2=1 (de 6*9 = 42)

Por LordPato March 31st, 2011, Categorías: Varios

6*9=42

  • Queremos Probar:

    6*9=42

  • Analicemos el 9…

    6*(8+1)=42

  • Como 1=2, 8+1=8+2=10

    6*10=42

  • 6*10=60=24+2*18, entonces

    24+2*18=42

  • pero 2=1

    24+18=42 :D

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Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Por LordPato February 22nd, 2011, Categorías: Varios

Publicado en la lista de MENSA Argentina. Quise mantener el idioma original.

Clever student:

I know!

x^{0} =  x^{1-1} = x^{1} x^{-1} = \frac{x}{x} = 1.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

0^{x} = 0^{1+x-1} = 0^{1} \times 0^{x-1} = 0 \times 0^{x-1}= 0

which is true since anything times 0 is 0. That means that

0^{0} = 0.

Cleverest student :

That doesn’t work either, because if x=0 then

0^{x-1} is 0^{-1} = \frac{1}{0}

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function x^{x} and see what happens as x>0 gets small. We have:

\lim_{x \to 0^{+}} x^{x} = \lim_{x \to 0^{+}} \exp(\log(x^{x}))

= \lim_{x \to 0^{+}} \exp(x \log(x))

= \exp( \lim_{x \to 0^{+} } x \log(x) )

= \exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )

= \exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )

= \exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )

= \exp( \lim_{x \to 0^{+} } -x )

= \exp( 0)

= 1

So, since  \lim_{x \to 0^{+}} x^{x} = 1, that means that 0^{0} = 1.

High School Teacher:

Showing that x^{x} approaches 1 as the positive value x gets arbitrarily close to zero does not prove that 0^{0} = 1. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that 0^{0} is undefined. 0^{0} does not have a value.

Calculus Teacher:

For all x>0, we have

0^{x} = 0.

Hence,

\lim_{x \to 0^{+}} 0^{x} = 0

That is, as x gets arbitrarily close to 0 (but remains positive), 0^{x} stays at 0.

On the other hand, for real numbers y such that y \ne 0, we have that

y^{0} = 1.

Hence,

\lim_{y \to 0} y^{0} = 1

That is, as y gets arbitrarily close to 0, y^{0} stays at 1.

Therefore, we see that the function f(x,y) = y^{x} has a discontinuity at the point (x,y) = (0,0). In particular, when we approach (0,0) along the line with x=0 we get

\lim_{y \to 0} f(0,y) = 1

but when we approach (0,0) along the line segment with y=0 and x>0 we get

\lim_{x \to 0^{+}} f(x,0) = 0.

Therefore, the value of \lim_{(x,y) \to (0,0)} y^{x} is going to depend on the direction that we take the limit. This means that there is no way to define 0^{0} that will make the function y^{x} continuous at the point (x,y) = (0,0).

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function f(x,y) = y^x for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

y^x := 1 \times y \times y \cdots \times y

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

y^{x} = 1 \times y.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

y^{0} = 1

which holds for any y. Hence, when y is zero, we have

0^0 = 1.

Look, we’ve just proved that 0^0 = 1! But this is only for one possible definition of y^x. What if we used another definition? For example, suppose that we decide to define y^x as

y^x := \lim_{z \to x^{+}} y^{z}.

In words, that means that the value of y^x is whatever y^z approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use y^z in our definition of y^x, which seems to be recursive. The reason it is okay is because we are working here only with z>0, and everyone agrees about what y^z equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

0^0 = \lim_{x \to 0^{+}} 0^{x} = \lim_{x \to 0^{+}} 0 = 0

Hence, we would find that 0^0 = 0 rather than 0^0 = 1. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what y^x means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is 0^0 really? Well, for x>0 and y>0 we know what we mean by y^x. But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of y^x is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what y^x means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that 0^0=1? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use 0^0=0 or if we say that 0^0 is undefined. For example, consider the binomial theorem, which says that:

(a+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}

where \binom{x}{k} means the binomial coefficients.

Now, setting a=0 on both sides and assuming b \ne 0 we get

b^x

= (0+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}

= \binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots

= \binom{x}{0} 0^0 b^{x}

= 0^0 b^{x}

where, I’ve used that 0^k = 0 for k>0, and that  \binom{x}{0} = 1. Now, it so happens that the right hand side has the magical factor 0^0. Hence, if we do not use 0^0 = 1 then the binomial theorem (as written) does not hold when a=0 because then b^x does not equal 0^0 b^{x}.

If mathematicians were to use 0^0 = 0, or to say that 0^0 is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using 0^0 = 1.

There are some further reasons why using 0^0 = 1 is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely helpful.

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MD5 y MySQL

Por Dracu December 1st, 2010, Categorías: Programación

Justo ayer me di cuenta que se puede hacer esto:

select * from tabla where md5(campo)=valor

Bastante util…

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Ruleta Rusa 1.0

Por LordPato November 23rd, 2010, Categorías: Humor, Linux

dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM

Ejecutar en produccion ;)

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Icono de una época

Por Dracu October 28th, 2010, Categorías: Varios

Lord Pato esta semana se recibió de:

Que el lo explique.

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El Humor …

Por LordPato October 26th, 2010, Categorías: Humor

Puede alguien explicarme la division celular?
o
0
8
oo

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Celulares Touch prohibidos para los Monk

Por Dracu October 21st, 2010, Categorías: Tecnología, Varios

Según un estudio de la Universidad de Stanford, Un celular táctil junta más gérmenes que un baño público.  Si usted es un fóbico de las bacterias, gérmenes y pestes malignas aléjese del IPhone (y si no también, quién en su sano juicio quiere uno?)

La nota de MinutoUno (Aguante CHICHE!!!!)

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Microsoft, creador de usuarios así:

Por Dracu October 20th, 2010, Categorías: Humor, Varios

Yo entiendo que Windows se enfoca al mercado de los usuarios sin conocimientos, pero esto es como mucho:

LINK

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Lacuna Expanse

Por LordPato October 20th, 2010, Categorías: Juegos

Un nuevo mmo web… la verdad se pasa para lo que estoy acosutmbrado, tiene tintes a otros conocidos pero es bastaaaante mas fluido que otros, no tenes que esperar tanto y podes administrar tus recursos de maneras bastante entretenidas para siempre tener algo para hacer.

Desde colonizar, o ser neutro, o estar a la defensiva… todo queda en ustedes… se los recomiendo.
Entren en www.lacunaexpanse.com y compruebenlo :D

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Entrando a Mensa (1)

Por LordPato October 11th, 2010, Categorías: Varios

No se por que… pero teniendo amigos mensos (miembros de Mensa) me tiraron la onda de tomar el examen de admision.
No es que crea en las diferencias entre gentes de diferente IQ (esta comprobado que la inteligencia abstracta no es lo mismo que la inteligencia emocional, u otras inteligencias que hay por ahi) … pero al enterarme la clase de “nerds” que hay , me picó la curiosidad.

Sin mas fui el pasado viernes y me di cuenta que dentro de los examenes online que hay pululando… si bien es del mismo estilo, no lo es su dificultad. Nada de dificultad creciente… todo igual de dificil (o sea dificiles todas).

Si me costó?. Si, me costo, no me esperaba que todo fuera igual en el nivel de dificultad, asi como tambien el hecho de que hubo un par de ejercicios que no los entendi (y probablemente no los llegue a entender).

Bueno… entre una y dos semanas me daran el resultado el cual publicare… nomas para ver si lo logre o no jaja, no voy a ser ni mas inteligente ni mas tonto por entrar o no. Fue divertido el examen mas alla de su dificultad.

Al ser por aplicacion, hay mucha gente que no se presenta nunca y tranquilamente podria entrar. Asi que… si tienen ganas… animense tambien.

Saludos

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