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Wikipedia:Reference desk/Archives/Mathematics/2010 March 8

Contents 1 March 9 1.1 Homework question 1.2 Divisibility In Digit-by-Digit Inter-Base Conversion 2 March 11 2.1 Quick question on integration by parts and non-integer values of the Gamma function 2.2 Length of a function 2.3 Why doesn't Cantor's diagonal argument apply to rationals? 2.4 What is the 98th percentile in a Normal distribution with μ=100 and σ=15? 3 March 12 3.1 little project 3.2 How many types of numbers are there? 3.3 Absolute value of a quaternion 3.4 Domain of a function 3.5 Statistical significance when discussing a population 3.6 feedback rating 4 March 13 4.1 Maclaurin expansion 4.2 Statistics - Conditioning 4.3 Stats -- algebraic manipulation of V(X) and E(X), etc. 4.4 Symmetric linear solve 4.5 simple question. 4.6 is there any proof that current compression couldn't be much better? 4.7 L'Hopital's rule 4.8 Approximations to roots 5 March 14 5.1 Approximate solutions to equation 5.2 Geometry problem 5.3 Terminology: equivalence of ordering or optima 5.4 Sparse Graph Colorings 5.5 Integration 5.6 L'Hopital vs. Landau

March 9

Homework question

For what values of x is the sequence Un = convergent to 0?

The answer is obviously 1 < x < 3 but I have no idea how to prove that. Could someone lead me in the right direction? --124.171.116.21 (talk) 04:38, 9 March 2010 (UTC)

Let such that , and let for each . Prove that . Once this has been established, write , and note that if and only if , and therefore, if , , , and thus .

To complete the above proof, you need to establish that , and that if , and we let for each , the sequence does not converge to zero. I have hidden the proof of the former below; if you feel that no furthur attempts at the former will be at least slightly productive, you may see the proof. However, even if you do see the proof of the former, I have not included a proof of the latter (which essentially follows from the line of thinking required to prove the former). If you feel that you cannot solve the latter, even after seeing the hidden proof of the former below, post here again, and I (or another volunteer), will give you additional hints. Hope this helps (and the proof of the former is hidden below). PST 05:36, 9 March 2010 (UTC)

PST 05:36, 9 March 2010 (UTC)

Alternatively, if you are OK with 1/n → 0 and with the Sandwich theorem, you may prove that if 0 ≤ a < 1 then 0 ≤ aⁿ ≤ C/n with C := a/(1-a). Start from the Bernoulli inequality with x := (1-a)/a .--pma 09:38, 9 March 2010 (UTC)

Divisibility In Digit-by-Digit Inter-Base Conversion

I've been trying to get my head around the meaning of some empirical results that don't seem to have an obvious explanation. Up to some very large number at least, there is no number that generates only numbers not divisible by any of {2, 3, 5} when one takes its representations in bases 2 through 5 and translates these into greater bases up to 6 (Ten numbers output for each input). Can anyone prove the impossibility of finding such a number, explain reasonably precisely why the first such number would be so inordinately large that it may not be computable, or demonstrate an example which does give ten numbers relatively prime to 30?Julzes (talk) 08:40, 9 March 2010 (UTC)

What's wrong with the number 1? Dmcq (talk) 09:45, 9 March 2010 (UTC)

20821 - outputs are 5321791, 285282577, 6348063151, 80542674037, 267781, 1973131, 10134727, 94531, 328147, 58633; outputs modulo 30 are 1, 7, 1, 7, 1, 1, 7, 1, 7, 13. Gandalf61 (talk) 11:51, 9 March 2010 (UTC)

Ok, that means my most recent program was in error. Anyway, can anyone find a number that generates 10 primes or suggest to me a stronger sieve than simply checking numbers congruent to 1 modulo 60? I know why the first such number is likely to be huge. In fact, just checking numbers like mentioned has another program that hopefully is not also defective up to around 190 billion without finding such a number.Julzes (talk) 16:36, 9 March 2010 (UTC)

Well, if anyone comes up with anything profound on this, please let me know.

Off-topic slightly, but here is a brief description of the granddaddy of coincidences that I discovered since the last time I posted anything of the nature here: List the primes that translate as primes twice in going from base 2 to base 10 and also translate twice as primes in going from base 3 to base 10. The 4th of these is the first to also generate a prime once in going from base 4, the 44th is the first to do it twice, both begin with the digits 234 in base 10, and the tenth on the list begins with 365. Furthermore, the 4th on the list is more special in that it translates twice as primes from base 5, does not translate once as a prime again until base 20 (allowing 'digits' greater than 9, and the tautology at base ten isn't counted) at which base it translates 5 times as a prime, and then it also does it 4 times at base 22 and twice at base 25 (not at bases 21, 23, or 24) to boot. Don't ask me how I found it.Julzes (talk) 03:42, 10 March 2010 (UTC)

Wikipedia:Reference desk/Archives/Mathematics/2010 March 10

March 11

Quick question on integration by parts and non-integer values of the Gamma function

Hi all,

Just briefly:

Could anyone explain to me why is non-singular? Or, more specifically, when I integrate by parts, I get ; but the first term, unless I'm being stupid here (most likely) goes to 0 at infinity, but not at 0. However, is a well-documented result (indeed on the Wiki page here), which would imply , no? Exp(0)=1, so the behaviour is like as . What have I done wrong?

Thanks very much! Otherlobby17 (talk) 02:21, 11 March 2010 (UTC)

The integral formula only works when the real part of the argument is greater than 0. So if you try to apply it with -1/2 then it's not surprising you get inconsistent results.--RDBury (talk) 02:53, 11 March 2010 (UTC)

Precisely, the integral formula holds for poisitive arguments of the gamma function, that is exponents strictly greater than -1; more generally, it gives Γ(z) for any complex z with Re(z)>0. --pma 16:16, 11 March 2010 (UTC)

Length of a function

Say I have a function f(x). If I know the lower and upper x bounds, is there a way, using calculus, I could find the length? I know how to do it for a linear function (distance formula), but what about a really curvy and irregular function? For example, if I have a function f(x)=3x³-6x²+x, how would I find how long it is between the x coordinates 7 and 15? Thanks 69.210.140.77 (talk) 03:26, 11 March 2010 (UTC)

I think you want Arc length. Algebraist 03:30, 11 March 2010 (UTC)

And the relevant formula is -- Meni Rosenfeld (talk) 08:35, 11 March 2010 (UTC)

Note that the linked article is somehow misleading, for it report the (correct) definition as total variation of f:[a,b]→X, ( X,d being a metric space) and then gives a (correct) integral formula for the curve , where f is a function (thus not the same role as the f of the definition!). That is, it is the "length of the graph curve". The integral formula corresponding to the definition given in the link, for the length (variation) of a C¹ curve f:[a,b]→Rⁿ with respect to any norm ||.||, is:

,

that also holds for Absolutely continuous f (the integral is then in the Lebesgue sense). I will rectify that thing (or somebody else) as soon as I'm more free.--pma 15:55, 11 March 2010 (UTC)

Why doesn't Cantor's diagonal argument apply to rationals?

Hi, I know that it is settled that Cantor's diagonal argument does not apply to integers, and that there exists a bijective map between integers and rationals numbers (so there is the "same number" of both). If it takes aleph₀ digits to represent all real in (0,1), and aleph₀ digits to represent all rationals in (0,1) then why doesn't the diagonal argument apply to rationals? 0¹⁸ (talk) 15:57, 11 March 2010 (UTC)

because, of course, not every infinite sequence of digits represents a rational number (try to follow the analogous argument, and it will stop in the middle before reaching any contradiction) --pma 16:00, 11 March 2010 (UTC)

Because the number constructed in the diagonalization argument (as it is usually presented) is guaranteed to be a real number, but not guaranteed to be a rational number. So the number constructed is a real number not in the initial sequence, but may not be a rational number not in the original sequence. The fact that the rationals are countable shows there is no way to patch up the argument to make it output a rational number regardless of what the input sequence is. — Carl (CBM · talk) 16:01, 11 March 2010 (UTC)

Okay, I follow your logic. I appreciate the prompt responses. I guess then what I wonder why this isn't an injection from the integers (indexed by j) into the decimals of base b:

d_i = [j mod_b bⁱ - j mod_b b^{i -1}] / b^i-1

where mod_b is the base b modulo (sorry, I don't recall how to make the minus sign without a math tag). 0¹⁸ (talk) 17:22, 11 March 2010 (UTC)

BTW, I think this is just saying, why can't I write the digits backwards in base b (which I realize I can not but I don't know why). 0¹⁸ (talk) 17:32, 11 March 2010 (UTC)

How can that produce an infinite decimal? —Bkell (talk) 17:58, 11 March 2010 (UTC)

Understanding your comment is what I want to do. How many digits can an integer have? If the answer is finite, I'd think you could tell me the length. I must be missing something. 0¹⁸ (talk) 18:44, 11 March 2010 (UTC)

For any finite length, there is an integer with that many digits. But there is no integer that has infinitely many digits. --Trovatore (talk) 18:55, 11 March 2010 (UTC)

Just to see if I understand: if there were an integer with infinitely many digits, it would be infinite and infinity is not in I. Is that right? 0¹⁸ (talk) 19:15, 11 March 2010 (UTC)

Well, almost. A string of infinitely many digits doesn't really represent "infinity", though. A string of infinitely many digits is pretty much just a string of digits — in the usual context, it doesn't represent anything but itself. (In the context of P-adic numbers, things are a bit different, but they're not relevant here.) --Trovatore (talk) 19:24, 11 March 2010 (UTC)

Your argument does prove that the set of real numbers (more precisely, those between 0 and 1) that have a finite decimal expansion is a countable set, which is true. —Bkell (talk) 19:10, 11 March 2010 (UTC)

Oh, and by the way, if I'm understanding it correctly, your map is an injection, as you claim, which shows that the set of real numbers is at least countably infinite. But it is not a surjection, because there are real numbers that are not the image of any integer under your map (for example, ). —Bkell (talk) 19:16, 11 March 2010 (UTC)

Thanks. This is not quite what I was originally asking but another related question I had was how do we know that pi and sqrt(2) have decimal representations (even allowing for infinite digits)? I can see how we can say that they are within delta for any delta larger than zero, but not that we really can write them down that way. 0¹⁸ (talk) 19:41, 11 March 2010 (UTC)

One first has to define "decimal expansion". But basically it's a function that takes n and returns the nth decimal digit. So your question comes down to, how do we know that if we fix n then the nth decimal digit of π exists? The answer is to put a grid on the real line starting at 0 and incrementing by a step size of 10^-n. Then π will be immediately between some pair of adjacent points in the grid, and the one on the left will tell you first nth decimal digits. — Carl (CBM · talk) 19:55, 11 March 2010 (UTC)

Okay, so what is n a member of, and how large does n have to be for the pi to not be arbitrarily close, but actually represented by by the decimal expansion? 0¹⁸ (talk) 20:11, 11 March 2010 (UTC)

n is any positive integer, and the decimal expansion has to match for any positive integer (the part before the decimal point needs to match too). -- Meni Rosenfeld (talk) 20:25, 11 March 2010 (UTC)

As for the second part of your question, there is no positive integer n such that pi is actually equal to the first n digits of the decimal expansion of pi. The decimal expansion of pi goes on forever; if you truncate it at some point, you get an approximation of pi, but not the true value. If you're familiar with infinite series, it's the same idea: Truncating a (convergent) infinite series after finitely many terms gives an approximation to the true value, but (generally) it is not exactly equal to the true value. —Bkell (talk) 20:34, 11 March 2010 (UTC)

(backdent) sorry, my question was how do we know we can represent an irrational real with an infinite string of decimals and not just get arbitrarily close. The answer I was told is that we can generate a string of digits until the cows come home but it will still just come close. I'm still confused. 0¹⁸ (talk) 02:33, 12 March 2010 (UTC)

If two real numbers are arbitrarily close to each other, then they are at distance 0, and so they are equal. You make a decimal expansion one digit at a time; the resulting infinite decimal expansion determines some real number, and that real number is arbitrarily close to the number you started with, so the infinite decimal expansion determines the same number you started with.

The most common mistake is to confuse the finite initial segments of the decimal expansion with the decimal expansion itself. To define the decimal expansion, all you have to do is define each of its initial segments. However, the overall decimal expansion of an irrational number will have an infinite number of digits. Thus each of the finite initial segments of the expansion will be unequal to the irrational number, but the entire infinite expansion gives back exactly the irrational number you started with. — Carl (CBM · talk) 02:42, 12 March 2010 (UTC)

I'd just add here that the reason it "gives that back" is that that's what it's defined to give back. The denotation of a decimal representation is, by definition, the limit of the denotations of the finite truncations.

It's not an arbitrary definition — it's the only reasonable definition that causes the decimal representations to denote real numbers. This is what those who refuse to believe that 0.999... is really exactly 1 are missing: The real numbers are the real, underlying Platonic objects we're trying to get at; the decimal representations are just names for them. If they didn't name real numbers, they wouldn't be doing the job we set them; the limit definition is the only reasonable one under which they do name real numbers. --Trovatore (talk) 02:50, 12 March 2010 (UTC)

Okay, so I asked about an injective map from the integers to the reals and was told it is not surjective because the algorithm could never generate enough digits--there simply are not enough. Then I asked you about generating the digits and you gave me a method that can generate any finite number of digits but clearly can not generate infinitely many digits (the bins are countable after all). So there must be an additional step that I am missing for how you get from an irrational number to an exact decimal representation. 0¹⁸ (talk) 04:35, 12 March 2010 (UTC)

As was mentioned, a decimal representation is a function which takes a positive integer as an argument and returns a digit (I'm focusing on the part after the dot for simplicity). Given any real x, the corresponding function is given by . So if you take and you get ; for you get . In practice we can only expect to calculate f for finitely many digits "until the cows come home", but in theory the function f is capable of producing a digit for any n, however large. In other words we have provided a "recipe" to obtain any digit of we want. Thus, the decimal expansion given by f describes exactly. -- Meni Rosenfeld (talk) 07:52, 12 March 2010 (UTC)

So when I build a number using the integers, I can say the same thing. For any fixed (finite) n, I can build the number, regardless of the size of n. There must be something that allows the decimal expansion to go past this. 0¹⁸ (talk) 17:15, 12 March 2010 (UTC)

You seem to be confusing two different functions:

• Given a real number, produce a decimal expansion
• Given an integer, use that as a finite decimal expansion of some real number

The first bullet is talking about a function whose domain is the entire set of real numbers, and whose range consists of decimal expansions. The second is talking about a function whose domain is the integers and whose range is included in the set of real numbers. These are not the same function. It is true that every real number has a decimal expansion (which may be infinite), but this is not the same as saying that there is a way to produce all these decimal expansions with a function whose domain is just the integers.— Carl (CBM · talk) 12:59, 12 March 2010 (UTC)

Carl, I agree with everything you say. I would especially emphasize that I am confused. My question is why are these different? See my above response to Meni Rosenfeld. 0¹⁸ (talk) 17:44, 12 March 2010 (UTC)

I don't know if what I'm about to say is going to make any more sense than what has already been said, but I'm going to say it anyway. Let's consider . Do you agree that this number is irrational? If so, then you must agree that it cannot have a finite decimal representation (for if it did we could multiply this finite decimal expansion by some sufficiently large power of 10 to get an integer, and then make a fraction that equals  by writing this integer over the power of 10 that we used). So there is no single integer that can be interpreted as the decimal expansion of , because every integer has only finitely many digits. Agreed? Therefore,  is an example of a real number that is not in the range of the map you defined.

That being said, it is possible to find arbitrarily good approximations of  that are in the range of your map—take the decimal expansion of  and truncate it after some number of digits. So the range of your map is dense in the real numbers, but that doesn't mean it actually includes all the real numbers (it doesn't). The rational numbers are dense in the real numbers (given any real number x and some allowable error ε > 0, no matter how small, we can find a rational number q that approximates x to within ε—though note that a different choice of ε may require a different choice for q), but there are many real numbers that are not rational. It is the same with the range of your map. Okay so far?

What we have said so far is that not all real numbers can be represented with finite decimal expansions (for example, , and in fact even simple fractions like 1/3 if we are using base 10), but every real number can be approximated arbitrarily well by a finite decimal. More precisely, if there is an allowable error ε > 0 specified ahead of time, we can find a finite-decimal approximation q to any real number x so that |x − q| < ε.

So what hope do we have of representing a real number like 1/3 with a decimal expansion? We must resort to an infinite decimal expansion, but we need to define what such a thing means. Let's assume we're working in base 10, for simplicity (this argument easily generalizes to any base), and let's consider just the part of the decimal expansion after the decimal point. Suppose an infinite decimal expansion has digits 0.d₁d₂d₃…. The value of this decimal expansion is defined to be

which is an infinite series that is in turn defined as the limit of the partial sums:

For example, 1/3 is 0.333…. If you replace d_i by 3 in the series above, you get a Geometric series, and you can calculate that the sum of the geometric series is indeed 1/3.

Now, how do we know that every infinite decimal expansion results in a real number, and that every real number can be represented with an infinite decimal expansion (i.e., as the sum of an infinite series like these)? The answer to the first question is that the real numbers are complete: every Cauchy sequence (roughly, every sequence of points that are getting closer and closer together) converges to a limit, i.e., a real number. The answer to the second question is essentially that this procedure is how the real numbers are constructed from the rationals (actually, usually the construction involves something a little subtler, like Dedekind cuts; but it is safe to think of the real numbers as precisely the set of "Limit points" of Cauchy sequences of rational numbers).

So, finally, here is the distinction between Carl's two bullet points above. The first idea is producing a decimal expansion from a real number. Unless the real number happens to have a finite decimal expansion, this process will produce an infinite sequence of digits, and the decimal expansion of the real number is the entire infinite sequence—not the finite pieces at the beginning, but the entire thing. (If you don't have all of the digits of the decimal expansion of 1/3, you don't have the decimal expansion of 1/3.) The infinite sequence of digits is an exact representation of the real number, but you need all infinity of them for it to be exact. The second idea is using a given integer as the (finite) decimal representation of some real number. This works fine for every real number that can be represented with a finite decimal expansion, but it cannot give you 1/3. It can give you real numbers that are very close to 1/3, but not 1/3 itself.

One last comment: Carl mentioned earlier that "If two real numbers are arbitrarily close to each other, then they are at distance 0, and so they are equal." This is true, but it requires careful interpretation. What it means is this: If two given real numbers x₁ and x₂ are such that for every ε > 0 the inequality |x₁ − x₂| < ε is satisfied, then x₁ = x₂. The important thing to notice is that x₁ and x₂ are fixed at the beginning—we aren't changing them for different values of ε—and they are real numbers, not sequences of real numbers. For example, let's consider the number 1/3 again, and try to find a number q in the image of your map that is "arbitrarily close" to 1/3. Well, q = 0.33333 is pretty close, and it will satisfy |1/3 − 0.33333| < ε for any ε ≥ 0.000003333…; but any value of ε less than this will cause the inequality to be false. So 0.33333 is not "arbitrarily close" to 1/3 in the sense that Carl meant. You can try to get closer by adding some more 3's to the end of q, but no matter how many 3's you add (as long as you add only finitely many), there will be some value of ε that is small enough (yet still positive) to make |1/3 − q| ≥ ε. So there is no such thing as a single number "arbitrarily close" to 1/3 in the image of your map, and therefore 1/3 is not in the image of your map.

Was this helpful? —Bkell (talk) 19:05, 12 March 2010 (UTC)

Bkell, that was very helpful. I think you answered my question but I am not sure. Lets see if you agree with my understanding. Consider these statements:

(1) Using the map I proposed, one can generate a string of digits of any finite length or a string larger than any finite length.

(2) Using the algorithm proposed above for coming up with digits for reals one can produce a string of digits of any finite length or a strong larger than any finite length.

(3) Any method that can produce only finite length strings of digits will never get exactly to a map to the reals.

(4) There is some map (not mentioned above) that gives infinite length strings of digits given a real value in its domain.

(5) statement 4 is false. Instead, reals are the set of all values generated by 1 or 2 plus the limit points of all Cauchy sequences that can be generated from this set.

I am quite sure of (1) and (3). If one of these is wrong, I'd really like to know. I think you are saying 1,2,3, and 5 are true. Is that right? If not, I am not sure about 2,4, and 5 and commenting on them would be very helpful for me. 0¹⁸ (talk) 19:31, 12 March 2010 (UTC)

Here are my comments:

1. It is certainly true that your map can generate a string of digits of any finite length. The second half of your statement depends heavily on interpretation. On the one hand, it is true that if I give you any finite length (say, n), then your map can produce a string of digits that is longer than that (for example, of length n + 1). But it is not true that your map can produce a string S of digits for which it can be said that "S is longer than any finite length." If S is finite, then it has some length n, and then it is not longer than length n + 1. So saying that "S is longer than any finite length" is the same as saying that "S has infinite length." Your map cannot produce strings of infinite length, because no integer has infinitely many digits. Most mathematicians would interpret "longer than any finite length" in the second way, that is, to mean "of infinite length"; under this interpretation, the second half of your statement is false.
2. It is true that the decimal expansion of a real number may be a finite sequence of digits or an infinite sequence of digits (that is to say, a sequence longer than any finite length). (It is often useful to consider the decimal expansion of any real number to be an infinite sequence of digits, by adding infinitely many zeroes at the end of the finite sequences. This way we don't have to consider two different cases. So, when I refer to an infinite decimal expansion below, you should consider this to include finite decimal expansions as a special case.) However, in your statement you need to be careful about what you mean by "algorithm" and "produce" (or "generate"). An Algorithm, by definition, must terminate in a finite amount of time, so it is impossible for an algorithm to produce infinitely much output (such as the entire infinite decimal expansion of ). Now, there is an algorithm that will produce, in a finite amount of time, any desired digit in the decimal expansion of , and for most practical purposes this is just as good as having the entire decimal expansion (since, in a finite amount of time, we could only use finitely much of the infinite decimal expansion anyway). But, strictly speaking, this algorithm does not actually generate the infinite decimal expansion itself. The idea of an infinite decimal expansion is a purely conceptual thing, just like any infinite set. There is no way to "produce" or "generate" an infinite set in "the real world." But, if you accept the idea of the natural numbers, for example, as an infinite set (an infinite set that exists conceptually in its entirety, rather than a more "real-world," "algorithmic" view of them as an unending list that is "produced" by a "process"), you should also accept the idea of an infinite decimal expansion. We accept that real numbers have infinite decimal expansions on a conceptual level, and there are algorithmic ways to symbolically manipulate these decimal expansions (or at least finite portions of them), but I can't actually present you with an infinite decimal expansion itself. The existence of the infinite decimal expansion of a real number does not depend on the existence of an algorithm to produce this expansion.
3. Yes, it is true that any function (I am using the word "function" instead of your word "method" to avoid the algorithmic connotations) that produces only finite strings of digits cannot produce the decimal expansions of all real numbers, because some (in a certain sense, most) real numbers do not have a finite decimal expansion.
4. Yes, there does exist a map from the set of real numbers to the set of infinite sequences of digits, such that every real number is mapped to its decimal expansion. There are some sticky points here, though. The first annoyance is that real numbers with finite decimal expansions can be represented in two ways (for example, 0.999… is exactly equal to 1.000…), so this map is not unique without some additional restrictions. The second difficulty is that this map may not be Computable—there are Uncomputable numbers for which no algorithm exists to compute their decimal expansions. (But their decimal expansions still do exist—remember, the existence of the decimal expansion does not require the existence of an algorithm to produce it). The third snag, from an algorithmic point of view, is the question of how to specify an arbitrary real number as "input" to this function without simply specifying its infinite decimal expansion in the first place. So the map you refer to in statement (4) exists, but it exists only in the mathematical sense of existence; it doesn't exist in a tangible or implementable sense.
5. Well, as I understand statement (2), you can say directly that the reals are the set of numbers whose decimal expansions are produced by (2), without having to say anything about Cauchy sequences or limit points at all; but that's a rather circular statement, because I understand (2) to refer directly to the set of decimal expansions of the real numbers. So here's what I think is a better statement: The set of real numbers is the set of all rational numbers that have finite decimal expansions, plus the limit points of Cauchy sequences of such numbers. (Often the reals are thought of as limit points of rationals, but the subset of rationals having finite decimal expansions is also dense in the reals.) This does not imply that statement (4) is false.

In summary, there is a difference between saying "every real number has an infinite decimal expansion" and saying "there is an algorithm to produce the infinite decimal expansion of any real number." Perhaps this is the core of the issue? —Bkell (talk) 23:15, 12 March 2010 (UTC)

Bkell, in the last paragraph you hit the nail on the head. Thank you. 0¹⁸ (talk) 23:48, 12 March 2010 (UTC)

What is the 98th percentile in a Normal distribution with μ=100 and σ=15?

What is the 98th percentile in a Normal distribution with μ=100 and σ=15?
It has been years since I went to school, so I have almost completely forgotten that which I knew about mathematics.

If the above question does not make sense, then please tell me how I should have said it, and then answer. ;-) --89.8.104.253 (talk) 17:22, 11 March 2010 (UTC)

Do you want the value such that 98 percent fall below that or the value such that 98 percent should not be that far from the mean, regardless of if they are too high or too low? 0¹⁸ (talk) 17:24, 11 March 2010 (UTC)

I want the value such that 98 percent fall below that value. --89.8.104.253 (talk) 17:32, 11 March 2010 (UTC)

Okay, then use a normal-table and look for about 0.98 in the table. You will find that it is between two points. This will tell you how many SD from the mean this value is. In this case, it is between 2.05 and 2.06. So lets say it is at about 2.055. Then the mean is 100, and the SD is 15, so this value is at 100 + 15 * 2.055 = 130.825, which is close to the correct answer (within 0.1). If you want the (more) exactly right answer, you have to use some software that has exact tables. 0¹⁸ (talk) 17:37, 11 March 2010 (UTC)

No, your answer was precisely what I wanted! Thank you! :-) --89.8.104.253 (talk) 18:31, 11 March 2010 (UTC)

Software typically doesn't "have" exact tables, it computes answers on the fly. -- Meni Rosenfeld (talk) 21:25, 11 March 2010 (UTC)

There is a formula that gives results to within about machine epsilon for all IEEE doubles. It's half way between a table (huge file) and a calculation (say, something based on Newton's method or numerical quadrature). 0¹⁸ (talk) 02:20, 12 March 2010 (UTC)

"98 Percentile" means the value which 98 Percent of the observations fall. This is a typical highschool math question. See our article on Standard score. You can also use one of calcultors listed in the links. --Kvasir (talk) 17:39, 11 March 2010 (UTC)

"Standard score" was helpful. Thank you! :-) --89.8.104.253 (talk) 18:39, 11 March 2010 (UTC)

A follow-up question: I seem to remember that in highschool and college, we used to have a small booklet with all the formulas, distributions and tables, that we would ever need in mathematics, probability and statistics. We were required to bring the booklet at exams (so it was not a cheat sheet). Q: What is this kind of booklet called in english? And is there such a booklet available in one of the Wikimedia Foundation projects? --89.8.104.253 (talk) 18:30, 11 March 2010 (UTC)

"Mathematical handbook" comes to mind, but the items I've seen with that name (such as this) are heavier than you describe. "Formula booklet" should do fine.

Wikibooks has Collection of Mathematical Formulas ("work in progress" is an understatement) and Engineering Tables. I also found this. Don't waste ink on the numerical tables - they do it all with computers now. This Wolfram Alpha query is a fairly impressive example.

Wikipedia also has many lists, such as List of probability distributions and Lists of integrals. -- Meni Rosenfeld (talk) 20:20, 11 March 2010 (UTC)

Thank you! :-) --89.8.104.253 (talk) 20:48, 11 March 2010 (UTC)

March 12

little project

I've been studding acase related to prime numbers and how to generate them and iam very convinced it is not possible to follow acertain pattern to generate infinte set of aprime numbers, so i came up with this postulate.Lets, p0, be aprime number≥7 and c(n) are constants generated by acertain pattern or afunction where; {P0+c(n)}; is aset of prime numbers,Now,the maximum prime number, [p0+c(max)]<[P0×P0].It means that any pattern will fail to exceed[p0×p0].I did lots of experiments about this.I want to know if my project is rite and if it can be proven?Husseinshimaljasimdini (talk) 00:05, 12 March 2010 (UTC)

You're going to have to be more precise about what you mean by "a certain pattern or a function." For example, I could define a function c(n) to be

c(n) = [the (n + 5)th prime number] − 7.

Then, for example, c(0) = 11 − 7 = 4, c(1) = 13 − 7 = 6, and so on. This is a perfectly reasonable definition of a function. If I do this, and let p0 = 7, then of course p0 + c(n) is a prime number for every nonnegative integer n. —Bkell (talk) 00:59, 12 March 2010 (UTC).

MY point here is that every pattern no matter what it was; or how it defined would fail under this condition, —Preceding unsigned comment added by Husseinshimaljasimdini (talk • contribs) 10:19, 12 March 2010 (UTC) As amatter of fact to be more specific ,let the set p0+c(0)=p1,.....p0+c(n)=pn,are positve prime numbers < [P0×P0].and my example is 41+2=43+4=47+6=53+.......this pattern fails until 1601+80=41×41.Husseinshimaljasimdini (talk) 10:09, 12 March 2010 (UTC)

The problem is still that you haven't clarified what you consider to be a "pattern". Your example is apparently Euler's trinomial n² - n + 41 which gives distinct primes for n = 1 to 40: 41, 43, 47, 53, ..., 1601. It's mentioned at Formula for primes#Prime formulas and polynomial functions. Most primes from 41 to 1601 are not included in the sequence so it seems you don't require consecutive primes. A sequence with larger jumps can easily end at a larger prime. 41 + 420n is prime for n = 0 to 5: 41, 461, 881, 1301, 1721, 2141. For an example where the number of primes is as large as the initial prime, 5 + 6n is prime for n = 0 to 4: 5, 11, 17, 23, 29. These were simple linear functions. If we allow arbitrary functions then there is no limit as Bkell showed. PrimeHunter (talk) 13:37, 12 March 2010 (UTC).

O!i see. i like these answers i appreciate you. I have also noticed something i`d like to include here,can we say or state that prime numbers can be devided generally into tow types, the first one is the prime numbers that satisfy the following way;for istance; 43.17=731, now if we reverse this number we will get 137 which is aprime;or 17.2=34,which leades to 43 and so on.The second type is the primes that stay primes when we reverse the digits such as 113,101..and so on.Husseinshimaljasimdini (talk) 14:58, 12 March 2010 (UTC)

That is one way to pick a subset of the primes. See List of prime numbers for many others. Palindromic primes are those which read the same backwards (usually with Base-10 implied) such as your 101. Emirps (also called reversible primes) are primes which become another prime when read backwards (again, base-10 usually implied) such as your 113. I (Jens Kruse Andersen) happen to be the discoverer of the largest known emirps [2] and several other computational emirp results.[3][4][5][6][7][8] Primality of a number is not a base-dependent property. Studying special properties of the decimal digits of numbers is usually considered Recreational mathematics. PrimeHunter (talk) 16:25, 12 March 2010 (UTC)

How many types of numbers are there?

I've known about Complex numbers for a while, but I recently discovered the article on Dual numbers. I was wondering, are there any other types of numbers? If so, what are they and how many are there? Furthermore, is it possible to have a "complex dual" number (i.e. a number in the form a + bi + cε?--203.22.23.9 (talk) 02:57, 12 March 2010 (UTC)

Well, the thing is, there's no agreed general definition of what it means to be a "number". So you have things like Cayley numbers that may not really strike you as all that "numerical". That makes this really a language question — how many classes of mathematical objects are called such-and-such numbers?

Ones you might be interested in looking at: Hyperreal numbers, Surreal numbers, transfinite Ordinal numbers and Cardinal numbers. --Trovatore (talk) 03:09, 12 March 2010 (UTC)

(edit conflict)The concept of number is a moving target and so it makes no sense to count how many types of number there are. See Natural number, Positive number, Rational number, Real number, Prime number, Square number. Lots of mathematical constructions generalize numbers, and some of them are even called numbers. Elements of groups, rings, and fields generalize numbers but are often not called numbers. Bo Jacoby (talk) 03:20, 12 March 2010 (UTC).

Another thought - Is there a set of numbers which can be used to solve equations such as |x| = -1 , in the same sense that the complex numbers are used to solve equations such as x² = -1?--203.22.23.9 (talk) 03:33, 12 March 2010 (UTC)

I can't make any sense of it, but that doesn't prove you shouldn't think about it. (Actually I did think about it, when I was a kid -- a point closer to a given point than that point is! But absolute values are not really analogous to quadratics -- they don't have algebraic properties that obviously generalize. Again, this is not a proof. Feel free to work on it; let us know if you get anywhere!) --Trovatore (talk) 03:41, 12 March 2010 (UTC)

Try Split-complex number. Staecker (talk) 12:23, 12 March 2010 (UTC)

(First off, no one has mentioned the Quaternions yet.) If you're going to allow |x| = −1, then your "absolute value" operation is no longer a norm. Norms have very useful properties—you can think of a norm as a generalized absolute value that can be applied to objects other than standard numbers. But I suppose there could be such a thing as a "signed norm" that is allowed to be negative; there are already concepts like Signed measures and Complex measures, which extend the idea of a measure (essentially a length, area, or volume) by allowing it to be negative or complex, respectively. —Bkell (talk) 03:55, 12 March 2010 (UTC)

Something similar is considering spaces that have "norms" that aren't positive-definite, like Minkowski space. In that context some vectors will satisfy |x|² = -1. But that also breaks from the standard definition of what a norm is. Rckrone (talk) 06:13, 12 March 2010 (UTC)

If |x|² = -1 then |x| = i (or -i), what value for x leads to that? --203.22.23.9 (talk)

Minkowski space describes events in space and time. We can choose some coordinates and describe a point in the space as w = (t,x,y,z) where t describes the point in time and x, y, z describe the position in space. The Minkowski norm of this point is |w|² = -t² + x² + y² + z² (compare this to the usual Euclidean norm which would be |w|² = t² + x² + y² + z²). So for example the point w = (1,0,0,0) has |w|² = -1. The Minkowski norm is useful in Special relativity because the value remains unchanged when we change our reference frame with a Lorentz transformation (similar to how the Euclidean norm doesn't change when we rotate the coordinate system), even though it doesn't meet all the criteria of the definition of a norm since |w|² can be negative. Rckrone (talk) 07:16, 12 March 2010 (UTC)

Absolute value of a quaternion

What's the absolute value of a quaternion?
Absolute value of a complex number:

Absolute value of a quaternion:
--220.253.247.165 (talk) 06:55, 12 March 2010 (UTC)

In general the abs of a Hypercomplex number is the sqrt of the number multiplied by its conjugate. Zunaid 07:16, 12 March 2010 (UTC)

Domain of a function

f(x)=√(x-4)÷x-2 then find the domain of a given function —Preceding unsigned comment added by 121.52.145.148 (talk) 07:26, 12 March 2010 (UTC)

I think it's {do, your, own, homework}. --Trovatore (talk) 08:17, 12 March 2010 (UTC)

Also, please be more clear when defining your function. It is difficult to determine whether the function f is defined by , , , or . If you use brackets to clarify the order of operation, we may be able to guide you towards the solution. We cannot give you the solution, of course, as this will not help you to solve any other problems of a similar nature. PST 08:36, 12 March 2010 (UTC)

How about assuming there is already a proper number of parens in appropriate places? May be there is no need to complicate things, if they are not obviously simplified too much. --CiaPan (talk) 12:12, 12 March 2010 (UTC)

Statistical significance when discussing a population

I understand the concept of statistical significance when dealing with sampling. E.g., if you draw a sample of 1000 American household's incomes for two years you can get a probability that the population income differs between the two years, and hence determine whether any statistically significant change in household income can be found.

But often social scientists use the concept of statistical significance when discussing a population, not a sample drawn from a population. For example, it can be claimed that there is a significant downward trend in GDP growth, or that there is a significant correlation between health and GDP per capita. But how can you talk of significance when you already know the population? I understand that you can use R2 to determine the fit of a regression, but is statistical significance a meaningful concept in this case? What's the intuition? Jacob Lundberg (talk) 17:51, 12 March 2010 (UTC)

Actually, you don't know the population. For example, I may have data on GDP from the beginning of time to the present, but that is not the whole population because it does not include the future. Wikiant (talk) 17:57, 12 March 2010 (UTC)

Jacob Lundberg, you have to postulate that the population is a draw from an infinite set of unrealized populations. Deming has a paper on this topic from his time at the Census. 0¹⁸ (talk) 18:34, 12 March 2010 (UTC)

Here is a link to the article at jstor [9]. 0¹⁸ (talk) 19:06, 12 March 2010 (UTC)

feedback rating

Feedback ratings tend to show how many buyers were satisfied with their purchase and not only do buyers rely on this statistic but auction houses rely on it too in deciding whether an item was defective that was returned. While 4,999 buyers that were satisfied with a food product versus only one that died will still have a 99.98% (4,999/5,000) rating while the risk of death may be ignored. What type of feedback rating calculation regardless of complexity can overcome this dilemma and give a much lower feedback rating to reflect the significance of a defective product so that potential buyers will be appropriately warned? 71.100.11.118 (talk) 22:59, 12 March 2010 (UTC)

Feedback systems tend to be game-able toys (and it's in the venue's interest for them to be game-able) but the article Trust metric and its references might give you some ideas. It's silly in general to try to compress complex multi-dimensional data into a single numerical score, although that doesn't stop many from trying. 66.127.52.47 (talk) 23:54, 12 March 2010 (UTC)

Any rating system is going to have a lot of subjectivity from the raters. But let's ignore some of that; there is still a lot of subjectivity from whoever designed the rating system. From a purely mathematical point of view, you're probably going to want some kind of Weighted average, but what weights do you assign? If "satisfied" is given a positive weight of 10, say, what weight should "death" be given? If it's given a weight of, say, −1000, then mathematically you're saying that 100 satisfied customers balance out one dead customer. Is that the correct value to put on death? I don't know; it doesn't seem right. On the other hand, if death is given a weight of −∞, then (under typical rules for arithmetic with ∞) your weighted average will be −∞ even if you have six billion satisfied customers and just one customer who died in a freak accident. That doesn't seem right either. So, if you're going to use a weighted-average approach, you're going to have to place numerical values on the relative worths of satisfaction and death, which is certainly going to involve a subjective judgement. —Bkell (talk) 00:03, 13 March 2010 (UTC)

Feedback ratings aren't usually used to assess safety. There are usually government standards to satisfy for that sort of thing, eg. the Kitemark system in the UK. You are right that giving one number like that doesn't account for different levels of satisfaction. Often the survey will let people choose from "Very poor, poor, satisfactory, good, very good" or similar and then the number you see is everyone that said "satisfactory" or better and they don't give a breakdown. This kind of loss of information is inevitable if you want to reduce things to a single number. --Tango (talk) 00:22, 13 March 2010 (UTC)

So you want to depend upon the lightning speed government to tell you it might be best if you did not stick that food in your mouth instead of having a rating system that would alert you to the danger if the person who eat the food before your just dropped dead? 71.100.11.118 (talk) 13:04, 14 March 2010 (UTC)

I've discovered an interesting problem with weighted feedback. Many people will give the maximum rating or the minimum rating, but very few choose any value in between. I suspect that people who would otherwise give a low rating actually give the minimum rating, to maximize their impact on the overall rating, and others do something similar at the high end. To deal with this, I would look at previous ratings by each responder, and "normalize" them. Do they always give a 0 or 100 ? Then maybe those ratings should actually count as a 30 or 70. StuRat (talk) 02:30, 13 March 2010 (UTC)

Actually lots of people give responses in the middle. It is (on a scale of 1-9) numbers around 3 and 7 that tend to be ignored. --Tango (talk) 02:58, 13 March 2010 (UTC)

You might be interested in the idea of 'quality adjusted life year' by NICE which has to deal with the problem of putting a price onto people's lives. If one in 5000 people dies who would otherwise live a healthy life your talking about a cost of more than a million pounds, so you'd only need a gain of a few hundred pounds for everyone else to easily offset it according to that. However in fact a death rate of 1 in 5000 would get people in a tizzy if you announced a new product a free HD television and cable for life and said there was a downside that unfortunately one in 5000 of those given the free HD would die of an electric shock. Dmcq (talk) 08:41, 13 March 2010 (UTC)

March 13

Maclaurin expansion

Question 6 on the homework is in two parts

(a) Obtain the first four terms of the Maclarin expansion for ln(1+x)
(b) Use the expansion from part (a) to show that:


Part (a) is a piece of cake:
Part (b) would also be a piece of cake if I worked it out the typical way (i.e. finding all the derivatives, powers and factorials etc.) and indeed I have already solved it that way, however the question specifically says Use the expansion from part (a) to solve it. It's possibly a mistake in the question. This is what I have so far:


I could probably expand the brackets to continue but you just know I'm never going to rearrange it to the correct form (after all, where the is that ln(c) going to come from?!)--220.253.247.165 (talk) 00:45, 13 March 2010 (UTC)

—Bkell (talk) 01:12, 13 March 2010 (UTC)

That's only helpful if the OP is allowed to use the rule for logs of products. That's always the problem with this type of question - guessing that you are and aren't allowed to take as given. --Tango (talk) 01:24, 13 March 2010 (UTC)

Thanks Bkell - That doesn't help the slightest bit :P Seriously, how is that useful?--220.253.247.165 (talk) 02:52, 13 March 2010 (UTC)

log(c+x)=log(c(1+x/c))=log(c)+log(1+x/c), you can then use your expression for log(1+x) and change all the x's to x/c's. Job done. --Tango (talk) 02:55, 13 March 2010 (UTC)

THANK YOU!!! Problem solved :)--220.253.247.165 (talk) 04:08, 13 March 2010 (UTC)

Statistics - Conditioning

My stats course recently covered conditioning for finding an expectation, probability, etc., but I really don't understand it. When do we use it and why and how is it useful? Please recommend any good resources. I don't understand my prof or textbook. —Preceding unsigned comment added by 142.58.129.94 (talk) 02:36, 13 March 2010 (UTC)

Have you ever watched a televised Poker tournament? They commonly have little cameras mounted around the edge of the table or somewhere so that the television audience can peek at the players' hands. Often there is a chart on the side of the screen listing the players and the percentage probabilities that each one will hold the best hand at the end. For example, the chart might say Jones 42%, Smith 9%, Brown 31%, Fernandez 18%, indicating that Jones has a 42% probability of holding the best hand at the end, etc. These probabilities are calculated based on the knowledge of the cards the players hold and (in a variant like Texas hold-em) the community cards that have been revealed. (The probabilities ignore such aspects of the game as bluffing; a player may hold the best hand but still lose if he decides to fold.)

In the hypothetical situation above with four players, before any cards are dealt or revealed, each player has a 25% probability of holding the best hand at the end. But as more knowledge is obtained about the cards, these probabilities change—if a player is dealt three aces, her probability of holding the best hand at the end will probably rise, whereas a player who is dealt a two, a six, and a nine of different suits will probably see his probability fall.

This is the idea of conditional probability. The probability that a person in a four-player poker game will be dealt the best hand, given no other information, is 25%. But the probability that she will be dealt the best hand, given that she has been dealt three aces, is quite a bit higher than 25%.

Of course, an example like poker is an overly complicated example. It's simpler to think about, say, the roll of two fair six-sided dice. Let's say one of the dice is white and the other is red, so that we can tell them apart. There are 36 possible rolls of the two dice; each of these rolls is equally likely. Let's shake the dice in an opaque plastic cup and then slam the cup upside-down on the table—we've rolled the dice, but we can't see what we rolled yet. What is the probability we rolled at least an 11? Well, there are three possible ways that could happen: white 5, red 6; white 6, red 5; or white 6, red 6. So the probability we rolled at least an 11 is 3/36, or 1/12. Now let's say we reveal only the white die, and we see it's a 5. Given that information, what is the probability that we rolled at least an 11? Well, there are only six possibilities for what we rolled (we know the white die was a 5, but the red die could be anything from 1 to 6). Out of those possibilities, two of them give us at least 11. So, given the fact that the white die was a 5, the probability that we rolled at least 11 is 2/6, or 1/3, which is significantly higher than 1/12. Our partial knowledge about the situation changed the probability—that's conditional probability. Now suppose instead that when we revealed the white die we saw that it was a 3. Then the conditional probability that we rolled at least 11, given the fact that the white die is a 3, is 0; do you see why? —Bkell (talk) 04:16, 13 March 2010 (UTC)

Sorry, but I already understand contional probability. I was asking specifically about using conditioning to find expectation or probability.. —Preceding unsigned comment added by 142.58.129.94 (talk) 02:36, 14 March 2010 (UTC) —Preceding unsigned comment added by 70.68.120.162 (talk)

I don't understand. What do you mean by "conditioning to find probability" if you don't mean "conditional probability"?

If you understand the idea of conditional probability, then conditional expectation should be an easy step to take: conditional expectation is to conditional probability as expectation is to probability. For example, if I roll two fair six-sided dice, what is the expected value of their sum? It's 7. Do you understand this? Then suppose I add more information: If I roll two fair six-sided dice, one red and one white, what is the expected value of their sum, given that the white die shows 5? The conditional expectation in this case is 8.5: There are six possible rolls (the white die is 5, but the red die could be anything from 1 to 6), each equally likely; the results of these six possible rolls are 6, 7, 8, 9, 10, and 11; so the expected value is (6 + 7 + 8 + 9 + 10 + 11)/6 = 8.5.

If something is still unclear, please explain in more detail what you are having trouble understanding. —Bkell (talk) 02:14, 14 March 2010 (UTC)

I think the OP is asking about Law of total probability and Law of total expectation. In that case their usefulness is, naturally, that they allow you to calculate the probabilityexpectation when you know the conditional probabilitiesexpectations. -- Meni Rosenfeld (talk) 10:55, 14 March 2010 (UTC)

Stats -- algebraic manipulation of V(X) and E(X), etc.

I can do simple things but can't solve them when it's really complicated. Could anyone please recommend any resources that have tips or practice problems for doing manipulations with V(X), E(X), etc.? —Preceding unsigned comment added by 142.58.129.94 (talk) 02:38, 13 March 2010 (UTC)

Are you talking about functions, or something else? 70.171.224.249 (talk) 16:27, 13 March 2010 (UTC)

Do Variance#Properties and Expected value#Properties help? --Tardis (talk) 02:30, 14 March 2010 (UTC)

Symmetric linear solve

Given with all three matrices symmetric, is there an explicit form for E? (If it and A aren't known to commute, of course.) --Tardis (talk) 03:55, 13 March 2010 (UTC)

You want to invert the linear endomorphism defined on the linear space of all symmetric square matrices of order . One case where is certainly invertible (both from matrices to matrices and from symmetric matrices to symmetric matrices), and you have an inversion formula is when A is a perturbation of the identity. Note that (because is the sum of a left multiplication operator and a right multiplication operator, and they commute so you can apply the binomial expansion formula). From you can compute the spectral radius of as . Therefore if and you have , with and has the geometric series expansion. This gives E as a double series, summing over all pairs (j,n) (with 0≤j≤n).

PS: Also note that the spectral theory for the above T_A is linked to that of A. For instance, if λ and μ are eigenvalues of A with eigenvectors resp. u and v, then the rank-one endomorphism u⊗v s.t. (u⊗v)x=u(v·x) is an eigenvector of T_A with eigenvalue λ+μ. In particular, since your A is symmetric, hence diagonalizable, T_A is also diagonalizable (because you get that way n² linearly independent eigenvectors), and if A has no pairs of opposite eigenvalues then T_A is invertible.

In other words, if A=diag(λ_i) in the base (e_i), then T_A=diag(λ_i+λ_j) in the base e_i⊗e_i (the endomorphism with matrix E_ij, which has an entry 1 at the place (i,j) and 0 otherwise)--pma 10:21, 13 March 2010 (UTC)

simple question.

Is there away to find out how many prime numbers before agiven one?like there are 4 prime numbers before 11.Husseinshimaljasimdini (talk) 06:45, 13 March 2010 (UTC)

Just counting them is one way. Prime-counting function#Algorithms for evaluating π(x) mentions some more sophisticated methods. Algebraist 06:59, 13 March 2010 (UTC)

the function Algebraist cited, π(x), is called the prime-counting function. So if you want to find out how many primes there are up to a certain number, all you need to do is put it into the parantheses. For example, if you are interested in knowing how many primes there are up to a thousand billion, you just need to put it in like this: π(1000000000000). 82.113.121.167 (talk) 11:17, 13 March 2010 (UTC)

Just because the function has a name doesn't mean it's easy to compute. "π(x)" is a just a concise way to represent the number that we are trying to find. Rckrone (talk) 21:59, 13 March 2010 (UTC)

http://primes.utm.edu/nthprime/ has an online tool to compute π(x) for x < 3×10¹³. PrimeHunter (talk) 14:44, 13 March 2010 (UTC)

If you're content with a simple approximation, will do. -- Meni Rosenfeld (talk) 20:26, 14 March 2010 (UTC)

is there any proof that current compression couldn't be much better?

Is there anything to show that in the future, people won't have computing power coupled with algorithms that will allow them to collapse something that is a 9 megabyte compressed file today, and actually just have it as a 50 kilobyte compressed file?

I don't mean compression of white noise, of random data - obviously impossible (pigeonhole principle). I mean the kinds of data people actually use and compress.

Thank you. 82.113.121.167 (talk) 11:06, 13 March 2010 (UTC)

Can you be more specific about what is being compressed ? Text files ? Digital photographs ? Movies ? And are we talking about lossless compression only ? StuRat (talk) 14:08, 13 March 2010 (UTC)

Yes. We are talking about lossless compression. I'm talking about the most common contents of zip files, these can be programs, documents, digital photographs, whatever. Is there anything to say that they are being compressed anywhere near the optimal level? 82.113.121.167 (talk) 14:23, 13 March 2010 (UTC)

There is a theoretical result that if you view the messages to be encoded as a random variables, under the correct hypotheses, then the minimum possible average compression ratio is determined by the entropy of the random variables. See Shannon's source coding theorem.

For specific types of messages, this entropy can be measured experimentally. For example, Shannon measured that the entropy of English is about 1 bit per letter (plus of minus). Assuming that experimental result is correct, it means that no lossless compression algorithm for English texts will be able to compress them smaller than 1 bit of output per letter of input, on average.

However, if the language is restricted, this could change the entropy. For example, if I only need to compress sentences that consist of the word "buffalo" repeated over and over, the entropy will be much lower than 1 bit per letter. So the best possible algorithm is only defined relative to a particular set of input messages. — Carl (CBM · talk) 14:35, 13 March 2010 (UTC)

Also, in general, if we consider binary files with some fixed number k bits, there are 2^k such files, and there are

1 + 2 + 2² + ... + 2^k-1

smaller files (adding up the number of files of every possible size starting with 0). But

1 + 2 + 2² + ... + 2^k-1 = 2^k − 1

Therefore there are more files of length k then the total number of files of lengths shorter than k. This means there is no lossless compression algorithm whatsoever that can compress every file of length k to a shorter file. This is a different sort of theoretical limitation on lossless compression algorithms for arbitrary binary data. — Carl (CBM · talk) 14:44, 13 March 2010 (UTC)

uh, thanks, but : JESUS CHRIST! I specifically said the data wasn't random (not "random variables" at all!) but instead the examples of data that we typically zip up. Moreover, when you ARE talking about "random data" then it can't be compressed - on average - at ALL. It's very easy to prove. If you take data of 1 bit, then it is either: 0 or 1 Now if you compress that down to less than 1 bit (ie 0 bits) then you are left with no alternatives - you can't differentiate a.zipped which is supposed to uncompress to a 1 and b.zipped, which is supposed to uncompress to a 0, if they are each 0 bytes long. Moving on, for 2-bit source data, you can have the following files: 00 01 10 or 11 now the only way to "compress" it is to go down to 1 bit compressed files or 0 bit compressed files. Now there is only one 0-bit compressed file, then two possible one-bit compressed files (0 and 1), leaving you three compressed files, A, B, or C. But if there are only three possible results, and four starting states, then obviously you can't represent all of them. One of them must compress to something LARGER than 1. And this is true all over the board. You can't compress white noise, as it has a complete sample space of every possible source input (all bit combinations) and so they must all be represented. Trying to "compress" any random data just can't be done.

Note: despite the above I do have a methodology that is able to compress random data; however it is at least 75 years ahead of current science, and so I am not eager to take on the abuse that I would get on this reference desk if i were to share it. 82.113.121.167 (talk) 14:55, 13 March 2010 (UTC)

The first post I made was about compressing English; the second is about binary data. The theoretical limitation is that you cannot compress English document tighter than about 1 bit per letter, on average. — Carl (CBM · talk) 15:04, 13 March 2010 (UTC)

The analysis in terms of random variables and information content is just that: an analysis. It addresses the question of "how many 90 kiB English text files are there"? Certainly the upper limit is , because that's the total number of 90 kiB files. But most of those won't look anything like English; Shannon's experimental result is that the "effective number of bits" is 1 per letter, so there are "just" 90 kiB English text files. This isn't really a count, but a mathematical estimate based on concepts from random variable theory. Given that many possible files to compress (even with a program that knew that it would never have to compress anything else), you obviously can't do much better than compressing them to about 90(128)=11520 bytes on average — on average because of course you could choose one of them to compress to 0 bytes. --Tardis (talk) 16:23, 13 March 2010 (UTC)

There is a difference between "uniformly random", which as you say cannot be compressed, and the kind of randomness that Carl talked about. Anything - be it English text, source code, images - can be modeled as a stochastic process. English text for example can be seen as a sequence of words from a dictionary, where the probability for each word to occur depends on those that came before it. If you have a sufficiently accurate model for those probabilities, you can 1) Design an optimal compression scheme for documents following the model (which may be computationally intractable) and 2) Estimate the entropy of the model, giving you the expected size of the compressed file. -- Meni Rosenfeld (talk) 17:02, 13 March 2010 (UTC)

Doesn't the unsolvability of the Busy beaver problem imply there will always be specific cases in which compression is possible but where you can't tell that it's possible until you figure out how to do it, and that there's no algorithm for that? (This of course relies on the Church–Turing thesis and might cease to apply if radically new concepts of algorithm were discovered.) Michael Hardy (talk) 19:36, 13 March 2010 (UTC)

I don't think so. Usually the way people formalize compression is by talking about an injective function from the set of strings on some finite alphabet back to the set of strings on that alphabet. Each such function can be viewed as a "decompression algorithm" in a general sense. Now for any particular string there is a decompression algorithm that returns that string when given any string of length 1 as input, but this is not particularly interesting. However, it does show that compression of a fixed string to a string of length 1 is always "possible" if you allow free choice of the compression algorithm. So if you are thinking about particular instances of the busy beaver problem as corresponding to incompressible strings, I don't think that will work. But maybe I am misreading what you said. — Carl (CBM · talk) 19:44, 13 March 2010 (UTC)

On the other hand, the busy beaver problem/halting problem is exactly the reason why the Kolmogorov complexity is not a computable function. Let f be a partial computable decompression function as in my previous paragraph, and let s be a string. Then K_f(s) is the smallest n such that there is a string d of length n with f(d) = s. (Usually, a "universality" assumption is made on f, which ensures among other things that K_f is a total function). For many choices of f the function K_f is not computable. Is that what you meant? (If f is taken to be total computable then we can enumerate the values of f on longer and longer strings, and so K_f would be computable). — Carl (CBM · talk) 19:54, 13 March 2010 (UTC)

You could think of the decompressor as a universal interpreter of fixed size, and the compressed text as a program for the interpreter. The question then becomes, what is the minimal sized program that generates the uncompressed text when run by the decompressor? The minimal size is the Kolmogorov complexity, which is uncomputable. So you might find a compressed text of size N, but in general you can't tell whether an even smaller one exists.

You could go even further and say we don't know how much entropy there really is in the whole Universe. Maybe the Big Bang started with a subatomic particle that could be completely described by a bunch of quantum numbers that would fit on a 3-by-5 card, and we now live in a giant Mandelbrot-like fractal that evolved deterministically from that small initial configuration, and all the weather patterns, biological organisms, galaxies, novels, sequences of coin flips or radioactive decay events, etc., all could in principle be predicted from that tiny set of parameters; and therefore, any plaintext that can occur in the real world can be compressed to a small constant size. That, too, is a statement about Kolmogorov complexity. There is just no way to know. 66.127.52.47 (talk) 00:02, 14 March 2010 (UTC)

L'Hopital's rule

In l'Hopital's rule, can x tend to 0, i.e., is ? —Preceding unsigned comment added by 76.230.230.42 (talk) 21:36, 13 March 2010 (UTC)

Yes, x can tend to anything. The rule applies if the limit gives 0/0 or infinity/infinity and the derivatives all exist (and are continuous?). --Tango (talk) 21:38, 13 March 2010 (UTC)

Oh, and the latter limit has to exist, too. --Tango (talk) 21:40, 13 March 2010 (UTC)

Question - According to the article, you only need an indeterminate former limit and for the latter limit to exist. If the latter limit exists, does that mean that continuous derivatives exist? Zain Ebrahim (talk) 22:40, 13 March 2010 (UTC)

No, the derivatives need not exist at the point you're taking the limit to (in this case at x=0). Rckrone (talk) 22:47, 13 March 2010 (UTC)

They do, however, need to exist around that point (otherwise the whole thing makes no sense). --Tango (talk) 22:53, 13 March 2010 (UTC)

Aren't we interested in the limit of the ratio of the derivatives only? Why do we need f' and g' to exist around 0 if f'/g' exists around 0? Zain Ebrahim (talk) 23:04, 13 March 2010 (UTC)

If at some point either f' or g' doesn't exist, what could f'/g' possibly mean there? Rckrone (talk) 23:20, 13 March 2010 (UTC)

f' and g' have to exist in arbitrarily small neighborhoods of x, but not necessarily at x itself. That's the whole concept of a limit. 66.127.52.47 (talk) 23:30, 13 March 2010 (UTC)

Ah, I see it now. Thanks. Zain Ebrahim (talk) 07:48, 14 March 2010 (UTC)

Btw, my suggestion, as an analyst, is: forget about de l'Hopital rule. It's a not-successfull tentative of a theorem, that still survives for some unclear unmathematical reason. Learn the Landau notation instead. --pma 22:37, 13 March 2010 (UTC)

I agree. I don't think I have ever used l'Hopital's rule outside of questions on l'Hopital's rule. There are almost always better ways to calculate limits. I think l'Hopital's rule is still taught because a lot of people prefer to just memorise algorithms than actually learn about a subject. If you actually understand what limits are and how they work, then you don't need l'Hopital's rule. --Tango (talk) 22:53, 13 March 2010 (UTC)

Yeah, it enjoyed some popularity in old times, where it was successfully employed as a quick tool to kick out students in examinations. But today we have the opposite problem, and try everything to let them survive. So de L'Hopital rule is going to go in the attic together with moustache nets.--pma 23:22, 13 March 2010 (UTC)

Approximations to roots

For large n, x^(1/n) ~= 1 + log(x)/n. What is the logical way of continuing the expression on the right-hand side to higher-order terms (and correspondingly greater accuracy)? 86.136.194.148 (talk) 22:42, 13 March 2010 (UTC).

If x>0 is fixed and you let n diverge, since x^1/n=exp(log(x)/n), you have all asymptotics you want from the exponential series, together with bounds on the remainder. For instance,

x^1/n=1+log(x)/n+ log(x)²/2n²+o(n^-2).--pma 23:02, 13 March 2010 (UTC)

Thanks pma. Do you mean "... +o(n^-2)" or should that be "... +o(n^-3)"? Do you know what the next few terms should be? Is there any pattern? 86.136.194.148 (talk) 00:14, 14 March 2010 (UTC). Don't worry, I see how to do it now!

March 14

Approximate solutions to equation

I'm trying to solve this for x:

1 + (n + 1)*x^(1/(n + 1)) = n*x^(1/n)

where n is some given positive number.

I can find numerical solutions to any desired accuracy using standard methods, but this doesn't tell me what I actually want to know, which is how the solution for x mathematically grows with n as n becomes very large. What I'm hoping for is a closed-form formula like

x ~= some expression involving n

which is asymptotically correct as n tends to infinity (and hopefully also gives good numerical approximations for reasonable-sized values of n). I feel this ought to be possible, but I don't think my math skills are good enough to figure it out. Any ideas? 86.136.194.148 (talk) 01:13, 14 March 2010 (UTC)

Is this homework? Try raising both sides to the (n+1)th power and seeing if you can solve it from there. 66.127.52.47 (talk) 02:48, 14 March 2010 (UTC)

No, this is not homework, so there's no need to be coy about the answer. At the moment I don't see how your proposal works. Can you be more explicit? 86.136.194.148 (talk) 03:04, 14 March 2010 (UTC).

My guess is where c is such that (Igny (talk) 06:00, 14 March 2010 (UTC))

Which is . Also, . -- Meni Rosenfeld (talk) 12:05, 14 March 2010 (UTC)

• Thank you both. Those answers look very plausible as far as they match the numerical results that I can calculate. If you have the time, and it's not going to take pages and pages to explain, I'd like to know how you arrived at this answer. 86.165.23.35 (talk) 20:04, 14 March 2010 (UTC).

What I did (which only resulted in the approximate solution I gave) is let . Then the equation is . Assuming the higher-order derivatives are small, this means (note - the derivative is wrt to n). The solution then follows without much effort. -- Meni Rosenfeld (talk) 20:16, 14 March 2010 (UTC)

I had a similar approach. I considered the solution to your equation, , and looked at equations for or Eventually that gave me insight to consider and it is very easy to check that satisfies and you can see that asymptotically introduced earlier. (Igny (talk) 21:16, 14 March 2010 (UTC))

Geometry problem

Let ABCD be a quadrilateral with AD = BC. Extend AD and BC to intersect at X. Let M be the midpoint of AB and N the midpoint of CD. Prove that MN is parallel to the bisector of angle X.--RDBury (talk) 05:14, 14 March 2010 (UTC)

This sounds like a homework question, so you will need to tell us what you have tried so far. --Tango (talk) 05:25, 14 March 2010 (UTC)

Use Law of sines lots of times. Rckrone (talk) 07:24, 14 March 2010 (UTC)

I'd first think about sliding BC along the line in which it lies, to the place where it is just as far from the intersection point as AD is. Then ask whether the line through the two midpoints has the same slope as it did before that move. If so, then you've done most of the work. Michael Hardy (talk) 11:45, 14 March 2010 (UTC)

It's not a homework question. I was hoping it was a special case of theorem I could reference, or maybe someone could see a relatively simple proof. If not I'm not going to worry about it.--RDBury (talk) 13:17, 14 March 2010 (UTC)

Here's what I meant before with law of sines. Suppose wolog that AB < CD and say AD intersects MN at E and BC intersects MN at F. Call angle AEM θ and angle BFM φ. Use law of sines on triangles AEM and BFM to show that AE/BF = sinθ/sinφ. Use it again on triangles DEN and CFN to show that DE/CF = sinθ/sinφ. Then AE = BF and so sinθ = sinφ. Either θ = φ or they are supplementary, but they can't be supplementary since AD and BC are not parallel. Rckrone (talk) 18:39, 14 March 2010 (UTC)

That makes no sense. AE and BF are not in general equal. And θ and φ are not angles in a common triangle, so applying the law of sines won't work directly. Michael Hardy (talk) 19:11, 14 March 2010 (UTC)

AE and BF are in fact necessarily equal (note: not AX and BX). Obviously I omitted steps but as I said you use law of sines on pairs of triangles to show the relationships between θ and φ. AM/sinθ = AE/sin(angleAME) and BN/sinφ = BF/sin(angleBNF). sin(angleBNF) = sin(angleAME) and AM = BN, so then AE/BF = sinθ/sinφ, etc. Does that make it more clear? Rckrone (talk) 20:13, 14 March 2010 (UTC)

Simple solution Let u and v be unit vectors pointing out from the point of intersection so that

A = ''xu

D = (x + a)u

B = ''yv

C = (y + a)v

Then

M = (''xu + yv)/2

N = (x + a)u/2 + (y + a)v/2

The vector from M to N is therefore

(''au + av)/2

and that does not depend on x or y! Therefore if it works for some x and some y, it works for all. Now just think about the case where x = y and use symmetry. Michael Hardy (talk) 22:27, 14 March 2010 (UTC)

Terminology: equivalence of ordering or optima

I'm looking for terminology related to the following idea. I'm sure this must have been studied and discussed by someone, but I'm having trouble finding the relevant phrases, so I'm getting nowhere googling it. Let's say you have two functions f and g, both (or something similar). I'm looking for a kind of property (equivalence?) that states that , and similar for other inequalities. It seems to me that this kind of thing would be very important in fields like optimization, so it's likely to have a name. My guess is that a far more general statement of this does have a name, but don't recognize it as such. Any insights? risk (talk) 13:34, 14 March 2010 (UTC)

Functions that preserve (or reverse) order are called Monotonic functions. Gandalf61 (talk) 14:51, 14 March 2010 (UTC)

I did get that far, but unfortunately, that doesn't quite capture what I wrote above. I'm looking for the phenomenon where two functions induce the same ordering on a given set (independent of any ordering that set may have had to begin with. Monotonocity may be relevant to this, but I haven't quite worked out how. risk (talk) 15:06, 14 March 2010 (UTC)

What is needed is not that they are monotonic functions of their argument. If they are monotonic functions of each other, that is enough. But even if they are not functions of each other, I think in some cases the proposed inequality may hold. Michael Hardy (talk) 19:03, 14 March 2010 (UTC)

Sorry to keep shooting everybody down, but I think you misunderstand what I'm after. I don't want to prove this for specific functions, rather I'm after terminology. If this property holds for two functions, what would you call that? Would that be some kind of equivalence? Isomorphism? Duality?

As an example of why this is relevant: consider an optimization problem where f(x) is the objective function we're trying to maximize. Any kind of hill-climbing or gradient ascent algorithm will then find maxima of f. If the above property holds for f and another function g, then we can replace g with f and the algorithm will still behave basically the same (in some sense). This can be helpful, for instance, if g is easier to compute. risk (talk) 19:55, 14 March 2010 (UTC)

I think Michael's answer above is what you are looking for. Your property certainly holds when f and g are monotonic functions of each other. I believe this is also the only case. No need for new terminology then. Maximising g=ln(f) instead of f (possible since ln is monotone) is a standard trick.213.160.108.26 (talk) 22:52, 14 March 2010 (UTC)

Oh crap, I just noticed I made a mistake (which explains a lot). The function definition above should actually read , so the functions are actually scalar fields. In any case, I can assume that there is no really obvious answer which I should've known about, which is good enough for now. risk (talk) 23:14, 14 March 2010 (UTC)

Sparse Graph Colorings

I'm trying to find examples of sparse graphs with relatively high chromatic number. More specifically, graphs with chromatic number 6, 7, or 8, which have the property that their total number of edges is less than four times their number of vertices (E<4V). The obvious examples are K6 and K7, but I'm having trouble finding others. Could you point me towards any resources? Black Carrot (talk) 16:37, 14 March 2010 (UTC)

The obvious thing to do is to take a graph of suitably high chromatic number and add loads of isolated vertices (or a long path, if you want connectivity) to bring down E/V. Algebraist 17:43, 14 March 2010 (UTC)

I was thinking more along the lines of minimal graphs, where removing an edge reduces the chromatic number. (And yes, connected.) Black Carrot (talk) 18:46, 14 March 2010 (UTC)

Integration

How would I integrate (where a=-1 and b=1)? I know it forms a half-circle and I could solve really easily by just пr², but I'm curious for a way that works for not just circles. THanks 68.76.147.34 (talk) 17:13, 14 March 2010 (UTC)

This kind of integrals are usually done using substitution. works in this case. -- Meni Rosenfeld (talk) 17:24, 14 March 2010 (UTC)

Look at Trigonometric substitution. Michael Hardy (talk) 18:56, 14 March 2010 (UTC)

L'Hopital vs. Landau

In a previous question by someone else (entitled L'Hopital's rule) there was mention that Landau notation (Big O notation) was superior to L'Hopital's rule in finding limits of indeterminate forms. I find this hard to comprehend, given that Big O notation throws away constants. e.g. For f(x) = 3e^x-3 = O(e^x) and g(x) = 2e^x-2 = O(e^x). However both and (both by L'Hopital's rule and by inspection), and both values depend critically on the very multiplicative constants Big O notation explicitly discards. Is it true that Big O notation is more applicable in finding limits of indeterminate forms than L'Hopital's rule, and if so, how does one accomplish it? Yes, I'm aware that the ratio f(x)/g(x) can be trivially simplified. I apologize for not having time to find a more complicated example. My key point is that Big O notation discards multiplicative constants, which may play a role in the final limit. i.e. given that it fails in this trivial example, how can you be confident that it would work in a more complicated one? -- 174.21.235.250 (talk) 19:49, 14 March 2010 (UTC)

You have a choice what to use the Biglittle O notation for. Taking for example , it's when . And it's also . And . In your example what you want is when . Then . If you mess something up (absorb too many terms in the Landau notation) you'll get an ambiguous result - e.g., for you get . -- Meni Rosenfeld (talk) 20:03, 14 March 2010 (UTC)