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I'm with doG here: 99. The universal answer, 42, just doesn't seem to fit. The question as written in the text would be nice.

What is the definition used in that book for upper bound? If it is something along the lines of given an ordered structure [math](E,\le)[/math] and a set [math]A\subset E[/math], then an element [math]c\in E[/math] is an upper bound of [math]A[/math] if [math]\{x \in E:x > c\} = \Phi[/math], then it should be obvious that every element of [math]E[/math] is an upper (and lower) bound of the empty set. If the set [math]E[/math] is itself unbounded (e.g., the reals), then there is no least upper bound (greatest lower bound).

Those reports are based on the mistaken assumption that just because recently observed sunspot #1030 had the opposite-than-expected polarity meant that it is a solar cycle 23 (or 25) sunspot. Sunspots of the opposite polarity do occur, and they do not necessarily mean they are a member of some other cycle.

The latest prediction (November 2, 2009) from the Solar Physics Group at NASA's Marshall Space Flight Center is for a weak cycle 24 (78 ± 18 sunspot count at solar max), with solar max in April or May 2013. The Sun has started to show signs of activity in the last month. With this slight uptick, the spotless count for 2009 will probably be just a bit less than the count for 2008. This corresponds well with a minimum in December 2008, and that in turn corresponds well with a max in early to mid 2013. We shall see ...

For crying out loud, triclino. Your obsession with "solid proof" has hit a new low (literally). This obsession is severely impeding your ability to learn. I strongly suggest you get over it.

Exactly. Given any real x, there is no member of the empty set that is larger than (or smaller than) x. It is a mathematical equivalent of the sound of one hand clapping.

http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture More interesting, to me, is that the identity [math]27^5 + 84^5 + 110^5 + 133^5 = 144^5[/math] was not found until nearly 200 years after Euler made his conjecture, even though the numbers involved are rather small.

http://en.wikipedia.org/wiki/Empty_set#Extended_real_numbers

Correct. That is the work energy theorem. (1) Incorrect, and (2) non sequitur. Regarding (1) incorrect: The relevant equation here is the Tsiolkovsky rocket equation, [math]\frac{\Delta v}{v_e} = \ln\left(\frac {m_r+m_f}{m_r}\right)[/math] where [math]\Delta v[/math] is the change in velocity experience by the rocket, [math]v_e[/math] is the exhaust velocity, [math]m_r[/math] is the mass of the rocket after achieving the specified delta-V, and [math]m_f[/math] is the mass of the fuel expended in achieving that delta-V. The problem at hand: Given that propelling a rocket of final mass [math]m_1[/math] to some velocity [math]\Delta v_1[/math] requires a quantity of fuel [math]m_{f1}[/math], how much fuel [math]m_{f2}[/math] is needed to make a rocket with a mass [math]m_1/4[/math] to twice the final velocity? The answer is [math]m_{f2} = \frac 1 2\,m_{f1} + \frac 1 4\,\frac{m_{f1}^2}{m_1}[/math] Regarding (2) non sequitur: Novak is ignoring the energy in the exhaust. Wrong math => wrong conclusion. Pure nonsense. mv has units of momentum, not energy. You quoted a crackpot, Kurt. I wrote a tutorial on rocket dynamics for another physics site. See http://www.physicsforums.com/showthread.php?t=199087. Bottom line: Someone saying along the lines of "About ninety percent of physics is corrupted by the error" is almost certainly a crackpot. Don't get sucked in by their nonsense.

Assume [math]c>0[/math] has two distinct positive real roots, call them [math]a[/math] and [math]b[/math]: [math]a^2=b^2=c[/math]. Since these roots are distinct, one will be smaller than the other. Denote this lesser root [math]a[/math]: [math]0<a<b[/math]. Multiplying both sides of an inequality by the same positive number does not change the nature of the inequality. Multiplying both sides of [math]a<b[/math] by [math]a[/math] yields [math]a^2<ab[/math] while multiplying both sides by [math]b[/math] yields [math]ab<b^2[/math]. As less than is a transitive relationship, [math]a^2<b^2[/math], and this contradicts the initial assumption that [math]a^2=b^2[/math].

??? I would argue the other way around: We haven't made any significant advances in symbolic AI in the past 30 years. It is GOFAI, not computational intelligence, that is at a dead end.

Isn't it more than a bit self-evident in this case? Some kindly advice: If you wish to progress in your mathematics education (self-education?) it would behoove to drop this obsessive-compulsive behavior of yours regarding formality. It is heeding your progress.

That derivation is fine. What is your objection?

The name of the image file, MKeclipse_mukensnable.jpg, pretty much gives it away. The photographer's name is Alex Mukensnable. The MK stands for Mauna Kea. What's left? Eclipse. The picture is of a lunar eclipse as seen from the top Mauna Kea.

One can pick any point on an object to be the center of rotation. Given that fixed point, the combined translational and rotational motion of the object can always be described in terms of a translation of this fixed point plus a rotation about some axis passing through this fixed point. (For that matter, the fixed point does not even a point not on an object!) This is a direct consequence of Euler's rotation theorem. In other words, there is nothing wrong per se with choosing to look at the system as continuing to rotate about the pre-release pivot point. There is however a very good reason for choosing the center of mass as that special fixed point. For an unconstrained, constant mass rigid body, the translational and rotational equations of motion decouple when the motion is viewed as a translation of the center of mass plus a rotation about some axis passing through the center of mass. This is not the case for any other point. For example, looking at things with the pivot point viewed as the fixed point is going to lead to a mess. If no external forces act on the system, the center of mass will move in a straight line once released. The motion of the pivot point is much more complex: it circles about a straight line path. Add in external forces and things become really messy. That said, there are cases where choosing the center of mass leads to a mess. When the pin is attached the mathematics are much simpler with the pivot point chosen to be the fixed point. The pin constrains the motion, so this does not qualify as an unconstrained, constant mass rigid body. Another example is a launch escape system that takes a capsule away from the exploding rocket underneath the capsule. The launch escape system rockets burn at an incredible rate to make the capsule accelerate away from (translational acceleration) and out of the way (rotational acceleration) of the main rocket. The launch escape system plus capsule is not a constant mass rigid body. In fact, the location of the center of mass of system moves within the system. The inertia tensor is not constant. No matter how you slice it, the translational and rotational equations of motion are tightly coupled. Some prefer to analyze this system from the perspective of a fixed point on the system as opposed to from the perspective of the center of mass.

The thing you must do is to pick one reference frame and stick with it throughout. In terms of the diagram you recently added to the original post, I will use the following reference frame: The origin is at the pivot point, The [math]\hat x[/math] axis points to the left, The [math]\hat y[/math] axis points toward the top of the screen, and The [math]\hat z[/math] axis is coming out of the screen. In this frame, the linear momentum, angular momentum and kinetic energy before the pin is released are [math]\aligned \vec p_- &= -m\omega_-x\hat y + m\omega_-(l-x)\hat y \\ &= m\omega_-(l-2x)\hat y \\ \vec L_- &= mx^2\omega_-\hat z + m(l-x)^2\omega_-\hat z \\ &= m(l^2-2lx+2x^2)\omega_-\hat z \\ E_- &= \frac 1 2 mx^2\omega_-^2+\frac 1 2m(l-x)^2\omega_-^2 \\ &= \frac 1 2m(l^2-2lx+2x^2)\omega_-^2 \endaligned[/math] After the pin is released, the bar+masses system will move up or down the screen, depending on the location of the center of mass with respect to the pivot point. Denoting the velocity of the center of mass after the release as [math]\vec v_{cm}[/math], the linear momentum of the bar+masses system after the release is [math]\vec p_+ = 2m\vec v_{cm}[/math] To conserve linear momentum, [math]\vec p_+ = \vec p_-[/math] and thus [math]\vec v_{cm} = \frac{\vec p_+}{2m} = (l/2-x)\omega_-\hat y[/math] The angular momentum after the release is the angular momentum due to the motion of the system as a whole plus the angular momentum due to rotation about the center of mass: [math] \vec L_+ = 2m\vec r_{cm}\times \vec v_{cm} + \frac 1 2 ml^2\omega_+\hat z [/math] The center of mass of the bar+masses system is [math]\vec r_{cm} = \frac 1{2m}(-mx\hat x + m(l-x)\hat x) = (l/2-x)\hat x[/math] Applying this to the expression for the post-release angular momentum yields [math]\aligned \vec L_+ &= 2m((l/2-x)\hat x)\times ((l/2-x)\omega_-\hat y) + \frac 1 2 ml^2\omega_+\hat z \\ &= \left(2m(l/2-x)^2\omega_- + \frac 1 2 ml^2\omega_+\right) \hat z \endaligned [/math] Equating this to the pre-prelease angular momentum, [math]\left(2m(l/2-x)^2\omega_- + \frac 1 2 ml^2\omega_+\right) \hat z = m(l^2-2lx+2x^2)\omega_-\hat z[/math] Solving for [math]\omega_+[/math] [math]\aligned \frac 1 2 ml^2\omega_+ &= m\left((l^2-2lx+2x^2)-2m(l/2-x)^2\right)\omega_- \\ &= \frac 1 2 ml^2\omega_- \endaligned[/math] Thus [math]\omega_+ = \omega_-[/math]. In other words, the angular velocity does not change. Is energy conserved? The post-release kinetic energy is the kinetic energy due to the translation of the center of mass plus the rotational kinetic energy. Since there is no change in angular velocity, I'll use [math]\omega[/math] sans the + and - subscripts. [math] \aligned E_+ &= \frac 1 2 \left(2m v_{cm}^2 + \frac 1 2 ml^2 \omega^2\right) \\ &= \frac 1 2 m\left(2 (l/2-x)^2+\frac 1 2 l^2\right)\omega^2 \\ &= \frac 1 2 m(l^2-2lx+2x^2)\omega^2 \endaligned[/math] This is the pre-release kinetic energy, so kinetic energy is conserved, as expected.

You are switching reference frames here. In particular, the origin of the frame in which you computed L' is shifted from the origin of the frame in which you computed L by x-l/2. These are two different frames of reference, so angular momentum is not the same in the two frames.

Emphasis mine: I see three possible meanings to phrase "the system spontaneously has the axis shifted to the centre." The masses somehow magically teleport to some other location so that "the axis is shifted to the centre." In this case, you are dealing with a unreal situation. Conserved quantities aren't conserved, but big deal. The situation is not real. You switch the frame of reference so that "the axis is shifted to the centre." Linear momentum, angular momentum, and energy are frame dependent quantities. Think of it this way. Consider a single particle moving at some velocity with respect to an observer. To this observer, the particle will have non-zero linear momentum and non-zero kinetic energy. To an observer who is co-moving with the particle, the particle has zero linear momentum and zero kinetic energy. Something entirely different from the above two interpretations. What exactly do you mean by the highlighted phrase?

Not in this case. It was the OP who misconstrued the article to being about silk. From the cited article, "The ultra-strong glue that spiders use to trap their prey has given up some of its genetic secrets, raising the hope that similar substances could one day be synthesised to produce surgical adhesives." The article does mention the spider web, but only in the context of spiders secrete this glue onto the web.

Read what ajb said again: ajb was nice. I'll be blunt. Nobody is going to read that huge block of text. It is unreadable.

There are about 1079 atoms in the visible universe (see here and here). Suppose that every atom in the observable universe is magically converted to 14C. How many atoms of 14C would you expect to find after 65 million years of decay? 65 million years is 11,343 times the half life of 14C. As [math]10^{79} \approx 2^{263}[/math], after 65 million years of decay, one would expect to see 2-11080 atoms of 14C in this magically converted universe. In short, after 65 million years of decay, there is absolutely nothing left.

Trurl: Here's a clue as to how to solve Bhavin's question: Don't bother. Whatever answer you say it is, Bhavin will say: Nope, wrong answer. He will then demonstrate that yours is not the answer he had in mind by producing a triangle in which one side has a length of 30, another has a length of 8, and third side has a length different from what you answered.

C-14 has a half-life of 5730 years. It's all gone in less than 100,000 years.

triclino, this looks like homework. Are you asking how to prove this, or are you posing this as a puzzle to which you already know the answer?

How do you know that doing this preserves the inequality? Yes. Multiplying both sides of an inequality of the form x>0 by any positive number b does preserve the inequality. In other words, if x>0 and b>0 then xb>0. Similarly, if x>0 and b≥0, then xb≥0. There is a particular non-negative number that transforms 1/a > 0 to a ≥ 0. Now all you have to deal with is the nasty case a=0. a must be non-zero because 1/0 is undefined.