Nehushtan

You can also try: \begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix} [latex]\begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix}[/latex]

The domain contains only two elements. This makes the problem extremely simple. If you let the domain be [latex]\{a,b\}[/latex] you can test any statement, say [latex]P(x)[/latex], by considering the statements [latex]P(a)[/latex] and [latex]P(b)[/latex]. If both of them are true, [latex](\forall x)\left(P(x)\right)[/latex] is true; if at least one of them is true, [latex](\exists x)\left(P(x)\right)[/latex] is true; and so on.

Hint (hope I’m allowed to post one). Let the speed of the bead at the given position be [latex]v[/latex]. Then you can calculate [latex]v^2[/latex] in two ways: Using energy considerations, you can express the kinetic energy [latex]\frac12mv^2[/latex] in terms of [latex]m,g,r,\omega,u[/latex]. The centripetal force [latex]\frac{mv^2}r[/latex] can be expressed in terms of [latex]m,g,r,\omega,R[/latex] where [latex]R[/latex] is the reaction of the wire on the bead. These equations should allow you to eliminate [latex]v[/latex] and solve for [latex]R[/latex].

Did you read what I wrote? I said: The IVT is about continuous functions, not uniformly continuous functions. You are trying to apply the definition of uniformly continuous function to the IVT – which is not what the theorem is about. That’s why you’re not getting anywhere closer towards understanding the proof.

That’s the definition of uniform continuity, not continuity. For continuity, [latex]\delta[/latex] depends not only on [latex]\epsilon[/latex] but also on each point at which the function is continuous. The intermediate-value theorem applies to continuous rather than uniformly continuous functions.

Why? [latex]\delta[/latex] is something we can choose here (due to the continuity of the funciton at [latex]c[/latex]) and we choose it to be less than [latex]b-c[/latex].

If [latex]c-\delta[/latex] is not an upper bound of [latex]S[/latex] there would be an element [latex]x\in S[/latex] such that [latex]c-\delta<x[/latex]. Now [latex]x\in S[/latex] implies [latex]x\leqslant c[/latex] (because [latex]c[/latex] is an upper bound) and [latex]f(x)\leqslant u[/latex] (definition of the set [latex]S[/latex]). But that part of the proof has shown that [latex]f(x)>u[/latex] for all [latex]x\in(c-\delta,\,c+\delta)[/latex], in particular for all [latex]x\in(c-\delta,\,c][/latex]. Hence [latex]c-\delta[/latex] must be an upper bound. Yeah, some details have been skipped here. The assumption here is that [latex]f(c )<u<f(b)[/latex]. Hence [latex]c<b[/latex] so [latex]\delta[/latex] can be taken small enough such that [latex]c+\delta<b[/latex]. For all [latex]x\in S[/latex], [latex]a\leqslant x\leqslant b[/latex]. So [latex]b[/latex] is an upper bound of [latex]S[/latex]; since [latex]c[/latex] is the least upper bound, [latex]c\leqslant b[/latex]. Hence [latex]a\leqslant x\leqslant c\leqslant b[/latex], i.e. [latex]a\leqslant c\leqslant b[/latex]. The strictness of the inequality signs follows from the fact that [latex]u=f( c)[/latex] is strictly between [latex]f(a)[/latex] and [latex]f(b)[/latex]. (NB: It is enough to have [latex]a\leqslant c\leqslant b[/latex] for the proof to proceed. When the proof is complete, it will follow that [latex]a<c<b[/latex].)

Where did you get [latex]\mathrm{450\ N}[/latex] from? The weight of the object (force exerted on it due to gravity) is [latex]\mathrm{500\ N}[/latex] taking [latex]g=\mathrm{10\ m\,s^{-2}}[/latex]; this is NOT the force the object would exert on the ground on impact. The latter depends (for a given object) on the object’s final velocity, which in turn depends on the height from which it is dropped. Hope this is clear.

Listening on the radio right now to Beethoven’s String Quartet in E Flat, Op 127, performed by the Leipzig String Quartet.

Correct.

Wrong definition to use. The principle that pressure increases with depth applies to the pressure exerted by a fluid on an object immersed in the fluid. Blood pressure on the other hand is the pressure exerted by blood on the walls of blood vessels due to the circulation of blood. Try the other physics definition of pressure: pressure = force/area. Thus blood pressure depends not only on the pumping action of the heart (force) but also on the size of the blood vessels: blood pressure is greater in blood vessels with smaller cross-sectional areas. This is why people with high blood pressure are given vasodilator drugs: these relax the walls of blood vessels, increasing their cross-sectional areas and lowering blood pressure.

But won’t you be blowing carbon dioxide rather than oxygen onto the flame?

If you have proved that [latex]\lim_{x\to0}\frac{\ln(1+x)}x=1[/latex] (which can be done by L’Hôpital’s rule) then you can use it to deduce the given limit. Hint: Let [latex]y=3x[/latex] and write the denominator in terms of [latex]y[/latex].

The group being considered is most likely [latex]C_3\times C_5\times C_{11}[/latex] (or a group that contains this group as a subgroup). An element in the group is not a vector; it’s just represented by an ordered triple. The subgroups [latex]\{(c,0,0):c\in C_3\}[/latex], [latex]\{(0,c,0):c\in C_5\}[/latex] and [latex]\{(0,0,c):c\in C_{11}\}[/latex] are isomorphic to [latex]C_3[/latex], [latex]C_5[/latex] and [latex]C_{11}[/latex] respectively, so the individual cyclic groups can be thought of as being embedded in the group in question.

William Shakespeare certainly thought so. In Act 3 Scene 1 of Measure for Measure he wrote: And the poor beetle that we tread upon In corporal sufferance finds a pang as great As when a giant dies. But then Shakespeare was no biologist …

Well, it looks like your substitution works perfectly after all. I must have made a careless mistake and didn’t get the [latex]v^2x[/latex] to cancel, ending up by going through a slightly more complicated path to solve the ODE. Just as well you didn’t take my advice.

Or hypervolume.

The answer to the second question is yes, provided the mass is constant. But the answer to the first question is no. The integral of force with respect to time is impulse, not work. To find work, you must express force as a function of distance and integrate with respect to distance.

No, [latex]y=vx[/latex] doesn’t work here. Did you find the integrating factor I told you to find earlier in the thread? In other words, multiply the expression [latex]2yy'-\frac2xy^2[/latex] by a function [latex]u(x)[/latex] of [latex]x[/latex] only such that [latex]u(x)\cdot2yy'-u(x)\frac2xy^2\,=\,\frac{\mathrm d}{\mathrm dx}\!\left(u(x)y^2\right)[/latex] [latex]u(x)[/latex] is called an integrating factor. (Hint for finding it: Clearly you want [latex]u'=-\frac2xu[/latex].) When you’ve found [latex]u(x)[/latex], the substitution you require for the original problem should be obvious.

Hopping is anything but an unpractical method of locomotion for a kangaroo – especially given the animal’s evolutionary adaptation in this respect: http://en.wikipedia.org/wiki/Kangaroo#Adaptations Compared with running, hopping enables kangaroos to cover large distances quickly and without expending too much energy. And I’ve never heard of waves “hopping” before.

You’re welcome. Solving differential equations often requires thinking outside the box. If a particular equation does not fit into any classifed category, it doesn’t mean it can’t be solved – you may need to try some other method.

Okay, suppose point C is a fixed distance [latex]d[/latex] from point A, which has co-ordinates [latex](a_1,a_2,a_3)[/latex]. Point C will lie on the surface of the sphere [latex](x-a_1)^2+(y-a_2)^2+(z-a_3)^3=d^2[/latex]. The line on which C lies will be described by the equation of a line in [latex]\mathbb R^3[/latex]. So you just need to solve the two equations to find the point(s) of intersection. (Note: Depending on the actual values, there may be two, one, or no points of intersection.)

Some time ago I read Antimatter by Frank Close. Excellent book on the topic. I’m also trying to read The Beat of a Different Drum, a biography of Richard Feynman by Jagdish Mehra.

Well, the most obvious special property is that the function [latex]e^x[/latex] is its own derivative and antiderivative (indeed the only such functions are given by [latex]f(x)=Ae^x[/latex] where [latex]A[/latex] is a constant). This property makes it extremely useful in many areas of mathematics, for example techniques for solving differential equations. The trigometric and hyperbolic functions for complex numbers are related to those for real numbers via [latex]e[/latex]. Many other important properties of [latex]e[/latex] are listed in the Wikipedia article on it.

The differential equation is clearly not homogeneous and I don’t see what homogeneity has got to do with the solution. Hint: Write the equation as [latex]2yy'-\frac{2}xy^2=2x^3[/latex] and find an integrating factor for the left-hand side. This will tell you exactly what substitution you need to use.