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March 12

Least Squares Question

Let $v_{1}={\begin{bmatrix}1\\1\\0\\1\end{bmatrix}}$ and $v_{2}={\begin{bmatrix}0\\0\\1\\0\end{bmatrix}}$ and $b={\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}}$. Let $V=span\{v_{1},v_{2}\}$. I am trying to find the vector in V that is closest to the vector b. Now, my questions is that I am trying to solve the system Ax=b and I know that b is my b above and $x={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}$ is a four dimensional vector but what would my matrix A be? Would it be just $A={\begin{bmatrix}1&1&0&1\\0&0&1&0\end{bmatrix}}$? But then I can't multiply both sides by transpose of A and them multiply by the inverse of (Transpose(A)*A)? Basically I want to have that $x=(A^{*}A)^{-1}*A^{*}$ but I can't figure out what my A will be. Any help would be appreciated! Thanks!A Real Kaiser (talk) 05:27, 12 March 2008 (UTC)

Because you're searching for the best approximation in V, you want the range of A to be V. The range of a matrix is the span of its columns, and so A is the 4 x 2 matrix with $v_{1}{\mbox{ and }}v_{2}$ as columns. Your x is actually a length 2 vector, representing the coefficients of $v_{1}{\mbox{ and }}v_{2}$. In this case, $A^{*}A$ will be square, and invertible. Check out Moore-Penrose inverse. 134.173.93.127 (talk) 07:46, 12 March 2008 (UTC)

Ironically, I can see by inspection that b is orthogonal to the vs. So the closest vector is just the zero vector. Baccyak4H (Yak!) 18:41, 12 March 2008 (UTC)

Actually, that was my intuition also that zero would be the answer (because that is the one and only point both spaces share so obviously it is the closest one). My I just couldn't understand how to setup my matrix A. And when I tried to do it (the way you basically said), I was wondering as to why my matrix x was only two dimensional. Now it makes sense the the entries of matrix x represent the coefficients for linear combination of v1 and v2. So this means that the x1v1+x2v2 would be the vector that I am looking for, right? In span{v1,v2}?A Real Kaiser (talk) 04:50, 13 March 2008 (UTC)

That's right. 134.173.93.127 (talk) 07:41, 13 March 2008 (UTC)

Which two spaces? There's the space V, and the vector b which is not in it. You may have meant span{b}, which indeed intersects V only at 0. But this is not enough for the closest vector to be zero - for this you need orthogonality. Consider v1 = {1, 0, 0}, v2 = {0, 1, 0} and b = {1, 1, 1}. Only {0, 0, 0} is common to span {v1, v2} and span {b}, but the closest vector in V to b is {1, 1, 0}. -- Meni Rosenfeld (talk) 14:36, 13 March 2008 (UTC)

Probabilities and this year's Champions League quarter finals

OK, I wasn't too shabby at Maths when I last studied it, but I always struggled with Probabilities. I thought of a nice bunch of Probabilities questions based on the forthcoming Champs Lg draw. Can you help me understand how to solve them, and thereby make up for either a duff teacher all those years ago, or, more likely, my inattention in class at crucial moments:

• There are 8 remaining teams in the competition
• 4 are English
• The draw will put them into 4 pairings that will decide the semi-final line up
• Assume all teams are of equal ability and AGF that UEFA don't rig their draws!

Q1 What's the probability of exactly 1 all-English quarter-final?
Q2 What's the probability of exactly 2 all-English quarter-finals?
Q3 What's the probability of at least 1 all-English quarter-final?
Q4 What's the probability of exactly 1 all-English semi-final?
Q5 What's the probability of exactly 2 all-English semi-finals?
Q6 What's the probability of at least 1 all-English semi-final?
Q7 What's the probability of an all-English final?

Oh, and leaving my probability of inattention in class aside (which, frankly is nearly 100%), if my maths teacher had used problems like this, instead of boys choosing socks in the dark (when they could just switch a ruddy light on) I might be able to do this without your help!

Cheers, --Dweller (talk) 12:23, 12 March 2008 (UTC)

Well, I will try to provide some of the answers, but I don't guarrantee that they are right :)

First, let's count how many combinations of quarter-final ties there are (only accounting that Team1 will be drawn against Team2, without accounting who is drawn to host the first leg). It should be $C_{2}^{8}={\frac {8!}{2!(8-2)!}}=28$.

A1. Out of these there is a $C_{2}^{4}={\frac {4!}{2!(4-2)!}}=6$ possibilities that exactly 2 English clubs are drawn together, thus making a probability of ${\frac {6}{28}}={\frac {3}{14}}\approx 0.21$, i.e. 21%.

A2. I suppose, therefore, that the probability of 2 all-English quarter-finals should be $({\frac {3}{14}})^{2}={\frac {9}{196}}\approx 0.05$, i.e. 5%.

A3. And, therefore, the probability of at least 1 all-English quarter-final should be the sum of above two probabilities, i.e. ${\frac {3}{14}}+{\frac {9}{196}}={\frac {51}{196}}\approx 0.26$, i.e. 26%.

It's hard to go on from here because we have to account for different cases.

A5. Of course, there can't be exactly 2 all-English semi-finals if there's at least 1 all-English quarterfinal. The probability that no English teams will be drawn together in the quarter-finals should be, according to the above information, $1-{\frac {51}{196}}={\frac {145}{196}}\approx 0.74$, i.e. 74%. Assuming that all clubs are of equal ability to qualify for the semi-finals, the probability that all English teams will win their ties in this case is $({\frac {1}{2}})^{4}={\frac {1}{16}}=0.0625$, i.e. 6.25%. Multiplying these two would give us the answer: ${\frac {145}{196}}\times {\frac {1}{16}}={\frac {145}{3136}}\approx 0.05$, i.e. 5%.

Well, the other questions are more complicated than these, maybe I will come back to them later... maybe not :). Again, I'm not sure if what I did above is right, but it sure makes sense. To me at least! Hope that helps!  ARTYOM  13:18, 12 March 2008 (UTC)

Urk. Sorry to be picky after you've done all that work, but you'll need to slow down. I think I remember what the ! is (but can't remember what it's called - is it "factorial" or something?) but haven't a clue what the big C thingy is, nor how/why you populate the first equation that leads to 28. Remember, I want to understand how to do this, not just learn the answers! --Dweller (talk) 13:47, 12 March 2008 (UTC)

For the benefit of people similarly mathematically illiterate to me who come here (I advertised this thread at WT:FOOTY) the ! is indeed Factorial --Dweller (talk) 13:52, 12 March 2008 (UTC)

Yes, it is, and the C stands for choose. $C_{2}^{8}$ is the number of ways of choosing 2 members from a set of 8 elements. --Tango (talk) 14:43, 12 March 2008 (UTC)

Thanks for that. So remembering about ! and learning that C notation means I've already learned something. --Dweller (talk) 15:10, 12 March 2008 (UTC)

I don't think that's right - only a 26% chance of an all-England matchup doesn't sound right at all. Your ${\frac {6}{28}}$ is the probability that any given tie, taken independently of the others, will be an all-England clash. But you can't just square that to get the probability of two, because the ties are not independent.

By my reasoning the probability that there are no all-English quarter-finals is given by placing the English teams into the draw first; so the first team can be placed in any of the 8 spots, the second in 6 of the remaining 7 spots (i.e. not against the first), the third in 4 of 6, and the last in 2 of 5, giving ${\frac {8}{8}}\times {\frac {6}{7}}\times {\frac {4}{6}}\times {\frac {2}{5}}={\frac {8}{35}}\approx 22.86\%$. So the probability of at least one all-English tie is the remainder, $\approx 77.14\%$. I can't think how to do the rest right now but I'll get back to you! — sjorford++ 14:53, 12 March 2008 (UTC)

In these things, it is often helpful to do the extreme cases first:

• Question 3 is easiest to answer, since we just want to find the probability of NOT having an english QF. As above, this is ${\dfrac {8}{8}}\times {\dfrac {6}{7}}\times {\dfrac {4}{6}}\times {\dfrac {2}{5}}={\dfrac {8}{35}}$, and thus P(at least 1 english QF) = ${\dfrac {27}{35}}$.
• Question 2. EXACTLY TWO english QF. Let us pick the english teams first. We can choose any position for the first team, so that is irrelevant. Let the 2nd team be a match (1/7). Then 6 choices for the 3rd, and 1 for the 4th. Thus we have ${\dfrac {8}{8}}\times {\dfrac {1}{7}}\times {\dfrac {6}{6}}\times {\dfrac {1}{5}}={\dfrac {1}{35}}$ chance of this. Now, suppose the 2nd team does not get a match (6/7), then the 3rd team has 2 options, and the 4th team just 1. Thus chances of 2 matches is ${\dfrac {1}{35}}+{\dfrac {8}{8}}\times {\dfrac {6}{7}}\times {\dfrac {2}{6}}\times {\dfrac {1}{5}}={\dfrac {1}{35}}+{\dfrac {2}{35}}={\dfrac {3}{35}}$
• Question 1. Now, there were 3 possible cases with 4 teams - 1 match, 2 matches, or no matches. So, now $P(1)=1-{\dfrac {3+8}{35}}={\dfrac {24}{35}}.$
• Question 5. For 2 english semi-finals, we require that the english teams all have different matches (8/35), and that they all win (1/2). Thus, P(2 E SF) = ${\dfrac {8}{35}}\left({\dfrac {1}{2}}\right)^{4}={\dfrac {8}{35}}\times {\dfrac {1}{16}}={\dfrac {1}{70}}$.
• Question 4. Here it gets difficult, as we need to consider many cases.
  • The easiest is if there are 2 english QFs (3/35) - then we can guarantee exactly two english teams in the SFs. The chance of them facing each other is ${\dfrac {1}{3}}$, which when combined with the probability of 2 english QFs is ${\dfrac {1}{35}}$.
  • Now suppose that there are no english QFs (8/35). Then to get 1 english QF we require at least 2 teams to win.
    • Suppose exactly 2 do win - there are $^{4}C_{2}=6$ ways to do this, each of which has a probaility of $\left({\dfrac {1}{2}}\right)^{4}={\dfrac {1}{16}}$ (2 teams must win, 2 lose). Now, we require that the 2nd team face the 1st team. This has a chance of 1/3 as above. Thus here we have ${\dfrac {8}{35}}\times {\dfrac {6}{16}}\times {\dfrac {1}{3}}={\dfrac {1}{35}}$
    • Suppose 3 English teams win - $^{4}C_{3}=4$ ways, each with P = 1/16. Any draw we are bound to have 2 english teams facing off, thus we have ${\dfrac {8}{35}}\times {\dfrac {4}{16}}\times {\dfrac {1}{1}}={\dfrac {2}{35}}$
  • Now, suppose there is one english QF (24/35). We are bound to have 1 english team in the SF.
    • Suppose both other english teams win (P = (1/2)^2 = 1/4). Then we have 3 teams and 4 spots - there's bound to be an english SF, with $P={\dfrac {24}{35}}\times {\dfrac {1}{4}}={\dfrac {6}{35}}$
    • Now, suppose that only one of the other enlish teams win. There are 2 ways to do this, each with probability 1/4. Now the chances of them facing each other in the SF is 1/3. Thus $P={\dfrac {24}{35}}\times {\dfrac {2}{4}}\times {\dfrac {1}{3}}={\dfrac {4}{35}}$
  • Thus, we now sum these probabilities: $P={\dfrac {1+1+2+6+4}{35}}={\dfrac {14}{35}}={\dfrac {2}{5}}$
• Question 6 - this is now easy, we just sum the probabilities of 1 and 2 english SFs. $P={\dfrac {2}{5}}+{\dfrac {1}{70}}={\dfrac {29}{70}}$
• Question 7 - Almost there, but again several cases to consider.
  • The most obvious place to start is with 2 all-english SFs. We know this had probability 1/70. We must have an all-english final here.
  • Next consider there are exactly 3 english teams in the SFs. This happens if either there is one english QF and both other teams win (P = 6/35), or no english QFs and 3 teams win (P=2/35). Thus we have a total P=8/35. We're bound to have 1 english team in the final, and the other has a 0.5 chance. Thus we have a 4/35 chance from this case.
  • We could take forever finding the probability of 2 english teams in the semis. However, note that there are an equal number of english and foreign teams, and that they all have equal chances of winning. Thus, the chances of two no-english SFs is 1/70, and the chance of only one english team in the SFs is 8/35. Thus, we have used up 17/35 options, and the remaining 18 must be exactly 2 teams facing each other, of which 1/3 are going to have the english teams facing each other and thus can be discarded. So we have a 12/35 chance of having exactly 2 english teams in different matches. Now, we require that both win. This has P = 1/4. Thus the chance of an english final here is ${\dfrac {12}{35}}\times {\dfrac {1}{4}}={\dfrac {3}{35}}$.
  • Now, sum these: $P={\dfrac {1}{70}}+{\dfrac {4}{35}}+{\dfrac {3}{35}}={\dfrac {3}{14}}$

Damn that was a lot of work. We'd better fkn win. -mattbuck (Talk) 16:16, 12 March 2008 (UTC)

Simpler approach for Q7: Forget the route to the final - there are (8x7)/2=28 possible pairs of teams in the final, of which (4x3)/2=6 consist of a pair of English teams. So probability of an all-English final is 6/28, which is 3/14. Gandalf61 (talk) 16:32, 12 March 2008 (UTC)

Damn my looking for the complicated route. I thought there should be an easy way to do it. -mattbuck (Talk) 16:46, 12 March 2008 (UTC)

Math: real-life examples of polynomial division?

what are some examples from real life in which you i might use polynomial division? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 13:38, 12 March 2008 (UTC)

Finding roots of polynomials is the obvious one. It comes up all the time in various disciplines, since lots of things are described (at least approximately) by polynomials and it's often useful to know when they are zero. --Tango (talk) 13:45, 12 March 2008 (UTC)

See Linear response function for one example. It is used in radio and television design. Bo Jacoby (talk) 13:56, 12 March 2008 (UTC).

I saw your previous question and felt I should answer it, but it's hard to come up with an answer at what I think is your level of understanding. As others have mentioned, it's used to find roots of polynomials (e.g. to find more roots after the first is known). I doubt you are familiar with eigenvalues or root locus plots, but those require polynomial roots. Let's say I'm designing a control system for an aircraft. I've taught this using the roll control system of an F-18 as one example. First, you write out the equations of motion of the aircraft. The result is a system of differential equations. You then apply a Laplace transform to that system, so you get a system of polynomial equations in 's' (the Laplace variable). If you can find one root, by numerical or other means, you can use polynomial division to simplify the equation, getting the entire set of real and complex roots. Depending on those roots (are they real or complex? are their real parts negative or positive? how close to the origin are they?) you can tell a lot about the behaviour of the aircraft. For instance, if the real parts are negative, the aircraft is stable, otherwise it is unstable. Adding a control system to the loop modifies the set of differential equations and allows us to set the roots where we want them, ensuring the behaviour of the aircraft is how we want it, and that the aircraft is stable, maneuverable, and other desirable properties. moink (talk) 14:37, 12 March 2008 (UTC)

And the reference desk serves its function of helping us improve articles: Root locus needs some serious work. moink (talk) 14:44, 12 March 2008 (UTC)

One thing is for sure - polynomial division is a much less important mathematical skill for an average individual than others which are unfortunately not studied enough (or at all) in school, such as logic and probabilistic thinking. -- Meni Rosenfeld (talk) 15:08, 12 March 2008 (UTC)

That's for sure -- I agree 100%. Logic especially needs to be taught more. It is completely foreign to most students, unfortunately. (Joseph A. Spadaro (talk) 08:18, 13 March 2008 (UTC))

Chi-square test

When calculating a chi-square test what are the steps between calculating the differences between the expected and observed values and obtaining the chi-square value? Then, knowing the degrees of freedom, how is the p-value of the chi-square obtained (other than by looking it up in a table)? Thanks. The chi-square test article does not explain these points. Itsmejudith (talk) 15:47, 12 March 2008 (UTC)

If you are talking about a contingency table, then the expected value for each entry is

(column total / table total) × (row total / table total) × table total.

I know this expression can be simplified, but in this form it is easier to see conceptually what is being estimated. Subtract this from the observed value, square the difference, and divide by this expected value again. The sum over all the table is the value of the χ² statistic. Call that value X. Then the p-value is just the probability that a χ² distribution with the appropriate degrees of freedom is at least X (that is, one minus the CDF of the χ² distribution, at X). Baccyak4H (Yak!) 18:33, 12 March 2008 (UTC)

Can this detail be added to the worked example in the article? Thanks. Itsmejudith (talk) 08:50, 13 March 2008 (UTC)

OK, done. Although I can see the point someone might make that it is too much "How to...". Baccyak4H (Yak!)

March 13

Generating random numbers

How would I generate random numbers with a distribution matching Zipf's law? --Carnildo (talk) 04:18, 13 March 2008 (UTC)

Could you start with a probability distribution that is linear ie each value occurs equally often and scale that to the zipf distribution? Would you need help scaling the fnuction?

Can it be assumed that you already have a method of generating random numbers in general?87.102.8.240 (talk) 09:10, 13 March 2008 (UTC)

According to the classic version of Zipf's law, the number of occurrences of the word that has rank k in a large corpus of words is proportional to 1/k, which means it is of the form c/k for some large constant c. But actually the value of c depends on the size of the corpus, where c is the number of different words in the corpus, and the size n of the corpus is roughly n = c·log c. Therefore it makes more sense to generate a sequence of numbers for which each initial segment obeys Zipf's law.

In pseudocode:

Let z₁, z₂, ... be the sequence to be generated.

Set c to 1

For n from 1 to whatever:

Set ν to 1/((log c)+1)

Select one alternative from:

(A) with probability ν:

Set z_n to c

Set c to c+1

(B) with probability 1 − ν:

Pick a random number uniformly from the range {1 ... (n−1)}

Set z_n to z_r

Obviously the beginning will have few different numbers, so you may want to discard a large initial part. It is then not actually necessary to store the segment to be discarded itself, as long as the number of occurrences of each value is kept track of. --Lambiam 09:58, 13 March 2008 (UTC)

The above is surely right - but looked a little mysterious to me - here's another version of 'psuedo-code' - in this case I don't explicitly normalise the probabilities but instead generate a sum "Total" of 1/k and find the position of a random number between 0 and "Total" ie the first 'word' is between 0 and 1/1, the second between 1/1 and 1/1+1/2 etc

For different functions simply change all instances of 1/value to the new thing eg 1/value^s I've marked these positions with a ^*

n=number of words

Total = Sum to n of 1/k ^* !ie 1/1+1/2+1/3 etc

LOOP 1

R=Total x RND() ! ie Generate a random number between 0 and Sum total called R {RND() is a random function between 0 and 1} !

Length=0

m=1 !a loop counter

LOOP 2

Length=Length+1/m ^*

if R<=Length then print/output "m"  ! m is the m^th word ! : GOTO LOOP 1 !start again!

increase m by 1 : goto LOOP 2

Just added in case others found the first example a little confusing.. apologies to any offended by the GOTOS.. the !!'s are comments. not factorials!87.102.8.240 (talk) 10:57, 13 March 2008 (UTC)

A counting variable can be added so that the program would stop after a certain number of random values had been produced. Would suggest adding this inside LOOP 1 eg Count=Count+1:If Count=10000 then END/STOP>87.102.8.240 (talk) 11:03, 13 March 2008 (UTC)

Note - what I've done here shares some similarities with Arithmetic coding or Entropy encoding - at least in small part.87.102.8.240 (talk) 11:15, 13 March 2008 (UTC)

(it could be extended to non-zipf distributions that have finite states from a single event, also the above could be speeded up slightly computer wise if that was neccessary - ask if you want further details on these.)83.100.138.116 (talk) 16:28, 13 March 2008 (UTC)

Financial calculator question

I have a financial calculator question. I have the APY for a Certificate of Deposit; I am looking for a calculation that will give me its dividend rate if the dividends are paid Quarterly or if they are paid monthly.

For example, for a 12 month Certificate the APY is 3.25% what is the calculation to find out what the dividends are? —Preceding unsigned comment added by 207.109.247.177 (talk) 17:58, 13 March 2008 (UTC)

Comment: The poster is referring to the annual percentage yield. Pallida  Mors 19:16, 13 March 2008 (UTC)

Different institutions may use different definitions of the Annual Percentage Yield, which may be different from the effective rate, the Annual percentage rate, because certain costs have not been accounted for yet. Assuming the formula in our article holds for the definition actually used, and assuming dividend rate may be equated with the nominal interest rate, then:

• for monthly dividend rate (12 periods) you get over a year:

  $12\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{12}}-1\right)\times 100\%=12\times \left(1.0325^{\frac {1}{12}}-1\right)\times 100\%=12\times 0.00267\times 100\%=3.20\%;$

• for quarterly dividend rate (4 periods) you get over a year:

  $4\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{4}}-1\right)\times 100\%=4\times \left(1.0325^{\frac {1}{4}}-1\right)\times 100\%=4\times 0.00803\times 100\%=3.21\%.$

Note the caveats. Your mileage may vary.  --Lambiam 19:18, 13 March 2008 (UTC)

PDE -> ODE

Are there a set of "standard" methods one could try to use to transform a PDE in several variables to a set of coupled ODEs? —Preceding unsigned comment added by 12.196.44.226 (talk) 20:22, 13 March 2008 (UTC)

maybe separation of variables#Partial differential equations? --Spoon! (talk) 03:49, 14 March 2008 (UTC)

I need help with pi & circumference

If you calculated the circumference of a circle the size of the known universe, requiring that the answer be accurate to within the radius of one proton, how many decimal places of pi would you need to use?

 A. Two million
 B. 39
 C. 48,000
 D. Six billion

Thanks in advance.207.69.140.24 (talk) 21:48, 13 March 2008 (UTC)

I seem to recall the answer is 39, though I don't remember why. Strad (talk) 21:56, 13 March 2008 (UTC)

Assuming you’re calculating it from the radius of the known universe I think you could find an upper bound by dividing the smallest distance in question (radius of a proton) by twice the largest distance in question (radius of the universe), and using pi with as many decimal places as this number: because then would be accurate to the radius of a proton when multiplied by twice the radius of the known universe. GromXXVII (talk) 22:11, 13 March 2008 (UTC)

[ec] According to Observable universe, its diameter is $9\cdot 10^{10}$ ly, so its circumference is $3\cdot 10^{11}$ ly or $3\cdot 10^{24}$ m. The radius of a proton is roughly $10^{-15}$ m. The ratio is roughly $10^{39}$, thus roughly 39 digits of π are required. But don't let anyone fool you into thinking that this means that more digits are not required for science and applications. -- Meni Rosenfeld (talk) 22:18, 13 March 2008 (UTC)

A question similar to this has been asked before. It was to find the circumference of the known universe to the accuracy of a planck length.

And the answer is that 62 digits of Pi was required, preferably 64 digits. I'll see if I can dig up a link.

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Science/2006_September_27#How_many_digits_of_pi_for_the_known_universe.3F

202.168.50.40 (talk) 00:08, 14 March 2008 (UTC)

We have no data to base the calculation on, so knowing the decimal expansion of π to many places is not going to help. Even if we knew the radius of the observable universe to within the radius of a proton, and we could halt the universe so that it stopped expanding while we are doing the calculation, we don't even know if the universe is flat.  --Lambiam 08:40, 14 March 2008 (UTC)

Well, the question didn't ask about the circumference of the observable universe, but rather of a circle the size of it. I guess what this means exactly is open to interpretation, but I think "fix a number of arbitrarily high precision and magnitude on the same order as the observable universe, and use it as a radius" is a valid one. -- Meni Rosenfeld (talk) 10:38, 14 March 2008 (UTC)

See Cosmic View: The Universe in 40 Jumps by Kees Boeke - a classic. Gandalf61 (talk) 14:21, 14 March 2008 (UTC)

March 14

Happy Pi Day

Just wishing all of you a happy Pi Day! Please celebrate responsibly. --Kinu ^t/_c 04:12, 14 March 2008 (UTC)

What should I do at 1:59:26? I've only got an hour! HYENASTE 04:22, 14 March 2008 (UTC)

Um... run around in a circle? —Keenan Pepper 05:54, 14 March 2008 (UTC)

Just returned from my circle! Happy Pi Day!!! HYENASTE 05:59, 14 March 2008 (UTC)

getting dizzy! this is so exciting87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Damn, I've just fallen over.--88.109.224.4 (talk) 13:41, 14 March 2008 (UTC)

Happy 3.14... Day! --Mayfare (talk) 15:19, 14 March 2008 (UTC)

I've just drank 3.1 pina coladas oh goddd.... Damien Karras (talk) 15:23, 14 March 2008 (UTC)

I've never liked the idea of having a celebration based on an arbitrary and unnatural number system. What say we move pi day to the third of July (or the seventh of March for transatlantic types)? Algebraist 15:48, 14 March 2008 (UTC)

Let's just make every day pi-day see Big yellow disc in sky87.102.83.204 (talk) 16:49, 14 March 2008 (UTC)

You may be an algebraist, but is that an excuse for taking a rational approach to pi day?  --Lambiam 16:56, 14 March 2008 (UTC)

How are continued fractions less transcendental than decimals? And that username's out of date: I'm currently applying for a DPhil in logic. Algebraist 20:54, 15 March 2008 (UTC)

That explains a lot.  --Lambiam 07:27, 16 March 2008 (UTC)

"Give three point one four one six cheers for the Science Officer with pointed ears." —Tamfang (talk) 00:24, 19 March 2008 (UTC)

Percentages

Here I am at a very advanced age totally ashamed to say that I cannot work out percentages! Please advise me how to work out the percentage of one sum against the other; for example what percentage has my pension increased from last year's income to this year's. How to I do that ? Thanks in anticipation.--88.109.224.4 (talk) 08:47, 14 March 2008 (UTC)

Simple a percentage is one over another times 100.

So if income was 110 this year, and 95 last year this gives (110/95)*100 = 115.8%

So the increase is 15.8% (since 100% equals 1:1 ratio or no change I subtract 100%)

Or if my wage (77) increases by 2% my new wage is 77 + ( 2/100 x 77 ) = 77+1.54 = 78.54 —Preceding unsigned comment added by 87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Did you check out the article Percentage?  --Lambiam 16:11, 14 March 2008 (UTC)

Thank you both, esp. Lambiam. The article is excellent and brings me up to speed. Thanks again.--88.109.224.4 (talk) 16:57, 14 March 2008 (UTC)

Also if you're struggling to do it 'mentally' (i.e. without a calculator/spreadsheet) I find it easiest to break things int0 10% and 1% then just do simple addition...E.g. If I want to know what 22% of 721 is I just go.... ok 10% is 72.1 and 1% is 7.21...I have 2 x 72.1 = 144.2 and then 2 x 7.21 so that's 14.42 add the two together and I get 158.62 which a quick google search (http://www.google.co.uk/search?hl=en&q=22%25+of+721&btnG=Search&meta= suggest is correct. This is particularly useful as a technique when you just want to get a ball-park amount really quite rapidly. ny156uk (talk) 17:50, 14 March 2008 (UTC)

Another thing. A lot of people get confused or thrown off by the percent symbol (%). Whenever you see that symbol, you can erase it altogether ... and simply replace it with the fraction 1/100 or, if you prefer, the decimal 0.01. That is, the % symbol simply means to take whatever number you are dealing with and to multiply it by the fraction 1/100 or (equivalently) the decimal 0.01 -- whichever you prefer or whichever is easier in a given situation. Thus:

• 7% means ( 7 * 1/100 ) = 7/100 --- if you need / want the fraction version of the answer
• 7% means ( 7 * 0.01 ) = 0.07 --- if you need / want the decimal version of the answer

Thanks. (Joseph A. Spadaro (talk) 22:25, 14 March 2008 (UTC))

Group Theory

The theory of groups interest me. I have a knowledge of mathematics up to:

a) Complex number arithmetic

b) Basic differentiation / integration

c) Basic knowledge of matrices (inverse, determinants)

d) Basic Logic and set theory

I understand the brief introductions and historical accounts of its development; and I'm tempted to pick up a book- although I don't know if I'm qualified to handle even the simplest introductory text.

It's basically a crude engineering mathematics background. Can you suggest a path of study to lead me up to investigations into group theory? Can I step into it with the basic knowledge I have or would you suggest deep study into which particular areas of mathematics? —Preceding unsigned comment added by 81.187.252.174 (talk) 14:12, 14 March 2008 (UTC)

Basic set theory is all you need, really. Find a book called something like "A first course in groups" and see how you find it. At a basic level, groups are pretty easy to understand. They are just sets with a way of combining 2 elements to get another element in a well-behaved way (for example, adding 2 integers to get a 3rd integer, or reflecting a polygon and then rotating it and realising that it's the same as having just reflected it in a different axis). --Tango (talk) 14:26, 14 March 2008 (UTC)

A textbook that is not expensive (used) is An introduction to the theory of groups by George W. Polites.[1] I don't know the book, so I can't vouch for it, but it is thin (80 pages), and if it is a mismatch to your needs, at least you did not waste lots of money.  --Lambiam 16:05, 14 March 2008 (UTC)

To hell with "not expensive"; here are three free online books for you to choose from: [2] [3] [4] —Keenan Pepper 21:59, 14 March 2008 (UTC)

Origin of asymmetry in conformation of symmetrical shape in a higher dimension (+ origami!)

Ok, this one has me totally stumped, and bear with me, as I'm not sure I'll explain it wonderfully well. There is a certain "mathematical" origami model, called a hyperbolic parabola (or paraboloid). Which is folded using a crease-pattern that is four-way symmetric under both reflectional and rotational transformations. (See: http://www.math.lsu.edu/~verrill/origami/parabola/ )

Now, here's the weirdness: when the crease-pattern is concertina-ed (fanned-up), the model assumes the three-dimensional shape for which it is named -- which is not completely symmetrical! The shape once conformed possesses chirality, or handedness, which is not possessed by the tetrahedron into which it will fit perfectly. Where does this rotational asymmetry come from? It is not there in the two-dimensional crease-pattern, but somehow emerges from the conformation into three dimensions, despite there being no way to distinguish between what is done to any four-way division of the model.

Can someone explain this puzzling phenomenon? (Preferably without using arcane technical terms). Many thanks, 85.194.245.82 (talk) 20:47, 14 March 2008 (UTC)

(original poster): Still be not-overly-confident of how well I explained my confusion, here's an addendum I just thought of: from the perspective of the paper as it is twisting into three-dimensions, both of its (indistinguishable) diagonals curve into circular arc-segments, one "up", the other "down" -- the puzzle is what logic does the paper use in conspiring with three-dimensional space to decide which of these diagonals goes up and which goes down?! 85.194.245.82 (talk) 20:53, 14 March 2008 (UTC)

I must admit I had no idea what you were talking about until I actually made one of these. Do you have one on hand? If so, try this: grab two opposite sides of the square piece of paper (which are two opposite edges of the tetrahedron) and twist in such a way as to flatten out the square. Then keep twisting in the same direction far past that point. Try to do it in one smooth motion. If you do it right, the thing will flip inside out and become an identical shape, but this time the diagonals go the opposite directions. So really it's not that a symmetrical pattern leads to an asymmetrical shape, it's that a symmetrical pattern leads to two possible stable shapes which are opposites, and they're symmetrical if you consider them as a pair. In between them is an unstable equilibrium which is a corrugated flat square (it's unstable because you effectively give the paper a negative Gaussian curvature, so it doesn't like to be flat anymore). Neat. —Keenan Pepper 21:41, 14 March 2008 (UTC)

Technical nitpick with one of your statements: The shape it makes is not chiral. It has point group $D_{2d}$, which is a proper subgroup of the point group of the flat square you start out with ($D_{4h}$), so it does have "less symmetry" in a specific sense, but it still has mirror symmetry, so it's not chiral. —Keenan Pepper 21:51, 14 March 2008 (UTC)

Hah, just realized I used many "arcane technical terms". Sorry about that. I'll be happy to explain them to you. —Keenan Pepper 21:53, 14 March 2008 (UTC)

Aye, empirical experimentation shortly after framing the question lead me to the "the paper does it" conclusion you point to, which at first made me think, "awww, reality's boring", as I was hoping for some kind of crazy metaphysical implication about the non-independence of spacial dimensions relative to one another (I have an overactive imagination that way). But then, I thought about it some more and had a little discussion on IRC, and decided it was still a very interesting phenomenon despite the "choice" of axis-curving resulting from environmental factors. Interesting because a purely geometrical consideration lead to an amplification of information. That is to say, although really only one bit of information is added to the system in choosing the subset of the point group, the effect is to cause a large difference in eventual topology. Someone said this was an example, or at least analogous, to the theoretical-physical concept of spontaneous symmetry breaking (I personally think the term "spontaneous" is a terrible misnomer, as it implies that the information arrives randomly, ex nihilo, from nowhere, which is not how mathematics or the universe works, no matter how hard we try to pretend otherwise). 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

(thanks for the terminology info, btw. I had a feeling chirality wasn't right, but couldn't think of any other asymmetry. Is there a specific term for it? Will look at the article you linked to.) 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

The phenomenon is an instance of spontaneous symmetry breaking. An even much simpler example is when you put a small ball (as from a ball bearing) exactly in the middle on top of a larger ball. Perfect symmetry. Turn around and the little ball will have dropped off, in an asymmetric way (since all ways of dropping off are asymmetric). If it does not drop off instantaneously, you may have to wait for a butterfly or Heisenberg to assist the process.  --Lambiam 23:43, 14 March 2008 (UTC)

March 15

A measurable set?

Let $(X,\mu )$ and $(Y,\nu )$ be measure spaces and $\{A_{k}\},\{B_{k}\}$ sequences of sets of finite measure in X and Y respectively. Let the "rectangles" $R_{k}=A_{k}\times B_{k}$, and assume that

$\sum _{k=1}^{\infty }(\mu \otimes \nu )(R_{k})<\infty .$

Let $R_{k,x}=\{y\ |\ (x,y)\in R_{k}\}$ and

$T_{n}=\{x\ |\ 1/n\leq \sum _{k}\nu (R_{k,x})\}.$

Why is it obvious that T_n is measurable?  — merge 17:01, 15 March 2008 (UTC)

Well, $R_{k,x}$ is just B_k if x is in A_k, and empty otherwise. So $\sum _{k}\nu (R_{k,x})$ is the some of the measures of the B_k such that x is in A_k. Thus whether x is in T_n is determined by which of the A_ks x is in, and T_n is a union of intersections of the A_ks. Algebraist 17:45, 15 March 2008 (UTC)

Oh, I think I see how it works out. If $\{f_{k}\}$ is a sequence of nonnegative measurable real-valued functions and α is a real number, the sets

$V_{k}=\{x\ |\ f_{1}(x)+\cdots +f_{k}(x)>\alpha \}$

are measurable, and so are

$W_{\alpha }=\bigcup _{k}V_{k}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)>\alpha \}$

and

$\bigcap _{j}W_{\alpha -1/j}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)\geq \alpha \}$.

In this case $f_{k}(x)=\nu (R_{k,x})=\chi _{A_{k}}(x)\nu (B_{k})$ and $\alpha =1/n$.  — merge 22:30, 16 March 2008 (UTC)

March 16

grasping math to an intermediate level

hi, I like to get to an intermediate level of understanding of different subjects and I want to get a decent grasp of mathematics, but I'm having a hard time with it. I have a BS in math and an MS in statistics. I feel as if I'm wandering blindly in a mathematical universe, feeling the obvious, stumbling over the less-obvious, but without confidence in my travels. When I started to learn typography, say, I had that blind feeling at first but with study soon found the fundamentals and understood the basic ideas and questions in that field.

I want to get to that level of confidence with math, but I'm not sure how to get there. Obviously it's a big world out there, but is there a field that I should concentrate study on (analysis maybe?) to get an intermediate grasp (i.e., not expert and not blind beginner) ? thanks 72.150.136.11 (talk) 14:24, 16 March 2008 (UTC)

In all honesty, it's pretty much impossible to get an intermediate grasp of everything, since there is so much. I mean, I am doing an MMath, and I'd say that by doing so I was intermediate in the subjects I do, but there are others I simply have no clue about - topology, fluid mechanics, relativity, anything related to statistics... maths is so vast, and a lot of it is so remote from anything you do at school as to be an entirely different subject. Group theory is probably a useful thing to try and understand, as is basic vector calculus and mathematical analysis. I personally like complex analysis and combinatorics. Of course, combinatorial game theory is also fun to try. -mattbuck (Talk) 15:00, 16 March 2008 (UTC)

Linear algebra, while rather boring, is also a useful subject to have a working knowledge of - it seems to pop up all over the place. However, if you've got a BS in Maths I would expect you've already covered the basics of all these areas. If you want to do more Maths, you need to specialise, there's no way to have a better than undergraduate understanding of more than one or two areas. --Tango (talk) 15:19, 16 March 2008 (UTC)

I disagree with the last statement, there are many researchers who are active in several quite distinct areas. It's not easy, of course.

Mathematical logic is very important to know the basics of (say, up to the level of understanding Godel's theorems). Likewise for topology, which at the very least one needs to know to the level of feeling the difference between metric features and topological ones. I recall that when I have been applying for a math MSc, virtually all schools frowned upon me not having taken a topology course in my BA. Fortunately I had learnt the material independently - my peers who hadn't had some trouble with many of the courses.

Of course, there's the obvious stuff, like analysis (real, complex, functional, differential geometry...), algebra (groups, rings, fields, Galois theory...), discrete (combinatorics, graphs, ...), and of course computer science (computational complexity, information theory, machine learning, ...) and applied math (numerical methods, mathematical physics, game theory...). Disclaimer: This categorization is based on my own personal view and the subjects are not equally divided. -- Meni Rosenfeld (talk) 17:37, 16 March 2008 (UTC)

Well, I guess you run into trouble defining an "area" of maths. There are plenty of people that do research in what would usually be considered distinct areas, but they're usually working in some kind of overlap between them - pretty much all branches of maths overlap somewhere. Are there many people that have a solid understanding of the whole of several areas? I would say most understand just the parts of those areas that are relevant to their research. It does depend very much on how large you consider an area to be (is algebra a single area or is ring theory a distinct area to Galois theory? - obviously, you can understand both ring theory and Galois theory to a high level, but understanding the whole of algebra to a high level is pretty challenging without even trying to start on any non-algebraic areas). Put simply: I think we're both right for appropriate definitions of "area". --Tango (talk) 18:12, 16 March 2008 (UTC)

Thanks for these answers. I thought it might be impossible for a not-particularly-gifted person to walk with confidence in the math spaces. I just hate that feeling of flailing around blindly, holding only to those proofs I really really understand. Sometimes I feel like I'm missing a sense that some others have; which I suppose is true for everyone on some level. I'll explore mathematical logic and topology and see how that goes. thanks everyone 72.150.136.11 (talk) 19:00, 16 March 2008 (UTC)

While we all strive for perfection, nobody gets there. However far you go, there's always going to be more you don't understand, so you'll probably always be walking blindly, you'll just be walking in more advanced parts of the world of maths. I think the important thing is to enjoy it - study the areas of maths that interest you and that you find fun. There's little point studying anything else. --Tango (talk) 19:31, 16 March 2008 (UTC)

From what you say, I'm not sure that what you're looking for is so much a subject as a teaching style. You need some reintroductions to subjects you already know, from a different viewpoint. At the moment, I don't know of any better way to find good sources than to read things randomly and hope to hit gold (as I have occasionally), but someone else on the board might. If you want to grasp the material, try some popularizers, avoiding as best you can those who dumb it down instead of actually making it clearer. I was just reading Indra's Pearls, for instance, which gives some solid visual meaning to some pretty abstract techniques from several fields (analysis, group theory, linear algebra, complex analysis, maybe some others). If you want to understand the motivation, that is the goals, behind something, look into its history. Even the original papers on the subject, if you can get them. For instance, I've heard over and over again this same fable about complex numbers being the solution to the equation x^2+1=0, which ties in nicely to some other equations, like x^2-2=0 and x+1=0. It wasn't until fairly recently that I found out that complex numbers were motivated for the first time, after thousands of years of people intentionally ignoring their effects on quadratic equations, by the solution to the cubic equation. There are a lot of nice books that give the story behind that, not least the Ars Magna itself. Also, something that will always help you grasp a subject is to use it. Set your sights on a number of small, random goals, and see what you can do with them. Abstraction is bad on an empty stomach - you need a lot of bulk to soak it up. Once you can navigate real situations with confidence, the patterns you find among them will motivate and support abstraction. The exercizes in textbooks are a start, but don't rely on them for originality, follow your own curiosity. I'm trying to keep my studies broad-based, since I like them that way, but if you want to focus on a single subject for awhile, there's certainly plenty to choose from. It doesn't sound like you plan to use this study immediately, so focus your decision on how much you can learn about learning. It may be best to relearn something concrete as an introduction. Euclid's geometry, for instance, moving through some historical works like the Arithmetica Infinitorum or Newton's papers, then up through the crisis a century ago with Cauchy and Weierstrass et al, to get to analysis. An interesting stop you might take on the way is in Algebraland - most of calculus, as it applies to polynomials, can be derived without hardly touching limits. Black Carrot (talk) 02:28, 17 March 2008 (UTC)

Local Sidereal Time Conversion

How do I convert local sidereal time to something that is more commonly used? Is this actually a measure of time? I read the article about sidereal time and it appeared to be an angle measurement, a measure of place. So when someone tells me "do this action at 13:20" local sidereal time, is this something that makes sense, or is it just astrological mumbo jumbo? And if it makes sense, how do I convert it to Eastern Standard Time or Hawaii Standard Time or GMT or anything else that is more commonly used? —Preceding unsigned comment added by RastaNancy (talk • contribs) 20:45, 16 March 2008 (UTC)

Sidereal time is measure of time, but it's not easy to convert to normal time. Regular time is determined by watching the Sun move across the celestial sphere, sidereal time is determined by watching the stars move - the stars move only due to Earth's rotation on it's axis, the Sun moves because of that and because of Earth orbiting it. Therefore, a solar day differs from a sidereal day (by about 4 minutes). This means that to convert from solar to sidereal time you need to know the date in order to work out how much they differ by. The easiest way to convert would be to find a converter online, I'm sure there are plenty. --Tango (talk) 21:42, 16 March 2008 (UTC)

Possibility of solving this equation

I have the following equation:

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left|-1+R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right|{ds}$

I can choose any values for ${\it {\phi _{i}}}$, and get the corresponding values of $\lambda _{i}$ through an experiment. The aim is to find the function $\theta \left(s\right)$ as accurately as possible. I am wondering if it is possible at all to do this. I was trying to expand $\theta \left(s\right)$ in a Fourier series with n terms (and n undetermined coefficients) and then using n different values of ${\it {\phi _{i}}}$ to get a system of n equations. However, it turns out that all these equations are linearly dependent. Note that ${\it {\phi _{i}}}$ is not a function of s. Any ideas on how best we can find $\theta \left(s\right)$ will be highly appreciated. Regards, deeptrivia (talk) 21:40, 16 March 2008 (UTC)

Do you have any additional knowledge about $\theta '$? If it is known to be monotonic then I have an idea which might work in some cases. The essence is to start with $\phi \approx \pi /2$, for which the absolute value can be solved; then decrease $\phi$ until you see a deviation from normal. The rate of deviation should give you the ratio $\theta '/\theta ''$ at the x-intercept. This might then be solvable. -- Meni Rosenfeld (talk) 22:56, 16 March 2008 (UTC)

Thanks, Meni. Although I can choose any values for ${\it {\phi _{i}}}$, I can't choose too many. Practically, n is two or three, at most six. $\theta '$ is smooth but not monotonous. Also, I don't need to find $\theta$ exactly, but just as best as I can. How best do you think can I find it by finding $\lambda _{i}$ for, say, n wisely chosen values of ${\it {\phi _{i}}}$ ? Regards, deeptrivia (talk) 23:13, 16 March 2008 (UTC)

PS: I think in my case it is safe to assume that

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left(1-R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right){ds}$

Then, since R and ${\it {\phi _{i}}}$ are not functions of s, we can directly integrate and get:

${\it {\lambda _{i}}}=c\left(L-R\cos \left({\it {\phi _{i}}}\right)\left(\theta \left(L\right)-\theta \left(0\right)\right)\right)$

Consequently, it appears that no matter how we vary φ, all we can find about θ is the difference between its values at 0 and L. Now, if we remove the assumption I just made regarding the absolute value, we can still break the integral up into a countable number of parts that can be directly integrated, and using a similar logic, we can argue that we cannot know anything more than θ(L) - θ(0) by varying φ. Am I right? Thanks, deeptrivia (talk) 23:33, 16 March 2008 (UTC)

It's certainly very hard to deduce information about $\theta$ in the general case, but it's not theoretically impossible. If you break up the interval, in some segments you will have $\theta (a)-\theta (b)$ and in some $\theta (b)-\theta (a)$, and when you add them up they don't cancel. Since changing $\theta$ can change the integral, knowledge of the integral can at least eliminate some possibilities. -- Meni Rosenfeld (talk) 00:15, 17 March 2008 (UTC)

Thanks. Am I at least right for the case where there is no sign change, becaause $R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\ll 1$ and no need to break the interval? Regards, deeptrivia (talk) 00:47, 17 March 2008 (UTC)

Yes, indeed. That is what I called the "normal" case, where the integral is reduced to a multiple of $\theta (L)-\theta (0)$. Deviations from this tells you something about $\theta$, but how to extract anything useful from it, especially with limited observations, is a mystery. -- Meni Rosenfeld (talk) 06:32, 17 March 2008 (UTC)

Am I missing something? If $\theta \left(s\right)$ solves the equation, then so does $\theta \left(s\right)+C$, so you can only determine solutions up to an additive constant.  --Lambiam 17:17, 17 March 2008 (UTC)

Certainly, but it appears that it's rather difficult, if not impossible, to even determine it that far. --Tango (talk) 17:28, 17 March 2008 (UTC)

March 17

is mathematics subjective?

So my understanding is that math isn't falsifiable, because it doesn't make physical predictions. Is it then subjective?

For example, whether "something has been proved", is that a question of viewpoint, a subjective thing, or is it objective? —Preceding unsigned comment added by 79.122.42.52 (talk) 01:41, 17 March 2008 (UTC)

I can't speak for all of mathematics but probability is subjective.

The probability of an event depends on what the observer/questioner knows. So two different observers can assign two different probability to the same event if they have different knowledge. 202.168.50.40 (talk) 02:26, 17 March 2008 (UTC)

Mathematics is objective in the sense that all proofs are derived from a set of axioms, so unlike the physical sciences where different experiments can lead to different results, whenever you start from the same axioms you'll get the same result for your proof. That said, the choice of axioms is somewhat subjective, since there is nothing to say that axiom set A is any "better" than axiom set B, although set A may result in something more familiar such as the natural numbers. Confusing Manifestation(Say hi!) 03:26, 17 March 2008 (UTC)

Sets of axioms are necessarily completely different, according to Gödel's incompleteness theorems. Because there are an infinite number of sets, subjectivity can not exist. Mac Davis (talk) 05:06, 17 March 2008 (UTC)

Mathematics is falsifiable, in that guesses can be checked. That's what falsifiability, as a doctrine, is for in the sciences - make sure your clever guesses aren't dead wrong. In mathematics, what's called proof serves roughly the same purpose. We can check our guesses. Whether we can check the system we use to check our guesses is open to philosophical debate, as the system of empirical validation used in the sciences is. Maybe it works, but maybe we live in god's big toe and he's playing games with us. That's a different level of fact-checking, and one that even falsifiable theories can't pass. So, to answer your question, mathematics is subjective to roughly the same extent that science is. Black Carrot (talk) 06:14, 17 March 2008 (UTC)

Mathematics differs in subjectivity from other sciences in a quantitative, not qualitative, way. Mathematics certainly has a higher standard of derivation - while in science, repeating an experiment several times passes as "proof" for its result, in mathematics we prefer to formulate our statements in an abstract way and derive them in a way we consider "logical". But don't be fooled into thinking that mathematics is completely rigorous, objective and devoid of human weaknesses. Not so much because of the choice of axioms (since it is understood that establishing a theorem doesn't mean that it's absolute, but rather that it follows from the specified axioms), but because of the choice of phrasing and logical inference. We have rules like "if we know A and we know A->B then we can deduce B", but there is nothing objectively "true" about this rule, other than that humanity's experience over the eons seems to support it. To this extent, mathematics relies on empirical evidence the same way as other sciences, the only difference being the amount of expected evidence. -- Meni Rosenfeld (talk) 06:49, 17 March 2008 (UTC)

The axioms of the logic in use are, to me, just as much part of the "choice of axioms" as the more mathematical axioms. It turns out that these are deeply intertwined anyway; a lovely result I saw once was that DeMorgan's Laws as logical axioms are equivalent to axiomatically stating that maximal ideals of commutative rings are necessarily prime -- one statement appears purely logical, the other most definitely in the realm of math not logic, yet in practice they are equivalent axioms. What I'm getting at with this little digression is that I don't really see the distinction you make between "choice of axioms" and axioms of logic that we assume. -- Leland McInnes (talk) 14:19, 17 March 2008 (UTC)

As I understand it - something that is 'subjective' depends on who is examining it. So a good maths proof should not be subjective.

Really though 'subjective' and 'objective' have no meaning here ie all mathematical knowledge should be 'objective'. Subjective examples of mathematical things include: imperfect models of phyical behaviour, best estimates of statisical behaviour/probabilities - eg things that are (educated) guesse.

I suggest reading about 'subjective and objective' and forming your own opinion. Picked these definitions 'at random' from the net:

Subjective: characteristic of or belonging to reality as perceived rather than as independent of mind : phenomenal

Objective: of, relating to, or being an object, phenomenon, or condition in the realm of sensible experience independent of individual thought and perceptible by all observers : having reality independent of the minds objective reality

Emphasis mine.87.102.13.144 (talk) 11:59, 17 March 2008 (UTC)

Answer - when something has been proved mathematically that should be objective, and not subjective knowledge. (It's work noting that peoples perception of what the words 'subjective' and 'objective' mean can be a bit subjective...)87.102.13.144 (talk) 14:07, 17 March 2008 (UTC)

Yes, something that has been proved should be objective, in the sense that we would want it to be. Unfortunately, it is not, per the responses above. -- Meni Rosenfeld (talk) 15:06, 17 March 2008 (UTC)

Our choice of what to try and prove is very subjective, but the actual proof is objective. We don't prove XYZ, we prove that our choice of axioms implies XYZ. That choice of axioms is very subjective, but the implication is entirely objective (give or take Godel, anyway - even then, I don't think it's subjective, it's just not wholly reliable). So, I guess the answer to the question "Is maths subjective?" is that maths is a subjective collection of objective facts (there is more to maths than a collection of facts, of course, but for the sake of this discussion I think that definition will do). --Tango (talk) 15:59, 17 March 2008 (UTC)

I disagree. First, nobody proves theorems in a completely formal language. People use natural language to describe their proofs, and hope that others will interpret it as they meant and agree that each of their implications is indeed valid. Sometimes it works, but there is still a lot variance in people's interpretation of arguments in a language such as used in articles. Second, even if the proof was written in a completely formal language, it still relies on the person's ability to validate each step. This validation is also composed of steps that can only be described in a natural language ("look at this line, see what's written here, compare it to what you see there..."). Especially for a long and complicated proof (like most proofs will be when written in a formal language), the possibility of human error is huge. It's also not unthinkable for the same person to make the same error repeatedly because of a certain mindset he is in. Thus different people will report different views on the correctness of a proof. In fact, when given a certain proof, there is no way for us to tell if it is truly valid (whatever that means) or is it just that every person who has ever examined it happened to mistakenly accept it. Thus I reiterate my claim that everything in mathematics is subjective, but to a much lesser extent than some other sciences. -- Meni Rosenfeld (talk) 16:15, 17 March 2008 (UTC)

What you say is certainly true, but I'm not sure if that's really a form of subjectivity - if it is, then nothing is truly objective and the question is moot. --Tango (talk) 17:16, 17 March 2008 (UTC)

Personally, I do believe there is an "objective truth", but that anything a human can ever discover is only a crude approximation of it, and thus subjective. The original question is not moot, as a priori one might think that modern mathematics has, at least in some parts, achieved the holy grail of true objectiveness - my argument is that this is not the case. It's probably as close as we'll be for a while, though. -- Meni Rosenfeld (talk) 18:38, 17 March 2008 (UTC)

Simultaneous Equation

how to solve this simultaneous equation 4x+y+8=x²+x-y=2 —Preceding unsigned comment added by 60.48.198.225 (talk) 07:32, 17 March 2008 (UTC)

You have the two equations $4x+y+8=2\;\!$ and $x^{2}+x-y=2\;\!$. There is a simple operation you can take to get rid of the y's and have an equation with only x. Can you see it? -- Meni Rosenfeld (talk) 07:35, 17 March 2008 (UTC)

Infinitely Many Factors

Consider the ring of formal power series over something like the integers. Is there any power series that can be split into infinitely many (non-unit) factors? That is, is there one where the process of pulling off factors will never stop with some number of irreducible factors? Black Carrot (talk) 10:02, 17 March 2008 (UTC)

Maybe I'm missing some subtlety in the question, but isn't

$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =x\prod _{n=-\infty }^{\infty }\left(1-{\frac {x}{\pi n}}\right)$

an example of a power series with an infinite number of irreducible factors ? Gandalf61 (talk) 10:44, 17 March 2008 (UTC)

Another example (unless I, too, misunderstand the question) is $\prod _{n=0}^{\infty }\left(1+2^{-n}x\right)=\sum _{n=0}^{\infty }{\frac {2^{n}}{\prod _{k=1}^{n}(2^{k}-1)}}x^{n}$. -- Meni Rosenfeld (talk) 10:50, 17 March 2008 (UTC)

Those are units: (1-ax)(1+ax+a^2*x^2+a^3*x^3...) = 1. You can always pull off infinitely many unit factors, even from something irreducible. Black Carrot (talk) 11:12, 17 March 2008 (UTC)

Oh, I see. No idea then. -- Meni Rosenfeld (talk) 11:22, 17 March 2008 (UTC)

Let's see if I have got this straight. You want an infinite number of factors each of which has, say, integer coefficients, is irreducible and is not a unit when considered as a formal power series over the integers ? Let's take out the largest power of x that we can, so we have an initial factor of x^k for some k (possibly 0). Doesn't this mean that the constant term of every other factor cannot be 0 (otherwise it is reducible) or +/-1 (otherwise it is a unit when considered as a formal power series) ? So you have x^k times an infinite number of factors each of which has constant term with absolute value greater than or equal to 2 - which seems to imply that the coefficient of the lowest power of x in the product is not finite, so the product is not a power series. Gandalf61 (talk) 11:29, 17 March 2008 (UTC)

What is meant by "non unit factor"?87.102.13.144 (talk) 11:47, 17 March 2008 (UTC)

In this context, a unit is something you can divide by without leaving the number system. For instance, in the whole numbers 1 is a unit, but anything bigger isn't. That's why 1 isn't considered prime - every number would have infinitely many prime factorizations, eg 5 = 5*1 = 5*1*1 = 5*1*1*1... You're right Gandalf, I hadn't thought of looking at it like that. So, does that mean that power series over the rationals also have finitely many factors? Since everything with a constant term would be a unit, and everything without could be divided by x. Black Carrot (talk) 12:04, 17 March 2008 (UTC)

Meni, should that be 2^n at the top of your last formula? Black Carrot (talk) 12:09, 17 March 2008 (UTC)

Yes, indeed. Fixed. -- Meni Rosenfeld (talk) 14:55, 17 March 2008 (UTC)

Was the first example given not an example? eg a power series over rational fractions expressed, but made of factors which are irrational (contain 1/pi)?87.102.13.144 (talk) 14:12, 17 March 2008 (UTC)

The original question needs to be made more precise. What do you mean by "something like the integers"? For example, are algebraic integers something like the integers? And wat about the complex numbers? Should the factors be over the same ring as the power series? Can the factors again be formal power series, or should they be polynomials?  --Lambiam 14:49, 17 March 2008 (UTC)

Gandalf has solved one form of the question: I believe his proof shows that if R is a Noetherian integral domain, then any element of R[[X]] can be expressed as a product of finitely many irreducibles in R[[X]]. Algebraist 16:09, 17 March 2008 (UTC)

math: polynomial as the difference of two squares

How do you determine if a polynomial is the difference of two squares? —Preceding unsigned comment added by 72.94.230.129 (talk) 14:08, 17 March 2008 (UTC)

Assuming that by "square" you mean the square of a polynomial, and that the coefficients are from a number system that is closed under division by 2, such as the rationals and the reals, any polynomial P can be rewritten as:

$\left({\frac {P+1}{2}}\right)^{2}-\left({\frac {P-1}{2}}\right)^{2}.$

 --Lambiam 14:33, 17 March 2008 (UTC)

I'm not sure that's what the OP means. We studied polynomials last year in algebra I. We learned the difference of two squares as $(x+S)(x-S)\,\!$, in which S is any number. This then takes the form $x^{2}-S^{2}\,\!$. So, to determine if a polynomial is the difference of two squares, just look for two square terms (such as $x^{2}-64$). The solution to a polynomial of the form $(x+S)(x-S)$ is $x=\pm {S}$. So the solution to $x^{2}-64$ would be $x=\pm {8}$.

Disclaimer: I learned these things last year so they are subject to possible error.

Zrs 12 (talk) 19:48, 17 March 2008 (UTC)

That's right, but the principle can be used for more complicated polynomials, as well. For example $x^{2}-y^{2}+8x+4y+12=x^{2}+8x+16-y^{2}+4y-4=(x+4)^{2}-(y-2)^{2}=(x+4+y-2)(x+4-y+2)=(x+y+2)(x-y+6)$. If you want everything to have integer coefficients, then I'm not sure there is a general algorithm to find out if you can write a polynomial as the difference of two squares, you just have to fiddle around with it and see what happens (you can try completing the square as a starting point). --Tango (talk) 20:08, 17 March 2008 (UTC)

We also have an article on this. See difference of two squares. Zrs 12 (talk) 20:06, 17 March 2008 (UTC)

Tango: A polynomial (over Z, say) is the difference of two squares iff it can be factorised with the sum of the (two) factors being divisible by 2 (i.e. corresponding coefficients in the factors have to be of the same parity). Thus a poly is a difference of two squares iff its reduction mod 2 is a square. Algebraist 23:25, 17 March 2008 (UTC)

Or rather not. We get an algorithm, anyway. There's an algorithm that factors a polynomial over Z into its irreducible factors, so we can just test all possible factorizations. I've no idea if this can be improved on. Algebraist 23:28, 17 March 2008 (UTC)

need a more indepth determination

Bold textHow do you determine if a polynomial is the difference of two roots? Thats how the question was brought to me.I don't quite get it either. —Preceding unsigned comment added by 72.94.241.142 (talk) 19:36, 17 March 2008 (UTC)

The difference of two roots? Well, assuming you mean square root, any number is a square root because it is able to be squared. Just go down a number line and mulitiply each term by itself. For example $S$ is the square root of $S^{2}$; 4 is the square root of 16. However, every number is not a square because not every number has an integer square root. I think the answer to your previous question above may be what you seek. Zrs 12 (talk) 19:55, 17 March 2008 (UTC)

Solve for y: x = y - 1/y

How can you solve the equation $x=y-{\frac {1}{y}}$ for y in terms of x? I know that the solution is $y={\frac {x\pm {\sqrt {x^{2}+4}}}{2}}$, and this works when you substitute it back into the original equation, but how can this be found analytically? (I used a calculator). --BrainInAVat (talk) 20:59, 17 March 2008 (UTC)

• Try multiplying both sides by $y$... you should then have a quadratic in terms of $y$ which can be solved using the "usual method". --Kinu ^t/_c 21:14, 17 March 2008 (UTC)
  • I'm at a loss, sorry. You get $xy=y^{2}-1$, but then what?--BrainInAVat (talk) 01:29, 18 March 2008 (UTC)

Take it all over to one side and you have a quadratic in y (just treat x like a number), you can then solve it using the quadratic formula and you'll get the answer you gave above. --Tango (talk) 01:53, 18 March 2008 (UTC)

If you are not familiar with it, you can take a look at out Quadratic equation article. -- Meni Rosenfeld (talk) 07:19, 18 March 2008 (UTC)

This can be a little confusing the first time you see it - here:

x=y - 1/y

xy=y² - 1 (equation 1)

y² - xy - 1 = 0

Now complete the square

(y-x/2)² - x²/4 - 1 = 0

As you should be able to see multilpling by y gives a quadratic in y (equation 1)

The solve the quadratic by completing the square - you can see that x is a parameter of the quadratic. Hopefully from the last equation you should be able to get the result.

Once you've got it, keep an eye out for equations of this type as it's a useful thing to remember.87.102.74.53 (talk) 11:35, 18 March 2008 (UTC)

March 18

Proof of Chain Rule

Let

$y=f(u)$ (1)

$u=g(x)$ (2)

where $f(u)$ and $g(x)$ are both differentiable functions. Then

${\frac {dy}{dx}}$ (3)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{\Delta x}}$ (4)

$=\lim _{\Delta x\rightarrow 0}\left[{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot {\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right]$ (5)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}$ (6)

Treat the arrow $\rightarrow$ as equal sign $=$. We can do the same operation on both sides of the arrow without changing the relationship

$\Delta x\rightarrow 0$ (7)

$x+\Delta x\rightarrow x$ (8)

Function $g(x)$ is continuous since it is differentiable. Apply $g(\cdot )$ to both sides of (8)

$g(x+\Delta x)\rightarrow g(x)$ (9)

$g(x+\Delta x)-g(x)\rightarrow 0$ (10)

Let

$\Delta u=g(x+\Delta x)-g(x)$ (11)

Replace (11) into (10)

$\Delta u\rightarrow 0$ (12)

Therefore

$\lim _{\Delta x\rightarrow 0}\cdot$  implies $\lim _{\Delta u\rightarrow 0}\cdot$ (13)

Replace (2) into (11)

$g(x+\Delta x)=u+\Delta u$ (14)

Replace (2), (11), (13) and (14) into (6)

${\frac {dy}{dx}}=\lim _{\Delta u\rightarrow 0}{\frac {f(u+\Delta u)-f(u)}{\Delta u}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}={\frac {dy}{du}}{\frac {du}{dx}}$ (15)

Q.E.D.

Is the proof of chain rule above correct and rigorous? - Justin545 (talk) 06:25, 18 March 2008 (UTC)

There are some questionable details. First, if we want a proof we can consider "rigorous", we would want to avoid treating functions as quantities (e.g., u instead of $g(x)$) and using Leibniz notation (${\tfrac {du}{dx}}$). So as a first step you should try formulating the proof without using u or y, only f, g and their composition $h=f\circ g$ (equivalently, $h(x)=f(g(x))$. Second, the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own. This would be acceptable for a handwaving proof, but not for a rigorous one. -- Meni Rosenfeld (talk) 07:37, 18 March 2008 (UTC)

>> "the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own."

I think you mean the result of (13) is incorrect or not rigorous. Does it mean the whole proof should be re-derived in a completely different way or we can somehow fix the problem so that we don't have to re-derive the whole proof? If (13) is not rigorous, is there any example which opposes it? Thanks! - Justin545 (talk) 09:00, 18 March 2008 (UTC)

(13) and the derivations that lead to it are "not even wrong" in the sense that in the standard framework of calculus they are pretty much meaningless - if you look at the standard rigorous definitions of limits, you will see that they do not allow a function to be used as a variable. It is "correct" in the sense that intuitively, the limit of a function "when" the variable approaches some value is equal to the limit when some othee function approaches its appropriate limit value. However, this "when" business lacks a rigorous interpretation and is haunted by Bishop Berkeley's ghosts.

I have thought about how one might amend the proof, and realized that you also have a mistake much earlier. Step (5), dividing and multiplying by $g(x+\Delta x)-g(x)$, is only valid if $g(x+\Delta x)-g(x)\neq 0$, but there is no reason to assume that should be the case. Take, for example

$g(x)=\left\{{\begin{array}{ll}x^{2}\sin {\tfrac {1}{x}}&x\neq 0\\0&x=0\end{array}}\right.$

- a perfectly differentiable function at 0, and yet $g(x)=0=g(0)$ infinitely many times in any neighborhood of 0. Thus your proof will not work for it. Those kinds of pathological counterexamples are one of the things that separates rigorous proofs from not-so-rigorous ones. -- Meni Rosenfeld (talk) 10:58, 18 March 2008 (UTC)

If you want a similar proof that really works, one way would be to apply the mean value theorem to f at (4). This allows you to replace $f(g(x+h))-f(g(x))$ 163.1.148.158 (talk) 12:54, 18 March 2008 (UTC)

Calculating inflation by consumer price index statistics

Hello,

I'm currently working on HMAS Melbourne (R21), which is currently at Featured Article Candidates. One of the reviewers has requested that more recent financial figures be provided for the various sums of money (listed at User talk:Saberwyn/HMAS Melbourne (R21)) mentioned in the history of the ship. I've been pointed to Australian consumer price index statistics (available at [5] from 1969 to 2007, and [6] from 1949 to 1997 - Both open directly as Excel spreadsheets).

I've looked at them, and realise I don't have the first idea as to using these statistics to convert, for example, a AU$1.4 million figure from 1985 to a AU$ figure in 2007. Any help, at least a point in the right direction to an online tutorial or something, would be muchly appreciated. -- saberwyn 10:32, 18 March 2008 (UTC)

Well, the table says that AU$69.7 in 06/1985 are equivalent to AU$157.5 in 06/2007 in terms of overall purchasing power, so 1.4m in 1985 AU$ would roughly amount to 1.4m*157.5/69.7 = 3.2m in 2007 AU$. Bikasuishin (talk) 12:21, 18 March 2008 (UTC)

Area Between Curves

I'm having a little trouble with an area between curves problem. The functions are f(x)=x^3-x and g(x)=3x. What I'm trying to figure out is if I should add the two areas (the one bellow and the one above the curve) or if i should subtract the area from underneath the X axis from the one above it. Thanks RedStateV (talk) 14:28, 18 March 2008 (UTC)

The distance between two points is the absolute value of one minus the other. So long as you keep your signs straight, it doesn't matter where either curve is in relation to the x-axis or to the other. — Lomn 14:42, 18 March 2008 (UTC)

So should I be getting a answer of 8 then? RedStateV (talk) 14:52, 18 March 2008 (UTC)