Wikipedia:Reference desk/Archives/Mathematics/2011 April 23

= April 23 =

Log help
This is homework help BUT I have done everthing i could to give myself aid, so now Im asking wikipedia How do you find log(500^500)?Accdude92 (talk) 00:12, 23 April 2011 (UTC)


 * Assuming you can compute logs of smaller numbers, look for a useful formula at Logarithm. PrimeHunter (talk) 00:58, 23 April 2011 (UTC)


 * Still stuckAccdude92 (talk) 01:01, 23 April 2011 (UTC)

log(500^500)=500 log(500) = 500 log(5*100) = 500 [log(5) + log(100)] = 500 [log(5) + log(10^2)] = 500 [log(5) + 2log(10)] = 500 [log(5) + 2] = 500log(5) + 1000

No further simplification is possible without the use of a calculator —Preceding unsigned comment added by 180.216.2.24 (talk) 01:32, 23 April 2011 (UTC)
 * Could you approximate it by means of a taylor series? Widener (talk) 09:16, 23 April 2011 (UTC)
 * Yes, you have $$\ln x = 2\sum_{k=0}^{\infty}\left(\frac{x-1}{x+1}\right)^{2k+1}/(2k+1)$$. It's more efficient to rescale x to be close to 1. -- Meni Rosenfeld (talk) 07:04, 24 April 2011 (UTC)
 * That is, of course, assuming the OP means base 10, and not the natural logarithm; log(5)=log(10/2)=1-log2. =>500log(5) + 1000 = 500(1-log(2)) + 1000 = 1500-500log(2) [or 500{3-log(2)}], not really simpler but probably easier to approximate. Grandiose (me, talk, contribs) 09:35, 23 April 2011 (UTC)
 * So, just to make it clear, still assuming base 10 logarithms, log(500^500) = 500 log(500) = 500·2.7 = 1350, right? If you need more significant digits, then look up log 5 in a table or compute it using a computer, finding that log 5 = 0.69897, so 500 log(500) = 500·2.69897 = 1349.48  &#x2013; b_jonas 16:15, 25 April 2011 (UTC)

0 vector and orthogonality
My understanding of orthogonality roughly matches up to the definition Wikipedia gives: "Two vectors, $$x$$ and $$y$$, in an inner product space, $$V$$, are orthogonal if their inner product, $$\langle x, y \rangle$$, is zero." If $$x$$ is the zero vector, the dot product of x and y would appear to be zero no matter what y is (given of course that x and y are both in the same vector space). Therefore, x and y would be perpendicular and orthogonal. How exactly does this work geometrically / how would one visualize this? There aren't really two right angles formed between y, -y, and x as there normally is with two orthogonal vectors, is there? NW ( Talk ) 03:19, 23 April 2011 (UTC)
 * Good question. Technically the 0 vector is orthogonal to every vector, but I don't think it forms a right angle with anything since it doesn't have direction. Normally a vector has length and direction but for the 0 vector the length is 0 and the direction is undefined. Geometrically, you can think of it as a single point or an arrow from a point to itself. The arrow doesn't go anywhere so it has no direction. In a way this makes the concept orthogonality more convenient that that of right angles since there are fewer exceptions to allow the 0 vector. Also, keep in mind that Euclidean space makes a rather poor model of a general inner product space because in the latter the inner product of a non-zero vector with itself can be zero or negative, so it becomes difficult to define the length of a vector. Anyway, that's just my interpretation so I'm sure there are differing ones.--RDBury (talk) 15:05, 23 April 2011 (UTC)
 * Generally inner products are defined to be positive-definite (i.e. $$\langle v,v\rangle > 0$$ for all v ≠ 0). Rckrone (talk) 01:43, 24 April 2011 (UTC)

simplicial mapping
1) Let |K| be polyhedra with triangulation K.whether this triangulation K and the complex of which |K| is the geometric carrier are same.                                                 2)Let φ be a simlicial mapping from acomplex K into another complex L.Then whether it's extension from the geometric carrier of K to that  of L is a simplicial mapping?If so, please give me the proofMathematics2011 (talk) 08:29, 23 April 2011 (UTC)

Minimum Value
Given that $$ c=\frac{(2m(V_0-E))^{1/2}}{\hbar}$$, that $$k=\frac{(2mE)^{1/2}}{\hbar}$$ and that $$E<V_0$$ how do I go about finding the minimum value of $$V_0$$ such that the equation $$c+k\cot(ka)=0$$? I have tried making $$V_0$$ the subject of the equation and then differentiating with respect to k but this leads only to approximate roots of the transcendtal equation $$1=ka\cot(ka)$$, and my question not only forbids me from using a calculator as well as wording itself so as to suggest an exact answer can be found, and the spurious result of 0, which is rather unhelpful when plugged into the original equation. Any suggestions? Thanks. meromorphic  [talk to me]  14:04, 23 April 2011 (UTC)
 * You say that you seeking the smallest $$V_0$$ that is a root of $$c+k\cot(ka)=0$$; what do you get when you solve for $$V_0$$, and what besides the wording of the question makes you think that there are multiple roots? -- 110.49.251.224 (talk) 17:06, 23 April 2011 (UTC)


 * Solving for $$V_0$$ gives $$V_0=\frac{{\hbar}^2}{2m}k^2{\csc}^2(ka)$$ and differentiating with respect to k gives, I think, $$\frac{dV_0}{dk}=\frac{{\hbar}^2}{m}k{\csc}^2(ka)(1-ka{\cot}(ka))$$. From this, I deduce that either k=0, which doesn't satisfy the original equation, cosec is zero for some value, which it isn't, or else $$1=ka\cot(ka)$$ for which Wolfram Alpha gives more than one solution, none of which are exact. What am I missing? meromorphic   [talk to me]  20:10, 23 April 2011 (UTC)=
 * I'm confused. What is the unknown in the equation $$c+k\cot(ka)=0$$? Why do you want to differentiate wrt k? Is the question perhaps "what is the minimum value of $$V_0$$ for which the equation has a solution in E?" If so, you need to solve for E to get $$E^*$$, and then solve for $$V_0$$ in $$E^*=V_0$$ (and check for the true solution to the problem). -- Meni Rosenfeld (talk) 07:27, 24 April 2011 (UTC)
 * To be honest, I'm confused by the question as well. It may help to know that this is a question in relation to quantum mechanics, specifically solving the Schrödinger equation, where m is a particle's mass, $$\hbar$$ is Plank's reduced constant, $$V_0$$ is the, constant, value of the potential in some range and E is the particle's energy. The question simply states 'find the minimum value of $$V_0$$ such that the equation $$c+k\cot(ka)=0$$ has a solution'. It makes no mention of E, although that does seem like the most likely culprit. I differentiated k initially because I had k squared, which is clearly just a constant multiple of E, and so I thought I could consider k first, find a solution and then just rearrange to find E but I have just realised the square root of E in the expression for k means I would need to differentiate wrt k squared not k. I'll do as you say now. Thank you. meromorphic   [talk to me]  09:49, 24 April 2011 (UTC)
 * I still don't know what differentiation has to do with it. You're not looking for an extremum of $$c+k\cot(ka)$$, you just need it to be 0. -- Meni Rosenfeld (talk) 10:01, 24 April 2011 (UTC)
 * Well, on rearranging, I have $$V_0=\frac{{\hbar}^2k^2{\cot}^2(ka)}{4m}+E$$. Do I now need to let E be the energy eigenvalue of least energy then? Or am I there? Surely I can't be because I haven't identidfied a 'least' value, merely a value? meromorphic   [talk to me]  11:02, 24 April 2011 (UTC)
 * Substitute the value of k, and solve for E in terms of V_0. Then equate the result to V_0. -- Meni Rosenfeld (talk) 14:29, 24 April 2011 (UTC)
 * OK, if that's the way to do it then fine but do you have any idea why they might have used the idea minimum? meromorphic   [talk to me]  14:38, 24 April 2011 (UTC)
 * The procedure I outlined is for finding the minimum. For every value of V_0 there will be a solution E* (I'll assume for simplicity there's just one). For some values of V_0 you will have E* < V_0, and for others you have E* > V_0. You want the value V* which is the infimal value of V_0 for which E* < V_0. By continuity, it is necessary, but not sufficient, that for this value you will have E* = V_0. So you solve E*=V_0 and you get some candidate solutions. You will need to check which of these is really the infimal value of V_0 for which E*R a smooth real valued function on some open set U containing x. Is there a smooth g:M->R that's equal to f on some small neighborhood of x? Money is tight (talk) 22:47, 23 April 2011 (UTC)
 * Yes. In fact you can require that f=g in any open neighborhood V of x whose closure is contained in U. The proof uses a smooth partition of unity subordinate to the open cover U, $$M-\overline{V}$$.  If you just want some neighborhood, you can instead use a cutoff function supported in a coordinate ball around x.  Sławomir Biały  (talk) 23:16, 23 April 2011 (UTC)