Mathspp: mathematics and programming

Pt En 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...

Pt En 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...

Pt En 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...

Pt En 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 join the mathspp mailing list