Wikipedia:Reference desk/Archives/Mathematics/2009 November 16

= November 16 =

Exponential Sum
Given the constants a, b and c, is there a general solution to the equation: ax + bx = cx ? If not, I ask why. —Preceding unsigned comment added by 189.24.43.141 (talk) 01:55, 16 November 2009 (UTC)


 * For x an integer greater than two, there are no (non-trivial) solutions. That is Fermat's Last Theorem. For x=2 there are infinitely many solutions, they are called Pythagorean triples. I don't know about non-integer x's. --Tango (talk) 02:34, 16 November 2009 (UTC)


 * Given that a,b,c are also integer. If a,b,c,x are all real then there is usually (always?) a solution. -- SGBailey (talk) 15:21, 16 November 2009 (UTC)


 * But what is given (as parameters) and what is unknown in this equation? (Igny (talk) 02:45, 16 November 2009 (UTC))
 * It's unknown for x>2 and coprime a,b,c even if a different x is allowed on each term. See Beal's conjecture. 69.228.171.150 (talk) 03:20, 16 November 2009 (UTC)


 * Following Igny's lead, it could be that a, b, c are constants, and x is unknown. For a, b, c positive and x real, then for the case where a < c and b < c, and the case where a > c and b > c there will be one real solution for x, but I'm not sure there's a closed form for it.  Otherwise there will be no real solutions since ax + bx > cx for all x. Rckrone (talk) 04:22, 16 November 2009 (UTC)


 * Or I am pretty sure that $$c=(a^x+b^x)^{1/x}$$ would solve the equation for any reasonable a,b,x. (Igny (talk) 05:52, 16 November 2009 (UTC))

I think what the question is saying is:

Given the complex numbers a,b and c, find the value of x.

That is the question in the MOST general meaning. 203.41.81.1 (talk) 21:54, 16 November 2009 (UTC)


 * Divide the equation through by $$\textstyle b^x$$, substitute $$\textstyle \left( \frac{c}{b} \right)^x = y$$ and $$\textstyle \log_{\frac{c}{b}} \frac{a}{b} = N$$. You get $$\textstyle y^N-y+1=0$$, which for non-negative integer $$\scriptstyle N$$ is solvable in terms of hypergeometric functions, as Glasser's derivation shows. Those, in turn, can sometimes be simplified to elementary functions, see e. g. the (downloadable) book "A=B". I wonder if there are any similar results for non-integer $$\scriptstyle N$$?undefined&mdash;undefinedPt(T) 16:33, 19 November 2009 (UTC)


 * Actually MathWorld (eq. (42)) claims any equation of the form $$x^p + b x^q + c = 0$$ to be solvable in closed form (by hypergeometric functions), using a similar technique. I haven't looked for the original cited papers of Cockle (1860) and Harley (1862), but they should shed some more light here.undefined&mdash;undefinedPt(T) 20:06, 19 November 2009 (UTC)

How to solve
Hello all.

I'm looking for a way to isolate x in this formula. My own mathematical skills look overwhelmed. I searched the web for some time. But, those kind of thing look hard to find (search engines gets nuts). I tought one of you could have the answer at hand.

$$y=x*\left|n^x-1\right|$$

$$x=?$$

I don't want to give anybody headach, just see if someone have the answer or a good pointer readyly. Thanks. --Iluvalar (talk) 02:48, 16 November 2009 (UTC)
 * First I went to Wikipedia talk:WikiProject Mathematics. They told me the question was more apropriate here. However noted that it might have some link with the Lambert W function. The graph there look pretty much as a rotation of my fonction. Thanks for your help. --Iluvalar (talk) 04:15, 16 November 2009 (UTC)
 * This equation probably cannot be solved in the way you want. If it was a little simpler - for example, $$y=x\cdot n^x$$ - then Lambert's W function would indeed come to the rescue.
 * What you can do is find solutions numerically for specific values of n and y. For example, if $$n=2,\ y=2$$ then the solution is $$x=1.32642766...$$. -- Meni Rosenfeld (talk) 05:41, 16 November 2009 (UTC)
 * Break down the original problem. Look at the two cases - one where $$n^x-1\geq0$$ and one where $$n^x-1<0$$. Then you get two easier equations to solve. Readro (talk) 09:31, 16 November 2009 (UTC)
 * There is no closed-form solution to these equations either, so they're not exactly easier. -- Meni Rosenfeld (talk) 19:49, 16 November 2009 (UTC)

The problem lacks an explicit closed-form solution. Your options are: (1) treat x as an implicit function of y (this is ordinarily good enough for most analytical purposes), (2) use a numerical root-finding method like Newton's method if you are interested in finding x for a particular value of y. 72.77.62.38 (talk) 19:35, 16 November 2009 (UTC)
 * I'm happy I've asked you. I would have prefered a simple formula, but since you say it's unsolvable, I have no choice but to make an iterative method. I must confess, I don't clearly undestand how to use that Newton's method. I've tried this thing :
 * $$x_n=x_{n-2}+\dfrac{(x_{n-1}-x_{n-2})*(y-f(x_{n-2}))}{f(x_{n-1})-f(x_{n-2})}$$
 * Which I guessed by myself. It's work well apparently until $$n^x+1>e$$ but then it goes nut. I'm a bit confused. can someone gives me a pointer ?
 * In anyway, thanks to all of you. --Iluvalar (talk) 18:05, 17 November 2009 (UTC)
 * What you've described is known as the secant method, which is one of the alternatives to Newton's. As you have noticed, numeric methods can fail for some choices of initial values. Can you describe what values you used for y, n, $$x_1$$ and $$x_2$$?
 * As for using Newton's method - first follow Readro's suggestion and consider two cases: $$y>0$$, for which the solution must have $$x>0$$, and $$y<0$$, for which the solution must have $$x<0$$. I'll cover the former. Your equation becomes $$y=x(n^x-1)$$, which is the same as $$y-x(n^x-1)=0$$. If we write $$f(x)=y-x(n^x-1)$$, we are looking for $$f(x)=0$$. The derivative of f with respect to x (y and n are constants) is $$f'(x)=1-n^x-n^xx\ln n\,\!$$, and you need to iterate $$x_{n}=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1}))}$$. -- Meni Rosenfeld (talk) 20:39, 17 November 2009 (UTC)

Old-fashioned solution without using computer. Multiply the equation $$\scriptstyle y=x(n^x-1)$$ by $$\scriptstyle \ln n$$ to get  $$\scriptstyle y \ln n=x \ln n(e^{x \ln n}-1).$$ Set $$\scriptstyle y \ln n=a^2$$ and  $$\scriptstyle x \ln n = z$$ to get  $$\scriptstyle a^2  = z(e^z-1).$$ This equation is a little nicer to work with, having only one parameter, a. The approximation  $$\scriptstyle e^z\approx 1+z$$ gives the approximate solution $$\scriptstyle z\approx a$$ and thus
 * $$\scriptstyle x \approx \sqrt {\frac y {\ln n}}.$$

This approximate solution may be further improved by substituting $$\scriptstyle z=a+w$$ into $$\scriptstyle a^2 = z(e^z-1)$$ giving
 * $$\scriptstyle a^2 = (a+w)(e^{a+w}-1) $$
 * $$\scriptstyle = (a+w)(e^ae^w-1) $$
 * $$\scriptstyle \approx (a+w)(e^a(1+w)-1) $$
 * $$\scriptstyle = (a+w)(e^a+e^aw-1) $$
 * $$\scriptstyle = ae^a+ae^aw-a+we^a+w^2e^a-w $$
 * $$\scriptstyle \approx ae^a+ae^aw-a+we^a-w$$
 * $$\scriptstyle = ae^a-a+w(e^a(a+1)-1)$$

So:
 * $$\scriptstyle x = \frac z{ln n}$$
 * $$\scriptstyle a = \sqrt{y \ln n}$$
 * $$\scriptstyle z = a+w $$
 * $$\scriptstyle w \approx ae^a\frac{1-(a+1)e^{-a}}{1-(a+1)e^a}$$

Perhaps this is what the OP was looking for. Bo Jacoby (talk) 12:36, 19 November 2009 (UTC).
 * I really appreciate all you've done for me. Many thanks. While I was trying to implement your methods in my more complex full fonction, I realised that a simple "divide by 2 check and do it again method" was good enough for what I'm trying to do. Plus ! it spare me the trouble to even know the exact value of n and it work as well for my entire fonction without more implementation. Look like it's called the Bisection method. It work perfectly for any value I will never have to bother with.
 * I feel a bit guilty about the trouble you undertook to help me. Sorry for that. You helped me a lot ! I knew that there was no absolute answer to my question at the first place, and it helped me to search for viable solutions.
 * Bonus question: I know my english skill at writing is a bit rusty... Does all this was clear enough ? --Iluvalar (talk) 21:22, 19 November 2009 (UTC)

CW complex
I'm a bit confused about the construction of CW complexes. Can someone give me the exact partitioning of the disjoint union of X^n-1 with all the $$D^n$$$$\alpha$$ to form quotient space $$X^n$$. $$D^n$$$$\alpha$$ means the collection of closed n unit disks. You can use any symbol for the attaching maps. Please define it properly in terms of sets instead of saying "make the following identifications". Thanks Money is tight (talk) 11:17, 16 November 2009 (UTC)
 * I'll give it a try. Let $$ \phi_\alpha : S^{n-1}_\alpha \to X^{n-1} $$ be the attaching maps. Then we partition $$ X^{n-1} \coprod_\alpha D^n_\alpha $$ as follows: Every point in the interior of a $$ D^n_\alpha $$ is its own class. And for each point $$ x \in X^{n-1} $$ we get the class $$ \{x\} \cup \bigcup_\alpha \{ s \in S^{n-1}_\alpha \mid \phi_\alpha(s) = x \} $$. Aenar (talk) 18:02, 16 November 2009 (UTC)
 * Thanks! That helps a lot. I find it really troublesome learning these CW complex / simplicial complex constructions. For example in Hatcher's book appendix under the heading "topology of cell complexes", he says "a set A of X = $$ \bigcup_\alpha \ X^n\ $$ is open iff its intersection with each $$X^n$$ is open. Replace $$\alpha$$ with n it says error to parse when I use n. From what I understand elements of the disjoint union are ordered pairs, and elements of the quotient space are sets containing order pairs; it's when we find a homeomorphism of a complex to, e.g. a torus, can we talk about the $$X^1$$ (the two arc skeleton of the torus) being a subspace of the entire torus $$X^2$$. So elements of $$X^n$$ are not on the same level as elements of X^n-1, hence the intersection might even be always empty on a set theoretic ground. If someone can tell me the hidden method behind these constructions I would really appreciate it. Money is tight (talk) 02:38, 17 November 2009 (UTC)
 * No, they're not ordered pairs. I think you are getting confused between disjoint unions and Cartesian products. A disjoint product is just the two spaces put next to each other without touching. A point will either be in one space or the other, you don't pick a point from each. I'm not sure why the TeX didn't work with the n - usually that happens to me because I've forgotten to put a space in-between a command and the letter, but that shouldn't be a problem here - there should be an underscore between them, not a space. I do notice you have some extra backslashes floating around - the last two don't need to be there. --Tango (talk) 03:03, 17 November 2009 (UTC)

How can that be? Given a collection $$X \alpha$$ of spaces, isn't the underlying set of their disjoint union defined to be all the (x,$$ \alpha$$), where x belongs to $$X \alpha$$? I mean the whole point of disjoint union is to make the spaces disjoint if they are not already, so for example when we say "two copies of the real line" we mean the disjoint union of the real line with itself. Am I misinterpreting the symbol $$ X^{n-1} \coprod_\alpha D^n_\alpha $$ as meaning the disjoint union of $$X^{n-1}$$ with all the $$D^n_\alpha$$? Money is tight (talk) 03:45, 17 November 2009 (UTC)
 * Formally speaking, $$X^n$$ does not contain any elements of $$X^{n-1}$$. However, for each $$x\in X^{n-1}$$ there is exactly one point in $$X^n$$ which includes $$x$$ in its equivalence class.  This defines an inclusion from $$X^{n-1}$$ into $$X^n$$.  People are normally going to be pretty cavalier about identifying $$X^{n-1}$$ with the image of this inclusion inside $$X^n$$.  You need to do that to make sense of what Hatcher means when he writes $$\bigcup X^n$$.Ctourneur (talk) 06:45, 17 November 2009 (UTC)
 * You're right that the elements of $$ \coprod_\alpha D^n_\alpha $$ formally are pairs $$ (d,\alpha) $$ with $$ d \in D^n $$, but in most cases the $$ \alpha $$ is suppressed, so that $$ (d,\alpha) $$ is just written d. Regarding the union $$ \bigcup_n X^n $$ I remember myself being annoyed when reading Hatcher for the first time, that he doesn't say more about it. Formally the union would usually be defined as $$ (\coprod_n X^n)/(x \sim i(x)) $$ where $$ x \in X^n $$ and $$ i\colon X^n \to X^{n+1} $$ is the inclusion. (This union is a colimit in the category of topological spaces.) Aenar (talk) 10:47, 17 November 2009 (UTC)
 * Thank you guys! I was thinking along the lines of Ctourneur last night and not-rigorously showed the inclusion is actually an imbedding. With restricted range to its image, bijectiveness and continuity follows easily; showing that it takes open sets to open sets was the bit I rushed. Aenar, your identification for $$ (\coprod_n X^n)/(x \sim i(x)) $$ is really helpful I think I have the tinge of understanding now. Do you know any books where they do these constructions formally? I find Hatcher's handwaving really annoying. I mean he has good parts and good intuition but the only way I understand something is to read all the details. Money is tight (talk) 02:14, 18 November 2009 (UTC)
 * I think that CW complexes are treated formally in various textbooks in the GTM series. Unfortunately, few of these have free versions available online. Alternatively, another excellent book, A Concise Course in Algebraic Topology, is available online for free. -- PS T  08:58, 18 November 2009 (UTC)
 * Can you give me some of the names in the GTM series? I can always borrow from my uni library. Thanks, and if you said "few of those are free" because of my name, well it's just a joke my parents give me plenty of money :D Money is tight (talk) 07:20, 19 November 2009 (UTC)
 * I took a look at limits and colimits yesterday, but since I'm not at all familiar with category theory, I had no idea what Aenar really meant. Then I thought about the imbeddings from $$X^n$$ to X^n+1 and knew it was like approximating some space as n tends to infinity (had that thought because of the word limit), but couldn't quite pin point the exact equivalence relation / partition down. Today I looked at the page "Direct limit" under the heading algebraic objects and whoaa, everything made sense now; just replace the index set I with the natural numbers, the homomorphisms f(nm) being the composition of imbeddings from n to n+1 to n+2 ... to m, the equivalence relation handles the rest. It seems to me like category theory makes everything easy, would it be better to learn it first? I certainly don't wish to... Money is tight (talk) 08:16, 19 November 2009 (UTC)
 * See Graduate texts in mathematics for every title in the GTM series (which I linked in my previous post). With regards to algebraic topology, it is one branch of mathematics which requires a tremendous breadth of knowledge (which only the best mathematicians attain). For instance, although most authors like to say (and do say) that only a basic familiarity with groups is required for the subject in terms of algebraic background, this is certainly an exaggeration of the truth. Modern algebraic topologists are required to have an understanding of algebra as if it were almost their specialist research interest, especially if they wish to be prominent in their fields. Likewise, by doing homological algebra, you may be able to develop insights into algebraic topology which would not be possible with the basic background. There are some rather concise category theory books available should you wish to attain the necessary background. However, I would not recommend you to become a category theory specialist, unless you plan to do algebraic topology in the long term (that is, try to do as much algebraic topology with as little category theory as possible). Rather, you may read textbooks in algebraic topology which are advanced, but yet do not require a huge amount of background. Another point to make is that life outside textbooks is rather diverse. As you may have noticed, Wikipedia articles almost always give information about concepts which you may have not seen within your textbook. This is good in that it provides you with a path by which you can learn particular concepts. Lastly (!), the GTM series consists of a number of good books in mathematics but do not forget the existence of others. The PDF to which I linked, is another excellent textbook. -- PS T  09:09, 19 November 2009 (UTC)
 * Another point to remember is that you are really doing topology; not algebra. Algebra is used to encapsulate topological information and to develop insights. However, the topological intuition is, of course, the most important. When you attain a slightly detailed understanding of particular concepts, algebra begins to take its place (you might remember, for instance, the connection between "conjugacy class", "fundamental group" and "covering space"). -- PS T  09:15, 19 November 2009 (UTC)
 * I'm a bit hesitant with giving advice on how best to learn algebraic topology (I don't think I'm sufficiently qualified). Personally I first really felt like I understood CW complexes after I learned about limits and colimits. But this happened after I had had courses where I had trained my geometric/topological intuition of these spaces. So just learning about limits and colimits might not be enough. And of course we are all different in how we learn best. Aenar (talk) 20:41, 19 November 2009 (UTC)