## Posts

Showing posts from November, 2018

### Twitter proof: the roots go hand in hand

PtEn< change language In this twitter proof we will have a look at a rather curious, yet simple, property of real polynomials.

Claim: if $p(x) = \sum_{i=0}^n a_ix^i$ is a polynomial with real coefficients, then for all complex numbers $z$, $$p(z) = 0 \iff p(\bar{z}) = 0$$ which means that the complex roots of $p(x)$ come in conjugate pairs.

Twitter proof: it suffices to show that $p(z) = 0 \implies p(\bar{z}) = 0$. Assume that $p(z) = 0$ and recall that $a_i = \bar{a_i}$: \begin{align} p(\bar{z}) &= \sum_{i=0}^n a_i\bar{z}^i \\ &= \sum_{i=0}^n \overline{a_iz^i} \\ &= \overline{\sum_{i=0}^n a_iz^i} = \overline{p(z)} = 0 \end {align} Neste post vamos dar uma olhadela a uma propriedade curiosa, mas simples, dos polinómios com coeficientes reais.

Proposição: se $p(x) = \sum_{i=0}^n a_ix^i$ é um polinómio com coeficientes reais, então para qualquer número complexo $z$ vem $$p(z) = 0 \iff p(\bar{z}) = 0$$ o que significa que as raízes complexas de $p(x)$ vêm em…

PtEn< change language In this post I will show the existence of a family of polynomials that are very useful for interpolation. For that I will use what are known as Lagrange polynomials.

Claim: given $n+1$ pairs $(x_i, y_i)$ with $0\leq i \leq n$ and with $x_i \neq x_j$ whenever $i\neq j$, there exists a polynomial $p(x)$ of degree at most $n$ such that $$p(x_i) = y_i,\ i = 0, \cdots, n$$
Twitter proof: consider the polynomial $$l_i(x) = \prod_{j\neq i} \frac{x - x_j}{x_i - x_j}$$ with $l_i(x_i) = 1$ and $l_i(x_j) = 0$ whenever $j \neq i$. Define $p(x)$ to be $$p(x) = \sum_{i=0}^{n} y_i l_i(x)$$ $p(x)$ has degree at most $n$ because so do the $l_i(x)$ and $p(x_k) = \sum_i y_i l_i(x_k) = y_k$.

In a future post I will show the uniqueness of the polynomial satisfying the constraints in the claim. Neste post vou mostrar a existência de uma família interessante de polinómios, muito útil em interpolação. Para isso vou usar uns polinómios chamados polinómios de Lagrange.…

### Twitter proof: can't touch this (exponential)

PtEn< change language In this twitter proof we will see that no polynomial grows faster than the exponential function.

Claim: the ratio $\frac{x^p}{e^x}$ tends to $0$ as $x$ tends to infinity.

Twitter proof: recall that by the Taylor expansion of $e^x$ we have $$e^x = \sum_{i=0}^\infty \frac{x^i}{i!}$$ and for $x > 0$ we have $$\frac{x^p}{e^x} \leq \frac{x^p}{\sum_{i=0}^{p+1} a_ix^i} \to 0$$ where $a_{p+1} \neq 0$ thus proving our claim.

The way I like to look at this is "if the exponential has a bit of every polynomial inside, then it will grow faster than any fixed polynomial $p(x)$" (because, in particular, the exponential "has a bit" of all other polynomials that have degree higher than that of $p(x)$. Neste post vamos ver que a função exponencial cresce mais depressa que qualquer função polinomial.

Proposição: o rácio $\frac{x^p}{e^x}$ tende para $0$ quando $x$ tende para infinito.

Prova num tweet: sabemos pela expansão de Taylor de $e^x$ qu…

### Twitter proof: irrational high-order roots of 2

PtEn< change language For this twitter proof we will be using a piece of mathematics straight from the 17th century.

Claim: the number $\sqrt[n]{2}$ is irrational for $n \geq 3$.

Twitter proof: suppose that $n \geq 3$ and $\sqrt[n]{2}$ is rational, i.e. $\sqrt[n]{2} = \frac{a}{b}$ for some integers $a, b$. Then taking the $n$-th power of both sides we get $2 = \frac{a^n}{b^n} \iff b^n + b^n = a^n$, contradicting the well-known Fermat's Last Theorem. Para esta prova num tweet vamos usar um pedaço de matemática do século 17.

Proposição: o número $\sqrt[n]{2}$ é irracional para $n \geq 3$.

Prova num tweet: suponhamos que $n\geq 3$ e que $\sqrt[n]{2}$ é racional, i.e. $\sqrt[n]{2} = \frac{a}{b}$ para alguns inteiros $a, b$. Se for esse o caso, elevando os dois lados da igualdade a $n$, obtemos $2 = \frac{a^n}{b^n} \iff b^n + b^n = a^n$, contrariando o famoso Último Teorema de Fermat.
&nbsp&nbsp- RGS

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