Wikipedia:Reference desk/Archives/Mathematics/2009 January 13

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 13

the complex plane

Please give a basic description of the complex plane, I'm having trouble understanding the article [1]. I am a student with knowledge through algebra 2 and geometry. Thanks. —Preceding unsigned comment added by 69.136.118.113 (talk) 01:46, 13 January 2009 (UTC)

A complex number has the from a+ib, where i is the square root of -1 and a and b are real numbers. Therefore, you can think of complex numbers of being points on a plane, a+ib would be the point (a,b). We call the set of all complex numbers, thought of as points in that manner, the complex plane. Does that help at all? If not, it would help if you explained in more detail what you are struggling with. --Tango (talk) 01:57, 13 January 2009 (UTC)

Please respond to Tango's comment or this question may be removed. PST (Point-set topologist)

If you remove this question, I will restore it. I do not understand why you are so anxious to delete perfectly valid threads recently. Algebraist 21:08, 13 January 2009 (UTC)

Likewise, why on Earth would we remove a perfectly reasonable question? --Tango (talk) 22:11, 13 January 2009 (UTC)

Well, the person who has asked the question (probably) has not read your response. Unless, in a particular thread, we get a reply from the OP, it is unlikely that he has read it (and hence the thread no longer has any purpose). When I wrote "may", I used italics to indicate that I myself would not remove it. I thought that it was a policy to remove such threads and therefore the use of "may" (and passive voice: I could have written "I will remove it" but I did not). PST

And just to clarify the relevant "policy" that I referred to, I don't see the point in having trivial threads here. PST

(ec)I was misled by your edit summary 'will remove question if OP does not show that he has read Tango's response', which seems to clearly state that you intended to remove the thread. No such policy exists; if you want one, suggest it on the talk page, but I have to say I think it very unlikely you will get any support. Add: your failure to understand a refdesk practice does not constitute a policy. Algebraist 22:20, 13 January 2009 (UTC)

It will be archived in a few days, that's the only removal that happens for good faith questions that don't involve medical or legal advice (removing medical/legal questions is controversial, but it does happen). --Tango (talk) 22:26, 13 January 2009 (UTC)

To Algebraist: "Your failure to understand a refdesk practice does not constitute a policy". That is why I used " " when I referred to "policy" (you probably did not pick this up because of the edit conflict). I don't have such a strong view that such threads should be removed so I won't say that I will remove threads in future (unless, for a different reason, I truly feel that the thread should be removed but in that case I will consult first). But if someone asks what is 5/6, is that a thread that should be removed (as I explained, I don't see the purpose in having trivial threads)?

Another questioner that run away forever... PST's removing policy is so strict, but also so nice, that doesn't matter! :-) --PMajer (talk) 23:46, 13 January 2009 (UTC)

Forever? It's been less than 24 hours... --Tango (talk) 23:47, 13 January 2009 (UTC)

No need to remove, just tell them to use a calculator, or give a link to long division or something. If you remove the thread you risk them not understanding why and putting the question back, which gets us nowhere. --Tango (talk) 23:47, 13 January 2009 (UTC)

PST, maybe they are right... after all "triviality" is a relative concept. Usually a question is not trivial for the questioner, otherwise he or she would answer by himself or herself... and of course it is very likely that it is trivial for you. But let's be patient with everybody ;-) --PMajer (talk) 23:57, 13 January 2009 (UTC)

Yes, I might as well go with their rules since I am not too concerned if my policy is not accepted. But is just seems that 80% of people post a question here but forget that they have done so! This wastes considerable time for us. The laughable thing is that some of their (more or less) trivial questions start a "heated debate" here (by us) when they are never going to see it (this is one such instance). :) PST

Yes consensus is a good idea with polcy. Four tildas ~~~~ to sign a posting is standard policy. Dmcq (talk) 10:33, 14 January 2009 (UTC)

You don't know how many people read the question, are satisfied with the answer, and just don't say anything. While a "thank you" is nice, I see little point in removing questions from people that don't say it. This debate is about policy, it has little to do with the question. When we get into debates about a question it's usually because it's something we find interesting and we enjoy the debate regardless of whether or not it is useful to the OP. --Tango (talk) 16:12, 14 January 2009 (UTC)

Thank you for your help, sorry it took a few days to respond. —Preceding unsigned comment added by 69.136.118.113 (talk) 23:13, 15 January 2009 (UTC)

Don't be. Despite PST's comment, you are in no way obliged to respond here. Algebraist 23:18, 15 January 2009 (UTC)

is a straight line a right angle triangle?

Technically speaking, can I consider a straight line to be a right angle triangle? If one of the sides of a right angle triangle is reduced to zero, then the other side is the same as the hypotenuse. So the Pythagorean theorem would say, $a^{2}+0^{2}=a^{2}$ which seems to hold up.

Duomillia (talk) 02:26, 13 January 2009 (UTC)

I suppose if you really wanted to you could consider a line segment to be a degenerate right triangle with two vertices at the same point. What do you want to do this for, exactly? Algebraist 02:29, 13 January 2009 (UTC)

Well, if you must know I want to use a degenerate triangle for degenerate purposes... no, wait, pretend you didn't hear that...

Following the same logic, could a point be considered an extremely degenerate triangle where all three vertices are the same point? Or for that matter, ditto for a rectangle, hexagon, ect? Not that there is any practical uses for it, but it's interesting to realise. Duomillia (talk) 02:46, 13 January 2009 (UTC)

I don't see why not. --Tango (talk) 03:06, 13 January 2009 (UTC)

Because then the side lines are undefined, and internal angles are undefined (you can't say if it is regular etc.), and diagonals either. Because generally a polygon
'is traditionally a plane figure that is bounded by a closed path or circuit, composed of a finite sequence of straight line segments (i.e., by a closed polygonal chain)',
and a line segment is in turn defined as
'a part of a line that is bounded by two distinct end points'.
CiaPan (talk) 06:58, 13 January 2009 (UTC)

Surely it is better not to allow that a line segment "is" a triangle. What sort of an "is" could that be, after all? Better to say that there are triangles with equal perimeter whose angles are 180°, 0°, and 0° (for the case of 90°, 90°, and 0°, see later), and that they are not all congruent with each other. Suppose that triangle T₁ has sides of length L (non-zero, let us say), 0.6L, and 0.4L, and triangle T₂ has sides of length L, 0.8L, and 0.2L. While T₁ is superimposable on T₂ in the sense that all points constituting T₁ can be "covered" by T₂, the two are not congruent because their vertices are not superimposable. Now, if there is a special relation of "coverage" that holds between T₁ and T₂, surely that same relation holds between any line segment of length L and each of T₁ and T₂. So I propose this instead: for every line segment of non-zero length L there is a class of triangles that it covers and that cover it: all triangles with largest side of length L and with perimeter 2L.

Clearly for every such line segment there are in fact infinitely many classes of polygons, of number of sides 2 (a very special case, only possible in degeneracy) to ∞. Clearly also each of the polygons covered by any such line segment also covers every other such polygon. Clearly also, for every such line segment there are infinitely many classes of polyhedra that it covers, of number of faces 2 (again a very special degenerate case) to ∞ (perhaps excluding 3, or allowing it as a stranger sort of degenerate case; and I assume that coverage is a matter of points on edges only, not all points on surfaces). And clearly for every n-sided polygon and for every natural number m there is an infinite number of classes of (n+m)-sided polygons – and (n+m)-faced polyhedra – that it covers and that cover it and each other. A similar extension may be made associating line segments with "polyhedra" of arbitrarily many dimensions above 3, and with ellipsoids and their higher-dimensional equivalents.

Returning to triangles, there seems to be no special difficulty in allowing triangles with angles 90°, 90°, and 0°. Triangles with sides of length L (non-zero), L, and 0 are all congruent with each other; and they too cover and are covered by a line segment of length L, and so with all of the polygons, polyhedra, and so on that cover and are covered by that line segment.

Excuse the extreme informality of all that. Not my field. And as for L equal to zero, I say nothing at this stage.

Note that not all such "collinear" polygons must have perimeter 2L to be coverable by a line segment of length L. That restriction applies only to triangles. For example, a polygon might have "internal" angles 0°, 360°, 0°, and 0°, and four equal sides of length L. Another polygon might have the same angles but sides of length L, 0.6L, 0.6L, and L. And so on. Every such polygon must have perimeter equal to or greater than 2L, of course.

–_⊥^{¡ɐɔıʇǝo}Noetica!^T– 09:50, 13 January 2009 (UTC)

Usually it wouldn't be considered a triangle, it depends on the circumstances though. For instance if the points are all on a line then there is no interior and one can't say which way the triangle is oriented. Of course it is always possible to assign an orientation as if the figure has an infinitesimal side. It is like asking what is the longitude at the north pole. A degenerate case is as good a description as any. Dmcq (talk) 12:32, 13 January 2009 (UTC)

Lefschetz numbers

I'm trying to understand the Deligne–Lusztig theory for characters of groups of Lie type. It appears that most of the l-adic cohomology stuff can be axiomatized out, but it still leaves a fair number of places where it would be convenient to understand what a Lefschetz number was. I think there are probably some differences from our article Lefschetz number and these, since the field has changed from C to some l-adic field, but since it does not appear to depend on l, I have some hope it is pretty close to the C theory. At any rate, I don't care whether one answers in terms of topology or algebraic geometry (C or Ql), but please specify which, just in case they really are different.

• Must the Lefschetz number always be an integer?

• The Lefschetz number of the identity is the Euler characteristic in the topology case; for other automorphisms is it like the Euler characteristic of the quotient?

• Is it easy to classify all the possible Lefshetz numbers for compact real surfaces in the topology case?

Thanks for any help. JackSchmidt (talk) 18:38, 13 January 2009 (UTC)

In the topological case, one can define (Hatcher does) the Lefschetz number in terms of the traces of the induced maps on integral homology mod torsion. In this case, the number is obviously an integer (when it is defined at all), and I think it is obviously the same as the Lefschetz number defined from Q- or C- homology. Algebraist 20:11, 13 January 2009 (UTC)

Cool. I would be very happy to use more of an integer version, both to have it obviously be an integer, and also to have some hope of computing on a computer (though so far I've got no idea how to write down the space I'm taking l-adic cohomology of yet). Would Hatcher be a good place to learn about the Lefschetz number? Is this Hatcher's Algebraic Topology text? Do you happen to think my "all surfaces" question could possibly have an answer (even if it looks like a pain to do it in general; I mean maybe I could do it for spheres and a 1-holed tori)? JackSchmidt (talk) 22:28, 13 January 2009 (UTC)

I doubt Hatcher would be a useful place to look (there's very little there), but you can always see for yourself. I think calculating Lefschetz numbers shouldn't be too hard, at least for sensible spaces and maps (for spheres it's just a matter of computing degree, which is simple enough), but this isn't my field and I'm just going on undergraduate memories here. Algebraist 22:31, 13 January 2009 (UTC)

Yeah Hatcher (p179-189) is pretty brief and focused on fixed points. The worked example was helpful, but I have trouble figuring out "all maps" (even up to homotopy) of a space. I'll see if I can remember what the degree of a self-map of a sphere is (I remember the degree of a circle map, and remember being confused by the Hopf fibration at some point). I'll reread the D-L theory stuff to see if there is any reason I should care about fixed points. I think there sort of is, but at a much more detailed level than just whether or not the Lefschetz number is 0 or not (the actual value matters). Thanks for the help again. JackSchmidt (talk) 22:56, 13 January 2009 (UTC)

Here's the general fixed point interpretation, at least over C: $L(f)=\sum _{x\in \mathrm {Fix} (f)}\mathrm {ind} (x)$. Here ind(x) is the index of a fixed point (see Fixed_point_index). kfgauss (talk) 12:42, 18 January 2009 (UTC)

In the article of Kleiman, "Algebraic cycles and the Weil conjectures", the Lefschetz fixed point formula is proved abstractly for any Weil cohomology theory. It gives the intersection number of two algebraic cycles (more generally $\langle v.{}^{t}w\rangle$ for any correspondences of certain degree), which lives in a field of characteristic 0. Intersection numbers in Chow theory are always integers, and probably one can deduce it from this, but I have to think about this. You could also look in Milne, Étale cohomology (there is a book and an online script). What is the map you are the Lefschetz number computing of? If it is a power of Frobenius, count the number of points of the variety over the fixed field. Ringspectrum (talk) 17:51, 18 January 2009 (UTC)

Interest calculation (discontinuous)

I want to calculate the end-of-year interest for my savings account, and moreover, I want to understand what I'm actually doing when calculating it. Maths isn't my strongest point, although I am quite interested in it, so I figure asking this here will benefit me in more than one way.

Without further ado, here's the "problem":

• I have a base interest rate of 3.0% per year, calculated based on the daily balance.
• I receive a further 0.6% bonus interest per year based on the lowest balance per quarter. I'm not quite sure how they calculate it, but I think they just take .6% of each quarterly low and let the mean of that be the bonus interest at the end of the year.
• As of today, my balance is €2116.56. My base interest built up until today is €2.02. Assume that the lowest balance this quarter is €2000.00.
• I will deposit €100 each month; I have already deposited the €100 for January.
• Assume I will not make any withdrawals.

Based on the above criteria, (a) what will the cumulative interest be on 1 January 2010 and (b) how do I calculate this?

Feel free to make your own assumptions as to precise dates for my deposits and calculation of the quarterly bonus interest, etcetera.

This kind of maths is quite a bit over my head, especially with the discontinuous aspect (depositing €100 each month) and the quarterly bonus, but I'd nevertheless I'd still like to learn to understand it. Thanks in advance!

P.S. sorry for stretching the page - if someone wants to fix my wiki-markup, be my guest. --Link ^(t•c•m) 20:34, 13 January 2009 (UTC)

I think the easiest way to do with would be to make a spreadsheet (you can do it with pen and paper if you aren't comfortable with computer spreadsheets). An annual interest rate of 3%, compounded daily, corresponds to a monthly rate of $(1+0.03)^{\frac {1}{12}}-1=0.0025$, or 0.25%. To get from one month to the next, do (last month)*1.0025+100. You can then work out the bonus manually, since it's only added on at the end of the year so won't affect the rest of the calculation. --Tango (talk) 20:50, 13 January 2009 (UTC)

If the interest is compounded daily, we probably want 1/365 of the "base interest rate" as the daily rate (0.008219%), and then that gets applied as a multiplier $d\approx 1.00008219$ every day. So, over the course of the year, you get multipliers of $d^{31}\approx 1.002551$, then $d^{28}\approx 1.002304$, $d^{31}$, $d^{30}\approx 1.0024689$, etc. Let us suppose that you deposit before the balance is checked for interest for the first of each month: then at the end of month i (January is 1) you have $b_{i}=(b_{i-1}+100)d^{n_{i}}$ where $n_{i}$ is the number of days in month i and $b_{0}=2116.56-100-2.02=2014.54$. The bonus, supposing it to be added at the end of the year (treating each quarter as precisely 1/4 year) rather than at the end of each quarter, is $0.006{\frac {b_{0}+b_{3}+b_{6}+b_{9}}{4}}$, taking the quarterly low to not include its first month's deposit. (You can use precisely 2000 instead of $b_{0}$ if you wish.) Writing down the full expression for the final balance (before January 1 2010's presumptive deposit) would take up a lot of useless space, but the calculation is trivial: I get $b_{3}=2330.98,b_{6}=2649.98,b_{9}=2971.60$, for a grand total of €3295.67. This isn't guaranteed to be precise, not only because of the question about $b_{0}$, but also because I just used floating-point math and the bank will presumably round (to eurocents, perhaps) as it goes. The interest is of course trivial to calculate since you know what you started with and what you deposited; it's much easier to get it this way than to try to only count interest as you go along. (If, however, the interest is not compounded at all, you just want to average the balance every day and multiply by 3%, so you get $(d-1)\sum _{i=1}^{12}n_{i}(100i+b_{0})$ as the base interest.) --Tardis (talk) 00:23, 14 January 2009 (UTC)

Convolution

Hi, I take a lecture in Evolutionary Dynamics (I am a biologists, thats why I am asking in the first place) and there was a lecture about Branching processes, specifically the Galton–Watson process. In the lecture script is say this about P(i,j) which is the probability to go from a state i to state j:

${\begin{aligned}P(i,j)=p_{j}^{*i}=\sum \limits _{k_{1}+\cdots +k_{i}=j}p_{k_{1}}\cdots p_{k_{i}}\end{aligned}}$

Then it says:

${\begin{aligned}\{p_{k}^{*i}\}_{k\geq 0}\end{aligned}}$ is the i-fold convolution of ${\begin{aligned}\{p_{k}\}_{k\geq 0}\end{aligned}}$

My problem is with that last sentence since I couldnt find anything in the article Convolution that relates to that statement. I think I understand that ${\begin{aligned}\{p_{k}\}_{k\geq 0}=\{p_{0},p_{1},p_{2}\cdots \}\end{aligned}}$ and thus per analogy that ${\begin{aligned}\{p_{k}^{*i}\}_{k\geq 0}\end{aligned}}$ denotes the set of this expression for all k bigger than zero. But I have trouble to see how they relate or how you can generate the latter from the first by a convolution. Maybe somebody here can help out? Did I just miss this part in the article or is there something strange in that formalism the professor uses? Thanks a lot. Greetings --hroest 22:08, 13 January 2009 (UTC)

This is a discrete convolution, in fact a one-sided discrete convolution, which just means that the $\textstyle p_{k}$ you are considering are zero for all k<0 (or if you prefer, you may think them defined for all k putting $\textstyle p_{k}=0$ for k<0). The one-sided discrete convolution is the same thing as the Cauchy product of power series, in the following sense. If you have two sequences $a:=\{a_{k}\}_{k\geq 0}$ and $b:=\{b_{k}\}_{k\geq 0}$ you can consider their (ordinary) generating functions: then the product of the two generating functions is the generating function of the convolution a*b. In your case you convolute $\{p_{k}\}_{k\geq 0}$ with itself i times, thus:

$\textstyle \sum _{j\geq 0}P(i,j)x^{j}=\left(\sum _{k\geq 0}p_{k}x^{k}\right)^{i}$

I hope this is clear enough --PMajer (talk) 22:58, 13 January 2009 (UTC)

This is the same discrete convolution defined in the article titled convolution, where it says

For complex-valued functions ƒ, g defined on the set of integers, the discrete convolution of ƒ and g is given by:

$(f*g)[n]\ {\stackrel {\mathrm {def} }{=}}\ \sum _{m=-\infty }^{\infty }f[m]\cdot g[n-m]\,$

Simply observe that the sum in the definition above is the same as

$\sum _{k_{1}+k_{2}=n}f[k_{1}]\cdot g[k_{2}].$

Michael Hardy (talk) 03:17, 14 January 2009 (UTC)

.... which is hroest's case, provided f[k] and g[k] vanish for k<0 (as I wrote) --PMajer (talk) 09:34, 14 January 2009 (UTC)

In the Galton–Watson process, Galton was modelling the distribution of surnames, which he assumed to be passed from father to son. Each male in one generation has n sons in the next generation, and we assume $\{p_{0},p_{1},p_{2}\cdots \}$, the probabilities that n=0, 1, 2 etc., are the same for all males and all generations. So the probability that there will be j males in one generation given that there were i males in the previous generation, denoted by $P(i,j)$, is found by summing $p_{k_{1}}p_{k_{2}}\cdots p_{k_{i}}$ across all ordered sets of i non-negative integers $\{k_{1},k_{2}\cdots k_{i}\}$ that sum to j. The set of values $\{P(i,j)\}_{j\geq 0}$ is the "i-fold convolution" of $\{p_{k}\}$. Because of the connection between convolution and multiplying generating functions mentioned above, this way of formulating the problem is particularly useful in cases where the generating function of $\{p_{k}\}$ itself has a simple closed form. Gandalf61 (talk) 10:57, 14 January 2009 (UTC)

Thanks a lot everybody, you really helped me a lot! So I guess I had a problem seeing what the i fold convolution means, but it is just a convolution with itself done i-1 times? As I understand, Gandalf61, if you have for example a Poisson distribution then you could just plug in for $p_{k}$ the value of $POI(k,\lambda )$ for a given parameter and "directly" calculate $P(i,j)$? I understand that you can basically think about this on the level of ${\begin{aligned}\{p_{k}\}_{k\geq 0}\end{aligned}}$ which you convolute i times or on the level of the generating functions which you multiply i times with itself. Thank you all very much --hroest 15:02, 14 January 2009 (UTC)

Yes, I think you have got it. And if $\{p_{k}\}$ follows a Poisson distribution then you get an especially tidy result because the sum of a set of independent random variables, each with a Poisson distribution, also has a Poisson distribution. Gandalf61 (talk) 15:29, 14 January 2009 (UTC)