^{+ \infty} F^k.$$ Substituting in $(\sharp)$, we get $$e^x=(I-F)^{-1}(1)=\left(\sum_{k=0}^{+ \infty} F^k \right) (1) = \sum_{k=0}^{+ \infty} F^k(1) = \sum_{k=0} ^{+ \infty}\frac{x^k}{k!},$$ which is the well-known Taylor series expansion of $e^x$.