Comments on: How to use LaTeX in comments http://polymathprojects.org Massively collaborative mathematical projects Sat, 28 Apr 2012 14:00:10 +0000 hourly 1 http://wordpress.com/ By: sporepigfish http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5343 Fri, 30 Mar 2012 15:11:40 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5343 parallel and perp does not work

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5239 Thu, 08 Mar 2012 01:57:30 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5239 functional of the probability of pulling a jewel of value v, is

I = \int_0^\infty dv \, p(v) \, ln \, p(v)

The constraints are that the probabilities must sum to one,

1 = \int_0^\infty dv \, p(v)

and that the expected value is 10,

10 = \int_0^\infty dv \, v \, p(v)

Using Lagrange multipliers, we solve for the probability distribution extremizing

\int_0^\infty dv \{ p \, ln \, p + \lambda p + \mu v p \}

The variation of this is

\int_0^\infty dv \, \delta p \{ ln \, p + 1 + \lambda + \mu v \}

which vanishes for

p(v) = p_0 e^{-\mu v}

We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the precise probability distribution consistent with our knowledge:

p(v) = \frac{1}{10} e^{- \frac{v}{10}}

What’s the probability of pulling a jeweled coin valued at least $100?

P(v \geq 100) = \int_{100}^\infty dv p(v) = \int_{100}^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5238 Thu, 08 Mar 2012 01:53:35 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5238 Heh, fail. value v, is

I = \int_0^\infty dv p(v) ln p(v)

The constraints are that the probabilities must sum to one,

1 = \int_0^\infty dv p(v)

and that the expected value is 10,

10 = \int_0^\infty dv v p(v)

Using Lagrange multipliers, we solve for the probability distribution extremizing

\int_0^\infty dv \{ p ln p + \lambda p + \mu v p \}

The variation of this is

\int_0^\infty dv \delta p \{ ln p + 1 + \lambda + \mu v \}

which vanishes for

p(v) = p_0 e^{-mu v}

We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the probability distribution consistent with our knowledge,

p(v) = \frac{1}{10} e^{- \frac{v}{10}}

which we can use to answer questions. What’s the probability of pulling a jeweled coin valued at least $100?

P(v \geq 100) = \int_100^\infty dv p(v) = \int_100^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3690 Mon, 17 Oct 2011 14:41:16 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3690 The task is to find a vector \mathbf{w} which is not parallel to v, since then \mathbf{w} \times \mathbf{v} is orthogonal to \mathbf{v}.

Clearly, any parameter-family which is not contained in the line \mathbb{R} \mathbf{v} will contain such a vector (this is just rephrasing the condition of being not parallel). Therefore, if we manage to find a one-parameter family which is not contained in any line through the origin then this construction will work for arbitrary vectors \mathbf{v}.

The simplest such construction is to simply fix two of the three components of your vector with at least one of them being non-zero (i.e., to choose as your one-parameter family a one-dimensional affine space which does not contain the origin), e.g.,

\mathbf{w} \in \begin{pmatrix}0\\w\\1\end{pmatrix}.

Then given a specific vector \mathbf{v} we only need to choose w in such a way that the last two components of \mathbf{w} are not parallel to those of \mathbf{v}. Suppose that we had failed to do so and the two 2-vectors are in fact parallel. Then

\mathrm{sign}(yz) = \mathrm{sign}(w).

It follows that we can simply choose

w = -\mathrm{sign}(yz)

to ensure that the two vectors \mathbf{v} and \mathbf{w} are non-parallel (one quickly verifies that this also works for yz = 0 as long as one defines \mathrm{sign}(0) = 1).

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3689 Mon, 17 Oct 2011 14:18:24 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3689 \mathbb{R} v

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3688 Mon, 17 Oct 2011 14:17:54 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3688 the line \mathbb{R} v

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3687 Mon, 17 Oct 2011 14:15:01 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3687 the line \mathbf{R} v

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3686 Mon, 17 Oct 2011 14:14:20 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3686 f(x)^2

]]> By: Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3685 Mon, 17 Oct 2011 14:13:58 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3685 $f(x)^2$

]]> By: Mike http://polymathprojects.org/how-to-use-latex-in-comments/#comment-3625 Fri, 30 Sep 2011 20:43:20 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-3625 oops…

$\frac{\partial \phi}{\partial t} = \nabla \cdot \vec{A}$

\frac{\partial \phi}{\partial t} = \nabla \cdot \vec{A}

]]>