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June 17

Arithmetic

shanu 07:42, 17 June 2008 (UTC)Suppose than x,y,z are real numbers not equal to zero. a & b are two negative real nos. Is it always possible to find another real no. c such that x^a + y^b = z^c . For example consider 2^-1 + 3^-1 = 1.2^-1 . I am thinking of this because fermat's last theorem doesen't allows it for natural nos. greater than 2.

For any given x, y, a and b values the left side of your equation is some constant value: v = x^a + y^b. Then the equation v = z^c with given z has a solution defined by logarithm: c = log_zv = log v / log z. --CiaPan (talk) 07:59, 17 June 2008 (UTC)

Cube Root

How can we find cube root of any no. ,like 2 without using calculator? —Preceding unsigned comment added by Rohit max (talk • contribs) 07:59, 17 June 2008 (UTC)

We can use a paper and a pencil. Possibly Tables of logarithms, too... --CiaPan (talk) 08:02, 17 June 2008 (UTC)

Without a log table it's a little harder. A good way to do it would be to start at some estimate $x_{0}$ and repeatedly iterate $x_{n+1}={\frac {2}{3}}x_{n}+{\frac {a}{3x_{n}^{2}}}$. -- Meni Rosenfeld (talk) 08:54, 17 June 2008 (UTC)

Perhaps this will also be interesting. Not very effective for pen and paper, though. -- Meni Rosenfeld (talk) 09:10, 17 June 2008 (UTC)

Apparently, there are a couple of "paper and pencil" algorithms for finding cube roots. This is one example (haven't checked it out, to be honest). Pallida  Mors 01:41, 18 June 2008 (UTC)

The simplest way is by the bisection method. Of course, if you actually mean *any* number, you'd have to consider calculating sines and cosines as well...--Fangz (talk) 01:55, 18 June 2008 (UTC)

-3^2 = ?

I brain farted with the initial question, asking 9^9. This is why Meni answered 'wrongly' - it's my fault -- 88.217.28.51 (talk) 08:54, 17 June 2008 (UTC)

Hi all,

-3^2 = 9

vs

-3^2 = -9

Which is correct? We can't agree, so I am asking here.. -- 88.217.28.51 (talk) 08:43, 17 June 2008 (UTC)

See Order of operations. Exponentiation takes precedence over negation. Thus to evaluate -9^2, we first replace 9^2 with 81, giving -81. Thus -9^2 = -81. If we wanted to do the negation first, we would use parentheses: (-9)^2 = (-9)*(-9) = 81. -- Meni Rosenfeld (talk) 08:49, 17 June 2008 (UTC)

^ vs **

Now, they claim that ^ (not the XOR kind) and ** are different operators. Opinions on that one? -- 88.217.28.51 (talk) 10:38, 17 June 2008 (UTC)

What is the context? In what way do "they" claim the operators are different? ** is not used in mathematical writing, only in some programming languages. Some languages use ^ to mean exponentiation, and some use **. I do not know of any other difference. -- Meni Rosenfeld (talk) 10:47, 17 June 2008 (UTC)

My thoughts exactly. Thanks -- 88.217.28.51 (talk) 10:52, 17 June 2008 (UTC)

The only language I know of that distinguishes between them is Haskell. ^ raises any number to an non-negative integer power (thus the result can have the same type as the base even in the case of integers, and it can be computed by repeated multiplication). ** raises numbers to floating-point powers and gives a floating-point result (in this case we can make no guarantee about whether the result is an integer or even rational, and we have to use a different algorithm to calculate the value). --Taejo|대조 11:51, 17 June 2008 (UTC)

See also Exponentiation#Exponentiation in programming languages. PrimeHunter (talk) 12:16, 17 June 2008 (UTC)

Python is another example. ** is exponentiation and ^ is XOR. Dragons flight (talk) 16:04, 18 June 2008 (UTC)

OP said "not the XOR kind" --Taejo|대조 17:44, 22 June 2008 (UTC)

June 18

Constant rule of differentiation

Constant rule states that derivative of any constant is zero:

let y=c.

${\frac {dy}{dx}}=\lim _{\Delta \ x\to 0}{\frac {\Delta y}{\Delta x}}=\lim _{\Delta \ x\to 0}{\frac {c-c}{\Delta x}}=\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}={\frac {0}{0}}=0$

Why should 0/0 be 0 rather than undefined?--218.102.124.108 (talk) 05:24, 18 June 2008 (UTC)

Because limit of the ratio of two functions is not necessarily equal the ratio of limits of those functions. Anyway check the limit's definition, and see that values of 0/Δx are arbitrarily close to zero when Δx approaches zero (in fact they are all equal zero), so they satisfy conditions in the limit definition. --CiaPan (talk) 05:47, 18 June 2008 (UTC)

Your mistake is in the step $\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}={\frac {0}{0}}$. Taking a limit is not the same as plugging in a value for the function; that is, $\lim _{x\to a}f(x)$ is not necessarily $f(a)\;\!$ (if it was, we wouldn't invent limits in the first place). The correct steps are $\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}=\lim _{\Delta \ x\to 0}0=0$. If you make this mistake it will happen when you calculate any derivative, not just for a constant. Do read up on the definition of limits. -- Meni Rosenfeld (talk) 08:28, 18 June 2008 (UTC)

More specifically, you can 'plug in' limits like that, provided the denominator does not equal zero, as it does in your case.--Fangz (talk) 11:51, 18 June 2008 (UTC)

You can plug in values if the function is continuous at the point. It is true that for rational functions, this happens exactly when the denominator is not zero at the point. Not all functions are rational, though. -- Meni Rosenfeld (talk) 12:54, 18 June 2008 (UTC)

Thanks everyone. I think I've known what's wrong with me. But what I intended to solve is 0/0:

${\frac {0}{0}}={\frac {0^{1}}{0^{1}}}=0^{1-1}=0^{0}=1?$

So what is the result of 0/0?--218.102.124.108 (talk) 15:34, 18 June 2008 (UTC)

0/0 is undefined in all common structures. See division by zero. -- Meni Rosenfeld (talk) 15:38, 18 June 2008 (UTC)

${\frac {0}{0}}=x\iff 0=x\cdot 0$, and that's true for any number x (it can also be true for infinity in some circumstances), so basically 0/0 can equal anything. If you end up with a 0/0 somewhere, you have to go back a step and work it out a different way, because it could come out to be absolutely anything. For example, if f(x)=0/x, then, if you want f to be continuous, you have to set f(0)=0, but if f(x)=x/x then f(0)=1, or if f(x)=2x/x, then f(0)=2. (Those are pretty trivial examples, but they should give you an idea of what can happen.) --Tango (talk) 15:52, 18 June 2008 (UTC)

The string of characters "0/0" can mean two different things. It can mean an attempt to divide 0 by 0, in which case it can't equal anything as it is undefined. It can also mean a certain Indeterminate form, in which case it could indeed evaluate to anything, but is only a mnemonic, not an actual mathematical entity. -- Meni Rosenfeld (talk) 16:05, 18 June 2008 (UTC)

Yes, I should have been clear I was speaking very informally, just to give an idea of why it is undefined. --Tango (talk) 16:18, 18 June 2008 (UTC)

cubic metres, cubic kilometers

A recent article about the Niger Delta talked about "10 trillion cubic meters of natural gas reserves" in one region there, and "28,800 cubic km" (kilometres, we assume) of natural gas reserves is Russia. Too many zeroes...will 1000 cubic meters be 1 cubic kilometre, and therefore 28,800 cubic km, will be 28,800,000 cubic meters...which is only 28.8 million cubic meters...which must be way less than 10 trillion (10,000,000,000,000 (is a trillion a thousand billion?)...So the Niger Delta reserves in this one region are millions of times more than the reserves in Russia? Surely we are confused over the math, but maybe the press was wrong. Can anyone confirm or correct our understanding? Thanks if you can help. —Preceding unsigned comment added by 213.84.41.211 (talk) 14:20, 18 June 2008 (UTC)

There are 1000 meters in a kilometer, but 1000^3 = 1,000,000,000 cubic meters in a cubic kilometer. Wikiant (talk) 14:27, 18 June 2008 (UTC)

So there are in fact 29 trillion cubic meters of gas in Russia. I think mass is a much more useful measure for the quantity of a gas then volume, though. -- Meni Rosenfeld (talk) 15:36, 18 June 2008 (UTC)

Agreed. I don't think Nigeria is the same temperature as Russia. Algebraist 15:42, 18 June 2008 (UTC)

It might be underground, though. At least, there is probably less of a difference. --Tango (talk) 15:48, 18 June 2008 (UTC)

I think such measures are based on what the real volume would be under certain standardized conditions. See Natural gas#Energy content, statistics and pricing. PrimeHunter (talk) 00:04, 19 June 2008 (UTC)

I figured as much, but I still find it unnecessarily confusing. -- Meni Rosenfeld (talk) 11:49, 19 June 2008 (UTC)

In answer to the side question, most people writing about numbers that big in English use a billion to mean 1,000,000,000 (10⁹) and a trillion to mean 1,000,000,000,000 (10¹²). But until recent decades, it was usual in English outside of North America for a billion to mean 1,000,000,000,000 (10¹²) and a trillion 1,000,000,000,000,000,000 (10¹⁸). The cognate words in most other languages have the larger meanings, and some people still prefer them for use in English. See long and short scales (and please do not debate them here). --Anonymous, 02:27 UTC, June 19, 2008.

A billion being a thousand million is pretty much universal these days. There's no need to worry about possible alternatives. --Tango (talk) 11:58, 19 June 2008 (UTC)

See billion (word). Bo Jacoby (talk) 22:35, 20 June 2008 (UTC).

Random numbers

Random numbers may be generated on computers, graphing calculators, and even Excel spreadsheets. (Although these are technically referred to as "pseudo" random numbers, that distinction is irrelevant for purposes of this question.) The following scenario came to mind, from which this question arises. Let's say that a teacher asks his students, as a homework practice exercise, to generate 5 random numbers, with each random number ranging from 0 to 9 inclusively. So, for example, one student might generate 3, 7, 2, 1, 9. Another student might generate 5, 6, 3, 5, 2. And so forth. Now, say there is a lazy student who did not do the homework and, when asked, lies to the teacher and fabricates that he generated 7, 7, 7, 7, 7. (Also, indicating a none-too-bright student, as he could just as easily have lied with a less "suspicious" result of another fictitious string of numbers such as, say, 4, 1, 8, 3, 0 or so.) So, the teacher gets suspicious at the result of 7, 7, 7, 7, 7, and suspects that the student cheated and did not, in fact, actually do the homework assignment. So, here is my question. Should the teacher, in fact, get suspicious at such a result? Is a result such as 7, 7, 7, 7, 7 equally likely as a series of 5 different digits (say, 3, 8, 0, 2, 6) … or is such a result not as equally likely in the generating of random numbers? My understanding is that any (and all) of the 5 separate digits is randomly generated and, thus, a result such as 7, 7, 7, 7, 7 should not be a surprising result. But, my gut instinct also tells me that it is highly unlikely to generate a result such as 7, 7, 7, 7, 7 and that a mixture of different digits is much more likely / probable. What is the truth of the matter? If all 5 digits are randomly generated, why does a result like 7, 7, 7, 7, 7 seem so improbable? Any insights? Thanks. (Joseph A. Spadaro (talk) 23:23, 18 June 2008 (UTC))

There are more than one way to answer this; I'll pick the one that seems most in line with your motivation. It all comes down to Bayes' law. Let A be the event "student cheated", B be "Student chose 38026", C be "student chose 77777". Assume P(A)=10% and that a cheating student can only come up with a constant sequence, thus P(C|A)=10%. Then $P(A|C)={\frac {P(A)P(C|A)}{P(C)}}={\frac {0.01}{0.010009}}\approx 1$, while $P(A|B)=0$. Of course you can replace the assumptions with whatever you find reasonable to get more realist estimates. -- Meni Rosenfeld (talk) 23:36, 18 June 2008 (UTC)

Theoretically a computer can generate a random number, not just a pseudorandom number. Although in my opinion and ignorance of physics, I am under the belief that of what’s on the market, only the quantum random number generators can be truly random, and those will cost as much as the computer to run them.

As for the actual question… I would try to find some way of measuring how random a number “looks”. Very subjective, but 2 4 9 5 2, 9 5 2 3 1, and 7 7 7 7 7 are all equally likely, but like most normal people my first reaction wouldn’t be to question the first two, but would be the third: because there are a lot of numbers that “look like” the first two, but only a few if any numbers that “look like” the third. GromXXVII (talk) 23:57, 18 June 2008 (UTC)

(e/c)Random is as random does - but human brains are very good at spotting patterns, even when they're spurious. Write down what you think would be 50 perfectly random coin tosses, and you're not likely to write down any sequence with more than 3 or 4 heads in a row. And yet, you're actually quite likely to get a run of 5, which you will find if you actually do the coin tosses a few times. The run of 5 heads is a pattern, and since we don't expect to see patterns in "random" data, it throws us.

That said, if you see something "odd", then you can perform a hypothesis test to see if it's statistically unlikely. First, you get an idea of what the results would look like if they were selected randomly - say, with each digit uniformly distributed between 0 and 9. Then you quantify just what is "odd" about Johnny's 7, 7, 7, 7, 7 (i.e. "it's the same digit 5 times). Finally, you calculate the probability that you would find a result equally as odd as or more odd than Johnny's, and from that probability determine whether it still seems unlikely (5% is a common cutoff, 1% if you're being more lenient). The probability of getting 5 digits the same is 1/1000, or 0.1%, so Johnny might be looking a little nervous right now.

And for the silly answer, both students fail. The perfectly random selection would be 4, 4, 4, 4, 4[1]. Confusing Manifestation(Say hi!) 00:06, 19 June 2008 (UTC)

[ec] Some more thoughts:

1. 77777 has a lower Kolmogorov complexity than 38026. This is why a cheater with no access to an adequate RNG is more likely to generate 77777 than 38026.
2. While 77777 is as likely as 38026, there are simple associated variables that are not equilikely. For example, you can define a variable "how many different digits are in the sequence". This will be 5 with probability ≈30%, and 1 with probability 0.01%.
3. Occam's razor says the simplest explanation to an observation tends to be the correct one. The sequence 38026 is not likely to occur, but if it does occur, the only explanation is that it happened randomly (again assuming that without a RNG it's difficult to come up with it). If, however, we are presented with 77777, the simplest explanation is that the student cheated.

-- Meni Rosenfeld (talk) 00:14, 19 June 2008 (UTC)

Here are some approximate odds (I didn't want to use a calculator to find the exact odds):

Chances of a given student picking 7 7 7 7 7 exactly: 1 in 100,000.

Chances of a given student picking any number 5 times in a row: 1 in 10,000.

Chances of any student in a class of 25 picking any number 5 times in a row: 25 in 10,000 or 1 in 400.

Chances of any student in a 8 classes of 25 each picking any number 5 times in a row: 200 in 10,000 or 1 in 50.

Chances of any student in a 8 classes of 25 each picking any number 5 times in a row or any sequence of 5 numbers in a row (0-4, 1-5, 2-6, 3-7, 4-8, 5-9, 6-0, 7-1, 8-2, 9-3): 200 in 5,000 or 1 in 25.

Add in reverse sequences like 9 8 7 6 5 and even sequences like 0 2 4 6 8 and odd sequences like 1 3 5 7 9 and now reverse those, too, and there's a good chance that one student would hit one of those. StuRat (talk) 00:17, 19 June 2008 (UTC)

I don't think that's a good argument. There's a lot of reason to think that a student without a RNG can just make up a number like 38026, and it's arguable that a number like 77777 is unlikely to arise out of a student told to work out a random number generation and thus *more* likely to be random. (In fact, IIRC, there's good evidence that in human generated sequences, usually runs of numbers are *less* common than in actual random numbers, especially in the binary case.) Trying to find ways that 77777 is special is also very dangerous, since the arguments previously put forward are equally applicable to any finite sequence. Even something like Kolmogorov complexity isn't helpful, because you can easily define a post-facto language where 38026 is easy to generate while 77777 is not. (E.g. 38026 can be someone's birthday, while 77777 cannot)

I think the probably best way to deal with the problem is a bayesian approach. You need to pre-specify an impression of what a human generated sequence is like, and what a good rng is like, and set some prior probabilities. Then just update your beliefs. Specifically, there isn't a good way to just say a probability - your view of whether the sequence is random or not can only be stated relative to a well (and in advance, especially not based on perceived specialness of the sequence) defined alternative hypothesis and/or a set of priors.--Fangz (talk) 00:48, 19 June 2008 (UTC)

1. You have raised the psychological warfare issue - "the student knows I will suspect 77777, and thus will avoid it if he cheats". The problem with this is that it can always be taken one step further - "The student knows I know he knows, and thus will choose 77777 if he cheats in order to trick me". It's thus not very reliable reasoning.

It is not reliable reasoning, but it does serve to point out a flaw in your argument. Implicit in any attempt at discriminating one hypothesis from another is a model - and indeed, a psychological model of how the student is likely to react. It is impossible to argue broadly that out of *all* possible algorithms the student may use instead of a random number generator, certain sequences are objectively more likely than according to random chance.

1. You say that trying to find projections from the space of sequences to a lower-dimensional space in which 77777 is projected to a salient outcome is dangerous. However, it is only dangerous if the projection is complicated and arbitrary. For example, we could define "$p(X)$ is 0 if $X=38026$, otherwise $p(X)=1$". Then $p(X)=0$ will have a very low probability, so we will reject an answer of 38026. However, this projection is very arbitrary and silly. If we stick to something simple and natural, like "$p(X)$ is the number of distinct digits in $X$", we should be good.

But the issue is that there is no rigorous mathematical framework around that definition of 'simple and natural'. I stress, the difference is that you need to pre-define your specialness criteria. Not choose it based on the number itself. (For example, regarding 38026, 38026 is 2 * 19013, which is a very rare 5 digit centered square prime number. 77777 is not a multiple of two primes like that. Are prime numbers more special and non-arbitary than repeating digits? Who can tell?) --Fangz (talk) 15:57, 21 June 2008 (UTC)

1. In your Bayesian framework, you say we need "an impression of what a... sequence is like". But this can only be formalized in terms of some projection like above. The essence of your suggestion is that instead of arbitrarily choosing the probability of a student cheating and the distribution of $p(X)$ given that he is, we should calculate them based on observations. However, such observations are hard to come by. Cheating students will generally not admit to doing so, and asking random students to invent their own sequences may not give you a lot of insight into the mind of the cheater.

-- Meni Rosenfeld (talk) 11:45, 19 June 2008 (UTC)

If you want to get psychological you also need to account for the fact that if a student genuinely gets 7,7,7,7,7 out of a RNG, they are likely to think something has gone wrong and do it again. --Tango (talk) 15:04, 19 June 2008 (UTC)

Basically, to give a sensible assignment like that, you should ask for enough numbers that a sensible randomness test will give a clear answer. Say you ask for 200 numbers and you get the following output from a student:

1549248135779511892489513549165798

7691249876216984798416498765194327

9165498732498490987951979164984691

9498496519496195498795195746249832

4981951976249519762495462498746579

5162195762169795194957264916

Did he do this by tapping on a keyboard or by using a RNG? (Hint - I am lazy!) So maybe you can do the chi-square test and the runs test and get a pretty good answer. --Slashme (talk) 13:52, 19 June 2008 (UTC)

Maybe the runs test isn't very sensitive! For example, for the numbers I gave (monkey style), I got a runs test value of -1.2535104, giving a P-value of 0.10601 and for 200 random numbers from Excel, I got a runs test result of -1.5420651, giving a P-value of 0.06258. (done using a javascript applet found on the web, so maybe there's a bug? I'm too lazy to check the code for correctness) --Slashme (talk) 14:40, 19 June 2008 (UTC)

Follow up

Thanks for all the responses above. They were all very helpful. What makes the most sense to me lies within Grom's reply, which essentially states that some data will "look" randomly generated and some data will not "look" randomly generated. (Thus, arousing the teacher's suspicions.) So, let me re-phrase my original question. Let me use "39710" as a value that does look randomly generated, and let me use "77777" as a value that does not look randomly generated. I understand that the definition of "look" is fuzzy, but I assume that people generally know what I mean. (For example, a value such as "12345" does not look randomly generated, but a value such as "83016" does.) So, to re-phrase my original question … in terms of mathematics and random number generation and probability … is a value that does look randomly generated (e.g., 39710) equally like / probable as a value that does not look randomly generated (e.g., 77777)? If you want to clarify the fuzzy definition of "look", I can ask it this way: is a random number which has no repeating digits (e.g., 39710) equally likely / probable as a random number which has all 5 digits repeated (e.g., 77777)? In a true random generator, will both of those values appear with equal likelihood / equal probability? Thanks. (Joseph A. Spadaro (talk) 16:24, 19 June 2008 (UTC))

The chance of a specific sequence coming up is the same for any sequence (assume a truly uniform and independent random number generator). So 83016 is just as likely as 77777. The real question is about how many "random looking" sequences there are and how many "non-random looking" sequences there are, because those are what determine the chance of the sequence you generate being random looking or not, because you don't actually care which (non-)random sequence you get, you just care if it looks random or not (ie. you could replace 77777 with 55555 and it wouldn't make any difference). There are 10⁵=100,000 possible sequences of 5 digits. How many of those don't look random depends on your definitions (for example, does 12121 look random or not?), but you even with quite restrictive definitions you can find quite a few. Also, it's important to note, as someone pointed out above, that even quite unlikely occurrences end up being quite likely to occur if you do the experiment enough times. --Tango (talk) 16:37, 19 June 2008 (UTC)

The problem isn't in the randomness, but in the prejudice of the reader. When selecting randomly from the set of integers from 1 to 100,000, the numbers 77777, 12345, 39710, and 83016 all have the same probability of occurring (1 out of 100,000). It is only the reader's misplaced "feeling" that 77777 is less likely than 83016 that is responsible for the suspicion that 77777 was not randomly selected. (In fact, one might exploit that prejudice as follows: A (large) list of numbers that contains no numbers such as 77777, 12345, 54321, etc. is likely "faked.") Wikiant (talk) 16:35, 19 June 2008 (UTC)

If I was the teacher then 77777 would make me suspicious and 83016 would not. Not because 77777 seems less likely than 83016 to be the result of a random process, but because 77777 seems more likely than 83016 to be the result of a non-random process. It could for example be cheating to avoid home work, or a cheeky student testing how the teacher or other students would react, or the result of using a poor method to generate the sequence, or simply a stuttering student. But without more information I don't think there is a meaningful way to assign percentages to these and other possibilities. PrimeHunter (talk) 17:21, 19 June 2008 (UTC)

I'm not sure that your reasoning is clear. Let F be the event "student has fabricated the data". Let 7 be the event that the student reports 77777, and let 8 be the event that the student reports 83016. What I heard you say is, "Pr(7|F) > Pr(8|F)." I believe that the pertinent question is, "is Pr(F|7) > Pr(F|8)?" Wikiant (talk) 21:13, 19 June 2008 (UTC)

I estimate Pr(7|not F) = Pr(8|not F), and Pr(7|F) > Pr(8|F). Based on this (and 0 < F < 1), I conclude Pr(F|7) > Pr(F|8). So I think 77777 gives rational reason to suspicion even though a random number would have the same chance of being 83016. PrimeHunter (talk) 22:19, 19 June 2008 (UTC)

From Bayes' theorem, we have: Pr(F|7) = Pr(7|F) Pr(F) / Pr(7) and Pr(F|8) = Pr(8|F) Pr(F) / Pr(8). If you are correct that Pr(F|7) > Pr(F|8), then (from Bayes) it must be true that Pr(7|F) / Pr(7) > Pr(8|F) / Pr(8). You claim that Pr(7|F) > Pr(8|F). But if the cheater is more likely to use 7 than he is to use 8, then the unconditional probabilities must be such that Pr(7) > Pr(8). This, however, leaves us with no conclusion as to the comparison of Pr(F|7) and Pr(F|8). Wikiant (talk) 22:53, 19 June 2008 (UTC)

P(F|7) > P(F|8) with my assumptions is intuitively clear to me and can probably be shown easier than the following.

P(8) = P(F)P(8|F) + (1-P(F))P(8|not F).
P(7) = P(F)P(7|F) + (1-P(F))P(7|not F).

I assume 0 < P(F) < 1, and 0 < P(7|not F) = P(8|not F).
Let a = (1-P(F))P(7|not F) = (1-P(F))P(8|not F). Then 0 < a < 1.
P(8) / P(7) = [P(F)P(8|F) + a] / [P(F)P(7|F) + a].

From Bayes' theorem:
P(F|7) = P(7|F)P(F)/P(7), and P(F|8) = P(8|F)P(F)/P(8).
Then P(F|7) / P(F|8) = P(7|F)/P(8|F) * P(8)/P(7).
Substituting the earlier expression for P(8)/P(7) gives:
P(F|7) / P(F|8) = P(7|F)/P(8|F) * [P(F)P(8|F) + a] / [P(F)P(7|F) + a]
                = [P(F) + a/P(8|F)] / [P(F) + a/P(7|F)].
I estimated P(8|F) < P(7|F), so (remembering a > 0):
P(F) + a/P(8|F) > P(F) + a/P(7|F), and then P(F|7) > P(F|8).

PrimeHunter (talk) 01:02, 20 June 2008 (UTC)

follow on (from another person)

(please excuse my bad manners)
Seeing as we've now had the discussion I would like to point out that "the teacher asks for five random numbers from his students, and then proceeds to mark them." does represent some sort of 'false hypothesis' - only the teacher is wrong here... Clearly though the sequence "1","7","banana","3.4","8","0","4" is wrong in at least two ways... Can you spot them?

Here's a 'spin' on the question - you ask me to generate 5 random numbers ignoring the original "only 0-9" qualification)- I present "pi","e","sqrt(2)","1" and "0" .. is this sequence any less random than eg "2","6","5","7","1" ? Do I gain or lose marks for 2i ?87.102.86.73 (talk) 17:16, 19 June 2008 (UTC)

Well, say the teacher asks for 2000 random digits and proceeds to mark them, and one set of numbers, according to one of the standard randomness tests has a probability of 0.0000001% of having been generated by a random process (because the student just banged on the numeric keypad 100 times and cut and pasted the resulting numbers 20 times) should he fail the student?. --Slashme (talk) 13:13, 20 June 2008 (UTC)

My opinion

I think the answer depends on whether or not the order of the numbers matters. If order does matter then I say yes something like "77777" is more unlikely than say "12345". If order does not matter then it is just as likely as anything else.

When order does matter, then the second you get something thats not a 7 you're screwed so to speak. So the chance of "77777" being the result is (1/10)^5 because you have to get a 7 each time. But with "12345" you could, for example, get any of the five digits, then any of the four remaining, then three etc so the probability can be worked out with combinations and it will be more likely than "77777"

When order does not matter, the probability of any number sequence (of 5 digits in our case) is always (1/10)^5 --212.120.247.132 (talk) 17:43, 20 June 2008 (UTC)

Yes, that sounds right, 5 different specific digits should be 120 (5!) times more likely than one specific digit 5 times. --Tango (talk) 18:15, 20 June 2008 (UTC)

Thanks to all for the above discussion. You have all been helpful. And your mathematical insights have been very elucidating for me, on this topic. Thanks. I really appreciate the input and the discussion. (Joseph A. Spadaro (talk) 01:13, 20 June 2008 (UTC))

Seems to me that the fault lies with the teacher for not specifying clearly what they meant by "5 random numbers". If the teacher wanted 5 different random numbers then they should have said so. Without this qualification, it is unfair to penalise the 7 7 7 7 7 response unless you also penalise every other response that might have been generated some non-random process (or not even pseudo-random) process. What about 1 2 4 8 6, or 1 3 9 7 1, or 1 4 9 6 5, or 1 1 2 3 5, or 3 1 4 1 5, or 2 7 1 8 2, or 5 7 7 2 1, or 1 6 4 4 9 ? Each of those sequences has a pattern, and so should also be penalised. Of course, the teacher's ambiguity could be deliberate, and the responses could be a springboard for a discussion of how assigning a meaning to the term "random number" is not as straightforward as it appears at first sight. Gandalf61 (talk) 12:07, 21 June 2008 (UTC)

Exactly - a single sample from any distribution tells us precisely nothing..87.102.86.73 (talk) 14:12, 21 June 2008 (UTC)

To clarify for Gandalf61 ... the teacher's instructions would have been something like: "Generate a random number from 0 to 9 inclusive. Then do so a second time. Then do so a third time. Then do so a fourth time. Then do so a fifth time." Or such. Thanks. (Joseph A. Spadaro (talk) 00:28, 22 June 2008 (UTC))

In that case 7 7 7 7 7 is a valid response - at any rate, it is just as valid as 1 2 4 8 6 or 1 3 9 7 1 or 1 4 9 6 5 etc. etc. Gandalf61 (talk) 17:29, 22 June 2008 (UTC)

Agreed. The question (and ensuing discussion above) was ... is the teacher's suspicion justified? (Joseph A. Spadaro (talk) 19:13, 22 June 2008 (UTC))

We're drifting into the realms of education psychology rather than maths here ... is the student who gives 7 7 7 7 7 as an answer simply correct, stupid, naively right, trying to prove a point? etc etc .. hypothetical teacher, hypothetical students - difficult to give answers - too many unknowns.87.102.86.73 (talk) 19:42, 22 June 2008 (UTC)

Not difficult at all ... and the answer was, in fact, given ... namely, that "77777" is equally likely to occur as any other 5-digit string of numbers such as "13974". (Joseph A. Spadaro (talk) 21:57, 22 June 2008 (UTC))

June 19

Infinite Series Involving Prime Numbers

I came across the following series while studying some Fourier Analysis. Given the infinite series

$S=1+{\frac {1}{5}}-{\frac {1}{7}}-{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{17}}-{\frac {1}{19}}-{\frac {1}{23}}+\cdots$

After writing it in the following closed form

$S={\frac {6}{5}}+\sum _{k=2}^{\infty }\left((-1)^{k-1}\right)\left({\frac {1}{p(2k)}}+{\frac {1}{p(2k+1)}}\right)$

where $p(n)$ returns the n-th prime number. Now this series seems to be converging and in fact it seems to be converging to a numerical value of $S=1.046932...$. My question is, how can we tell for sure if this series is converging or not and if it is converging, what is it converging to? Is there anyway to find out what the exact value is? Is this a well-known fact, a well-known constant?A Real Kaiser (talk) 01:49, 19 June 2008 (UTC)

The series certainly converges, by the alternating series test. Have you tried Plouffe's inverter for the limit (I don't have a precise enough estimate)? Algebraist 01:13, 19 June 2008 (UTC)

Actually no, I had not tried looking up in the inverter. I want to but the problem is that this series converges so slowly, after adding up the first 100,000 terms, I still only had 1.04693... and I am not even sure about the last digit 2 yet. So, first I have to write up a more efficient program to calculate more digits (faster) so that I can obtain more decimal places. Meanwhile, does anyone have any idea what this number is?A Real Kaiser (talk) 01:49, 19 June 2008 (UTC)

I'm fairly certain it starts 1.046932971... (with the last 1 possibly a 0). Plouffe's returns no hits. -- Meni Rosenfeld (talk) 09:56, 19 June 2008 (UTC)

See Proof that the sum of the reciprocals of the primes diverges -- Q Chris (talk) 10:32, 19 June 2008 (UTC)

How is this relevant? -- Meni Rosenfeld (talk) 10:47, 19 June 2008 (UTC)

I think I've found it on OEIS: [2] - note that their formulation differs from yours, because you've omitted 2 and 3, and included 1. Oops, didn't notice the oddity in your thing. It isn't the same after all. Well, maybe that's a start, at least.--Fangz (talk) 11:35, 19 June 2008 (UTC)

Maths (Indian mathematicians)

How do I get the clips of Sridhara on wikipedia?Minintelligent (talk) 10:52, 19 June 2008 (UTC)

What do you mean by "clips"? You can add pieces of information to the Sridhara article if you have anything suitable. If you are having trouble with editing then you should try reading Wikipedia:How to edit a page, which gives lots of assistance. --_{tiny plastic} Grey Knight ⊖ 11:55, 19 June 2008 (UTC)

Meaning of Exponent In Trigonometric Expressions?

What does $\cos ^{2}(x)$ mean? And $\sin ^{2}(x)?$ And $\tan ^{2}(x)$? Etc.
The Piano Man (talk) 14:14, 19 June 2008 (UTC)

• $\sin ^{2}(x)$ means $(\sin(x))^{2}$ or equivently $\sin(x)*\sin(x)$. It is used to distinguish easily from $\sin(x^{2}))$. Similarly the same rule applies for the other trig functions. Rambo's Revenge (talk) 14:32, 19 June 2008 (UTC)

It's important to note that for most other functions, the same notation means to apply the function twice, $f^{2}(x)=f\circ f(x)=f(f(x)$. The way it's used for trig function is an abuse of notation, but it's a very convenient one. --Tango (talk) 14:39, 19 June 2008 (UTC)

Be careful though, as $\sin ^{-1}(x)\neq {\dfrac {1}{\sin(x)}}$, but is instead equal to $\arcsin(x)$. -mattbuck (Talk) 15:20, 19 June 2008 (UTC)

And if you see $\sin ^{-2}(x)$, it's best just to run away and hide! --Tango (talk) 15:27, 19 June 2008 (UTC)

Are you sure about that? I’ve only seen this notation used as composition when $f$ is being considered as an operator. It then only naturally makes sense because operator spaces have their multiplication operation as composition, while function spaces have their multiplication operation as pointwise multiplication. Thus if $f$ is being considered an operator, it is unambiguous $f^{2}(x)=(f\cdot f)(x)=f(f(x))$, while if $f$ is being considered in a typical function space it is still unambiguous, $f^{2}(x)=(f\cdot f)(x)=(f(x))^{2}$ GromXXVII (talk) 00:02, 20 June 2008 (UTC)

The notation is sometimes unambiguous for trivial reasons: if the image of f is not contained in its domain, then the first interpretation is impossible, while if the range of f has no multiplication operation, then the second is impossible. Algebraist 00:09, 20 June 2008 (UTC)

Yeah, pretty sure. That's the way I've generally seen it used. The most obvious example is f^-1 - that almost universally means the inverse of f, never it's reciprocal, and other exponents generally work the same way. I'm not sure what you mean by a "function space" - I would usually that term to refer to a vector space of functions, which doesn't have a multiplication (other than scalar multiplication, of course). The closest you get to a multiplication would be an inner product, but that gives you a scalar, not a function. --Tango (talk) 12:14, 20 June 2008 (UTC)

I think he was referring to when you have an algebra of functions. --137.205.233.134 (talk) 13:06, 20 June 2008 (UTC)

savant mathématicien

Bonjour, Pourquoi M.Jan MIKUSINSKI né le 3 avril 1913 à Stanislawow et décédé le 27 juillet 1987 à KATOWICE ne figure pas dans la liste des grands mathématiciens connu pour son travail de pionnier dans l'analyse mathématique.M.Mikusinski a développé un calcul opérationnel 44a40 calculus of Mikusinski,qui sont pertinentes pour résoudre les équations différentielles.Son calcul opérationnel est basé sur un calcul de la convolution des fonctions à l'égard de la transformés de Fourier.De la convolution produit qu'il va à définir ce que dans d'autres contextes est appelé le domaine de fractions ou un quotient domaine.Ces couples de fonctions M.Mikusinski appels opérateurs-Mikusinski opérateur,44a40 il est également bien connu pour Cube et Antosik-Mikusinski Théorem,Mikusinski convolution algèbre etc. Récupéré "de http://en.wikipedia.org/wiki/Jan_Mikusinski" Plusieurs grands savants polonais figurent aux États-Unis en Suède et font des conférences partout dans le monde. La Pologne peut être très fière d'avoir des savant d'une telle renommée dans le monde entier. —Preceding unsigned comment added by 90.54.102.79 (talk) 14:41, 19 June 2008 (UTC)

Sir, this is not a question, but a statement. You might have also read that he died twenty years ago... Monsieur, ce n'est pas une question, mais une déclaration. Vous avez peut-être également lire qu'il est mort il ya vingt ans. 86.152.76.21 (talk) 17:02, 19 June 2008 (UTC)

Ne pleure pas, Alfred - j'ai besoin tout mon courage pour mourir á vingt ans! Damien Karras (talk) 21:05, 19 June 2008 (UTC)

I don't understand this language. Can anyone make a translation?--218.102.124.108 (talk) 17:03, 20 June 2008 (UTC)

It's french. He's complaining that a certain man did various work in mathematical analysis, but does not figure in some list of great mathematicians.--Fangz (talk) 23:13, 20 June 2008 (UTC)

Translation: Hello, Why does Mr. Jan MIKUSINSKI born April 3 1913 in Stanislawow, deceased July 27 1987 in KATOWICE and known for his pioneering work in mathematical analysis not figure in the list of great mathematicians[?] Mr. Mikusinski developed an operational calculus [which helps] resolve differential equations. His calculation...is based on the convolution of Fourier functions. The convolution produces what is defined in other contexts as the domain of fractions or a quotient domain. Mr. Mikusinski calls this set of operators mikusinski operater, it is known equally well as Cube and antosik-mikusinski theorem, Mikusinski convolution algebra, etc. Recovered from http://en.wikipedia.org/wiki/Jan_Mikusinsk . Several great Polish scholars figure in the United States [and] in Sweden and give lectures all over the world. Poland can be very proud to have such world famous scholars. Cuddlyable3 (talk) 14:30, 21 June 2008 (UTC)

June 20

No questions this day

Awwwww... — Kieff | Talk 04:25, 23 June 2008 (UTC)

June 21

Differentiation

I'm looking at the assumed knowledge for a financial mathematics course, and I'm ok with all of it except this one question:

Differential Calculus

As an indication of the level of calculus required, students should be able to interpret the following equation:

$D={-}{(1+y) \over P}{dP \over dy}$

I only know very basic differentiation (e.g. polynomials, basic trig functions). I've spent some time trying to figure it out but it's got me stumped. I can't tell if P is meant to be a function of y, or if D is meant to be a constant... I always thought capital letters in calculus just referred to the integral of an expression.

I understand if you don't want to tell me the answer, but can someone please identify the specific area of differential calculus I should be researching? And even better, a website/chapter in a book that can help me with this?

Thanks,

60.242.124.184 (talk) 12:19, 21 June 2008 (UTC)

P has to be a function of y, or differentiating it with respect to y is meaningless. D could be a constant or a function of y. If I had to guess, I think I'd go for a constant, since that's a common use for capitals from the beginning of the alphabet (though one normally starts with either A or C). Algebraist 12:27, 21 June 2008 (UTC)

This looks like a formula for bond duration (that is a really poor article). P is the price of the bond and y is the yield to maturity. So P is definitely a function of y. For small interest rate moves, D is typically treated as a constant but, in theory, it will change if the other variables change.

Duration is a measure of the sensitivity of the price of a fixed income asset to changes in interest rates. The Macaulay Duration is calculated as the average time to each cashflow weighted by the PV of that cashflow. Express the value as

$P=\sum _{i}PV(CF_{i})=\sum _{i}CF_{i}(1+y)^{-t_{i}}$

and find the derivative wrt y to see how the formula above gives the Macaulay Duration. Zain Ebrahim (talk) 13:25, 21 June 2008 (UTC)

Given that we are expected to use dP/dy I'd automatically assume that P is a function of y, and (again) in the absence of further information that D is a constant - since the alternatives (eg D is also a function of y) are just too complex..

So I'd rearrange to get

dP/dy = -D P/(1+y)

ie

d fn(y)/dy = k fn(y)/(1+y) (k is a constant equal to -D)

SPOILER BELOW ?? to avoid just read the bolded and underlined bits first

Since there is a equation that has the derivative of a function to be the function times another function of y , my first guess would be to try fn(y)=e^g(y) since

d(e^g(y))/dy = g'(y) e^g(y)

ie if P=e^g(y) then dP/dy = g'(y) P (g'(y) is the derivative of g(y) = d g(y)/dy )

Therefor I get one solution if g'(y) = k/(1+y), and because I already know that d (ln(1+x))/dx = 1/(1+x) I get g(y)=k ln(1+y)

So I think g(y)=k ln(1+y)

This gives P=e^{k ln(1+y)} = (e^ln(1+y))^k = (1+y)^k

So P=(1+y)^-D

Hopefully that will help, please ask about any specific parts, and hppefully others will pick me up on any mistakes I've made.

That should be one type of solution - I've no idea if there may be others as well...

As general advice - if you get a differential equation of the form d fn(x)/dx = fn(x) times another function of x it's worth trying the substituion fn(x)=e^g(x) due the properties of the derivative of e^g(x)87.102.86.73 (talk) 14:35, 21 June 2008 (UTC)

The thing is, it didn't say to solve it, it said to interpret it. That could mean any number of things, but probably nothing as complicated as actually solving for P. --Tango (talk) 14:57, 21 June 2008 (UTC)

The area to look at would be "first order differential equations", assuming you know how to differentiate (and integrate) polynomials, logs, and exponentials the equation should be ok..

Even if D is a function of y it may be possible to still solve it with the above knowledge.

I assumed that an answer would be a simple way to show the type of knowledge required..

In general capital letters don't mean the integral of an expression. They simply refer to a variable (which may or may not be a function) in the same way lower case letters do.87.102.86.73 (talk) 15:14, 21 June 2008 (UTC)

Indeed, however D(x) is often used for the integral of d(x), which is probably what the OP was thinking. I would never assume that's what was meant unless it was actually stated, though - it's a common choice of symbol, it's not a standard notation. --Tango (talk) 15:17, 21 June 2008 (UTC)

To the original poster: In financial mathematics, the formula you wrote is a very common way to express Macaulay Duration. You should check with the course convenor at the institution (they're normally quite responsive to emails) whether you need to know how to solve that differential equation but I rather doubt it since they clearly specified that you need to interpret it. The solution calculated by 87 above represents a special case: here, P represents the price of a zero coupon bond (ZCB) that pays one dollar D years from now. For ZCBs, the duration is merely the term to maturity. Zain Ebrahim (talk) 15:27, 21 June 2008 (UTC)

Thanks a lot everyone 60.242.124.184 (talk) 02:19, 22 June 2008 (UTC)

June 22

Brahmagupta's formula

Brahmagupta#Geometry has a "Reference diagram" marked ABCDpqrs and text using abcd and s. I think abcd are really pqrs and the text s is something else. Could someone please explain what the text all means - I find it unclear - and better yet edit the article or diagram to make them match. -- SGBailey (talk) 09:37, 22 June 2008 (UTC)

S is half the sum of the sides and I think the picture should be replaced with one that shows the lengths of the sides as a, b, c and d. So, in the text, the lowercase variables are the lengths of the four sides which are shown as p, q, r and s in the picture. Zain Ebrahim (talk) 09:43, 22 June 2008 (UTC)

It's easier to edit the text, so I've done that. —Ilmari Karonen (talk) 11:25, 22 June 2008 (UTC)

Carrying on down to the next figure we have "root(ac+bd)" without abcd being defined. I think these are some of pqrs in the diagram above, but which is which. Should this be changed to "root((AB)(CD)+(BC)(AD))"? -- SGBailey (talk) 13:48, 22 June 2008 (UTC)

June 23

Probability

I have a question concerning probability:

Given: 57% Breakfast eating {P(A)}, 80% Teeth flossing {P(B)}, and 46% doing both. I figured that (Teeth + Breakfast) - Both = Probability of doing either activity. However I am not sure how to find the probability of those who eat breakfast, but do NOT floss. Vivio Testarossa^{Talk Who} 03:27, 23 June 2008 (UTC)

Try drawing a Venn diagram, and use that to see what areas you could add or subtract to get the one you want. Confusing Manifestation(Say hi!) 04:00, 23 June 2008 (UTC)

Ultimately, you want to find the percentage of people who only eat breakfast. This is given by (Percent breakfast eating) - (Percent doing both). 76.238.91.253 (talk) 04:34, 23 June 2008 (UTC)

Or, alternatively, since they do seem to be assuming these are independent events, you can multiply the portion who eat breakfast (0.57) by the portion NOT brushing (0.20). StuRat (talk) 11:51, 23 June 2008 (UTC)

Is "but do not floss" equivalent to "and do not floss"? I think you've assumed so, StuRat. Zain Ebrahim (talk) 12:07, 23 June 2008 (UTC)

Yes, "but" and "and" mean effectively the same thing, "but" is used when there is some kind of contradiction between the two statements (that is, given the first you would expect the opposite of the second). --Tango (talk) 14:54, 23 June 2008 (UTC)

Polynomials

Who uses polynomials in the real world for real life situations?67.101.98.121 (talk) 04:45, 23 June 2008 (UTC) —Preceding unsigned comment added by 67.101.98.121 (talk) 04:30, 23 June 2008 (UTC)

Just about everyone? Calculators normally use polynomial approximations to calculate many functions like cosine or logarithms. The path of a thrown ball is a polynomial - a quadratic - to a reasonable approximation. Computer programmers think in terms of polynomial versus non-polynomial complexity, and some encryption algorithms use an extension of polynomials (or at least polynomials over fields other than the real numbers). Complicated functions are also sometimes approximated using polynomial interpolation, although care needs to be taken sometimes (using a high order polynomial to approximate the path your Mars lander needs to take may result in one that dips below ground level). Confusing Manifestation(Say hi!) 07:00, 23 June 2008 (UTC)

Chi-squared question

Given a set of measurements with uncertainty which are presumably linearly distributed, I'm comfortable using the chi-squared parameter to find a line of best fit. However, what if I want to fit a line to data that has uncertainties in both the independent and dependent variables? How would I define a $\chi ^{2}$ in that case? Thanks! — gogobera (talk) 05:35, 23 June 2008 (UTC)

Regarding the Wiki article on Markov chains

I just edited and added some content to a section (reproduced below) of the Markov chain article, and I have some questions about the math and terminology. All of my questions and such pertain only to Markov processes with invariant transition matrices that are stochastic.

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Markov chains with a finite state space

If the state space is finite, the transition probability distribution can be represented by a matrix, called the transition matrix, with the (i, j)'th element of P equal to

$p_{ij}=\Pr(X_{n+1}=j\mid X_{n}=i).\,$

P is a stochastic matrix, which is an important fact to keep in mind for the rest of this discussion. When the Markov chain is a time-homogeneous Markov chain, so that the transition matrix P always remains the same at each step, then the k-step transition probability can be computed as the k'th power of the transition matrix, P^k.

The stationary distribution π is a (row) vector that satisfies the equation

$\pi =\pi \mathbf {P} .\,$

In other words, the stationary distribution π is a normalized left eigenvector of the transition matrix associated with the eigenvalue 1.

Alternatively, π can be viewed as a fixed point of the linear (hence continuous) transformation on the unit simplex associated to the matrix P. As any continuous transformation in the unit simplex has a fixed point, a stationary distribution always exists, but is not guaranteed to be unique, in general. However, if the Markov chain is irreducible and aperiodic, then there is a unique stationary distribution π. Additionally, in this case P^k converges to a rank-one matrix in which each row is the stationary distribution π, that is,

$\lim _{k\rightarrow \infty }\mathbf {P} ^{k}=\mathbf {1} \pi$

where 1 is the column vector with all entries equal to 1. This is stated by the Perron-Frobenius theorem. If, by whatever means, $\scriptstyle \lim _{k\to \infty }\mathbf {P} ^{k}$ is found, then the stationary distribution of the Markov chain in question can be easily determined for any starting distribution, as will be explained below.

Since P is a stochastic matrix, $\scriptstyle \lim _{k\to \infty }\mathbf {P} ^{k}$ always exists. Because there are a number of different special cases to consider, the process of finding this limit can be a lengthy task. All the same, there are several general rules and guidelines to keep in mind. Let P be an n×n matrix, and define $\scriptstyle \mathbf {Q} =\lim _{k\to \infty }\mathbf {P} ^{k}.$

It is always true that

$\mathbf {QP} =\mathbf {Q} .$

Subtracting Q from both sides and factoring then yields

$\mathbf {Q} (\mathbf {P} -\mathbf {I} _{n})=\mathbf {0} _{n,n}$

where I_n is the identity matrix of size n, and 0_n,n is the zero matrix of size n×n. Multiplying together stochastic matrices always yields another stochastic matrix, so Q must be a stochastic matrix. It is sometimes sufficient to use the matrix equation above and the fact that Q is a stochastic matrix to solve for Q.

Here is one method for doing so: first, define the function f(A) to return the matrix A with its right-most column replaced with all 1's. Then evaluate the following equation:

$\lim _{k\rightarrow \infty }\mathbf {P} ^{k}=f(\mathbf {0} _{n,n})[f(\mathbf {P} -\mathbf {I} _{n})]^{-1}.$

This equation does not work when [f(P – I_n)]^–1 does not exist. If this is the case, then it is necessary to take into account more information in order to find Q. One thing to notice is that if P has an element P_ii on its main diagonal that is equal to 1 and the ith row or column is otherwise filled with 0's, then that row or column will remain unchanged in all of the subsequent powers P^k. Hence, the ith row or column of Q will have the 1 and the 0's in the same positions as in P.

In most cases, P^k approaches but never actually equals its limit. There are numerous exceptions to this, however, such as the case in which

$\mathbf {P} ={\begin{bmatrix}0&0&1\\0&0&1\\0&0&1\end{bmatrix}}.$

If A₀ (which is a row vector) represents the starting distribution, then the stationary distribution is equal to A₀Q. Note that any distribution, regardless of the number of steps it is away from the starting distribution, can be used in place of A₀ without affecting the result for the stationary distribution.

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I added all the information in the lower half of this section about how to find $\scriptstyle \lim _{k\to \infty }\mathbf {P} ^{k}$. First of all, I was wondering whether what I added would be more appropriately placed in the stochastic matrix article. While I was editing, I just figured that some people might want to know how to find $\scriptstyle \lim _{k\to \infty }\mathbf {P} ^{k}$ while they are looking up information about topics like these.

Anyway, what exactly is a "distribution"? Specifically, am I correct in believing that a distribution is represented by a row vector and shows the status of a Markov process after some number of transition steps? Are distributions linked to specific starting conditions, or are they only dependent on the transition matrix P? Do all the elements in a distribution vector have to sum to 1? As a related matter, I know that a Markov chain of the type described above always converges to a particular distribution after a large number of steps, and I'd like to know the proper term for this distribution that it converges to.

In the part of the section that I did not write, it was stated that the stationary distribution π is a row vector that satisfies the equation πP = π. Does that constitute the entire definition for π? After all, any scalar multiple of some original π would also satisfy the equation.

It was also stated that π is a normalized vector, which means to me that π is appropriately scaled to have a vector length of 1. However, earlier in the article I believe it stated that all the elements of π sum to 1. These two statements are contradictory. I noticed that the latter statement coincides with the equation (as quoted above) :$\lim _{k\rightarrow \infty }\mathbf {P} ^{k}=\mathbf {1} \pi$.

I have a number of other questions about the proper terminology to use for describing Markov chains, but I'll ask them later. 76.238.91.253 (talk) 05:44, 23 June 2008 (UTC)

In this context a distribution is a probability measure. Yes, in the case of a finite space (finite state space) the distribution is sometimes described by a vector. Yes, the elements are non-negative and sum to 1. A stationary distribution is a distribution that is P-invariant. For some P, there might be more than one stationary distributions. But there are a large class of Markov chains for which there is a unique stationary distribution. Don't you think it would be more appropriate to have this discussion in the talk page of the article? (Not that I particularly mind.) Oded (talk) 05:50, 23 June 2008 (UTC)

I asked one question on the talk page (the title of the question starts "Just added a formula..."), but then I realized that the reference desk gets more user traffic. I would continue the discussion at the talk page if I could get at least one person knowledgeable in the subject to notice what I'm doing/asking there. I haven't taken a class about Markov chains proper (I will soon, though), but I have worked with them informally for a long time. Thus, for the most part I just need someone to help me with the terminology so that I can appropriately describe what I know about this subject. Do you think you could help me with that over at the talk page (I don't have that many questions left)? —Preceding unsigned comment added by 76.224.118.12 (talk) 00:35, 24 June 2008 (UTC)

square roots and cube roots.

Is the amount of numbers, which are both squares and cube, (eg 64) infinite? And how does one prove it? I imagine that it would be some sort of proof by contradiction, eg "Assume there is some some number x such that the square root is an integer and the cube root is an integer and that there is no larger such number." Now, if I think about it, (this is the part I'm not 100% clear on) x^6 should also satisfy the two initial conditions, because if the square root of x is an integer then the square root of x^6 is also an integer, ditto for cube roots, therefore there are an infinite amount of numbers that are both squares and cube.

Am I way off base???

Duomillia (talk) 19:50, 23 June 2008 (UTC)

After thought.

Can the same proof be used to demonstrate the same thing for higher degrees? Example, an number that is both x^10 and y^6 just to pick random numbers. 65.110.174.74 Duomillia (talk) 20:09, 23 June 2008 (UTC)

Your proof is valid (assuming you have shown at least one x exists to start a chain), but it can be simplified. Regardless of x (just assuming it's an integer), what can you say about the square root of x^6 and the cube root of x^6? Remember x^6 = x*x*x*x*x*x. PrimeHunter (talk) 20:18, 23 June 2008 (UTC)

Indeed, and the same approach works for any other powers you may choose. Just remember that there are an infinite number of integers. --Tango (talk) 20:20, 23 June 2008 (UTC)

Divergent Series

Studying some analysis, I came across these two questions in my book. Give an example of a series that does not diverge to infinity but whose partial sums are increasing. One easy answer is just $\sum _{n=0}^{\infty }\left({\frac {1}{2}}\right)^{n}$ which converges to 1.5. The second question, the converse of the first one, asks for a series which diverges to positive infinity but whose partial sums do not form an increasing sequence. What would such a series look like?76.79.202.34 (talk) 00:44, 24 June 2008 (UTC)