The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
Does anybody know an elementary geometric argument (slicing and reattaching or the like) that shows that the area under x2 from 0 to 1 is 1/3? I tried for a good 20 minutes and couldn't come up with anything. Equivalently you could try to show the area between x2 and is 1/3, which doesn't seem any easier. Staecker (talk) 15:17, 8 December 2009 (UTC)[reply]
Archimedes investigated this problem in The Method of Mechanical Theorems and found that the area is 1/3. He used an argument that imagined the parabola on a lever, balancing a triangle on the other side of the fulcrum. I don't know if this is as "elementary" as you want—it does use the idea of infinitesimals—but it's different from the typical calculus approach at least. —Bkell (talk) 15:26, 8 December 2009 (UTC)[reply]
Interesting- certainly not something I've seen before. Not really as elementary as I wanted, but I suspect that I'm asking for too much. Staecker (talk) 15:54, 8 December 2009 (UTC)[reply]
The standard way of deriving the definite integral of from the definition of Riemann integral uses the identity The (upper) Riemann sum for with respect to the uniform partition is
PS: As far as I recall Archimedes computes the area of a segment of parabola by the exhaustion method, filling it with triangles; he has a simple lemma about the area of an inscribed triangle, and then the final result is equivalent to summing a geometric series with ratio 1/4. --pma (talk) 20:57, 8 December 2009 (UTC)[reply]
Thanks pma- I'm familiar with Riemann sums. (I taught it today to my freshman calc class.) The pyramid volume sounds close to what I'm looking for, but I can't see how that works. Can you explain more? Staecker (talk) 22:06, 8 December 2009 (UTC)[reply]
OK, that's pma's "PS" method. Those links are good, but this is still much more complicated than what I was looking for (still using an infinitesimal-type argument in the summing). I guess maybe there just isn't a simple argument. But I was intrigued by pma's comment about the volume of a pyramid. Are you (PST) saying that your links explain that too? (If so I need to read them more carefully because I didn't see anything about pyramids.) Staecker (talk) 02:21, 9 December 2009 (UTC)[reply]
Explanation of pyramid volume: Consider a pyramid with unit square base and height 1. It's made of infinitesimal cubic bricks. Count them layer by layer, numbering each layer by its distance from the tip. We find that the area in layer x is x2, and the volume of the entire pyramid is therefore . But that is exactly the same as the area under the parabola you're asking for. You probably need to appeal to infinitesimals to make this identification, but it should be possible to derive the volume of the pyramid without infinitesimals. –Henning Makholm (talk) 03:02, 9 December 2009 (UTC)[reply]
Pyramid of KhafraPyramids of GüímarAgreed. To identify the volume of the pyramid (on the left) with the integral of x2 one may use the following exhaustion argument. First interpret the upper and lower Riemann sums as the volume of the outer and inner terraced pyramids (on the right); this equality volume=Riemann sum is a direct verification. Then, we compute the difference between the volume of the inner and outer terraced pyramids with layers of width 1/n, and note that it is exactly 1/n (it is made by n co-centric square frames that put together form a rectangular parallelepiped of size 1 × 1 × 1/n), which allows to extend the equality volume = integral to the Egyptian fashion pyramid. --pma (talk) 11:27, 9 December 2009 (UTC)[reply]
Great- thanks for those descriptions. I see it now. (Though I still don't see it in PST's link.) So it still relies on infinitesimals, but in a much simpler way than the two Archimedes methods referenced above. I guess that's as simple as it's going to get. Thanks- Staecker (talk) 12:48, 9 December 2009 (UTC)[reply]
Talking about buildings, there's also a nice way to compute the sum of the first n squares. Consider 6 rectangular parallelepipeds of size 1 × k × k, (you can think each of them as made by k2 unit-cube bricks), and arrange them to form the walls of a room (with no floor and no ceiling) of outer size (2k+1) × (k+1) × k and inner size (2k-1) × (k-1) × k. Note that the inner size of the k-th object coincides with the outer size of the (k-1)-th object. Therefore, you can put a collection of them, from k=1 to n, one inside the other like matryoshka dolls and form a full parallelepiped of size (2n+1) × (n+1) × n, which explains the formula.--pma (talk) (talk) 11:27, 9 December 2009 (UTC)[reply]
Here's another more algebraic argument, that may be of interest for your class. Call α the integral of x2 over [0,1]. The integral over [0,2] is then 8α: you may justify it to your class by considering the linear transformation (x,y)↦(2x,4y), that expands all areas by a factor 8 (that's true for rectangles, hence also true for the area of the parabola), or alternatively, by a linear change of variable formula for the Riemann integral, to be proven separately. But the integral of x2 over [0,2] is also the sum of the integral over [0,1] plus the integral over [1,2]; the latter may be written as the integral of (x+1)2 over [0,1], and expanded by linearity of the integral into a sum of three integrals. Assuming they already know what's the integral of x and of 1, we end up with the linear equation 8α=2α+2 (I didn't try to solve it but I'm quite confident that it will give α=1/3 ).--pma (talk) (talk) 13:01, 9 December 2009 (UTC)[reply]
Here's an elementary argument, that turns the integral into a geometric problem. I'll write it out in a lot of detail, but many steps are obvious and can be skipped.
We can interpret the integral of as half of the center of mass of the line from 0 to 1 where the point at position x has mass xdx:
Equivalently, M can be interpreted as the x-coordinate of the center of mass (i.e., the centroid) of the triangle T with vertices at (0, 0), (1, 0), and (1, 1). So it remains to be shown that the x-coordinate of the centroid of T is 2/3.
Unfortunately the direct way to compute the centroid of T involves an integral, which just gets us back to where we started. However, by appealing to physical intuition, one can argue that the centroid lies on the medians of a triangle (use the fact that the area of a triangle is half of base times height and maybe some hand-waving). Thus one needs to show (the well-known fact) that the intersection of the medians divides the medians into segments whose lengths have ratio 2:1. One way to do that is, show that it is true for equilateral triangles (divide an equilateral triangle into its medians and use similar triangles), and then argue that the ratio 2:1 is preserved by linear transformations (scaling, skewing, rotating), and that any triangle can be gotten from an equilateral triangle via linear transformations (just skew the sides until they're all the right length).
Hopefully the above is clear; most of it should generalize to higher dimensions, too. (I put a diagram of the linear transformation thing here.)
though i m fond of solving sudokus i m unable 2 complete them. i feel very irritated when i get struck in the last step. so pls provide me the details of how to solve a sudoku, the tricks behind it. thanx —Preceding unsigned comment added by Srividhyaathreya (talk • contribs) 06:46, 10 December 2009 (UTC)[reply]
See Sudoku. One good techinique is to complete the puzzle square by square; in each empty square, neatly write the possible numbers that could be within the square at the particular stage you are at in solving the puzzle. Once you progress furthur, you will get more information which will allow you to eliminate numbers you had thought could be in a square at an earlier stage, eventually finding only one possibility for that particuar square. Continue in this manner to solve the puzzle. (Wikipedia:Reference Desk/Miscellaneous would be more appropriate; puzzles like this have nothing to do with proper mathematics, since mathematics is not about arranging numbers in a 9x9 grid). --PST07:18, 10 December 2009 (UTC)[reply]
Wow, PST, that's a bit of a sweeping statement. I would say that there is a lot of very nice combinatorial mathematics to be found by "arranging numbers in a 9×9 grid." I think that the Mathematics of Sudoku article shows exactly that. In fact this subsection of the main Sudoku article says that "A completed Sudoku grid is a special type of Latin square". From the Latin Square article: "The name Latin square originates from Leonhard Euler, who used Latin characters as symbols." ~~ Dr Dec(Talk) ~~19:04, 10 December 2009 (UTC)[reply]
I am aware of all that but I would not call someone who can solve sudoku puzzles a mathematician. One could argue that there is a lot of mathematics in chess, but a chess player alone does not constitute a mathematician. Both these traits (chess and sudoku) require (admirable) intelligence but I doubt that they constitute a mathematical intelligence. --PST02:25, 11 December 2009 (UTC)[reply]
But nobody has ever claimed that someone who can solve sudoku puzzles is automatically a "mathematician", so what is it you're so vehemently denying? Of course sudoku puzzles involve mathematical concepts, whether the solver realises it or not. Sheesh, making a cake involves mathematical concepts. -- Jack of Oz ... speak! ... 11:00, 11 December 2009 (UTC)[reply]
I never denied anything; I just suggested that perhaps the miscellaneous reference desk would be more appropriate for this sort of question (...since there may be people there who are used to solving sudoku puzzles; that is what "miscellaneous" is for! On the other hand, people here are used to responding to mathematics questions and may not have experience in sudoku (and arranging numbers in a 9x9 grid...)). When Dr. Dec commented on my post (as he usually does to promote his superiority (only joking!)), I responded to justify the reasoning behind my post. Thus, I never really denied anything said by anyone. On a different note, I would not say that making a cake involves mathematics; rather it involves following the instructions given on the packet. ;) --PST12:50, 11 December 2009 (UTC)[reply]
This is a near-universal convention, which, oddly enough, doesn't seem to be mentioned at Sudoku. It's discussed at Mathematics of Sudoku. A puzzle with a non-unique solution would require guessing at some step (or at least some arbitrary choice), which most people regard as lame. (Most people who care at all, that is.) Staecker (talk) 02:33, 11 December 2009 (UTC)[reply]
Technically sudoku is not math, it is logic, since one could substitute something else for the numbers, letters, simple animal drawings, other symbols etc... 65.121.141.34 (talk) 16:18, 15 December 2009 (UTC)[reply]
Integral estimate
Hello, I'm having a lot of problems while trying to prove that
for small enought (it seems to be true from my numerical approximations).
I'd begin by noting that the integrand is negative, thus it can be negated and the absolute value dropped. Then I'd use the substitution to make the integral more manageable. The Taylor polynomial of around may be useful for bounding the integrand. -- Meni Rosenfeld (talk) 13:53, 10 December 2009 (UTC)[reply]
These things usually involve the triangle inequality. It says that |x + y| ≤ |x| + |y|. Replacing y with −y we have |x − y| ≤ |x| + |y|. We know that | ∫ ƒ(x) dx | ≤ ∫ |ƒ(x)| dx. Thus:
Be careful, you must exploit the fact that x is close to y. You'll loose everything writing |x-y|≤|x|+|y| (think to x=1,000,000 and y=1,000,001). Can you see that the RHS you wrote is O(ε-2)? pma
I think you can argue that since the magnitude of the derivative of x-3/2 is decreasing. Then you want to show that as z goes to zero, that expression grows no faster than 1/z2 which is not so bad once the ε's are gone. Rckrone (talk) 23:46, 10 December 2009 (UTC)[reply]
Well, sounds good. Here I'm using more TeX than thought, so it could be possibly made shorter. Let's start with Meni's remark on the sign, and factor out the first term of the integrand
To bound the term into square brakets use the convexity of for all we have so we can go on with:
Now change variable: and bound the resulting ugly integral:
I have a sample of data points xi, each with an associated error Δxi, now i would like to average the xi's , weighteing the data by the its Δxi (ie the data with the smallest error contributes the most to the average. Currently i have my weighting factors as wi=1-(abs(Δxi/ (sum of Δxi))
Now, is this
a) Something that would actualy give me an meaningful result? (wi will sum to 1 so i think it can just do sum of wi * xi)
b) Is there a better way of doing this kind of problem? (surely someones has done a theory or two on this problem)
c) If i wanted to calculate the error in the average, could i do something with error propagation/adding in quadrature (weighted somehow) or would i be confined to only being able to use the standard deviation as a measure of the error?
Thanks for any help--137.205.21.58 (talk) 14:47, 11 December 2009 (UTC)[reply]
If you know the variances of the errors, then the standard thing is to make the weigths proportional to the reciprocals of the variances. That minimizes the mean squared error of estimation. But in just what sense you "know the errors" is not clear from your posting. Michael Hardy (talk) 21:28, 11 December 2009 (UTC)[reply]
original op here, the errors for each data point are a combination of the systematic/statistical errors associated with measuring the value, i'm using some one elses data set and they've quoted x with an error Δx--86.27.192.94 (talk) 23:03, 11 December 2009 (UTC)[reply]
Are the data points all supposed to approximate the same value? The reciprocal of variances is fairly reasonable in that case. Or is it more like something like a problem I was looking where the points themselves are all different and I had to give some average value to their aggregate? With that I used the variance of the points plus their intrinsic variance to weigh them which probably didn't distringuish the better points as much as theoretically one should but worked well in practice. Dmcq (talk) 21:51, 13 December 2009 (UTC)[reply]
Original op here, yes the data should be approximately the same, just for clarity for working this out using the reciprocal of the variances:
Calculate (average of errors)
Calculate (get the sum of the variances)
Calculate (normalise)
op here i think what i just wrote wasnt worth the electrons its displayed on, and thinking about this, doesnt using the reciprocal variance mean were giving more weight to the points closest to the mean, rather than the smallest points? which if i'm weighting by the errors (small error good, big error bad) what i want? —Preceding unsigned comment added by 86.27.192.94 (talk) 19:33, 14 December 2009 (UTC)[reply]
What was being said is to use the following formulae:
Personally I think this tends to accentuate the ones with a small variance too much - perhaps there is some way of taking into account that the variance has its own variation but I've not thought about that too much. Dmcq (talk) 00:59, 15 December 2009 (UTC)[reply]
Non-linear ODE question
Hi guys, I've been trying to find an analytical solution to for about half a day now, to no avail. What's screwing with me is the non-linear term in there; even if I use the fact that to try to simplify it to a linear ODE, I get a nasty elliptical integral that I can't evaluate. Any ideas how to tackle this one? Titoxd(?!? - cool stuff)17:25, 11 December 2009 (UTC)[reply]
As you said: 2F·dF/dψ = d(F2)/dψ, so your equation becomes d3F/dψ3 − d(F2)/dψ = 0. It follows that d2F/dψ2 − F2 = k for some constant k. You should be able to solve this in terms of the Weierstrass ℘-function. ~~ Dr Dec(Talk) ~~17:41, 11 December 2009 (UTC)[reply]
I'm back with another question related to the ones I've mentioned lately. I'll be brief, but not assume the reader has been following what I've said. The specific question I have involves a tree search in number theory. I doubt my own ability to resolve the question, which is whether my computer is ever going to cease the search it's doing right now.
The question is one that I mentioned, the base-ten question, except that I thought I'd start with the first interesting base for the question, which turns out to be base nine. So, the problem is to find the largest number all of whose rightmost segments (including itself) are composite and relatively prime to each other. What we have in base nine is a choice of two final digits--4 & 8--an odd digit preceding it that generates a composite number, and then a string of digits of some fairly large length chosen from the even digits. The constraint that the numbers not be prime creates some uniformity for the problem and cuts down the tree significantly in the final digits, but doesn't affect the higher order digits very much. Every terminal node in the tree is more likely for the larger numbers to be final on account of small prime factors arising for all four choices of digit. There are really effectively only three digits available in any branch where a number divisible by 5 has cropped up, and for every small prime that shows up, there is a significant factor reduction in the size of the part of the tree emanating from it. One can be reasonably certain that there is a finite number to be found, but at present I have no clue how long a computer running optimally will take in finding it (another hour, two days, five months, or 10^87 times the age of the Universe). Would someone mind taking a look at whether I should shut this down?Julzes (talk) 06:26, 12 December 2009 (UTC)[reply]
I've concluded that I don't actually need help on this after all. A simple modification of the program is to count the number of numbers of each length up to a reasonable point and then I can rely on some empirical analysis to get the length of time I can expect this to take. One thing I have noticed is that there is significant slowing due to primality testing. Only actual primes don't have to be checked for relative primality, so numbers with only large prime factors are taking a double hit in terms of time. What this may mean is that the problem without the restriction to composites may actually be as tractable or more so than the original, so I'll have to investigate both. The problem skirts the boundaries of tractability also, so the question may have a different outcome for different bases. A lot to work on, but I should be able to get my own answers.Julzes (talk) 00:21, 13 December 2009 (UTC)[reply]
I've definitely settled the question of whether the problem allowing primes is tractable--it certainly is not. While not having to check primality dramatically speeds things up, in the smaller to moderate size numbers, the fact that a number is relatively prime to all of its predecessors so often means the number is prime; and by the time the ratio between successive counts of how many numbers fill the conditions is relatively close between the two cases, the size of the tree with primes allowed is certain to be many, many orders of magnitude larger than in the case without primes. Already at twelve digits, with primes allowed, just counting the number takes ten minutes. That amount of time for counting, even with all of the slow primality testing, gets the other smaller tree up around thirty digits. One thing has crossed my mind in the interest of getting a final result, and that is that the primality testing need only be done up to a certain depth until candidate final values are presented. Cutting out all the time spent on numbers that only have large prime factors and aren't part of candidate numbers for the final result is bound, though more complicated to implement, to be the most efficient route. Once candidate numbers are complete is the time for primality testing.Julzes (talk) 07:28, 13 December 2009 (UTC)[reply]
Since my last post here I have done all that is possible with the smaller bases vis a vis allowing primes. Base 2 is trivial: 1112=7. Base 3 gives nine terminal nodes and a value of 18709 in base ten. Base 4 has 10608 terminal nodes and gives a 44-digit base-ten number. Base 5 has 490 terminal nodes, with evaluation 7444858551025390541 in base ten. Base six is intractable with values exceeding 101500; and base seven is the last solvable case, with 99 base-ten digits for the solution and 29735375 maximal numbers.Julzes (talk) 05:38, 14 December 2009 (UTC)[reply]
Hi, I would be very very grateful if anyone can tell me how I might get past STEP papers and their mark schemes... the more the better! Desperately need them! Thanks LOADS!!! :)
Convergence of the boundary for this series
Hi all,
how would I go about investigating the convergence of the series on the boundary of the R.O.C? I calculated the R.O.C as e, so on the boundary |z|=e, we could write z as , but that doesn't really help me.
If, as you say, z is an element of the real numbers, there are exactly two points on the boundary of the region of convergence, z=±e. (I haven't verified your solution that |z| = e, but I'm trusting you here...) In this case, the solution is trivially checked with just two cases. Are you sure that you don't mean to say z is a complex number? Nimur (talk) 16:51, 12 December 2009 (UTC)[reply]
Yes sorry, I do of course mean , it seems my brain's not working today (changed now)! So yes sorry, for |z|=e in the complex plane, what's the best way to approach the problem? Thanks, 86.26.6.36 (talk) 17:18, 12 December 2009 (UTC)[reply]
Hi, I would be very very grateful if anyone can tell me how I might get past STEP papers and their mark schemes... the more the better! Desperately need them! Thanks LOADS!!! :) —Preceding unsigned comment added by 121.202.252.24 (talk) 16:12, 12 December 2009 (UTC)[reply]
How do I solve 2^x - x =2 for x. I know the solution (at least one of them) is x = 2. But how do I arrive at it through the general process?--71.91.58.62 (talk) 19:13, 12 December 2009 (UTC)[reply]
This equation is harder than it may appear. To solve it you need some familiarity with the Lambert W function - you'll find that the substitution solves it. The other real solution is -1.69009..., and there are infinitely many complex solutions. -- Meni Rosenfeld (talk) 19:39, 12 December 2009 (UTC)[reply]
Staring at the graph it's instantly obvious that one solution is 2, and another is between −1 and −2. Reasonable numerical methods like Newton's method should approximate the negative solution quite well after a few steps, but if you want a closed form, that's more problematic. Simple algebra alone won't do it, but notice that you've got x in an exponent in one term and as a base in another term. When that happens, it seems algebra often reduces it to something expressible in terms of Lambert's W. Whether you consider that closed form is probably not worth arguing about for most purposes. Michael Hardy (talk) 22:52, 12 December 2009 (UTC)[reply]
Looking at the article on the Mayan civilisation, there is a subsection on their mathematics, see here. Looking at the article on the Babylonian civilisation, there is a subsection on their mathematics, see here. My guess would be that the Babylonians used zero first; they are a much older people than the Maya. The Maya were using zero by 36 BC, "the earliest inscriptions in an identifiably-Maya script date back to 200–300 BC". But there is a well edited mathematical text from the Old Babylonian period (1830-1531 BC). ~~ Dr Dec(Talk) ~~15:24, 13 December 2009 (UTC)[reply]
More on finite differences
Thanks to everyone who answered my previous question, I now know the finite difference for n! (based on a difference of 1), and from that I can calculate the finite difference for (n-1)! or (n-j)! etc. But to turn things upside-down, what please is the finite difference (based on a difference of 1) for "1/n!"? For "1/((n-j)!(n!))"? —Preceding unsigned comment added by ImJustAsking (talk • contribs) 19:54, 13 December 2009 (UTC)[reply]
It's pretty basic algebra to work it out - have you tried to do it yourself? You have the difference of two fractions, so your first step should be to put them over a common denominator (this is really easy) and then simplify it (which is also easy). --Tango (talk) 20:36, 13 December 2009 (UTC)[reply]
AS in your previous question, it's not clear if you are just happy with the first difference, or you want in general the k-th iterated difference (in which case the answer is to be expressed in terms of the k-th Charlier polynomial). --pma (talk) 23:12, 13 December 2009 (UTC)[reply]
Scratch the question about 1/(n!(n-j)!) - it was downright careless of me not to realize that the two terms can separated using partial fractions.
Acccording to an example I worked out, the first difference of 1/n! is "1/(n+1)! - 1/n!", which seems almost too simple to be true. Is it?
Finally, I am happy with just the first difference. —Preceding unsigned comment added by ImJustAsking (talk • contribs) 13:14, 14 December 2009 (UTC)[reply]
That is the first difference, by definition. You then need to simplify it. I suspect you don't really understand what finite differences are - I suggest you learn about them a little more before trying to solve your own problems. --Tango (talk) 14:46, 14 December 2009 (UTC)[reply]
Anyway, if you just want the first difference, your answer is correct. You may also like to write it -n/(n+1)! as Tango suggests. --pma (talk) 15:49, 14 December 2009 (UTC)[reply]
One thousand, four hundred and thirty seven = 1437. Seems left to right to me. Do you not read and think about numbers starting with the largest (i.e. "most significant") digit first? Dragons flight (talk) 02:23, 14 December 2009 (UTC)[reply]
(Edit Conflict) I think that most people write a number from left to right (for instance, if you write 568, you will naturally start at the 5 and end at the 8). However, you are right that (some) people write the units digit of a number only after writing each and every other digit. Perhaps that is just the way the convention was adopted. However, most of mathematics is not concerned with how one denotes a number; in fact, the whole of mathematics would run smoothly if we swapped the names of numbers.
Some mathematicians, particularly algebraists, believe that function composition should be written from left to right. Other mathematicians, however, believe that function composition should be written from right to left. Irrespective of this, many mathematicians agree that notational conventions should not lie at the real heart of mathematics; one can worry about them up to a certain point, but then mathematical ideas become of greater value. --PST02:26, 14 December 2009 (UTC)[reply]
Hmm... neither Hindu-Arabic numeral system nor Positional notation seem to discuss the historical development of our big-endian convention. What I've learned (but cannot source right now) is that the European convention was taken over from Arabic practice which also had the most significant digit on the left; but as Arabic is written right-to-left, this actually amounted to little-endian notation in their context.
Not to mention a discussion of the confusion brought about by computer representation of Arabic text containing decimal numbers; here (at least in Unicode) numbers are stored in big-endian (Western) order in memory, which must be reversed when rendering the text for display or printing... –Henning Makholm (talk) 02:38, 14 December 2009 (UTC)[reply]
There is some (unsourced) discussion about endianness in Eastern Arabic numerals, which probably ought to be moved to some more general location. 02:42, 14 December 2009 (UTC)
Strangely that article says some European languages traditionally read from the small end like the Indians, however the Egyptians, Babylonians Greeks an Romans all started with the big end just like western languages today as far as I'm aware. Dmcq (talk) 10:41, 14 December 2009 (UTC)[reply]
That used to be pretty common in English too as in four and twenty blackbirds, but is there evidence of people writing numbers that way as opposed to writing speech in longhand, for instance in accounts or anything like that? Dmcq (talk) 23:17, 14 December 2009 (UTC)[reply]
Another thing. Were the original numbers from India written left to right? Weren't their languages also written left to right so the Arabs would have reversed the numbers when taking them direct and then they got reverse back to the original in Europe? What order was the original in? Dmcq (talk) 11:00, 14 December 2009 (UTC)[reply]
Cardinal numbers measure the size of a set: how many things there are in it. Ordinal numbers measure the position of a term in sequence. For finite numbers, there is (as long as you adopt the logician's and computer scientist's convention that a sequence starts with the zeroth term, not the first) a direct correspondence between cardinals (zero, one, two, ...) and ordinals (zeroth, first, second, ...), so the distinction is not very important mathematically and is not often made. Specifically, the correspondence is that the nth (ordinal) term in a sequence has n (cardinal) terms coming before it.
In the infinite case, this breaks down and the two notions become very different. Ordinals continue in the obvious way: zeroth, first, second, third, ..., infinityeth (strictly ωth), infinity-plus-oneth, infinity-plus-twoth, ..., but if we work try to use our correspondence, we find that the infinityeth term and the infinity-plus-oneth term both have the same (cardinal) number of terms before them (this cardinal number is called aleph-null). In fact, it turns out that each infinite cardinal number has a great many ordinal numbers corresponding to it (in contrast to the finite case, where there is always exactly one). This leads to the theories of infinite cardinal and ordinal numbers, such as cardinal arithmetic and ordinal arithmetic, being very different from one another. Algebraist16:51, 14 December 2009 (UTC)[reply]
So, in the finite case, the English and Mathish meanings are the same. Cardinal numbers include zero (for the empty set) while ordinal numbers do not. The 'ordinal' zeroth is confusing and distasteful, both in English and Mathish. Bo Jacoby (talk) 21:02, 14 December 2009 (UTC)[reply]
Starting your indices with zero makes a lot of sense in maths. For example, a general expression for a degree n polynomial will have n+1 constants, it makes far more sense to call them a0 to an than a1 to an+1 since then the index of the constant and the degree of the term can be equal. There are lots of other cases where such notation is also clearer. --Tango (talk) 22:36, 14 December 2009 (UTC)[reply]
The distinction in ordinary language arguably has no referent — it's just a convention that makes things easier to parse, like agreement between verb and pronoun. (This has actually been a bit of a problem, in figuring out how to disambiguate the search term ordinal; the current solution is a bit awkward, but no one has proposed one that isn't.) On the other hand, in mathematics, there's a real distinction, at least in the infinite case. I don't know that there's any real distinction between finite ordinals and finite cardinals. --Trovatore (talk) 21:12, 14 December 2009 (UTC)[reply]
Do the words exist in ordinary language? I know them as technical mathematical terms and technical linguistic terms (and they do agree, really - cardinals are used for counting, ordinals are used for ordering and there is a 1-to-1 correspondence in the finite case [which is the only case linguistics covers]). --Tango (talk) 22:36, 14 December 2009 (UTC)[reply]
Indistinguishable balls in distinguishable boxes
Resolved
What is the quickest route for answering the following sort of question, and what is the proper nomenclature surrounding it: In how many ways can fifteen indistinguishable objects be placed in six labeled boxes (any number of which may remain empty, though the general problems are identical with an offset)? This is something I believe I should have learned to do generally, but my present situation is that I would tackle it on an ad hoc basis. It's a little different from partitions.Julzes (talk) 22:44, 14 December 2009 (UTC)[reply]
I just wrote a 1-line program to just count them in the specific case mentioned, but what I got looks smaller than I would have expected: 15504. Just chime in if it happens to be wrong.Julzes (talk) 22:57, 14 December 2009 (UTC)[reply]
It sounds exactly like partitions to me. The answer should be 20 Choose 5, which is 15,504, so your computer program seems to be correct. The easiest way I know to think about it is that you have 20 objects in a row, 15 of them are balls and 5 of them are dividers and you want to decide where to put the dividers. --Tango (talk) 23:38, 14 December 2009 (UTC)[reply]
"The management of Home Depot has 1600 feet of fencing to fence in a rectangular storage yard using the building as one side of the yard. If the fencing is used for the remaining 3 sides, find the area of the largest possible yard."
Okay, I'm having a bit of difficulty with this problem. I have attempted to set up a primary equation and a secondary equation.
Primary: A=xy
Secondary: 2x+y=1600
How am I doing so far?
Solving for y yields y=1600/2x and substituting the answer into the primary equation gives A=x(1600/2x) which simplifies to A=1600/x
Oh my gosh! Thanks so much, I misread my own equation! I got the maximum area to be 320,000 ft squared. Thanks again Henning I appreciate it :) —Preceding unsigned comment added by 98.108.33.148 (talk) 01:13, 15 December 2009 (UTC)[reply]
rotated figure
A triangle is rotated in plane through an angle (clock wise or anti clock wise ), point P as centre of rotation.? if only object triangle and image triangle is given .How can we find out centre of rotation only by observation.( I know the method of point of intersection of perpendicular bisectors ).--True path finder (talk) 02:05, 15 December 2009 (UTC)mks[reply]
Resolved
![{\displaystyle 0\leq \int _{\epsilon }^{1}\left({\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}-{\frac {1}{\left(\left(z+1\right)^{2}z^{2}+\epsilon ^{2}\right)^{3/2}}}\right)dz\leq \int _{\epsilon }^{1}{\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}\left[1-\left({\frac {z^{2}+\epsilon ^{2}}{(z+1)^{2}z^{2}+\epsilon ^{2}}}\right)^{3/2}\right]dz}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b2f6276bf3d97e024bdb77dda394d0e92f1df5f)
