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{"id":59364,"date":"2024-09-20T02:46:38","date_gmt":"2024-09-20T11:16:38","guid":{"rendered":"https:\/\/radiocubalibre.live\/los-desconcertantes-numeros-anomalos-ciencia\/"},"modified":"2024-09-20T02:46:38","modified_gmt":"2024-09-20T11:16:38","slug":"los-desconcertantes-numeros-anomalos-ciencia","status":"publish","type":"post","link":"https:\/\/radiocubalibre.live\/los-desconcertantes-numeros-anomalos-ciencia\/","title":{"rendered":"Los desconcertantes n\u00fameros an\u00f3malos | Ciencia"},"content":{"rendered":"


\n<\/p>\n

\n
\"N\u00fameros<\/span>
Simon Newcomb (1835-1909).<\/span>Alamy Stock Photo<\/span><\/figcaption><\/figure>\n

Nuestro comentarista habitual Rafael Granero da las siguientes respuestas a las preguntas de la semana pasada<\/a>:<\/p>\n

Los n\u00fameros cuyos logaritmos decimales est\u00e1n comprendidos entre 0 y 2 son aquellos que se encuentran entre 1 y 100. Esto se debe a que:<\/p>\n

logd(1) = 0<\/p>\n

logd(100) = 2<\/p>\n

Por lo tanto, cualquier n\u00famero x que cumpla 1 < x < 100 tendr\u00e1 un logaritmo decimal entre 0 y 2.<\/p>\n

Logaritmo decimal de 0,01<\/p>\n

El logaritmo decimal de 0,01 es -2.<\/p>\n

Esto se puede entender de la siguiente manera: logd(0,01) = logd(1\/100) = logd(1) \u2013 logd(100) = -2<\/p>\n

Logaritmos decimales de 9, 30 y 1\/3<\/p>\n

Sabiendo que logd(3) = 0,477, podemos calcular los dem\u00e1s:<\/p>\n

Logaritmo decimal de 9: logd(9) = logd(3^2) = 2 * logd(3) = 2 * 0,477 = 0,954<\/p>\n

Logaritmo decimal de 30: logd(30) = logd(3 * 10) = logd(3) + logd(10) = 0,477 + 1 = 1,477<\/p>\n

Logaritmo decimal de 1\/3: logd(1\/3) = logd(1) \u2013 logd(3) = -0,477<\/p>\n

Por su parte, Manuel Amor\u00f3s encuentra de esta ingeniosa manera el valor de x cuando x elevado a la potencia x3 es igual a 3:<\/p>\n

x^(x^3) = 3<\/p>\n

(x^(x^3))^3 = 3^3<\/p>\n

(x^3)^(x^3) = 3^3<\/p>\n

x^3 = y<\/p>\n

y^y = 3^3<\/p>\n

y = 3, ergo x = ra\u00edz c\u00fabica de 3<\/p>\n

Y Bretos Burs\u00f3 propone una interesante interpretaci\u00f3n del n\u00famero e, base de los logaritmos neperianos que, aunque solo es apta para personas con ciertos conocimientos matem\u00e1ticos, no he resistido la tentaci\u00f3n de incluirla:<\/p>\n

Supongamos que vemos c\u00f3mo se forma al azar una fila de personas arbitrariamente larga, y que somos capaces de distinguir siempre, dadas dos personas, cu\u00e1l es la m\u00e1s alta (por peque\u00f1\u00edsima que sea la diferencia). Contamos la cantidad de personas que llegan hasta que la \u00faltima en hacerlo es m\u00e1s alta que la pen\u00faltima (dicha cantidad ser\u00e1 siempre mayor o igual a 2). Entonces:<\/p>\n

– el valor esperado o promedio de esta cantidad variable es el n\u00famero e.<\/p>\n

– la probabilidad de que esa \u00faltima persona sea adem\u00e1s m\u00e1s alta que todas las anteriores es e-2.<\/p>\n

(Cada una de las dos afirmaciones anteriores equivale a que la suma de la serie de 1\/n! para n = 0, 1, 2, 3\u2026 es el n\u00famero e).<\/p>\n