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Nehushtan

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About Nehushtan

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  1. 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]
  2. 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.
  3. 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].
  4. 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.
  5. 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.
  6. 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].
  7. 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].)
  8. 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.
  9. Listening on the radio right now to Beethoven’s String Quartet in E Flat, Op 127, performed by the Leipzig String Quartet.
  10. 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.
  11. But won’t you be blowing carbon dioxide rather than oxygen onto the flame?
  12. 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].
  13. 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.
  14. Nehushtan

    pain

    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 …
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