Wikipedia:Reference desk/Archives/Mathematics/2010 March 4

= March 4 =

Simplifying an expression involving radicals
Hello Math Reference Desk. I'm continuing to wrangle with algebraic numbers of small degree. I've got this horrible function:

$$f\left(x\right)=\frac{x^2-2x+4+\sqrt{2\left(-4x^2+13x+14\right)}}{10}$$,

and I know that $$f\left(\sqrt[3]{2}\right) = 1$$. I just can't figure out how to carry out the simplification. I can equivalently get the same result with:

$$f\left(x\right)=\frac{x^2-2x+4+2\sqrt{13x^{-2}+7-2x^2}}{10}$$,

but I don't know if that helps. It still doesn't look square under that radical, even with even powers. Maple simplifies it to 1, but doesn't say how. This feels like a dumb question, but oh well. Thanks in advance for any insights. -GTBacchus(talk) 04:30, 4 March 2010 (UTC)
 * If what you want is to prove that $$f\left(\sqrt[3]{2}\right) = 1$$, you can write $$f(x)=a$$ and solve for a (by moving to one side and squaring). You'll get an equation for a for which one solution is 1 and the other you need to show is invalid (more detail: You get to $$\mathrm{stuff}=\sqrt{\mathrm{radicand}}$$ which is true iff $$\mathrm{stuff}\ge0$$ and $$\mathrm{stuff}^2=\mathrm{radicand}$$. The latter turns out to be $$-1-4a-5a^2-2r+2ar+r^2-ar^2=0$$. $$a=1$$ satisfies both conditions).
 * A technique that might generalize a bit better is to let $$\sqrt{2\left(-4r^2+13r+14\right)} = ar^2+br+c$$, where $$r=\sqrt[3]{2}$$, and solve for a, b and c. You get $$2ac+b^2=-8,\ a^2+bc=13,\ 4ab+c^2=28$$ which should be solvable with some work. -- Meni Rosenfeld (talk) 08:40, 4 March 2010 (UTC)


 * I equivalenced the sides so I got the radical should equal $$6+2x-x^2$$. Squaring that and substituting for x3 gave something near to what was under the radical sign except I got $$2(-4x^2+7x+14)$$. Dmcq (talk) 09:40, 4 March 2010 (UTC)
 * I had to rush off and did my sums wrong. It does come out to the original expression under the radical. I somehow missed out a factor of two in the product for the 6 and 2x when I should have got 24x. Rushing is not good idea for doing things right. Dmcq (talk) 14:06, 4 March 2010 (UTC)
 * That is the right approach. Starting from
 * $$(x^2-2x-6)^2 = x^4 - 4x^3 - 8x^2 + 24x + 36$$
 * when $$x = \sqrt[3]{2}$$ we have
 * $$x^4 - 4x^3 - 8x^2 + 24x + 36 = 2x - 8 - 8x^2 + 24x + 36 = -8x^2 + 26x + 28$$
 * so
 * $$\sqrt{2\left(-4x^2+13x+14\right)} = \pm(x^2 - 2x - 6) \text{ for } x = \sqrt[3]{2}$$
 * and, taking the positive valued root, we have
 * $$\frac{x^2-2x+4+\sqrt{2\left(-4x^2+13x+14\right)}}{10} = \frac{(x^2-2x+4) - (x^2-2x-6)}{10}=1 \text{ for } x = \sqrt[3]{2}$$
 * Gandalf61 (talk) 11:05, 4 March 2010 (UTC)

Awesome. Proving that $$x=\sqrt[3]{2}$$ is a solution, I could already do. This technique for getting there makes perfect sense, and I can generalize it to similar questions. Thank you all very much! -GTBacchus(talk) 15:39, 4 March 2010 (UTC)

Vectors
I originally asked this question in the science reference desk, but they suggested that I repost it here.

To my current understanding, a "quantity with direction" isn't necessarily a vector. It must also satisfy certain criteria, such as commutativity under addition and rotational invariance (are these the only ones?). With this in mind, how would one prove that angular velocity is a vector? I was already able to show that infinitesimal rotations commute under addition, implying that ω = dθ_x/dt i + dθ_y/dt j + dθ_z/dt k should likewise commute under addition. My problem is showing that it's invariant under rotation. Secondly, how could one prove that the cross product is a vector? Thanks. —Preceding unsigned comment added by 173.179.59.66 (talk) 15:40, 4 March 2010 (UTC)


 * If you're trying to prove that angular velocity is a vector, you need to start by stating your definition of what angular velocity is. If the angular velocity of an object is defined so that it is an element of a vector space, then angular velocity is a vector. If it is defined in some other way, it is not a vector. Everything will come down to the definition. If you are asking about the physics definition of a "vector" (which is an entirely different concept), you may want to look at Pseudovector. &mdash; Carl (CBM · talk) 16:34, 4 March 2010 (UTC)


 * I'm asking about the physics definition of a vector. The article discusses the issues I'm talking about (covariance and such), but I don't know how to apply that to the problem at hand... —Preceding unsigned comment added by 173.179.59.66 (talk) 20:21, 4 March 2010 (UTC)


 * Start with the formula in terms of the cross product
 * $$\boldsymbol\omega=\frac{\mathbf{r}\times\mathbf{v}}{|\mathrm{\mathbf{r}}|^2}$$
 * Applying a proper rotation is the same as multiplying the velocity and position vectors by some matrix R of determinant 1. It seems to me that
 * $$\frac{R\mathbf{r}\times R\mathbf{v}}{|\mathrm{R\mathbf{r}}|^2}

=\frac{\mathrm{det}(R)\cdot R(\mathbf{r}\times\mathbf{v})}{\mathrm{det}(R)^2|\mathrm{\mathbf{r}}|^2} = \frac{R(\mathbf{r}\times\mathbf{v})}{|\mathrm{\mathbf{r}}|^2} = R(\boldsymbol\omega)$$
 * However, I have only thought about this for 30 seconds, and I am not very knowledgeable about physics. Try asking your professor. &mdash; Carl (CBM · talk) 13:43, 5 March 2010 (UTC)


 * I'm not too sure that would work, because that equation isn't a definition; it's derived from the assumption that ω is a vector. —Preceding unsigned comment added by 173.179.59.66 (talk) 20:54, 5 March 2010 (UTC)


 * That takes you back to my original comment: If you're trying to prove that angular velocity is a vector, you need to start by stating your definition of what angular velocity is. &mdash; Carl (CBM · talk) 21:17, 5 March 2010 (UTC)


 * Well, it's clear that a scalar is inadequate to describe angular velocity. We could have rotation in any combination of three independent planes (call them x = 0, y = 0, z = 0 in a Cartesian coordinate system if you wish).  That implies that we need three numbers to describe angular velocity; that is, a vector.  Alternatively, the more intuitive view of it is nicer—we define it such that the vector points out of the plane of rotation.  Of course, there's an ambiguity in this (you could pick the vector pointing in the opposite direction), so by convention we pick one of them.  This is why angular velocity is a pseudovector, and can pick up obnoxious sign changes upon reflections. — Zazou 00:05, 6 March 2010 (UTC)

Small lie brackets and commutators
Is there an easy way to prove that if the Lie bracket of two vector fields $$[X,Y]$$ is small then also the commutator of the correspondig flows $$\phi_X^t\phi_Y^s-\phi_Y^s\phi_X^t$$ is small for fixed s and t?--Pokipsy76 (talk) 18:36, 4 March 2010 (UTC)

Random walk
Given a simple random walk, s0 = 0, and with probability of 1/2 either sn+1 = sn + 1 or sn+1 = sn - 1, is there a simple expression, or an asymptotic expression (for large n) for the probability that no negative number occurred after n steps? I don't really know how to count the paths in an efficient way. Thanks in advance. Icek (talk) 18:57, 4 March 2010 (UTC)
 * The combinatorial argument escapes me as well, so I went with a more analytical approach: Denote $$p_{i,n}=\mathrm{Pr}(s_0,\ldots,s_n\ge0|s_0=i)$$. Then $$p_{-1,n}=0,\ p_{i,0}=1, p_{i,n}=(p_{i+1,n-1}+p_{i-1,n-1})/2$$. Numerically it can be observed that for $$n=2m,2m-1\,\!$$ we have $$p_{0,n}=4^{-m}\binom{2m}{m}$$. With Stirling you get $$p_{0,n} \sim \sqrt{\frac{2}{\pi n}}$$. -- Meni Rosenfeld (talk) 19:21, 4 March 2010 (UTC)


 * Check Bertrand's ballot theorem.--Pokipsy76 (talk) 19:25, 4 March 2010 (UTC)