Wikipedia:Reference desk/Archives/Mathematics/2010 November 15

= November 15 =

(delta, epsilon)
How could I show that $$\lim_{x \to 0} \frac{e^x-1}{x}=1$$ using delta-epsilon? L'Hopital's Rule is out of the question. 24.92.78.167 (talk) 02:16, 15 November 2010 (UTC)


 * What definition of e^x are you allowed to use? There are quite a few--see characterizations of the exponential function for several. 67.158.43.41 (talk) 04:21, 15 November 2010 (UTC)


 * Exactly. It's quite a tricky one this. If the OP wants a nice εδ-proof then s/he needs to show that for each real ε > 0 there exists a real δ > 0 such that for all x with &thinsp;x&thinsp; we have
 * $$ \left| \frac{e^x - 1}{x} - 1 \right| < \varepsilon . $$
 * If you try to solve this inequality explicitly to give yourself an ε then you need the Lambert W function which, in this case, is as useful as a chocolate fire guard. The exponential function is an entire function, so we might as well jump to the power series instead:
 * $$ e^z = \sum_{k=0}^{\infty} \frac{z^k}{k!} = 1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \cdots . $$
 * (Notice I changed from x to z, meaning I am working over the complex plane and not just the real line.) Doing a little bit of algebra shows us that
 * $$ \frac{e^z - 1}{z} = \sum_{k=1}^{\infty} \frac{z^{k-1}}{k!} = 1 + \frac{1}{2}z + \frac{1}{6}z^2 + \frac{1}{24}z^3 + \cdots . $$
 * The rest is straight forward, we find that
 * $$ \lim_{z \to 0} \left( \frac{e^z - 1}{z} \right) = \lim_{z \to 0} \left( \sum_{k=1}^{\infty} \frac{z^{k-1}}{k!} \right) = 1 . $$
 * One could apply the εδ-argument to the above power series, but there really is no point. — Fly by Night  ( talk )  21:01, 15 November 2010 (UTC)
 * It should be mentioned that this argument relies on the fact that $$\sum_{k=1}^{\infty} \frac{z^{k-1}}{k!}$$ converges for all z in a neighborhood around 0. It's fairly clear that this is true since (ez-1)/z is defined and finite everywhere except at 0, but in general it's something to be careful about of when you want to reverse the order of infinite summation and taking a limit. Rckrone (talk) 01:46, 16 November 2010 (UTC)
 * Thanks for mentioning that. I didn't mention it because the series has an infinite radius of convergence. But it was worth mentioning. Thanks Rckrone. — Fly by Night  ( talk )  16:15, 16 November 2010 (UTC)


 * It follows quite easily from the differential equation definition of the exponential function, i.e. $$\frac{d}{dx} e^x = e^x$$ and e^0 = 1. I didn't suggest this at first since the equivalences can be more involved than your question, but now that another proof has been given, I'll suggest mine. With the differential equation definition, we can take a 3-term Taylor series about 0 to get
 * $$e^x = 1 + x + \frac{x^2}{2} + \frac{e^{\xi_x}}{6} x^3$$
 * for each x for some $$0 < |\xi_x| < |x|$$ using the Lagrange form of the remainder term as discussed in Taylor's theorem. The equation is then
 * $$\lim_{x \to 0} \frac{e^x-1}{x} = \lim_{x \to 0} \frac{x + \frac{x^2}{2} + \frac{e^{\xi_x}}{6} x^3}{x} = \lim_{x \to 0} \left(1 + \frac{x}{2} + \frac{e^{\xi_x}}{6} x^2 \right)$$
 * Since e^x is differentiable, it is continuous, so $$e^{\xi_x}/6$$ can be taken arbitrarily close to e^0 = 1 by taking x small enough to force $$\xi_x$$ small enough. So we may split up this three-term limit into a sum of three limits, and we may further split the final term into the product of two limits. This gives
 * $$\lim_{x \to 0} 1 + \lim_{x \to 0} \frac{x}{2} + \frac{1}{6} \lim_{x \to 0} e^{\xi_x} \lim_{x \to 0} x^2$$
 * $$ = 1 + 0 + 1*0 = 1$$.
 * If you're sadistic enough to use epsilon-delta the entire way instead of relying on other properties, you can unwind the epsilons and deltas used in the proofs of splitting limits over sums and products and starting the entire process with a delta found with the continuity of e^x which bounds $$e^{\xi_x}$$. It doesn't seem at all helpful, though. 67.158.43.41 (talk) 03:40, 16 November 2010 (UTC)

Limit of [sin (x - sin x)] / x^3
Hi Reference desk

I tried solving lim x-->0 [sin (x - sin x)] / x^3 and I think I ended up over complicating things. I checked with a calculator the limit is supposed to be 1/6 but I keep on getting 1/2. Can anyone teach me how to do this?

Thanks! —Preceding unsigned comment added by 169.232.246.218 (talk) 07:55, 15 November 2010 (UTC)


 * Do you know the MacLaurin series expansion for sin(x)? Since the denominator is x-cubed, you only need to expand the top up to x-cubed to see the answer.  The other method is to use L'Hospital's rule, taking derivatives of the top and bottom until you get a limit you can evaluate.  (In this case, that would be 3 derivatives.)  I think the series expansion is a better method, and it certainly is faster than L'Hospital's rule for this problem.  If you're stuck on any particular step, explain your work so far.140.114.81.55 (talk) 08:28, 15 November 2010 (UTC)


 * The series for sin(x) is readily available; that for sin(x-sin(x)) not so much. To get it means differentiating, so one may as well use L'Hôpital's rule (by which I do get 1/6). —Tamfang (talk) 08:46, 15 November 2010 (UTC)
 * No, you can find it with substitution. $$x-\sin x = \frac{x^3}6+O(x^5)$$, so
 * $$\sin(x-\sin x) = (x-\sin x) +O((x-\sin x)^3) = \frac{x^3}{6}+O(x^5)$$.
 * -- Meni Rosenfeld (talk) 17:25, 15 November 2010 (UTC)

http://www.wolframalpha.com/input/?i=sin(x-sin(x))+%2Fx^3 gives you the power series. If you want to do it by hand, substitute sin(x)≈x-x3/6, x-sin(x)≈x-(x-x3/6)=x3/6, sin(x-sin(x))≈sin(x3/6)≈x3/6, sin(x-sin(x)))/x3≈(x3/6)/x3=1/6. Bo Jacoby (talk) 10:39, 15 November 2010 (UTC).

How to know is a system of linear equations has an infinite number of solutions or none
I know that if a system of linear equations is expressed in the form Ax = y, where A is a square matrix, x is a column vector of variables and y is a column vector, then if det(A)=0 the system of linear equations has either an infinite number of solutions, or it has none. How do you know which one of these is the case (i.e. whether there are an infinite number of solutions, or no solutions)? 220.253.217.130 (talk) 08:35, 15 November 2010 (UTC)


 * Hi, you can find the answer under the section "Determining the Number of Solutions of a Nonhomogeneous System of Equations" with some examples. Hope this helps. ~ Elitropia (talk) 11:44, 15 November 2010 (UTC)


 * Use Gaussian elimination to solve the system. If it fails to find a solution, there is no solution. If it finds a solution, you know that there is at least one solution, and therefore (because of det(A) = 0) there are infinitely many.—Emil J. 14:54, 15 November 2010 (UTC)


 * If det(A)=0, I think Gaussian elimination leaves you with an identity (like 0=0) if there are infinitely many solutions and a contradiction (like 0=1) if there are none. Gandalf61 (talk) 15:06, 15 November 2010 (UTC)
 * You'll only end up with a straight identity if everything is a solution. Otherwise you get a system containing identities and nonidentities, such as x-y=0, 0=0, giving the solution set {(t,t)}. Algebraist 15:12, 15 November 2010 (UTC)


 * (e/c) Not at all. Regardless of regularity of the system, Gaussian elimination gives you a modified linear system A'x = y' where A' is in a reduced row-echelon form. Then the system is solvable iff the entries in y' corresponding to the zero rows of A' are zero, and the algorithm tells you how to find the solution. This is the whole point of Gaussian elimination, that it works even for non-regular (or non-square, for that matter) linear systems, unlike e.g. Cramer's rule.—Emil J. 15:15, 15 November 2010 (UTC)
 * An elegant characterization is that in this case, the system has solutions iff the augmented matrix, $$[A\ y]$$, has the same rank as A. Though I think the best way to actually find this out is to do Gaussian elimination as above. -- Meni Rosenfeld (talk) 17:21, 15 November 2010 (UTC)

Group theory questions
I have two questions: 1. If H is a subgroup of G containing all squares then H is normal. 2. If gcd(m,n)=1 and mth powers of G commute with each other and so do nth powers, then G is abelian. Thanks-Shahab (talk) 16:16, 15 November 2010 (UTC)
 * Ad 1: hg = (g−1)2(gh)2h−1.—Emil J. 16:35, 15 November 2010 (UTC)


 * Ad 2: let H be the subgroup generated by mth powers, and K the subgroup generated by nth powers. It follows easily from the assumptions that H and K are both abelian. Also, both are normal (in fact, characteristic) subgroups, and Bézout's identity implies that HK = KH = G. Thus, $$G/(H\cap K)=H/(H\cap K)\times K/(H\cap K)$$ is abelian. Since $$H\cap K\subseteq Z(G)$$, G is class-2 nilpotent (and, in particular, metabelian). I don't know how to continue the argument. I'm probably missing something basic.—Emil J. 17:48, 15 November 2010 (UTC)

A question on Riemann surfaces and spaces of germs
Hello everyone, I would really, -really- appreciate some detailed help on this question: I am taking a lecture course on Riemann Surfaces, and the lecturer has failed to explain how to approach a question like this at all unfortunately - plenty of theorems (monodromy, existence of lifts etc.) but nothing in the way of examples on how to approach a question like this.

Once I'm done with this problem I've got 4 or 5 more along very similar lines, and given that I'd like to try and get those done completely by myself following this, I would really appreciate as much detail as you can possibly give me. Obviously I will be using my own mathematical knowledge/intuition, but the more detail & help I can get from you for this one question, the more likely I am to grasp the concept for subsequent ones.

The question is as follows: "Show that the component of the space of germs over $$\mathbb{C}^{*}$$ corresponding to the complex logarithm is analytically isomorphic to the Riemann surface constructed by gluing, and hence also analytically isomorphic to $$\mathbb{C}.$$"

Now I believe I can choose a 'base point' somewhere in the surface constructed by the gluing and then use analytic continuation to extend this to any given point - I am also fairly sure that whatever path we choose to extend it, we should get the same end result presumably, because we need the map to be well-defined - I suspect this follows from the Monodromy theorem.

However, I don't know where to go from here. Obviously we can project the Riemann surface down onto $$\mathbb{C}^{*}$$ by 'flattening it', but at the same time I believe the surface is simply connected (its complement is just {0} which is connected), and my thoughts are something along the lines of mapping from a point on the surface to the germ given by analytic continuation of the logarithm from our arbitrary fixed basepoint to the point on the surface - however, I'm uncertain as to how to formalize this argument, among other issues.

For one thing, how can a map be analytically isomorphic onto the space of germs? Aren't germs (as far as we've got in the course) considered to be essentially functions at a point under the equivalence relation f~g at the point x iff they are identical on some open neighbourhood of x? So how can something be analytic onto such a space? I think part of the issue is that I am being expected (sadly!) to hand this work in before the lecturer is completely done covering the topic - so how would you go about a problem like this? As much as the mathematical content, actually knowing -what- to write is unclear to me too, so if you would be able to provide me with an example of how something like this should be properly attacked, that would be incredibly helpful. (The next question, for example, discusses the situation with the function (z3-z)(1/2), and I would like to give that a go ASAP, so please respond if you can.) I hugely appreciate your response in advance, it is desperately needed! ;-) Typeships17 (talk) 20:26, 15 November 2010 (UTC)
 * I've a feeling the question is referring to the fact that the Complex logarithm is not uniquely defined. So the space of Germ corresponding to the Complex logarithm is going to consist multiple elements, hence you can construct a space of all those germs. The question is a little unclear, it seems some words a re missing so its hard to tell. The picture lower down in Complex logarithm might also help.--Salix (talk): 23:46, 15 November 2010 (UTC)
 * Yes, I believe that's exactly what it refers to, since we construct the Riemann surface by identifying the distinct branches of the logarithm and gluing them together - and I agree, the wording of the question is poor, but unfortunately that's how it was stated. I believe each point in $$\mathbb{C}^{*}$$ should have a set of germs corresponding to $$\mathbb{Z}$$, one for each 'sheet' of the construction. So how would you actually construct the analytic isomorphism, and show it is analytic and isomorphic then? The way I picture the glued surface is actually -already- as $$\mathbb{Z}$$ copies of $$\mathbb{C}^{*}$$, or rather as $$\mathbb{C}^{*} \times \mathbb{Z}$$, but now it appears that is the structure of the space of germs - however, I find it hard to believe that my map is from (z,n) to the germ corresponding to $$\log(|z|)+2n\pi i +i\theta$$, where $$0 \leq \theta = arg(z) < 2\pi $$ - that seems almost -too- obvious in a way, and I suspect I've made an incorrect assumption which has trivialized a non-trivial question. Typeships17 (talk) 00:37, 16 November 2010 (UTC)
 * I often find that the simplest questions are often the hardest. How about this: we have the space $$\mathbb{C}^{*} \times \mathbb{Z}$$ you can parameterise it as (r,θ) with $$r \in (0,\infty)$$, $$\theta \in (-\infty,\infty)$$ which is easy to show is just $$\mathbb{C}$$ .--Salix (talk): 06:11, 16 November 2010 (UTC)
 * But surely if $$r \in (0,\infty)$$, we can never hit zero, so that's a map onto the punctured complex plane $$\mathbb{C}^{*}$$ instead of $$\mathbb{C}$$? Typeships17 (talk) 10:45, 16 November 2010 (UTC)
 * You could use real log function to rescale the the domain $$\log :(0,\infty) \to (-\infty,\infty)$$. --Salix (talk): 14:37, 16 November 2010 (UTC)